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checked that these do not depend on the choices of representatives for the equivalence classes, and that we obtain in this way an S 1A-module S 1M D f m s j m 2 M; s 2 S g and a homomorphism m 7! m 1 W M iS! S 1M of A-modules whose kernel is fa 2 M j sa D 0 for some s 2 S g: PROPOSITION 1.17. The elements of S act invertibly on S 1M , and every homomorphism from M to an A-module N with this property factors uniquely through iS , iS M S 1M 9Š N: PROOF. Similar to the proof of 1.10. PROPOSITION 1.18. The functor M S 1M is exact. In other words, if the sequence of A-modules ˇ! M 00 is exact, then so also is the sequence of S 1A-modules M 0 ˛! M S 1M 0 S 1˛! S 1M S 1ˇ! S 1M 00: PROOF. Because ˇ ı˛ D 0, we have 0 D S 1.ˇ ı˛/ D S 1ˇ ıS 1˛. Therefore Im.S 1˛/ 2 Ker.S 1ˇ/, where m 2 M and s 2 S. Then Ker.S 1ˇ/. For the reverse inclusion, let m s D 0 and so, for some t 2 S, we have tˇ.m/ D 0. Then ˇ.tm/ D 0, and so tm D ˛.m0/ ˇ.m/ s for some m0 2 M 0. Now m s D tm ts D ˛.m0/ ts 2 Im.S 1˛/: PROPOSITION 1.19. Let A be a ring, and let M be an A-module. The canonical map M ! YfMm j m a maximal ideal in Ag is injective. 22 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let m 2 M map to zero in all Mm. The annihilator a D fa 2 A j am D 0g of m is an ideal in A. Because m maps to zero Mm, there exists an s 2 A X m such that sm D 0. Therefore a is not contained in m. Since this is true for all maximal ideals m, a D A, and so it contains 1. Now m D 1m D 0. COROLLARY 1.20. An A-module M D 0 if Mm D 0 for all maximal ideals m in A. PROOF. Immediate consequence of the lemma. PROPOSITION 1.21. Let A be a ring. A sequence of A-modules is exact if and only if is exact for all maximal ideals m. M 0 ˛! M ˇ! M 00 M 0 m ˛m! Mm ˛m! M 00 m () () PROOF. The necessity is a special case of Proposition 1.18. For the sufficiency, let N D Ker.ˇ/= Im.˛/. Because the functor M Mm is exact, Nm D Ker.ˇm/= Im.˛m/: If () is exact for all m, then Nm D 0 for all m, and so N D 0 (by 1.20). But this means that () is exact. COROLLARY 1.22. A homomorphism M ! N of A-modules is injective (resp. surjective) if and only if Mm ! Nm is injective (resp. surjective) for all maximal ideals m: PROOF. Apply the proposition to 0 ! M ! N (resp. M ! N ! 0). Direct limits A directed set is a pair .I; / consisting of a set I and a preorder4 on I such that for all i; j 2 I , there exists a k 2 I with i; j k. Let .I; / be a directed set, and let A be a ring. A direct system of A-modules indexed W Mi ! Mj /i j of k all i j k.5 An A-module M j for all i j is the direct by .I; / is a family .Mi /i 2I of A-modules together with a family .˛i j D idMi and ˛j A-linear maps such that ˛i i together with A-linear maps ˛i W Mi ! M such that ˛i D ˛j ı ˛i limit of the system .Mi ; ˛j D ˛i ı ˛i j k (a) M D S (b) mi 2 Mi maps to zero in M if and only if it maps to zero in Mj for some j i . i / if i 2I ˛i .Mi /, and Direct limits of A-algebras are defined similarly. PROPOSITION 1.23. For every multiplicative subset S of A, S 1A ' lim! Ah, where h runs over the elements of S (partially ordered by division). 4A preorder is a reflexive transitive binary relation. 5Regard I as a category with Hom.a; b/ empty unless a b, in which case it contains a single element. Then a direct system is a functor from I to the category of A-modules. c. Unique factorization 23 PROOF. When hjh0, say, h0 D hg, there is a canonical homomorphism a h0 W Ah ! h Ah0, and so the rings Ah form a direct system indexed by the set S. When h 2 S , the W Ah ! S 1A homomorphism A ! S 1A extends uniquely to a homomorphism a h (1.10), and these homomorphisms are compatible with the maps in the direct system. Now it is easy to see that S 1A satisfies the conditions to be the direct limit of the Ah. 7! a h 7! ag c. Unique factorization Let A be an integral domain. An element a of A is irreducible if it is not zero, not a unit, and admits only trivial factorizations, i.e., a D bc H) b or c is a unit. An element a is said to be prime if .a/ is a prime ideal, i.e., ajbc H) ajb or ajc. An integral domain A is called a unique factorization domain (or a factorial domain) if every nonzero nonunit in A can be written as a finite product of irreducible elements in exactly one way up to units and the order of the factors: Principal ideal domains, for example, Z and kŒX, are unique factorization domains, PROPOSITION 1.24. Let A be an integral domain, and let a be an element of A that is neither zero nor a unit. If a is prime, then a is irreducible, and the converse holds when A is a unique factorization domain. PROOF. Assume that a is prime. If a D bc, then a divides bc and so a divides b or c. Suppose the first, and write b D aq. Now a D bc D aqc, which implies that qc D 1 because A is an integral domain, and so c is a unit. Therefore a is irreducible. For the converse, assume that a is irreducible and that A is a unique factorization domain. If ajbc, then bc D aq, some q 2 A: On writing each of b, c, and q as a product of irreducible elements, and using the uniqueness of factorizations, we see that a differs from one of the irreducible factors of b or c by a unit. Therefore a divides b or c. COROLLARY 1.25. Let A be an integral domain. If A is a unique factorization domain, then every prime ideal of height 1 is principal. PROOF. Let p be a prime ideal of height 1. Then p contains a nonzero element, and hence an irreducible element a. We have p .a/ .0/. As .a/ is prime and p has height 1, we must have p D .a/. PROPOSITION 1.26. Let A be an integral domain in which every nonzero nonunit element is a finite product of irreducible elements. If every irreducible element of A is prime, then A is a unique factorization domain. PROOF. Suppose that a1 am D b1 bn (9) 24 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA with the ai and bi irreducible elements in A. As a1 is prime, it divides one of the bi , which we may suppose to be b1. As b1 is irreducible, b1 D ua1 for some unit u. On cancelling a1 from both sides of (9), we obtain the equality a2 am D .ub2/b3 bn: Continuing in this fashion, we find that the two factorizations are the same up to units and the order of the factors. PROPOSITION 1.27 (GAUSS’S LEMMA). Let A be a unique factorization domain with field of fractions F . If f .X/ 2 AŒX factors into the product of two nonconstant polynomials in F ŒX, then it factors into the product of two nonconstant polynomials in AŒX . PROOF. Let f D gh in F ŒX. For suitable c; d 2 A, the polynomials g1 D cg and h1 D dh have coefficients in A, and so we have a factorization cdf D g1h1 in AŒX. If an irreducible element p of A divides cd , then, looking modulo .p/, we see that 0 D g1 h1 in .A=.p// ŒX. According to Proposition 1.24, .p/ is prime, and so .A=.p// ŒX is an integral domain. Therefore, p divides all the coefficients of at least one of the polynomials g1; h1, say g1, so that g1 D pg2 for some g2 2 AŒX. Thus, we have a factorization .cd=p/f D g2h1 in AŒX . Continuing in this fashion, we can remove all the irreducible factors of cd , and so obtain a factorization of f in AŒX. Let A be a unique factorization domain. A nonzero polynomial f D a0 C a1X C C amX m in AŒX is said to be primitive if the coefficients ai have no common factor (other than units). Every polynomial f in F ŒX can be written f D c.f / f1 with c.f / 2 F and f1 primitive. The element c.f /, which is well-defined up to multiplication by a unit, is called the content of f . Note that f 2 AŒX if and only if c.f / 2 A. LEMMA 1.28. The product of two primitive polynomials is primitive. PROOF. Let f D a0 C a1X C C amX m g D b0 C b1X C C bnX n; be primitive polynomials, and let p be an irreducible element of A. Let ai0, i0 m, be the first coefficient of f not divisible by p, and let bj0, j0 n, the first coefficient of g not divisible by p. Then all the terms in the sum P i Cj Di0Cj0 ai bj are divisible by p, except ai0bj0, which is not divisible by p. Therefore, p doesn’t divide the .i0 C j0/th-coefficient of fg. We have shown that no irreducible element of A divides all the coefficients of fg, which must therefore be primitive. c. Unique factorization 25 PROPOSITION 1.29. Let A be a unique factorization domain with field of fractions F . For polynomials f; g 2 F ŒX, c.fg/ D c.f / c.g/I hence every factor in AŒX of a primitive polynomial is primitive. PROOF. Let f D c.f / f1 and g D c.g/ g1 with f1 and g1 primitive. Then fg D c.f / c.g/ f1g1 with f1g1 primitive, and so c.fg/ D c.f /c.g/. COROLLARY 1.30. An element f 2 AŒX is irreducible if and only if either (a) f is constant, say f D a, with a an irreducible element of A, or (b) f is a nonconstant primitive polynomial that is irreducible in F ŒX . PROOF. (: If f is as in (a) and f D gh in AŒX, then g and h both lie in A and one must be a unit in A, and hence a unit in AŒX. If f is as in (b) and f D gh, then one of g or h must be constant because otherwise f would be reducible in F ŒX. If it is g that is constant, then, because f is primitive, g must be a unit in A, hence in AŒX. ): Let f 2 AŒX be irreducible. If f is a constant polynomial, say f D a, then a is obviously irreducible in A. If f nonconstant, then it must be primitive because otherwise f D c.f / f1 would be a nontrivial factorization in AŒX . It must also be irreducible in F ŒX, because otherwise it would have a nontrivial factorization in AŒX (by 1.27). PROPOSITION 1.31. If A is a unique factorization domain, then so also is AŒX. PROOF. We shall check that A satisfies the conditions of Proposition 1.26. Let f 2 AŒX, and write f D c.f /f1. Then c.f / is a product of irreducible elements in A, and f1 is a product of irreducible primitive polynomials. This shows that f is a product of irreducible elements in AŒX. Let a be an irreducible element of A. If a divides fg, then it divides c.fg/ D c.f /c.g/. As a is prime (1.24), it divides c.f / or c.g/, and hence also f or g. Let f be an irreducible primitive polynomial in AŒX . Then f is irreducible in F ŒX , and so if f divide
s the product gh of g; h 2 AŒX, then it divides g or h in F ŒX. Suppose the first, and write f q D g with q 2 F ŒX. Then c.q/ D c.f /c.q/ D c.f q/ D c.g/ 2 A, and so q 2 AŒX. Therefore f divides g in AŒX. We have shown that every element of AŒX is a product of irreducible elements and that every irreducible element of AŒX is prime, and so AŒX is a unique factorization domain (1.26). Polynomial rings Let k be a field. The elements of the polynomial ring kŒX1; : : : ; Xn are finite sums X ca1anX a1 1 X an n ; ca1an 2 k; aj 2 N; with the obvious notions of equality, addition, and multiplication. In particular, the monomials form a basis for kŒX1; : : : ; Xn as a k-vector space. The degree, deg.f /, of a nonzero polynomial f is the largest total degree of a monomial occurring in f with nonzero coefficient. Since deg.fg/ D deg.f / C deg.g/, kŒX1; : : : ; Xn is an integral domain and kŒX1; : : : ; Xn D k. An element f of kŒX1; : : : ; Xn is irreducible if it is nonconstant and f D gh H) g or h is constant. 26 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA THEOREM 1.32. The ring kŒX1; : : : ; Xn is a unique factorization domain. PROOF. Note that kŒX1; : : : ; Xn1ŒXn D kŒX1; : : : ; Xn: This simply says that every polynomial f in n symbols X1; : : : ; Xn can be expressed uniquely as a polynomial in Xn with coefficients in kŒX1; : : : ; Xn1, f .X1; : : : ; Xn/ D a0.X1; : : : ; Xn1/X r n C C ar .X1; : : : ; Xn1/: Since, as we noted, kŒX is a unique factorization domain, the theorem follows by induction from Proposition 1.31. COROLLARY 1.33. A nonzero proper principal ideal .f / in kŒX1; : : : ; Xn is prime if and only if f is irreducible. PROOF. Special case of Proposition 1.24. d. Integral dependence Let A be a subring of a ring B. An element ˛ of B is said to be6 integral over A if it is a root of a monic7 polynomial with coefficients in A, i.e., if it satisfies an equation ˛n C a1˛n1 C C an D 0; ai 2 A: If every element of B is integral over A, then B is said to be integral over A. In the next proof, we shall need to apply a variant of Cramer’s rule: if x1; : : : ; xm is a solution to the system of linear equations m X j D1 cij xj D 0; i D 1; : : : ; m; with coefficients in a ring A, then det.C / xj D 0; j D 1; : : : ; m; (10) where C is the matrix of coefficients. To prove this, expand out the left hand side of 0 B @ det c11 ::: cm1 : : : : : : c1 j 1 ::: cm j 1 P P i c1i xi ::: i cmi xi c1 j C1 ::: cm j C1 : : : : : : c1m ::: cmm 1 C A D 0 using standard properties of determinants. An A-module M is faithful if aM D 0, a 2 A, implies that a D 0. PROPOSITION 1.34. Let A be a subring of a ring B. An element ˛ of B is integral over A if and only if there exists a faithful AŒ˛-submodule M of B that is finitely generated as an A-module. 6More generally, if f W A ! B is an A-algebra, an element ˛ of B is integral over A if it satisfies an equation ˛n C f .a1/˛n1 C C f .an/ D 0; ai 2 A: Thus, ˛ is integral over A if and only if it is integral over the subring f .A/ of B. 7A polynomial is monic if its leading coefficient is 1, i.e., f .X/ D X nC terms of degree less than n. d. Integral dependence 27 PROOF. )W Suppose that ˛n C a1˛n1 C C an D 0; ai 2 A: Then the A-submodule M of B generated by 1, ˛, ..., ˛n1 has the property that ˛M M , and it is faithful because it contains 1. (W Let M be a faithful AŒ˛-submodule of B admitting a finite set fe1; : : : ; eng of generators as an A-module. Then, for each i , ˛ei D P aij ej , some aij 2 A: We can rewrite this system of equations as .˛ a11/e1 a12e2 a13e3 D 0 a21e1 C .˛ a22/e2 a23e3 D 0 D 0: Let C be the matrix of coefficients on the left-hand side. Then Cramer’s formula tells us that det.C / ei D 0 for all i. As M is faithful and the ei generate M , this implies that det.C / D 0. On expanding out the determinant, we obtain an equation ˛n C c1˛n1 C c2˛n2 C C cn D 0; ci 2 A: PROPOSITION 1.35. An A-algebra B is finite if it is generated as an A-algebra by a finite set of elements each of which is integral over A. PROOF. Suppose that B D AŒ˛1; : : : ; ˛m and that C ai1˛ni 1 C C ai ni D 0; ˛ni i i aij 2 A; i D 1; : : : ; m. Any monomial in the ˛i divisible by some ˛ni is equal (in B) to a linear combination of i monomials of lower degree. Therefore, B is generated as an A-module by the finite set of monomials ˛r1 1 m , 1 ri < ni . ˛rm COROLLARY 1.36. An A-algebra B is finite if and only if it is finitely generated and integral over A. PROOF. (: Immediate consequence of 1.35. ): We may replace A with its image in B. Then B is a faithful AŒ˛-module for all ˛ 2 B (because 1B 2 B), and so 1.34 shows that every element of B is integral over A. As B is finitely generated as an A-module, it is certainly finitely generated as an A-algebra. PROPOSITION 1.37. Consider rings A B C . If B is integral over A and C is integral over B, then C is integral over A. PROOF. Let 2 C . Then n C b1 n1 C C bn D 0 for some bi 2 B. Now AŒb1; : : : ; bn is finite over A (see 1.35), and AŒb1; : : : ; bnŒ is finite over AŒb1; : : : ; bn, and so it is finite over A. Therefore is integral over A by 1.34. THEOREM 1.38. Let A be a subring of a ring B. The elements of B integral over A form an A-subalgebra of B. 28 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let ˛ and ˇ be two elements of B integral over A. Then AŒ˛; ˇ is finitely generated as an A-module (1.35). It is stable under multiplication by ˛ ˙ ˇ and ˛ˇ and it is faithful as an AŒ˛ ˙ ˇ-module and as an AŒ˛ˇ-module (because it contains 1A). Therefore 1.34 shows that ˛ ˙ ˇ and ˛ˇ are integral over A. DEFINITION 1.39. Let A be a subring of the ring B. The integral closure of A in B is the subring of B consisting of the elements integral over A. PROPOSITION 1.40. Let A be an integral domain with field of fractions F , and let ˛ be an element of some field containing F . If ˛ is algebraic over F , then there exists a d 2 A such that d˛ is integral over A. PROOF. By assumption, ˛ satisfies an equation ˛m C a1˛m1 C C am D 0; ai 2 F: Let d be a common denominator for the ai , so that dai 2 A for all i , and multiply through the equation by d m: .d˛/m C a1d.d˛/m1 C C amd m D 0: As a1d; : : : ; amd m 2 A, this shows that d˛ is integral over A. COROLLARY 1.41. Let A be an integral domain and let E be an algebraic extension of the field of fractions of A. Then E is the field of fractions of the integral closure of A in E. PROOF. In fact, the proposition shows that every element of E is a quotient ˇ=d with ˇ integral over A and d 2 A. DEFINITION 1.42. An integral domain A is said to be integrally closed if it is equal to its integral closure in its field of fractions F , i.e., if ˛ 2 F; ˛ integral over A H) ˛ 2 A: An integrally closed integral domain is called an integrally closed domain or normal domain. PROPOSITION 1.43. Unique factorization domains are integrally closed. PROOF. Let A be a unique factorization domain, and let a=b be an element of its field of fractions. If a=b … A, then b divisible by some prime element p not dividing a. If a=b is integral over A, then it satisfies an equation .a=b/n C a1.a=b/n1 C C an D 0; ai 2 A: On multiplying through by bn, we obtain the equation an C a1an1b C C anbn D 0: The element p then divides every term on the left except an, and hence divides an. Since it doesn’t divide a, this is a contradiction. d. Integral dependence 29 Let F E be fields, and let ˛ 2 E be algebraic over F . The minimal polynomial of ˛ over F is the monic polynomial of smallest degree in F ŒX having ˛ as a root. If f is the minimal polynomial of ˛, then the homomorphism X 7! ˛W F ŒX ! F Œ˛ defines an isomorphism F ŒX =.f / ! F Œ˛, i.e., F Œx ' F Œ˛, x $ ˛. PROPOSITION 1.44. Let A be an integrally closed domain, and let E be a finite extension of the field of fractions F of A. An element of E is integral over A if and only if its minimal polynomial over F has coefficients in A. PROOF. Let ˛ 2 E be integral over A, so that ˛m C a1˛m1 C C am D 0; some ai 2 A; m > 0. Let f .X / be the minimal polynomial of ˛ over F , and let ˛0 be a conjugate of ˛, i.e., a root of f in some splitting field of f . Then f is also the minimal polynomial of ˛0 over F , and so (see above), there is an F -isomorphism W F Œ˛ ! F Œ˛0; .˛/ D ˛0: On applying to the above equation we obtain the equation ˛0m C a1˛0m1 C C am D 0; which shows that ˛0 is integral over A. As the coefficients of f are polynomials in the conjugates of ˛, it follows from (1.38) that the coefficients of f .X/ are integral over A. They lie in F , and A is integrally closed, and so they lie in A. This proves the “only if” part of the statement, and the “if” part is obvious. COROLLARY 1.45. Let A F E be as in the proposition, and let ˛ be an element of E integral over A. Then NmE=F .˛/ 2 A, and ˛ divides NmE=F .˛/ in AŒ˛. PROOF. Let f .X/ D X m C a1X m1 C C am be the minimal polynomial of ˛ over F . Then Nm.˛/ D .1/mnan (FT 5.45), and so Nm.˛/ 2 A. Because f .˛/ D 0, m, where n D ŒEW F Œ˛ 0 D an1 m .˛m C a1˛m1 C C am/ m ˛m1 C C an1 D ˛.an1 m am1/ C .1/mn Nm.˛/; and so ˛ divides NmE=F .˛/ in AŒ˛. COROLLARY 1.46. Let A be an integrally closed domain with field of fractions F , and let f .X/ be a monic polynomial in AŒX. Then every monic factor of f .X/ in F ŒX has coefficients in A. PROOF. It suffices to prove this for an irreducible monic factor g of f in F ŒX . Let ˛ be a root of g in some extension field of F . Then g is the minimal polynomial of ˛. As ˛ is a root of f , it is integral over A, and so g has coefficients in A. PROPOSITION 1.47. Let A B be rings, and let A0 be the integral closure of A in B. For any multiplicative subset S of A, S 1A0 is the integral closure of S 1A in S 1B. 30 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let b s 2 S 1A0 with b 2 A0 and s 2 S . Then bn C a1bn1 C C an D 0 for some ai 2 A, and so n b s C a1 s b s n1 C C an sn D 0: Therefore b=s is integral over S 1A. This shows that S 1A0 is contained in the integral closure of S 1B. For the converse, let b=s (b 2 B, s 2 S ) be int
egral over S 1A. Then n b s C a1 s1 b s n1 C C an sn D 0: for some ai 2 A and si 2 S . On multiplying this equation by sns1 sn, we find that s1 snb 2 A0, and therefore that b s 2 S 1A0. D s1snb ss1sn COROLLARY 1.48. Let A B be rings, and let S be a multiplicative subset of A. If A is integrally closed in B, then S 1A is integrally closed in S 1B. PROOF. Special case of the proposition in which A0 D A. PROPOSITION 1.49. The following conditions on an integral domain A are equivalent: (a) A is integrally closed; (b) Ap is integrally closed for all prime ideals p; (c) Am is integrally closed for all maximal ideals m. PROOF. The implication (a))(b) follows from 1.48, and (b))(c) is obvious. It remains to prove (c))(a). If c is integral over A, then it is integral over each Am, and hence lies in each Am. It follows that the ideal consisting of the a 2 A such that ac 2 A is not contained in any maximal ideal m, and therefore equals A. Hence 1 c 2 A. Let E=F be a finite extension of fields. Then .˛; ˇ/ 7! TrE=F .˛ˇ/W E E ! F (11) is a symmetric bilinear form on E regarded as a vector space over F . LEMMA 1.50. If E=F is separable, then the trace pairing (11) is nondegenerate. PROOF. Let ˇ1; :::; ˇm be a basis for E as an F -vector space. We have to show that the discriminant det.Tr.ˇi ˇj // of the trace pairing is nonzero. Let 1; :::; m be the distinct F -homomorphisms of E into some large Galois extension ˝ of F . Recall (FT 5.45) that TrL=K .ˇ/ D 1ˇ C C mˇ (12) By direct calculation, we have det.Tr.ˇi ˇj // D det.P k k.ˇi ˇj // D det.P k k.ˇi / k.ˇj // D det.k.ˇi // det.k.ˇj // D det.k.ˇi //2: (by 12) d. Integral dependence 31 Suppose that det.i ˇj / D 0. Then there exist c1; :::; cm 2 ˝ such that ci i .ˇj / D 0 all j: X i By linearity, it follows that P theorem on the independence of characters (FT 5.14). i ci i .ˇ/ D 0 for all ˇ 2 E, but this contradicts Dedekind’s PROPOSITION 1.51. Let A be an integrally closed domain with field of fractions F , and let B be the integral closure of A in a separable extension E of F of degree m. There exist free A-submodules M and M 0 of E such that M B M 0. (13) If A is noetherian, then B is a finite A-algebra. PROOF. Let fˇ1; :::; ˇmg be a basis for E over F . According to Proposition 1.40, there exists a d 2 A such that d ˇi 2 B for all i . Clearly fd ˇ1; : : : ; d ˇmg is still a basis for E as a vector space over F , and so we may assume to begin with that each ˇi 2 B. Because the trace pairing is nondegenerate, there is a dual basis fˇ0 g of E over F with the property that Tr.ˇi ˇ0 j / D ıij for all i; j . We shall show that 1; :::; ˇ0 m Aˇ1 C Aˇ2 C C Aˇm B Aˇ0 1 C Aˇ0 2 C C Aˇ0 m: Only the second inclusion requires proof. Let ˇ 2 B. Then ˇ can be written uniquely as a linear combination ˇ D P bj ˇ0 j with coefficients bj 2 F , and we have to show that each bj 2 A. As ˇi and ˇ are in B, so also is ˇ ˇi , and so Tr.ˇ ˇi / 2 A (1.44). But j of the ˇ0 Tr.ˇ ˇi / D Tr. X bj ˇ0 j ˇi / D X bj Tr.ˇ0 j ˇi / D X bj ıij D bi : j j j Hence bi 2 A. If A is Noetherian, then M 0 is a Noetherian A-module, and so B is finitely generated as an A-module. LEMMA 1.52. Let A be a subring of a field K. If K is integral over A, then A is also a field. PROOF. Let a be a nonzero element of A. Then a1 2 K, and it is integral over A: .a1/n C a1.a1/n1 C C an D 0; ai 2 A: On multiplying through by an1, we find that a1 C a1 C C anan1 D 0; from which it follows that a1 2 A. THEOREM 1.53 (GOING-UP THEOREM). Let A B be rings with B integral over A. (a) For every prime ideal p of A, there is a prime ideal q of B such that q \ A D p. (b) Let p D q \ A; then p is maximal if and only if q is maximal. 32 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. (a) If S is a multiplicative subset of a ring A, then the prime ideals of S 1A are in one-to-one correspondence with the prime ideals of A not meeting S (see 1.14). It therefore suffices to prove (a) after A and B have been replaced by S 1A and S 1B, where S D A p. Thus we may assume that A is local, and that p is its unique maximal ideal. In this case, for all proper ideals b of B, b \ A p (otherwise b A 3 1/. To complete the proof of (a), we shall show that for all maximal ideals n of B, n \ A D p. Consider B=n A=.n \ A/. Here B=n is a field, which is integral over its subring A=.n \ A/, and n \ A will be equal to p if and only if A=.n \ A/ is a field. This follows from Lemma 1.52. (b) The ring B=q contains A=p, and it is integral over A=p. If q is maximal, then Lemma 1.52 shows that p is also. For the converse, note that any integral domain integral over a field is a field because it is a union of integral domains finite over the field, which are automatically fields (left multiplication by an element is injective, and hence surjective, being a linear map of a finite-dimensional vector space). COROLLARY 1.54. Let A B be rings with B integral over A. Let p p0 be prime ideals of A, and let q be a prime ideal of B such that q \ A D p. Then there exists a prime ideal q0 of B containing q and such that q0 \ A D p0, B A q q0 p p0: PROOF. We have A=p B=q, and B=q is integral over A=p. According to the (1.53), there exists a prime ideal q00 in B=q such that q00 \ .A=p/ D p0=p. The inverse image q0 of q00 in B has the required properties. ASIDE 1.55. Let A be a noetherian integral domain, and let B be the integral closure of A in a finite extension E of the field of fractions F of A. Is B always a finite A-algebra? When A is integrally closed and E is separable over F , or A is a finitely generated k-algebra, then the answer is yes (1.51, 8.3). However, in 1935, Akizuki found an example of a noetherian integral domain whose integral closure in its field of fractions is not finite (according to Matsumura 1986, finding the example cost him a year’s hard struggle). F.K. Schmidt found another example at about the same time.8 e. Tensor Products Tensor products of modules Let A be a ring, and let M , N , and P be A-modules. A map W M N ! P of A-modules is said to be A-bilinear if .x C x0; y/ D .x; y/ C .x0; y/; .x; y C y0/ D .x; y/ C .x; y0/; .ax; y/ D a.x; y/; .x; ay/ D a.x; y/; x; x0 2 M; y 2 N x 2 M; y; y0 2 N a 2 A; a 2 A; x 2 M; y 2 N x 2 M; y 2 N; i.e., if is A-linear in each variable. 8For a discussion of the examples Akizuki and Schmidt and generalizations, see Olberding, Bruce, One- dimensional bad Noetherian domains. Trans. Amer. Math. Soc. 366 (2014), no.8, 4067–4095. e. Tensor Products 33 An A-module T together with an A-bilinear map W M N ! T is called the tensor product of M and N over A if it has the following universal property: every A-bilinear map 0W M N ! T 0 factors uniquely through . M N T 0 9Š linear T 0: As usual, the universal property determines the tensor product uniquely up to a unique isomorphism. We write it M ˝A N . Note that HomA-bilinear.M N; T / ' HomA-linear.M ˝A N; T /: CONSTRUCTION Let M and N be A-modules, and let A.M N / be the free A-module with basis M N . Thus each element A.M N / can be expressed uniquely as a finite sum X ai .xi ; yi /; ai 2 A; xi 2 M; yi 2 N: Let P be the submodule of A.M N / generated by the following elements .x C x0; y/ .x; y/ .x0; y/; x; x0 2 M; y 2 N .x; y C y0/ .x; y/ .x; y0/; x 2 M; y; y0 2 N .ax; y/ a.x; y/; a 2 A; .x; ay/ a.x; y/; a 2 A; x 2 M; y 2 N x 2 M; y 2 N; and define M ˝A N D A.M N /=P: Write x ˝ y for the class of .x; y/ in M ˝A N . Then .x; y/ 7! x ˝ yW M N ! M ˝A N is A-bilinear — we have imposed the fewest relations necessary to ensure this. Every element of M ˝A N can be written as a finite sum9 X ai .xi ˝ yi /; ai 2 A; xi 2 M; yi 2 N; and all relations among these symbols are generated by the following relations .x C x0/ ˝ y D x ˝ y C x0 ˝ y x ˝ .y C y0/ D x ˝ y C x ˝ y0 a.x ˝ y/ D .ax/ ˝ y D x ˝ ay: The pair .M ˝A N; .x; y/ 7! x ˝ y/ has the correct universal property because any bilinear map 0W M N ! T 0 defines an A-linear map A.M N / ! T 0, which factors through A.M N /=K, and gives a commutative triangle. 9“An element of the tensor product of two vector spaces is not necessarily a tensor product of two vectors, but sometimes a sum of such. This might be considered a mathematical shenanigan but if you start with the state vectors of two quantum systems it exactly corresponds to the notorious notion of entanglement which so displeased Einstein.” Georges Elencwajg on mathoverflow.net. 34 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA Tensor products of algebras Let A and B be k-algebras. A k-algebra C together with homomorphisms i W A ! C and j W B ! C is called the tensor product of A and B if it has the following universal property: for every pair of homomorphisms (of k-algebras) ˛W A ! R and ˇW B ! R, there is a unique homomorphism W C ! R such that ı i D ˛ and ı j D ˇ: A i C j B ˛ 9Š ˇ R: If it exists, the tensor product, is uniquely determined up to a unique isomorphism by this property. We write it A ˝k B. Note that Homk.A ˝k B; R/ ' Homk.A; R/ Homk.B; R/ (homomorphisms of k-algebras). CONSTRUCTION Form the tensor product A ˝k B of A and B regarded as k-vector spaces. There is a multiplication map A ˝k B A ˝k B ! A ˝k B for which .a ˝ b/.a0 ˝ b0/ D aa0 ˝ bb0. This makes A ˝k B into a ring, and the homomorphism c 7! c.1 ˝ 1/ D c ˝ 1 D 1 ˝ c makes it into a k-algebra. The maps a 7! a ˝ 1W A ! C and b 7! 1 ˝ bW B ! C are homomorphisms, and they make A ˝k B into the tensor product of A and B in the above sense. EXAMPLE 1.56. The algebra B, equipped with the given map k ! B and the identity map B ! B, has the universal property characterizing k ˝k B, so k ˝k B ' B. In terms of the constructive definition of tensor products, the isomorphism is c ˝ b 7! cbW k ˝k B ! B. EXAMPLE 1.57. The ring kŒX1; : : : ; Xm; XmC1; : : : ; XmCn, equipped with the obvious inclusions kŒX1; : : : ; Xm ,! kŒX1; : : : ; XmCn - kŒXmC1; : : : ; XmCn is the tensor product of kŒX1; : : : ; Xm and kŒXmC1; : : : ; XmCn. To verify this we only have to check that, for every k-algebra R, the map Homk-alg.kŒX1; : : : ; XmCn; R/
! Homk-alg.kŒX1; : : :; R/ Homk-alg.kŒXmC1; : : :; R/ induced by the inclusions is a bijection. But this map can be identified with the obvious bijection RmCn ! Rm Rn: In terms of the constructive definition of tensor products, the isomorphism is f ˝ g 7! fgW kŒX1; : : : ; Xm ˝k kŒXmC1; : : : ; XmCn ! kŒX1; : : : ; XmCn. f. Transcendence bases 35 REMARK 1.58. (a) If .b˛/ is a family of generators (resp. basis) for B as a k-vector space, then .1 ˝ b˛/ is a family of generators (resp. basis) for A ˝k B as an A-module. (b) Let k ,! ˝ be fields. Then ˝ ˝k kŒX1; : : : ; Xn ' ˝Œ1 ˝ X1; : : : ; 1 ˝ Xn ' ˝ŒX1; : : : ; Xn: If A D kŒX1; : : : ; Xn=.g1; : : : ; gm/, then ˝ ˝k A ' ˝ŒX1; : : : ; Xn=.g1; : : : ; gm/: (c) If A and B are algebras of k-valued functions on sets S and T respectively, then .f ˝ g/.x; y/ D f .x/g.y/ realizes A ˝k B as an algebra of k-valued functions on S T . f. Transcendence bases We review the theory of transcendence bases. For the proofs, see Chapter 9 of FT. 1.59. Elements ˛1; :::; ˛n of a k-algebra A are said to be algebraically dependent over k there exists a nonzero polynomial f .X1; :::; Xn/ 2 kŒX1; :::; Xn such that f .˛1; :::; ˛n/ D 0. Otherwise, the ˛i are said to be algebraically independent over k. Now let ˝ be a field containing k. 1.60. For a subset A of ˝, we let k.A/ denote the smallest subfield of ˝ containing k and A. For example, if A D fx1; : : : ; xmg, then k.A/ consists of the quotients f .x1;:::;xmg g.x1;:::;xmg with f; g 2 kŒX1; : : : ; Xm. A subset B of ˝ is algebraically dependent on A if each element of B is algebraic over k.A/. 1.61 (FUNDAMENTAL THEOREM). Let A D f˛1; :::; ˛mg and B D fˇ1; :::; ˇng be two subsets of ˝. Assume that (a) A is algebraically independent (over k), and (b) A is algebraically dependent on B (over k). Then m n. The reader should note the similarity of this to the statement in linear algebra with “algebraically” replaced by “linearly”. 1.62. A transcendence basis for ˝ over k is an algebraically independent set A such that ˝ is algebraic over k.A/: 1.63. Assume that there is a finite subset A ˝ such that ˝ is algebraic over k.A/. Then (a) every maximal algebraically independent subset of ˝ is a transcendence basis; (b) every subset A minimal among those such that ˝ is algebraic over k.A/ is a transcen- dence basis; (c) all transcendence bases for ˝ over k have the same finite number of elements (called the transcendence degree, tr degk˝, of ˝ over k). 1.64. Let k L ˝ be fields. Then tr degk˝ D tr degkL C tr degL˝. More precisely, if A is a transcendence basis for L=k and B is a transcendence basis for ˝=L, then A [ B is a transcendence basis for ˝=k. 36 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA Exercises 1-1. Let k be an infinite field (not necessarily algebraically closed). Show that an f 2 kŒX1; : : : ; Xn that is identically zero on kn is the zero polynomial (i.e., has all its coefficients zero). 1-2. Find a minimal set of generators for the ideal .X C 2Y; 3X C 6Y C 3Z; 2X C 4Y C 3Z/ in kŒX; Y; Z. What standard algorithm in linear algebra will allow you to answer this question for any ideal generated by homogeneous linear polynomials? Find a minimal set of generators for the ideal .X C 2Y C 1; 3X C 6Y C 3X C 2; 2X C 4Y C 3Z C 3/: 1-3. A ring A is said to be normal if Ap is a normal integral domain for all prime ideals p in A. Show that a noetherian ring is normal if and only if it is a finite product of normal integral domains. 1-4. Prove the statement in 1.64. CHAPTER 2 Algebraic Sets a. Definition of an algebraic set An algebraic subset V .S/ of kn is the set of common zeros of some collection S of polynomials in kŒX1; : : : ; Xn, V .S/ D f.a1; : : : ; an/ 2 kn j f .a1; : : : ; an/ D 0 all f 2 Sg: We refer to V .S/ as the zero set of S. Note that S S 0 H) V .S/ V .S 0/I — more equations means fewer solutions. Recall that the ideal a generated by a set S consists of the finite sums X fi gi ; fi 2 kŒX1; : : : ; Xn; gi 2 S: Such a sum P fi gi is zero at every point at which the gi are all zero, and so V .S/ V .a/, but the reverse conclusion is also true because S a. Thus V .S/ D V .a/ — the zero set of S is the same as the zero set of the ideal generated by S. Therefore the algebraic subsets of kn can also be described as the zero sets of ideals in kŒX1; : : : ; Xn. An empty set of polynomials imposes no conditions, and so V .;/ D kn. Therefore kn is n for kn regarded an algebraic subset. It is also the zero set of the zero ideal .0/. We write A as an algebraic set. Examples 2.1. If S is a set of homogeneous linear equations, ai1X1 C C ai nXn D 0; i D 1; : : : ; m; then V .S/ is a subspace of kn. If S is a set of nonhomogeneous linear equations, ai1X1 C C ai nXn D di ; i D 1; : : : ; m; then V .S/ is either empty or is the translate of a subspace of kn. 37 38 2. ALGEBRAIC SETS 2.2. If S consists of the single equation Y 2 D X 3 C aX C b; 4a3 C 27b2 ¤ 0; then V .S/ is an elliptic curve. For example.X 2 1/ We generally visualize algebraic sets as though the field k were R, i.e., we draw the real locus of the curve. However, this can be misleading — see the examples 4.11 and 4.17 below. 2.3. If S consists of the single equation Z2 D X 2 C Y 2; then V .S/ is a cone. 2.4. A nonzero constant polynomial has no zeros, and so the empty set is algebraic. 2.5. The proper algebraic subsets of k are the finite subsets, because a polynomial f .X/ in one variable X has only finitely many roots. 2.6. Some generating sets for an ideal will be more useful than others for determining what the algebraic set is. For example, the ideal a D .X 2 C Y 2 C Z2 1; X 2 C Y 2 Y; X Z/ can be generated by1 X Z; Y 2 2Y C 1; Z2 1 C Y: The middle polynomial has (double) root 1, from which it follows that V .a/ consists of the single point .0; 1; 0/. b. The Hilbert basis theorem In our definition of an algebraic set, we didn’t require the set S of polynomials to be finite, but the Hilbert basis theorem shows that, in fact, every algebraic set is the zero set of a finite set of polynomials. More precisely, the theorem states that every ideal in kŒX1; : : : ; Xn can be generated by a finite set of elements, and we have already observed that a set of generators of an ideal has the same zero set as the ideal. 1This is, in fact, a Gr¨obner basis for the ideal. c. The Zariski topology 39 THEOREM 2.7 (HILBERT BASIS THEOREM). The ring kŒX1; : : : ; Xn is noetherian. As we noted in the proof of 1.32, kŒX1; : : : ; Xn D kŒX1; : : : ; Xn1ŒXn: Thus an induction argument shows that the theorem follows from the next statement. THEOREM 2.8. If A is noetherian, then so also is AŒX . PROOF. We shall show that every ideal in AŒX is finitely generated. Recall that for a polynomial f .X/ D a0X r C a1X r1 C C ar ; ai 2 A; a0 ¤ 0; a0 is called the leading coefficient of f . Let a be a proper ideal in AŒX, and let a.i / denote the set of elements of A that occur as the leading coefficient of a polynomial in a of degree i (we also include 0). Clearly, a.i/ is an ideal in A, and a.i/ a.i C 1/ because, if cX i C 2 a, then X.cX i C / 2 a. Let b be an ideal of AŒX contained in a. Then b.i/ a.i/, and if equality holds for all i , then b D a. To see this, let f be a polynomial in a. Because b.deg f / D a.deg f /, there exists a g 2 b such that deg.f g/ < deg.f /. In other words, f D g C f1 with g 2 b and deg.f1/ < deg.f /. Similarly, f1 D g1 C f2 with g1 2 b and deg.f2/ < deg.f1/. Continuing in this fashion, we find that f D g C g1 C g2 C 2 b.. As A is noetherian, the sequence a.1/ a.2/ a.i/ eventually becomes constant, say a.d / D a.d C 1/ D : : : (and then a.d / contains the leading coefficient of every polynomial in a). For each i d , there exists a finite generating set fai1; ai2; : : : ; ai ni g of a.i/, and for each .i; j /, there exists an fij 2 a with leading coefficient aij . The ideal b of AŒX generated by the (finitely many) fij is contained in a and has the property that b.i/ D a.i / for all i . Therefore b D a, and a is finitely generated. ASIDE 2.9. One may ask how many elements are needed to generate a given ideal a in kŒX1; : : : ; Xn, or, what is not quite the same thing, how many equations are needed to define a given algebraic set V . For n D 1, the ring kŒX is a principal ideal domain, which means that every ideal is generated by a single element. Also, if V is a linear subspace of kn, then linear algebra shows that it is the zero set of n dim.V / polynomials. All one can say in general, is that at least n dim.V / polynomials are needed to define V (see 3.45), but often more are required. Determining exactly how many is an area of active research — see 3.55. c. The Zariski topology Recall that, for ideals a and b in kŒX1; : : : ; Xn, a b H) V .a/ V .b/. PROPOSITION 2.10. There are the following relations: (a) V .0/ D kn; V .kŒX1; : : : ; Xn/ D ;I (b) V .ab/ D V .a \ b/ D V .a/ [ V .b/I (c) V .P i 2I ai / D T i 2I V .ai / for every family of ideals .ai /i 2I . 40 2. ALGEBRAIC SETS PROOF. (a) This is obvious. (b) Note that ab a \ b a; b H) V .ab/ V .a \ b/ V .a/ [ V .b/: For the reverse inclusions, observe that if a … V .a/ [ V .b/, then there exist f 2 a, g 2 b such that f .a/ ¤ 0, g.a/ ¤ 0; but then .fg/.a/ ¤ 0, and so a … V .ab/. (c) Recall that, by definition, P ai consists of all finite sums of the form P fi , fi 2 ai . Thus (c) is obvious. Proposition 2.10 shows that the algebraic subsets of A n satisfy the axioms to be the n: both the whole space and the empty set are algebraic; closed subsets for a topology on A a finite union of algebraic sets is algebraic; an arbitrary intersection of algebraic sets is n for which the closed subsets are exactly the algebraic. Thus, there is a topology on A n. The induced topology on a algebraic subsets — this is called the Zariski topology on A subset V of A n is called the Zariski topology on V . The Zariski topology has many strange properties, but it is nevertheless of great importance. For the Zariski topology on k, the closed subsets are jus
t the finite sets and the whole space, and so the topology is not Hausdorff (in fact, there are no disjoint nonempty open subsets at all). We shall see in 2.68 below that the proper closed subsets of k2 are finite 2 are much coarser unions of points and curves. Note that the Zariski topologies on C and C (have fewer open sets) than the complex topologies. d. The Hilbert Nullstellensatz n and the ideals of We wish to examine the relation between the algebraic subsets of A kŒX1; : : : ; Xn more closely, but first we must answer the question of when a collection S of polynomials has a common zero, i.e., when the system of equations g.X1; : : : ; Xn/ D 0; g 2 S; is “consistent”. Obviously, equations gi .X1; : : : ; Xn/ D 0; i D 1; : : : ; m are inconsistent if there exist fi 2 kŒX1; : : : ; Xn such that P fi gi D 1, i.e., if 1 2 .g1; : : : ; gm/ or, equivalently, .g1; : : : ; gm/ D kŒX1; : : : ; Xn. The next theorem provides a converse to this. THEOREM 2.11 (HILBERT NULLSTELLENSATZ). 2 Every proper ideal a in kŒX1; : : : ; Xn has a zero in kn. A point P D .a1; : : : ; an/ in kn defines a homomorphism “evaluate at P ” kŒX1; : : : ; Xn ! k; f .X1; : : : ; Xn/ 7! f .a1; : : : ; an/; whose kernel contains a if P 2 V .a/. Conversely, from a homomorphism 'W kŒX1; : : : ; Xn ! k of k-algebras whose kernel contains a, we obtain a point P in V .a/, namely, 2Nullstellensatz = zero-points-theorem. P D .'.X1/; : : : ; '.Xn//: e. The correspondence between algebraic sets and radical ideals 41 Thus, to prove the theorem, we have to show that there exists a k-algebra homomorphism kŒX1; : : : ; Xn=a ! k. Since every proper ideal is contained in a maximal ideal (see p. 16), it suffices to prove this for a maximal ideal m. Then K defD kŒX1; : : : ; Xn=m is a field, and it is finitely generated as an algebra over k (with generators X1 C m; : : : ; Xn C m/. To complete the proof, we must show that K D k. The next lemma accomplishes this. In the next lemma, we need to allow k to be arbitrary in order to make the induction in the proof work. We shall also need to use that kŒX has infinitely many distinct monic irreducible polynomials. When k is infinite, the polynomials X a, a 2 k, are distinct and irreducible. When k is finite, we can adapt Euclid’s argument: if p1; : : : ; pr are monic irreducible polynomials in kŒX, then p1 pr C 1 is divisible by a monic irreducible polynomial distinct from p1; : : : ; pr . LEMMA 2.12 (ZARISKI’S LEMMA). Let k K be fields, not necessarily algebraically closed. If K is finitely generated as an algebra over k, then K is algebraic over k. (Hence K D k if k is algebraically closed.) In other words, if K is finitely generated as a ring over k, then it is finitely generated as a module. PROOF. We shall prove this by induction on r, the minimum number of elements required to generate K as a k-algebra. The case r D 0 being trivial, we may suppose that K D kŒx1; : : : ; xr ; r 1: If K is not algebraic over k, then at least one xi , say x1, is not algebraic over k. Then, kŒx1 is a polynomial ring in one symbol over k, and its field of fractions k.x1/ is a subfield of K. Clearly K is generated as a k.x1/-algebra by x2; : : : ; xr , and so the induction hypothesis implies that x2; : : : ; xr are algebraic over k.x1/. From 1.40, we see that there exists a c 2 kŒx1 such that cx2; : : : ; cxr are integral over kŒx1. Let f 2 k.x1/. Then f 2 K D kŒx1; : : : ; xr and so, for a sufficiently large N , cN f 2 kŒx1; cx2; : : : ; cxr . Therefore cN f is integral over kŒx1 by 1.38, which implies that cN f 2 kŒx1 because kŒx1 is integrally closed in k.x1/ (1.43). But this contradicts the fact that that kŒx1 has infinitely many distinct monic irreducible polynomials that can occur as the denominator of an f in k.x1/. e. The correspondence between algebraic sets and radical ideals The ideal attached to a subset of kn For a subset W of kn, we write I.W / for the set of polynomials that are zero on W : I.W / D ff 2 kŒX1; : : : ; Xn j f .P / D 0 all P 2 W g: Clearly, it is an ideal in kŒX1; : : : ; Xn. There are the following relations: (a) V W H) I.V / I.W /I (b) I.;/ D kŒX1; : : : ; Xn; I.kn/ D 0I (c) I.S Wi / D T I.Wi /. Only the statement I.kn/ D 0 is (perhaps) not obvious. It says that every nonzero polynomial in kŒX1; : : : ; Xn is nonzero at some point of kn. This is true for any infinite field k (see Exercise 1-1). Alternatively, it follows from the strong Hilbert Nullstellensatz (2.19 below). 42 2. ALGEBRAIC SETS EXAMPLE 2.13. Let P be the point .a1; : : : ; an/, and let mP D .X1 a1; : : : ; Xn an/. Clearly I.P / mP , but mP is a maximal ideal, because “evaluation at .a1; : : : ; an/” defines an isomorphism kŒX1; : : : ; Xn=.X1 a1; : : : ; Xn an/ ! k: As I.P / is a proper ideal, it must equal mP : PROPOSITION 2.14. Let W be a subset of kn. Then V I.W / is the smallest algebraic subset of kn containing W . In particular, V I.W / D W if W is an algebraic set. PROOF. Certainly V I.W / is an algebraic set containing W . Let V D V .a/ be another algebraic set containing W . Then a I.W /, and so V .a/ V I.W /. Radicals of ideals The radical of an ideal a in a ring A is rad.a/ defD ff j f r 2 a, some r 2 Ng: PROPOSITION 2.15. Let a be an ideal in a ring A. (a) The radical of a is an ideal. (b) rad.rad.a// D rad.a/. PROOF. (a) If a 2 rad.a/, then clearly f a 2 rad.a/ for all f 2 A. Suppose that a; b 2 rad.a/, with say ar 2 a and bs 2 a. When we expand .a C b/rCs using the binomial theorem, we find that every term has a factor ar or bs, and so lies in a. (b) If ar 2 rad.a/, then ars D .ar /s 2 a for some s. An ideal is said to be radical if it equals its radical. Thus a is radical if and only if the ring A=a is reduced, i.e., without nonzero nilpotent elements. Since integral domains are reduced, prime ideals (a fortiori, maximal ideals) are radical. Note that rad.a/ is radical (2.15b), and hence is the smallest radical ideal containing a. If a and b are radical, then a \ b is radical, but a C b need not be: consider, for example, a D .X 2 Y / and b D .X 2 C Y /; they are both prime ideals in kŒX; Y , but X 2 2 a C b, X … a C b. (See 2.22 below.) The strong Nullstellensatz For a polynomial f and point P 2 kn, f r .P / D f .P /r . Therefore f r is zero on the same set as f , and it follows that the ideal I.W / is radical for every subset W kn. In particular, I V .a/ rad.a/. The next theorem states that these two ideals are equal. THEOREM 2.16 (STRONG NULLSTELLENSATZ). For every ideal a in kŒX1; : : : ; Xn, in particular, I V .a/ D a if a is a radical ideal. I V .a/ D rad.a/I e. The correspondence between algebraic sets and radical ideals 43 PROOF. We have already noted that I V .a/ rad.a/. For the reverse inclusion, we have to show that if a polynomial h vanishes on V .a/, then hN 2 a for some N > 0. We may assume h ¤ 0. Let g1; : : : ; gm generate a, and consider the system of m C 1 equations in n C 1 symbols, gi .X1; : : : ; Xn/ D 0; 1 Y h.X1; : : : ; Xn/ D 0: i D 1; : : : ; m; If .a1; : : : ; an; b/ satisfies the first m equations, then .a1; : : : ; an/ 2 V .a/; consequently, h.a1; : : : ; an/ D 0, and .a1; : : : ; an; b/ doesn’t satisfy the last equation. Therefore, the equations are inconsistent, and so, according to the original Nullstellensatz, there exist fi 2 kŒX1; : : : ; Xn; Y such that 1 D m X i D1 fi gi C fmC1 .1 Y h/ (in the ring kŒX1; : : : ; Xn; Y ). On applying the homomorphism Xi 7! Xi Y 7! h1 W kŒX1; : : : ; Xn; Y ! k.X1; : : : ; Xn/ to the above equality, we obtain the identity 1 D m X i D1 fi .X1; : : : ; Xn; h1/ gi .X1; : : : ; Xn/ () in k.X1; : : : ; Xn/. Clearly fi .X1; : : : ; Xn; h1/ D polynomial in X1; : : : ; Xn hNi for some Ni . Let N be the largest of the Ni . On multiplying () by hN we obtain an equation hN D m X i D1 (polynomial in X1; : : : ; Xn/ gi .X1; : : : ; Xn/; which shows that hN 2 a. COROLLARY 2.17. The map a 7! V .a/ defines a one-to-one correspondence between the set of radical ideals in kŒX1; : : : ; Xn and the set of algebraic subsets of kn; its inverse is I . PROOF. We know that I V .a/ D a if a is a radical ideal (2.16), and that V I.W / D W if W is an algebraic set (2.14). Therefore, I and V are inverse bijections. COROLLARY 2.18. The radical of an ideal in kŒX1; : : : ; Xn is equal to the intersection of the maximal ideals containing it. PROOF. Let a be an ideal in kŒX1; : : : ; Xn. Because maximal ideals are radical, every maximal ideal containing a also contains rad.a/, and so rad.a/ \ ma m. 44 2. ALGEBRAIC SETS For each P D .a1; : : : ; an/ 2 kn, the ideal mP D .X1 a1; : : : ; Xn an/ is maximal in kŒX1; : : : ; Xn, and (see 2.13). Thus f 2 mP ” f .P / D 0 mP a ” P 2 V .a/. If f 2 mP for all P 2 V .a/, then f is zero on V .a/, and so f 2 I V .a/ D rad.a/. We have shown that rad.a/ \ P 2V .a/ mP \ ma m. Remarks 2.19. Because V .0/ D kn, I.kn/ D I V .0/ D rad.0/ D 0I in other words, only the zero polynomial is zero on the whole of kn. In fact, this holds whenever k is infinite (Exercise 1-1). 2.20. The one-to-one correspondence in Corollary 2.17 is order reversing. Therefore the maximal proper radical ideals correspond to the minimal nonempty algebraic sets. But the maximal proper radical ideals are simply the maximal ideals in kŒX1; : : : ; Xn, and the minimal nonempty algebraic sets are the one-point sets. As I..a1; : : : ; an// D .X1 a1; : : : ; Xn an/ (see 2.13), this shows that the maximal ideals of kŒX1; : : : ; Xn are exactly the ideals .X1 a1; : : : ; Xn an/ with .a1; : : : ; an/ 2 kn. 2.21. The algebraic set V .a/ is empty if and only if a D kŒX1; : : : ; Xn (Nullstellensatz, 2.11). 2.22. Let W and W 0 be algebraic sets. As W \ W 0 is the largest algebraic subset contained in both W and W 0, I.W \ W 0/ must be the smallest radical ideal containing both I.W / and I.W 0/: I.W \ W 0/ D rad.I.W / C I.W 0//: For example, let W D V .X 2 Y / and W 0 D V .X 2 C Y /; then I.W \ W 0/ D rad.X 2; Y / D .X; Y / (assuming characteristic ¤ 2/. Note that W \ W 0 D f.0; 0/g, but when rea
lized as the intersection of Y D X 2 and Y D X 2, it has “multiplicity 2”. P Q ! P ! and V W be the set of subsets of kn and let 2.23. Let Then I W and (see FT 7.19). It follows that I and V define a one-to-one correspondence between I . and V . ideals, and (by definition) V . 2.17. Q / P / consists exactly of the radical / consists of the algebraic subsets. Thus we recover Corollary define a simple Galois correspondence between /. But the strong Nullstellensatz shows that I . be the set of subsets of kŒX1; : : : ; Xn. Q Q Q Q P P P •V(X2−Y)V(X2+Y) f. Finding the radical of an ideal 45 n capture only part of the ideal theory of kŒX1; : : : ; Xn ASIDE 2.24. The algebraic subsets of A because two ideals with the same radical correspond to the same algebraic subset. There is a finer n are in notion of an algebraic scheme over k for which the closed algebraic subschemes of A one-to-one correspondence with the ideals in kŒX1; : : : ; Xn (see Chapter 11 on my website). f. Finding the radical of an ideal Typically, an algebraic set V is defined by a finite set of polynomials fg1; : : : ; gsg, and we need to find I.V / D rad.g1; : : : ; gs/. PROPOSITION 2.25. A polynomial h 2 rad.a/ if and only if 1 2 .a; 1 Y h/ (the ideal in kŒX1; : : : ; Xn; Y generated by the elements of a and 1 Y h). PROOF. We saw that 1 2 .a; 1 Y h/ implies h 2 rad.a/ in the course of proving 2.16. Conversely, from the identities 1 D Y N hN C .1 Y N hN / D Y N hN C .1 Y h/ .1 C Y h C C Y N 1hN 1/ we see that, if hN 2 a, then 1 2 a C .1 Y h/. Given a set of generators of an ideal, there is an algorithm for deciding whether or not a polynomial belongs to the ideal, and hence an algorithm for deciding whether or not a polynomial belongs to the radical of the ideal. There are even algorithms for finding a set of generators for the radical. These algorithms have been implemented in the computer algebra systems CoCoA and Macaulay 2. g. Properties of the Zariski topology n. n and on an algebraic subset of A We now examine more closely the Zariski topology on A n, V I.W / is the closure of W , and 2.17 says Proposition 2.14 says that, for a subset W of A n and the radical that there is a one-to-one correspondence between the closed subsets of A ideals of kŒX1; : : : ; Xn. Under this correspondence, the closed subsets of an algebraic set V correspond to the radical ideals of kŒX1; : : : ; Xn containing I.V /. PROPOSITION 2.26. Let V be an algebraic subset of A n. (a) The points of V are closed for the Zariski topology. (b) Every ascending chain of open subsets U1 U2 of V eventually becomes constant. Equivalently, every descending chain of closed subsets of V eventually becomes constant. (c) Every open covering of V has a finite subcovering. PROOF. (a) We have seen that f.a1; : : : ; an/g is the algebraic set defined by the ideal .X1 a1; : : : ; Xn an/. (b) We prove the second statement. A sequence V1 V2 of closed subsets of V gives rise to a sequence of radical ideals I.V1/ I.V2/ : : :, which eventually becomes constant because kŒX1; : : : ; Xn is noetherian. (c) Given an open covering of V , let can be expressed as a finite union of sets in the covering. If every element of ascending chain of sets in be the collection of open subsets of V that does not contain V , then is properly contained in another element, and so there exists an infinite (axiom of dependent choice), contradicting (b). U U U U 46 2. ALGEBRAIC SETS A topological space whose points are closed is said to be T1; the condition means that, for any pair of distinct points, each has an open neighbourhood not containing the other. A topological space having the property (b) is said to be noetherian. The condition is equivalent to the following: every nonempty set of closed subsets of V has a minimal element. A topological space having property (c) is said to be quasicompact (by Bourbaki at least; others call it compact, but Bourbaki requires a compact space to be Hausdorff). The proof of (c) shows that every noetherian space is quasicompact. Since an open subset of a noetherian space is again noetherian, it is also quasicompact. h. Decomposition of an algebraic set into irreducible algebraic sets A topological space is said to be irreducible if it is not the union of two proper closed subsets. Equivalent conditions: every pair of nonempty open subsets has nonempty intersection; every nonempty open subset is dense. By convention, the empty space is not irreducible. Obviously, every nonempty open subset of an irreducible space is irreducible. In a Hausdorff topological space, any two points have disjoint open neighbourhoods. Therefore, the only irreducible Hausdorff spaces are those consisting of a single point. PROPOSITION 2.27. An algebraic set W is irreducible if and only if I.W / is prime. PROOF. Let W be an irreducible algebraic set, and let fg 2 I.W / — we have to show that either f or g is in I.W /. At each point of W , either f is zero or g is zero, and so W V .f / [ V .g/. Hence W D .W \ V .f // [ .W \ V .g//: As W is irreducible, one of these sets, say W \ V .f /, must equal W . But then f 2 I.W /. Let W be an algebraic set such that I.W / is prime, and let W D V .a/ [ V .b/ with a and b radical ideals — we have to show that W equals V .a/ or V .b/. The ideal a \ b is radical, and V .a \ b/ D V .a/ [ V .b/ (2.10); hence I.W / D a \ b. If W ¤ V .a/, then there exists an f 2 a X I.W /. Let g 2 b. Then fg 2 a \ b D I.W /, and so g 2 I.W / (because I.W / is prime). We conclude that b I.W /, and so V .b/ V .I.W // D W . SUMMARY 2.28. There are one-to-one correspondences, radical ideals in kŒX1; : : : ; Xn $ algebraic subsets of A prime ideals in kŒX1; : : : ; Xn $ irreducible algebraic subsets of A n n maximal ideals in kŒX1; : : : ; Xn $ one-point sets of A n: EXAMPLE 2.29. Let f 2 kŒX1; : : : ; Xn. We saw (1.32) that kŒX1; : : : ; Xn is a unique factorization domain, and so .f / is a prime ideal if and only if f is irreducible (1.33). Thus f is irreducible H) V .f / is irreducible. On the other hand, suppose f factors as f D Y f mi i ; fi distinct irreducible polynomials. h. Decomposition of an algebraic set into irreducible algebraic sets 47 Then .f / D T f mi i rad.f / D T .fi / V .f / D S V .fi / f mi i distinct ideals .fi / distinct prime ideals V .fi / distinct irreducible algebraic sets. LEMMA 2.30. Let W be an irreducible topological space. If W D W1 [ : : : [ Wr with each Wi closed, then W is equal to one of the Wi . PROOF. When r D 2, the statement is the definition of “irreducible”. Suppose that r > 2. Then W D W1 [ .W2 [ : : : [ Wr /, and so W D W1 or W D .W2 [ : : : [ Wr /; if the latter, then W D W2 or W3 [ : : : [ Wr , etc. PROPOSITION 2.31. Let V be a noetherian topological space. Then V is a finite union of irreducible closed subsets, V D V1 [ : : : [ Vm. If the decomposition is irredundant in the sense that there are no inclusions among the Vi , then the Vi are uniquely determined up to order. PROOF. Suppose that V cannot be written as a finite union of irreducible closed subsets. Then, because V is noetherian, there will be a nonempty closed subset W of V that is minimal among those that cannot be written in this way. But W itself cannot be irreducible, and so W D W1 [ W2, with W1 and W2 proper closed subsets of W . Because W was minimal, each Wi is a finite union of irreducible closed subsets. Hence W is also, which is a contradiction. Suppose that V D V1 [ : : : [ Vm D W1 [ : : : [ Wn are two irredundant decompositions of V . Then Vi D S j .Vi \ Wj /, and so, because Vi is irreducible, Vi D Vi \ Wj for some j . Consequently, there is a function f W f1; : : : ; mg ! f1; : : : ; ng such that Vi Wf .i/ for each i . Similarly, there is a function gW f1; : : : ; ng ! f1; : : : ; mg such that Wj Vg.j / for each j . Since Vi Wf .i / Vgf .i /, we must have gf .i/ D i and Vi D Wf .i/; similarly fg D id. Thus f and g are bijections, and the decompositions differ only in the numbering of the sets. The Vi given uniquely by the proposition are called the irreducible components of V . They are exactly the maximal irreducible closed subsets of V .3 In Example 2.29, the V .fi / are the irreducible components of V .f /. An algebraic set with two irreducible components. 3In fact, they are exactly the maximal irreducible subsets of V because the closure of an irreducible subset is also irreducible. 48 2. ALGEBRAIC SETS COROLLARY 2.32. The radical of an ideal a in kŒX1; : : : ; Xn is a finite intersection of prime ideals, a D p1 \ : : : \ pn. If there are no inclusions among the pi , then the pi are uniquely determined up to order (and they are exactly the minimal prime ideals containing a). PROOF. Write V .a/ as a union of its irreducible components, V .a/ D Sn pi D I.Vi /. Then rad.a/ D p1 \ : : : \ pn because they are both radical ideals and i D1 Vi , and let V .rad.a// D V .a/ D [ V .pi / 2.10bD V . \ p/: i The uniqueness similarly follows from the proposition. Remarks 2.33. An irreducible topological space is connected, but a connected topological space 2, need not be irreducible. For example, V .X1X2/ is the union of the coordinate axes in A n is disconnected if and which is connected but not irreducible. An algebraic subset V of A only if there exist radical ideals a and b such that V is the disjoint union of V .a/ and V .b/, that is, V D V .a/ [ V .b/ D V .a \ b/ ” a \ b D I.V / ; D V .a/ \ V .b/ D V .a C b/ ” a C b D kŒX1; : : : ; Xn: Note that then kŒV ' kŒX1; : : : ; Xn a kŒX1; : : : ; Xn b (Chinese remainder theorem, 1.1). 2.34. A Hausdorff space is noetherian if and only if it is finite, in which case its irreducible components are the one-point sets. 2.35. In kŒX1; : : : ; Xn, a principal ideal .f / is radical if and only if f is square-free, in which case f is a product of distinct irreducible polynomials, f D f1 : : : fr , and .f / D .f1/ \ : : : \ .fr /. 2.36. In a noetherian ring, every proper ideal a has a decomposition into primary ideals: a D T qi (see CA 19). For ra
dical ideals, this becomes a simpler decomposition into prime ideals, as in the corollary. For an ideal .f / with f D Q f mi , the primary decomposition is the decomposition .f / D T.f mi / in Example 2.29. i i i. Regular functions; the coordinate ring of an algebraic set Let V be an algebraic subset of A n, and let I.V / D a. The coordinate ring of V is kŒV defD kŒX1; : : : ; Xn=a. This is a finitely generated k-algebra. It is reduced (because a is radical), but not necessarily an integral domain. An f 2 kŒX1; : : : ; Xn defines a function P 7! f .P /W V ! k: i. Regular functions; the coordinate ring of an algebraic set 49 Functions of this form are said to be regular. Two polynomials f; g 2 kŒX1; : : : ; Xn define the same regular function on V if and only if they define the same element of kŒV , and so kŒV is the ring of regular functions on V . The coordinate function xi W V ! k; .a1; : : : ; an/ 7! ai is regular, and kŒV D kŒx1; : : : ; xn. In other words, the coordinate ring of an algebraic set V is the k-algebra generated by the coordinate functions on V . For an ideal b in kŒV , set V .b/ D fP 2 V j f .P / D 0, all f 2 bg — it is a closed subset of V . Let W D V .b/. The quotient maps kŒX1; : : : ; Xn kŒV D kŒX1; : : : ; Xn a kŒW D kŒV b send a regular function on kn to its restriction to V and then to its restriction to W . Write for the quotient map kŒX1; : : : ; Xn kŒV . Then b 7! 1.b/ is a bijection from the set of ideals of kŒV to the set of ideals of kŒX1; : : : ; Xn containing a, under which radical, prime, and maximal ideals correspond to radical, prime, and maximal ideals (because each of these conditions can be checked on the quotient ring, and kŒX1; : : : ; Xn= 1.b/ ' kŒV =b). Clearly V . 1.b// D V .b/; and so b 7! V .b/ is a bijection from the set of radical ideals in kŒV to the set of algebraic sets contained in V . Now 2.28 holds for ideals in kŒV and algebraic subsets of V , radical ideals in kŒV $ algebraic subsets of V prime ideals in kŒV $ irreducible algebraic subsets of V maximal ideals in kŒV $ one-point sets of V: Moreover (see 2.33), the decompositions of a closed subset W of V into a disjoint union of closed subsets correspond to pairs of radical ideals a; b 2 kŒV such that kŒW D kŒV =a \ b ' kŒV =a kŒV =b: For h 2 kŒV , set D.h/ D fa 2 V j h.a/ ¤ 0g: It is an open subset of V , because its complement is the closed set V ..h//. It is empty if and only if h is zero (2.19). PROPOSITION 2.37. The sets D.h/, h 2 kŒV , are a base for the topology on V , i.e., each D.h/ is open, and every open set is a (finite) union of this form. PROOF. We have already observed that D.h/ is open. Every open subset U V is the complement of a set of the form V .b/, with b an ideal in kŒV . If f1; : : : ; fm generate b, then U D S D.fi /. 50 2. ALGEBRAIC SETS The D.h/ are called the basic (or principal) open subsets of V . We sometimes write Vh for D.h/. Note that D.h/ D.h0/ ” V .h/ V .h0/ ” rad..h// rad..h0// ” hr 2 .h0/ some r ” hr D h0g, some g: Some of this should look familiar: if V is a topological space, then the zero set of a family of continuous functions f W V ! R is closed, and the set where a continuous function is nonzero is open. Let V be an irreducible algebraic set. Then I.V / is a prime ideal, and so kŒV is an integral domain. Let k.V / be its field of fractions — k.V / is called the function field of V or the field of rational functions on V . j. Regular maps Let W km and V kn be algebraic sets. Let xi denote the ith coordinate function .b1; : : : ; bn/ 7! bi W V ! k on V . The i th component of a map 'W W ! V is 'i D xi ı '. Thus, ' is the map P 7! '.P / D .'1.P /; : : : ; 'n.P //W W ! V kn: DEFINITION 2.38. A continuous map 'W W ! V of algebraic sets is regular if each of its components 'i is a regular function. As the coordinate functions generate kŒV , a continuous map ' is regular if and only if f ı ' is a regular function on W for every regular function f on V . Thus a regular map 'W W ! V of algebraic sets defines a homomorphism f 7! f ı 'W kŒV ! kŒW of k-algebras, which we sometimes denote by '. k. Hypersurfaces; finite and quasi-finite maps A hypersurface in A polynomial, nC1 is the algebraic set H defined by a single nonzero nonconstant We examine the regular map H ! A n defined by the projection H W f .T1; : : : ; Tn; X/ D 0. .t1; : : : ; tn; x/ 7! .t1; : : : ; tn/: We can write f in the form f D a0X m C a1X m1 C C am; ai 2 kŒX1; : : : ; Xm; a0 ¤ 0; m 2 N: k. Hypersurfaces; finite and quasi-finite maps 51 We assume that m ¤ 0, i.e., that X occurs in f (otherwise, H is a cylinder over a hypersurface n over .t1; : : : ; tn/ 2 kn is the set of points .t1; : : : ; tn; c/ n). The fibre of the map H ! A in A such that c is a root of the polynomial a0.t /X m C a1.t/X m1 C C am.t/; ai .t/ defD ai .t1; : : : ; tn/ 2 k: Suppose first that a0 2 k, so that a0.t/ is a nonzero constant independent of t . Then the fibre over t consists of the roots of the polynomial a0X m C a1.t/X m1 C C am.t/; (14) in kŒX. Counting multiplicities, there are exactly m of these. More precisely, let D be the discriminant of the polynomial a0X m C a1X m1 C C am: Then D 2 kŒX1; : : : ; Xm, and the fibre has exactly m points over the open subset D ¤ 0, and fewer then m points over the closed subset D D 0.4 We can picture it schematically as follows (m D 3): Now drop the condition that a0 is constant. For certain t , the degree of (14) may drop, which means that some roots have “disappeared off to infinity”. Consider, for example, f .T; X/ D TX 1; for each t ¤ 0, there is one point .t; 1=t/, but there is no point with t D 0 (see the figure p. 71). Worse, for certain t all coefficients may be zero, in which case n such that the number the fibre is a line. There is a nested collection of closed subsets of A of points in the fibre (counting multiplicities) drops as you pass to a smaller subset, except that over the smallest subset the fibre may be a full line. DEFINITION 2.39. Let 'W W ! V be a regular map of algebraic subsets, and let 'W kŒV ! kŒW be the map f 7! f ı '. (a) The map ' is dominant if '.W / is dense in V . (b) The map ' is quasi-finite if '1.P / is finite for all P 2 V . (c) The map ' is finite if kŒW is a finite kŒV -algebra. As we shall see (8.28), finite maps are indeed quasi-finite. As kŒW is finitely generated as a k-algebra, a fortiori as a kŒV -algebra, to say that kŒW is a finite kŒV -algebra means that it is integral over kŒV (1.36). The map H ! A n considered above is finite if and only if a0 is constant, and quasi-finite if and only if the polynomials a0; : : : ; am have no common zero in kn. 4I’m ignoring the possibility that D is identically zero. Then the open set where D ¤ 0 is empty. This case occurs when the characteristic is p ¤ 0, and f is a polynomial in T1; : : : ; Tn; and X p. HAn 52 2. ALGEBRAIC SETS PROPOSITION 2.40. A regular map 'W W ! V is dominant if and only if 'W kŒV ! kŒW is injective. PROOF. Let f 2 kŒV . If the image of ' is dense, then f ı ' D 0 H) f D 0: On the other hand, if the image of ' is not dense, then its closure Z is a proper closed subset of V , and so there exists a nonzero regular function f zero on Z. Then f ı ' D 0. PROPOSITION 2.41. A dominant finite map is surjective. PROOF. Let 'W W ! V be dominant and finite. Then 'W kŒV ! kŒW is injective, and kŒW is integral over the image of kŒV . According to the going-up theorem (1.53), for every maximal ideal m of kŒV there exists a maximal ideal n of kŒW such that m D n \ kŒV . Because of the correspondence between points and maximal ideals, this implies that ' is surjective. l. Noether normalization theorem Let H be a hypersurface in A projection map .x1; : : : ; xnC1/ 7! .x1; : : : ; xn/W A nC1. We show that, after a linear change of coordinates, the n. n defines a finite map H ! A nC1 ! A PROPOSITION 2.42. Let H W f .X1; : : : ; XnC1/ D 0 be a hypersurface in A nC1. There exist c1; : : : ; cn 2 k such that the map H ! A n defined by .x1; : : : ; xnC1/ 7! .x1 c1xnC1; : : : ; xn cnxnC1/ is finite. PROOF. Let c1; : : : ; cn 2 k. In terms of the coordinates x0 i H is the zero set of D xi ci xnC1, the hyperplane f .X1 C c1XnC1; : : : ; Xn C cnXnC1; XnC1/ D a0X m nC1 C a1X m1 nC1 C : The next lemma shows that the ci can be chosen so that a0 is a nonzero constant. This n defined by .x1; : : : ; xnC1/ 7! .x0 implies that the map H ! A n/ is finite. 1; : : : ; x0 LEMMA 2.43. Let k be an infinite field (not necessarily algebraically closed), and let f 2 kŒX1; : : : ; Xn; T . There exist c1; : : : ; cn 2 k such that f .X1 C c1T; : : : ; Xn C cnT; T / D a0T m C a1T m1 C C am with a0 2 k and all ai 2 kŒX1; : : : ; Xn. PROOF. Let F be the homogeneous part of highest degree of f and let r D deg.F /. Then F .X1 C c1T; : : : ; Xn C cnT; T / D F .c1; : : : ; cn; 1/T r C terms of degree < r in T , because the polynomial F .X1 C c1T; : : : ; Xn C cnT; T / is still homogeneous of degree r in X1; : : : ; Xn; T , and so the coefficient of the monomial T r can be obtained by setting each Xi equal to zero in F and T to 1. As F .X1; : : : ; Xn; T / is a nonzero homogeneous l. Noether normalization theorem 53 polynomial, F .X1; : : : ; Xn; 1/ is a nonzero polynomial, and so we can choose the ci so that F .c1; : : : ; cn; 1/ ¤ 0 (Exercise 1-1). Now f .X1 C c1T; : : : ; Xn C cnT; T / D F .c1; : : : ; cn; 1/T r C terms of degree < r in T; with F .c1; : : : ; cn; 1/ 2 k, as required. In fact, every algebraic set V admits a finite surjective map to A d for some d . THEOREM 2.44. Let V be an algebraic set. For some natural number d , there exists a finite surjective map 'W V ! A d . This follows from the next statement applied to A D kŒV : the regular functions d , which is finite and surjective because kŒx1; : : : ; xd ! A x1; : : : ; xd define a map V ! A is finite and injective. THEOREM 2.45 (NOETHER NORMALIZATION THEOREM). Let A be a finitely generated k-algebra. There exist elements x1; : : : ; xd 2 A that are algebraically indepen
dent over k, and such that A is finite over kŒx1; : : : ; xd . It is not necessary to assume that A is reduced in Theorem 2.45, nor that k is algebraically closed, although the proof we give requires it to be infinite (for the general proof, see CA 8.1). Let A D kŒx1; : : : ; xn. We prove the theorem by induction on n. If the xi are algebraically independent, there is nothing to prove. Otherwise, the next lemma shows that A is finite over a subring B D kŒy1; : : : ; yn1. By induction, B is finite over a subring C D kŒz1; : : : ; zd with z1; : : : ; zd algebraically independent, and A is finite over C . LEMMA 2.46. Let A D kŒx1; : : : ; xn be a finitely generated k-algebra, and let fx1; : : : ; xd g be a maximal algebraically independent subset of fx1; : : : ; xng. If n > d , then there exist c1; : : : ; cd 2 k such that A is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1. PROOF. By assumption, the set fx1; : : : ; xd ; xng is algebraically dependent, and so there exists a nonzero f 2 kŒX1; : : : ; Xd ; T such that f .x1; : : : ; xd ; xn/ D 0: (15) Because fx1; : : : ; xd g is algebraically independent, T occurs in f , and so f .X1; : : : ; Xd ; T / D a0T m C a1T m1 C C am with ai 2 kŒX1; : : : ; Xd , a0 ¤ 0, and m > 0. If a0 2 k, then (15) shows that xn is integral over kŒx1; : : : ; xd . Hence x1; : : : ; xn are integral over kŒx1; : : : ; xn1, and so A is finite over kŒx1; : : : ; xn1. If a0 … k, then, for a suitable choice of .c1; : : : ; cd / 2 k, the polynomial g.X1; : : : ; Xd ; T / defD f .X1 C c1T; : : : ; Xd C cd T; T / takes the form g.X1; : : : ; Xd ; T / D bT r C b1T C C br with b 2 k (see 2.43). As g.x1 c1xn; : : : ; xd cd xn; xn/ D 0 this shows that xn is integral over kŒx1 c1xn; : : : ; xd cd xn, and so A is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1 as before. (16) 54 Remarks 2. ALGEBRAIC SETS 2.47. For an irreducible algebraic subset V of A prove the following more precise statement: n, the above argument can be modified to Let x1; : : : ; xn be the coordinate functions on V ; after possibly renumbering the coordinates, we may suppose that fx1; : : : ; xd g is a maximal algebraically independent subset of fx1; : : : ; xng; then there exist cij 2 k such that the map .x1; : : : ; xn/ 7! x1 n X j Dd C1 c1j xj ; : : : ; xd n X W A n ! A d cdj xj j Dd C1 induces a finite surjective map V ! A d : Indeed, Lemma 2.46 shows that there exist c1; : : : ; cn 2 k such that kŒV is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1. Now fx1; : : : ; xd g is algebraically dependent on fx1 c1xn; : : : ; xd cd xng. If the second set were not algebraically independent, we could drop one of its elements, but this would contradict 1.61. Therefore fx1 c1xn; : : : ; xd cd xng is a maximal algebraically independent subset of fx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1g and we can repeat the argument. m. Dimension The dimension of a topological space Let V be a noetherian topological space whose points are closed. DEFINITION 2.48. The dimension of V is the supremum of the lengths of the chains V0 V1 Vd of distinct irreducible closed subsets (the length of the displayed chain is d ). 2.49. Let V1; : : : ; Vm be the irreducible components of V . Then (obviously) dim.V / D max i .dim.Vi //: 2.50. Assume that V is irreducible, and let W be a proper closed subspace of V . Then every chain W0 W1 in W extends to a chain V W0 , and so dim.W / C 1 dim.V /. If dim.V / < 1, then dim.W / < dim.V /. Thus an irreducible topological space V has dimension 0 if and only if it is a point; it has dimension 1 if and only if every proper closed subset is a point; and, inductively, V has dimension n if and only if every proper closed subset has dimension n 1. The dimension of an algebraic set DEFINITION 2.51. The dimension of an algebraic set is its dimension as a topological space. Because of the correspondence between the prime ideals in kŒV and irreducible closed subsets of V , dim.V / D Krull dimension of kŒV : m. Dimension 55 Note that, if V1; : : : ; Vm are the irreducible components of V , then dim V D max i dim.Vi /: When the Vi all have the same dimension d , we say that V has pure dimension d . A one-dimensional algebraic set is called a curve; a two-dimensional algebraic set is called a surface; and an n-dimensional algebraic set is called an n-fold. Let V be an irreducible algebraic set and W an algebraic subset of V . If W is irreducible, then its codimension in V is codimV W D dim V dim W: Dimension and transcendent degree THEOREM 2.52. Let V be an irreducible algebraic set. Then dim.V / D tr degkk.V /: The proof will occupy the rest of this subsection. Let A be an arbitrary commutative ring. Let x 2 A, and let Sfxg denote the multiplicative subset of A consisting of the elements of the form xn.1 ax/; n 2 N; a 2 A: The boundary Afxg of A at x is defined to be the ring of fractions S 1 fxgA. We write dim.A/ for the Krull dimension of A. PROPOSITION 2.53. Let A be a ring and let n 2 N. Then dim.A/ n ” for all x 2 A, dim.Afxg/ n 1: PROOF. We shall use (1.14) that there is a one-to-one correspondence between the prime ideals of S 1A and the prime ideals of A disjoint from S . We begin with two observations. (a) For every x 2 A and maximal ideal m A, m \ Sfxg ¤ ;. Indeed, if x 2 m, then certainly x 2 m \ Sfxg. On the other hand, if x … m, then it is invertible modulo m, and so there exists an a 2 A such that 1 ax 2 m (hence also m \ Sfxg). (b) Let m be a maximal ideal, and let p be a prime ideal contained in m; for every x 2 m X p, we have p \ Sfxg D ;. Indeed, if xn.1 ax/ 2 p, then 1 ax 2 p (as x … p/; hence 1 ax 2 m, and so 1 2 m, which is a contradiction. Statement (a) shows that every chain of prime ideals beginning with a maximal ideal is shortened when passing from A to Afxg, while statement (b) shows that a maximal chain of length n is shortened only to n 1 when x is chosen appropriately. From this, the proposition is follows. PROPOSITION 2.54. Let A be an integral domain, and let k be a subfield of A. Then dim.A/ tr degkF .A/; where F .A/ is the field of fractions of A. 56 2. ALGEBRAIC SETS PROOF. If tr degkF .A/ D 1, there is nothing to prove, and so we suppose that tr degkF .A/ D n 2 N. We argue by induction on n. We can replace k with its algebraic closure in A without changing tr degkF .A/. Let x 2 A. If x … k, then it is transcendental over k, and so tr degk.x/F .A/ D n 1 by 1.64; since k.x/ Afxg, this implies (by induction) that dim.Afxg/ n 1. If x 2 k, then 0 D 1 x1x 2 Sfxg, and so Afxg D 0; again dim.Afxg/ n 1. We deduce from 2.53 that dim.A/ n. COROLLARY 2.55. The polynomial ring kŒX1; : : : ; Xn has Krull dimension n. PROOF. The existence of the sequence of prime ideals .X1; : : : ; Xn/ .X1; : : : ; Xn1/ .X1/ .0/ shows that kŒX1; : : : ; Xn has Krull dimension at least n. Now 2.54 completes the proof. COROLLARY 2.56. Let A be an integral domain and let k be a subfield of A. If A is finitely generated as a k-algebra, then tr degkF .A/ D dim.A/: PROOF. According to the Noether normalization theorem (2.45), A is integral over a polynomial subring kŒx1; : : : ; xn. Clearly n D tr degkF .A/. The going up theorem (1.54), implies that a chain of prime ideals in kŒx1; : : : ; xn lifts to a chain in A, and so dim.A/ dim.kŒx1; : : : ; xn/ D n. Now 2.54 shows that dim.A/ D n. COROLLARY 2.57. Let V be an irreducible algebraic set. Then V has dimension n if and only if there exists a finite surjective map V ! A n. PROOF. The d in Theorem 2.44 is the dimension of V . ASIDE 2.58. In linear algebra, we justify saying that a vector space V has dimension n by proving that its elements are parametrized by n-tuples. It is not true in general that the points of an algebraic set of dimension n are parametrized by n-tuples. All we can say is Corollary 2.57. ASIDE 2.59. The inequality in Proposition 2.54 may be strict; for example, A D k.x/ has dimension 0 but its field of fractions k.x/ has transcendence degree 1 over k. It is possible to deduce 2.54 from 2.56 — see mo79959. NOTES. The above proof of 2.55 is based on that in Coquand and Lombardi, Amer. Math. Monthly 112 (2005), no. 9, 826–829. Examples EXAMPLE 2.60. Let V D A X1; : : : ; Xn over k, and so dim.V / D n. n. Then k.V / D k.X1; : : : ; Xn/, which has transcendence basis m. Dimension 57 EXAMPLE 2.61. If V is a linear subspace of kn (or a translate of a linear subspace), then the dimension of V as an algebraic set is the same as its dimension in the sense of linear algebra — in fact, kŒV is canonically isomorphic to kŒXi1; : : : ; Xid , where the Xij are the “free” variables in the system of linear equations defining V . More specifically, let c be an ideal in kŒX1; : : : ; Xn generated by linear forms `1; : : : ; `r , which we may assume to be linearly independent. Let Xi1; : : : ; Xinr be such that f`1; : : : ; `r ; Xi1; : : : ; Xinr g is a basis for the linear forms in X1; : : : ; Xn. Then kŒX1; : : : ; Xn=c ' kŒXi1; : : : ; Xinr : This is obvious if the forms are X1; : : : ; Xr . In the general case, because fX1; : : : ; Xng g are both bases for the linear forms, each element of one set and f`1; : : : ; `r ; Xi1; : : : ; Xinr can be expressed as a linear combination of the elements of the other. Therefore, kŒX1; : : : ; Xn D kŒ`1; : : : ; `r ; Xi1; : : : ; Xinr ; and so kŒX1; : : : ; Xn=c D kŒ`1; : : : ; `r ; Xi1; : : : ; Xinr =c ' kŒXi1; : : : ; Xinr : EXAMPLE 2.62. If W is a proper algebraic subset of an irreducible algebraic set V , then dim.W / < dim.V / (see 2.50). EXAMPLE 2.63. Every nonempty algebraic set contains a point, which is a closed irreducible subset. Therefore an irreducible algebraic set has dimension 0 if and only if it consists of a single point. n has dimension n 1. It suffices to prove this for an EXAMPLE 2.64. A hypersurface in A irreducible hypersurface H . Such an H is the zero set of an irreducible polynomial f (see 2.29). Let kŒx1; : : : ; xn D kŒX1; : : : ; Xn=.f /; xi D Xi C .f /; and let k.x1; : : :
; xn/ be the field of fractions of kŒx1; : : : ; xn. As f is not the zero polynomial, some Xi , say, Xn, occurs in it. Then Xn occurs in every nonzero multiple of f , and so no nonzero polynomial in X1; : : : ; Xn1 belongs to .f /. This means that x1; : : : ; xn1 are algebraically independent. On the other hand, xn is algebraic over k.x1; : : : ; xn1/, and so fx1; : : : ; xn1g is a transcendence basis for k.x1; : : : ; xn/ over k. (Alternatively, use 2.57.) EXAMPLE 2.65. Let F .X; Y / and G.X; Y / be nonconstant polynomials with no common factor. Then V .F .X; Y // has dimension 1 by 2.64, and so V .F .X; Y // \ V .G.X; Y // must have dimension zero; it is therefore a finite set. PROPOSITION 2.66. Let W be a closed set of codimension 1 in an algebraic set V ; if kŒV is a unique factorization domain, then I.W / D .f / for some f 2 kŒV . PROOF. Let W1; : : : ; Ws be the irreducible components of W ; then I.W / D T I.Wi /, and so if we can prove I.Wi / D .fi /, then I.W / D .f1 fr /. Thus we may suppose that W is irreducible. Let p D I.W /; it is a prime ideal, and it is not zero because otherwise 58 2. ALGEBRAIC SETS dim.W / D dim.V /. Therefore it contains an irreducible polynomial f . From (1.33) we know .f / is prime. If .f / ¤ p , then we have p .f / .0/ (distinct prime ideals) and hence W D V .p/ V .f / V (distinct irreducible closed subsets). But then (2.62) dim.W / < dim.V .f // < dim V , which contradicts the hypothesis. COROLLARY 2.67. The closed sets of codimension 1 in A n are exactly the hypersurfaces. PROOF. Combine 2.64 and 2.66. 2. If V has dimension EXAMPLE 2.68. We classify the irreducible algebraic sets V of A 2. If V has dimension 1, then 2, then (by 2.62) it can’t be a proper subset of A V D V .f /, where f is any irreducible polynomial in I.V / (see 2.66 and its proof). Finally, if V has dimension zero, then it is a point. Correspondingly, the following is a complete list of the prime ideals in kŒX; Y : 2, so it is A .0/; .f / with f irreducible, .X a; Y b/ with a; b 2 k: Exercises 2-1. Find I.W /, where W D .X 2; XY 2/. Check that it is the radical of .X 2; X Y 2/. 2-2. Identify kmn with the set of m n matrices, and let r 2 N. Show that the set of matrices with rank r is an algebraic subset of kmn. 2-3. Let V D f.t; t 2; : : : ; t n/ j t 2 kg. Show that V is an algebraic subset of kn, and that kŒV kŒX (polynomial ring in one variable). (Assume k has characteristic zero.) 2-4. Let f1; : : : ; fm 2 QŒX1; : : : ; Xn. If the fi have no common zero in C, prove that there exist g1; : : : ; gm 2 QŒX1; : : : ; Xn such that f1g1 C C fmgm D 1. (Hint: linear algebra). 2-5. Let k K be algebraically closed fields, and let a be an ideal in kŒX1; : : : ; Xn. Show that if f 2 KŒX1; : : : ; Xn vanishes on V .a/, then it vanishes on VK.a/. Deduce that the zero set V .a/ of a in kn is dense in the zero set VK.a/ of a in Kn. [Hint: Choose a basis .ei /i2I for K as a k-vector space, and write f D P ei fi (finite sum) with fi 2 kŒX1; : : : ; Xn.] 2-6. Let A and B be (not necessarily commutative) Q-algebras of finite dimension over Q, and let Qal be the algebraic closure of Q in C. Show that if there exists a C-algebra homomorphism C ˝ Q B, then there exists a Qal-algebra homomorphism Qal ˝ Q A ! Qal ˝ Q B. (Hint: The proof takes only a few lines.) Q A ! C ˝ 2-7. Let A be finite dimensional k-algebra, where k is an infinite field, and let M and N be A-modules. Show that if kal ˝k M and kal ˝k N are isomorphic kal ˝k A-modules, then M and N are isomorphic A-modules. 2-8. Show that the subset f.z; ez/ j z 2 Cg is not an algebraic subset of C 2. CHAPTER 3 Affine Algebraic Varieties In this chapter, we define the structure of a ringed space on an algebraic set. In this way, we are led to the notion of an affine algebraic variety — roughly speaking, this is an algebraic set n. This is in preparation for Chapter 5, where we define with no preferred embedding into A an algebraic variety to be a ringed space that is a finite union of affine algebraic varieties satisfying a natural separation axiom. a. Sheaves Let k be a field (in this section 3a, k need not be algebraically closed). DEFINITION 3.1. Let V be a topological space, and suppose that, for every open subset U V .U / is a sheaf of k-algebras of V we have a set if the following statements hold for every open U in V : V .U / of functions U ! k. Then U O O (a) V .U / is a k-subalgebra of the algebra of all k-valued functions on U , i.e., V .U / V .U /, then so also do f C g and O O contains the constant functions and, if f; g lie in fg; O (b) the restriction of an f in O (c) a function f W U ! k lies in V .U / to an open subset U 0 of U is in V .U 0/I V .U / if there exists an open covering U D S O i 2I Ui of U such that f jUi 2 O V .Ui / for all i 2 I . O V is a sheaf if, for all U , In other words, lies in lies in V .U / is a k-subalgebra and a function f W U ! k V .U / if and only if every point P of U has a neighbourhood UP such that f jUP V .UP / (so the condition for f to lie in V .U / is local). O O Note that, for disjoint open subsets Ui of V , condition (c) says that O O O V .U / ' Q O V .Ui /. i O Examples 3.2. Let V be a topological space, and for each open subset U of V let of all continuous real-valued functions on U . Then V is a sheaf of R-algebras. O V .U / be the set O 3.3. Recall that a function f W U ! R on an open subset U of R n is said to be smooth (or infinitely differentiable) if its partial derivatives of all orders exist and are continuous. Let n, and for each open subset U of V , let V be an open subset of R V .U / be the set of all smooth functions on U . Then V is a sheaf of R-algebras. O O 59 60 3. AFFINE ALGEBRAIC VARIETIES 3.4. Recall that a function f W U ! C on an open subset U of C n, is said to be analytic (or holomorphic) if it is described by a convergent power series in a neighbourhood of each V .U / be point of U . Let V be an open subset of C the set of all analytic functions on U . Then n, and for each open subset U of V , let V is a sheaf of C-algebras. O O 3.5. Let V be a topological space, and, for each open subset U of V , let V .U / be the set of all constant functions U ! k. If V is not connected, then V is not a sheaf: let U1 and U2 be disjoint open subsets of V , and let f be the function on U1 [ U2 that takes the V .U1 [ U2/, and constant value 0 on U1 and the constant value 1 on U2; then f is not in so condition (3.1c) fails. When “constant” is replaced with “locally constant”, V becomes a sheaf of k-algebras (in fact, the smallest such sheaf). O O O O 3.6. Let V be a topological space, and, for each open subset U of V , let of all functions U ! k. The k-algebras are subsheaves of this one. V .U / be the set V is a sheaf of k-algebras. By definition, all our sheaves of O O b. Ringed spaces O A pair .V; V / consisting of a topological space V and a sheaf of k-algebras on V will be called a k-ringed space (or just a ringed space when the k is understood). For historical reasons, we sometimes write .U; over U . V .U / and call its elements the sections of V / for O O O V O Let .V; V / be a k-ringed space. For each open subset U of V , the restriction the collection of open subsets of U is a sheaf of k-algebras on U . O Let .V; V / be a k-ringed space, and let P 2 V . A germ of a function at P is an equivalence class of pairs .U; f / with U an open neighbourhood of P and f 2 V .U /; two pairs .U; f / and .U 0; f 0/ are equivalent if the functions f and f 0 agree on some open neighbourhood of P in U \ U 0. The germs of functions at P form a k-algebra V;P , called the stalk of O O V to O V at P . In other words, O V;P D lim! O In the interesting cases, are zero at P . We often write P for O O O V;P . O V .U / (direct limit over open neighbourhoods U of P ). V;P is a local ring with maximal ideal mP the set of germs that O V be the sheaf of holomorphic functions on V D C, and let c 2 C. EXAMPLE 3.7. Let A power series P n0 an.z c/n, an 2 C, is said to be convergent if it converges on some open neighbourhood of c. The set of such power series is a C-algebra, and I claim that it is canonically isomorphic to the stalk O To prove this, let f be a holomorphic function on a neighbourhood U of c. Then f has a unique power series expansion f D P an.z c/n in some (possibly smaller) open neighbourhood of c (Cartan 19631, II 2.6). Moreover, another holomorphic function f 0 on a neighbourhood U 0 of c defines the same power series if and only if f and f 0 agree on some neighbourhood of c contained in U \ U 0 (ibid. I, 4.3). Thus we have a well-defined injective map from the ring of germs of holomorphic functions at c to the ring of convergent power series, which is obviously surjective. V at c. V;c of O 1Cartan, Henri. Elementary theory of analytic functions of one or several complex variables. Hermann, Paris; Addison-Wesley; 1963. c. The ringed space structure on an algebraic set 61 c. The ringed space structure on an algebraic set Let V be an algebraic subset of kn. Recall that the basic open subsets of V are those of the form D.h/ D fQ j h.Q/ ¤ 0g; h 2 kŒV : A pair g; h 2 kŒV with h ¤ 0 defines a function Q 7! g.Q/ h.Q/ W D.h/ ! k. A function on an open subset of V is regular if it is locally of this form. More formally: DEFINITION 3.8. Let U be an open subset of V . A function f W U ! k is regular at P 2 U if there exist g; h 2 kŒV with h.P / ¤ 0 such that f .Q/ D g.Q/= h.Q/ for all Q in some neighbourhood of P . A function f W U ! k is regular if it is regular at every P 2 U . Let O V .U / denote the set of regular functions on an open subset U of V . PROPOSITION 3.9. The map U V .U / is a sheaf of k-algebras on V . O PROOF. We have to check the conditions of (3.1). (a) Clearly, a constant function is regular. Suppose f and f 0 are regular on U , and let P 2 U . By assumption, there exist g; g0; h; h0 2 kŒV , with h.P / ¤ 0 ¤ h0.P / such that f h0 respectively on a neighbourhood U 0 of P . Then f C f 0 agrees and f 0 agree wit
h g with gh0Cg 0h hh0 on U 0, and so is regular at P . on U 0, and so f C f 0 is regular at P . Similarly, ff 0 agrees with gg 0 h and g 0 hh0 (b,c) The definition is local. We next determine O V .U / when U is a basic open subset of V . LEMMA 3.10. Let g; h 2 kŒV with h ¤ 0. The function P 7! g.P /= h.P /mW D.h/ ! k is zero if and only if and only if gh D 0 in kŒV . PROOF. If g= hm is zero on D.h/, then gh is zero on V because h is zero on the complement of D.h/. Therefore gh is zero in kŒV . Conversely, if gh D 0, then g.P /h.P / D 0 for all P 2 V , and so g.P / D 0 for all P 2 D.h/. Let kŒV h denote the ring kŒV with h inverted (see 1.11). The lemma shows that V .D.h// sending g= hm to the regular function P 7! g.P /= h.P /m is the map kŒV h ! well-defined and injective. O PROPOSITION 3.11. The above map kŒV h ! V .D.h// is an isomorphism of k-algebras. O PROOF. It remains to show that every regular function f on D.h/ arises from an element of kŒV h. By definition, we know that there is an open covering D.h/ D S Vi and elements gi , hi 2 kŒV with hi nowhere zero on Vi such that f jVi D gi . We may assume that each hi set Vi is basic, say, Vi D D.ai / for some ai 2 kŒV . By assumption D.ai / D.hi /, and so aN i 2 kŒV (see p. 50). On D.ai /, i for some N 2 N and g0 D hi g0 i f D gi hi D gi g0 i hi g0 i D gi g0 i aN i : 62 3. AFFINE ALGEBRAIC VARIETIES Note that D.aN assume that Vi D D.hi /. i / D D.ai /. Therefore, after replacing gi with gi g0 i and hi with aN i , we can We now have that D.h/ D S D.hi / and that f jD.hi / D gi . Because D.h/ is quasicomhi on D.hi / \ D.hj / D D.hi hj /, pact, we can assume that the covering is finite. As gi hi D gj hj hi hj .gi hj gj hi / D 0, i.e., hi h2 () — this follows from Lemma 3.10 if hi hj ¤ 0 and is obvious otherwise. Because D.h/ D S D.hi / D S D.h2 i /, j gi D h2 i hj gj V ..h// D V ..h2 1; : : : ; h2 m//; and so h lies in rad.h2 1; : : : ; h2 m/: there exist ai 2 kŒV such that hN D m X i D1 ai h2 i : for some N . I claim that f is the function on D.h/ defined by (*) : P ai gi hi hN Let P be a point of D.h/. Then P will be in one of the D.hi /, say D.hj /. We have the following equalities in kŒV : h2 j m X i D1 ai gi hi ./D m X i D1 ai gj h2 i hj ./D gj hj hN . But f jD.hj / D gj hj following equality of functions on D.hj /: , i.e., f hj and gj agree as functions on D.hj /. Therefore we have the h2 j m X i D1 ai gi hi D f h2 j hN : Since h2 on D.hj / equals that defined by P ai gi hi . j is never zero on D.hj /, we can cancel it, to find that, as claimed, the function f hN On taking h D 1 in the proposition, we see that the definition of a regular function on V in this section agrees with that in Section 2i. COROLLARY 3.12. For every P 2 V , there is a canonical isomorphism where mP is the maximal ideal I.P /. P ! kŒV mP , O PROOF. In the definition of the germs of a sheaf at P , it suffices to consider pairs .f; U / with U lying in a some basis for the neighbourhoods of P , for example, the basis provided by the basic open subsets. Therefore, P D lim! O h.P /¤0 .D.h/; V / O (3.11)' lim! h…mP kŒV h (1.23)' kŒV mP : c. The ringed space structure on an algebraic set 63 Remarks 3.13. Let V be an algebraic set and let P be a point on V . Proposition 1.14 shows that there is a one-to-one correspondence between the prime ideals of kŒV contained in mP and the P . In geometric terms, this says that there is a one-to-one correspondence prime ideals of between the irreducible closed subsets of V passing through P and the prime ideals in P . The irreducible components of V passing through P correspond to the minimal prime ideals P . The ideal p corresponding to an irreducible closed subset Z consists of the elements in P represented by a pair .U; f / with f jZ\U D 0. of O O O O O V;P depends only on .U; V;P is an integral domain if P lies on a single irreducible component 3.14. The local ring V jU / for U an open neighbourhood of P , we may of V . As suppose that V itself is irreducible, in which case the statement follows from 3.12. On the other hand, if P lies on more than one irreducible component of V , then P contains more than one minimal prime ideal 3.13, and so the ideal .0/ can’t be prime. O O O 3.15. Let V be an algebraic subset of kn, and let A D kŒV . Propositions 2.37 and 3.11 allow us to describe .V; V / purely in terms of A: ˘ V is the set of maximal ideals in A. ˘ For each f 2 A, let D.f / D fm j f … mg; the topology on V is that for which the O sets D.f / form a base. ˘ For f 2 Ah and m 2 D.h/, let f .m/ denote the image of f in Ah=mAh ' k; in this V is the V / D Ah for all h 2 A. way Ah becomes identified with a k-algebra of functions D.h/ ! k, and unique sheaf of k-valued functions on V such that .D.h/; O O 3.16. When V is irreducible, all the rings attached to it can be identified with subrings of its function field k.V /. For example, .D.h/; V / D O .U hN n g h 2 k.V / ˇ 2 k.V / ˇ o ˇ g 2 kŒV ; N 2 N o ˇ h.P / ¤ 0 P P 2U O .D.hi /; V / if U D [ D.hi /: O Note that every element of k.V / defines a function on some dense open subset of V . Following tradition, we call the elements of k.V / rational functions on V .2 Examples 3.17. The ring of regular functions on A h.X1; : : : ; Xn/, the ring of regular functions on D.h/ is n is kŒX1; : : : ; Xn. For a nonzero polynomial n g hN 2 k.X1; : : : ; Xn/ ˇ ˇ g 2 kŒX1; : : : ; Xn; N 2 N ˇ o : For a point P D .a1; : : : ; an/, the local ring at P is O P D ˚ g 2 k.X1; : : : ; Xn/ j h.P / ¤ 0 D kŒX1; : : : ; Xn.X1a1;:::;Xnan/; h and its maximal ideal consists of those g= h with g.P / D 0: 2The terminology is similar to that of “meromorphic function”, which is also not a function on the whole space. 64 3. AFFINE ALGEBRAIC VARIETIES 3.18. Let U D A 2, but it is not a basic open subset because its complement f.0; 0/g has dimension 0, and therefore can’t be of the form V ..f // (see 2.64). Since U D D.X/ [ D.Y /, the ring of regular functions on U is 2 X f.0; 0/g. It is an open subset of A .U; OA2/ D kŒX; Y X \ kŒX; Y Y (intersection inside k.X; Y /). Thus, a regular function f on U can be expressed f D g.X; Y / X N D h.X; Y / Y M : We may assume that X − g and Y − h. On multiplying through by X N Y M , we find that g.X; Y /Y M D h.X; Y /X N : Because X doesn’t divide the left hand side, it can’t divide the right hand side either, and so N D 0. Similarly, M D 0, and so f 2 kŒX; Y . We have shown that every regular function on U extends uniquely to a regular function on A 2: .U; 2; OA2/ D kŒX; Y D .A OA2/: d. Morphisms of ringed spaces A morphism of k-ringed spaces .V; that O V / ! .W; O W / is a continuous map 'W V ! W such f 2 W .U / H) f ı ' 2 V .'1U / O O for all open subsets U of W . Then, for every pair of open subsets U W and U 0 V with '.U 0/ U , we get a homomorphism of k-algebras f 7! f ı 'W W .U 0/ ! O V .U /; O and these homomorphisms are compatible with restriction to smaller open subsets. Sometimes we write '.f / for f ı '. A morphism of ringed spaces is an isomorphism if it is bijective and its inverse is also a morphism of ringed spaces (in particular, it is a homeomorphism). If U is an open subset of V , then the inclusion U ,! V is a morphism of k-ringed spaces .U; V jU / ! .V; V /. O A morphism of ringed spaces maps germs of functions to germs of functions. More O precisely, a morphism 'W .V; V / ! .W; W / induces a k-algebra homomorphism O O OW;'.P / ! V;P O for each P 2 V , which sends the germ represented by .U; f / to the germ represented by .'1.U /; f ı '/. In the interesting cases, V;P is a local ring with maximal ideal mP O consisting of the germs represented by pairs .U; f / with f .P / D 0. Therefore, the homoV;P defined by ' maps m'.P / into mP : it is a local homomorphism morphism of local rings. OW ;'.P / ! O e. Affine algebraic varieties 65 Examples 3.19. Let V and W be topological spaces endowed with their sheaves W of continuous real valued functions (3.2). Every continuous map 'W V ! W is a morphism of ringed structures .V; V and V / ! .W; O 3.20. Let V and W be open subsets of R function .a1; : : : ; an/ 7! ai W V ! R. Recall from advanced calculus that a map n and R m respectively, and let xi be the coordinate W /. O O O 'W V ! W R m defD xi ı 'W V ! R is smooth. If is said to be smooth if each of its component functions 'i ' is smooth, then f ı ' is smooth for every smooth function f W W ! R. Since a similar statement is true for functions f on open subsets of W , we see that a continuous map 'W V ! W is smooth if and only if it is a morphism of the associated ringed spaces (3.3). 3.21. Same as 3.20, but replace R with C and “smooth” with “analytic”. e. Affine algebraic varieties We have just seen that every algebraic set V kn gives rise to a k-ringed space .V; V /. A k-ringed space isomorphic to one of this form is called an affine algebraic variety over k. We usually denote an affine algebraic variety .V; V / by V . Let .V; V / and .W; W / be affine algebraic varieties. A map 'W V ! W is regular (or a morphism of affine algebraic varieties) if it is a morphism of k-ringed spaces. With these definitions, the affine algebraic varieties become a category. We usually shorten “affine algebraic variety” to “affine variety”. O O O O In particular, the regular functions define the structure of an affine variety on every n as an affine variety. The affine varieties we have n. We now explain how to construct affine algebraic set. We usually regard A constructed so far have all been embedded in A varieties with no preferred embedding. An affine k-algebra is a reduced finitely generated k-algebra. For such an algebra A, there exist xi 2 A such that A D kŒx1; : : : ; xn, and the kernel of the homomorphism Xi 7! xi W kŒX1; : : : ; Xn ! A is a radical ideal. Therefore 2.18 implies that the intersection of the maximal ideals in A is 0. Moreover, 2.12 implies that, for every maximal ideal m A, the map k ! A ! A=m is an isomorphism. Thus we can identify A=m with k. For f 2 A, we write f .m/ for th
e image of f in A=m D k, i.e., f .m/ D f (mod m/. This allows us to identify elements of A with functions spm.A/ ! k. We attach a ringed space .V; For f 2 A, let O V / to A by letting V be the set of maximal ideals in A. D.f / D fm j f .m/ ¤ 0g D fm j f … mg: Since D.fg/ D D.f / \ D.g/, there is a topology on V for which the D.f / form a base. A pair of elements g; h 2 A, h ¤ 0, defines a function For U an open subset of V , we define of this form in some neighbourhood of each point of U . O W D.h/ ! k: m 7! g.m/ h.m/ V .U / to be the set of functions f W U ! k that are 66 3. AFFINE ALGEBRAIC VARIETIES PROPOSITION 3.22. The pair .V; Ah for each h 2 A X f0g. O V / is an affine algebraic variety with .D.h/; V / ' O PROOF. Represent A as a quotient kŒX1; : : : ; Xn=a D kŒx1; : : : ; xn. Then .V; morphic to the k-ringed space attached to the algebraic set V .a/ (see 3.15). V / is iso- O We write spm.A/ for the topological space V , and Spm.A/ for the k-ringed space .V; V /. O ASIDE 3.23. We have attached to an affine k-algebra A an affine variety whose underlying topological space is the set of maximal ideals in A. It may seem strange to be describing a topological space in terms of maximal ideals in a ring, but the analysts have been doing this for more than 70 years. Gel’fand and Kolmogorov in 19393 proved that if S and T are compact topological spaces, and the rings of real-valued continuous functions on S and T are isomorphic (just as rings), then S and T are homeomorphic. The proof begins by showing that, for such a space S , the map P 7! mP defD ff W S ! R j f .P / D 0g is one-to-one correspondence between the points in the space and maximal ideals in the ring. f. The category of affine algebraic varieties For each affine k-algebra A, we have an affine variety Spm.A/, and conversely, for each V /. We now make this affine variety .V; correspondence into an equivalence of categories. V /, we have an affine k-algebra kŒV D .V; O O Let ˛W A ! B be a homomorphism of affine k-algebras. For every h 2 A, ˛.h/ is invertible in B˛.h/, and so the homomorphism A ! B ! B˛.h/ extends to a homomorphism g hm 7! ˛.g/ ˛.h/m W Ah ! B˛.h/: For every maximal ideal n of B, m D ˛1.n/ is maximal in A because A=m ! B=n D k is an injective map of k-algebras which implies that A=m D k. Thus ˛ defines a map 'W spm B ! spm A; '.n/ D ˛1.n/ D m: For m D ˛1.n/ D '.n/, we have a commutative diagram: A A=m ˛ ' B B=n: Recall that the image of an element f of A in A=m ' k is denoted f .m/. Therefore, the commutativity of the diagram means that, for f 2 A, f .'.n// D ˛.f /.n/, i.e., f ı ' D ˛ ı f: () Since '1D.f / D D.f ı '/ (obviously), it follows from () that '1.D.f // D D.˛.f //; 3On rings of continuous functions on topological spaces, Doklady 22, 11-15. See also Allen Shields, Banach Algebras, 1939–1989, Math. Intelligencer, Vol 11, no. 3, p15. g. Explicit description of morphisms of affine varieties 67 and so ' is continuous. Let f be a regular function on D.h/, and write f D g= hm, g 2 A. Then, from (*) we see that f ı ' is the function on D.˛.h// defined by ˛.g/=˛.h/m. In particular, it is regular, and so f 7! f ı ' maps regular functions on D.h/ to regular functions on D.˛.h//. It follows that f 7! f ı ' sends regular functions on any open subset of spm.A/ to regular functions on the inverse image of the open subset. Thus ˛ defines a morphism of ringed spaces Spm.B/ ! Spm.A/. Conversely, by definition, a morphism of 'W .V; W / of affine algebraic varieties defines a homomorphism of the associated affine k-algebras kŒW ! kŒV . Since these maps are inverse, we have shown: V / ! .W; O O PROPOSITION 3.24. For all affine algebras A and B, Homk-alg.A; B/ ' ! Mor.Spm.B/; Spm.A//I for all affine varieties V and W , Mor.V; W / ' ! Homk-alg.kŒW ; kŒV /: In terms of categories, Proposition 3.24 can be restated as: PROPOSITION 3.25. The functor A Spm A is a (contravariant) equivalence from the category of affine k-algebras to the category of affine algebraic varieties over k, with quasi-inverse .V; V / .V; V /. O O g. Explicit description of morphisms of affine varieties PROPOSITION 3.26. Let V km and W kn be algebraic subsets. The following conditions on a continuous map 'W V ! W are equivalent: (a) ' is a morphism of ringed spaces .V; V / ! .W; W /; O O (b) the components '1; : : : ; 'm of ' are regular functions on V ; (c) f 2 kŒW H) f ı ' 2 kŒV . PROOF. (a) H) (b). By definition 'i D yi ı ', where yi is the coordinate function .b1; : : : ; bn/ 7! bi W W ! k: Hence this implication follows directly from the definition of a regular map. (b) H) (c). The map f 7! f ı ' is a k-algebra homomorphism from the ring of all functions W ! k to the ring of all functions V ! k, and (b) says that the map sends the coordinate functions yi on W into kŒV . Since the yi generate kŒW as a k-algebra, this implies that it sends kŒW into kŒV . (c) H) (a). The map f 7! f ı ' is a homomorphism ˛W kŒW ! kŒV . It therefore defines a map spm .kŒV / ! spm .kŒW /, and it remains to show that this coincides with ' when we identify spm .kŒV / with V and spm .kŒW / with W . Let P 2 V , let Q D '.P /, and let mP and mQ be the ideals of elements of kŒV and kŒW that are zero at P and Q respectively. Then, for f 2 kŒW , ˛.f / 2 mP ” f .'.P // D 0 ” f .Q/ D 0 ” f 2 mQ: Therefore ˛1.mP / D mQ, which is what we needed to show. 68 3. AFFINE ALGEBRAIC VARIETIES The equivalence of (a) and (b) means that 'W V ! W is a regular map of algebraic sets (in the sense of Chapter 2) if and only if it is a regular map of the associated affine algebraic varieties. Now consider equations Y1 D f1.X1; : : : ; Xm/ : : : Yn D fn.X1; : : : ; Xm/: On the one hand, they define a regular map 'W A m ! A n, namely, .a1; : : : ; am/ 7! .f1.a1; : : : ; am/; : : : ; fn.a1; : : : ; am//: On the other hand, they define a homomorphism ˛W kŒY1; : : : ; Yn ! kŒX1; : : : ; Xm of kalgebras, namely, that sending Yi to fi .X1; : : : ; Xm/. This map coincides with g 7! g ı ', because ˛.g/.P / D g.: : : ; fi .P /; : : :/ D g.'.P //: Now consider closed subsets V .a/ A m and V .b/ A n with a and b radical ideals. I claim that ' maps V .a/ into V .b/ if and only if ˛.b/ a. Indeed, suppose '.V .a// V .b/, and let g 2 b; for Q 2 V .a/, ˛.g/.Q/ D g.'.Q// D 0; and so ˛.g/ 2 I V .a/ D a. Conversely, suppose ˛.b/ a, and let P 2 V .a/; for f 2 b, f .'.P // D ˛.f /.P / D 0; and so '.P / 2 V .b/. When these conditions hold, ' is the morphism of affine varieties V .a/ ! V .b/ corresponding to the homomorphism kŒY1; : : : ; Yn=b ! kŒX1; : : : ; Xm=a defined by ˛. Thus, we see that the regular maps are all of the form V .a/ ! V .b/ P 7! .f1.P /; : : : ; fn.P //; fi 2 kŒX1; : : : ; Xm: In particular, they all extend to regular maps A m ! A n. Examples of regular maps 3.27. Let R be a k-algebra. To give a k-algebra homomorphism kŒX ! R is the same as giving an element (the image of X under the homomorphism): Homk-alg.kŒX; R/ ' R: Therefore Mor.V; A 1/ 3.24' Homk-alg.kŒX; kŒV / ' kŒV : In other words, the regular maps V ! A would hope). 1 are simply the regular functions on V (as we g. Explicit description of morphisms of affine varieties 69 0 D Spm k. Then A 0 consists of a single point and .A 0; 0 ! V from A 0 to an affine variety, sends A 0 to a point of V , and so Mor.A OA0/ D k. Every 0; V / ' V . 3.28. Let A map A Alternatively, Mor.A 0; V / ' Homk-alg.kŒV ; k/ ' V; where the last map sends ˛W kŒV ! k to the point corresponding to the maximal ideal Ker.˛/. 3.29. Consider the regular map t 7! .t 2; t 3/W A bijective onto its image, 1 ! A 2. This is V W Y 2 D X 3; but it is not an isomorphism onto its image because the inverse map is not regular. In view of 3.25, to prove this it suffices to show that t 7! .t 2; t 3/ does not induce an isomorphism on the 1 D kŒT and kŒV D rings of regular functions. We have kŒA kŒX; Y =.Y 2 X 3/ D kŒx; y. The map on rings is x 7! T 2; y 7! T 3; kŒx; y ! kŒT ; which is injective, but its image is kŒT 2; T 3 ¤ kŒT . In fact, kŒx; y is not integrally closed: .y=x/2 x D 0, and so .y=x/ is integral over kŒx; y, but y=x … kŒx; y (it maps to T under the inclusion k.x; y/ ,! k.T //. 3.30. Let k have characteristic p ¤ 0, and consider the regular map .a1; : : : ; an/ 7! .ap 1 ; : : : ; ap n /W A n ! A n: This is a bijection, but it is not an isomorphism because the corresponding map on rings, Xi 7! X p i W kŒX1; : : : ; Xn ! kŒX1; : : : ; Xn; is not surjective. This is the famous Frobenius map. Take k to be the algebraic closure of Fp, and write F for the map. Recall that for each m 1 there is a unique subfield Fpm of k of degree m over Fp, and that its elements are the solutions of X pm D X (FT 4.23). The fixed points of F m n with coordinates in Fpm. Let f .X1; : : : ; Xn/ be a polynomial are precisely the points of A with coefficients in Fpm, say, f D X ci1inX i1 1 X in n ; ci1in 2 Fpm: If f .a1; : : : ; an/ D 0, then X 0 D c˛ai1 1 ain n pm D X c˛apmi1 1 apmin n ; 1 ; : : : ; apm and so f .apm form .X C Y /pm D X pm C Y pm its fixed points are the solutions of n / D 0. Here we have used that the binomial theorem takes the simple in characteristic p. Thus F m maps V .f / into itself, and in Fpm. f .X1; : : : ; Xn/ D 0 70 3. AFFINE ALGEBRAIC VARIETIES ASIDE 3.31. In one of the most beautiful pieces of mathematics of the second half of the twentieth century, Grothendieck defined a cohomology theory (´etale cohomology) and proved a fixed point formula that allowed him to express the number of solutions of a system of polynomial equations with coordinates in Fpm as an alternating sum of traces of operators on finite-dimensional vector spaces, and Deligne used this to obtain very precise estimates for the number of solutions. See my article The Riemann hypothesis over finite fields: from Weil to the present day and my notes Lectures on ´Etale Cohomology. h. Subvarieties Let A be an affine k-algebra. For any ideal a in A, we define V .a/ D fP 2 spm.A/ j f .P / D 0 all f 2 ag D fm maximal ideal in A j a mg: This
is a closed subset of spm.A/, and every closed subset is of this form. Now let a be a radical ideal in A, so that A=a is again reduced. Corresponding to the homomorphism A ! A=a, we get a regular map Spm.A=a/ ! Spm.A/: The image is V .a/, and spm.A=a/ ! V .a/ is a homeomorphism. Thus every closed subset of spm.A/ has a natural ringed structure making it into an affine algebraic variety. We call V .a/ with this structure a closed subvariety of V: PROPOSITION 3.32. Let .V; kŒV . Then O V / be an affine variety and let h be a nonzero element of in particular, it is an affine variety. .D.h/; O V jD.h// ' Spm.Ah/I PROOF. The map A ! Ah defines a morphism spm.Ah/ ! spm.A/. The image is D.h/, and it is routine (using (1.13)) to verify the rest of the statement. If V D V .a/ A n, then .a1; : : : ; an/ 7! .a1; : : : ; an; h.a1; : : : ; an/1/W D.h/ ! A nC1; defines an isomorphism of D.h/ onto V .a; 1 hXnC1/. For example, there is an isomorphism of affine varieties a 7! .a; 1=a/W A 1 X f0g ! V A 2; i. Properties of the regular map Spm.˛/ 71 with V equal to the subvariety XY D 1 of A 2, By an open affine (subset) U of an affine algebraic variety V , we mean an open subset V jU / is an affine algebraic variety. Thus, the proposition says that, for U such that .U; all nonzero h 2 .V; V /, the open subset of V , where h is nonzero is an open affine. An open affine subset of an irreducible affine algebraic variety V is irreducible with the same dimension as V (2.52). O O REMARK 3.33. We have seen that all closed subsets and all basic open subsets of an affine variety V are again affine varieties with their natural ringed structure, but this is not true for all open subsets of V . For an open affine subset U , the natural map U ! spm .U; V / is a bijection. However, for O U defD A 2 X f.0; 0/g D D.X/ [ D.Y / A 2; OA2/ D kŒX; Y (see 3.18), but U ! spm kŒX; Y is not a bijection, we know that .U; OA2jU / is a union of affine because the ideal .X; Y / is not in the image. Clearly .U; algebraic varieties, and in Chapter 5 we shall recognize it as a (nonaffine) algebraic variety. i. Properties of the regular map Spm.˛/ PROPOSITION 3.34. Let ˛W A ! B be a homomorphism of affine k-algebras, and let 'W Spm.B/ ! Spm.A/ be the corresponding morphism of affine varieties. (a) The image of ' is dense for the Zariski topology if and only if ˛ is injective. (b) The morphism ' is an isomorphism from Spm.B/ onto a closed subvariety of Spm.A/ if and only if ˛ is surjective. PROOF. (a) Let f 2 A. If the image of ' is dense, then f ı ' D 0 H) f D 0: XY=1 72 3. AFFINE ALGEBRAIC VARIETIES On the other hand, if the image of ' is not dense, then the closure of its image is a proper closed subset of Spm.A/, and so there is a nonzero function f 2 A that is zero on it. Then f ı ' D 0. (See 2.40.) (b) If ˛ is surjective, then it defines an isomorphism A=a ! B, where a is the kernel of ˛. This induces an isomorphism of Spm.B/ with its image in Spm.A/. The converse follows from the description of the closed subvarieties of Spm.A/ in the last section. A regular map 'W V ! W of affine algebraic varieties is said to be a dominant if its image is dense in W . The proposition then says that: ' is dominant ” f 7! f ı 'W .W; W / ! .V; O V / is injective. O A regular map 'W V ! W of affine algebraic varieties is said to be a closed immersion if it is an isomorphism of V onto a closed subvariety of W . The proposition then says that ' is a closed immersion ” f 7! f ı 'W .W; W / ! .V; O O V / is surjective. j. Affine space without coordinates Let E be a vector space over k of dimension n. The set A.E/ of points of E has a natural structure of an algebraic variety: the choice of a basis for E defines a bijection A.E/ ! A n, and the inherited structure of an affine algebraic variety on A.E/ is independent of the choice of the basis (because the bijections defined by two different bases differ by an automorphism of A n). We now give an intrinsic definition of the affine variety A.E/. Let V be a finite- dimensional vector space over a field k. The tensor algebra of V is T V D M i 0 V ˝i D k ˚ V ˚ .V ˝ V / ˚ .V ˝ V ˝ V / ˚ with multiplication defined by .v1 ˝ ˝ vi / .v0 1 ˝ ˝ v0 j / D v1 ˝ ˝ vi ˝ v0 1 ˝ ˝ v0 j : It is a noncommutative k-algebra, and the choice of a basis e1; : : : ; en for V defines an isomorphism e1 ei 7! e1 ˝ ˝ ei W kfe1; : : : ; eng ! T .V / to T V from the k-algebra of noncommuting polynomials in the symbols e1; : : : ; en. The symmetric algebra S .V / of V is defined to be the quotient of T V by the two- sided ideal generated by the elements v ˝ w w ˝ v; v; w 2 V: This algebra is generated as a k-algebra by commuting elements (namely, the elements of V D V ˝1), and so is commutative. The choice of a basis e1; : : : ; en for V defines an isomorphism e1 ei 7! e1 ˝ ˝ ei W kŒe1; : : : ; en ! S .V / to S .V / from the commutative polynomial ring in the symbols e1; : : : ; en. This shows that S .V / is an affine k-algebra. The pair .S .V /; i/ consisting of S .V / and the natural k. Birational equivalence 73 k-linear map i W V ! S .V / has the following universal property: every k-linear map V ! A from V into a k-algebra A extends uniquely to a k-algebra homomorphism S .V / ! A: V i S .V / k-linear 9Š k-algebra A: (17) As usual, this universal property determines the pair .S .V /; i/ uniquely up to a unique isomorphism. We now define A.E/ to be Spm.S .E_//, where E_ is the dual vector space. For an affine k-algebra A, Mor.Spm.A/; A.E// ' Homk-algebra.S .E_/; A/ ' Homk-linear.E_; A/ ' E ˝k A .3.24/ .17/ .linear algebra/: In particular, A.E/.k/ ' E: Moreover, the choice of a basis e1; : : : ; en for E determines a (dual) basis f1; : : : ; fn of E_, and hence an isomorphism of k-algebras kŒf1; : : : ; fn ! S .E_/. The map of algebraic varieties defined by this homomorphism is the isomorphism n A.E/ ! A whose map on the underlying sets is the isomorphism E ! kn defined by the basis of E. k. Birational equivalence Recall that if V is irreducible, then kŒV is an integral domain, and we write k.V / for its field of fractions. If U is an open affine subvariety of V , then kŒV kŒU k.V /, and so k.V / is also the field of fractions of kŒU . DEFINITION 3.35. Two irreducible affine algebraic varieties over k are birationally equivalent if their function fields are isomorphic over k. PROPOSITION 3.36. Irreducible affine varieties V and W are birationally equivalent if and only if there exist open affine subvarieties UV and UW of V and W respectively such that UV UW . PROOF. Let V and W be birationally equivalent irreducible affine varieties, and let A D kŒV and B D kŒW . We use the isomorphism to identify k.V / and k.W /. This allows us to suppose that A and B have a common field of fractions K. Let x1; : : : ; xn generate B as k-algebra. As K is the field of fractions of A, there exists a d 2 A such that dxi 2 A for all i ; then B Ad . The same argument shows that there exists an e 2 B such that Ad Be. Now B Ad Be H) Be Ade .Be/e D Be; and so Ade D Be. This shows that the open subvarieties D.de/ V and D.e/ W are isomorphic. We have proved the “only if” part, and the “if” part is obvious. 74 3. AFFINE ALGEBRAIC VARIETIES THEOREM 3.37. Every irreducible affine algebraic variety of dimension d is birationally equivalent to a hypersurface in A d C1. PROOF. Let V be an irreducible variety of dimension d . According to (3.38) below, there exist rational functions x1; : : : ; xd C1 on V such that k.V / D k.x1; : : : ; xd ; xd C1/. Let f 2 kŒX1; : : : ; Xd C1 be an irreducible polynomial satisfied by the xi , and let H be the hypersurface f D 0. Then k.V / k.H / and so V and H are birationally equivalent. We review some definitions from FT, Chapter 2. Let F be a field. A polynomial f 2 F ŒX is separable if it has no multiple roots. Equivalent condition: gcd.f; df dX / D 1. When f is irreducible, this just says that df dX < deg f . An element dX of an algebraic extension E of F is separable over F if its minimal polynomial over F is separable, and E is separable over F if all its elements are separable over F . ¤ 0 because deg df PROPOSITION 3.38. Let ˝ be a finitely generated field extension of k of transcendence degree d . If k is perfect, then there exist x1; : : : ; xd C1 2 ˝ such that ˝ D k.x1; : : : ; xd C1/. After renumbering, fx1; : : : ; xd g will be a transcendence basis for ˝ over k and xd C1 will be separable over k.x1; : : : ; xd /. PROOF. Let ˝ D k.x1; : : : ; xn/. After renumbering, we may suppose that x1; : : : ; xd are algebraically independent, hence a transcendence basis (1.63). If F has characteristic zero, then xd C1; : : : ; xn are separable over k.x1; : : : xd /, and so the primitive element theorem (FT 5.1) shows that there exists a y 2 ˝ for which ˝ D k.x1; : : : ; xd ; y/. Thus, we may suppose that k has characteristic p ¤ 0. Because k is perfect, every polynomial in with coefficients in k is a pth power in kŒX1; : : : ; Xn: X ai1inX i1p 1 : : : X inp n D X 1 p i1in a X i1 1 : : : X in n p : (18) Let .x1; : : : ; xn/ be a generating set for ˝ over k with the fewest elements. We shall assume that n > d C 1 and obtain a contradiction. As before, we may suppose that x1; : : : ; xd are algebraically independent. Then f .x1; : : : ; xd C1/ D 0 for some nonzero irreducible polynomial f .X1; : : : ; Xd C1/ with coefficients in k. Not all polynomials @f =@Xi are zero, for otherwise f would be a polynomial in X p d C1, and hence a pth power. After renumbering, we may suppose that @f =@Xd C1 ¤ 0. Now xd C1 is separably algebraic over k.x1; : : : ; xd / and xd C2 is algebraic over k.x1; : : : ; xd C1/ (hence over k.x1; : : : ; xd /). According to the primitive element theorem (FT 5.1), there exists a y 2 ˝ such that k.x1; : : : ; xd C2/ D k.x1; : : : ; xd ; y/. Now ˝ D k.x1; : : : ; xd ; y; xd C3; : : : ; xn/, contradicting the minimality of n. 1 ; : : : ; X p We have shown that ˝ D k.z1; : : : ; zd C1/ for some zi 2 ˝. The argument in the last paragraph shows
that, after renumbering, zd C1 will be separably algebraic over k.z1; : : : ; zd /, and this implies that fz1; : : : ; zd g is a transcendence basis for ˝ over k (1.63). l. Noether Normalization Theorem DEFINITION 3.39. The dimension of an affine algebraic variety is the dimension of the underlying topological space (2.48). DEFINITION 3.40. A regular map 'W W ! V of affine algebraic varieties is finite if the map 'W kŒV ! kŒW makes kŒW a finite kŒV -algebra. m. Dimension 75 THEOREM 3.41. Let V be an affine algebraic variety of dimension n. Then there exists a finite map V ! A n. PROOF. Immediate consequence of (2.45). m. Dimension By definition, the dimension d of an affine variety V is the maximum length of a chain V0 V1 of distinct closed irreducible affine subvarieties. In this section, we prove that it is the length of every maximal chain of such subvarieties. THEOREM 3.42. Let V be an irreducible affine variety, and let f be a nonzero regular function on V . If f has a zero in V , then its zero set is of pure codimension 1. The Noether normalization theorem allows us to deduce this from the special case V D A proved in 2.64. n, PROOF. 4Let Z1; : : : ; Zn be the irreducible components of V .f /. We have to show that dim Zi D dim V 1 for each i. There exists a point P 2 Zi not contained in any other Zj . Because the Zj are closed, there exists an open affine neighbourhood U of P in V not meeting any Zj with j ¤ i . Now V .f jU / D Zi \ U , which is irreducible. Therefore, on replacing V with U , we may assume that V .f / is irreducible. As V .f / is irreducible, the radical of .f / is a prime ideal p in kŒV . According to the d ,! kŒV realizing d and f divides d . We claim that, in Noether normalization theorem (2.45), there exists an inclusion kŒA kŒV as a finite kŒA f0 in kŒV (see 1.45). Hence f0 2 .f / p, and so rad.f0/ p \ kŒA fact, d -algebra. Let f0 D Nmk.V /=k.Ad / f . Then f0 2 kŒA d . d . Then g 2 p defD rad.f /, and so gm D f h for some h 2 kŒV , m 2 N. rad.f0/ D p \ kŒA Let g 2 p \ kŒA Taking norms, we find that gme D Nm.f h/ D f0 Nm.h/ 2 .f0/; where e D Œk.V / W k.A The inclusion kŒA n/, and so g 2 rad.f0/, as claimed. d ,! kŒV therefore induces an inclusion kŒA d = rad.f0/ ,! kŒV =p: This makes kŒV =p into a finite algebra over kŒA of these two k-algebras have the same transcendence degree: d = rad.f0/, and so the fields of fractions dim V .p/ D dim V .f0/: Clearly f ¤ 0 ) f0 ¤ 0, and f0 2 p ) f0 is nonconstant. Therefore dim V .f0/ D d 1 by (2.64). 4This proof was found by John Tate. 76 3. AFFINE ALGEBRAIC VARIETIES We can restate Theorem 3.42 as follows: let V be a closed irreducible subvariety of A and let F 2 kŒX1; : : : ; Xn; then n V \ V .F / D 8 < : V ; pure codimension 1 otherwise. if F is identically zero on V if F has no zeros on V COROLLARY 3.43. Let V be an irreducible affine variety, and let Z be a maximal proper irreducible closed subset of V . Then dim.Z/ D dim.V / 1. PROOF. Because Z is a proper closed subset of V , there exists a nonzero regular function f on V vanishing on Z. Let V .f / be the zero set of f in V . Then Z V .f / V , and Z must be an irreducible component of V .f / for otherwise it wouldn’t be maximal in V . Thus Theorem 3.42 shows that dim Z D dim V 1. COROLLARY 3.44. Let V be an irreducible affine variety. Every maximal (i.e., nonrefinable) chain V D V0 V1 Vd (19) of distinct irreducible closed subsets of V has length d D dim.V /. PROOF. The last set Vd must be a point and each Vi must be maximal in Vi 1, and so, from 3.43, we find that dim V0 D dim V1 C 1 D dim V2 C 2 D D dim Vd C d D d: COROLLARY 3.45. Let V be an irreducible affine variety, and let f1; : : : ; fr be regular functions on V . Every irreducible component Z of V .f1; : : : fr / has codimension at most r: codim.Z/ r: For example, if the fi have no common zero on V , so that V .f1; : : : ; fr / is empty, then there are no irreducible components, and the statement is vacuously true. PROOF. We use induction on r. Because Z is an irreducible closed subset of V .f1; : : : ; fr1/, it is contained in some irreducible component Z0 of V .f1; : : : fr1/. By induction, codim.Z0/ r 1. Also Z is an irreducible component of Z0 \ V .fr / because Z Z0 \ V .fr / V .f1; : : : ; fr / and Z is a maximal irreducible closed subset of V .f1; : : : ; fr /. If fr vanishes identically on Z0, then Z D Z0 and codim.Z/ D codim.Z0/ r 1; otherwise, the theorem shows that Z has codimension 1 in Z0, and codim.Z/ D codim.Z0/ C 1 r. EXAMPLE 3.46. In the setting of 3.45, the components of V .f1; : : : ; fr / need not all have the same dimension, and it is possible for all of them to have codimension < r without any of the fi being redundant. For example, let V be the cone in A 4. Then V .X1/ \ V is the union of two planes: X1X4 X2X3 D 0 V .X1/ \ V D f.0; 0; ; /g [ f.0; ; 0; /g: m. Dimension 77 Both of these have codimension 1 in V (as required by 3.42). Similarly, V .X2/ \ V is the union of two planes, V .X2/ \ V D f.0; 0; ; /g [ f.; 0; ; 0/g: However V .X1; X2/ \ V consists of a single plane f.0; 0; ; /g: it still has codimension 1 in V , but it requires both X1 and X2 to define it. PROPOSITION 3.47. Let Z be an irreducible closed subvariety of codimension r in an affine variety V . Then there exist regular functions f1; : : : ; fr on V such that Z is an irreducible component of V .f1; : : : ; fr / and all irreducible components of V .f1; : : : ; fr / have codimension r. PROOF. We know that there exists a chain of irreducible closed subsets V Z1 Zr D Z with codim Zi D i . We shall show that there exist f1; : : : ; fr 2 kŒV such that, for all s r, Zs is an irreducible component of V .f1; : : : ; fs/ and all irreducible components of V .f1; : : : ; fs/ have codimension s. We prove this by induction on s. For s D 1, take any f1 2 I.Z1/, f1 ¤ 0, and apply Theorem 3.42. Suppose f1; : : : ; fs1 have been chosen, and let Y1; Y2; : : : ; Ym, be the irreducible components of V .f1; : : : ; fs1/, numbered so that Zs1 D Y1. We seek an element fs that is identically zero on Zs but is not identically zero on any Yi — for such an fs, all irreducible components of Yi \ V .fs/ will have codimension s, and Zs will be an irreducible component of Y1 \ V .fs/. But no Yi is contained in Zs because Zs has smaller dimension than Yi , and so I.Zs/ is not contained in any of the ideals I.Yi /. Now the prime i I.Yi /, and this is avoidance lemma (see below) tells us that there exist an fs 2 I.Zs/ X S the function we want. LEMMA 3.48 (PRIME AVOIDANCE LEMMA). If an ideal a of a ring A is not contained in any of the prime ideals p1; : : : ; pr , then it is not contained in their union. PROOF. We may assume that none of the prime ideals pi is contained in a second, because then we could omit it. For a fixed i , choose an fi 2 a X pi and, for each j ¤ i , choose an j D1 fj lies in each pj with j ¤ i and a, but not in pi (here we fj 2 pj X pi . Then hi use that pi is prime). The element Pr i D1 hi is therefore in a but not in any pi . defD Qr EXAMPLE 3.49. When V is an affine variety whose coordinate ring is a unique factorization domain, every closed subset Z of codimension 1 is of the form V .f / for some f 2 kŒV (see 2.66). The condition that kŒV be a unique factorization domain is definitely needed. Again consider the cone, in A 4 and let Z and Z0 be the planes V W X1X4 X2X3 D 0 Z D f.; 0; ; 0/g Z0 D f.0; ; 0; /g: Then Z \ Z0 D f.0; 0; 0; 0/g, which has codimension 2 in Z0. If Z D V .f / for some regular function f on V , then V .f jZ0/ D f.0; : : : ; 0/g, which has codimension 2, in violation of 3.42. Thus Z is not of the form V .f /, and so kŒX1; X2; X3; X4=.X1X4 X2X3/ is not a unique factorization domain. 78 3. AFFINE ALGEBRAIC VARIETIES Restatement in terms of affine algebras We restate some of these results in terms of affine algebras. 3.50. Theorem 3.42 says the following: let A be an affine k-algebra; if A is an integral domain and f 2 A is neither zero nor a unit, then every prime ideal p minimal among those containing .f / has height 1 (principal ideal theorem). 3.51. Corollary 3.44 says the following: let A be an affine k-algebra; if A is integral domain, then every maximal chain pd pd 1 p0 of distinct prime ideals has length equal to the Krull dimension of A. In particular, every maximal ideal in A has height dim.A/. 3.52. Let A be an affine k-algebra; if A is an integral domain and every prime ideal of height 1 in A is principal, then A is a unique factorization domain. In order to prove this, it suffices to show that every irreducible element f of A is prime (1.26). Let p be minimal among the prime ideals containing .f /. According to 3.50, p has height 1, and so it is principal, say p D .g/. As .f / .g/, f D gq for some q 2 A. Because f is irreducible, q is a unit, and so .f / D .g/ D p — the element f is prime. 3.53. Proposition 3.47 says the following: let A be an affine k-algebra, and let p be a prime ideal in A. If p has height r, then there exist elements f1; : : : ; fr 2 A such that p is minimal among the prime ideals containing .f1; : : : ; fr /. ASIDE 3.54. Statements 3.50 and 3.53 are true for all noetherian rings (CA 21.3, 21.8). However, 3.51 may fail. For example, as we noted on p. 16 a noetherian ring may have infinite Krull dimension. Moreover, a noetherian ring may have finite Krull dimension d without all of its maximal ideals having height d . For example, let A D RŒX, where R D kŒt.t/ is a discrete valuation ring with maximal ideal .t/. The Krull dimension of A is 2, and .t; X/ .t/ .0/ is a maximal chain of prime ideals, but the ideal .tX 1/ is maximal (because A=.tX 1/ ' Rt , see 1.13) and of height 1 (because it is in kŒt; X and A is obtained from kŒt; X by inverting the elements of kŒt X .t/). 3 is an irreducible component of V .f1; f2/ ASIDE 3.55. Proposition 3.47 shows that a curve C in A for some f1, f2 2 kŒX; Y; Z. In fact C D V .f1; f2; f3/ for suitable polynomials f1; f2, and f3 — this is an exercise in Shafarevich
1994 (I 6, Exercise 8; see also Hartshorne 1977, I, Exercise 2.17). 3 — see Apparently, it is not known whether two polynomials always suffice to define a curve in A 3 can’t be defined by two polynomials (ibid. Kunz 1985, p136.5 The union of two skew lines in P 3 can be defined by two polynomials. p. 140), but it is unknown whether all connected curves in P Macaulay (the man, not the program) showed that for every r 1, there is a curve C in A 3 such that I.C / requires at least r generators (see the same exercise in Hartshorne for a curve whose ideal can’t be generated by 2 elements).6 5Kunz, Ernst Introduction to commutative algebra and algebraic geometry. Birkh¨auser Boston, Inc., Boston, MA 6In 1882 Kronecker proved that every algebraic subset in P n can be cut out by n C 1 polynomial equations. 3 which he claimed was not In 1891 Vahlen asserted that the result was best possible by exhibiting a curve in P the zero locus of 3 equations. It was only 50 years later, in 1941, that Perron gave 3 equations defining Vahlen’s curve, thus refuting Vahlen’s claim which had been accepted for half a century. Finally, in 1973 Eisenbud and n (mo35468 Evans proved that n equations always suffice to describe (set-theoretically) an algebraic subset of P Georges Elencwajg). m. Dimension 79 In general, a closed variety V of codimension r in A n/ is said to be a set-theoretic complete intersection if there exist r polynomials fi 2 kŒX1; : : : ; Xn (resp. homogeneous polynomials fi 2 kŒX0; : : : ; Xn/ such that n (resp. P V D V .f1; : : : ; fr /: Such a variety is said to be an ideal-theoretic complete intersection if the fi can be chosen so that I.V / D .f1; : : : ; fr /. Chapter V of Kunz’s book is concerned with the question of when a variety is a complete intersection. Obviously there are many ideal-theoretic complete intersections, but most of the varieties one happens to be interested in turn out not to be. For example, no abelian variety of dimension > 1 is an ideal-theoretic complete intersection (being an ideal-theoretic complete intersection imposes constraints on the cohomology of the variety, which are not fulfilled in the case of abelian varieties). Let P be a point on an irreducible variety V A n. Then 3.47 shows that there is a neighbourhood n and functions f1; : : : ; fr on U such that U \ V D V .f1; : : : ; fr / (zero set in U /. Thus U of P in A U \ V is a set-theoretic complete intersection in U . One says that V is a local complete intersection n such that the ideal I.V \ U / can be at P 2 V if there is an open affine neighbourhood U of P in A generated by r regular functions on U . Note that ideal-theoretic complete intersection ) local complete intersection at all p: It is not difficult to show that a variety is a local complete intersection at every nonsingular point (cf. 4.36). Exercises 3-1. Show that a map between affine varieties can be continuous for the Zariski topology without being regular. 3-2. Let V D Spm.A/, and let Z D Spm.A=a/ Spm.A/. Show that a function f on an open subset U of Z is regular if and only if, for each P 2 U , there exists a regular function f 0 on an open neighbourhood U 0 of P in V such that f and f 0 agree on U 0 \ U . 3-3. Find the image of the regular map .x; y/ 7! .x; xy/W A 2 ! A 2 and verify that it is neither open nor closed. 3-4. Show that the circle X 2 C Y 2 D 1 is isomorphic (as an affine variety) to the hyperbola XY D 1, but that neither is isomorphic to A 1. (Assume char.k/ ¤ 2:/ 3-5. Let C be the curve Y 2 D X 2 C X 3, and let ' be the regular map t 7! .t 2 1; t.t 2 1//W A 1 ! C: Is ' an isomorphism? CHAPTER 4 Local Study Geometry is the art of drawing correct conclusions from incorrect figures. (La g´eom´etrie est l’art de raisonner juste sur des figures fausses.) Descartes In this chapter, we examine the structure of an affine algebraic variety near a point. We begin with the case of a plane curve, since the ideas in the general case are the same but the proofs are more complicated. a. Tangent spaces to plane curves Consider the curve V in the plane defined by a nonconstant polynomial F .X; Y /, V W F .X; Y / D 0: We assume that F .X; Y / has no multiple factors, so that .F .X; Y // is a radical ideal and I.V / D .F .X; Y //. We can factor F into a product of irreducible polynomials, F .X; Y / D Q Fi .X; Y /, and then V D S V .Fi / expresses V as a union of its irreducible components (see 2.29). Each component V .Fi / has dimension 1 (by 2.64) and so V has pure dimension 1. If F .X; Y / itself is irreducible, then kŒV D kŒX; Y =.F .X; Y // D kŒx; y is an integral domain. Moreover, if F ¤ X c, then x is transcendental over k and y is algebraic over k.x/, and so x is a transcendence basis for k.V / over k. Similarly, if F ¤ Y c, then y is a transcendence basis for k.V / over k. Let .a; b/ be a point on V . If we were doing calculus, we would say that the tangent space at P D .a; b/ is defined by the equation .a; b/.X a/ C @F @Y @F @X .a; b/.Y b/ D 0: (20) This is the equation of a line unless both @F the equation of a plane. @X .a; b/ and @F @Y .a; b/ are zero, in which case it is We are not doing calculus, but we can define @ X aij X i Y j D X iaij X i 1Y j ; @ @X 81 @ @Y @X and @ X @Y by aij X i Y j D X jaij X i Y j 1, 82 4. LOCAL STUDY and make the same definition. DEFINITION 4.1. The tangent space TP V to V at P D .a; b/ is the algebraic subset defined by equation (20). If @F @X .a; b/ and @F @Y .a; b/ are not both zero, then TP .V / is a line through .a; b/, and we say that P is a nonsingular or smooth point of V . Otherwise, TP .V / has dimension 2, and we say that P is singular or multiple. The curve V is said to be nonsingular or smooth if all its points are nonsingular. Examples For each of the following examples, the reader is invited to sketch the curve. Assume that char.k/ ¤ 2; 3. 4.2. X m C Y m D 1. The tangent space at .a; b/ is given by the equation mam1.X a/ C mbm1.Y b/ D 0: All points on the curve are nonsingular unless the characteristic of k divides m, in which case X m C Y m 1 has multiple factors, X m C Y m 1 D X m0p C Y m0p 1 D .X m0 C Y m0 1/p: 4.3. Y 2 D X 3 (sketched in 4.12 below). The tangent space at .a; b/ is given by the equation 3a2.X a/ C 2b.Y b/ D 0: The only singular point is .0; 0/. 4.4. Y 2 D X 2.X C 1/ (sketched in 4.10 below). Here again only .0; 0/ is singular. 4.5. Y 2 D X 3 C aX C b. In 2.2 we sketched two nonsingular examples of such curves, and in 4.10 and 4.11 we sketch two singular examples. The singular points of the curve are the common zeros of the polynomials Y 2 X 3 aX b; 2Y; 3X 2 C a, which consist of the points .c; 0/ with c a common zero of X 3 C aX C b; 3X 2 C a. As 3X 2 C a is the derivative of X 3 C aX C b, we see that V is singular if and only if X 3 C aX C b has a multiple root. 4.6. V D V .F G/ where F G has no multiple factors (so F and G are coprime). Then V D V .F / [ V .G/, and a point .a; b/ is singular if and only if it is ˘ a singular point of V .F /, ˘ a singular point of V .G/, or ˘ a point of V .F / \ V .G/. This follows immediately from the product rule: @.F G/ @X D F @G @X C @F @X G; @.F G/ @Y D F @G @Y C @F @Y G: b. Tangent cones to plane curves 83 The singular locus PROPOSITION 4.7. The nonsingular points of a plane curve form a dense open subset of the curve. PROOF. Let V D V .F /, where F is a nonconstant polynomial in kŒX; Y without multiple factors. It suffices to show that the nonsingular points form a dense open subset of each irreducible component of V , and so we may assume that V (hence F ) is irreducible. It suffices to show that the set of singular points is a proper closed subset. Since it is the set of common zeros of the polynomials F ; @F @X ; @F @Y ; it is obviously closed. It will be proper unless @F=@X and @F=@Y are both identically zero on V , and hence both multiples of F , but, as they have lower degree than F , this is impossible unless they are both zero. Clearly @F=@X D 0 if and only if F is a polynomial in Y (k of characteristic zero) or is a polynomial in X p and Y (k of characteristic p/. A similar remark applies to @F=@Y . Thus if @F=@X and @F=@Y are both zero, then F is constant (characteristic zero) or a polynomial in X p, Y p, and hence a pth power (characteristic p, see (18)). These are contrary to our assumptions. Thus the singular points form a proper closed subset, called the singular locus. ASIDE 4.8. In common usage, “singular” means uncommon or extraordinary as in “he spoke with singular shrewdness”. Thus the proposition says that singular points (mathematical sense) are singular (usual sense). b. Tangent cones to plane curves A polynomial F .X; Y / can be written (uniquely) as a finite sum F D F0 C F1 C F2 C C Fm C (21) with each Fm a homogeneous polynomial of degree m. The term F1 will be denoted F` and called the linear form of F , and the first nonzero term on the right of (21) (the homogeneous summand of F of least degree) will be denoted F and called the leading form of F . If P D .0; 0/ is on the curve V defined by F , then F0 D 0 and (21) becomes F D aX C bY C higher degree terms, and the equation of the tangent space is aX C bY D 0: DEFINITION 4.9. Let F .X; Y / be a polynomial without square factors, and let V be the curve defined by F . If .0; 0/ 2 V , then the geometric tangent cone to V at .0; 0/ is the zero set of F. The tangent cone is the pair .V .F/; kŒX; Y =F/. To obtain the tangent cone at any other point, translate to the origin, and then translate back. Note that the geometric tangent cone at a point on a curve always has dimension 1. While the tangent space tells you whether a point is nonsingular or not, the tangent cone also gives you information on the nature of a singularity. 84 4. LOCAL STUDY In general we can factor F as F.X; Y / D cX r0 Y .Y ai X/ri : i Then deg F D P ri is called the multiplicity of the singularity, multP .V /. A multiple point is ordinary if its tangents are nonmultiple, i.e., ri D 1 all i . An ordinary
double point is called a node. There are many names for special types of singularities — see any book, especially an old book, on algebraic curves. Examples The following examples are adapted from Walker, Robert J., Algebraic Curves. Princeton Mathematical Series, vol. 13. Princeton University Press, Princeton, N. J., 1950 (reprinted by Dover 1962). 4.10. F .X. The tangent cone at .0; 0/ is defined by Y 2 X 2. It is the pair of lines Y D ˙X, and the singularity is a node. 4.11. F .X. The origin is an isolated point of the real locus. It is again a node, but the tangent cone is defined by Y 2 C X 2, which is the pair of lines Y D ˙iX. In this case, the real locus of the tangent cone is just the point (0,0). 4.12. F .X; Y / D X 3 Y 2. Here the origin is a cusp. The tangent cone is defined by Y 2, which is the X-axis (doubled). 4.13. F .X; Y / D 2X 4 3X 2Y C Y 2 2Y 3 C Y 4. The origin is again a double point, but this time it is a tacnode. The tangent cone is again defined by Y 2. 4.14. F .X; Y / D X 4 C X 2Y 2 2X 2Y XY 2 Y 2. The origin is again a double point, but this time it is a ramphoid cusp. The tangent cone is again defined by Y 2. c. The local ring at a point on a curve 85 4.15. F .X; Y / D .X 2 C Y 2/2 C 3X 2Y Y 3. The origin is an ordinary triple point. The tangent cone is defined by 3X 2Y Y 3, which is the triple of lines Y D 0, Y D ˙ p 3X . 4.16. F .X; Y / D .X 2 C Y 2/3 4X 2Y 2. The origin has multiplicity 4. The tangent cone is defined by 4X 2Y 2, which is the union of the X and Y axes, each doubled. 4.17. F .X; Y / D X 6 X 2Y 3 Y 5. The tangent cone is defined by X 2Y 3 C Y 5, which consists of a triple line Y 3 D 0 and a pair of lines Y D ˙iX . ASIDE 4.18. Note that the real locus of the algebraic curve in 4.17 is smooth even though the curve itself is singular. Another example of such a curve is Y 3 C 2X 2Y X 4 D 0. This is singular at .0; 0/, but its real locus is the image of R under the analytic map t 7! .t 3 C 2t; t.t 3 C 2//, which is injective, 2 with closed image. See Milnor, J., Singular proper, and immersive, and hence an embedding into R points of complex hypersurfaces. PUP, 1968, or mo98366 (Elencwajg). c. The local ring at a point on a curve PROPOSITION 4.19. Let P be a point on a plane curve V , and let m be the corresponding maximal ideal in kŒV . If P is nonsingular, then dimk.m=m2/ D 1, and otherwise dimk.m=m2/ D 2. PROOF. Assume first that P D .0; 0/. Then m D .x; y/ in kŒV D kŒX; Y =.F .X; Y // D kŒx; y. Note that m2 D .x2; xy; y2/, and m=m2 D .X; Y /=.m2 C F .X; Y // D .X; Y /=.X 2; XY; Y 2; F .X; Y //: In this quotient, every element is represented by a linear polynomial cx C dy, and the only relation is F`.x; y/ D 0. Clearly dimk.m=m2/ D 1 if F` ¤ 0, and dimk.m=m2/ D 2 otherwise. Since F` D 0 is the equation of the tangent space, this proves the proposition in this case. The same argument works for an arbitrary point .a; b/ except that one uses the variables X 0 D X a and Y 0 D Y b; in essence, one translates the point to the origin. We explain what the condition dimk.m=m2/ D 1 means for the local ring P D kŒV m. Let n be the maximal ideal mkŒV m of this local ring. The map m ! n induces an isomorphism m=m2 ! n=n2 (see 1.15), and so we have O P nonsingular ” dimk m=m2 D 1 ” dimk n=n2 D 1: Nakayama’s lemma (1.3) shows that the last condition is equivalent to n being a principal ideal. As P has Krull dimension one (2.64), for its maximal ideal to be principal means that it is a regular local ring of dimension 1 (see 1.6). Thus, for a point P on a curve, O P nonsingular ” P regular. O 86 4. LOCAL STUDY PROPOSITION 4.20. Every regular local ring of dimension one is a principal ideal domain. PROOF. Let A be such a ring, and let m D ./ be its maximal ideal. According to the Krull intersection theorem (1.8), T r0 mr D .0/. Let a be a proper nonzero ideal in A. As a is finitely generated, there exists an r 2 N such that a mr but a 6 mrC1. Therefore, there exists an a D c r 2 a such that a … mrC1. The second condition implies that c … m, and so it is a unit. Therefore . r / D .a/ a . r /, and so a D . r / D mr . We have shown that all ideals in A are principal. By assumption, there exists a prime ideal p properly contained in m. Then A=p is an integral domain. As … p, it is not nilpotent in A=p, and hence not nilpotent in A. Let a and b be nonzero elements of A. There exist r; s 2 N such that a 2 mr X mrC1 and b 2 ms X msC1. Then a D u r and b D v s with u and v units, and ab D uv rCs ¤ 0. Hence A is an integral domain. It follows from the elementary theory of principal ideal domains that the following conditions on a principal ideal domain A are equivalent: (a) A has exactly one nonzero prime ideal; (b) A has exactly one prime element up to associates; (c) A is local and is not a field. A ring satisfying these conditions is called a discrete valuation ring. THEOREM 4.21. A point P on a plane algebraic curve is nonsingular if and only if regular, in which case it is a discrete valuation ring. P is O PROOF. The statement summarizes the above discussion. d. Tangent spaces to algebraic subsets of Am Before defining tangent spaces at points of an algebraic subset of A terminology from linear algebra (which should be familiar from advanced calculus). m we review some LINEAR ALGEBRA For a vector space km, let Xi be the i th coordinate function a 7! ai . Thus X1; : : : ; Xm is the dual basis to the standard basis for km. A linear form P ai Xi can be regarded as an element of the dual vector space .km/_ D Hom.km; k/. Let A D .aij / be an n m matrix. It defines a linear map ˛W km ! kn, by 0 B @ a1 ::: am 1 C A 7! A 1 C A D 0 B @ a1 ::: am 0 Pm B @ j D1 a1j aj ::: j D1 anj aj Pm 1 C A : Write X1; : : : ; Xm for the coordinate functions on km and Y1; : : : ; Yn for the coordinate functions on kn. Then Yi ı ˛ D m X j D1 aij Xj : This says that the i th coordinate of ˛.a/ is m X j D1 aij .Xj a/ D m X j D1 aij aj : d. Tangent spaces to algebraic subsets of A m 87 TANGENT SPACES DEFINITION 4.22. Let V km be an algebraic subset of km, and let a D I.V /. The tangent space Ta.V / to V at a point a D .a1; : : : ; am/ of V is the subspace of the vector space with origin a cut out by the linear equations m X i D1 @F @Xi ˇ ˇ ˇ ˇa .Xi ai / D 0; F 2 a. (22) In other words, Ta.A the subspace of Ta.A m/ defined by the equations (22). m/ is the vector space of dimension m with origin a, and Ta.V / is Write .dXi /a for .Xi ai /; then the .dXi /a form a basis for the dual vector space m/_. As in m/_ to Ta.A m/ — in fact, they are the coordinate functions on Ta.A Ta.A advanced calculus, we define the differential of a polynomial F 2 kŒX1; : : : ; Xm at a by the equation: .dF /a D .dXi /a: m X i D1 @F @Xi ˇ ˇ ˇ ˇa It is again a linear form on Ta.A m/ defined by the equations: Ta.A m/. In terms of differentials, Ta.V / is the subspace of .dF /a D 0; F 2 a: (23) I claim that, in (22) and (23), it suffices to take the F to lie in a generating subset for a. The product rule for differentiation shows that if G D P j Hj Fj , then .dG/a D X Hj .a/ .dFj /a C Fj .a/ .dHj /a: j If F1; : : : ; Fr generate a and a 2 V .a/, so that Fj .a/ D 0 for all j , then this equation becomes .dG/a D X Hj .a/ .dFj /a: j Thus .dF1/a; : : : ; .dFr /a generate the k-vector space f.dF /a j F 2 ag. DEFINITION 4.23. A point a on an algebraic set V is nonsingular (or smooth) if it lies on a single irreducible component W of V and the dimension of the tangent space at a is equal to the dimension of W ; otherwise it is singular (or multiple). Thus, a point a on an irreducible algebraic set V is nonsingular if and only if dim Ta.V / D dim V . As in the case of plane curves, a point on V is nonsingular if and only if it lies on a single irreducible component of V , and is nonsingular on it. Let a D .F1; : : : ; Fr /, and let J D Jac.F1; : : : ; Fr / D D @Fi @Xj F1 @X1 ::: @Fr @X1 ; : : : ; 1 C C A : @F1 @Xm ::: @Fr @Xm Then the equations defining Ta.V / as a subspace of Ta.A linear algebra shows that m/ have matrix J.a/. Therefore, dimk Ta.V / D m rank J.a/; 88 4. LOCAL STUDY and so a is nonsingular if and only if the rank of Jac.F1; : : : ; Fr /.a/ is equal to m dim.V /. For example, if V is a hypersurface, say I.V / D .F .X1; : : : ; Xm//, then Jac.F /.a/ D @F @X1 .a/; : : : ; .a/ ; @F @Xm and a is nonsingular if and only if not all of the partial derivatives @F @Xi We can regard J as a matrix of regular functions on V . For each r, vanish at a. fa 2 V j rank J.a/ rg is closed in V , because it is the set where certain determinants vanish. Therefore, there is an open subset U of V on which rank J.a/ attains its maximum value, and the rank jumps on closed subsets. Later (4.37) we shall show that the maximum value of rank J.a/ is m dim V , and so the nonsingular points of V form a nonempty open subset of V . e. The differential of a regular map Consider a regular map 'W A m ! A n; a 7! .P1.a1; : : : ; am/; : : : ; Pn.a1; : : : ; am//: We think of ' as being given by the equations Yi D Pi .X1; : : : ; Xm/; i D 1; : : : ; n: It corresponds to the map of rings 'W kŒY1; : : : ; Yn ! kŒX1; : : : ; Xm sending Yi to Pi .X1; : : : ; Xm/, i D 1; : : : ; n. Let a 2 A n/ to be the map such that m, and let b D '.a/. Define .d'/aW Ta.A .d Yi /b ı .d'/a D X @Pi @Xj m/ ! Tb.A ˇ ˇ ˇ ˇa .dXj /a; i.e., relative to the standard bases, .d'/a is the map with matrix Jac.P1; : : : ; Pn/.a/ D 0 B B @ @P1 @X1 @Pn @X1 .a/; ::: .a/; : : : ; .a/ @P1 @Xm ::: : : : ; @Pn @Xm .a/ 1 C C A : For example, suppose a D .0; : : : ; 0/ and b D .0; : : : ; 0/, so that Ta.A kn, and m/ D km and Tb.A n/ D Pi D cij Xj C .higher terms), i D 1; : : : ; n: m X j D1 Then Yi ı .d'/a D P i.e., it is simply t 7! .cij /t. j cij Xj , and the map on tangent spaces is given by the matrix .cij /, Let F 2 kŒX1; : : : ; Xm. We can regard F as a regular map A m ! A 1, whose differential will be a linear map m/ ! Tb.A .dF /aW Ta.A 1/ with k, we obtain an identification of the differential of F (F b D F .a/
: When we identify Tb.A regarded as a regular map) with the differential of F (F regarded as a regular function). 1/; f. Tangent spaces to affine algebraic varieties 89 LEMMA 4.24. Let 'W A km into W D V .b/ kn, then .d'/a maps Ta.V / into Tb.W /, b D '.a/. n be as at the start of this subsection. If ' maps V D V .a/ m ! A PROOF. We are given that and have to prove that f 2 b ) f ı ' 2 a; f 2 b ) .df /b ı .d'/a is zero on Ta.V /: The chain rule holds in our situation: @f @Xi D n X j D1 @f @Yj @Yj @Xi ; Yj D Pj .X1; : : : ; Xm/; f D f .Y1; : : : ; Yn/: If ' is the map given by the equations Yj D Pj .X1; : : : ; Xm/; j D 1; : : : ; n; then the chain rule implies d.f ı '/a D .df /b ı .d'/a; b D '.a/: Let t 2 Ta.V /; then .df /b ı .d'/a.t/ D d.f ı '/a.t/; which is zero if f 2 b because then f ı ' 2 a. Thus .d'/a.t/ 2 Tb.W /. We therefore get a map .d'/aW Ta.V / ! Tb.W /. The usual rules from advanced calculus hold. For example, .d /b ı .d'/a D d. ı '/a; b D '.a/: f. Tangent spaces to affine algebraic varieties The definition (4.22) of the tangent space at a point on an algebraic set uses the embedding n. In this section, we give an intrinsic definition of the tangent of the algebraic set into A space at a point of an affine algebraic variety that makes clear that it depends only on the local ring at the point. Dual numbers For an affine algebraic variety V and a k-algebra R (not necessarily an affine k-algebra), we define V .R/ to be Homk-alg.kŒV ; R/. For example, if V A n and a D I.V /, then V .R/ D f.a1; : : : ; an/ 2 Rn j f .a1; : : : ; an/ D 0 for all f 2 ag: A homomorphism R ! S of k-algebras defines a map V .R/ ! V .S/ of sets. The ring of dual numbers is kŒ" defD kŒX=.X 2/, where " D X C .X 2/. Thus kŒ" D k ˚ k" as a k-vector space, and .a C b"/.a0 C b0"/ D aa0 C .ab0 C a0b/"; a; b; a0; b0 2 k: Note that there is a k-algebra homomorphism " 7! 0W kŒ" ! k. 90 4. LOCAL STUDY DEFINITION 4.25. Let P be a point on an affine algebraic variety V over k. The tangent space to V at P is TP .V / D fP 0 2 V .kŒ"/ j P 0 7! P under V .kŒ"/ ! V .k/g: Thus an element of TP .V / is a homomorphism of k-algebras ˛W kŒV ! kŒ" whose "7!0! k is the point P . To say that kŒV ! k is the point P means that composite with kŒ" its kernel is mP , and so mP D ˛1.."//. PROPOSITION 4.26. Let V be an algebraic subset of A equipped with its canonical structure of an affine algebraic variety. Let P 2 V . Then O n, and let V 0 D .V; V / be V TP .V / (as defined in 4.22) ' TP .V 0/ (as defined in 4.25). PROOF. Let I.V / D a and let P D .a1; : : : ; an/. On rewriting a polynomial F .X1; : : : ; Xn/ in terms of the variables Xi ai , we obtain the (trivial Taylor) formula, F .X1; : : : ; Xn/ D F .a1; : : : ; an/ CX @F @Xi ˇ ˇ ˇ ˇa .Xi ai / C R with R a finite sum of products of at least two terms .Xi ai /. According to 4.25, TP .V 0/ consists of the elements a C "b of kŒ"n D kn ˚ kn" lying in V .kŒ"/. Let F 2 a. On setting Xi equal to ai C "bi in the above formula, we obtain: F .a1 C "b1; : : : ; an C "bn/ D " X @F @Xi ˇ ˇ ˇ ˇa bi : Thus, .a1 C "b1; : : : ; an C "bn/ lies in V .kŒ"/ if and only if .b1; : : : ; bn/ 2 Ta.V /. We can restate this as follows. Let V be an affine algebraic variety, and let P 2 V . Choose an embedding V ,! A n, and let P map to .a1; : : : ; an/. Then the point .a1; : : : ; an/ C .b1; : : : ; bn/" n.kŒ"/ is an element of TP .V / (definition 4.25) if and only if .b1; : : : ; bn/ is an element of A of TP .V / (definition 4.22). PROPOSITION 4.27. Let V be an affine variety, and let P 2 V . There is a canonical isomorphism TP .V / ' Hom. O P ; kŒ"/ (local homomorphisms of local k-algebras). PROOF. By definition, an element of TP .V / is a homomorphism ˛W kŒV ! kŒ" such that ˛1.."// D mP . Therefore ˛ maps elements of kŒV X mP into .kŒ" X ."// D kŒ", and so ˛ extends (uniquely) to a homomorphism ˛0W P ! kŒ". By construction, ˛0 is a local homomorphism of local k-algebras, and clearly every such homomorphism arises in this way from an element of TP .V /. O f. Tangent spaces to affine algebraic varieties 91 Derivations DEFINITION 4.28. Let A be a k-algebra and M an A-module. A k-derivation is a map DW A ! M such that (a) D.c/ D 0 for all c 2 k; (b) D.f C g/ D D.f / C D.g/; (c) D.fg/ D f Dg C g Df (Leibniz’s rule). Note that the conditions imply that D is k-linear (but not A-linear). We write Derk.A; M / for the k-vector space of all k-derivations A ! M . For example, let A be a local k-algebra with maximal ideal m, and assume that A=m D k. For f 2 A, let f .m/ denote the image of f in A=m. Then f f .m/ 2 m, and the map f 7! df defD f f .m/ mod m2 is a k-derivation A ! m=m2 because, mod m2, 0 D .f f .m//.g g.m// D fg C f .m/g.m/ C f .g g.m// C g.f f .m// D d.fg/ C f dg C g df: PROPOSITION 4.29. Let .A; m/ be as above. There are canonical isomorphisms Homlocal k-algebra.A; kŒ"/ ! Derk.A; k/ ! Homk-linear.m=m2; k/: c7!c! A PROOF. The composite k k-vector space, A decomposes into f 7!f .m/ ! k is the identity map, and so, when regarded as A D k ˚ m; f $ .f .m/; f f .m//: Let ˛W A ! kŒ" be a local homomorphism of k-algebras, and write ˛.a/ D a0 C D˛.a/". Because ˛ is a homomorphism of k-algebras, a0 D a.m/. We have ˛.ab/ D .ab/0 C D˛.ab/"; and ˛.a/˛.b/ D .a0 C D˛.a/"/.b0 C D˛.b/"/ D a0b0 C .a0D˛.b/ C b0D˛.a//": On comparing these expressions, we see that D˛ satisfies Leibniz’s rule, and therefore is a k-derivation A ! k. Conversely, if DW A ! k is a k-derivation, then ˛W a 7! a.m/ C D.a/" is a local homomorphism of k-algebras A ! kŒ", and all such homomorphisms arise in this way. A derivation DW A ! k is zero on k and on m2 (by Leibniz’s rule). It therefore defines a k-linear map m=m2 ! k. Conversely, a k-linear map m=m2 ! k defines a derivation by composition f 7!df! m=m2 ! k: A 92 4. LOCAL STUDY Tangent spaces and differentials We now summarize the above discussion in the context of affine algebraic varieties. 4.30. Let V be an affine algebraic variety, and let P be a point on V . Write mP for the corresponding maximal ideal in kŒV and nP for the maximal ideal mP V;P in the local ring at P . There are canonical isomorphisms O TP .V / Derk.kŒV ; k/ Homk-linear.mP =m2 P ; k/ (24) Homlocal k-algebra. P ; kŒ"/ Derk. P ; k/ O O In the middle term on the top row, kŒV acts on k through kŒV ! kŒV =mP ' k,1 and on P =nP ' k. The maps have the following the bottom row descriptions. P acts on k through P ! O O O Homk-linear.nP =n2 P ; k/: (a) By definition, TP .V / is the fibre of V .kŒ"/ ! V .k/ over P . To give an element of TP .V / amounts to giving a homomorphism ˛W kŒV ! kŒ" such that ˛1.."// D mP . (b) The homomorphism ˛ in (a) can be decomposed, ˛.f / D f .mP / ˚ D˛.f /"; f 2 kŒV , f .mP / 2 k, D˛.f / 2 k: ! k. The map D˛ is a k-derivation kŒV ! k, and D˛ induces a k-linear map mP =m2 P (c) The homomorphism ˛W kŒV ! kŒ" in (a) extends uniquely to a local homomorphism P ! kŒ". Similarly, a k-derivation kŒV ! k extends uniquely to a k-derivation P ! k. (d) The two right hand groups are related through the isomorphism mP =m2 P ! nP =n2 P O O of (1.15). 4.31. A regular map 'W V ! W defines a map '.kŒ"/W V .kŒ"/ ! W .kŒ"/. If Q D '.P /, then ' maps the fibre over P to the fibre over Q, i.e., it defines a map d'W TP .V / ! TQ.W /: This map of tangent spaces is called the differential of ' at P . (a) When V and W are embedded as closed subvarieties of A p. 89. TP .V / V .kŒ"/ d' ' TQ.W / W .kŒ"/ "7!0 "7!0 ' V .k/ W .k/ n, d' has the description in (b) As a map Hom. (c) As a map Hom.mP =m2 O P ; kŒ"/ ! Hom. Q; kŒ"/, d' is induced by '/ ! Hom.mQ=m2 Q; k/, d' is induced by the map mQ=m2 Q ! O mP =m2 P defined by 'W kŒW ! kŒV . EXAMPLE 4.32. Let E be a finite dimensional vector space over k. Then To.A.E// ' E: ASIDE 4.33. A map Spm.kŒ"/ ! V should be thought of as a curve in V but with only the first infinitesimal structure retained. Thus, the descriptions of the tangent space provided by the terms in the top row of (24) correspond to the three standard descriptions of the tangent space in differential geometry (Wikipedia: TANGENT SPACE). 1Thus, Derk.kŒV ; k/ depends on P . g. Tangent cones 93 g. Tangent cones Let V be an algebraic subset of km, and let a D I.V /. Assume that P D .0; : : : ; 0/ 2 V . Define a to be the ideal generated by the polynomials F for F 2 a, where F is the leading form of F (see p. 83). The geometric tangent cone at P , CP .V / is V .a/, and the tangent cone is the pair .V .a/; kŒX1; : : : ; Xn=a/. Obviously, CP .V / TP .V /. CAUTION. If a is principal, say a D .F /, then a D .F/, but if a D .F1; : : : ; Fr /, then it need not be true that a D .F1; : : : ; Fr/. Consider for example a D .XY; XZ C Z.Y 2 Z2//. One can show that this is an intersection of prime ideals, and hence is radical. As the polynomial Y Z.Y 2 Z2/ D Y .XZ C Z.Y 2 Z2// Z .XY / lies in a and is homogeneous, it lies in a, but it is not in the ideal generated by XY , XZ. In fact, a is the ideal generated by XY; XZ; Y Z.Y 2 Z2/: Let A be a local ring with maximal ideal n. The associated graded ring is gr.A/ D M ni =ni C1: i 0 Note that if A D Bm and n D mA, then gr.A/ D L mi =mi C1 (because of 1.15). PROPOSITION 4.34. The map kŒX1; : : : ; Xn=a ! gr. kŒX1; : : : ; Xn=a to the class of Xi in gr. P / is an isomorphism. O PROOF. Let m be the maximal ideal in kŒX1; : : : ; Xn=a corresponding to P . Then O P / sending the class of Xi in gr mi =mi C1 .X1; : : : ; Xn/i =.X1; : : : ; Xn/i C1 C a \ .X1; : : : ; Xn/i .X1; : : : ; Xn/i =.X1; : : : ; Xn/i C1 C ai ; where ai is the homogeneous piece of a of degree i (that is, the subspace of a consisting of homogeneous polynomials of degree i ). But .X1; : : : ; Xn/i =.X1; : : : ; Xn/i C1 C ai D i th homogeneous piece of kŒX1; : : : ; Xn=a: For an affine algebraic variety V and P 2 V , we define the geometric tangent cone P / by its P /red is the quotient of gr. CP .V / of V at P to be Spm.gr. nilradical, and we define the tangent cone to be .CP .V /; gr. P /red/, where gr. P //. O O O As in the case of a curve, the d
imension of the geometric tangent cone at P is the same as the dimension of V (because the Krull dimension of a noetherian local ring is equal to P / is a polynomial ring in dim V variables if and that of its graded ring). Moreover, gr. P / is a only if polynomial ring in d variables, in which case CP .V / D TP .V /. P is regular. Therefore, P is nonsingular (see below) if and only if gr. O O O O A regular map 'W V ! W sending P to Q induces a homomorphism gr. and hence a map CP .V / ! CQ.V / of the geometric tangent cones. CAUTION. The map on the rings kŒX1; : : : ; Xn=a defined by a map of algebraic varieties is not the obvious one, i.e., it is not necessarily induced by the same map on polynomial rings as the original map. To see what it is, it is necessary to use Proposition 4.34, i.e., it is necessary to work with the rings gr. Q/ ! gr. P /, O O P /. O 94 4. LOCAL STUDY h. Nonsingular points; the singular locus DEFINITION 4.35. A point P on an affine algebraic variety V is said to be nonsingular or smooth if it lies on a single irreducible component W of V and dim TP .V / D dim W ; otherwise the point is said to be singular. A variety is nonsingular if all of its points are nonsingular. The set of singular points of a variety is called its singular locus. Thus, on an irreducible variety V of dimension d , P is nonsingular ” dimk TP .V / D d ” dimk.nP =n2 P / D d . PROPOSITION 4.36. Let V be an irreducible variety of dimension d , and let P be a nonsingular point on V . Then there exist d regular functions f1; : : : ; fd defined in an open neighbourhood U of P such that P is the only common zero of the fi on U . PROOF. Suppose that P is nonsingular. Let f1; : : : ; fd generate the maximal ideal nP in P . Then f1; : : : ; fd are all defined on some open affine neighbourhood U of P , and I O claim that P is an irreducible component of the zero set V .f1; : : : ; fd / of f1; : : : ; fd in U . If not, there will be some irreducible component Z ¤ P of V .f1; : : : ; fd / passing through P . Write Z D V .p/ with p a prime ideal in kŒU . Because V .p/ V .f1; : : : ; fd / and because Z contains P and is not equal to it, we have .f1; : : : ; fd / p ¤ mP (ideals in kŒU /: On passing to the local ring P D kŒU mP , we find (using 1.14) that O .f1; : : : ; fd / p P ¤ nP O (ideals in P /: O This contradicts the assumption that the fi generate nP . Hence P is an irreducible component of V .f1; : : : ; fd /. On removing the remaining irreducible components of V .f1; : : : ; fd / from U , we obtain an open neighbourhood of P with the required property. Let P be a point on an irreducible variety V , and let f1; : : : ; fr generate the maximal ideal nP in P . The proof of the proposition shows that P is an irreducible component of V .f1; : : : ; fr /, and so r d (see 3.45). Nakayama’s lemma (1.3) shows that f1; : : : ; fr P span it. Thus dim TP .V / dim V , with generate nP if and only if their images in nP =n2 equality if and only if P is nonsingular. O A point P on V is nonsingular if and only if there exists an open affine neighbourhood U of P and functions f1; : : : fd on U such that .f1; : : : ; fd / is the ideal of all regular functions on U zero at P . THEOREM 4.37. The set of nonsingular points of an affine algebraic variety is dense and open. PROOF. Let V be an irreducible component of the variety. It suffices to show that the singular locus of V is a proper closed subset.2 2Let V1; : : : ; Vr be the irreducible components of V . Then Vi \ .T j ¤i Vj / is a proper closed subset of Vi . We show that .Vi /sing is a proper closed subset of Vi . It follows that Vi \ Vsing is the union of two proper closed subsets of Vi , and so it is proper and closed in Vi . Hence the points of Vi that are nonsingular on V form a nonempty open (hence dense) subset of Vi . h. Nonsingular points; the singular locus 95 We first show that it is closed. We may suppose that V D V .a/ A n. Let P1; : : : ; Pr generate a. Then the singular locus is the zero set of the ideal generated by the .n d / .n d / minors of the matrix Jac.P1; : : : ; Pr /.a/ D 0 B B @ @P1 @X1 ::: @Pr @X1 .a/ : : : .aa/ @P1 @Xn ::: @Pr @Xn .a/ which is closed. We now show that the singular locus is not equal to V . According to 3.36 and 3.37 some nonempty open affine subset of V is isomorphic to a nonempty open affine subset of an d C1, and so we may suppose that V itself is an irreducible irreducible hypersurface in A d C1, say, equal to the zero set of the nonconstant irreducible polynomial hypersurface in A F .X1; : : : ; Xd C1/. By 2.64, dim V D d . The singular locus is the set of common zeros of the polynomials F ; @F @X1 ; : : : ; @F @Xd C1 ; and so it will be proper unless the polynomials @F=@Xi are identically zero on V . As in the proof of 4.7, if @F=@Xi is identically zero on V .F /, then it is the zero polynomial, and so F is a polynomial in X1; : : : ; Xi 1; Xi C1; : : : Xd C1 (characteristic zero) or in X1; : : : ; X p i ; : : : ; Xd C1 (characteristic p). Therefore, if the singular locus equals V , then F is constant (characteristic 0) or a pth power (characteristic p), which contradicts the hypothesis. COROLLARY 4.38. If V is irreducible, then dim V D min P 2V dim TP .V /. PROOF. By definition dim TP .V / dim V , with equality if and only if P is nonsingular. As there exists a nonsingular P , dim V is the minimum value of dim TP .V /. This formula can be useful in computing the dimension of a variety. COROLLARY 4.39. An irreducible algebraic variety is nonsingular if and only if the tangent spaces TP .V /, P 2 V , have constant dimension. PROOF. The constant dimension is the dimension of V , and so all points are nonsingular. COROLLARY 4.40. Every variety on which a group acts transitively by regular maps is nonsingular. PROOF. The group must act by isomorphisms, and so the tangent spaces have constant dimension. In particular, every group variety (see p. 109) is nonsingular. 96 Examples 4. LOCAL STUDY 4.41. For the surface Z3 D XY , the only singular point is .0; 0; 0/. The tangent cone at .0; 0; 0/ has equation XY D 0, and so it is the union of two planes intersecting in the z-axis. 4.42. For the surface V W Z3 D X 2Y , the singular locus is the line X D 0 D Z (and the singularity at .0; 0/ is very bad: for example, it lies in the singular set of the singular set.3 The intersection of the surface with the surface Y D c is the cuspidal curve X 2 D Z3=c: 3, and let W be the zero set of 4.43. Let V be the union of the coordinate axes in A XY.X Y / in A 2. Each of V and W is a union of three lines meeting at the origin. Are they isomorphic as algebraic varieties? Obviously, the origin o is the only singular point on V or W . An isomorphism V ! W would have to send the singular point o to the singular point o and map To.V / isomorphically onto To.W /. But V D V .XY; Y Z; XZ/, and so To.V / has dimension 3, whereas ToW has dimension 2. Therefore, V and W are not isomorphic. i. Nonsingularity and regularity THEOREM 4.44. Let P be a point on an irreducible variety V . Every generating set for the maximal ideal nP of P has at least d elements, and there exists a generating set with d elements if and only if P is nonsingular. O PROOF. If f1; : : : ; fr generate nP , then the proof of 4.36 shows that P is an irreducible component of V .f1; : : : ; fr / in some open neighbourhood U of P . Therefore 3.45 shows that 0 d r, and so r d . The rest of the statement has already been noted. COROLLARY 4.45. A point P on an irreducible variety is nonsingular if and only if regular. P is O PROOF. This is a restatement of the second part of the theorem. According to CA 22.3, a regular local ring is an integral domain. If P lies on two P is not P is not an integral domain (3.14), and so irreducible components of a V , then regular. Therefore, the corollary holds also for reducible varieties. O O 3In fact, it belongs to the worst class of singularities (sx2848895, KReiser). 0.10.512y j. Examples of tangent spaces 97 j. Examples of tangent spaces The description of the tangent space in terms of dual numbers is particularly convenient when our variety is given to us in terms of its points functor. For example, let Mn be the set of n n matrices, and let I be the identity matrix. Write e for I when it is to be regarded as the identity element of GLn. 4.46. A matrix I C "A has inverse I "A in Mn.kŒ"/, and so lies in GLn.kŒ"/. In fact, Te.GLn/ D fI C "A j A 2 Mng ' Mn.k/: 4.47. On expanding det.I C "A/ as a sum of signed products and using that "2 D 0, we find that det.I C "A/ D I C "trace.A/: Hence Te.SLn/ D fI C "A j trace.A/ D 0g ' fA 2 Mn.k/ j trace.A/ D 0g: 4.48. Assume that the characteristic ¤ 2, and let On be the orthogonal group: On D fA 2 GLn j Atr A D I g: (Atr denotes the transpose of A). This is the group of matrices preserving the quadratic form n . The determinant defines a surjective regular homomorphism detW On ! f˙1g, X 2 1 whose kernel is defined to be the special orthogonal group SOn. For I C "A 2 Mn.kŒ"/, .I C "A/tr .I C "A/ D I C "Atr C "A; C C X 2 and so Te.On/ D Te.SOn/ D fI C "A 2 Mn.kŒ"/ j A is skew-symmetricg ' fA 2 Mn.k/ j A is skew-symmetricg: ASIDE 4.49. On the tangent space Te.GLn/ ' Mn of GLn, there is a bracket operation ŒM; N defD MN NM which makes Te.GLn/ into a Lie algebra. For any closed algebraic subgroup G of GLn, Te.G/ is stable under the bracket operation on Te.GLn/ and is a sub-Lie-algebra of Mn, which we denote Lie.G/. The Lie algebra structure on Lie.G/ is independent of the embedding of G into GLn (in fact, it has an intrinsic definition in terms of left invariant derivations), and G 7! Lie.G/ is a functor from the category of linear group varieties to that of Lie algebras. This functor is not fully faithful, for example, every ´etale homomorphism G ! G0 defines an isomorphism Lie.G/ ! Lie.G0/, but it is nevertheless very useful. Assume that k has characteristic zero. A connected algebraic group G is said to be semisimple if it has no close
d connected solvable normal subgroup (except feg). Such a group G may have a finite nontrivial centre Z.G/, and we call two semisimple groups G and G0 locally isomorphic if G=Z.G/ G0=Z.G0/. For example, SLn is semisimple, with centre n, the set of diagonal matrices diag.; : : : ; /, n D 1, and SLn =n D PSLn. A Lie algebra is semisimple if it has no commutative ideal (except f0g). One can prove that G is semisimple ” Lie.G/ is semisimple; and the map G 7! Lie.G/ defines a one-to-one correspondence between the set of local isomorphism classes of semisimple algebraic groups and the set of isomorphism classes of Lie algebras. The classification of semisimple algebraic groups can be deduced from that of semisimple Lie algebras and a study of the finite coverings of semisimple algebraic groups — this is quite similar to the relation between Lie groups and Lie algebras. 98 4. LOCAL STUDY Exercises 4-1. Find the singular points, and the tangent cones at the singular points, for each of (a 3Y 2X C 3X 2Y C 2XY I (b) X 4 C Y 4 X 2Y 2 (assume that the characteristic is not 2). 4-2. Let V A n be an irreducible affine variety, and let P be a nonsingular point on V . Let n (i.e., the subvariety defined by a linear equation P ai Xi D d with H be a hyperplane in A not all ai zero) passing through P but not containing TP .V /. Show that P is a nonsingular point on each irreducible component of V \ H on which it lies. (Each irreducible component has codimension 1 in V — you may assume this.) Give an example with H TP .V / and P singular on V \ H . Must P be singular on V \ H if H TP .V /? 4-3. Given a smooth point on a variety and a tangent vector at the point, show that there is a smooth curve passing through the point with the given vector as its tangent vector (see mo111467). 4-4. Let P and Q be points on varieties V and W . Show that T.P;Q/.V W / D TP .V / ˚ TQ.W /: 4-5. For each n, show that there is a curve C and a point P on C such that the tangent space to C at P has dimension n (hence C can’t be embedded in A 0 I 4-6. Let I be the n n identity matrix, and let J be the matrix . The symplectic group Spn is the group of 2n 2n matrices A with determinant 1 such that Atr J A D J . (It is the group of matrices fixing a nondegenerate skew-symmetric form.) Find the tangent space to Spn at its identity element, and also the dimension of Spn. n1 ). I 0 4-7. Find a regular map ˛W V ! W which induces an isomorphism on the geometric tangent cones CP .V / ! C˛.P /.W / but is not ´etale at P . 4-8. Show that the cone X 2 C Y 2 D Z2 is a normal variety, even though the origin is singular (characteristic ¤ 2). See p. 174. n. Suppose that a ¤ I.V /, and for a 2 V , let T 0 n/ defined by the equations .df /a D 0, f 2 a. Clearly, T 0 a a be the subspace Ta.V /, but need they 4-9. Let V D V .a/ A of Ta.A always be different? 4-10. Let W be a finite-dimensional k-vector space, and let RW D k ˚W endowed with the k-algebra structure for which W 2 D 0. Let V be an affine algebraic variety over k. Show that the elements of V .RW / defD Homk-algebra.kŒV ; RW / are in natural one-to-one correspondence with the pairs .P; t / with P 2 V and t 2 W ˝ TP .V / (cf. Mumford, Lectures on curves . . . , 1966, p25). CHAPTER 5 Algebraic Varieties An algebraic variety is a ringed space that is locally isomorphic to an affine algebraic variety, just as a topological manifold is a ringed space that is locally isomorphic to an open subset of R n. We require both to satisfy a separation axiom. a. Algebraic prevarieties As motivation, recall the following definitions. DEFINITION 5.1. (a) A topological manifold of dimension n is a ringed space .V; V / such that V is Hausdorff and every point of V has an open neighbourhood U for which V jU / is isomorphic to the ringed space of continuous functions on an open subset of .U; R n (cf. 3.2). O O (b) A differentiable manifold of dimension n is a ringed space such that V is Hausdorff V jU / is isomorphic to and every point of V has an open neighbourhood U for which .U; the ringed space of smooth functions on an open subset of R n (cf. 3.3). O (c) A complex manifold of dimension n is a ringed space such that V is Hausdorff and V jU / is isomorphic to the every point of V has an open neighbourhood U for which .U; ringed space of holomorphic functions on an open subset of C O n (cf. 3.4). These definitions are easily seen to be equivalent to the more classical definitions in terms of charts and atlases.1 Often one imposes additional conditions on V , for example, that it be connected or that it have a countable base of open subsets. DEFINITION 5.2. An algebraic prevariety over k is a k-ringed space .V; is quasicompact and every point of V has an open neighbourhood U for which .U; is isomorphic to the ringed space of regular functions on an algebraic set over k. V / such that V V jU / O O V / is an algebraic prevariety over k if there exists a finite open Thus, a ringed space .V; covering V D S Vi such that .Vi ; V jVi / is an affine algebraic variety over k for all i . An algebraic variety will be defined to be an algebraic prevariety satisfying a certain separation condition. O O An open subset U of an algebraic prevariety V such that .U , V jU / is an affine algebraic variety is called an open affine (subvariety) in V . Because V is a finite union of open affines, and in each open affine the open affines (in fact the basic open subsets) form a base for the topology, it follows that the open affines form a base for the topology on V . O 1Provided the latter are stated correctly, which is frequently not the case. 99 100 5. ALGEBRAIC VARIETIES Let .V; V / be an algebraic prevariety, and let U be an open subset of V . The functions O f W U ! k lying in .U; V / are called regular. Note that if .Ui / is an open covering of V by affine varieties, then f W U ! k is regular if and only if f jUi \ U is regular for all i (by 3.1(c)). Thus understanding the regular functions on open subsets of V amounts to understanding the regular functions on the open affine subvarieties and how these subvarieties fit together to form V . O EXAMPLE 5.3. (Projective space). Let P relation n denote knC1 X foriging modulo the equivalence .a0; : : : ; an/ .b0; : : : ; bn/ ” .a0; : : : ; an/ D .cb0; : : : ; cbn/ some c 2 k: Thus the equivalence classes are the lines through the origin in knC1 (with the origin omitted). Write .a0W : : : W an/ for the equivalence class containing .a0; : : : ; an/. For each i , let Ui D f.a0 W : : : W ai W : : : W an/ 2 P n j ai ¤ 0g: Then P n D S Ui , and the map .a0W : : : W an/ 7! a0 ai ; : : : ; bai ai ; : : : ; an ai W Ui ui! A n (the term ai =ai is omitted) is a bijection. In Chapter 6 we shall show that there is a unique n for which each Ui is an open affine structure of a (separated) algebraic variety on P subvariety of P n and each map ui is an isomorphism of algebraic varieties. b. Regular maps In each of the examples (5.1a,b,c), a morphism of manifolds (continuous map, smooth map, holomorphic map respectively) is just a morphism of ringed spaces. This motivates the following definition. O Let .V; V / and .W; W / be algebraic prevarieties. A map 'W V ! W is said to be regular if it is a morphism of k-ringed spaces. In other words, a continuous map 'W V ! W is regular if f 7! f ı ' sends a regular function on an open subset U of W to a regular function on '1.U /. A composite of regular maps is again regular (this is a general fact about morphisms of ringed spaces). O Note that we have three categories: (affine varieties) (algebraic prevarieties) (ringed spaces). Each subcategory is full, i.e., the morphisms Mor.V; W / are the same in the three categories. W / be prevarieties, and let 'W V ! W be a PROPOSITION 5.4. Let .V; O continuous map (of topological spaces). Let W D S Wj be a covering of W by open affines, and let '1.Wj / D S Vj i be a covering of '1.Wj / by open affines. Then ' is regular if and only if its restrictions V / and .W; O are regular for all i; j . 'jVj i W Vj i ! Wj PROOF. We assume that ' satisfies this condition, and prove that it is regular. Let f be a regular function on an open subset U of W . Then f jU \ Wj is regular for each Wj (sheaf condition 3.1(b)), and so f ı 'j'1.U / \ Vj i is regular for each j; i (this is our assumption). It follows that f ı ' is regular on '1.U / (sheaf condition 3.1(c)). Thus ' is regular. The converse is even easier. c. Algebraic varieties 101 ASIDE 5.5. A differentiable manifold of dimension n is locally isomorphic to an open subset of n. In particular, all manifolds of the same dimension are locally isomorphic. This is not true for R algebraic varieties, for two reasons: (a) We are not assuming our varieties are nonsingular (see Chapter 4). (b) The inverse function theorem fails in our context: a regular map that induces an isomorphism on the tangent space at a point P need not induce an isomorphism in a neighbourhood of P . However, see 5.55 below. c. Algebraic varieties In the study of topological manifolds, the Hausdorff condition eliminates such bizarre possibilities as the line with the origin doubled in which a sequence tending to the origin has two limits (see 5.10 below). It is not immediately obvious how to impose a separation axiom on our algebraic varieties, because even affine algebraic varieties are not Hausdorff. The key is to restate the Hausdorff condition. Intuitively, the significance of this condition is that it prevents a sequence in the space having more than one limit. Thus a continuous map into the space should be determined by its values on a dense subset, i.e., if '1 and '2 are continuous maps Z ! V that agree on a dense subset U of Z, then they should agree on the whole of Z.2 Equivalently, the set where two continuous maps '1; '2W Z U agree should be closed. Surprisingly, affine varieties have this property, provided '1 and '2 are required to be regular maps. LEMMA 5.6. Let '1; '2W Z V regular maps of affine algebraic varieties. Th
e subset of Z on which '1 and '2 agree is closed. PROOF. There are regular functions xi on V such that P 7! .x1.P /; : : : ; xn.P // identifies n (take the xi to be any set of generators for kŒV as a k-algebra). V with a closed subset of A Now xi ı '1 and xi ı '2 are regular functions on Z, and the set where '1 and '2 agree is Tn iD1 V .xi ı '1 xi ı '2/, which is closed. DEFINITION 5.7. An algebraic prevariety V is said to be separated if it satisfies the following additional condition: Separation axiom: for every pair of regular maps '1; '2W Z V with Z an affine algebraic variety, the set fz 2 Z j '1.z/ D '2.z/g is closed in Z. An algebraic variety over k is a separated algebraic prevariety over k.3 PROPOSITION 5.8. Let '1 and '2 be regular maps Z V from an algebraic prevariety Z to a separated prevariety V . The subset of Z on which '1 and '2 agree is closed. PROOF. Let W be the set on which '1 and '2 agree. For any open affine U of Z, W \ U is the subset of U on which '1jU and '2jU agree, and so W \ U is closed. This implies that W is closed because Z is a finite union of open affines. 2Let z 2 Z, and let z D lim un with un 2 U . Then '1.z/ D lim '1.un/ because '1 is continuous, and lim '1.un/ D lim '2.un/ D '2.z/. 3These are sometimes called “algebraic varieties in the sense of FAC” (Serre, Jean-Pierre. Faisceaux alg´ebriques coh´erents. Ann. of Math. (2) 61, (1955). 197–278; 34). In Grothendieck’s language, they are separated and reduced schemes of finite type over k (assumed to be algebraically closed), except that we omit the nonclosed points; cf. EGA IV, 10.10. Some authors use a more restrictive definition — they may require a variety to be connected, irreducible, or quasi-projective — usually because their foundations do not allow for a more flexible definition. 102 5. ALGEBRAIC VARIETIES EXAMPLE 5.9. The open subspace U D A when endowed with the sheaf OA2jU (cf. 3.33). A subvariety of an affine variety is said to be quasi-affine. For example, A 2 X f.0; 0/g of A 2 becomes an algebraic variety 2 X f.0; 0/g is quasi-affine but not affine. 1. EXAMPLE 5.10. (The affine line with the origin doubled.)4 Let V1 and V2 be copies of A Let V D V1 t V2 (disjoint union), and give it the obvious topology. Define an equivalence relation on V by x (in V1/ y (in V2/ ” x D y and x ¤ 0: Let V be the quotient space V D V = with the quotient topology (a set is open if and only if its inverse image in V is open): Then V1 and V2 are open subspaces of V , V D V1 [ V2, and V1 \ V2 D A 1 f0g. Define a function on an open subset to be regular if its restriction to each Vi is regular. This makes V into a prevariety, but not a variety: it fails the separation axiom because the two maps 1 D V1 ,! V ; A A 1 D V2 ,! V agree exactly on A 1 f0g, which is not closed in A 1. Let Vark denote the category of algebraic varieties over k and regular maps. The functor A Spm .A/ is a fully faithful contravariant functor Affk ! Vark, and defines an equivalence of the first category with the subcategory of the second whose objects are the affine algebraic varieties. 5.11. When V is irreducible, all the rings attached to it have a common field of fractions k.V / (see p. 113 below). Moreover, P D fg= h 2 k.V / j h.P / ¤ 0g O V .U / D \f O D \f V .U 0/ j U 0 U , U 0 open affineg P j P 2 U g: O O d. Maps from varieties to affine varieties O V / be an algebraic variety, and let ˛W A ! .V; Let .V; V / be a homomorphism from an affine k-algebra A to the k-algebra of regular functions on V . For any P 2 V , f 7! ˛.f /.P / is a k-algebra homomorphism A ! k, and so its kernel '.P / is a maximal ideal in A. In this way, we get a map O 'W V ! spm.A/ which is easily seen to be regular. Conversely, from a regular map 'W V ! Spm.A/, we get a k-algebra homomorphism f 7! f ı 'W A ! .V; V /. Since these maps are inverse, we have proved the following result. O 4This is the algebraic analogue of the standard example of a non Hausdorff topological space. Let R denote the real line with the origin removed but with two points a ¤ b added. The topology is generated by the open intervals in R together with the sets of the form .u; 0/ [ fag [ .0; v/ and .u; 0/ [ fbg [ .0; v/, where u < 0 < v. Then X is not Hausdorff because a and b cannot be separated by disjoint open sets. Every sequence that converges to a also converges to b; for example, 1=n converges to both a and b. e. Subvarieties 103 PROPOSITION 5.12. For an algebraic variety V and an affine k-algebra A, there is a canonical one-to-one correspondence Mor.V; Spm.A// ' Homk-algebra.A; .V; V //: O Let V be an algebraic variety such that .V; V / is an affine k-algebra. The proposition O shows that the regular map 'W V ! Spm. .V; V // defined by id .V;OV / has the following universal property: every regular map from V to an affine algebraic variety U factors uniquely through ': O ' V Spm. .V; V // O 9Š U: CAUTION 5.13. For a nonaffine algebraic variety V , .V; ated as a k-algebra. O V / need not be finitely gener- e. Subvarieties Let .V; O V / be an algebraic variety over k. Open subvarieties Let U be an open subset of V . Then U is a union of open affines, and it follows that V jU / is a variety, called an open subvariety of V . A regular map 'W W ! V is an open .U; immersion if '.W / is open in V and ' defines an isomorphism W ! '.W / of varieties. O Closed subvarieties Let Z be a closed subset of V . A function f on an open subset U of Z is regular if, for every P 2 U , there exists a germ .U 0; f 0/ of a regular function at P on V such that f 0jU 0 \ U D f jU 0 \ U . This defines a ringed structure Z/ is a variety it ZjU \ Z/ suffices to check that, for every open affine U V , the ringed space .U \ Z; is an affine algebraic variety, but this is only an exercise (Exercise 3-2 to be precise). Such Z/ is called a closed subvariety of V . A regular map 'W W ! V is a closed a pair .Z; immersion if '.W / is closed in V and ' defines an isomorphism W ! '.W / of varieties. Z on Z. To show that .Z; O O O O Subvarieties A subset W of a topological space V is said to be locally closed if every point P in W has an open neighbourhood U in V such that W \ U is closed in U . Equivalent conditions: W is the intersection of an open and a closed subset of V ; W is open in its closure. A locally closed subset W of a variety V acquires a natural structure as a variety: write it as the intersection W D U \ Z of an open and a closed subset; Z is a variety, and W (being open in Z/ therefore acquires the structure of a variety. This structure on W has the following characterization: the inclusion map W ,! V is regular, and a map 'W V 0 ! W with V 0 a variety is regular if and only if it is regular when regarded as a map into V . With this structure, W is called a subvariety of V . A regular map 'W W ! V is an immersion if it induces an isomorphism of W onto a subvariety of V . Every immersion is the composite of an open immersion with a closed immersion (in both orders). 104 5. ALGEBRAIC VARIETIES Application PROPOSITION 5.14. A prevariety V is separated if and only if two regular maps from a prevariety to V agree on the whole prevariety whenever they agree on a dense subset of it. PROOF. If V is separated, then the set on which a pair of regular maps '1; '2W Z V agree is closed, and so must be the whole of the Z. Conversely, consider a pair of maps '1; '2W Z V , and let S be the subset of Z on which they agree. We assume that V has the property in the statement of the proposition, and show that S is closed. Let NS be the closure of S in Z. According to the above discussion, NS has the structure of a closed prevariety of Z and the maps '1j NS and '2j NS are regular. Because they agree on a dense subset of NS they agree on the whole of NS, and so S D NS is closed. f. Prevarieties obtained by patching PROPOSITION 5.15. Suppose that the set V is a finite union V D S i 2I Vi of subsets Vi and that each Vi is equipped with ringed space structure. Assume that the following “patching” condition holds: for all i; j , Vi \ Vj is open in both Vi and Vj and jVi \ Vj D Vi O jVi \ Vj . Vj O Then there is a unique structure of a ringed space on V for which (a) each inclusion Vi ,! V is a homeomorphism of Vi onto an open set, and (b) for each i 2 I , V jVi D Vi . O O If every Vi is an algebraic prevariety, then so also is V , and to give a regular map from V to a prevariety W amounts to giving a family of regular maps 'i W Vi ! W such that 'i jVi \ Vj D 'j jVi \ Vj : PROOF. One checks easily that the subsets U V such that U \ Vi is open for all i are the open subsets for a topology on V satisfying (a), and that this is the only topology to satisfy Vi .U \ Vi / V .U / to be the set of functions f W U ! k such that f jU \ Vi 2 (a). Define for all i . Again, one checks easily that V is a sheaf of k-algebras satisfying (b), and that it is the only such sheaf. O O O For the final statement, if each .Vi ; Vi / is a finite union of open affines, so also V /. Moreover, to give a map 'W V ! W amounts to giving a family of maps is .V; 'i W Vi ! W such that 'i jVi \ Vj D 'j jVi \ Vj (obviously), and ' is regular if and only 'jVi is regular for each i . O O Clearly, the Vi may be separated without V being separated (see, for example, 5.10). In 5.29 below, we give a condition on an open affine covering of a prevariety sufficient to ensure that the prevariety is separated. g. Products of varieties Let V and W be objects in a category C. A triple .V W; pW V W ! V; qW V W ! W / g. Products of varieties 105 is said to be the product of V and W if it has the following universal property: for every pair of morphisms Z ! V , Z ! W in C, there exists a unique morphism Z ! V W making the diagram Z 9Š commute. In other words, the triple is a product if the map p V V W q W ' 7! .p ı '; q ı '/W Hom.Z; V W / ! Hom.Z; V / Hom.Z; W / is a bijection. The product, if it exists, is uniquely determined up to a unique isomorphism by its universal property. For example, the product
of two sets (in the category of sets) is the usual cartesian product of the sets, and the product of two topological spaces (in the category of topological spaces) is the product of the underlying sets endowed with the product topology. We shall show that products exist in the category of algebraic varieties. Suppose, for the 0; Z/ is the underlying set of Z; 0 ! Z with image z is regular, and these are all moment, that V W exists. For any prevariety Z, Mor.A more precisely, for any z 2 Z, the map A the regular maps (cf. 3.28). Thus, from the definition of products we have (underlying set of V W / ' Mor.A ' Mor.A ' (underlying set of V / (underlying set of W /: 0; V W / 0; V / Mor.A 0; W / Hence, our problem can be restated as follows: given two prevarieties V and W , define on the set V W the structure of a prevariety such that (a) the projection maps p; qW V W V; W are regular, and (b) a map 'W T ! V W of sets (with T an algebraic prevariety) is regular if its compo- nents p ı '; q ı ' are regular. There can be at most one such structure on the set V W . Products of affine varieties EXAMPLE 5.16. Let a and b be ideals in kŒX1; : : : ; Xm and kŒXmC1; : : : ; XmCn respectively, and let .a; b/ be the ideal in kŒX1; : : : ; XmCn generated by the elements of a and b. Then there is an isomorphism f ˝ g 7! fgW kŒX1; : : : ; Xm a ˝k kŒXmC1; : : : ; XmCn b ! kŒX1; : : : ; XmCn .a; b/ : Again this comes down to checking that the natural map Homk-alg.kŒX1; : : : ; XmCn=.a; b/; R/ Homk-alg.kŒX1; : : : ; Xm=a; R/ Homk-alg.kŒXmC1; : : : ; XmCn=b; R/ is a bijection. But the three sets are respectively 106 5. ALGEBRAIC VARIETIES V .a; b/ D zero set of .a; b/ in RmCn; V .a/ D zero set of a in Rm; V .b/ D zero set of b in Rn; and so this is obvious. The tensor product of two k-algebras A and B has the universal property to be a product in the category of k-algebras, but with the arrows reversed. Because of the category antiequivalence (3.25), this shows that Spm.A ˝k B/ will be the product of Spm A and Spm B in the category of affine algebraic varieties once we have shown that A ˝k B is an affine k-algebra. PROPOSITION 5.17. Let A and B be k-algebras with A finitely generated. (a) If A and B are reduced, then so also is A ˝k B. (b) If A and B are integral domains, then so also is A ˝k B. PROOF. Let ˛ 2 A ˝k B. Then ˛ D Pn is a linear combination of the remaining bi , say, bn D Pn1 bilinearity of ˝, we find that i D1 ai ˝ bi , some ai 2 A, bi 2 B. If one of the bj i D1 ci bi , ci 2 k, then, using the ˛ D n1 X i D1 ai ˝ bi C n1 X i D1 ci an ˝ bi D n1 X i D1 .ai C ci an/ ˝ bi : Thus we can suppose that in the original expression of ˛, the bi are linearly independent over k. Now assume A and B to be reduced, and suppose that ˛ is nilpotent. Let m be a maximal ideal of A. From a 7! NaW A ! A=m D k we obtain homomorphisms a ˝ b 7! Na ˝ b 7! NabW A ˝k B ! k ˝k B ' ! B: The image P Nai bi of ˛ under this homomorphism is a nilpotent element of B, and hence is zero (because B is reduced). As the bi are linearly independent over k, this means that the Nai are all zero. Thus, the ai lie in all maximal ideals m of A, and so are zero (see 2.18). Hence ˛ D 0, and we have shown that A ˝k B is reduced. Now assume that A and B are integral domains, and let ˛, ˛0 2 A ˝k B be such that i with the sets fb1; b2; : : :g 2; : : :g each linearly independent over k. For each maximal ideal m of A, we know i / D 0. Thus either all ˛˛0 D 0. As before, we can write ˛ D P ai ˝bi and ˛0 D P a0 i 1; b0 and fb0 .P Nai bi /.P Na0 the ai 2 m or all the a0 i i / D 0 in B, and so either .P Nai bi / D 0 or .P Na0 2 m. This shows that ˝b0 i b0 i b0 spm.A/ D V .a1; : : : ; am/ [ V .a0 1; : : : ; a0 n/: As spm.A/ is irreducible (see 2.27), it follows that spm.A/ equals either V .a1; : : : ; am/ or V .a0 n/. In the first case ˛ D 0, and in the second ˛0 D 0. 1; : : : ; a0 REMARK 5.18. The proof of 5.17 fails when k is not algebraically closed, because then A=m may be a finite extension of k over which the bi become linearly dependent (see sx599391). The following examples show that the statement of 5.17 also fails in this case. (a) Suppose that k is nonperfect of characteristic p, so that there exists an element ˛ in an algebraic closure of k such that ˛ … k but ˛p 2 k. Let k0 D kŒ˛, and let ˛p D a. Then g. Products of varieties 107 .˛ ˝ 1 1 ˝ ˛/ ¤ 0 in k0 ˝k k0 (in fact, the elements ˛i ˝ ˛j , 0 i; j p 1, form a basis for k0 ˝k k0 as a k-vector space), but .˛ ˝ 1 1 ˝ ˛/p D .a ˝ 1 1 ˝ a/ D .1 ˝ a 1 ˝ a/ D 0: .because a 2 k/ Thus k0 ˝k k0 is not reduced, even though k0 is a field. (b) Let K be a finite separable extension of k and let ˝ be a second field containing k. By the primitive element theorem (FT 5.1), K D kŒ˛ D kŒX=.f .X//; for some ˛ 2 K and its minimal polynomial f .X/. Assume that ˝ is large enough to split f , say, f .X/ D Q i .X ˛i / with ˛i 2 ˝. Because K=k is separable, the ˛i are distinct, and so ˝ ˝k K ' ˝ŒX=.f .X// ' Y ˝ŒX=.X ˛i /; (1.58(b)) (1.1) which is not an integral domain. For example, C ˝ R C ' CŒX=.X i/ CŒX=.X C i/ ' C C: The proposition allows us to make the following definition. DEFINITION 5.19. The product of the affine varieties V and W is .V W; O V W / D Spm.kŒV ˝k kŒW / with the projection maps p; qW V W ! V; W defined by the homomorphisms f 7! f ˝ 1W kŒV ! kŒV ˝k kŒW g 7! 1 ˝ gW kŒW ! kŒV ˝k kŒW : PROPOSITION 5.20. Let V and W be affine varieties. (a) The variety .V W; V W / is the product of .V; W / in the category of affine algebraic varieties; in particular, the set V W is the product of the sets V and W and p and q are the projection maps. (b) If V and W are irreducible, then so also is V W . V / and .W; O O O PROOF. (a) As noted at the start of the subsection, the first statement follows from 5.17(a), and the second statement then follows by the argument on p. 105. (b) This follows from 5.17(b) and 2.27. COROLLARY 5.21. Let V and W be affine varieties. For every prevariety T , a map 'W T ! V W is regular if p ı ' and q ı ' are regular. PROOF. If p ı ' and q ı ' are regular, then 5.20 implies that ' is regular when restricted to any open affine of T , which implies that it is regular on T . 108 5. ALGEBRAIC VARIETIES The corollary shows that V W is the product of V and W in the category of prevarieties (hence also in the categories of varieties). EXAMPLE 5.22. (a) It follows from 1.57 that A mCn endowed with the projection maps m p A mCn q! A n; A p.a1; : : : ; amCn/ D .a1; : : : ; am/ q.a1; : : : ; amCn/ D .amC1; : : : ; amCn/; is the product of A m and A n. (b) It follows from 5.16 that is the product of V .a/ and V .b/. V .a/ p V .a; b/ q! V .b/ CAUTION. When V and W have dimension > 0, the topology on V W is strictly finer 1, every than product topology. For example, for the product topology on A proper closed subset is contained in a finite union of vertical and horizontal lines, whereas A 2 has many more closed subsets (see 2.68). If V is affine, then the diagonal in V V is closed for the Zariski topology. Therefore, if the Zariski topology on V V is equal to the product topology, then V is Hausdorff. We deduce that the Zariski topology on V V is the product topology if and only if V is finite. 2 D A 1 A Products in general We now define the product of two algebraic prevarieties V and W . Write V as a union of open affines V D S Vi , and note that V can be regarded as the Vi /; in particular, this covering satisfies the patching variety obtained by patching the .Vi ; condition (5.15). Similarly, write W as a union of open affines W D S Wj . Then O V W D [ Vi Wj and the .Vi Wj ; W; O O V W / to be the variety obtained by patching the .Vi Wj ; Vi Wj /. O Vi Wj / satisfy the patching condition. Therefore, we can define .V V W just defined, V W becomes the PROPOSITION 5.23. With the sheaf of k-algebras product of V and W in the category of prevarieties. In particular, the structure of prevariety on V W defined by the coverings V D S Vi and W D S Wj are independent of the coverings. O PROOF. Let T be a prevariety, and let 'W T ! V W be a map of sets such that p ı ' and q ı ' are regular. Then 5.21 implies that the restriction of ' to '1.Vi Wj / is regular. As these open sets cover T , this shows that ' is regular. PROPOSITION 5.24. If V and W are separated, then so also is V W . PROOF. Let '1; '2 be two regular maps U ! V W . The set where '1; '2 agree is the intersection of the sets where p ı '1; p ı '2 and q ı '1; q ı '2 agree, which is closed. PROPOSITION 5.25. If V and W are connected, then so also is V W . h. The separation axiom revisited 109 PROOF. For v0 2 V , we have continuous maps W ' v0 W closed V W: Similarly, for w0 2 W , we have continuous maps V ' V w0 closed V W: The images of V and W in V W intersect in .v0; w0/ and are connected, which shows that .v0; w/ and and .v; w0/ lie in the same connected component of V W for all v 2 V and w 2 W . Since v0 and w0 were arbitrary, this shows that any two points lie in the same connected component. Group varieties A group variety is an algebraic variety G together with a group structure defined by regular maps mW G G ! G; invW G ! G; eW A 0 ! G. A homomorphism of group varieties is a regular map that is also a homomorphism of groups. The algebraic variety, 8 ˆ< ˆ: SLn D Spm kŒX11; X12; : : : ; Xnn .det.Xij / 1// SLn.k/ D fM 2 Mn.k/ j det M D 1g becomes a group variety when endowed with its usual group structures. Matrix multiplication .aij / .bij / D .cij /; cij D Pn lD1 ail blj ; is given by polynomials, and Cramer’s rule gives an explicit expression of the entries of A1 as polynomials in the entries of A. The only affine group varieties of dimension 1 over k are Gm D Spm kŒX; X 1 and Ga D Spm kŒX: Every finite group N can be made into a group variety by setting N D Spm.A/ with A the k-algebra of all maps f W N ! k. h. The separation axiom revisited By way of motivation, consider a topological space V and the diagonal V V , defD .x; x/ j x 2 V . If is closed for the produ
ct topology, then every pair of points .x; y/ … has an open neighbourhood U U 0 such that .U U 0/ \ D ¿. In other words, if x and y are distinct points in V , then there are open neighbourhoods U and U 0 of x and y respectively such that U \ U 0 D ¿. Thus V is Hausdorff. Conversely, if V is Hausdorff, the reverse argument shows that is closed. For a variety V , we let D V (the diagonal) be the subset f.v; v/ j v 2 V g of V V . 110 5. ALGEBRAIC VARIETIES PROPOSITION 5.26. An algebraic prevariety V is separated if and only if V is closed.5 PROOF. We shall use the criterion 5.8: V is separated if and only if, for every pair of regular maps '1; '2W Z V , the subset of Z on which '1 and '2 agree is closed. Suppose that V is closed. The map .'1; '2/W Z ! V V; z 7! .'1.z/; '2.z// is regular because its components '1 and '2 are regular (see p. 105). In particular, it is continuous, and so .'1; '2/1.V / is closed, but this is exactly the subset on which '1 and '2 agree. Conversely, V is the set on which the two projection maps V V ! V agree, and so it is closed if V is separated. COROLLARY 5.27. For any prevariety V , the diagonal is a locally closed subset of V V . PROOF. Let P 2 V , and let U be an open affine neighbourhood of P . Then U U is an open neighbourhood of .P; P / in V V , and V \ .U U / D U , which is closed in U U because U is separated (5.6). Thus V is always a subvariety of V V , and it is closed if and only if V is separated. The graph ' of a regular map 'W V ! W is defined to be f.v; '.v// 2 V W j v 2 V g: COROLLARY 5.28. For any morphism 'W V ! W of prevarieties, the graph ' of ' is locally closed in V W , and it is closed if W is separated. The map v 7! .v; '.v// is an isomorphism of V onto ' (as algebraic prevarieties). PROOF. The map .v; w/ 7! .'.v/; w/W V W ! W W is regular because its composites with the projections are ' and idW which are regular. In particular, it is continuous, and as ' is the inverse image of W under this map, this proves the first statement. The second statement follows from the fact that the regular map ' ,! V W p! V is an inverse to v 7! .v; '.v//W V ! '. THEOREM 5.29. The following three conditions on a prevariety V are equivalent: (a) V is separated; (b) for every pair of open affines U and U 0 in V , U \ U 0 is an open affine, and the map f ˝ g 7! f jU \U 0 gjU \U 0W kŒU ˝k kŒU 0 ! kŒU \ U 0 is surjective; 5Recall that the topology on V V is not the product topology. Thus the statement does not contradict the fact that V is not Hausdorff. VΓϕvWϕ(v)(v,ϕ(v)) i. Fibred products 111 (c) the condition in (b) holds for the sets in some open affine covering of V . PROOF. Let U and U 0 be open affines in V . We shall prove that (i) if is closed then U \ U 0 affine, (ii) when U \ U 0 is affine, .U U 0/ \ is closed ” kŒU ˝k kŒU 0 ! kŒU \ U 0 is surjective: Assume (a); then these statements imply (b). Assume that (b) holds for the sets in an open affine covering .Ui /i 2I of V . Then .Ui Uj /.i;j /2I I is an open affine covering of V V , and V \ .Ui Uj / is closed in Ui Uj for each pair .i; j /, which implies (a). Thus, the statements (i) and (ii) imply the theorem. Proof of (i): The graph of the inclusion U \ U 0 ,! V is the subset .U U 0/ \ of .U \ U 0/ V: If V is closed, then .U U 0/ \ V is a closed subvariety of an affine variety, and hence is affine. Now 5.28 implies that U \ U 0 is affine. Proof of (ii): Assume that U \ U 0 is affine. Then .U U 0/ \ V is closed in U U 0 ” v 7! .v; v/W U \ U 0 ! U U 0 is a closed immersion ” kŒU U 0 ! kŒU \ U 0 is surjective (3.34). Since kŒU U 0 D kŒU ˝k kŒU 0, this completes the proof of (ii). In more down-to-earth terms, condition (b) says that U \ U 0 is affine and every regular function on U \ U 0 is a sum of functions of the form P 7! f .P /g.P / with f and g regular functions on U and U 0. EXAMPLE 5.30. (a) Let V D P 5.3). Then U0 \ U1 D A U0 \ U1 ,! Ui are 1, and let U0 and U1 be the standard open subsets (see 1 X f0g, and the maps on rings corresponding to the inclusions f .X/ 7! f .X/W kŒX ! kŒX; X 1 f .X/ 7! f .X 1/W kŒX ! kŒX; X 1: Thus the sets U0 and U1 satisfy the condition in (b). (b) Let V be A lower copies of A inclusions U0 \ U1 ,! Ui are 1 with the origin doubled (see 5.10), and let U and U 0 be the upper and 1 in V . Then U \ U 0 is affine, but the maps on rings corresponding to the X 7! XW kŒX ! kŒX; X 1 X 7! XW kŒX ! kŒX; X 1: Thus the sets U0 and U1 fail the condition in (b). (c) Let V be A 2 in V . Then U \ U 0 is not affine (see 3.33). of A 2 with the origin doubled, and let U and U 0 be the upper and lower copies i. Fibred products Let 'W V ! S and W W ! S be regular maps of algebraic varieties. The set V S W defD f.v; w/ 2 V W j '.v/ D .w/g 112 5. ALGEBRAIC VARIETIES is closed in V W; because it is the set where ' ı p and ı q agree, and so it has a canonical structure of an algebraic variety (see p. 103). The algebraic variety V S W is called the fibred product of V and W over S . Note that if S consists of a single point, then V S W D V W . Write '0 for the map .v; w/ 7! wW V S W ! W and 0 for the map .v; w/ 7! vW V S W ! V . We then have a commutative diagram: '0 V S W 0 ' V W S: The system .V S W; '0; 0/ has the following universal property: for any regular maps ˛W T ! V , ˇW T ! W such that '˛ D ˇ, there is a unique regular map .˛; ˇ/W T ! V S W such that the following diagram T ˇ .˛; ˇ / V S W ˛ 0 V '0 ' W S commutes. In other words, Hom.T; V S W / ' Hom.T; V / Hom.T;S/ Hom.T; W /: Indeed, there is a unique such map of sets, namely, t 7! .˛.t/; ˇ.t//, which is regular because it is as a map into V W . The map '0 in the above diagrams is called the base change of ' with respect to . For any point P 2 S , the base change of 'W V ! S with respect to P ,! S is the map '1.P / ! P induced by ', which is called the fibre of V over P . EXAMPLE 5.31. If f W V ! S is a regular map and U is a subvariety of S , then V S U is the inverse image of U in V . Notes 5.32. Since a tensor product of rings A ˝R B has the opposite universal property to that of a fibred product, one might hope that Spm.A/ Spm.R/ Spm.B/ ‹‹D Spm.A ˝R B/: This is true if A ˝R B is an affine k-algebra, but in general it may have nonzero nilpotent elements. For example, let k have characteristic p, let R D kŒX, and consider the kŒXalgebras kŒX ! k; X 7! a kŒX ! kŒX; X 7! X p: j. Dimension 113 Then A ˝R B ' k ˝kŒX p kŒX ' kŒX=.X p a/; which contains the nilpotent element x a 1 p . The correct statement is Spm.A/ Spm.R/ Spm.B/ ' Spm.A ˝R B=N/; (25) where N is the ideal of nilpotent elements in A ˝R B. To prove this, note that for any algebraic variety T , Mor.T; Spm.A ˝R B=N// ' Hom.A ˝R B=N; T .T // O T .T // ' Hom.A ˝R B; ' Hom.A; O T .T // O ' Mor.T; Spm.A// Hom.R;OT .T // Mor.T;Spm.R// Hom.B; T .T // O Mor.T; Spm.B// (5.12) .5.12). For the second isomorphism we used that the ring isomorphism, we used the universal property of A ˝R B. O T .T / is reduced, and for the third 5.33. Fibred products may differ depending on whether we are working in the category of algebraic varieties or algebraic schemes. For example, Spec.A/ Spec.R/ Spec.B/ ' Spec.A ˝R B/ in the category of schemes. Consider the map x 7! x2W A 1 (see 5.49). The fibre '1.a/ consists of two points if a ¤ 0, and one point if a D 0. Thus '1.0/ D Spm.kŒX=.X//. However, the scheme-theoretic fibre is Spec.kŒX=.X 2//, which reflects the fact that 0 is “doubled” in the fibre over 0. 1 '! A 5.34. Fibred products exist also for prevarieties. In this case, V S W is only locally closed in V W . j. Dimension Recall p. 46 that, in an irreducible topological space, every nonempty open subset is dense and irreducible. Let V be an irreducible algebraic variety V , and let U and U 0 be nonempty open affines in V . Then U \ U 0 is also a nonempty open affine (5.29), which is dense in U , and so the V .U \ U 0/ is injective. Therefore restriction map V .U / ! O O kŒU kŒU \ U 0 k.U /; where k.U / is the field of fractions of kŒU , and so k.U / is also the field of fractions of kŒU \ U 0 and of kŒU 0. Thus, attached to V there is a field k.V /, called the function field of V or the field of rational functions on V , which is the field of fractions of kŒU for any open affine U in V . The dimension of V is defined to be the transcendence degree of k.V / over k. Note the dim.V / D dim.U / for any open subset U of V . In particular, dim.V / D dim.U / for U an open affine in V . It follows that some of the results in 2 carry over — for example, if Z is a proper closed subvariety of V , then dim.Z/ < dim.V /. 114 5. ALGEBRAIC VARIETIES PROPOSITION 5.35. Let V and W be irreducible varieties. Then dim.V W / D dim.V / C dim.W /: PROOF. We may suppose V and W to be affine. Write kŒV D kŒx1; : : : ; xm kŒW D kŒy1; : : : ; yn; where the x and y have been chosen so that fx1; : : : ; xd g and fy1; : : : ; yeg are maximal algebraically independent sets of elements of kŒV and kŒW . Then fx1; : : : ; xd g and fy1; : : : ; yeg are transcendence bases of k.V / and k.W / (see 1.63), and so dim.V / D d and dim.W / D e. Now6 kŒV W defD kŒV ˝k kŒW kŒx1; : : : ; xd ˝k kŒy1; : : : ; ye; which is a polynomial ring in the symbols x1 ˝1; : : : ; xd ˝1; 1˝y1; : : : ; 1˝ye (see 1.57). In particular, the elements x1 ˝1; : : : ; xd ˝1; 1˝y1; : : : ; 1˝ye are algebraically independent in kŒV ˝k kŒW . Obviously kŒV W is generated as a k-algebra by the elements xi ˝ 1, 1 ˝ yj , 1 i m, 1 j n, and all of them are algebraic over kŒx1; : : : ; xd ˝k kŒy1; : : : ; ye. Thus the transcendence degree of k.V W / is d C e. We extend the definition of dimension to an arbitrary variety V as follows. An algebraic variety is a finite union of noetherian topological spaces, and so is noetherian. Consequently (see 2.31), V is a finite union V D S Vi of its irreducible components, and we define dim.V / D max dim.Vi /. When all the irreducible components of V have dimension n; V is said to be pure of dimension n (or to be of pure dimension n). PROPOSITION 5.36.
Let V and W be closed subvarieties of A ducible component Z of V \ W , n; for any (nonempty) irre- that is, dim.Z/ dim.V / C dim.W / nI codim.Z/ codim.V / C codim.W /: PROOF. In the course of the proof of Theorem 5.29, we saw that V \ W is isomorphic to \ .V W /, and this is defined by the n equations Xi D Yi in V W . Thus the statement follows from 3.45. REMARK 5.37. (a) The subvariety X 2 C Y 2 D Z2 Z D 0 3 is the curve X 2 C Y 2 D 0, which is the pair of lines Y D ˙iX if k D C; in particular, of A the codimension is 2. Note however, that real locus is f.0; 0/g, which has codimension 3. Thus, Proposition 5.36 becomes false if one looks only at real points (and the pictures we draw can mislead). 6In general, it is not true that if M 0 and N 0 are R-submodules of M and N , then M 0 ˝R N 0 is an Rsubmodule of M ˝R N . However, this is true if R is a field, because then M 0 and N 0 will be direct summands of M and N , and tensor products preserve direct summands. k. Dominant maps 115 (b) Proposition 5.36 becomes false if A n is replaced by an arbitrary affine variety. Consider for example the affine cone V It contains the planes, X1X4 X2X3 D 0: Z W X2 D 0 D X4I Z0 W X1 D 0 D X3I Z D f.; 0; ; 0/g Z0 D f.0; ; 0; /g and Z \ Z0 D f.0; 0; 0; 0/g. Because V is a hypersurface in A of Z and Z0 has dimension 2. Thus 4, it has dimension 3, and each codim Z \ Z0 D 3 — 1 C 1 D codim Z C codim Z0: The proof of 5.36 fails because the diagonal in V V cannot be defined by 3 equations 4) — the diagonal is not a set-theoretic (it takes the same 4 that define the diagonal in A complete intersection. k. Dominant maps As in the affine case, a regular map 'W V ! W is said to be dominant if the image of ' is dense in W . Suppose V and W are irreducible. If V 0 and W 0 are open affine subsets of V and W such that '.V 0/ W 0, then 3.34 implies that the map f 7! f ı 'W kŒW 0 ! kŒV 0 is injective. Therefore it extends to a map on the fields of fractions, k.W / ! k.V /, and this map is independent of the choice of V 0 and W 0. l. Rational maps; birational equivalence Loosely speaking, a rational map from a variety V to a variety W is a regular map from a dense open subset of V to W , and a birational map is a rational map admitting a rational inverse. Let V and W be varieties over k, and consider pairs .U; 'U /, where U is a dense open subset of V and 'U is a regular map U ! W . Two such pairs .U; 'U / and .U 0; 'U 0/ are said to be equivalent if 'U and 'U 0 agree on U \ U 0. An equivalence class of pairs is called a rational map 'W V Ü W . A rational map ' is said to be defined at a point v of V if v 2 U for some .U; 'U / 2 '. The set U1 of v at which ' is defined is open, and there is a regular map '1W U1 ! W such that .U1; '1/ 2 ' — clearly, U1 D S .U;'U /2' U and we can define '1 to be the regular map such that '1jU D 'U for all .U; 'U / 2 '. Hence, in the equivalence class, there is always a pair .U; 'U / with U largest (and U is called “the open subvariety on which ' is defined”). PROPOSITION 5.38. Let V and V 0 be irreducible varieties over k. A regular map 'W U 0 ! U from an open subset U 0 of V 0 onto an open subset U of V defines a k-algebra homomorphism k.V / ! k.V 0/, and every such homomorphism arises in this way. PROOF. The first part of the statement is obvious, so let k.V / ,! k.V 0/ be a k-algebra homomorphism. We identify k.V / with a subfield of k.V 0/. Let U (resp. U 0) be an open affine subset of V (resp. U 0). Let kŒU D kŒx1; : : : ; xm. Each xi 2 k.V 0/, which is the field of fractions of kŒU 0, and so there exists a nonzero d 2 kŒU 0 such that dxi 2 kŒU 0 116 5. ALGEBRAIC VARIETIES for all i . After inverting d , i.e., replacing U 0 with basic open subset, we may suppose that kŒU kŒU 0. Thus, the inclusion k.V / ,! k.V 0/ is induced by a dominant regular map 'W U 0 ! U . According to Theorem 9.1 below, the image of ' contains an open subset U0 of U . Now '1.U0/ '! U0 is the required map. A rational (or regular) map 'W V Ü W is birational if there exists a rational map '0W W Ü V such that '0 ı ' D idV and ' ı '0 D idW as rational maps. Two varieties V and V 0 are birationally equivalent if there exists a birational map from one to the other. In this case, there exist dense open subsets U and U 0 of V and V 0 respectively such that U U 0. PROPOSITION 5.39. Two irreducible varieties V and V 0 are birationally equivalent if and only if their function fields are isomorphic over k. PROOF. Assume that k.V / k.V 0/. We may suppose that V and W are affine, in which case the existence of U U 0 is proved in 3.36. This proves the “if” part, and the “only if” part is obvious. PROPOSITION 5.40. Every irreducible algebraic variety of dimension d is birationally equivalent to a hypersurface in A d C1. PROOF. Let V be an irreducible variety of dimension d . According to Proposition 3.38, there exist x1; : : : ; xd ; xd C1 2 k.V / such that k.V / D k.x1; : : : ; xd ; xd C1/. Let f 2 kŒX1; : : : ; Xd C1 be an irreducible polynomial satisfied by the xi , and let H be the hypersurface f D 0. Then k.V / k.H /. m. Local study Everything in Chapter 4, being local, extends mutatis mutandis, to general algebraic varieties. 5.41. The tangent space TP .V / at a point P on an algebraic variety V is the fibre of V .kŒ"/ ! V .k/ over P . There are canonical isomorphisms TP .V / ' Derk. P ; k/ ' Homk-linear.nP =n2 P ; k/; where nP is the maximal ideal of O P . O 5.42. A point P on an algebraic variety V is nonsingular (or smooth) if it lies on a single irreducible component W and dim TP .V / D dim W . A point P is nonsingular if and only if the local ring P is regular. The singular points form a proper closed subvariety, called the O singular locus. 5.43. A variety is nonsingular (or smooth) if every point is nonsingular. n. ´Etale maps DEFINITION 5.44. A regular map 'W V ! W of smooth varieties is ´etale at a point P of V if the map .d'/P W TP .V / ! T'.P /.W / is an isomorphism; ' is ´etale if it is ´etale at all points of V . n. ´Etale maps 117 Examples 5.45. A regular map 'W A n ! A n, a 7! .P1.a1; : : : ; an/; : : : ; Pn.a1; : : : ; an// is ´etale at a if and only if rank Jac.P1; : : : ; Pn/.a/ D n, because the map on the tangent spaces has matrix Jac.P1; : : : ; Pn/.a/. Equivalent condition: det ¤ 0. .a/ @Pi @Xj 5.46. Let V D Spm.A/ be an affine variety, and let f D P ci X i 2 AŒX be such that AŒX=.f .X// is reduced. Let W D Spm.AŒX=.f .X///, and consider the map W ! V corresponding to the inclusion A ,! AŒX=.f /. Thus AŒX=.f / AŒX W A V A 1 V The points of W lying over a point a 2 V are the pairs .a; b/ 2 V A 1 such that b is a root of P ci .a/X i . I claim that the map W ! V is ´etale at .a; b/ if and only if b is a simple root of P ci .a/X i . To see this, write A D kŒX1; : : : ; Xn=a, a D .f1; : : : ; fr /, so that AŒX=.f / D kŒX1; : : : ; Xn=.f1; : : : ; fr ; f /: The tangent spaces to W and V at .a; b/ and a respectively are the null spaces of the matrices 0 B B B B @ .a/ : : : @f1 @X1 ::: @f1 @Xn ::: .a/ 0 @fr @X1 @f @X1 .a/ .a/ : : : : : : @fr @Xn @f @Xn .a/ .a/ 0 @f @X .a; bf1 @X1 ::: @fr @X1 .a/ : : : .a/ : : : 1 C C A .a/ @f1 @Xn ::: @fr @Xn .a/ and the map T.a;b/.W / ! Ta.V / is induced by the projection map knC1 ! kn omitting the last coordinate. This map is an isomorphism if and only if @f @X .a; b/¤ 0, because then every solution of the smaller set of equations extends uniquely to a solution of the larger set. But @f @X .a; b/ D d.P i ci .a/X i / dX which is zero if and only if b is a multiple root of P i ci .a/X i . The intuitive picture is that W ! V is a finite covering with deg.f / sheets, which is ramified exactly at the points where two or more sheets cross. .b/; 5.47. Consider a dominant map 'W W ! V of smooth affine varieties, corresponding to a map A ! B of rings. Suppose B can be written B D AŒY1; : : : ; Yn=.P1; : : : ; Pn/ (same number of polynomials as variables). A similar argument to the above shows that ' is ´etale if and only if det is never zero. .a/ @Pi @Xj 5.48. The example in 5.46 is typical; in fact every ´etale map is locally of this form, provided P is a normal domain for all P 2 V . More precisely, let 'W W ! V V is normal, i.e., O 118 5. ALGEBRAIC VARIETIES be ´etale at P 2 W , and assume V to be normal; then there exist a map '0W W 0 ! V 0 with kŒW 0 D kŒV 0ŒX=.f .X//, and a commutative diagram W ' V U1 ´etale U2 U 0 1 ´etale U 0 2 W 0 '0 V 0 with all the U open subvarieties and P 2 U1. The failure of the inverse function theorem for the Zariski topology 5.49. In advanced calculus (or differential topology, or complex analysis), the inverse function theorem says that a map ' that is ´etale at a point a is a local isomorphism there, i.e., there exist open neighbourhoods U and U 0 of a and '.a/ such that ' induces an isomorphism U ! U 0. This is not true in algebraic geometry, at least not for the Zariski topology: a map can be ´etale at a point without being a local isomorphism. Consider for example the map 'W A 1 X f0g ! A 1 X f0g; a 7! a2: This is ´etale if the characteristic is ¤ 2, because the Jacobian matrix is .2X/, which has rank one for all X ¤ 0 (alternatively, it is of the form 5.46 with f .X/ D X 2 T , where T is the 1, and X 2 c has distinct roots for c ¤ 0). Nevertheless, I claim coordinate function on A that there do not exist nonempty open subsets U and U 0 of A 1 f0g such that ' defines an isomorphism U ! U 0. If there did, then ' would define an isomorphism kŒU 0 ! kŒU 1/. But on the fields of and hence an isomorphism on the fields of fractions k.A fractions, ' defines the map k.X/ ! k.X/, X 7! X 2, which is not an isomorphism. 1/ ! k.A 5.50. Let V be the plane curve Y 2 D X and ' the map V ! A 2 W 1 except over 0, and so we may view it schematically as 1, .x; y/ 7! x. Then ' is However, when viewed as a Riemann surface, V .C/ consists of two sheets joined at a single point O. As a point on the surface moves around O, it shifts from one sheet to the other. Thus the true picture is mo
re complicated. To get a section to ', it is necessary to remove a line in C from 0 to infinity, which is not closed for the Zariski topology. \|0VA1ϕ n. ´Etale maps 119 It is not possible to fit the graph of the complex curve Y 2 D X into 3-space, but the picture at right is an early depiction of it (from Neumann, Carl, Vorlesungen ¨uber Riemann’s theorie der Abel’schen integrale, Leipzig: Teubner, 1865). ´Etale maps of singular varieties Using tangent cones, we can extend the notion of an ´etale morphism to singular varieties. Obviously, a regular map ˛W V ! W induces a homomorphism gr. P /. We say that ˛ is ´etale at P if this is an isomorphism. Note that then there is an isomorphism of the geometric tangent cones CP .V / ! C˛.P /.W /, but this map may be an isomorphism without ˛ being ´etale at P . Roughly speaking, to be ´etale at P , we need the map on geometric tangent cones to be an isomorphism and to preserve the “multiplicities” of the components. O˛.P // ! gr. O O O˛.P / ! O P ! kŒŒX1; : : : ; Xd . It is a fairly elementary result that a local homomorphism of local rings ˛W A ! B induces an isomorphism on the graded rings if and only if it induces an isomorphism on the completions (Atiyah-MacDonald 1969, 10.23).7 Thus ˛W V ! W is ´etale at P if and only if the map O P is an isomorphism. Hence 5.53 shows that the choice of a local system of parameters f1; : : : ; fd at a nonsingular point P determines an isomorphism O O nonsingular point P ; then there is a canonical isomorphism O O f 2 O O For example, let V D A let t1; : : : ; td be a local system of parameters at a P ! kŒŒt1; : : : ; td . For 1, and let P be the point a. Then t D X a is a local parameter P consists of quotients f .X/ D g.X/= h.X / with h.a/ ¤ 0, and the coefficients of n0 an.X a/n of f .X/ can be computed as in elementary calculus P , the image of f 2 kŒŒt1; : : : ; td can be regarded as the Taylor series of f . We can rewrite this as follows: O at a, the Taylor expansion P courses: an D f .n/.a/=nŠ. PROPOSITION 5.51. Let 'W W ! V be a map of irreducible affine varieties. If k.W / is a finite separable extension of k.V /, then ' is ´etale on a nonempty open subvariety of W . PROOF. After passing to open subvarieties, we may assume that W and V are nonsingular, and that kŒW D kŒV ŒX=.f .X//, where f .X/ is separable when considered as a polynomial in k.V /. Now the statement follows from 5.46. 7Atiyah, M. F.; Macdonald, I. G., Introduction to commutative algebra. Addison-Wesley Publishing Co., 1969. Die lliemaim'sche Windungsfläche erster Ordnung.Tergl. Seite 162 168, 213-214ml 218-2 21LiAdhsti\MSvyer, Zeinzu/.Generated for jmilne (University of Michigan) on 2014-06-05 00:08 GMT / http://hdl.handle.net/2027/ucm.530650220xPublic Domain, Google-digitized / http://www.hathitrust.org/access_use#pd-google 120 5. ALGEBRAIC VARIETIES ASIDE 5.52. There is an old conjecture that every ´etale map 'W A write ' D .P1; : : : ; Pn/, then this becomes the statement: n ! A n is an isomorphism. If we .a/ is never zero (for a 2 kn), then ' has an inverse. never zero, implies that det @Pi @Xj @Pi @Xj is a nonzero constant (by the Null- stellensatz 2.11 applied to the ideal generated by det ). This conjecture, which is known as the Jacobian conjecture, has not been settled even for k D C and n D 2, despite the existence of several published proofs and innumerable announced proofs. It has caused many mathematicians a good deal of grief. It is probably harder than it is interesting. See the Wikipedia: JACOBIAN CONJECTURE. if det @Pi @Xj .a/ @Pi @Xj The condition, det o. ´Etale neighbourhoods Recall that a regular map ˛W W ! V is said to be ´etale at a nonsingular point P of W if the map .d˛/P W TP .W / ! T˛.P /.V / is an isomorphism. O Let P be a nonsingular point on a variety V of dimension d . A local system of parameters at P is a family ff1; : : : ; fd g of germs of regular functions at P generating the maximal ideal nP P . Equivalent conditions: the images of f1; : : : ; fd in nP =n2 P generate it as a k-vector space (see 1.4); or .df1/P ; : : : ; .dfd /P is a basis for the dual space to TP .V /. PROPOSITION 5.53. Let ff1; : : : ; fd g be a local system of parameters at a nonsingular point P of V . Then there is a nonsingular open neighbourhood U of P such that f1; f2; : : : ; fd are represented by pairs . Qf1; U /; : : : ; . Qfd ; U / and the map . Qf1; : : : ; Qfd /W U ! A PROOF. Obviously, the fi are represented by regular functions Qfi defined on a single open neighbourhood U 0 of P , which, because of 4.37, we can choose to be nonsingular. The map d is ´etale at P , because the dual map to .d˛/a is .dXi /o 7! .d Qfi /a. ˛ D . Qf1; : : : ; Qfd /W U 0 ! A The next lemma then shows that ˛ is ´etale on an open neighbourhood U of P . d is ´etale. LEMMA 5.54. Let W and V be nonsingular varieties. If ˛W W ! V is ´etale at P , then it is ´etale at all points in an open neighbourhood of P . PROOF. The hypotheses imply that W and V have the same dimension d , and that their tangent spaces all have dimension d . We may assume W and V to be affine, say W A m and V A n, and that ˛ is given by polynomials P1.X1; : : : ; Xm/; : : : ; Pn.X1; : : : ; Xm/. Then m/ ! T˛.a/.A .d˛/aW Ta.A , and ˛ is not ´etale at a if and only if the kernel of this map contains a nonzero vector in the subspace Ta.V / of n/. Let f1; : : : ; fr generate I.W /. Then ˛ is not ´etale at a if and only if the matrix Ta.A n/ is a linear map with matrix @Pi @Xj .afi @Xj @Pi @Xj .a/ .a/ has rank less than m. This is a polynomial condition on a, and so it fails on a closed subset of W , which doesn’t contain P . Let V be a nonsingular variety, and let P 2 V . An ´etale neighbourhood of a point P of V is a pair .Q; W U ! V / with an ´etale map from a nonsingular variety U to V and Q a point of U such that .Q/ D P . o. ´Etale neighbourhoods 121 COROLLARY 5.55. Let V be a nonsingular variety of dimension d , and let P 2 V . There is an open Zariski neighbourhood U of P and a map W U ! A d realizing .P; U / as an ´etale neighbourhood of .0; : : : ; 0/ 2 A d . PROOF. This is a restatement of the Proposition. ASIDE 5.56. Note the similarity to the definition of a differentiable manifold: every point P on a nonsingular variety of dimension d has an open neighbourhood that is also a “neighbourhood” of d . There is a “topology” on algebraic varieties for which the “open neighbourhoods” the origin in A of a point are the ´etale neighbourhoods. Relative to this “topology”, any two nonsingular varieties are locally isomorphic (this is not true for the Zariski topology). The “topology” is called the ´etale topology — see my notes Lectures on ´Etale Cohomology. The inverse function theorem (for the ´etale topology) THEOREM 5.57 (INVERSE FUNCTION THEOREM). If a regular map of nonsingular varieties 'W V ! W is ´etale at P 2 V , then there exists a commutative diagram V ' W open UP '0 ´etale U'.P / with UP an open neighbourhood of P , Uf .P / an ´etale neighbourhood '.P /, and '0 an isomorphism. PROOF. According to 5.54, there exists an open neighbourhood U of P such that the restriction 'jU of ' to U is ´etale. To get the above diagram, we can take UP D U , U'.P / to be the ´etale neighbourhood 'jU W U ! W of '.P /, and '0 to be the identity map. The rank theorem For vector spaces, the rank theorem says the following: let ˛W V ! W be a linear map of k-vector spaces of rank r; then there exist bases for V and W relative to which ˛ has matrix Ir 0 . In other words, there is a commutative diagram 0 0 ˛ .x1;:::;xm/7!.x1;:::;xr ;0;:::/ V km W kn: A similar result holds locally for differentiable manifolds. In algebraic geometry, there is the following weaker analogue. THEOREM 5.58 (RANK THEOREM). Let 'W V ! W be a regular map of nonsingular varieties of dimensions m and n respectively, and let P 2 V . If rank.TP .'// D n, then there exists a commutative diagram 'jUP .x1;:::;xm/7!.x1;:::;xn/ UP ´etale m A U'.P / ´etale n A 122 5. ALGEBRAIC VARIETIES in which UP and U'.P / are open neighbourhoods of P and '.P / respectively and the vertical maps are ´etale. PROOF. Choose a local system of parameters g1; : : : ; gn at '.P /, and let f1 D g1 ı'; : : : ; fn D gn ı '. Then df1; : : : ; dfn are linearly independent forms on TP .V /, and there exist fnC1; : : : ; fm such df1; : : : ; dfm is a basis for TP .V /_. Then f1; : : : ; fm is a local system of parameters at P . According to 5.54, there exist open neighbourhoods UP of P and U'.P / of '.P / such that the maps .f1; : : : ; fm/W UP ! A .g1; : : : ; gn/W U'.P / ! A m n are ´etale. They give the vertical maps in the above diagram. ASIDE 5.59. Tangent vectors at a point P on a smooth manifold V can be defined to be certain equivalence classes of curves through P (Wikipedia: TANGENT SPACE). For V D A n, there is a similar description with a curve taken to be a regular map from an open neighbourhood U of 0 1 to V . In the general case there is a map from an open neighbourhood of the point P in X in A onto affine space sending P to 0 and inducing an isomorphism from tangent space at P to that at 0 (5.53). Unfortunately, the maps from U A n need not lift to X, and so it is necessary to allow maps from smooth curves into X (pull-backs of the covering X ! A n). There is a description of the tangent vectors at a point P on a smooth algebraic variety V as certain equivalence classes of regular maps from an ´etale neighbourhood U of 0 in A n by the maps from U into A 1 to A 1 to V . p. Smooth maps DEFINITION 5.60. A regular map 'W V ! W of nonsingular varieties is smooth at a point P of V if .d'/P W TP .V / ! T'.P /.W / is surjective; ' is smooth if it is smooth at all points of V . THEOREM 5.61. A map 'W V ! W is smooth at P 2 V if and only if there exist open neighbourhoods UP and U'.P / of P and '.P / respectively such that 'jUP factors into UP ´etale!A dim V dim W U'.P / q! U'.P /: PROOF. Certainly, if 'jUP factors in this way, it is smooth. Conversely, if ' is smooth at P
, then we get a diagram as in the rank theorem. From it we get maps UP ! A m An U'.P / ! U'.P /: The first is ´etale, and the second is the projection of A mn U'.P / onto U'.P /. COROLLARY 5.62. Let V and W be nonsingular varieties. If 'W V ! W is smooth at P , then it is smooth on an open neighbourhood of V . PROOF. In fact, it is smooth on the neighbourhood UP in the theorem. q. Algebraic varieties as functors 123 Separable maps A transcendence basis S of an extension E F of fields is separating if the algebraic extension E F .S/ is separable. A finitely generated extension E F of fields is separable if it admits a separating transcendence basis. DEFINITION 5.63. A dominant map 'W W ! V of irreducible algebraic varieties is separable if k.W / is a separable extension of k.V /. THEOREM 5.64. Let 'W W ! V be a map of irreducible varieties. (a) If there exists a nonsingular point P of W such that 'P is nonsingular and .d'/P is surjective, then ' is dominant and separable. (b) Conversely if ' is dominant and separable, then the set of P 2 W satisfying (a) is open and dense. PROOF. Replace W and V with their open subsets of nonsingular points. Then apply the rank theorem. q. Algebraic varieties as functors Let R be an affine k-algebra, and let V be an algebraic variety. We define a point of V with coordinates in R (or an R-point of V ) to be a regular map Spm.R/ ! V . For example, if V D V .a/ A n, then V .R/ D f.a1; : : : ; an/ 2 Rn j f .a1; : : : ; an/ D 0 all f 2 ag; which is what you should expect. In particular V .k/ D V (as a set), i.e., V (as a set) can be identified with the set of points of V with coordinates in k. Note that .V W /.R/ D V .R/ W .R/ (property of a product). CAUTION 5.65. If V is the union of two subvarieties, V D V1 [ V2, then it need not be true that V .R/ D V1.R/ [ V2.R/. For example, for any polynomial f .X1; : : : ; Xn/, n D Df [ V .f /; A where Df ' Spm.kŒX1; : : : ; Xn; T =.1 Tf // and V .f / is the zero set of f , but Rn ¤ fa 2 Rn j f .a/ 2 Rg [ fa 2 Rn j f .a/ D 0g in general. In fact, it need not be true even when V1 and V2 are open in V . Indeed, this would say that every regular map U ! V with U affine must factor through V1 or V2, which 2 n f.0; 0/g is the union of the open subsets is nonsense. For example, the variety V D A V1W X ¤ 0 and V2W Y ¤ 0, but the affine subvariety U W X C Y D 1 of V is not contained in V1 or V2. 124 5. ALGEBRAIC VARIETIES THEOREM 5.66. A regular map 'W V ! W of algebraic varieties defines a family of maps of sets, '.R/W V .R/ ! W .R/, one for each affine k-algebra R, such that for every homomorphism ˛W R ! S of affine k-algebras, rhe diagram '.R/ V .R/ W .R/ V .˛/ V .ˇ / '.S/ V .S/ W .S/ (*) commutes. Every family of maps with this property arises from a unique morphism of algebraic varieties. Let Vark (resp. Affk) denote the category of algebraic varieties over k (resp. affine V denote the functor sending an affine algebraic varieties over k). For a variety V , let haff variety T D Spm.R/ to V .R/ D Hom.T; V /. We can restate Theorem 5.66 as follows. THEOREM 5.67. The functor V haff V W Vark ! Fun.Affk; Sets/ if fully faithful. PROOF. For an algebraic variety V over k, let hV denote the functor T Hom.T; V /W Vark ! Set: According to the Yoneda lemma (q.v. Wikipedia) the functor V hV W Vark ! Fun.Vark; Sets/ is fully faithful. Let ' be a morphism of functors haff V 0, and let T be an algebraic V variety. Let .Ui /i 2I be a finite affine covering of T . Each intersection Ui \ Uj is affine (5.29), and so ' gives rise to a commutative diagram ! haff 0 0 hV .T / hV 0.T / Y i hV .Ui / '.Ui / Y i hV 0.Ui / Y i;j hV .Ui \ Uj // '.Ui \Uj / Y i;j hV 0.Ui \ Uj // in which the pairs of maps are defined by the inclusions Ui \ Uj ,! Ui ; Uj . As the rows are exact (5.15, last sentence), this shows that 'V extends uniquely to a functor hV ! hV 0, which (by the Yoneda lemma) arises from a unique regular map V ! V 0. COROLLARY 5.68. To give an affine group variety is the same as giving a functor GW Affk ! Grp such that for some n and some finite set S of polynomials in kŒX1; X2; : : : ; Xn, G is isomorphic to the functor sending R to the set of zeros of S in Rn. PROOF. Certainly an affine group variety defines such a functor. Conversely, the conditions imply that G D hV for an affine algebraic variety V (unique up to a unique isomorphism). The multiplication maps G.R/G.R/ ! G.R/ give a morphism of functors hV hV ! hV . As hV hV ' hV V (by definition of V V ), we see that they arise from a regular map V V ! V . Similarly, the inverse map and the identity-element map are regular. q. Algebraic varieties as functors 125 It is not unusual for a variety to be most naturally defined in terms of its points functor. For example: SLnW R fM 2 Mn.R/ j det.M / D 1g GLnW R fM 2 Mn.R/ j det.M / 2 Rg GaW R .R; C/: We now describe the essential image of h 7! hV W Vark ! Fun.Affk; Sets/. The fibred product of two maps ˛1W F1 ! F3, ˛2W F2 ! F3 of sets is the set F1 F3 F2 D f.x1; x2/ j ˛1.x1/ D ˛2.x2/g: When F1; F2; F3 are functors and ˛1; ˛2; ˛3 are morphisms of functors, there is a functor F D F1 F3 F2 such that .F1 F3 F2/.R/ D F1.R/ F3.R/ F2.R/ for all affine k-algebras R. To simplify the statement of the next proposition, we write U for hU when U is an affine variety. PROPOSITION 5.69. A functor F W Affk ! Sets is in the essential image of Vark if and only if there exists an affine variety U and a morphism U ! F such that (a) the functor R defD U F U is a closed affine subvariety of U U and the maps R U defined by the projections are open immersions; (b) the set R.k/ is an equivalence relation on U.k/, and the map U.k/ ! F .k/ realizes F .k/ as the quotient of U.k/ by R.k/. PROOF. Let F D hV for V an algebraic variety. Choose a finite open affine covering V D S Ui of V , and let U D F Ui . It is again an affine variety (Exercise 5-2). The functor R is hU 0, where U 0 is the disjoint union of the varieties Ui \ Uj . These are affine (5.29), and so U 0 is affine. As U 0 is the inverse image of V in U U , it is closed (5.26). This proves (a), and (b) is obvious. The converse is omitted for the present. ASIDE 5.70. A variety V defines a functor R V .R/ from the category of all k-algebras to Sets. Again, we call the elements of V .R/ the points of V with coordinates in R. For example, if V is affine, V .R/ D Homk-algebra.kŒV ; R/: More explicitly, if V kn and I.V / D .f1; : : : ; fm/, then V .R/ is the set of solutions in Rn of the system equations fi .X1; : : : ; Xn/ D 0; i D 1; : : : ; m: Note that, when we allow R to have nilpotent elements, it is important to choose the fi to generate I.V / (i.e., a radical ideal) and not just an ideal a such that V .a/ D V .8 For a general variety V , we write V as a finite union of open affines V D S i Vi , and we define V .R/ to be the set of families .˛i /i 2I 2 Q i 2I Vi .R/ such that ˛i agrees with ˛j on Vi \ Vj for all i; j 2 I . This is independent of the choice of the covering, and agrees with the previous definition when V is affine. 8Let a be an ideal in kŒX1; : : :. If A has no nonzero nilpotent elements, then every k-algebra homomorphism kŒX1; : : : ! A that is zero on a is also zero on rad.a/, and so Homk.kŒX1; : : :=a; A/ ' Homk.kŒX1; : : :=rad.a/; A/: This is not true if A has nonzero nilpotents. 126 5. ALGEBRAIC VARIETIES The functor defined by A.E/ (see p. 72) is R R ˝k E. A criterion for a functor to arise from an algebraic prevariety 5.71. By a functor we mean a functor from the category of affine k-algebras to sets. A subfunctor U of a functor X is open if, for all maps 'W hA ! X, the subfunctor '1.U / of hA is defined by an open subvariety of Spm.A/. A family .Ui /i 2I of open subfunctors of X is an open covering of X if each Ui is open in X and X.K/ D S Ui .K/ for every field K. A functor X is local if, for all k-algebras R and all finite families .fi /i of elements of A generating A as an ideal, the sequence of sets X.R/ ! Y i X.Rfi / Y X.Rfi fj / i;j is exact. Let A .U / D Hom.U; A 1 denote the functor sending a k-algebra R to its underlying set. For a functor U , .U / is an affine let k-algebra and the canonical map U ! hO.U / is an isomorphism. A local functor admitting a finite covering by open affines is representable by an algebraic variety over k. 1/ — it is a k-algebra.9 A functor U is affine if O O In the functorial approach to algebraic geometry, an algebraic prevariety over k is defined to be a functor satisfying this criterion. See, for example, I, 1, 3.11, p. 13, of Demazure and Gabriel, Groupes alg´ebriques: g´eom´etrie alg´ebrique, g´en´eralit´es, groupes commutatifs. 1970. r. Rational and unirational varieties DEFINITION 5.72. Let V be an algebraic variety over k. (a) V is unirational if there exists a dominant rational map P (b) V is rational if there exists a birational map P n Ü V: n Ü V . In more down-to-earth terms, V is rational if k.V / is a pure transcendental extension of k, and it is unirational if k.V / is contained in such an extension of k. In 1876 (over C), L¨uroth proved that every unirational curve is rational. For a proof over any field, see FT 9.19. The L¨uroth problem asks whether every unirational variety is rational. Already for surfaces, this is a difficult problem. In characteristic zero, Castelnuovo and Severi proved that all unirational surfaces are rational, but in characteristic p ¤ 0, Zariski showed that some surfaces of the form Zp D f .X; Y /; while obviously unirational, are not rational. Surfaces of this form are now called Zariski surfaces. Fano attempted to find counter-examples to the L¨uroth problem in dimension 3 among the so-called Fano varieties, but none of his attempted proofs satisfies modern standards. In 1971-72, three examples of nonrational unirational three-folds were found. For a description of them, and more discussion of the L¨uroth problem in characteristic zero, see: Arnaud Beauville, The L¨uroth problem, arXiv:1507.02476. 9Actually, one needs to be more c
areful to ensure that O.U / is a set; for example, restrict U and A category of k-algebras of the form kŒX0; X1; : : :=a for a fixed family of symbols .Xi / indexed by N. 1 to the r. Rational and unirational varieties 127 A little history In his first proof of the Riemann hypothesis for curves over finite fields, Weil made use of the Jacobian variety of the curve, but initially he was not able to construct this as a projective variety. This led him to introduce “abstract” algebraic varieties, neither affine nor projective (in 1946). Weil first made use of the Zariski topology when he introduced fibre spaces into algebraic geometry (in 1949). For more on this, see my article: The Riemann hypothesis over finite fields: from Weil to the present day. Exercises 1 is 5-1. Show that the only regular functions on P not affine. When k D C, P 1 is the Riemann sphere (as a set), and one knows from complex analysis that the only holomorphic functions on the Riemann sphere are constant. Since regular functions are holomorphic, this proves the statement in this case. The general case is easier.] 1 are the constant functions. [Thus P 5-2. Let V be the disjoint union of algebraic varieties V1; : : : ; Vn. This set has an obvious topology and ringed space structure for which it is an algebraic variety. Show that V is affine if and only if each Vi is affine. 5-3. Show that an algebraic variety G equipped with a group structure is an algebraic group if the map .x; y/ 7! x1yW G G ! G is regular. 5-4. Let G be an algebraic group. Show: (a) The neutral element e of G is contained in a unique irreducible component Gı of G, which is also the unique connected component of G containing e. (b) The subvariety Gı is a normal subgroup of G of finite index, and every algebraic subgroup of G of finite index contains Gı. 5-5. Show that every subgroup variety of a group variety is closed. 5-6. Show that a prevariety V is separated if and only if it satisfies the following condition: a regular map U X fP g ! V with U a curve and P a nonsingular point on U extends in at most one way to a regular map U ! V . 5-7. Prove the final statement in 5.71. CHAPTER 6 Projective Varieties Recall (5.3) that we defined P the relation n to be the set of equivalence classes in knC1 X foriging for .a0; : : : ; an/ .b0; : : : ; bn/ ” .a0; : : : ; an/ D c.b0; : : : ; bn/ for some c 2 k: Let .a0 W : : : W an/ denote the equivalence class of .a0; : : : ; an/, and let denote the map knC1 X f.0; : : : ; 0/g ! P n: Let Ui be the set of .a0 W : : : W an/ 2 P n such that ai ¤ 0, and let ui be the bijection .a0W : : : W an/ 7! a0 ai ; : : : ; bai ai ; : : : ; an ai W Ui ui! A n ( ai ai omitted). n has a unique structure of an algebraic variety for which In this chapter, we show that P these maps become isomorphisms of affine algebraic varieties. A variety isomorphic to n is called a projective variety, and a variety isomorphic to a a closed subvariety of P n is called a quasiprojective variety. Every affine variety is locally closed subvariety of P quasiprojective, but not all algebraic varieties are quasiprojective. We study morphisms between quasiprojective varieties. Projective varieties are important for the same reason compact manifolds are important: results are often simpler when stated for projective varieties, and the “part at infinity” often plays a role, even when we would like to ignore it. For example, a famous theorem of Bezout (see 6.37 below) says that a curve of degree m in the projective plane intersects a curve of degree n in exactly mn points (counting multiplicities). For affine curves, one has only an inequality. a. Algebraic subsets of Pn A polynomial F .X0; : : : ; Xn/ is said to be homogeneous of degree d if it is a sum of terms ai0;:::;inX i0 0 n with i0 C C in D d ; equivalently, X in F .tX0; : : : ; tXn/ D t d F .X0; : : : ; Xn/ for all t 2 k. The polynomials homogeneous of degree d form a subspace kŒX0; : : : ; Xnd of kŒX0; : : : ; Xn, and kŒX0; : : : ; Xn D M d 0 kŒX0; : : : ; Xnd I 129 130 6. PROJECTIVE VARIETIES in other words, every polynomial F can be written uniquely as a sum F D P Fd with Fd homogeneous of degree d . Let P D .a0 W : : : W an/ 2 P n. Then P also equals .ca0 W : : : W can/ for any c 2 k, and so we can’t speak of the value of a polynomial F .X0; : : : ; Xn/ at P . However, if F is homogeneous, then F .ca0; : : : ; can/ D cd F .a0; : : : ; an/, and so it does make sense to say n (or projective algebraic set) is the set that F is zero or not zero at P . An algebraic set in P of common zeros in P n of some set of homogeneous polynomials. EXAMPLE 6.1. Consider the projective algebraic subset of P equation 2 defined by the homogeneous E W Y 2Z D X 3 C aXZ2 C bZ3. It consists of the points .x W y W 1/ on the affine curve E \ U2 (26) Y 2 D X 3 C aX C b (see 2.2) together with the point “at infinity” .0 W 1 W 0/. Note that E \ U1 is the affine curve Z D X 3 C aXZ2 C bZ3; and that .0W 1W 0/ corresponds to the point .0; 0/ on E \ U1: Z D X 3 C XZ2 C Z3 As .0; 0/ is nonsingular on E \ U1, we deduce from (4.5) that E is nonsingular unless X 3 C aX C b has a multiple root. A nonsingular curve of the form (26) is called an elliptic curve. An elliptic curve has a unique structure of a group variety for which the point at infinity n a. Algebraic subsets of P 131 is the zero: Q P P C Q When a; b 2 Q, we can speak of the zeros of (26) with coordinates in Q. They also form a group E.Q/, which Mordell showed to be finitely generated. It is easy to compute the torsion subgroup of E.Q/, but there is at present no known algorithm for computing the rank of E.Q/. More precisely, there is an “algorithm” which works in practice, but which has not been proved to always terminate after a finite amount of time. There is a very beautiful theory surrounding elliptic curves over Q and other number fields, whose origins can be traced back almost 1,800 years to Diophantus. (See my book on Elliptic Curves for all of this.) An ideal a kŒX0; : : : ; Xn is said to be graded or homogeneous if it contains with any polynomial F all the homogeneous components of F , i.e., if F 2 a H) Fd 2 a, all d: It is straightforward to check that ˘ an ideal is graded if and only if it is generated by (a finite set of) homogeneous polynomials; ˘ the radical of a graded ideal is graded; ˘ an intersection, product, or sum of graded ideals is graded. For a graded ideal a, we let V .a/ denote the set of common zeros of the homogeneous polynomials in a. Clearly a b H) V .a/ V .b/. If F1; : : : ; Fr are homogeneous generators for a, then V .a/ is also the set of common zeros of the Fi . Clearly every polynomial in a is zero on every representative of a point in V .a/. We write V aff.a/ for the set of common zeros of a in knC1. It is a cone in knC1, i.e., together with any point P it contains the line through P and the origin, and V .a/ D V aff.a/ X f.0; : : : ; 0/g : The sets V .a/ in P n have similar properties to their namesakes in A n. 132 6. PROJECTIVE VARIETIES PROPOSITION 6.2. There are the following relations: nI V .a/ D ; ” rad.a/ .X0; : : : ; Xn/I (a) V .0/ D P (b) V .ab/ D V .a \ b/ D V .a/ [ V .b/I (c) V .P ai / D T V .ai /. PROOF. For the second statement in (a), note that V .a/ D ; ” V aff.a/ f.0; : : : ; 0/g ” rad.a/ .X0; : : : ; Xn/ (strong Nullstellensatz 2.16). The remaining statements can be proved directly, as in (2.10), or by using the relation between V .a/ and V aff.a/. on P Proposition 6.2 shows that the projective algebraic sets are the closed sets for a topology n. This topology is called the Zariski topology on P If C is a cone in knC1, then I.C / is a graded ideal in kŒX0; : : : ; Xn: if F .ca0; : : : ; can/ D n. 0 for all c 2 k, then Fd .a0; : : : ; an/ cd D F .ca0; : : : ; can/ D 0; X d for infinitely many c, and so P Fd .a0; : : : ; an/X d is the zero polynomial. For a subset S of n, we define the affine cone over S in knC1 to be P C D 1.S / [ foriging and we set I.S/ D I.C /. Note that if S is nonempty and closed, then C is the closure of 1.S/ ¤ ;, and that I.S/ is spanned by the homogeneous polynomials in kŒX0; : : : ; Xn that are zero on S. PROPOSITION 6.3. The maps V and I define inverse bijections between the set of algebraic n and the set of proper graded radical ideals of kŒX0; : : : ; Xn. An algebraic set subsets of P V in P n is irreducible if and only if I.V / is prime; in particular, P n is irreducible. PROOF. Note that we have bijections falgebraic subsets of P ng S 7! C fnonempty closed cones in knC1g V I fproper graded radical ideals in kŒX0; : : : ; Xng Here the top map sends S to the affine cone over S, and the maps V and I are in the sense of projective geometry and affine geometry respectively. The composite of any three of these maps is the identity map, which proves the first statement because the composite of the top map with I is I in the sense of projective geometry. Obviously, V is irreducible if and only if the closure of 1.V / is irreducible, which is true if and only if I.V / is a prime ideal. b. The Zariski topology on P n 133 Note that the graded ideals .X0; : : : ; Xn/ and kŒX0; : : : ; Xn are both radical, but V .X0; : : : ; Xn/ D ; D V .kŒX0; : : : ; Xn/ and so the correspondence between irreducible subsets of P quite one-to-one. n and radical graded ideals is not ASIDE 6.4. In English “homogeneous ideal” is more common than “graded ideal”, but we follow Bourbaki, Alg, II, 11. A graded ring is a pair .S; .Sd /d 2N/ consisting of a ring S and a family of additive subgroups Sd such that ( S D M Sd Sd Se Sd Ce, all d; e 2 N: d 2N An ideal a in S is graded if and only if a D M d 2N .a \ Sd /; this means that it is a graded submodule of .S; .Sd //. The quotient of a graded ring S by a graded ideal a is a graded ring S=a D L d Sd =.a \ Sd /. b. The Zariski topology on Pn For a graded polynomial F , let D.F / D fP 2 P n j F .P / ¤ 0g: Then, just as in the affine case, D.F / is open and the sets of this type form a base for the to
pology of P n. As in the opening paragraph of this chapter, we let Ui D D.Xi /. To each polynomial f .X1; : : : ; Xn/, we attach the homogeneous polynomial of the same degree f .X0; : : : ; Xn/ D X deg.f / 0 X1 X0 f ; : : : ; Xn X0 ; and to each homogeneous polynomial F .X0; : : : ; Xn/, we attach the polynomial F.X1; : : : ; Xn/ D F .1; X1; : : : ; Xn/: PROPOSITION 6.5. Each subset Ui of P we endow it with the induced topology, the bijection n is open in the Zariski topology on P n, and when Ui $ A n, .a0 an/ $ .a0; : : : ; ai1; ai C1; : : : ; an/ becomes a homeomorphism. PROOF. It suffices to prove this with i D 0. The set U0 D D.X0/, and so it is a basic open subset in P n. Clearly, for any homogeneous polynomial F 2 kŒX0; : : : ; Xn, D.F .X0; : : : ; Xn// \ U0 D D.F .1; X1; : : : ; Xn// D D.F/ and, for any polynomial f 2 kŒX1; : : : ; Xn, D.f / D D.f / \ U0: Thus, under the bijection U0 $ A tersections with Ui of the basic open subsets of P homeomorphism. n correspond to the inn, which proves that the bijection is a n, the basic open subsets of A 134 6. PROJECTIVE VARIETIES n is irreducible. We REMARK 6.6. It is possible to use this to give a different proof that P apply the criterion that a space is irreducible if and only if every nonempty open subset is dense (see p. 46). Note that each Ui is irreducible, and that Ui \ Uj is open and dense in each of Ui and Uj (as a subset of Ui , it is the set of points .a0 aj W : : : W an/ with aj ¤ 0/. Let U be a nonempty open subset of P n; then U \ Ui is open in Ui . For some i , U \ Ui is nonempty, and so must meet Ui \ Uj . Therefore U meets every Uj , and so is dense in every Uj . It follows that its closure is all of P n. c. Closed subsets of An and Pn We identify A n with U0, and examine the closures in P n of closed subsets of A n. Note that n D A P n t H1; H1 D V .X0/: generated by ff j f 2 ag. For a closed subset V of A With each ideal a in kŒX1; : : : ; Xn, we associate the graded ideal a in kŒX0; : : : ; Xn n, set V D V .a/ with a D I.V /. With each graded ideal a in kŒX0; X1; : : : ; Xn], we associate the ideal a in kŒX1; : : : ; Xn n, we set V D V .a/ with generated by fF j F 2 ag. When V is a closed subset of P a D I.V /. n, PROPOSITION 6.7. (a) Let V be a closed subset of A and .V / D V . If V D S Vi is the decomposition of V into its irreducible components, then V D S V i is the decomposition of V into its irreducible components. n. Then V D V \ A n. Then V is the closure of V in P (b) Let V be a closed subset of P n, and if no irreducible component n, and .V/ D V . of V lies in H1 or contains H1, then V is a proper subset of A PROOF. Straightforward. Examples 6.8. For we have and . aX C b; V W Y 2Z D X 3 C aXZ2 C bZ3; 6.9. Let V D V .f1; : : : ; fm/; then the closure of V in P components of V .f m/ not contained in H1. For example, let 1 ; : : : ; f n is the union of the irreducible V D V .X1; X 2 1 C X2/ D f.0; 0/gI then V .X0X1; X 2 1 (which is contained in H1).1 C X0X2/ consists of the two points .1W 0W 0/ (the closure of V ) and .0W 0W 1/ 6.10. For V D H1 D V .X0/, we have V D ; D V .1/ and .V/ D ; ¤ V . 1Of course, in this case a D .X1; X2/, a D .X1; X2/, and V D f.1W 0W 0/g, and so this example doesn’t contradict the proposition. d. The hyperplane at infinity 135 d. The hyperplane at infinity n as being A It is often convenient to think of P More precisely, we identify the set U0 with A n D U0 with a hyperplane added “at infinity”. n; the complement of U0 in P n is H1 D f.0 W a1 W : : : W an/ 2 P ng; which can be identified with P For example, P 1 D A n1. 2 D A P 2 [ H1 with H1 a projective line. Consider the line 1 t H1 (disjoint union), with H1 consisting of a single point, and in A 2. Its closure in P 2 is the line 1 C aX1 C bX2 D 0 X0 C aX1 C bX2 D 0: This line intersects the line H1 D V .X0/ at the point .0 W b W a/, which equals .0 W 1 W a=b/ when b ¤ 0. Note that a=b is the slope of the line 1 C aX1 C bX2 D 0, and so the point at which a line intersects H1 depends only on the slope of the line: parallel lines meet in one 2 with point at infinity. We can think of the projective plane P one point added at infinity for each “direction” in A 2 as being the affine plane A 2. Similarly, we can think of P n as being A direction in A one point at infinity for each equivalence class of lines. n — being parallel is an equivalence relation on the lines in A n with one point added at infinity for each n, and there is We can replace U0 with Un in the above discussion, and write P n D Un t H1 with H1 D f.a0W : : : W an1W 0/g, as in Example 6.1. Note that in this example the point at infinity on the elliptic curve Y 2 D X 3 C aX C b is the intersection of the closure of any vertical line with H1. e. Pn is an algebraic variety O For each i , write i for the sheaf on Ui P n defined by the homeomorphism ui W Ui ! A j jUij . When endowed with this sheaf; LEMMA 6.11. Let Uij D Ui \ Uj ; then i jUij D O O Uij is an affine algebraic variety; moreover, .Uij ; i / is generated as a k-algebra by the i /, g 2 .Uj ; functions .f jUij /.gjUij / with f 2 .Ui ; j /. O n. O O PROOF. It suffices to prove this for .i; j / D .0; 1/. All rings occurring in the proof will be identified with subrings of the field k.X0; X1; : : : ; Xn/. Recall that U0 D f.a0 W a1 W : : : W an/ j a0 ¤ 0g; .a0 W a1 W : : : W an/ $ . a1 a0 ; a2 a0 ; : : : ; an a0 / 2 A n: ; X2 X0 ; : : : ; Xn X0 Let kŒ X1 X0 — it is the polynomial ring in the n symbols X1 X0 kŒ X1 X0 be the subring of k.X0; X1; : : : ; Xn/ generated by the quotients Xi X0 ; : : : ; Xn / 2 X0 . An element f . X1 X0 defines a map ; : : : ; Xn X0 ; : : : ; Xn X0 .a0 W a1 W : : : W an/ 7! f . a1 a0 ; : : : ; an a0 /W U0 ! k; 136 6. PROJECTIVE VARIETIES and in this way kŒ X1 X0 U0; and U0 with Spm ; X2 X0 kŒ X1 X0 ; : : : ; Xn X0 ; : : : ; Xn X0 . becomes identified with the ring of regular functions on Next consider the open subset of U0; U01 D f.a0 W : : : W an/ j a0 ¤ 0, a1 ¤ 0g: /, and is therefore an affine subvariety of .U0; It is D. X1 X0 corresponds to the inclusion of rings kŒ X1 X0 f . X1 / of kŒ X1 X0 X0 on U01. ; : : : ; Xn X0 ; : : : ; Xn X0 ; X0 X1 ; X0 X1 O 0/. The inclusion U01 ,! U0 . An element ; : : : ; Xn X0 ,! kŒ X1 X0 defines the function .a0 W : : : W an/ 7! f . a1 a0 ; : : : ; Xn X0 ; X0 X1 Similarly, U1 D f.a0 W a1 W : : : W an/ j a1 ¤ 0g; .a0 W a1 W : : : W an/ $ . a0 a1 ; X2 and we identify U1 with Spm X0 defines the map .a0 W : : : W an/ 7! f . a0 a1 ; : : : ; Xn X1 ; : : : ; an a1 kŒ X0 X1 /W U1 ! k. . A polynomial f . X0 X1 / in kŒ X0 X1 ; : : : ; Xn X1 / 2 A ; : : : ; an a1 ; : : : ; Xn X1 ; : : : ; an a0 ; a0 a1 / n; ; X1 X0 ; X1 X0 ; X0 X1 ; : : : ; Xn X1 ; : : : ; Xn X1 ; : : : ; Xn X0 / of kŒ X0 X1 ; a1 a0 ; : : : ; Xn X1 The two subrings kŒ X1 X0 When regarded as an open subset of U1; U01 D D. X0 X1 /, and is therefore an affine 1/, and the inclusion U01 ,! U1 corresponds to the inclusion of rings ; X1 X0 subvariety of .U1; O kŒ X0 ,! kŒ X0 . An element f . X0 ; : : : ; Xn ; : : : ; Xn X1 X1 X1 X1 X1 defines the function .a0 W : : : W an/ 7! f . a0 ; : : : ; an / on U01. a1 a1 ; X1 and kŒ X0 of k.X0; X1; : : : ; Xn/ are X0 X1 equal, and an element of this ring defines the same function on U01 regardless of which of the two rings it is considered an element. Therefore, whether we regard U01 as a subvariety of U0 or of U1 it inherits the same structure as an affine algebraic variety (3.15). This proves the first two assertions, and the third is obvious: kŒ X1 is generated by X0 and kŒ X0 its subrings kŒ X1 X1 X0 ; X2 X1 PROPOSITION 6.12. There is a unique structure of an algebraic variety on P each Ui is an open affine subvariety of P varieties. Moreover, P n for which n and each map ui is an isomorphism of algebraic n is separated. ; : : : ; Xn X0 ; : : : ; Xn X1 ; : : : ; Xn X0 ; X0 X1 . PROOF. Endow each Ui with the structure of an affine algebraic variety for which ui is an n D S Ui , and the lemma shows that this covering satisfies the patching isomorphism. Then P n has a unique structure of a ringed space for which Ui ,! P n is a condition 5.15, and so P OPnjUi D n and homeomorphism onto an open subset of P Ui . Moreover, because each n into an algebraic prevariety. Finally, the Ui is an algebraic variety, this structure makes P lemma shows that P n satisfies the condition 5.29(c) to be separated. O EXAMPLE 6.13. Let C be the plane projective curve C W Y 2Z D X 3 and assume that char.k/ ¤ 2. For each a 2 k, there is an automorphism 'a! C: .x W y W z/ 7! .ax W y W a3z/W C Patch two copies of C A 1 together along C .A 1 f0g/ by identifying .P; a/ with .'a.P /; a1/, P 2 C , a 2 A 1 X f0g. One obtains in this way a singular surface that is not quasiprojective (see Hartshorne 1977, Exercise 7.13). It is even complete — see below — and so if it were quasiprojective, it would be projective. In Shafarevich 1994, VI 2.3, there is an example of a nonsingular complete variety of dimension 3 that is not projective. It is known that every irreducible separated curve is quasiprojective, and every nonsingular complete surface is projective, and so these examples are minimal. f. The homogeneous coordinate ring of a projective variety 137 f. The homogeneous coordinate ring of a projective variety Recall (p. 114) that attached to each irreducible variety V , there is a field k.V / with the property that k.V / is the field of fractions of kŒU for any open affine U V . We now describe this field in the case that V D P . We regard this as a subring of k.X0; : : : ; Xn/, and wish to identify the field of fractions of kŒU0 as a subfield of k.X0; : : : ; Xn/. Every nonzero F 2 kŒU0 can be written n. Recall that kŒU0 D kŒ X1 X0 ; : : : ; Xn X0 F . X1 X0 ; : : : ; Xn X0 / D F .X0; : : : ; Xn/ X deg.F / 0 with F homogeneous of degree deg.F /, and it follows that the field of fractions of kŒU0 is k.U0/ D G.X0; : : : ; Xn/ H.X0; : : : ; Xn/ ˇ ˇ ˇ ˇ G, H homogeneous of the same degree [ f0g: Write k.X0; : : : ; Xn/0 for
this field (the subscript 0 is short for “subfield of elements of n/ D k.X0; : : : ; Xn/0. Note that for F D G degree 0”), so that k.P H in k.X0; : : : ; Xn/0; .a0 W : : : W an/ 7! G.a0; : : : ; an/ H.a0; : : : ; an/ W D.H / ! k, is a well-defined function, which is obviously regular (look at its restriction to Ui /. We now extend this discussion to any irreducible projective variety V . Such a V can be written V D V .p/ with p a graded radical ideal in kŒX0; : : : ; Xn, and we define the homogeneous coordinate ring of V (with its given embedding) to be khomŒV D kŒX0; : : : ; Xn=p. Note that khomŒV is the ring of regular functions on the affine cone over V ; therefore its dimension is dim.V / C 1: It depends, not only on V , but on the embedding of V into P n, i.e., it is not intrinsic to V . For example, .a0 W a1/ 7! .a2 0 W a0a1 W a2 2 1 ! P 1/W P 1/W X0X2 D X 2 1 D kŒX0; X1, which is the affine coordinate ring of the smooth variety A 1 onto its image .P is an isomorphism from P khomŒP 1/ D kŒX0; X1; X2=.X0X2 X 2 khomŒ.P gular variety X0X2 X 2 1 . 1 (see 6.23 below), but 2, whereas 1 /, which is the affine coordinate ring of the sin- We say that a nonzero f 2 khomŒV is homogeneous of degree d if it can be represented by a homogeneous polynomial F of degree d in kŒX0; : : : ; Xn, and we say that 0 is homogeneous of degree 0. LEMMA 6.14. Each element of khomŒV can be written uniquely in the form with fi homogeneous of degree i. f D f0 C C fd PROOF. Let F represent f ; then F can be written F D F0 C C Fd with Fi homogeneous of degree i ; when read modulo p, this gives a decomposition of f of the required type. Suppose f also has a decomposition f D P gi , with gi represented by the homogeneous polynomial Gi of degree i . Then F G 2 p, and the homogeneity of p implies that Fi Gi D .F G/i 2 p. Therefore fi D gi . 138 6. PROJECTIVE VARIETIES It therefore makes sense to speak of homogeneous elements of kŒV . For such an element h, we define D.h/ D fP 2 V j h.P / ¤ 0g. Since khomŒV is an integral domain, we can form its field of fractions khom.V /. Define khom.V /0 D n g h 2 khom.V / ˇ o ˇ ˇ g and h homogeneous of the same degree [ f0g: PROPOSITION 6.15. The field of rational functions on V is k.V / defD khom.V /0. PROOF. Consider V0 of khomŒV , and then the field of fractions of kŒV0 becomes identified with khom.V /0. n, we can identify kŒV0 with a subring defD U0 \ V . As in the case of P g. Regular functions on a projective variety Let V be an irreducible projective variety, and let f 2 k.V /. By definition, we can write f D g h with g and h homogeneous of the same degree in khomŒV and h ¤ 0. For any P D .a0 W : : : W an/ with h.P / ¤ 0, f .P / defD g.a0; : : : ; an/ h.a0; : : : ; an/ is well-defined: if .a0; : : : ; an/ is replaced by .ca0; : : : ; can/, then both the numerator and denominator are multiplied by cdeg.g/ D cdeg.h/. We can write f in the form g h in many different ways,2 but if then f D g h D g0 h0 (in k.V /0), gh0 D g0h (in khomŒV ) and so g.a0; : : : ; an/ h0.a0; : : : ; an/ D g0.a0; : : : ; an/ h.a0; : : : ; an/: Thus, if h0.P / ¤ 0, the two representations give the same value for f .P /. PROPOSITION 6.16. For each f 2 k.V / defD khom.V /0, there is an open subset U of V , where f .P / is defined, and P 7! f .P / is a regular function on U ; every regular function on an open subset of V arises from a unique element of k.V /. PROOF. From the above discussion, we see that f defines a regular function on U D S D.h/, where h runs over the denominators of expressions f D g h with g and h homogeneous of the same degree in khomŒV . Conversely, let f be a regular function on an open subset U of V , and let P 2 U . Then P lies in the open affine subvariety V \ Ui for some i , and so f coincides with the function defined by some fP 2 k.V \ Ui / D k.V / on an open neighbourhood of P . If f coincides with the function defined by fQ 2 k.V / in a neighbourhood of a second point Q of U , then fP and fQ define the same function on some open affine U 0, and so fP D fQ as elements of kŒU 0 k.V /. This shows that f is the function defined by fP on the whole of U . 2Unless khomŒV is a unique factorization domain, there will be no preferred representation f D g h . h. Maps from projective varieties 139 REMARK 6.17. (a) The elements of k.V / D khom.V /0 should be regarded as the algebraic analogues of meromorphic functions on a complex manifold; the regular functions on an open subset U of V are the “meromorphic functions without poles” on U . [In fact, when k D C, this is more than an analogy: a nonsingular projective algebraic variety over C defines a complex manifold, and the meromorphic functions on the manifold are precisely the rational functions on the variety. For example, the meromorphic functions on the Riemann sphere are the rational functions in z.] (b) We shall see presently (6.24) that, for any nonzero homogeneous h 2 khomŒV , D.h/ is an open affine subset of V . The ring of regular functions on it is kŒD.h/ D fg= hm j g homogeneous of degree m deg.h/g [ f0g: We shall also see that the ring of regular functions on V itself is just k, i.e., any regular function on an irreducible (connected will do) projective variety is constant. However, if U V / of regular functions can be almost is an open nonaffine subset of V , then the ring .U; anything — it needn’t even be a finitely generated k-algebra! O h. Maps from projective varieties We describe the morphisms from a projective variety to another variety. PROPOSITION 6.18. The map W A nC1 X foriging ! P n, .a0; : : : ; an/ 7! .a0 W : : : W an/ is an open morphism of algebraic varieties. A map ˛W P if and only if ˛ ı is regular. n ! V with V a prevariety is regular PROOF. The restriction of to D.Xi / is the projection .a0; : : : ; an/ 7! . a0 ai W : : : W an ai /W knC1 X V .Xi / ! Ui ; which is the regular map of affine varieties corresponding to the map of k-algebras h X0 Xi k ; : : : ; Xn Xi i ! kŒX0; : : : ; XnŒX 1 i : (In the first algebra Xj Xi is regular. is to be thought of as a single symbol.) It now follows from (5.4) that Let U be an open subset of knC1 X foriging, and let U 0 be the union of all the lines through the origin that meet U , that is, U 0 D 1.U /. Then U 0 is again open in knC1 X foriging, because U 0 D S cU , c 2 k, and x 7! cx is an automorphism of knC1 X foriging. The complement Z of U 0 in knC1 X foriging is a closed cone, and the proof of (6.3) shows n; but .U / is the complement of .Z/. Thus sends open sets that its image is closed in P to open sets. The rest of the proof is straightforward. Thus, the regular maps P n (as maps of sets). through P n ! V are just the regular maps A nC1 X foriging ! V factoring 140 6. PROJECTIVE VARIETIES REMARK 6.19. Consider polynomials F0.X0; : : : ; Xm/; : : : ; Fn.X0; : : : ; Xm/ of the same degree. The map .a0 W : : : W am/ 7! .F0.a0; : : : ; am/ W : : : W Fn.a0; : : : ; am// obviously defines a regular map to P that is, on the set S D.Fi / D P will also be regular. It may be possible to extend the map to a larger set by representing it by different polynomials. Conversely, every such map arises in this way, at least locally. More precisely, there is the following result. n X V .F1; : : : ; Fn/. Its restriction to any subvariety V of P m, where not all Fi vanish, m n on the open subset of P PROPOSITION 6.20. Let V D V .a/ P regular if and only if, for every P 2 V , there exist polynomials m and W D V .b/ P n. A map 'W V ! W is F0.X0; : : : ; Xm/; : : : ; Fn.X0; : : : ; Xm/; homogeneous of the same degree, such that ' ..b0 W : : : W bn// D .F0.b0; : : : ; bm/ W : : : W Fn.b0; : : : ; bm// for all points .b0 W : : : W bm/ in some neighbourhood of P in V .a/. PROOF. Straightforward. EXAMPLE 6.21. We prove that the circle X 2 CY 2 D Z2 is isomorphic to P 1. This equation can be rewritten .X C iY /.X iY / D Z2, and so, after a change of variables, the equation of the circle becomes C W XZ D Y 2. Define 'W P 1 ! C , .a W b/ 7! .a2 W ab W b2/: For the inverse, define W C ! P 1 by .a W b W c/ 7! .a W b/ .a W b W c/ 7! .b W c/ if a ¤ 0 if b ¤ 0 : Note that, a ¤ 0 ¤ b; ac D b2 H) c b D b a and so the two maps agree on the set where they are both defined. Clearly, both ' and are regular, and one checks directly that they are inverse. i. Some classical maps of projective varieties We list some of the classic maps. HYPERPLANE SECTIONS AND COMPLEMENTS 6.22. Let L D P ci Xi be a nonzero linear form in n C 1 variables. Then the map .a0 W : : : W an/ 7! a0 L.a/ ; : : : ; an L.a/ i. Some classical maps of projective varieties 141 is a bijection of D.L/ P inverse n onto the hyperplane L.X0; X1; : : : ; Xn/ D 1 of A nC1, with .a0; : : : ; an/ 7! .a0 W : : : W an/: Both maps are regular — for example, the components of the first map are the regular functions Xj . As V .L 1/ is affine, so also is D.L/, and its ring of regular functions is kŒ X0 P ci Xi symbol, and P cj is to be thought of as a single D 1; thus it is a polynomial ring in n symbols; any one symbol : In this ring, each quotient Xj P ci Xi ; : : : ; Xn P ci Xi P ci Xi Xj P ci Xi Xj P ci Xi for which cj ¤ 0 can be omitted. For a fixed P D .a0W : : : W an/ 2 P n, the set of c D .c0W : : : W cn/ such that Lc.P / defD X ci ai ¤ 0 is a nonempty open subset of P n (n > 0). Therefore, for any finite set S of points of P n, fc 2 P n j S D.Lc/g n is irreducible). In particular, S is contained in is a nonempty open subset of P n. Moreover, if S V , where V is a closed subvariety of an open affine subset D.Lc/ of P n, then S V \ D.Lc/: any finite set of points of a projective variety is contained in an P open affine subvariety. n (because P THE VERONESE MAP; HYPERSURFACE SECTIONS 6.23. Let I D f.i0; : : : ; in/ 2 N nC1 j X ij D mg: elements3. Note that I indexes the monomials of degree m in n C 1 variables. It has mCn m Write n;m D mCn n;m whose coordinates are W : : :/. The Veronese mapping indexed by I ; thus a point of P is defined to be 1, and consider the pro
jective space P n;m can be written .: : : W bi0:::in m vW P n ! P n;m, .a0 W : : : W an/ 7! .: : : W bi0:::in W : : :/; bi0:::in D ai0 0 : : : ain n : In other words, the Veronese mapping sends an n C 1-tuple .a0W : : : W an/ to the set of monomials in the ai of degree m. For example, when n D 1 and m D 2, the Veronese map is 3This can be proved by induction on m C n. If m D 0 D n, then 0 0 homogeneous polynomial of degree m can be written uniquely as 1 ! P P 2, .a0 W a1/ 7! .a2 0 W a0a1 W a2 1/: D 1, which is correct. A general F .X0; X1; : : : ; Xn/ D F1.X1; : : : ; Xn/ C X0F2.X0; X1; : : : ; Xn/ with F1 homogeneous of degree m and F2 homogeneous of degree m 1. But mCn m D mCn1 m C mCn1 m1 because they are the coefficients of X m in .X C 1/mCn D .X C 1/.X C 1/mCn1; and this proves the induction. 142 6. PROJECTIVE VARIETIES Its image is the curve .P 1/ W X0X2 D X 2 .b2;0 W b1;1 W b0;2/ 7! 1 , and the map .b2;0 W b1;1/ if b2;0 ¤ 1 .b1;1 W b0;2/ if b0;2 ¤ 0 is an inverse .P 1/ ! P 1. (Cf. Example 6.22.) When n D 1 and m is general, the Veronese map is 1 ! P m, .a0 W a1/ 7! .am 0 P W am1 0 a1 W : : : W am 1 /: I claim that, in the general case, the image of is a closed subset of P defines an isomorphism of projective varieties W P n ! .P n/. n;m and that First note that the map has the following interpretation: if we regard the coordinates ai n as being the coefficients of a linear form L D P ai Xi (well-defined up of a point P of P to multiplication by nonzero scalar), then the coordinates of .P / are the coefficients of the homogeneous polynomial Lm with the binomial coefficients omitted. As L ¤ 0 ) Lm ¤ 0, the map is defined on the whole of P n, that is, .a0; : : : ; an/ ¤ .0; : : : ; 0/ ) .: : : ; bi0:::in; : : :/ ¤ .0; : : : ; 0/: Moreover, L1 ¤ cL2 ) Lm ¤ cLm 1 and so is injective. It is clear from its definition that is regular. 2 , because kŒX0; : : : ; Xn is a unique factorization domain, We shall see in the next chapter that the image of any projective variety under a regular n/ is defined by the system of map is closed, but in this case we can prove directly that .P equations: bi0:::inbj0:::jn D bk0:::knb`0:::`n; ih C jh D kh C `h, all h: (*) Obviously P n maps into the algebraic set defined by these equations. Conversely, let Vi D f.: : : : W bi0:::in W : : :/ j b0:::0m0:::0 ¤ 0g: Then .Ui / Vi and 1.Vi / D Ui . It is possible to write down a regular map Vi ! Ui inverse to jUi : for example, define V0 ! P n to be .: : : W bi0:::in W : : :/ 7! .bm;0;:::;0 W bm1;1;0;:::;0 W bm1;0;1;0;:::;0 W : : : W bm1;0;:::;0;1/: Finally, one checks that .P For any closed variety W P .W / of .P n/ P n;m. n/ S Vi . n, jW is an isomorphism of W onto a closed subvariety 6.24. The Veronese mapping has a very important property. If F is a nonzero homogeneous form of degree m 1, then V .F / P n is called a hypersurface of degree m and V .F / \ W is called a hypersurface section of the projective variety W . When m D 1, “surface” is replaced by “plane”. Now let H be the hypersurface in P n of degree m X ai0:::inX i0 0 X in n D 0, and let L be the hyperplane in P n;m defined by X ai0:::inXi0:::in: i. Some classical maps of projective varieties 143 Then .H / D .P n/ \ L, i.e., H.a/ D 0 ” L..a// D 0: n, defines an isomorphism of the hypersurface Thus for any closed subvariety W of P section W \ H of V onto the hyperplane section .W / \ L of .W /. This observation often allows one to reduce questions about hypersurface sections to questions about hyperplane sections. As one example of this, note that maps the complement of a hypersurface section of W isomorphically onto the complement of a hyperplane section of .W /, which we know to be affine. Thus the complement of any hypersurface section of a projective variety is an affine variety. AUTOMORPHISMS OF P 6.25. An element A D .aij / of GLnC1 defines an automorphism of P n n: .x0 W : : : W xn/ 7! .: : : W P aij xj W : : :/I clearly it is a regular map, and the inverse matrix gives the inverse map. Scalar matrices act as the identity map. Let PGLnC1 D GLnC1 =kI , where I is the identity matrix, that is, PGLnC1 is the .nC1/21 of the quotient of GLnC1 by its centre. Then PGLnC1 is the complement in P hypersurface det.Xij / D 0, and so it is an affine variety with ring of regular functions kŒPGLnC1 D fF .: : : ; Xij ; : : :/= det.Xij /m j deg.F / D m .n C 1/g [ f0g: It is an affine group variety. The homomorphism PGLnC1 ! Aut.P it is surjective.4 Consider a hypersurface n/ is obviously injective. We sketch a proof that in P n and a line H W F .X0; : : : ; Xn/ D 0 L D f.ta0 W : : : W tan/ j t 2 kg in P n. The points of H \ L are given by the solutions of F .ta0; : : : ; tan/ D 0, which is a polynomial of degree deg.F / in t unless L H . Therefore, H \ L contains deg.F / points, and it is not hard to show that for a fixed H and most L it will contain n exactly deg.F / points. Thus, the hyperplanes are exactly the closed subvarieties H of P such that (a) dim.H / D n 1; (b) #.H \ L/ D 1 for all lines L not contained in H . These are geometric conditions, and so any automorphism of P hyperplanes. But on an open subset of P n, such an automorphism takes the form n must map hyperplanes to .b0 W : : : W bn/ 7! .F0.b0; : : : ; bn/ W : : : W Fn.b0; : : : ; bn//; where the Fi are homogeneous of the same degree d (see 6.20). Such a map will take hyperplanes to hyperplanes if and only if d D 1. 4This is related to the fundamental theorem of projective geometry — see E. Artin, Geometric Algebra, Interscience, 1957, Theorem 2.26. 144 6. PROJECTIVE VARIETIES THE SEGRE MAP 6.26. This is the mapping ..a0 W : : : W am/; .b0 W : : : W bn// 7! ..: : : W ai bj W : : ://W P m P n ! P mnCmCn: mnCmCn is f.i; j / j 0 i m; 0 j ng. Note that if we interpret The index set for P the tuples on the left as being the coefficients of two linear forms L1 D P ai Xi and L2 D P bj Yj , then the image of the pair is the set of coefficients of the homogeneous form of degree 2, L1L2. From this observation, it is obvious that the map is defined on n .L1 ¤ 0 ¤ L2 ) L1L2 ¤ 0/ and is injective. On any subset of the the whole of P form Ui Uj it is defined by polynomials, and so it is regular. Again one can show that it mnCmCn defined by the is an isomorphism onto its image, which is the closed subset of P equations m P – see Shafarevich 1994, I 5.1. For example, the map wij wkl wil wkj D 0 ..a0 W a1/; .b0 W b1// 7! .a0b0 W a0b1 W a1b0 W a1b1/W P 1 P 1 ! P 3 has image the hypersurface H W W Z D XY: The map .w W x W y W z/ 7! ..w W y/; .w W x// 1 is not isomorphic to 2, because in the first variety there are closed curves, e.g., two vertical lines, that don’t is an inverse on the set where it is defined. [Incidentally, P P intersect.] 1 P If V and W are closed subvarieties of P isomorphically onto a closed subvariety of P are projective. n, then the Segre map sends V W m and P mnCmCn. Thus products of projective varieties The product P 1 P fore a finite disjoint union of copies of P union of projective varieties is projective. n contains many disjoint copies of P n as closed subvarieties. Theren is projective, which shows that a finite disjoint There is an explicit description of the topology on P m P n W the closed sets are the sets of common solutions of families of equations F .X0; : : : ; XmI Y0; : : : ; Yn/ D 0 with F separately homogeneous in the Xi and in the Yj . PROJECTIONS WITH GIVEN CENTRE 6.27. Let L1; : : : ; Lnd be linearly independent linear forms in n C 1 variables. Their zero set E in knC1 has dimension d C 1, and so their zero set in P n is a d -dimensional nd 1 by .a/ D .L1.a/ W : : : W Lnd .a//; such a map linear space. Define W P is called a projection with centre E. If V is a closed subvariety disjoint from E, then nd 1. More generally, if F1; : : : ; Fr are homogeneous forms defines a regular map V ! P of the same degree, and Z D V .F1; : : : ; Fr /, then a 7! .F1.a/ W : : : W Fr .a// is a morphism n Z ! P P n E ! P r1. By carefully choosing the centre E, it is possible to linearly project any smooth curve in 3, and nonisomorphically (but bijectively on an open n isomorphically onto a curve in P P j. Maps to projective space 145 3. To project to P 2 with only nodes as singularities.5 For example, suppose we have subset) onto a curve in P 2 we need three linear forms L0, L1, L2 and a nonsingular curve C in P the centre of the projection is the point P0 where all forms are zero. We can think of the map as projecting from the centre P0 onto some (projective) plane by sending the point P to the point where P0P intersects the plane. To project C to a curve with only ordinary nodes as singularities, one needs to choose P0 so that it doesn’t lie on any tangent to C , any trisecant (line crossing the curve in 3 points), or any chord at whose extremities the tangents are coplanar. See for example Samuel, P., Lectures on Old and New Results on Algebraic Curves, Tata Notes, 1966. Projecting a nonsingular variety in P n to a lower dimensional projective space usually introduces singularities. Hironaka proved that every singular variety arises in this way in characteristic zero. See Chapter 8. APPLICATION PROPOSITION 6.28. Every finite set S of points of a quasiprojective variety V is contained in an open affine subset of V . n, let NV be the closure of V in P n, and let Z D NV X V . PROOF. Regard V as a subvariety of P Because S \ Z D ;, for each P 2 S there exists a homogeneous polynomial FP 2 I.Z/ such that FP .P / ¤ 0. We may suppose that the FP have the same degree. An elementary argument shows that some linear combination F of the FP , P 2 S , is nonzero at each P . Then F is zero on Z, and so NV \ D.F / is an open affine of V , but F is nonzero at each P , and so NV \ D.F / contains S . j. Maps to projective space Under construction. NOTES. There is no nonconstant map P f0/ ! P maps A nC1 X n, namely, .x0; : : : ; xn/ 7! .x0W W xn). Somewhat surprisingly, there are surjective regular n ! P n. However, there is a
surjective regular map A n. Consider the map n ! A .x0W : : : W xn/ 7! .x2 0 W W x2 n/W P n ! P n: It is mW 1 with m > 1 except over the points .0W W 1W W 0/. If H is a general hyperplane avoiding these points, then P n. For example, when we take n still maps onto P n X H A we obtain the surjective map H W x0 C C xn D 0, .x1; : : : ; xn/ 7! .x2 1 W W x2 n W .1 x1 xn/2/W A n ! P n: k. Projective space without coordinates Let E be a vector space over k of dimension n. The set P.E/ of lines through zero in E has a natural structure of an algebraic variety: the choice of a basis for E defines a bijection P.E/ ! P n, and the inherited structure of an algebraic variety on P.E/ is independent of 5A nonsingular curve of degree d in P 2 has genus .d 1/.d 2/ 2 genus g can’t be realized as a nonsingular curve in P 2. . Thus, if g is not of this form, a curve of 146 6. PROJECTIVE VARIETIES the choice of the basis (because the bijections defined by two different bases differ by an n, which has n C 1 distinguished hyperplanes, automorphism of P namely, X0 D 0; : : : ; Xn D 0, no hyperplane in P.E/ is distinguished. n). Note that in contrast to P l. The functor defined by projective space Let R be a k-algebra. A submodule M of an R-module N is said to be a direct summand of N if there exists another submodule M 0 of M (a complement of M ) such that N D M ˚ M 0. Let M be a direct summand of a finitely generated projective R-module N . Then M is also finitely generated and projective, and so Mm is a free Rm-module of finite rank for every maximal ideal m in R. If Mm is of constant rank r, then we say that M has rank r. See CA 12. Let P n.R/ D fdirect summands of rank 1 of RnC1g. Then P n is a functor from k-algebras to sets. When K is a field, every K-subspace of KnC1 is a direct summand, and so P n.K/ consists of the lines through the origin in KnC1. Let Hi be the hyperplane Xi D 0 in knC1, and let Pi .R/ D fL 2 P n.R/ j L ˚ HiR D RnC1g: Let L 2 Pi .R/; then Now ei D ` CX j ¤i aj ej . L 7! .aj /j ¤i W Pi .R/ ! Ui .R/ ' Rn is a bijection. These combine to give an isomorphism P n.R/ ! P Y Y n.R/: P n.R/ Pi .R/ Pi .R/ \ Pj .R/ 0i n 0i;j n n.R/ P Y 0i n Ui .R/ Y 0i;j n Ui .R/ \ Uj .R/: More generally, to give a regular map from a variety V to P n is the same as giving an isomorphism class of pairs .L; .s0; : : : ; sn// where L is an invertible sheaf on V and s0; : : : ; sn are sections of L that generate it. m. Grassmann varieties Let E be a vector space over k of dimension n, and let Gd .E/ be the set of d -dimensional subspaces of E. When d D 0 or n, Gd .E/ has a single element, and so from now on we assume that 0 < d < n. Fix a basis for E, and let S 2 Gd .E/. The choice of a basis for S then determines a d n matrix A.S/ whose rows are the coordinates of the basis elements. Changing the basis for S multiplies A.S/ on the left by an invertible d d matrix. Thus, the family of d d minors of A.S/ is determined up to multiplication by a nonzero constant, and so defines a point P .S/ in P n d 1 . m. Grassmann varieties 147 PROPOSITION 6.29. The map S 7! P .S/W Gd .E/ ! P closed subset of P 1 n d . n d 1 is injective, with image a n d 1 We give the proof below. The maps P defined by different bases of E differ by an automorphism of P , and so the statement is independent of the choice of the basis — later (6.34) we shall give a “coordinate-free description” of the map. The map realizes Gd .E/ as a projective algebraic variety called the Grassmann variety of d -dimensional subspaces of E. EXAMPLE 6.30. The affine cone over a line in P Thus, G2.k4/ can be identified with the set of lines in P x D .x0 W x1 W x2 W x3/ and y D .y0 W y1 W y2 W y3/ be distinct points on L. Then 3 is a two-dimensional subspace of k4. 3, and let 3. Let L be a line in P P .L/ D .p01 W p02 W p03 W p12 W p13 W p23/ 2 P 5; pij defD ˇ ˇ ˇ ˇ xi xj yi yj ˇ ˇ ˇ ˇ ; depends only on L. The map L 7! P .L/ is a bijection from G2.k4/ onto the quadric ˘ W X01X23 X02X13 C X03X12 D 0 in P 5. For a direct elementary proof of this, see (9.41, 9.42) below. REMARK 6.31. Let S 0 be a subspace of E of complementary dimension n d , and let Gd .E/S 0 be the set of S 2 Gd .V / such that S \ S 0 D f0g. Fix an S0 2 Gd .E/S 0, so that E D S0 ˚ S 0. For any S 2 Gd .V /S 0, the projection S ! S0 given by this decomposition is an isomorphism, and so S is the graph of a homomorphism S0 ! S 0: s 7! s0 ” .s; s0/ 2 S: Conversely, the graph of any homomorphism S0 ! S 0 lies in Gd .V /S 0. Thus, Gd .V /S 0 Hom.S0; S 0/ Hom.E=S 0; S 0/: (27) The isomorphism Gd .V /S 0 Hom.E=S 0; S 0/ depends on the choice of S0 — it is the element of Gd .V /S 0 corresponding to 0 2 Hom.E=S 0; S 0/. The decomposition E D S0 ˚ S 0 gives a decomposition End.E/ D End.S0/ Hom.S0; S 0/ Hom.S 0; S0/ End.S 0/ ; and the bijections (27) show that the group Gd .E/S 0. 0 1 Hom.S0;S 0/ 1 acts simply transitively on REMARK 6.32. The bijection (27) identifies Gd .E/S 0 with the affine variety A.Hom.S0; S 0// defined by the vector space Hom.S0; S 0/ (cf. p. 72). Therefore, the tangent space to Gd .E/ at S0, TS0.Gd .E// ' Hom.S0; S 0/ ' Hom.S0; E=S0/: Since the dimension of this space doesn’t depend on the choice of S0, this shows that Gd .E/ is nonsingular (4.39). (28) 148 6. PROJECTIVE VARIETIES REMARK 6.33. Let B be the set of all bases of E. The choice of a basis for E identifies B with GLn, which is the principal open subset of A In particular, B has a natural structure as an irreducible algebraic variety. The map .e1; : : : ; en/ 7! he1; : : : ; ed iW B ! Gd .E/ is a surjective regular map, and so Gd .E/ is also irreducible. where det ¤ 0. n2 REMARK 6.34. The exterior algebra V E D L Vd E of E is the quotient of the tensor algebra by the ideal generated by all vectors e ˝ e, e 2 E. The elements of Vd E are called (exterior) d -vectors:The exterior algebra of E is a finite-dimensional graded algebra over k wedge with V0 E D k, V1 E D E; if e1; : : : ; en form an ordered basis for V , then the n products d 0 d ei1 ^ : : : ^ eid .i1 < < id / form an ordered basis for Vd E. In particular, Vn E has dimension 1. For a subspace S of E of dimension d , Vd S is the one-dimensional subspace of Vd E spanned by e1 ^ : : : ^ ed for any basis e1; : : : ; ed of S. Thus, there is a well-defined map S 7! ^d S W Gd .E/ ! P. ^d E/ (29) which the choice of a basis for E identifies with S 7! P .S/. Note that the subspace spanned by e1; : : : ; en can be recovered from the line through e1 ^ : : : ^ ed as the space of vectors v such that v ^ e1 ^ : : : ^ ed D 0 (cf. 6.35 below). FIRST PROOF OF PROPOSITION 6.29. Fix a basis e1; : : : ; en of E, and let S0 D he1; : : : ; ed i and S 0 D hed C1; : : : ; eni. Order the coordinates in P so that n d 1 P .S/ D .a0W : : : W aij W : : : W : : :/; where a0 is the left-most d d minor of A.S/, and aij , 1 i d , d < j n, is the minor obtained from the left-most d d minor by replacing the i th column with the j th column. Let U0 be the (“typical”) standard open subset of P consisting of the points with nonzero zeroth coordinate. Clearly,6 P .S/ 2 U0 if and only if S 2 Gd .E/S 0. We shall prove the proposition by showing that P W Gd .E/S 0 ! U0 is injective with closed image. For S 2 Gd .E/S 0, the projection S ! S0 is bijective. For each i , 1 i d , let n d 1 e0 i D ei C P d <j n aij ej (30) denote the unique element of S projecting to ei . Then e0 d is a basis for S. Conversely, for any .aij / 2 kd.nd /, the e0 i defined by (30) span an S 2 Gd .E/S 0 and project to the ei . Therefore, S $ .aij / gives a one-to-one correspondence Gd .E/S 0 $ kd.nd / (this is a restatement of (27) in terms of matrices). 1; : : : ; e0 Now, if S $ .aij /, then P .S/ D .1 W : : : W aij W : : : W : : : W fk.aij /W : : :/: 6If e 2 S 0 \ S is nonzero, we may choose it to be part of the basis for S, and then the left-most d d submatrix of A.S/ has a row of zeros. Conversely, if the left-most d d submatrix is singular, we can change the basis for S so that it has a row of zeros; then the basis element corresponding to the zero row lies in S 0 \ S . m. Grassmann varieties 149 where fk.aij / is a polynomial in the aij whose coefficients are independent of S. Thus, P .S/ determines .aij / and hence also S. Moreover, the image of P W Gd .E/S 0 ! U0 is the graph of the regular map .: : : ; aij ; : : :/ 7! .: : : ; fk.aij /; : : :/W A d.nd / ! A n d d.nd /1 ; which is closed (5.28). SECOND PROOF OF PROPOSITION 6.29. An exterior d -vector v is said to be pure (or decomposable) if there exist vectors e1; : : : ; ed 2 V such that v D e1 ^ : : : ^ ed . According to 6.34, the image of Gd .E/ in P.Vd E/ consists of the lines through the pure d -vectors. LEMMA 6.35. Let w be a nonzero d -vector and let M.w/ D fv 2 E j v ^ w D 0gI then dimk M.w/ d , with equality if and only if w is pure. PROOF. Let e1; : : : ; em be a basis of M.w/, and extend it to a basis e1; : : : ; em; : : : ; en of V . Write w D X ai1:::id ei1 ^ : : : ^ eid ; ai1:::id 2 k. 1i1<:::<id If there is a nonzero term in this sum in which ej does not occur, then ej ^ w ¤ 0. Therefore, each nonzero term in the sum is of the form ae1 ^ : : : ^ em ^ : : :. It follows that m d , and m D d if and only if w D ae1 ^ : : : ^ ed with a ¤ 0. For a nonzero d -vector w, let Œw denote the line through w. The lemma shows that Œw 2 Gd .E/ if and only if the linear map v 7! v ^ wW E 7! Vd C1 E has rank n d (in which case the rank is n d ). Thus Gd .E/ is defined by the vanishing of the minors of order n d C 1 of this map. 7 Flag varieties The discussion in the last subsection extends easily to chains of subspaces. Let d D .d1; : : : ; dr / be a sequence of integers with 0 < d1 < < dr < n, and let Gd.E/ be the set of flags F W E E1 Er 0 (31) 7In more detail, the map w 7! .v 7! v ^ w/W^d E ! Homk.E; ^d C1 E/ is injective and linear, and so defines an injective regular map ^d P. E/ ,! P.Homk.E; ^d C1 E//: The condition rank n d defines a closed subset W of P.Homk.E; Vd C1 E// (once a basis has been cho
sen for E, the condition becomes the vanishing of the minors of order n d C 1 of a linear map E ! Vd C1 E), and Gd .E/ D P.Vd E/ \ W: 150 6. PROJECTIVE VARIETIES with Ei a subspace of E of dimension di . The map Gd.E/ F 7!.E i / ! Q i Gdi .E/ Q i P.Vdi E/ realizes Gd.E/ as a closed subset8 Q i Gdi .E/, and so it is a projective variety, called a flag variety. The tangent space to Gd.E/ at the flag F consists of the families of homomorphisms 'i W Ei ! E=Ei ; 1 i r; (32) that are compatible in the sense that 'i jEi C1 'i C1 mod EiC1: ASIDE 6.36. A basis e1; : : : ; en for E is adapted to the flag F if it contains a basis e1; : : : ; eji for each E i . Clearly, every flag admits such a basis, and the basis then determines the flag. As in (6.33), this implies that Gd.E/ is irreducible. Because GL.E/ acts transitively on the set of bases for E, it acts transitively on Gd.E/. For a flag F , the subgroup P .F / stabilizing F is an algebraic subgroup of GL.E/, and the map g 7! gF0W GL.E/=P .F0/ ! Gd.E/ is an isomorphism of algebraic varieties. Because Gd.E/ is projective, this shows that P .F0/ is a parabolic subgroup of GL.E/. n. Bezout’s theorem n (that is, a closed subvariety of dimension n 1). For such a Let V be a hypersurface in P variety, I.V / D .F .X0; : : : ; Xn// with F a homogenous polynomial without repeated factors. We define the degree of V to be the degree of F . The next theorem is one of the oldest, and most famous, in algebraic geometry. 2 of degrees m and n respectively. If C and D THEOREM 6.37. Let C and D be curves in P have no irreducible component in common, then they intersect in exactly mn points, counted with appropriate multiplicities. PROOF. Decompose C and D into their irreducible components. Clearly it suffices to prove the theorem for each irreducible component of C and each irreducible component of D. We can therefore assume that C and D are themselves irreducible. We know from 2.62 that C \ D is of dimension zero, and so is finite. After a change of variables, we can assume that a ¤ 0 for all points .a W b W c/ 2 C \ D. Let F .X; Y; Z/ and G.X; Y; Z/ be the polynomials defining C and D, and write F D s0Zm C s1Zm1 C C sm; G D t0Zn C t1Zn1 C C tn with si and tj polynomials in X and Y of degrees i and j respectively. Clearly sm ¤ 0 ¤ tn, for otherwise F and G would have Z as a common factor. Let R be the resultant of F and G, regarded as polynomials in Z. It is a homogeneous polynomial of degree mn in X and 8For example, if ui is a pure di -vector and ui C1 is a pure diC1-vector, then it follows from (6.35) that M.ui / M.ui C1/ if and only if the map v 7! .v ^ ui ; v ^ ui C1/W E ! ^di C1 E ˚ ^diC1C1 E has rank n di (in which case it has rank n di ). Thus, Gd.E/ is defined by the vanishing of many minors. o. Hilbert polynomials (sketch) 151 Y , or else it is identically zero. If the latter occurs, then for every .a; b/ 2 k2, F .a; b; Z/ and G.a; b; Z/ have a common zero, which contradicts the finiteness of C \ D. Thus R is a nonzero polynomial of degree mn. Write R.X; Y / D X mnR. Y X /, where R.T / is a polynomial of degree mn in T D Y X . Suppose first that deg R D mn, and let ˛1; : : : ; ˛mn be the roots of R (some of them may be multiple). Each such root can be written ˛i D bi , and R.ai ; bi / D 0. According to ai 7.28 this means that the polynomials F .ai ; bi ; Z/ and G.ai ; bi ; Z/ have a common root ci . Thus .ai W bi W ci / is a point on C \ D, and conversely, if .a W b W c/ is a point on C \ D (so a is a root of R.T /. Thus we see in this case, that C \ D has precisely mn a ¤ 0/, then b points, provided we take the multiplicity of .a W b W c/ to be the multiplicity of b a as a root of R. Now suppose that R has degree r < mn. Then R.X; Y / D X mnr P .X; Y /, where P .X; Y / is a homogeneous polynomial of degree r not divisible by X . Obviously R.0; 1/ D 0, and so there is a point .0 W 1 W c/ in C \ D, in contradiction with our assumption. REMARK 6.38. The above proof has the defect that the notion of multiplicity has been too obviously chosen to make the theorem come out right. It is possible to show that the theorem holds with the following more natural definition of multiplicity. Let P be an isolated point of C \ D. There will be an affine neighbourhood U of P and regular functions f and g on U such that C \ U D V .f / and D \ U D V .g/. We can regard f and g as elements P , and clearly rad.f; g/ D m, the maximal ideal in P . It follows that of the local ring P =.f; g/ is finite-dimensional over k, and we define the multiplicity of P in C \ D to be P =.f; g//. For example, if C and D cross transversely at P , then f and g will form O dimk. a system of local parameters at P — .f; g/ D m — and so the multiplicity is one. O O O The attempt to find good notions of multiplicities in very general situations motivated much of the most interesting work in commutative algebra in the second half of the twentieth century. o. Hilbert polynomials (sketch) Recall that for a projective variety V P n, khomŒV D kŒX0; : : : ; Xn=b D kŒx0; : : : ; xn; where b D I.V /. We observed that b is graded, and therefore khomŒV is a graded ring: khomŒV D M m0 khomŒV m; where khomŒV m is the subspace generated by the monomials in the xi of degree m. Clearly khomŒV m is a finite-dimensional k-vector space. THEOREM 6.39. There is a unique polynomial P .V; T / such that P .V; m/ D dimk kŒV m for all m sufficiently large. PROOF. Omitted. EXAMPLE 6.40. For V D P 141), dim khomŒV m D mCn n n, khomŒV D kŒX0; : : : ; Xn, and (see the footnote on page D .mCn/.mC1/ , and so nŠ P .P n; T / D T Cn n D .T C n/ .T C 1/ nŠ : 152 6. PROJECTIVE VARIETIES The polynomial P .V; T / in the theorem is called the Hilbert polynomial of V . Despite the notation, it depends not just on V but also on its embedding in projective space. THEOREM 6.41. Let V be a projective variety of dimension d and degree ı; then P .V terms of lower degree. PROOF. Omitted. The degree of a projective variety is the number of points in the intersection of the variety and of a general linear variety of complementary dimension (see later). EXAMPLE 6.42. Let V be the image of the Veronese map .a0 W a1/ 7! .ad 0 W ad 1 0 a1 W : : : W ad 1 /W P 1 ! P d : Then khomŒV m can be identified with the set of homogeneous polynomials of degree m d d C1 given by the same equations), which is a in two variables (look at the map A space of dimension d m C 1, and so 2 ! A P .V; T / D d T C 1: Thus V has dimension 1 (which we certainly knew) and degree d . Macaulay knows how to compute Hilbert polynomials. REFERENCES: Hartshorne 1977, I.7; Harris 1992, Lecture 13. p. Dimensions The results for affine varieties extend to projective varieties with one important simplification: n and r C s n, then if V and W are projective varieties of dimensions r and s in P V \ W ¤ ;. n be a projective variety of dimension 1, and let THEOREM 6.43. Let V D V .a/ P f 2 kŒX0; : : : ; Xn be homogeneous, nonconstant, and … a; then V \ V .f / is nonempty and of pure codimension 1. PROOF. Since the dimension of a variety is equal to the dimension of any dense open affine subset, the only part that doesn’t follow immediately from 3.42 is the fact that V \ V .f / nC1 (that is, the affine cone over V /. is nonempty. Let V aff.a/ be the zero set of a in A Then V aff.a/ \ V aff.f / is nonempty (it contains .0; : : : ; 0/), and so it has codimension 1 in V aff.a/. Clearly V aff.a/ has dimension 2, and so V aff.a/ \ V aff.f / has dimension 1. This implies that the polynomials in a have a zero in common with f other than the origin, and so V .a/ \ V .f / ¤ ;. COROLLARY 6.44. Let f1; : : : ; fr be homogeneous nonconstant elements of kŒX0; : : : ; Xn; and let Z be an irreducible component of V \ V .f1; : : : fr /. Then codim.Z/ r, and if dim.V / r, then V \ V .f1; : : : fr / is nonempty. PROOF. Induction on r, as before. p. Dimensions 153 PROPOSITION 6.45. Let Z be an irreducible closed subvariety of V ; if codim.Z/ D r, then there exist homogeneous polynomials f1; : : : ; fr in kŒX0; : : : ; Xn such that Z is an irreducible component of V \ V .f1; : : : ; fr /. PROOF. Use the same argument as in the proof 3.47. PROPOSITION 6.46. Every pure closed subvariety Z of P i.e., I.Z/ D .f / for some f homogeneous element of kŒX0; : : : ; Xn. n of codimension one is principal, PROOF. Follows from the affine case. COROLLARY 6.47. Let V and W be closed subvarieties of P then V \W ¤ ;, and every irreducible component of it has codim.Z/ codim.V /Ccodim.W /. PROOF. Write V D V .a/ and W D V .b/, and consider the affine cones V 0 D V .a/ and W 0 D V .b/ over them. Then n; if dim.V / C dim.W / n, dim.V 0/ C dim.W 0/ D dim.V / C 1 C dim.W / C 1 n C 2: As V 0 \ W 0 ¤ ;, V 0 \ W 0 has dimension 1, and so it contains a point other than the origin. Therefore V \ W ¤ ;. The rest of the statement follows from the affine case. n of dimension r < n; then there is a PROPOSITION 6.48. Let V be a closed subvariety of P linear projective variety E of dimension n r 1 (that is, E is defined by r C 1 independent linear forms) such that E \ V D ;. PROOF. Induction on r. If r D 0, then V is a finite set, and the lemma below shows that there is a hyperplane in knC1 not meeting V . Suppose r > 0, and let V1; : : : ; Vs be the irreducible components of V . By assumption, they all have dimension r. The intersection Ei of all the linear projective varieties containing Vi is the smallest such variety. The lemma below shows that there is a hyperplane H containing none of the nonzero Ei ; consequently, H contains none of the irreducible components Vi of V , and so each Vi \ H is a pure variety of dimension r 1 (or is empty). By induction, there is an linear subvariety E0 not meeting V \ H . Take E D E0 \ H . LEMMA 6.49. Let W be a vector space of dimension d over an infinite field k, and let E1; : : : ; Er be a finite set of nonzero subspaces of W . Then there is a hyperplane H in W containing none of the Ei . PROOF. Pass to the dual space V of W
. The problem becomes that of showing V is not a finite union of proper subspaces E_ i by a hyperplane Hi containing it. Then Hi is defined by a nonzero linear form Li . We have to show that Q Lj is not identically zero on V . But this follows from the statement that a polynomial in n variables, with coefficients not all zero, cannot be identically zero on kn (Exercise 1-1). i . Replace each E_ Let V and E be as in Proposition 6.48. If E is defined by the linear forms L0; : : : ; Lr then the projection a 7! .L0.a/ W W Lr .a// defines a map V ! P r . We shall see later that this map is finite, and so it can be regarded as a projective version of the Noether normalization theorem. In general, a regular map from a variety V to P n corresponds to a line bundle on V n X foriging are trivial and a set of global sections of the line bundle. All line bundles on A (see, for example, Hartshorne II 7.1 and II 6.2), from which it follows that all regular maps nC1 X foriging ! P m are given by a family of homogeneous polynomials. Assuming this, A it is possible to prove the following result. 154 6. PROJECTIVE VARIETIES n ! P nC1 foriging ! P COROLLARY 6.50. Let ˛W P m be regular; if m < n, then ˛ is constant. n be the map .a0; : : : ; an/ 7! .a0 W : : : W an/. Then ˛ ı PROOF. Let W A is regular, and there exist polynomials F0; : : : ; Fm 2 kŒX0; : : : ; Xn such that ˛ ı is the map .a0; : : : ; an/ 7! .F0.a/ W : : : W Fm.a//: As ˛ ı factors through P n, the Fi must be homogeneous of the same degree. Note that ˛.a0 W : : : W an/ D .F0.a/ W : : : W Fm.a//: If m < n and the Fi are nonconstant, then 6.43 shows they have a common zero and so ˛ is not defined on all of P n. Hence the Fi must be constant. q. Products It is useful to have an explicit description of the topology on some product varieties. The topology on Pm Pn. Suppose we have a collection of polynomials Fi .X0; : : : ; XmI Y0; : : : ; Yn/, i 2 I , each of which is separately homogeneous in the Xi and Yj . Then the equations Fi .X0; : : : ; XmI Y0; : : : ; Yn/ D 0; i 2 I; m P define a closed subset of P from a (finite) set of polynomials. n, and every closed subset of P m P n arises in this way The topology on Am Pn m P The closed subsets of A n are exactly those defined by sets of equations Fi .X1; : : : ; XmI Y0; : : : ; Yn/ D 0; i 2 I; with each Fi homogeneous in the Yj . The topology on V Pn Let V be an irreducible affine algebraic variety. We look more closely at the topology n in terms of ideals. Let A D kŒV , and let B D AŒX0; : : : ; Xn. Note that B D on V P nC1: for A ˝k kŒX0; : : : ; Xn, and so we can view it as the ring of regular functions on V A f 2 A and g 2 kŒX0; : : : ; Xn, f ˝ g is the function .v; a/ 7! f .v/ g.a/W V A nC1 ! k: n , a 2 A, has degree P ij — The ring B has an obvious grading — a monomial aX i0 and so we have the notion of a graded ideal b B. It makes sense to speak of the zero set V .b/ V P n of such an ideal. For any ideal a A, aB is graded, and V .aB/ D V .a/ P n. LEMMA 6.51. (a) For each graded ideal b B, the set V .b/ is closed, and every closed subset of V P n is of this form. 0 : : : X in (b) The set V .b/ is empty if and only if rad.b/ .X0; : : : ; Xn/. (c) If V is irreducible, then V D V .b/ for some graded prime ideal b. q. Products 155 PROOF. (a) In the case that A D k, we proved this in 6.1 and 6.2, and similar arguments apply in the present more general situation. For example, to see that V .b/ is closed, cover n with the standard open affines Ui and show that V .b/ \ Ui is closed for all i . P The set V .b/ is empty if and only if the cone V aff.b/ V A nC1 defined by b is contained in V foriging. But X ai0:::inX i0 0 : : : X in n ; ai0:::in 2 kŒV ; is zero on V foriging if and only if its constant term is zero, and so I aff.V foriging/ D .X0; X1; : : : ; Xn/: Thus, the Nullstellensatz shows that V .b/ D ; ) rad.b/ D .X0; : : : ; Xn/. Conversely, if X N i 2 b for all i , then obviously V .b/ is empty. For (c), note that if V .b/ is irreducible, then the closure of its inverse image in V A nC1 is also irreducible, and so I V .b/ is prime. Exercises 6-1. Show that a point P on a projective curve F .X; Y; Z/ D 0 is singular if and only if @F=@X, @F=@Y , and @F=@Z are all zero at P . If P is nonsingular, show that the tangent line at P has the (homogeneous) equation .@F=@X/P X C .@F=@Y /P Y C .@F=@Z/P Z D 0. Verify that Y 2Z D X 3 C aXZ2 C bZ3 is nonsingular if X 3 C aX C b has no repeated root, and find the tangent line at the point at infinity on the curve. 6-2. Let L be a line in P by a homogeneous polynomial of degree 2). Show that either 2 and let C be a nonsingular conic in P 2 (i.e., a curve in P 2 defined (a) L intersects C in exactly 2 points, or (b) L intersects C in exactly 1 point, and it is the tangent at that point. 3. Prove 6-3. Let V D V .Y X 2; Z X 3/ A (a) I.V / D .Y X 2; Z X 3/; (b) ZW X Y 2 I.V / kŒW; X; Y; Z, but ZW XY … ..Y X 2/; .Z X 3//. r generate a, even (Thus, if F1; : : : ; Fr generate a, it does not follow that F if ais radical.) 1 ; : : : ; F 6-4. Let P0; : : : ; Pr be points in P through P0 but not passing through any of P1; : : : ; Pr . n. Show that there is a hyperplane H in P n passing 6-5. Is the subset of P 2 locally closed? f.a W b W c/ j a ¤ 0; b ¤ 0g [ f.1 W 0 W 0/g 6-6. Show that the image of the Segre map P in any hyperplane of P mnCmCn. m P n ! P mnCmCn (see 6.26) is not contained 156 6. PROJECTIVE VARIETIES 6-7. Write 0, 1, 1 for the points .0W 1/, .1W 1/, and .1W 0/ on P (a) Let ˛ be an automorphism of P 1 such that 1. ˛.0/ D 0; ˛.1/ D 1; ˛.1/ D 1: Show that ˛ is the identity map. (b) Let P0, P1, P2 be distinct points on P 1. Show that there exists an ˛ 2 PGL2.k/ such that ˛.0/ D P0; ˛.1/ D P1; ˛.1/ D P2: (c) Deduce that Aut.P 1/ ' PGL2.k/. 6-8. Show that the functor R P n.R/ D fdirect summands of rank 1 of RnC1g satisfies the criterion 5.71 to arise from an algebraic prevariety. (This gives an alternative definition of P n.) 6-9. (a) Let V A m be algebraic varieties and 'W V ! W a map. Show that ' is regular if and only if every point in V has an open neighbourhood U on which there are regular functions f0; : : : ; fm such that n and W P '.a1; : : : ; an/ D .f0.a1; : : : ; an/W : : : W fm.a1; : : : ; an// for all .a1; : : : ; an/ 2 U . (b) Show that, for a regular map ' as in (a), it may not be possible to take U D V . Hint: Let V A 4 be the complement of .0; 0; 0; 0/ in XY ZW D 0; and let 'W V ! P .wW y/. See sx4626969 (Mohan). 1 send .w; x; y; z/ to .xW z/ if one of x or z is nonzero and .w; 0; y; 0/ to CHAPTER 7 Complete Varieties Complete varieties are the analogues in the category of algebraic varieties of compact topological spaces in the category of Hausdorff topological spaces. Recall that the image of a compact space under a continuous map is compact, and hence is closed if the image space is Hausdorff. Moreover, a Hausdorff space V is compact if and only if, for all topological spaces T , the projection map qW V T ! T is closed, i.e., maps closed sets to closed sets (see Bourbaki, N., General Topology, I, 10.2, Corollary 1 to Theorem 1). a. Definition and basic properties Definition DEFINITION 7.1. An algebraic variety V is complete if for all algebraic varieties T , the projection map qW V T ! T is closed. Note that a complete variety is required to be separated — we really mean it to be a variety and not a prevariety. We shall see 7.22 that projective varieties are complete. EXAMPLE 7.2. Consider the projection map .x; y/ 7! yW A 1 A 1 ! A 1: This is not closed; for example, the variety V W XY D 1 is closed in A omits the origin. However, when we replace V with its closure in P 1. To see this, note that becomes the whole of A NV defD f..xW z/; y/ 2 P 1 j xy D z2g 1 A 2 but its image in A 1 1 A 1, its projection contains V as an open dense subset, and so must be its closure in P ..xW 0/; 0/ of NV maps to 0. 1 A 1. The point Properties 7.3. Closed subvarieties of complete varieties are complete. Let Z be a closed subvariety of a complete variety V . For any variety T , Z T is closed in V T , and so the restriction of the closed map qW V T ! T to Z T is also closed. 7.4. A variety is complete if and only if its irreducible components are complete. 157 158 7. COMPLETE VARIETIES Each irreducible component is closed, and hence complete if the variety is complete (7.3). Conversely, suppose that the irreducible components Vi of a variety V are complete. If Z is defD Z \ .Vi T / is closed in Vi T . Therefore, q.Zi / is closed in closed in V T , then Zi T , and so q.Z/ D S q.Zi / is also closed. 7.5. Products of complete varieties are complete. Let V1; : : : ; Vn be complete varieties, and let T be a variety. The projection Q is the composite of the projections i Vi T ! T V1 Vn T ! V2 Vn T ! ! Vn T ! T; all of which are closed. 7.6. If 'W W ! V is surjective and W is complete, then V is complete. Let T be a variety, and let Z be a closed subset of V T . Let Z0 be the inverse image of Z in W T . Then Z0 is closed, and its image in T equals that of Z. 7.7. Let 'W W ! V be a regular map of varieties. If W is complete, then '.W / is a complete closed subvariety of V . In particular, every complete subvariety of a variety is closed. defD f.w; '.w//g W V be the graph of '. It is a closed subset of W V (because Let ' V is a variety, see 5.28), and '.W / is the projection of ' into V . Therefore '.W / is closed, and 7.6 shows that it is complete. The second statement follows from the first applied to the identity map. 7.8. A regular map V ! P surjective. 1 from a complete connected variety V is either constant or 1 are the finite sets, and such a set is connected if and The only proper closed subsets of P only if it consists of a single point. Because '.V / is connected and closed, it must either be 1 (and ' is onto). a single point (and ' is constant) or P 7.9. The only regular functions on a complete connected variety are the constant functions. A regular function on a variety V is a
regular map f W V ! A 7.8. 1 P 1, to which we can apply 7.10. A regular map 'W V ! W from a complete connected variety to an affine variety has image equal to a point. In particular, every complete connected affine variety is a point. n, and write ' D .'1; : : : ; 'n/, where 'i is the composite Embed W as a closed subvariety of A of ' with the coordinate function xi W A 1. Each 'i is a regular function on V , and hence is constant. (Alternatively, apply 5.12.) This proves the first statement, and the second follows from the first applied to the identity map. n ! A 7.11. In order to show that a variety V is complete, it suffices to check that qW V T ! T is a closed mapping when T is affine (or even an affine space A Every variety T can be written as a finite union of open affine subvarieties T D S Ti . If Z is closed in V T , then Zi defD Z \ .V Ti / is closed in V Ti . Therefore, q.Zi / is closed in Ti for all i . As q.Zi / D q.Z/ \ Ti , this shows that q.Z/ is closed. This shows that it suffices to check that V T ! T is closed for all affine varieties T . But T can be realized as a closed subvariety of A n, and then V T ! T is closed if V A n is closed. n ! A n). b. Proper maps 159 Remarks 7.12. The statement that a complete variety V is closed in every larger variety W perhaps explains the name: if V is complete, W is connected, and dim V D dim W , then V D W . Contrast A n P n. 7.13. Here is another criterion: a variety V is complete if and only if every regular map C X fP g ! V extends uniquely to a regular map C ! V ; here P is a nonsingular point on a curve C . Intuitively, this says that all Cauchy sequences have limits in V and that the limits are unique. b. Proper maps DEFINITION 7.14. A regular map 'W V ! S of varieties is said to be proper if it is “universally closed”, that is, if for all regular maps T ! S , the base change '0W V S T ! T of ' is closed. 7.15. For example, a variety V is complete if and only if the map V ! fpointg is proper. 7.16. From its very definition, it is clear that the base change of a proper map is proper. In particular, (a) if V is complete, then V S ! S is proper, (b) if 'W V ! S is proper, then the fibre '1.P / over a point P of S is complete. 7.17. If 'W V ! S is proper, and W is a closed subvariety of V , then W '! S is proper. PROPOSITION 7.18. A composite of proper maps is proper. PROOF. Let V3 ! V2 ! V1 be proper maps, and let T be a variety. Consider the diagram V3 V2 V1 V3 V2 .V2 V1 T / ' V3 V1 T closed V2 V1 T closed T: Both smaller squares are cartesian, and hence so also is the outer square. The statement is now obvious from the fact that a composite of closed maps is closed. COROLLARY 7.19. If V ! S is proper and S is complete, then V is complete. PROOF. Special case of the proposition. COROLLARY 7.20. The inverse image of a complete variety under a proper map is complete. PROOF. Let 'W V ! S be proper, and let Z be a complete subvariety of S . Then V S Z ! Z is proper, and V S Z ' '1.Z/. 160 7. COMPLETE VARIETIES EXAMPLE 7.21. Let f 2 kŒT1; : : : ; Tn; X; Y be homogeneous of degree m in X and Y , and let H be the subvariety of A 1 defined by n P f .T1; : : : ; Tn; X; Y / D 0. The projection map A 7.15). The fibre over a point .t1; : : : ; tn/ 2 A mial 1 ! A n P n defines a regular map H ! A n is the subvariety of P n, which is proper (7.22, 1 defined by the polyno- f .t1; : : : ; tn; X; Y / D a0X m C a1X m1Y C C amY m; ai 2 k: Assume that not all ai are zero. Then this is a homogeneous of degree m and so the fibre always has m points counting multiplicities. The points that “disappeared off to infinity” when P 1 (see p. 51) have literally become the point at infinity on P 1 was taken to be A 1. c. Projective varieties are complete The reader may skip this section since the main theorem is given a more explicit proof in Theorem 7.31 below. THEOREM 7.22. A projective variety is complete. n itself; thus we n W ! W is a closed mapping in the case that W PROOF. After 7.3, it suffices to prove the Theorem for projective space P have to prove that the projection map P is an irreducible affine variety (7.11). Write p for the projection W P n ! W . We have to show that Z closed in W P implies that p.Z/ closed in W . If Z is empty, this is true, and so we can assume it to be nonempty. Then Z is a finite union of irreducible closed subsets Zi of W P n, and it suffices to show that each p.Zi / is closed. Thus we may assume that Z is irreducible, and hence that Z D V .b/ with b a graded prime ideal in B D AŒX0; : : : ; Xn (6.51). If p.Z/ is contained in some closed subvariety W 0 of W , then Z is contained in W 0 P n, and we can replace W with W 0. This allows us to assume that p.Z/ is dense in W , and we now have to show that p.Z/ D W . n Because p.Z/ is dense in W , the image of the cone V aff.b/ under the projection W nC1 ! W is also dense in W , and so (see 3.34a) the map A ! B=b is injective. Let w 2 W : we shall show that if w … p.Z/, i.e., if there does not exist a P 2 P n such A that .w; P / 2 Z, then p.Z/ is empty, which is a contradiction. Let m A be the maximal ideal corresponding to w. Then mB C b is a graded ideal, n/ \ V .b/, and so w will be in the image of Z and V .mB C b/ D V .mB/ \ V .b/ D .w P unless V .mB C b/ ¤ ;. But if V .mB C b/ D ;, then mB C b .X0; : : : ; Xn/N for some N (by 6.51b), and so mB C b contains the set BN of homogeneous polynomials of degree N . Because mB and b are graded ideals, BN mB C b H) BN D mBN C BN \ b: In detail: the first inclusion says that an f 2 BN can be written f D g C h with g 2 mB and h 2 b. On equating homogeneous components, we find that fN D gN C hN . Moreover: fN D f ; if g D P mi bi , mi 2 m, bi 2 B, then gN D P mi biN ; and hN 2 b because b is homogeneous. Together these show f 2 mBN C BN \ b. Let M D BN =BN \ b, regarded as an A-module. The displayed equation says that M D mM . The argument in the proof of Nakayama’s lemma (1.3) shows that .1 C m/M D 0 for some m 2 m. Because A ! B=b is injective, the image of 1 C m in B=b is nonzero. But d. Elimination theory 161 M D BN =BN \ b B=b, which is an integral domain, and so the equation .1 C m/M D 0 2 b for all i , which contradicts the implies that M D 0. Hence BN b, and so X N i assumption that Z D V .b/ is nonempty. Remarks 7.23. Every complete curve is projective. 7.24. Every nonsingular complete surface is projective (Zariski), but there exist singular complete surfaces that are not projective (Nagata). 7.25. There exist nonsingular complete three-dimensional varieties that are not projective (Nagata, Hironaka). 7.26. A nonsingular complete irreducible variety V is projective if and only if every finite set of points of V is contained in an open affine subset of V (Conjecture of Chevalley; proved by Kleiman1; see 6.22 for the necessity). d. Elimination theory When given a system of polynomial equations to solve, we first use some of the equations to eliminate some of the variables; we then find the solutions of the reduced system, and go back to find the solutions of the original system. Elimination theory does this more systematically. Note that the fact that P system of polynomial equations n is complete has the following explicit restatement: for each ./ 8 ˆ< ˆ: P1.X1; : : : ; XmI Y0; : : : ; Yn/ D 0 ::: Pr .X1; : : : ; XmI Y0; : : : ; Yn/ D 0 such that each Pi is homogeneous in the Yj , there exists a system of polynomial equations ./ 8 ˆ< ˆ: R1.X1; : : : ; Xm/ D 0 ::: Rs.X1; : : : ; Xm/ D 0 with the following property; an m-tuple .a1; : : : ; am/ is a solution of (*) if and only if there exists a nonzero n-tuple .b0; : : : ; bn/ such that .a1; : : : ; am; b0; : : : ; bn/ is a solution of (). In other words, the polynomials Pi .a1; : : : ; amI Y0; : : : ; Yn/ have a common zero if and only if Rj .a1; : : : ; am/ D 0 for all j . The polynomials Rj are said to have been obtained from the polynomials Pi by elimination of the variables Yi . Unfortunately, the proof we gave of the completeness of P n, while short and elegant, gives no indication of how to construct (*) from (). The purpose of elimination theory is to provide an algorithm for doing this. 1Kleiman, Steven L., Toward a numerical theory of ampleness. Ann. of Math. (2) 84 1966 293–344 (Theorem 3, p. 327, et seq.). See also, Hartshorne, Robin, Ample subvarieties of algebraic varieties. Lecture Notes in Mathematics, Vol. 156 Springer, 1970, I 9 p45. 162 7. COMPLETE VARIETIES Elimination theory: special case Let P D s0X m C s1X m1 C C sm and Q D t0X n C t1X n1 C C tn be polynomials. The resultant of P and Q is defined to be the determinant s0 t0 s1 s0 t1 t0 : : : : : : : : : : : : : : : : : : sm tn sm tn rows m rows There are n rows with s0 : : : sm and m rows with t0 : : : tn, so that the matrix is .m C n/ .m C n/; all blank spaces are to be filled with zeros. The resultant is a polynomial in the coefficients of P and Q. PROPOSITION 7.27. The resultant Res.P; Q/ D 0 if and only if (a) both s0 and t0 are zero; or (b) the two polynomials have a common root. PROOF. If (a) holds, then Res.P; Q/ D 0 because the first column is zero. Suppose that ˛ is a common root of P and Q, so that there exist polynomials P1 and Q1 of degrees m 1 and n 1 respectively such that P .X/ D .X ˛/P1.X/; Q.X/ D .X ˛/Q1.X/: Using these equalities, we find that P .X/Q1.X/ Q.X/P1.X/ D 0: (33) On equating the coefficients of X mCn1; : : : ; X; 1 in (33) to zero, we find that the coefficients of P1 and Q1 are the solutions of a system of m C n linear equations in m C n unknowns. The matrix of coefficients of the system is the transpose of the matrix s0 t0 0 B B B B B B @ s1 s0 t1 t0 sm tn sm tn : : : : : : : : : : : : : : : : : : 1 C C C C C C A : : : : : : The existence of the solution shows that this matrix has determinant zero, which implies that Res.P; Q/ D 0. Conversely, suppose that Res.P; Q/ D 0 but neither s0 nor t0 is zero. Because the above matrix has determinant zero, we can solve the linear equations to find polyno
mials P1 and Q1 satisfying (33). A root ˛ of P must be also be a root of P1 or of Q. If the former, cancel X ˛ from the left hand side of (33), and consider a root ˇ of P1=.X ˛/. As deg P1 < deg P , this argument eventually leads to a root of P that is not a root of P1, and so must be a root of Q. d. Elimination theory 163 The proposition can be restated in projective terms. We define the resultant of two homogeneous polynomials P .X; Y / D s0X m C s1X m1Y C C smY m; Q.X; Y / D t0X n C C tnY n; exactly as in the nonhomogeneous case. PROPOSITION 7.28. The resultant Res.P; Q/ D 0 if and only if P and Q have a common zero in P 1. PROOF. The zeros of P .X; Y / in P (a) .1 W 0/ in the case that s0 D 0; (b) .a W 1/ with a a root of P .X; 1/. 1 are of the form: Since a similar statement is true for Q.X; Y /, 7.28 is a restatement of 7.27. Now regard the coefficients of P and Q as indeterminates. The pairs of polynomials mCnC2. Consider the closed subset .P; Q/ are parametrized by the space A mCnC2 is the set mCnC2 P V .P; Q/ in A defined by Res.P; Q/ D 0. Thus, not only have we shown that the projection of V .P; Q/ is closed, but we have given an algorithm for passing from the polynomials defining the closed set to those defining its projection. 1. The proposition shows that its projection on A nC1 D A mC1 A Elimination theory does this in general. Given a family of polynomials Pi .T1; : : : ; TmI X0; : : : ; Xn/; homogeneous in the Xi , elimination theory gives an algorithm for finding polynomials Rj .T1; : : : ; Tm/ such that the Pi .a1; : : : ; amI X0; : : : ; Xn/ have a common zero if and only if Rj .a1; : : : ; am/ D 0 for all j . (Theorem 7.22 shows only that the Rj exist.) Maple can find the resultant of two polynomials in one variable: for example, entering “resultant..x C a/5; .x C b/5; x/” gives the answer .a C b/25. Explanation: the polynomials have a common root if and only if a D b, and this can happen in 25 ways. Macaulay doesn’t seem to know how to do more. Elimination theory: general case In this subsection, we give a proof of Theorem 7.22, following Cartier and Tate 1978,2 which is a more explicit proof than that given above. Throughout, k is a field (not necessarily algebraically closed) and K is an algebraically closed field containing k. THEOREM 7.29. For any graded ideal a in kŒX0; : : : ; Xn, exactly one of the following statements is true: (a) there exists an integer d0 0 such that a contains every homogeneous polynomial of degree d d0; (b) the ideal a has a nontrivial zero in KnC1. PROOF. Statement (a) says that the radical of a contains .X0; : : : ; Xn/, and so the theorem is a restatement of 6.2(a), which we deduced from the strong Nullstellensatz. For a direct proof of it, see the article of Cartier and Tate. 2Cartier, P., Tate, J., A simple proof of the main theorem of elimination theory in algebraic geometry. Enseign. Math. (2) 24 (1978), no. 3-4, 311–317. 164 7. COMPLETE VARIETIES THEOREM 7.30. Let R D L Rd be a graded k-algebra such that R0 D k, R is generated as a k-algebra by R1, and Rd is finite-dimensional for all d . Then exactly one of the following statements is true: d 2N (a) there exists an integer d0 0 such that Rd D 0 for all d d0; (b) no Rd D 0, and there exists a k-algebra homomorphism R ! K whose kernel is not equal to RC defD L d 1 Rd . PROOF. The hypotheses on R say that it is a quotient of kŒX0; : : : ; Xn by a graded ideal. Therefore 7.30 is a restatement of 7.29. Let P1; : : : ; Pr be polynomials in kŒT1; : : : ; TmI X0; : : : ; Xn with Pj homogeneous of degree dj in the variables X0; : : : ; Xn. Let J be the ideal .P1; : : : ; Pr / in kŒT1; : : : ; TmI X0; : : : ; Xn, and let A be the ideal of polynomials f in kŒT1; : : : ; Tm with the following property: there exists an integer N 1 such that all lie in J . n.K/ P n.K/. The projection of V into THEOREM 7.31. Let V be the zero set of J in A A n.K/ is the zero set of A. Consider the ring B D kŒT1; : : : ; TmI X0; : : : ; Xn and its subring B0 D kŒT1; : : : ; Tm. Then B is a graded B0-algebra with Bd the B0-submodule generated by the monomials of degree d in X0; : : : ; Xn, and J is a homogeneous (graded) ideal in B. Let A D L Ad be the quotient graded ring B=J D L Bd =.Bd \ J /. Let S be the ideal of elements a d 2N of A0 such that aAd D 0 for all sufficiently large d . d 2N THEOREM 7.32. A ring homomorphism 'W A0 ! K extends to a ring homomorphism W A ! K not annihilating the ideal AC defD L d 1 Ad if and only if '.S/ D 0. Following Cartier and Tate, we leave it to reader to check that 7.32 is equivalent to 7.31. Proof of Theorem 7.32 We shall prove 7.32 for any graded ring A D L tions: d 0 Ad satisfying the following two condi- (a) as an A0-algebra, A is generated by A1; (b) for every d 0, Ad is finitely generated as an A0-module. In the statement of the theorem, K is any algebraically closed field. The proof proceeds by replacing A with other graded rings with the properties (a) and (b) and also having the property that no Ad is zero. Let 'W A0 ! K be a homomorphism such that '.S/ D 0, and let P D Ker.'/. Then P is a prime ideal of A0 containing S. Step 1. Let J be the ideal of elements a of A for which there exists an s 2 A0 X P such that sa D 0. For every d 0, the annihilator of the A0-module Ad is contained in S, hence in P, and so J \ Ad ¤ Ad . The ideal J is graded, and the quotient ring A0 D A=J has the required properties. Step 2. Let A00 be the ring of fractions of A0 whose denominators are in ˙ defD A0 0 Let A00 L d be the set of fractions with numerator in A0 d 0 A00 d is a graded ring with the required properties, and A00 X P. d and denominator in ˙. Then A00 D 0 is a local ring with maximal ideal P00 defD P0 A0 0. e. The rigidity theorem; abelian varieties 165 Step 3. Let R be the quotient of A00 by the graded ideal P00 A00. As A00 finitely generated module over the local ring A00 Therefore R is graded ring with the required properties, and k D R0 0, Nakayama’s lemma shows that A00 defD A00 d is a nonzero ¤ P00A00 d . 0=P00 is a field. d Step 4. At this point R satisfies the hypotheses of Theorem 7.30. Let " be the composite of the natural maps A ! A0 ! A00 ! R. In degree 0, this is nothing but the natural map from A0 to k with kernel P. As ' has the same kernel, it factors through "0, making K into an algebraically closed extension of k. Now, by Theorem 7.30, there exists a k-algebra homomorphism f W R ! K such that f .RC/ ¤ 0. The composite map D f ı " has the required properties. For more on elimination theory, see Chapter 8, Section 5, of Cox, David A.; Little, John; O’Shea, Donal, Ideals, varieties, and algorithms. Springer, Cham, 2015. ASIDE 7.33. Elimination theory became unfashionable several decades ago — one prominent algebraic geometer went so far as to announce that Theorem 7.22 eliminated elimination theory from mathematics,3 provoking Abhyankar, who prefers equations to abstractions, to start the chant “eliminate the eliminators of elimination theory”. With the rise of computers, it has become fashionable again. e. The rigidity theorem; abelian varieties The paucity of maps between complete varieties has some interesting consequences. First an observation: for any point w 2 W , the projection map V W ! V defines an isomorphism V fwg ! V with inverse v 7! .v; w/W V ! V W (this map is regular because its components are). THEOREM 7.34 (RIGIDITY THEOREM). Let 'W V W ! T be a regular map, and assume that V is complete, V and W are irreducible, and T is separated. If '.v; w0/ is independent of v for one w0 2 W , then '.v; w/ D g.w/ with g a regular map gW PROOF. Choose a v0 2 V , and consider the regular map gW W ! T; w 7! '.v0; w/: We shall show that ' D g ı q. Because V is complete, the projection map qW V W ! W is closed. Let U be an open affine neighbourhood U of '.v0; w0/; then T X U is closed in T , '1.T X U / is closed in V W , and C defD q.'1.T X U // is closed in W . By definition, C consists of the w 2 W such that '.v; w/ … U for some v 2 V , and so W X C D fw 2 W j '.V fwg/ U g: 3Weil, A., Foundations of Algebraic Geometry, 1946/1962, p. 31: “The device that follows, which, it may be hoped, finally eliminates from algebraic geometry the last traces of elimination-theory, is borrowed from C. Chevalley’s Princeton lectures.” Demazure 2012 quotes Dieudonn´e as saying: “Il faut ´eliminer la th´eorie de l’´elimination.” 166 7. COMPLETE VARIETIES As '.V; w0/ D '.v0; w0/, we see that w0 2 W X C . Therefore W X C is nonempty, and so it is dense in W . As V fwg is complete and U is affine, '.V fwg/ must be a point whenever w 2 W X C (see 7.10); in fact '.V fwg/ D '.v0; w/ D g.w/: We have shown that ' and g ı q agree on the dense subset V .W X C / of V W , and therefore on the whole of V W . COROLLARY 7.35. Let 'W V W ! T be a regular map, and assume that V is complete, that V and W are irreducible, and that T is separated. If there exist points v0 2 V , w0 2 W , t0 2 T such that '.V fw0g/ D ft0g D '.fv0g W /, then '.V W / D ft0g. PROOF. With g as in the proof of the theorem, '.v; w/ D g.w/ D '.v0; w/ D t0: In more colloquial terms, the corollary says that if ' collapses a vertical and a horizontal slice to a point, then it collapses the whole of V W to a point, which must therefore be “rigid”. DEFINITION 7.36. An abelian variety is a complete connected group variety. THEOREM 7.37. Every regular map ˛W A ! B of abelian varieties is the composite of a homomorphism with a translation; in particular, a regular map ˛W A ! B such that ˛.0/ D 0 is a homomorphism. PROOF. After composing ˛ with a translation, we may suppose that ˛.0/ D 0. Consider the map 'W A A ! B; '.a; a0/ D ˛.a C a0/ ˛.a/ ˛.a0/: Then '.A 0/ D 0 D '.0 A/ and so ' D 0. This means that ˛ is a homomorphism. COROLLARY 7.38. The group law on an abelian variety is commutative. PROOF. Commutative groups are distinguished among all groups by the fact that the map taking an element to its inverse is a homomorphism: if .gh/1 D g1h1, then, on
taking inverses, we find that gh D hg. Since the negative map, a 7! aW A ! A, takes the identity element to itself, the preceding corollary shows that it is a homomorphism. f. Chow’s Lemma The next theorem is a useful tool in extending results from projective varieties to complete varieties. It shows that a complete variety is not far from a projective variety. THEOREM 7.39 (CHOW’S LEMMA). For every complete irreducible variety V , there exists a surjective regular map f W V 0 ! V from a projective algebraic variety V 0 to V such that, for some dense open subset U of V , f induces an isomorphism f 1.U / ! U (in particular, f is birational). f. Chow’s Lemma 167 Write V as a finite union of nonempty open affines, V D U1 [: : :[Un, and let U D T Ui . Because V is irreducible, U is a dense in V . Realize each Ui as a dense open subset of a projective variety Pi . Then P defD Q i Pi is a projective variety (6.26). We shall construct an algebraic variety V 0 and regular maps f W V 0 ! V and gW V 0 ! P such that (a) f is surjective and induces an isomorphism f 1.U / ! U ; (b) g is a closed immersion (hence V 0 is projective). Let '0 (resp. 'i ) denote the given inclusion of U into V (resp. into Pi ), and let ' D .'0; '1; : : : ; 'n/W U ! V P1 Pn; be the diagonal map. We set U 0 D '.U / and V 0 equal to the closure of U 0 in V P1 Pn. The projection maps pW V P ! V and qW V P ! P restrict to regular maps f W V 0 ! V and gW V 0 ! P . Thus, we have a commutative diagram ': (34) PROOF OF (a) In the upper-left triangle of the diagram (34), the maps ' and '0 are isomorphisms from U onto its images U 0 and U . Therefore f restricts to an isomorphism U 0 ! U . Note that U 0 D f.u; '1.u/; : : : ; 'n.u// j u 2 U g; which is the graph of the map .'1; : : : ; 'n/W U ! P . Therefore, U 0 is closed in U P (5.28), and so U 0 D V 0 \ .U P / D f 1.U /: The map f is dominant, and f .V 0/ D p.V /, which is closed because P is complete. Hence f is surjective. PROOF OF (b) We first show that g is an immersion. As this is a local condition, it suffices to find g! Vi is an open subsets Vi P such that S q1.Vi / V 0 and each map V 0 \ q1.Vi / immersion. We set Vi D p1 i .Ui / D P1 Ui Pn where pi is the projection map P ! Pi . We first show that the sets q1.Vi / cover V 0. The sets Ui cover V , hence the sets f 1.Ui / cover V 0, and so it suffices to show that q1.Vi / f 1.Ui / 168 7. COMPLETE VARIETIES for all i . Consider the diagrams q1.Vi / V P pi ıq Ui 'i Pi f 1.Ui / f V P pi ıq Ui 'i Pi '0 U ' V P pi ıq Ui 'i Pi : The diagram at left is cartesian, i.e., it realizes q1.Vi / as the fibred product q1.Vi / D .V P / Pi Ui ; and so it suffices to show that the middle diagram commutes. But U 0 is dense in V 0, hence in f 1.Ui /, and so it suffices to prove that the middle diagram commutes with f 1.Ui / replaced by U 0. But then it becomes the diagram at right, which obviously commutes. We next show that V 0 \ q1.Vi / g! Vi is an immersion for each i . Recall that and so Vi D Ui P i ; where P i D Y j ¤i Pj : q1.Vi / D V Ui P i V P: Let i denote the graph of the map Ui P i pi! Ui ,! V : Being a graph, i is closed in V Ui P i and the projection map V Ui P i ! Ui P i restricts to an isomorphism i ! Ui P i . In other words, i is closed in q1.Vi /, and the projection map q1.Vi / ! Vi restricts to an isomorphism i ! Vi . As i is closed in q1.Vi / and contains U 0, it contains V 0 \ q1.Vi /, and so the projection map q1.Vi / ! Vi restricts to an immersion V 0 \ q1.Vi / ! Vi . Finally, V P is complete because V and P are, and so V 0 is complete (7.3). Hence g.V / is closed (7.7), and so g is a closed immersion. Notes 7.40. Let V be a complete variety, and let V1; : : : ; Vs be the irreducible components of V . Each Vi is complete (7.4), and so there exists a surjective birational regular map V 0 ! Vi i with V 0 i projective (7.39). Now F V 0 i is projective 6.26, and the composite G ! G V 0 i Vi ! V is surjective and birational. 7.41. Chow (1956, Lemma 1)4 proved essentially the statement 7.42 by essentially the above argument. He used the lemma to prove that all homogeneous spaces are quasiprojective. See also EGA II, 5.6.1. 4Chow, Wei-Liang. On the projective embedding of homogeneous varieties. Algebraic geometry and topology. A symposium in honor of S. Lefschetz, pp. 122–128. Princeton University Press, Princeton, N. J., 1957. g. Analytic spaces; Chow’s theorem 169 g. Analytic spaces; Chow’s theorem We summarize a little of Serre, Jean-Pierre. G´eom´etrie alg´ebrique et g´eom´etrie analytique. Ann. Inst. Fourier, Grenoble 6 (1955–1956), 1–42, commonly referred to as GAGA. 7.42. The following is more general than Theorem 7.39: for every algebraic variety V , there exists a projective algebraic variety V 0 and a birational regular map ' from an open dense subset U of V 0 onto V whose graph is closed in V 0 V ; the subset U equals V 0 if and only if V is complete. Ibid. p. 12. A subset V of C n is analytic if every v 2 V admits an open neighbourhood U in C n such that V \ U is the zero set of a finite collection of holomorphic functions on U . An analytic subset is locally closed. Let V 0 be an open subset of an analytic set V . A function f W V 0 ! C is holomorphic if, for every v 2 V 0, there exists an open neighbourhood U of v in C n and a holomorphic function h on U such that f D h on V 0 \ U . The holomorphic functions on open subsets of V define on V the structure of a C-ringed space. DEFINITION 7.43. An analytic space is a C-ringed space .V; two conditions: O V / satisfying the following (a) there exists an open covering V D S Vi of V such that, for each i , the C-ringed space V jVi / is isomorphic to an analytic set equipped with its sheaf of holomorphic .Vi ; O functions; (b) the topological space V is Hausdorff. PROPOSITION 7.44. An algebraic variety V is complete if and only if V .C/ is compact in the complex topology. PROOF. The proof uses Chow’s lemma (ibid. Proposition 6, p. 12). There is a natural functor V V an from algebraic varieties over C to complex analytic spaces (ibid. 2). We omit the definition of a coherent sheaf of V -modules. O THEOREM 7.45. Let V be a projective variety over C. Then the functor equivalence from the category of coherent modules, under which locally free modules correspond. In particular, .V an; .V; V -modules to the category of coherent O O F F V /. 7! an is an V anO V an/ ' O PROOF. This summarizes the main results of GAGA (ibid. Th´eor´eme 2,3, p. 19, p. 20). THEOREM 7.46 (CHOW’S THEOREM). Every closed analytic subset of a projective variety is algebraic. PROOF. Let V be a projective space, and let Z be a closed analytic subset of V an. A theorem Zan is a coherent analytic sheaf on V an, and so there exists a of Henri Cartan states that coherent algebraic sheaf is Zariski closed, Zan. The support of F and equals Z (ibid. p. 29). O on V such that an D O F F THEOREM 7.47. Every compact analytic subset of an algebraic variety is algebraic. 170 7. COMPLETE VARIETIES PROOF. Let V be an algebraic variety, and let Z be a compact analytic subset. By Chow’s lemma (7.42), there exists a projective variety V 0, a dense open subset U of V 0, and a surjective regular map 'W U ! V whose graph is closed in V V 0. Let 0 D \.Z V 0/. As Z and V 0 are compact and is closed, 0 is compact, and so its projection V 00 on V 0 is also compact. On the other hand, V 00 D f 1.Z/, which shows that it is an analytic subset of U , and therefore also of V 0. According to Chow’s theorem, it is a Zariski closed subset of V 0 (hence an algebraic variety). Now Z D f .V 00/ is constructible (Zariski sense; see 9.7 below), and therefore its Zariski closure coincides with its closure for the complex topology, but (by assumption) it is closed. COROLLARY 7.48. Let V and W be algebraic varieties over C. If V is complete, then every holomorphic map f W V an ! W an is algebraic. PROOF. Apply the preceding theorem to the graph of f . EXAMPLE 7.49. The graph of z 7! ezW C ! C C is closed in C C but it is not Zariski closed. h. Nagata’s Embedding Theorem A necessary condition for a prevariety to be an open subvariety of a complete variety is that it be separated. An important theorem of Nagata says that this condition is also sufficient. THEOREM 7.50. Every variety V admits an open immersion V ,! W into a complete variety W . If V is affine, then one can embed V ,! A n. The proof in the general case is quite difficult. See: n ,! P in P n, and take W to be the closure of V Nagata, Masayoshi. Imbedding of an abstract variety in a complete variety. J. Math. Kyoto Univ. 2 1962 1–10; A generalization of the imbedding problem of an abstract variety in a complete variety. J. Math. Kyoto Univ. 3 1963 89–102. For a modern exposition, see: L¨utkebohmert, W. On compactification of schemes. Manuscripta Math. 80 (1993), no. 1, 95–111. In the 1970s, Deligne translated Nagata’s work into the language of schemes. His personal notes are available in three versions. Deligne, P., Le th´eor`eme de plongement de Nagata, Kyoto J. Math. 50, Number 4 (2010), 661-670. Conrad, B., Deligne’s notes on Nagata compactifications. J. Ramanujan Math. Soc. 22 (2007), no. 3, 205–257. Vojta, P., Nagata’s embedding theorem, 19pp., 2007, arXiv:0706.1907. See also: Temkin, Michael. Relative Riemann-Zariski spaces. Israel J. Math. 185 (2011), 1–42. A little history When he defined abstract algebraic varieties, Weil introduced the term “complete variety” to denote the algebraic geometer’s analogue of a compact manifold. h. Nagata’s Embedding Theorem 171 Exercises 7-1. Identify the set of homogeneous polynomials F .X; Y / D P aij X i Y j , 0 i; j m, with an affine space. Show that the subset of reducible polynomials is closed. 7-2. Let V and W be complete irreducible varieties, and let A be an abelian variety. Let P and Q be points of V and W . Show that any regular map hW V W ! A such that h.P; Q/ D 0 can be written h D f ı p C g ı q where f W V ! A and gW W ! A are regular maps carrying P and
Q to 0 and p and q are the projections V W ! V; W . CHAPTER 8 Normal Varieties; (Quasi-)finite maps; Zariski’s Main Theorem We begin by studying normal varieties. These varieties have some of the good properties of nonsingular varieties, and it is easy to show that every variety is birationally equivalent to a normal variety. After studying finite and quasi-finite maps, we discuss the celebrated Zariski’s Main Theorem (ZMT), which says that every quasi-finite map of algebraic varieties can be obtained from a finite map by removing a closed subset from the source variety. In its original form, the theorem says that a birational regular map to a normal algebraic variety fails to be a local isomorphism only at points where the fibre has dimension > 0. a. Normal varieties Recall (1.42) that an integrally closed domain is an integral domain that is integrally closed in its field of fractions. Moreover, that an integral domain A is normal if and only if Am is normal for every maximal ideal m in A (see 1.49). DEFINITION 8.1. A point P on an algebraic variety V is normal if closed domain. An algebraic variety is said to be normal if all of its points are normal. V;P is an integrally O Since the local ring at a point lying on two irreducible components can’t be an integral domain (see 3.14), a normal variety is a disjoint union of its irreducible components, which are therefore its connected components. PROPOSITION 8.2. The following conditions on an irreducible variety V are equivalent. (a) The variety V is normal. (b) For all open affine subsets U of V , the ring V .U / is an integrally closed domain. O (c) For all open subsets U of V , a rational function on V that satisfies a monic polynomial equation on U whose coefficients are regular on U is itself regular on U . PROOF. The equivalence of (a) and (b) follows from 1.49. (a) H) (c). Let U be an open subset of V , and let f 2 k.V / satisfy f n C a1f n1 C C an D 0; ai 2 V .U /; O (equality in k.V /). Then ai 2 This implies that f 2 O V .U / (5.11). V .U / O P for all P 2 U , and so f 2 O P for all P 2 U . O 173 174 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM (c) H) (b). The condition applied to an open affine subset U of V implies that integrally closed in k.V /. V .U / is O A regular local noetherian ring is normal — this is a difficult result that we don’t prove here (see CA 22.5 for references). Conversely, a normal local domain of dimension one is regular. Thus nonsingular varieties are normal, and normal curves are nonsingular. However, a normal surface need not be nonsingular: the cone X 2 C Y 2 Z2 D 0 is normal, but it is singular at the origin — the tangent space at the origin is k3. The singular locus of a normal variety V must have dimension dim V 2 (see 8.12 below). For example, a normal surface can only have isolated singularities — the singular locus can’t contain a curve. In particular, the surface Z3 D X 2Y (see 4.42) is not normal. The normalization of an algebraic variety Let E F be a finite extension of fields. The extension E=F is said to be normal if the minimal polynomial of every element of E splits in E. Let F al be an algebraic closure of F containing E. The composite in F al of the fields E, 2 Aut.E=F /, is normal over F (and is called the normal closure of F in F al). If E is normal over F , then E is Galois over EAut.E=F / (FT 3.10), and EAut.E=F / is purely inseparable over F (because HomF .EAut.E=F /; F al/ consists of a single element). PROPOSITION 8.3. Let A be a finitely generated k-algebra. Assume that A is an integral domain, and let E be a finite field extension of its field of fractions F . Then the integral closure A0 of A in E is a finite A-algebra (hence a finitely generated k-algebra). PROOF. According to the Noether normalization theorem (2.45), A contains a polynomial subalgebra A0 and is finite over A0. Now E is a finite extension of F .A0/ and A0 is the integral closure of A0 in E, and so we only need to consider the case that A is a polynomial ring kŒX1; : : : ; Xd . Let QE denote the normal closure of E in some algebraic closure of F containing E, and let QA denote the integral closure of A in QE. If QA is finitely generated as an A-module, then so is its submodule A0 (because A is noetherian). Therefore we only need to consider the case that E is normal over F . According to the above discussion, E E1 F with E Galois over E1 and E1 purely inseparable over F . Let A1 denote the integral closure of A in E1. Then A0 is a finite A1-algebra (1.51), and so it suffices to show that A1 is a finite A-algebra. Therefore we only need to consider the case that E is purely inseparable over F . In this case, k has characteristic p ¤ 0, and, for each x 2 E, there is a power q.x/ of p such that xq.x/ 2 F . As E is finitely generated over F , there is a single power q of p such that xq 2 F for all x 2 E. Let F al denote an algebraic closure of F containing E. For each i , there is a unique Yi 2 F al such that Y q i D Xi . Now F D k.X1; : : : ; Xd / E k.Y1; : : : ; Yd / and A D kŒX1; : : : ; Xd A0 kŒY1; : : : ; Yd because kŒY1; : : : ; Yd contains A and is integrally closed (1.32, 1.43). Obviously kŒY1; : : : ; Yd is a finite A-algebra, and this implies, as before, that A0 is a finite A-algebra. a. Normal varieties 175 COROLLARY 8.4. Let A be as in 8.3. If Am is normal for some maximal ideal m in A, then Ah is normal for some h 2 A X m. PROOF. Let A0 be the integral closure of A in its field of fractions. Then A0 D AŒf1; : : : ; fm 0 D Am, and so there exists an h 2 A X m such that, for some fi 2 A0. Now .A0/m for all i , hfi 2 A. Now A0 h D Ah, and so Ah is normal. 1.47D .Am/ The proposition shows that if A is an integral domain finitely generated over k, then the integral closure A0 of A in a finite extension E of F .A/ has the same properties. Therefore, Spm.A0/ is an irreducible algebraic variety, called the normalization of Spm.A/ in E. This construction extends without difficulty to nonaffine varieties. PROPOSITION 8.5. Let V be an irreducible algebraic variety, and let K be a finite field extension of k.V /. Then there exists an irreducible algebraic variety W with k.W / D K and a regular map 'W W ! V such that, for all open affines U in V , '1.U / is affine and kŒ'1.U / is the integral closure of kŒU in K. The map ' (or just W ) is called the normalization of V in K. PROOF. For each v 2 V , let W .v/ be the set of maximal ideals in the integral closure of in K. Let W D F For an open affine subset U of V , v O v2V W .v/, and let 'W W ! V be the map sending the points of W .v/ to v. '1.U / ' spm.kŒU 0/; where kŒU 0 is the integral closure of kŒU in K. We endow W with the k-ringed space structure for which .'1.U /; W j'1.U // ' Spm.kŒU 0/. O W / is an algebraic variety with the required properties. A routine argument shows that .W; O EXAMPLE 8.6. (a) The normalization of the cuspidal cubic V W Y 2 D X 3 in k.V / is the map A 1 ! V , t 7! .t 2; t 3/ (see 3.29). (b) The normalization of the nodal cubic 4.10) in k.V / is the map 1 ! V , t 7! .t 2 1; t 3 t/. A PROPOSITION 8.7. The normal points in an irreducible algebraic variety form a dense open subset. PROOF. Corollary 8.4 shows that the set of normal points is open, and it remains to show that it is nonempty. Let V be an irreducible algebraic variety. According to (3.37, 3.38), V is birationally equivalent to a hypersurface H in A d C1, d D dim V , H W a0X m C a1X m1 C C am; ai 2 kŒT1; : : : ; Td ; a0 ¤ 0; m 2 NI moreover, T1; : : : ; Td can be chosen to be a separating transcendence basis for k.V / over k. Therefore the discriminant D of the polynomial a0X m C C am is nonzero (it is an element of kŒT1; : : : ; Td ). Let A D kŒT1; : : : ; Td ; then kŒH D AŒX=.a0X m C C am/ D AŒx. Let y D c0 C C cm1xm1; ci 2 k.T1; : : : ; Td /; (35) be an element of k.H / integral over A. For each j 2 N, Trk.H /=F .A/.yxj / is a sum of conjugates of yxj , and hence is integral over A (cf. the proof of 1.44). As it lies in F .A/, it 176 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM is an element of A. On multiplying (35) with xj and taking traces, we get a system of linear equations c0 Tr.xj / C c1 Tr.x1Cj / C C cm1 Tr.xm1Cj / D Tr.yxj /; j D 0; : : : ; m 1: By Cramer’s rule (p. 26), det.Tr.xi Cj // cl 2 A; l D 0; : : : ; m 1: But det.Tr.xi Cj // D D,1 and so cl 2 AŒD1. Hence kŒH becomes normal once we invert the nonzero element D. We have shown that H contains a dense open normal subvariety, which implies that V does also. PROPOSITION 8.8. For every irreducible algebraic variety V , there exists a surjective regular map 'W V 0 ! V from a normal algebraic variety V 0 to V such that, for some dense open subset U of V; ' induces an isomorphism '1.U / ! U (in particular ' is birational). PROOF. Proposition 8.7 shows that the normalization of V in k.V / has this property. 8.9. More generally, for a dominant map 'W W ! V of irreducible algebraic varieties, there exists a normalization of V in W . For each open affine U in V we have kŒU .'1.U /; W / k.W /: O The integral closure kŒU 0 of .U; V / in .'1.U /; W / is a finite kŒU -algebra (because O it is a kŒU -submodule of the integral closure of kŒU in k.W /). The normalization of V in W is a regular map '0W V 0 ! V such that, for every open affine U in V , O .'01.U /; O V 0/ D Spm.kŒU 0/: In particular, '0 is an affine map. For example, if W and V are affine, then V 0 D Spm.kŒV 0/, where kŒV 0 is the integral closure of kŒV in kŒW . There is a commutative triangle W ' V 0 '0 j V: b. Regular functions on normal varieties DEFINITION 8.10. An algebraic variety V is factorial at a point P if domain. The variety V is factorial if it is factorial at all points P . P is a factorial O When V is factorial, it does not follow that V .U / is factorial for all open affines U in V . O A prime divisor Z on a variety V is a closed irreducible subvariety of codimension 1. Let Z be a prime divisor on V , and let P 2 V ; we say that
Z is locally principal at P if there exists an open affine neighbourhood U of P and an f 2 kŒU such that I.Z \ U / D .f /; the regular function f is then called a local equation for Z at P . If P … Z, then Z is locally principal at P because then we can choose U so that Z \ U D ;, and I.Z \ U / D .1/. 1See, for example, 2.34 of my notes Algebraic Number Theory. b. Regular functions on normal varieties 177 PROPOSITION 8.11. An irreducible variety V is factorial at a point P if and only if every prime divisor on V is locally principal at P . PROOF. Recall that an integral domain is factorial if and only if every prime ideal of height 1 is principal (1.24, 3.52). PROPOSITION 8.12. The codimension of the singular locus in a normal variety is at least 2. PROOF. Let V be a normal algebraic variety of dimension d , and suppose that its singular locus has an irreducible component W of codimension 1. After replacing V with an open subvariety, we may suppose that it is affine and that W is principal, say, W D .f / (see 8.11). There exists a nonsingular point P on W (4.37). Let .U; f1/; : : : ; .U; fd 1/ be germs of functions at P (on V ) whose restrictions to W generate the maximal ideal in W;P (cf. O V;P , and so P is 4.36). Then .U; f1/; : : : ; .U; fd 1/; .U; f / generate the maximal ideal in nonsingular on V . This contradicts the definition of W . O SUMMARY 8.13. For an algebraic variety V , nonsingular H) factorial H) normal H) singular locus has codimension 2. ˘ The variety X 2 C C X 2 1 ˘ The cone Z2 D XY in A ˘ The variety Spm.kŒX; XY; Y 2; Y 3/ is a surface in A 5 is factorial but singular. 3 is normal but not factorial (see 9.39 below). 4 with exactly one singular point, namely, the origin. Its singular locus has codimension 2, but the variety is not normal (the normalization kŒX; XY; Y 2; Y 3 is kŒX; Y ). ˘ Every singular curve has singular locus of codimension 1 (hence fails all conditions). ZEROS AND POLES OF RATIONAL FUNCTIONS ON NORMAL VARIETIES Let V be a normal irreducible variety. A divisor on V is an element of the free abelian group Div.V / generated by the prime divisors. Thus a divisor D can be written uniquely as a finite (formal) sum D D X ni Zi ; ni 2 Z; Zi a prime divisor on V: The support jDj of D is the union of the Zi corresponding to nonzero ni . A divisor is said to be effective (or positive) if ni 0 for all i . We get a partial ordering on the divisors by defining D D0 to mean D D0 0: O Because V is normal, there is associated with every prime divisor Z on V a discrete valuation ring Z. This can be defined, for example, by choosing an open affine subvariety U of V such that U \ Z ¤ ;; then U \ Z is a maximal proper closed subset of U , and so the ideal p corresponding to it is minimal among the nonzero ideals of R D .U; /I so Rp is an integrally closed domain with exactly one nonzero prime ideal pRp — it is therefore a Z. More intrinsically we can define discrete valuation ring (4.20), which is defined to be Z to be the set of rational functions on V that are defined an open subset U of V meeting O O O Z. Let ordZ be the valuation k.V / onto! Z with valuation ring element of Z, then O a D unit ordZ .a/: The divisor of a nonzero element f of k.V / is defined to be div.f / D X ordZ.f / Z: Z; thus, if is a prime O 178 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM The sum is over all the prime divisors of V , but in fact ordZ.f / D 0 for all but finitely many Z. In proving this, we can assume that V is affine (because it is a finite union of affines), say V D Spm.R/. Then k.V / is the field of fractions of R, and so we can write f D g= h with g; h 2 R, and div.f / D div.g/ div.h/. Therefore, we can assume f 2 R. The zero set of f , V .f / either is empty or is a finite union of prime divisors, V D S Zi (see 3.42) and ordZ.f / D 0 unless Z is one of the Zi . The map f 7! div.f /W k.V / ! Div.V / is a homomorphism. A divisor of the form div.f / is said to be principal, and two divisors are said to be linearly equivalent, denoted D D0, if they differ by a principal divisor. When V is nonsingular, the Picard group Pic.V / of V is defined to be the group of divisors on V modulo principal divisors. (The definition of the Picard group of a general algebraic variety agrees with this definition only for nonsingular varieties; it may differ for normal varieties.) THEOREM 8.14. Let V be a normal variety, and let f be rational function on V . If f has no zeros or poles on an open subset U of V , then f is regular on U . PROOF. We may assume that V is connected, hence irreducible. Now apply the following statement (proof omitted): a noetherian domain is normal if and only if Ap is a discrete valuation ring for all prime ideals p of height 1 and A D T ht.p/D1 Ap. COROLLARY 8.15. A rational function on a normal variety, regular outside a subset of codimension 2, is regular everywhere. PROOF. This is a restatement of the theorem. COROLLARY 8.16. Let V and W be affine varieties with V normal, and let 'W V X Z ! W be a regular map defined on the complement of a closed subset Z of V . If codim.Z/ 2, then ' extends to a regular map on the whole of V . PROOF. We may suppose that W is affine, and embed it as a closed subvariety of A map V X Z ! W ,! A V . Therefore V X Z ! A n. The n is given by n regular functions on V X Z, each of which extends to n, and its image is contained in W . n extends to A c. Finite and quasi-finite maps Finite maps DEFINITION 8.17. A regular map 'W W ! V of algebraic varieties is finite if there exists a finite covering V D S i Ui of V by open affines such that, for each i , the set '1.Ui / is affine and kŒ'1.Ui / is a finite kŒUi -algebra. EXAMPLE 8.18. Let V be an irreducible algebraic variety, and let 'W W ! V be the normalization of V in a finite extension of k.V /. Then ' is finite. This follows from the definition 8.5 and Proposition 8.3. The next lemma shows that, for maps of affine algebraic varieties, the above definition agrees with Definition 2.39. c. Finite and quasi-finite maps 179 LEMMA 8.19. A regular map 'W W ! V of affine algebraic varieties is finite if and only if kŒW is a finite kŒV -algebra. PROOF. The necessity being obvious, we prove the sufficiency. For simplicity, we shall assume in the proof that V and W are irreducible. Let .Ui /i be a finite family of open affines covering V and such that, for each i , the set '1.Ui / is affine and kŒ'1.Ui / is a finite kŒUi -algebra. Each Ui is a finite union of basic open subsets of V . These are also basic open subsets of Ui , because D.f / \ Ui D D.f jUi /, and so we may assume that the original Ui are basic open subsets of V , say, Ui D D.fi / with fi 2 A. Let A D kŒV and B D kŒW . We are given that .f1; : : : ; fn/ D A and that Bfi is a finite Afi -algebra for each i . We have to show that B is a finite A-algebra. Let fbi1; : : : ; bimi g generate Bfi as an Afi -module. After multiplying through by a power of fi , we may assume that the bij lie in B. We shall show that the family of all bij generate B as an A-module. Let b 2 B. Then b=1 2 Bfi , and so b D ai1 f ri i n / D A because any maximal ideal containing .f r1 1 ; : : : ; f rn bimi , some aij 2 A and ri 2 N: bi1 C C aimi f ri i The ideal .f r1 have to contain .f1; : : : ; fn/ D A. Therefore, 1 ; : : : ; f rn n / would 1 D h1f r1 1 C C hnf rn n , some hi 2 A: Now b D b 1 D h1 bf r1 1 C C hn bf rn n D h1.a11b11 C C a1m1b1m1/ C C hn.an1bn1 C C anmnbnmn/, as required. LEMMA 8.20. Let 'W W ! V be a regular map with V affine, and let U be an open affine in V . There is a canonical isomorphism of k-algebras PROOF. Let U 0 D '1.U /. The map is defined by the kŒV -bilinear pairing .W; W / ˝kŒV kŒU ! .'1.U /; O W /: O .f; g/ 7! .f jU 0; g ı 'jU 0/W .W; W / kŒU ! .U 0; O W /: O When W is also affine, it is an isomorphism (see 5.31, 5.32). Let W D S Wi be a finite open affine covering of W , and consider the commutative diagram: 0 0 .W; O W / ˝kŒV kŒU Q .Wi ; i O W / ˝kŒV kŒU Q .Wij ; i;j W / ˝kŒV kŒU O .U 0; W / O Q .U 0 \ Wi ; i W / O .U \ Wij ; W /. O O Here Wij D Wi \ Wj . The bottom row is exact because W is a sheaf, and the top row is W is a sheaf and kŒU is flat over kŒV .2 The varieties Wi and Wi \ Wj are exact because all affine, and so the two vertical arrows at right are products of isomorphisms. This implies that the first is also an isomorphism. 2A sequence 0 ! M 0 ! M ! M 00 is exact if and only if 0 ! Am ˝A M 0 ! Am ˝A M ! Am ˝A M 00 is ' OU;P ' OV;P ' kŒV mP exact for all maximal ideals m of A (1.21). This implies the claim because kŒU mP for all P 2 U . O 180 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM PROPOSITION 8.21. Let 'W W ! V be a regular map of algebraic varieties. If ' is finite, then, for every open affine U in V , '1.U / is affine and kŒ'1.U / is a finite kŒU -algebra. PROOF. Let Vi be an open affine covering of V (which we may suppose to be finite) such defD '1.Vi / is an affine subvariety of W for all i and kŒWi is a finite kŒVi -algebra. that Wi Let U be an open affine in V , and let U 0 D '1.U /. Then .U 0; W / is a subalgebra of Q W /, and so it is an affine k-algebra finite over kŒU .3 We have a morphism O i .U 0 \ Wi ; O of varieties over V U 0 canonical Spm. .U 0; W // O (36) V which we shall show to be an isomorphism. We know that each of the maps U 0 \ Wi ! Spm. .U 0 \ Wi ; W // O is an isomorphism. But Spm. .U 0 \Wi ; Therefore the canonical morphism is an isomorphism over each Vi , and so it is an isomorphism. W // is the inverse image of Vi in Spm. .U 0; O O W //. SUMMARY 8.22. Let 'W W ! V be a regular map, and consider the following condition on an open affine subset U of V : () '1.U / is affine and kŒ'1.U / is a finite over kŒU . The map ' is finite if () holds for the open affines in some covering of V , in which case (*) holds for all open affines of V . PROPOSITION 8.23. (a) Closed immersions are finite. (b) The composite of two finite morphisms is finite. (c) The product of two finite
morphisms is finite. PROOF. (a) Let Z be a closed subvariety of a variety V , and let U be an open affine subvariety of V . Then Z \ U is a closed subvariety of U . It is therefore affine, and the map Z \ U ! U corresponds to a map A ! A=a of rings, which is obviously finite. This proves (a). As to be finite is a local condition, it suffices to prove (a) and (b) for maps of affine varieties. Then the statements become statements in commutative algebra. (b) If B is a finite A-algebra and C is a finite B-algebra, then C is a finite A-algebra. To see this, note that if fbi g is a set of generators for B as an A-module, and fcj g is a set of generators for C as a B-module, then fbi cj g is a set of generators for C as an A-module. (c) If B and B 0 are respectively finite A and A0-algebras, then B ˝k B 0 is a finite A ˝k A0-algebra. To see this, note that if fbi g is a set of generators for B as an A-module, and fb0 g is a set of generators j for B ˝A B 0 as an A ˝ A0-module. g is a set of generators for B 0 as an A0-module, then fbi ˝ b0 j 3Recall that a module over a noetherian ring is noetherian if and only if it is finitely generated, and that a submodule of a noetherian module is noetherian. Therefore, a submodule of a finitely generated module over a noetherian ring is finitely generated. c. Finite and quasi-finite maps 181 1 X f0g ,! A By way of contrast, open immersions are rarely finite. For example, the inclusion 1 is not finite because the ring kŒT; T 1 is not finitely generated as a kŒT A module (any finitely generated kŒT -submodule of kŒT; T 1 is contained in T nkŒT for some n). THEOREM 8.24. Finite maps of algebraic varieties are closed. PROOF. It suffices to prove this for affine varieties. Let 'W W ! V be a finite map of affine varieties, and let Z be a closed subset of W . The restriction of ' to Z is finite (by 8.23a and b), and so we can replace W with Z; we then have to show that Im.'/ is closed. The map corresponds to a finite map of rings A ! B. This will factors as A ! A=a ,! B, from which we obtain maps Spm.B/ ! Spm.A=a/ ,! Spm.A/: The second map identifies Spm.A=a/ with the closed subvariety V .a/ of Spm.A/, and so it remains to show that the first map is surjective. This is a consequence of the going-up theorem (1.53). The base change of a finite map Recall that the base change of a regular map 'W V ! S is the map '0 in the diagram: V S W '0 W 0 V ' S: PROPOSITION 8.25. The base change of a finite map is finite. PROOF. We may assume that all the varieties concerned are affine. Then the statement becomes: if A is a finite R-algebra, then A ˝R B=N is a finite B-algebra, which is obvious. PROPOSITION 8.26. Finite maps of algebraic varieties are proper. PROOF. The base change of a finite map is finite, and hence closed. COROLLARY 8.27. Let 'W V ! S be finite; if S is complete, then so also is V . PROOF. Combine 7.19 and 8.26. Quasi-finite maps Recall that the fibres of a regular map 'W W ! V are the closed subvarieties '1.P / of W for P 2 V . As for affine varieties (2.39), we say that a regular map of algebraic varieties is quasi-finite if all of its fibres are finite. PROPOSITION 8.28. A finite map 'W W ! V is quasi-finite. PROOF. Let P 2 V ; we wish to show '1.P / is finite. After replacing V with an affine neighbourhood of P , we can suppose that it is affine, and then W will be affine also. The map ' then corresponds to a map ˛W A ! B of affine k-algebras, and a point Q of W maps to P if and only ˛1.mQ/ D mP . But this holds if and only if mQ ˛.mP /, and so the points of W mapping to P are in one-to-one correspondence with the maximal ideals of 182 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM B=˛.mP /B. Clearly B=˛.mP /B is generated as a k-vector space by the image of any generating set for B as an A-module, and so it is a finite k-algebra. The next lemma shows that it has only finitely many maximal ideals. LEMMA 8.29. A finite k-algebra A has only finitely many maximal ideals. PROOF. Let m1; : : : ; mn be maximal ideals in A. They are obviously coprime in pairs, and so the Chinese Remainder Theorem (1.1) shows that the map A ! A=m1 A=mn; a 7! .: : : ; ai mod mi ; : : :/; is surjective. It follows that dimk A X dimk.A=mi / n — here dimk means dimension as a k-vector space. Finite and quasi-finite maps of prevarieties are defined as for varieties. Examples 8.30. The projection from the curve XY D 1 onto the X axis (see p. 71) is quasi-finite but not finite — its image is not closed in A 1, and kŒX; X 1 is not finite over kŒX. 8.31. The map t 7! .t 2; t 3/W A 1 ! V .Y 2 X 3/ A 2 from the line to the cuspidal cubic is finite because the image of kŒX; Y in kŒT is kŒT 2; T 3, and f1; T g is a set of generators for kŒT as a kŒT 2; T 3-module (see 3.29). 8.32. The map A 1 ! A 1, a 7! am is finite. 8.33. The obvious map .A 1 with the origin doubled / ! A 1 is quasi-finite but not finite (the inverse image of A 1 is not affine). 8.34. The map A of A 2 is not affine (see 3.33). The map 2 X foriging ,! A 2 is quasi-finite but not finite, because the inverse image 2 X f.0; 0/g t fOg ! A 2 A sending O to .0; 0/ is bijective but not finite (here fOg D Spm.k/ D A 0). 8.35. The map in 8.31, and the Frobenius map .t1; : : : ; tn/ 7! .W A n ! A n in characteristic p ¤ 0; are examples of finite bijective regular maps that are not isomorphisms. c. Finite and quasi-finite maps 183 8.36. Let V D A 2 D Spm.kŒX; Y / and let f be the map defined on the ring level by X 7! X D A Y 7! XY 2 C Y C 1 D B: Then f is (obviously) quasi-finite, but it is not finite. For this we have to show that kŒX; Y is not integral over its subring kŒA; B. The minimal polynomial of Y over kŒA; B is AY 2 C Y C 1 B D 0; which shows that it is not integral over kŒA; B (see 1.44). Alternatively, one can show directly that Y can never satisfy an equation Y s C g1.A; B/Y s1 C C gs.A; B/ D 0; gi .A; B/ 2 kŒA; B; by multiplying the equation by A. 8.37. Let V be the hyperplane X n C T1X n1 C C Tn D 0 in A nC1, and consider the projection map .a1; : : : ; an; x/ 7! .a1; : : : ; an/W V ! A n: The fibre over a point .a1; : : : ; an/ 2 A n is the set of solutions of X n C a1X n1 C C an D 0; and so it has exactly n points, counted with multiplicities. The map is certainly quasi-finite; it is also finite because it corresponds to the finite map of k-algebras, kŒT1; : : : ; Tn ! kŒT1; : : : ; Tn; X=.X n C T1X n1 C C Tn/: See also the more general example p. 51. 8.38. Let V be the hyperplane T0X n C T1X n1 C C Tn D 0 in A nC2. The projection map .a0; : : : ; an; x/ 7! .a0; : : : ; an/W V '! A nC1 has finite fibres except for the fibre above o D .0; : : : ; 0/, which is A 1. Its restriction to V X '1.o/ is quasi-finite, but not finite. Above points of the form .0; : : : ; 0; ; : : : ; / some of the roots “vanish off to 1”. (Example 8.30 is a special case of this.) See also the more general example p. 51. 8.39. Let P .X; Y / D T0X n C T1X n1Y C C TnY n; and let V be its zero set in P A nC1 X fog is finite. 1 .A nC1 X fog/. In this case, the projection map V ! 184 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM d. The fibres of finite maps Let 'W W ! V be a finite dominant morphism of irreducible varieties. Then dim.W / D dim.V /, and so k.W / is a finite field extension of k.V /. Its degree is called the degree of the map '. The map ' is said to be separable if the field k.W / is separable over k.V /. Recall that jS j denotes the number of elements in a finite set S . THEOREM 8.40. Let 'W W ! V be a finite surjective regular map of irreducible varieties, and assume that V is normal. ˇ'1.P /ˇ ˇ'1.P /ˇ ˇ D deg.'/ is an open subset of V , and it (a) For all P 2 V , ˇ ˇ deg.'/. (b) The set of points P of V such that ˇ is nonempty if ' is separable. Before proving the theorem, we give examples to show that we need W to be separated and V to be normal in (a), and that we need k.W / to be separable over k.V / for the second part of (b). EXAMPLE 8.41. (a) The map fA 1 with origin doubled g ! A 1 has degree one and is one-to-one except over the origin where it is two-to-one. (b) Let C be the curve Y 2 D X 3 C X 2, and consider the map t 7! .t 2 1; t.t 2 1//W A 1 ! C . It is one-to-one except that the points t D ˙1 both map to 0. On coordinate rings, it corresponds to the inclusion kŒx; y ,! kŒT , x 7! T 2 1 y 7! T .T 2 1/ , and so is of degree one. The ring kŒx; y is not integrally closed — in fact kŒT is the integral closure of kŒx; y in its field of fractions k.x; y/ D k.T /. (c) The Frobenius map .a1; : : : ; an/ 7! .ap 1 ; : : : ; ap n /W A n ! A n in characteristic p ¤ 0 is bijective on points, but has degree pn. The field extension corresponding to the map is k.X1; : : : ; Xn/ k. / which is purely inseparable. LEMMA 8.42. Let Q1; : : : ; Qr be distinct points on an affine variety V . Then there is a regular function f on V taking distinct values at the Qi . PROOF. We can embed V as closed subvariety of A statement with V D A n — almost any linear form will do. n, and then it suffices to prove the d. The fibres of finite maps 185 PROOF (OF 8.40). In proving (a) of the theorem, we may assume that V and W are affine, and so the map corresponds to a finite map of k-algebras, kŒV ! kŒW . Let '1.P / D fQ1; : : : ; Qr g. According to the lemma, there exists an f 2 kŒW taking distinct values at the Qi . Let F .T / D T m C a1T m1 C C am be the minimal polynomial of f over k.V /. It has degree m Œk.W / W k.V / D deg ', and it has coefficients in kŒV because V is normal (see 1.44). Now F .f / D 0 implies F .f .Qi // D 0, i.e., f .Qi /m C a1.P / f .Qi /m1 C C am.P / D 0: Therefore the f .Qi / are all roots of a single polynomial of degree m, and so r m deg.'/. In order to prove the first part of (b), we show that, if there is a point P 2 V such that '1.P / has deg.'/ elements, then the same is true for all points in an open neighbourhood of P . Choose f as in the last paragraph corresponding to such a P . Then the polynomial T m
C a1.P / T m1 C C am.P / D 0 () has r D deg ' distinct roots, and so m D r. Consider the discriminant disc F of F . Because () has distinct roots, disc.F /.P / ¤ 0, and so disc.F / is nonzero on an open neighbourhood U of P . The factorization kŒV ! kŒV ŒT =.F / T 7!f! kŒW gives a factorization W ! Spm.kŒV ŒT =.F // ! V: Each point P 0 2 U has exactly m inverse images under the second map, and the first map is finite and dominant, and therefore surjective (recall that a finite map is closed). This proves that '1.P 0/ has at least deg.'/ points for P 0 2 U , and part (a) of the theorem then implies that it has exactly deg.'/ points. We now show that if the field extension is separable, then there exists a point such that '1.P / has deg ' elements. Because k.W / is separable over k.V /, there exists an f 2 kŒW such that k.V /Œf D k.W /. Its minimal polynomial F has degree deg.'/ and its discriminant is a nonzero element of kŒV . The diagram W ! Spm.kŒV ŒT =.F // ! V shows that j'1.P /j deg.'/ for P a point such that disc.f /.P / ¤ 0. Let E F be a finite extension of fields. The elements of E separable over F form a subfield F sep of E, and the separable degree of E over F is defined to be the degree of F sep over F . The separable degree of a finite surjective map 'W W ! V of irreducible varieties is the separable degree of k.W / over k.V /. THEOREM 8.43. Let 'W W ! V be a finite surjective regular map of irreducible varieties, and assume that V is normal. ˇ'1.P /ˇ ˇ sepdeg.'/, with equality holding on a dense open subset. (a) For all P 2 V , ˇ (b) For all i , is closed in V . Vi D fP 2 V j ˇ ˇ'1.P /ˇ ˇ i g 186 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM PROOF. If ' is separable, this was proved in 8.40. If ' is purely inseparable, then ' is one-to-one because, for some q, the Frobenius map V .q1/ F! V factors through '. To prove the general case, factor ' as the composite of a purely inseparable map with a separable map. ASIDE 8.44. A finite map from a variety onto a normal variety is open (hence both open and closed). For an elementary proof, see Theorem 63.12 of Musili, C., Algebraic geometry for beginners. Texts and Readings in Mathematics, 20. Hindustan Book Agency, New Delhi, 2001. e. Zariski’s main theorem In this section, we explain a fundamental theorem of Zariski. Statement and proof One obvious way to construct a nonfinite quasi-finite map is to take a finite map W ! V and remove a closed subset of W . Zariski’s Main Theorem (ZMT) shows that, for algebraic varieties, every quasi-finite map arises in this way. THEOREM 8.45 (ZARISKI’S MAIN THEOREM). Every quasi-finite map of algebraic varieties 'W W ! V factors into W ! V with '0 finite and j an open immersion: ,! V 0 '0 j W open immersion V 0 quasi-finite finite V: When ' is a dominant map of irreducible varieties, the statement is true with '0W V 0 ! V equal to the normalization of V in W (in the sense of 8.9). The key result needed to prove 8.45 is the following statement from commutative algebra. For a ring A and a prime ideal p in A, .p/ denotes the field of fractions of A=p. THEOREM 8.46 (LOCAL VERSION OF ZMT). Let A be a commutative ring, and let i W A ! B be a finitely generated A-algebra. Let q be a prime ideal of B, and let p D i 1.q/. Finally, let A0 denote the integral closure of A in B. If Bq=pBq is a finite .p/-algebra, then there exists an f 2 A0 not in q such that the map A0 f ! Bf is an isomorphism. PROOF. The proof is quite elementary, but intricate — see 17 of my notes CA. Recall that a point v in a topological space V is isolated if fvg is an open subset of V . The isolated points v of an algebraic variety V are those such that fvg is both open and closed. Thus they are the irreducible components of V of dimension 0. Let 'W W ! V be a continuous map of topological spaces. We say that w 2 W is isolated in its fibre if it is isolated in the subspace '1.'.w// of W . Let 'W A ! B be a homomorphism of finitely generated k-algebras, and consider spm.'/W spm.B/ ! spm.A/; then n 2 spm.B/ is isolated in its fibre if and only if Bn=mBn is a finite k-algebra; here m D '1.n/. PROPOSITION 8.47. Let 'W W ! V be a regular map of algebraic varieties. The set W 0 of points of W isolated in their fibres is open in W . e. Zariski’s main theorem 187 PROOF. Let w 2 W 0. Let Ww and Vv be open affine neighbourhoods of w and v D '.w/ such that '.Ww / Vv, and let A D kŒVv and B D kŒWw . Let n D ff 2 B j f .w/ D 0g — it is the maximal ideal in B corresponding to w. Let A0 be the integral closure of A in B. Theorem 8.46 shows that there exists an f 2 A0 ' Bf . Write A0 as the union of the finitely generated A-subalgebras not in m such that A0 f Ai of A0 containing f : A0 D [ i Ai : Because A0 is integral over A, each Ai is finite over A (see 1.35). We have Bf ' A0 f D [ Aif : i Because Bf is a finitely generated A-algebra, Bf D Aif for all sufficiently large Ai . As the Ai are finite over A, Bf is quasi-finite over A, and spm.Bf / is an open neighbourhood of w consisting of quasi-finite points. PROPOSITION 8.48. Every quasi-finite map of affine algebraic varieties 'W W ! V factors into W j! V 0 '0 ! V with j a dominant open immersion and '0 finite. PROOF. Let A D kŒV and B D kŒW . Because ' is quasi-finite, Theorem 8.46 shows that there exist fi 2 A0 such that the sets spm.Bfi / form an open covering of W and A0 ' Bfi for all i. As W quasicompact, finitely many sets spm.Bfi / suffice to cover fi W . The argument in the proof of (8.47) shows that there exists an A-subalgebra A00 of A0, finite over A, which contains f1; : : : ; fn and is such that Bfi for all i . Now the map W D Spm.B/ ! Spm.A00/ is an open immersion because it is when restricted to Spm.Bfi / for each i . As Spm.A00/ ! Spm.A/ D V is finite, we can take V 0 D Spm.A00/. ' A00 fi Recall (Exercise 8-3) that a regular map 'W W ! V is affine if '1.U / is affine whenever U is an open affine subset of V . PROPOSITION 8.49. Let 'W W ! V be an affine map of irreducible algebraic varieties. Then the map j W W ! V 0 from W into the normalization V 0 of V in W (8.9) is an open immersion. PROOF. Let U be an open affine in V . Let A D kŒU and B D kŒ'1.U /. In this case, the normalization A0 of A in B is finite over A (because it is contained in the normalization of A in k.W /, which is finite over A (8.3)). Thus, in the proof of 8.48 we can take A00 D A0, and then '1.U / ! Spm.A0/ is an open immersion. As Spm.A0/ is an open subvariety of V 0 and the sets '1.U / cover W , this implies that j W W ! V 0 is an open immersion. As V 0 ! V is finite, this proves Theorem 8.45 in the case that ' is an affine map of irreducible varieties. To deduce the general case of Theorem 8.45 from 8.44 requires an additional argument. See Theorem 12.83 of G¨ortz, U. and Wedhorn, T., Algebraic Geometry I., Springer Spektrum, Wiesbaden, 2020. NOTES 8.50. Let 'W W ! V be a quasi-finite map of algebraic varieties. In 8.45, we may replace V 0 with the closure of the image of j . Thus, there is a factorization ' D '0 ı j with '0 finite and j a dominant open immersion. 188 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM 8.51. Theorem 8.45 is false for prevarieties (see 8.33). However, it is true for separated maps of prevarieties. (A regular map 'W V ! S of algebraic prevarieties is separated if the image V =S of the map v 7! .v; v/W V ! V S V is closed; the map ' is separated if V is separated.) 8.52. Assume that V is normal in 8.45. Then '0 is open (8.44), and so ' is open. Thus, every quasi-finite map from an algebraic variety to a normal algebraic variety is open. Applications to finite maps Zariski’s main theorem allows us to give a geometric criteria for a regular map to be finite. PROPOSITION 8.53. Every quasi-finite regular map 'W W ! V of algebraic varieties with W complete is finite. PROOF. The map j W W ,! V 0 in 8.45 is an isomorphism of W onto its image j.W / in V 0. If W is complete, then j.W / is closed (7.7), and so the restriction of '0 to j.W / is finite. PROPOSITION 8.54. Every proper quasi-finite map 'W W ! V of algebraic varieties is finite. PROOF. Factor ' into W j ,! W 0 ˛! V with ˛ finite and j an open immersion. Factor j into W w7!.w;jw/ ! W V W 0 .w;w 0/7!w 0 The image of the first map is j , which is closed because W 0 is a variety (see 5.28; W 0 is separated because it is finite over a variety — exercise). Because ' is proper, the second map is closed. Hence j is an open immersion with closed image. It follows that its image is a connected component of W 0, and that W is isomorphic to that connected component. ! W 0: NOTES 8.55. When W and V are curves, every surjective map W ! V is closed. Thus it is easy to give examples of closed surjective quasi-finite, but nonfinite, maps. Consider, for example, the map A 1 X f0g t A 0 ! A 1; sending each a 2 A because the map is only closed, not universally closed. 1 X f0g to a and O 2 A 0 to 0. This doesn’t violate the Proposition 8.54, Applications to birational maps Recall (p. 116) that a regular map 'W W ! V of irreducible varieties is said to be birational if it induces an isomorphism k.V / ! k.W / on the fields of rational functions. 8.56. One may ask how a birational regular map 'W W ! V can fail to be an isomorphism. Here are three examples. (a) The inclusion of an open subset into a variety is birational. e. Zariski’s main theorem 189 (b) The map (8.31) from A 1 to the cuspidal cubic, 1 ! C; A t 7! .t 2; t 3/; is birational. Here C is the cubic Y 2 D X 3, and the map kŒC ! kŒA 1 D kŒT identifies kŒC with the subring kŒT 2; T 3 of kŒT . Both rings have k.T / as their fields of fractions. (c) For any smooth variety V and point P 2 V , there is a regular birational map 'W V 0 ! V such that the restriction of ' to V 0 X '1.P / is an isomorphism onto V X P , but '1.P / is the projective space attached to the vector space TP .V /. See the section on blow-ups below. The next result says that, if we require the target variety to b
e normal (thereby excluding example (b)), and we require the map to be quasi-finite (thereby excluding example (c)), then we are left with (a). PROPOSITION 8.57. Let 'W W ! V be a birational regular map of irreducible varieties. If V is normal and the map ' is quasi-finite, then ' is an isomorphism from W onto an open subvariety of V . PROOF. Factor ' as in the Theorem 8.45 (so, in particular, '0W V 0 ! V is the normalization of V in W ). For each open affine subset U of V , kŒ'01.U / is the integral closure of kŒU in k.W /. Because ' is birational, the inclusion k.V / k.V 0/ D k.W / is an equality. Now kŒU is integrally closed in k.V / (because V is normal), and so U D '01.U / (as varieties). We have shown that '0W V 0 ! V is an isomorphism locally on the base V , and hence an isomorphism. 8.58. In topology, a continuous bijective map 'W W ! V need not be a homeomorphism, but it is if W is compact and V is Hausdorff. Similarly, a bijective regular map of algebraic varieties need not be an isomorphism. Here are three examples: (a) In characteristic p, the Frobenius map .x1; : : : ; xn/ 7! .xp 1 ; : : : ; xp n /W A n ! A n is bijective and regular, but it is not an isomorphism even though A n is normal. (b) The map t 7! .t 2; t 3/ from A 1 to the cuspidal cubic (see 8.56b) is bijective, but not an isomorphism. (c) Consider the regular map A 1 sending x to 1=x for x ¤ 0 and 0 to 0. Its graph is the union of .0; 0/ and the hyperbola xy D 1, which is a closed subvariety of 1 is a bijective, regular, birational map, A but it is not an isomorphism even though A 1. The projection .x; y/ 7! xW ! A 1 is normal. 1 ! A 1 A If we require the map to be birational (thereby excluding example (a)), V to be normal (thereby excluding example (b)), and the varieties to be irreducible (thereby excluding example (c)), then the map is an isomorphism. PROPOSITION 8.59. Let 'W W ! V be a bijective regular map of irreducible algebraic varieties. If the map ' is birational and V is normal, then ' is an isomorphism. PROOF. The hypotheses imply that ' is an isomorphism of W onto an open subset of V (8.57). Because ' is bijective, the open subset must be the whole of V . 190 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM In fact, example (a) can be excluded by requiring that ' be generically separable (instead of birational). PROPOSITION 8.60. Let 'W W ! V be a bijective regular map of irreducible varieties. If V is normal and k.W / is separably generated over k.V /, then ' is an isomorphism. PROOF. Because ' is bijective, dim.W / D dim.V / (see Theorem 9.9 below) and the separable degree of k.W / over k.V / is 1 (apply 8.40 to the variety V 0 in 8.45). Hence ' is birational, and we may apply 8.59. 8.61. In functional analysis, the closed graph theorem states that, if a linear map 'W W ! V between two Banach spaces has a closed graph defD f.w; 'w/ j w 2 W g, then ' is continuous (q.v. Wikipedia). One can ask (cf. mo113858) whether a similar statement is true in algebraic geometry. Specifically, if 'W W ! V is a map (in the set-theoretic sense) of algebraic varieties V; W whose graph is closed (for the Zariski topology), then is ' a regular map? The answer is no in general. For example, even in characteristic zero, the map .t 2; t 3/ ! tW C ! A 1 inverse to that in 8.56(b) has closed graph but is not regular. In characteristic p, the inverse of the Frobenius map x 7! xp provides another counterexample. For a third counterexample, see 8.58(c). The projection from to W is a bijective regular map, and so ' will be regular if is an isomorphism. According to 8.60, is an isomorphism if the varieties are irreducible, W is normal, and is generically separable. In particular, a map between irreducible normal algebraic varieties in characteristic zero is regular if its graph is closed. A condition for an algebraic monoid to be a group A monoid variety is an algebraic variety G together with the structure of a monoid defined by regular maps mW G G ! G; eW A 0 ! G: LEMMA 8.62. Let .G; m; e/ be an algebraic monoid. The map TeG ˚ TeG ' T.e;e/.G G/ .d m/.e;e/ ! Te.G/ is addition. PROOF. The first isomorphism is .X; Y / 7! .d˛/e.X/ C .dˇ/e.Y /, where ˛ is the map x 7! .x; e/W G ! G G and ˇ is x 7! .e; x/. To compute .d m/.e;e/..dˇ/e.X/C.d˛/e.Y //; note that m ı ˛ D idG D m ı ˇ. PROPOSITION 8.63. Let .G; m; e/ be an algebraic monoid over k. If .G.k/; m.k// is a group with identity element e, then .G; m/ is an algebraic group, that is, the map a 7! a1 is regular. PROOF. Let a 2 G.k/. The translation map LaW x 7! ax is an isomorphism G ! G because it has an inverse La1. Therefore G is homogeneous as an algebraic variety: for any two points in jGj, there is an isomorphism G ! G mapping one to the other. It follows that G is nonsingular, in particular, normal. The map .x; y/ 7! .x; xy/W G G ! G G f. Stein factorization 191 is regular, a bijection on k-points, and induces an isomorphism on the tangent spaces at .e; e/ (apply the lemma). It is therefore an isomorphism of algebraic varieties over k. Therefore, its inverse .x; y/ 7! .x; x1y/ is regular, and so .x; y/ 7! x1yW G G ! G is regular. This implies that .G; m/ is an algebraic group. Note that it is necessary in the proposition that G be reduced: consider G D Spec kŒT =.T n/, n > 1, with the trivial monoid structure G G ! e ! G. Variants of Zariski’s main theorem Mumford, 1966,4 III, 9, lists the following variants of ZMT. Original form (8.57) Let 'W W ! V be a birational regular map of irreducible varieties. If V is normal and ' is quasi-finite, then ' is an isomorphism of W onto an open subvariety of V . Topological form Let V be a normal variety over C, and let v 2 V . Let S be the singular locus of V . Then the complex neighbourhoods U of v such that U X U \ S is connected form a base for the system of complex neighbourhoods of v. Power series form Let V be a normal variety, and let an irreducible closed subset of V (cf. p. 177). If domain, then so also is its completion. O V;Z be the local ring attached to O V;Z is an integrally closed integral Grothendieck’s form (8.45) Every quasi-finite map of algebraic varieties factors as the composite of an open immersion with a finite map. Connectedness theorem Let 'W W ! V be a proper birational map, and let v be a (closed) normal point of V . The '1.v/ is a connected set (in the Zariski topology). The original form of the theorem was proved by Zariski using a fairly direct argument whose method doesn’t seem to generalize.5 The power series form was also proved by Zariski, who showed that it implied the original form. The last two forms are much deeper and were proved by Grothendieck. See the discussion in Mumford 1966. NOTES. The original form of the theorem (8.57) is the “Main theorem” of Zariski, O., Foundations of a general theory of birational correspondences. Trans. Amer. Math. Soc. 53, (1943). 490–542. f. Stein factorization The following important theorem shows that the fibres of a proper map are disconnected only because the fibres of finite maps are disconnected. THEOREM 8.64 (STEIN FACTORIZATION). Every proper map 'W W ! V of algebraic varieties factors into W '1! W 0 '2! V with '1 proper with connected fibres and '2 finite. 4Introduction to Algebraic Geometry, Harvard notes. Reprinted as “The Red Book of Varieties and Schemes” (with the introduction of misprints) by Springer 1999. 5See Lang, S., Introduction to Algebraic Geometry, 1958, V 2, for Zariski’s original statement and proof of this theorem. See Springer, T.A., Linear Algebraic Groups, 1998, 5.2.8, for a direct proof of (8.59). 192 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM When V is affine, this is the factorization W ! Spm. W .W // ! V: O The first major step in the proof of the theorem is to show that ' on V . Here ' O W is the sheaf of V -algebras on V , O U W .'1.U //: O W is a coherent sheaf O O To say that ' of regular map '2W Spm.' O attached by Spm to the map of k-algebras kŒU ! W is coherent means that, on every open affine subset U of V , it is the sheaf U -algebras defined by a finite kŒU -algebra. This, in turn, means that there exists a W / ! V that, over every open affine subset U of V , is the map W .'1.U //: O The Stein factorization is then O '1! W 0 defD Spm.' W '2! V: W / O By construction, '2 is finite and '1W W ! W 0 has the property that W is an isomorphism. That its fibres are connected is a consequence of the following extension of Zariski’s connectedness theorem to non birational maps. W 0 ! '1 O O THEOREM 8.65. Let 'W W ! V be a proper map such that the map isomorphism. Then the fibres of ' are connected. O V ! ' W is an O See Hartshorne 1977, III, 11. NOTES. The Stein factorization was originally proved by Stein for complex spaces (q.v. Wikipedia). g. Blow-ups Under construction. Let P be a nonsingular point on an algebraic variety V , and let Tp.V / be the tangent space at P . The blow-up of V at P is a regular map QV ! V that replaces P with the projective space P.TP .V //. More generally, the blow-up at P replaces P with P.CP .V //, where CP .V / is the geometric tangent cone at P . Blowing up the origin in An Let O be the origin in A .a1W : : : W an/. Let be the graph of , and let fA The map W fA n defined by the projection map A at O. n, and let W A n ! A n X fOg ! P n be the closure of in A n1 be the map .a1; : : : ; an/ 7! n1. n n is the blow-up of A n1 ! A n P n P Blowing up a point on a variety Examples 8.66. The nodal cubic 8.67. The cuspidal cubic h. Resolution of singularities 193 h. Resolution of singularities Let V be an algebraic variety. A desingularization of V is birational regular map W W ! V such that W is nonsingular and is proper; if V is projective, then W should also be projective, and should induce an isomorphism W X 1.Sing.V // ! V X Sing.V /: In other words, the nonsingular variety W is the same as V except over the singular locus of V . When a variety admits a desingularization, then we say tha
t resolution of singularities holds for V . Note that with “nonsingular” replaced by “normalization”, the normalization of V (see 8.5) provides such a map (resolution of abnormalities). Nagata’s embedding theorem 7.50 shows that it suffices to prove resolution of singularities for complete varieties, and Chow’s lemma 7.39 then shows that it suffices to prove resolution of singularities for projective varieties. From now on, we shall consider only projective varieties. Resolution of singularities for curves was first obtained using blow-ups (see Chapter 7 of Fulton’s book, Algebraic Curves). Zariski introduced the notion of the normalization of a variety, and observed that the normalization W QV ! V of a curve V in k.V / is a desingularization of V . There were several proofs of resolution of singularities for surfaces over C, but the first to be accepted as rigorous is that of Walker (patching Jung’s local arguments; 1935). For a surface V , normalization gives a surface with only point singularities (8.12), which can then be blown up. Zariski showed that the desingularization of a surface in characteristic zero can be obtained by alternating normalizations and blow-ups. The resolution of singularities for three-folds in characteristic zero is much more difficult, and was first achieved by Zariski (Ann. of Math. 1944). His result was extended to nonzero characteristic by his student Abhyankar and to all varieties in characteristic zero by his student Hironaka. The resolution of singularities for higher dimensional varieties in nonzero characteristic is one of the most important outstanding problems in algebraic geometry. In 1996, de Jong proved a weaker result in which, instead of the map being birational, k.W / is allowed to be a finite extension of k.V /. A little history Normal varieties were introduced by Zariski in a paper, Amer. J. Math. 61, 1939, p. 249–194. There he noted that the singular locus of a normal variety has codimension at least 2 and that the system of hyperplane sections of a normal variety relative to a projective embedding is complete (i.e., is a complete rational equivalence class). Zariski’s introduction of the notion of a normal variety and of the normalization of a variety was an important insertion of commutative algebra into algebraic geometry. It is not easy to give a geometric intuition for “normal”. One criterion is that a variety is normal if and only if every surjective finite birational map onto it is an isomorphism (8.57). See mo109395 for a discussion of this question. 194 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM Exercises 8-1. Prove that a finite map is an isomorphism if and only if it is bijective and ´etale. (Cf. Harris 1992, 14.9.) 8-2. Give an example of a surjective quasi-finite regular map that is not finite (different from any in the notes). 8-3. Let 'W W ! V be a regular map with the property that '1.U / is an open affine subset of W whenever U is an open affine subset of V (such a map is said to be affine). Show that if V is separated, then so also is W . 8-4. For every n 1, find a finite map 'W W ! V with the following property: for all 1 i n, defD fP 2 V j '1.P / has i pointsg Vi is a nonempty closed subvariety of dimension i . CHAPTER 9 Regular Maps and Their Fibres Consider again the regular map 'W A 2, .x; y/ 7! .x; xy/ (Exercise 3-3). The line Y D c maps to the line Y D cX . As c runs over the elements of k, this line sweeps out the whole x; y-plane except for the y-axis, and so the image of ' is 2 ! A which is neither open nor closed, and, in fact, is not even locally closed. The fibre C D .A 2 X fy-axisg/ [ f.0; 0/g; '1.a; b/ D 8 < : point .a; b=a/ Y -axis ; if a ¤ 0 if .a; b/ D .0; 0/ if a D 0, b ¤ 0: From this unpromising example, it would appear that it is not possible to say anything about the image of a regular map or its fibres. However, it turns out that almost everything that can go wrong already goes wrong in this example. We shall show: (a) the image of a regular map is a finite union of locally closed sets; (b) the dimensions of the fibres can jump only over closed subsets; (c) the number of elements (if finite) in the fibres can drop only on closed subsets, provided the map is finite, the target variety is normal, and k has characteristic zero. a. The constructibility theorem THEOREM 9.1. Let 'W W ! V be a dominant regular map of irreducible affine algebraic varieties. Then '.W / contains a dense open subset of V . PROOF. Because ' is dominant, the map f 7! f ı 'W kŒV ! kŒW is injective (3.34). According to Lemma 9.4 below, there exists a nonzero a 2 kŒV such that every homomorphism ˛W kŒV ! k such that ˛.a/ ¤ 0 extends to a homomorphism ˇW kŒW ! k with ˇ.1/ ¤ 0. In particular, for P 2 D.a/, the homomorphism g 7! g.P /W kŒV ! k extends to a nonzero homomorphism ˇW kŒW ! k. The kernel of ˇ is a maximal ideal of kŒW whose zero set is a point Q of W such that '.Q/ D P . Before beginning the proof of Lemma 9.4, we should look at an example. EXAMPLE 9.2. Let A be an affine k-algebra, and let B D AŒT =.f / with f D amT m C C a0. When does a homomorphism ˛W A ! k extend to B? The extensions of ˛ correspond to roots of the polynomial ˛.am/T m C C˛.a0/ in k, and so there exists an extension unless this is a nonzero constant polynomial. In particular, ˛ extends if ˛.am/ ¤ 0. 195 196 9. REGULAR MAPS AND THEIR FIBRES LEMMA 9.3. Let A B be finitely generated k-algebras. Assume that A and B are integral domains, and that B is generated by a single element, say, B D AŒt ' AŒT =a. Let c A be the set of leading coefficients of the polynomials in a. Then every homomorphism ˛W A ! k such that ˛.c/ ¤ 0 extends to a homomorphism B ! k. PROOF. Note that c is an ideal in A. If a D 0, then every homomorphism ˛ extends. Thus we may assume that a ¤ 0. Let f D amT m C C a0 be a nonzero polynomial of minimum degree in a such that ˛.am/ ¤ 0. Because B ¤ 0, we have that m 1. Extend ˛ to a homomorphism Q˛W AŒT ! kŒT by sending T to T . The k-submodule of kŒT generated by Q˛.a/ is an ideal (because T P ci Q˛.gi / D P ci Q˛.gi T //. Unless Q˛.a/ contains a nonzero constant, it generates a proper ideal in kŒT , which will have a zero c in k (2.11). The homomorphism AŒT Q˛! kŒT h7!h.c/! k; T 7! T 7! c then factors through AŒT =a D B and extends ˛. In the contrary case, a contains a polynomial g.T / D bnT n C C b0; ˛.bi / D 0 .i > 0/; ˛.b0/ ¤ 0: On dividing f .T / into g.T /, we find that ad mg.T / D q.T /f .T / C r.T /; d 2 N; q; r 2 AŒT ; deg r < m: On applying Q˛ to this equation, we obtain ˛.am/d ˛.b0/ D Q˛.q/ Q˛.f / C Q˛.r/: Because Q˛.f / has degree m > 0, we must have Q˛.q/ D 0, and so Q˛.r/ is a nonzero constant. After replacing g.T / with r.T /, we may assume n < m. If m D 1, such a g.T / can’t exist, and so we may suppose m > 1 and (by induction) that the lemma holds for smaller values of m. For h.T / D cr T r C cr1T r1 C C c0, let h0.T / D cr C C c0T r . Then the Amodule generated by the polynomials T sh0.T /, s 0, h 2 a, is an ideal a0 in AŒT . Moreover, a0 contains a nonzero constant if and only if a contains a nonzero polynomial cT r , which implies t D 0 and A D B (since B is an integral domain). If a0 does not contain nonzero constants, then set B 0 D AŒT =a0 D AŒt 0. Then a0 contains the polynomial g0 D bn C C b0T n, and ˛.b0/¤ 0. Because deg g0 < m, the induction hypothesis implies that ˛ extends to a homomorphism B 0 ! k. Therefore, there is a c 2 k such that, for all h.T / D cr T r C cr1T r1 C C c0 2 a, h0.c/ D ˛.cr / C ˛.cr1/c C C c0cr D 0: On taking h D g, we see that c D 0, and on taking h D f , we obtain the contradiction ˛.am/ D 0. LEMMA 9.4. Let A B be finitely generated k-algebras. Assume that A and B are integral domains, and let b be a nonzero element of B. Then there exists a nonzero a 2 A with the following property: every homomorphism ˛W A ! k from A into k such that ˛.a/ ¤ 0 extends to a homomorphism ˇW B ! k such that ˇ.b/ ¤ 0. a. The constructibility theorem 197 PROOF Suppose that we know the proposition in the case that B is generated by a single element, and write B D AŒx1; : : : ; xn. Then there exists an element bn1 2 AŒx1; : : : ; xn1 with the following property: every homomorphism ˛W AŒx1; : : : ; xn1 ! k such that ˛.bn1/ ¤ 0 extends to a homomorphism ˇW B ! k such that ˇ.b/ ¤ 0. Then there exists a bn2 2 AŒx1; : : : ; xn2 etc. Continuing in this fashion, we obtain an element a 2 A with the required property. Thus we may assume B D AŒx. Let a be the kernel of the homomorphism T 7! x, AŒT ! AŒx. Case (i). The ideal a D .0/. Write b D f .x/ D a0xn C a1xn1 C C an; ai 2 A; and take a D a0. If ˛W A ! k is such that ˛.a0/ ¤ 0, then there exists a c 2 k such that f .c/ ¤ 0, and we can take ˇ to be the homomorphism P di xi 7! P ˛.di /ci . Case (ii). The ideal a ¤ .0/. Let f .T / D amT m C C a0; am ¤ 0; be an element of a of minimum degree. Let h.T / 2 AŒT represent b. As b is nonzero, h … a. Because f is irreducible over the field of fractions of A, it and h are coprime over that field. Hence there exist u; v 2 AŒT and c 2 A X f0g such that uh C vf D c: It follows now that cam satisfies our requirements, for if ˛.cam/ ¤ 0, then ˛ can be extended to ˇW B ! k by the preceding lemma, and ˇ.u.x/ b/ D ˇ.c/ ¤ 0, and so ˇ.b/ ¤ 0. ASIDE 9.5. It is also possible to deduce Theorem 9.1 from the generic freeness theorem (CA 21.11). In order to generalize 9.1 to arbitrary maps of arbitrary varieties, we need the notion of a constructible set. Let W be a topological space. A subset C of W is said to constructible if it is a finite union of sets of the form U \ Z with U open and Z closed. Obviously, if C is constructible in W and V W , then C \ V is constructible in V , and it is constructible in W if V is open or closed. A constructible subset of A n is one that is definable by a finite number of polynomials. More precisely, it is defined by a finite number of statements of the form f .X1; : : : ; Xn/ D 0; g.X1; : : : ; Xn/ ¤ 0 1 that o
mits an infinite set. combined using only “and” and “or” (or, better, statements of the form f D 0 combined using “and”, “or”, and “not”). The next proposition shows that a constructible set C that is dense in an irreducible variety V must contain a nonempty open subset of V . Contrast Q, which is dense in R (real topology), but does not contain an open subset of R, or an infinite subset of A PROPOSITION 9.6. Let C be a constructible set whose closure NC is irreducible. Then C contains a nonempty open subset of its closure NC . PROOF. We are given that C D S.Ui \ Zi / with each Ui open and each Zi closed. We may assume that each set Ui \ Zi in this decomposition is nonempty. Clearly NC S Zi , and as NC is irreducible, it must be contained in one of the Zi . For this i C Ui \ Zi Ui \ NC Ui \ C Ui \ .Ui \ Zi / D Ui \ Zi : Thus Ui \ Zi D Ui \ NC is a nonempty open subset of NC contained in C . 198 9. REGULAR MAPS AND THEIR FIBRES THEOREM 9.7. Every regular map 'W W ! V sends constructible sets to constructible sets. PROOF We first show that it suffices to prove the theorem with W and V affine. Write V as a finite union of open affines, and then write the inverse image of each of the affines as a finite union of open affines. In this way, we get W D S i 2I Wi with each Wi open affine and '.Wi / contained in an open affine of V . If C is a constructible subset of W , then '.C / D S i 2I '.C \Wi /, and so '.C / is constructible if each set '.C \Wi / is constructible. Now assume that W and V are affine, and let C be a constructible subset of W . Let Wi be the irreducible components of W . They are closed in W , and so C \ Wi is constructible in W . As '.W / D S '.C \ Wi /, it is constructible if the '.C \ Wi / are. Hence we may suppose that W is irreducible. Moreover, C is a finite union of its irreducible components. As these are closed in C , they are constructible in W . We may therefore assume that C is also irreducible; NC is then an irreducible closed subvariety of W . We prove the theorem by induction on the dimension of W . If dim.W / D 0, then the statement is obvious because W is a point. If NC ¤ W , then dim. NC / < dim.W /, and '.C / is '! V . We may therefore assume constructible by the induction hypothesis applied to NC that NC D W . Replace V with '.C /. According to Proposition 9.6, C contains a dense open subset U 0 of W , and Theorem 9.1 applied to U 0 '! V shows that '.C / contains a dense open subset U of V . Write '.C / D U [ '.C \ '1.V U //: Then '1.V U / is a proper closed subset of W (the complement of V U is dense in V and ' is dominant). As C \ '1.V U / is constructible in '1.V U /, the set '.C \ '1.V U // is constructible in V by induction, which completes the proof. ASIDE 9.8. Let X be a subset of C closure of X for the Zariski topology is equal to its closure for the complex topology. n. If X is constructible for the Zariski topology on C n, then the b. The fibres of morphisms We wish to examine the fibres of a regular map 'W W ! V . We can replace V by the closure of '.W / in V and so assume that ' is dominant. THEOREM 9.9. Let 'W W ! V be a dominant regular map of irreducible varieties. Then (a) dim.W / dim.V /; (b) if P 2 '.W /, then dim.'1.P // dim.W / dim.V / for every P 2 V , with equality holding exactly on a nonempty open subset U of V . (c) The sets are closed in '.W /. Vi D fP 2 V j dim.'1.P // i g In other words, for P on a dense open subset U of V , the fibre '1.P / has the expected dimension dim.W / dim.V /. On the closed complement of U (possibly empty), the dimension of the fibre is > dim.W / dim.V /, and it may jump further on closed subsets. Before proving the theorem, we should look at an example. b. The fibres of morphisms 199 EXAMPLE 9.10. Consider the subvariety W V A m defined by r linear equations m X j D1 aij Xj D 0; aij 2 kŒV ; i D 1; : : : ; r; and let ' be the projection W ! V . For P 2 V , '1.P / is the set of solutions of system of equations m X aij .P /Xj D 0; aij .P / 2 k; i D 1; : : : ; r; j D1 and so its dimension is m rank.aij .P //. Since the rank of the matrix .aij .P // drops on closed subsets, the dimension of the fibre jumps on closed subsets. More precisely, for each r 2 N, fP 2 V j rank.aij .P // rg is a closed subset of V (see Exercise 2-2); hence, for each r 0 2 N, fP 2 V j dim '1.P / r 0g is closed in V . PROOF. (a) Because the map is dominant, there is a homomorphism k.V / ,! k.W /, and obviously tr degkk.V / tr degkk.W / (an algebraically independent subset of k.V / remains algebraically independent in k.W /). (b) In proving the first part of (b), we may replace V by any open neighbourhood of P . In particular, we can assume V to be affine. Let m be the dimension of V . From (3.47) we know that there exist regular functions f1; : : : ; fm such that P is an irreducible component of V .f1; : : : ; fm/. After replacing V by a smaller neighbourhood of P , we can suppose that P D V .f1; : : : ; fm/. Then '1.P / is the zero set of the regular functions f1 ı '; : : : ; fm ı ', and so (if nonempty) has codimension m in W (see 3.45). Hence dim '1.P / dim W m D dim.W / dim.V /: In proving the second part of (b), we can replace both W and V with open affine subsets. Since ' is dominant, kŒV ! kŒW is injective, and we may regard it as an inclusion (we identify a function x on V with x ı ' on W /. Then k.V / k.W /. Write kŒV D kŒx1; : : : ; xM and kŒW D kŒy1; : : : ; yN , and suppose V and W have dimensions m and n respectively. Then k.W / has transcendence degree n m over k.V /, and we may suppose that y1; : : : ; ynm are algebraically independent over kŒx1; : : : ; xm, and that the remaining yi are algebraic over kŒx1; : : : ; xm; y1; : : : ; ynm. There are therefore relations Fi .x1; : : : ; xm; y1; : : : ; ynm; yi / D 0; i D n m C 1; : : : ; N; (37) with Fi .X1; : : : ; Xm; Y1; : : : ; Ynm; Yi / a nonzero polynomial. We write Nyi for the restriction of yi to '1.P /. Then kŒ'1.P / D kŒ Ny1; : : : ; NyN : The equations (37) give an algebraic relation among the functions x1; : : : ; yi on W . When we restrict them to '1.P /, they become equations: Fi .x1.P /; : : : ; xm.P /; Ny1; : : : ; Nynm; Nyi / D 0; i D n m C 1; : : : ; N: 200 9. REGULAR MAPS AND THEIR FIBRES If these are nontrivial algebraic relations, i.e., if none of the polynomials Fi .x1.P /; : : : ; xm.P /; Y1; : : : ; Ynm; Yi / is identically zero, then the transcendence degree of k. Ny1; : : : ; NyN / over k will be n m. Thus, regard Fi .x1; : : : ; xm; Y1; : : : ; Ynm; Yi / as a polynomial in the Y ’s with coefficients polynomials in the x’s. Let Vi be the closed subvariety of V defined by the simultaneous vanishing of the coefficients of this polynomial — it is a proper closed subset of V . Let U D V X S Vi — it is a nonempty open subset of V . If P 2 U , then none of the polynomials Fi .x1.P /; : : : ; xm.P /; Y1; : : : ; Ynm; Yi / is identically zero, and so for P 2 U , the dimension of '1.P / is n m, and hence D n m by (a). Finally, if for a particular point P , dim '1.P / D n m, then we can modify the above argument to show that the same is true for all points in an open neighbourhood of P . (c) We prove this by induction on the dimension of V — it is obviously true if dim V D 0. We know from (b) that there is an open subset U of V such that dim '1.P / D n m ” P 2 U: Let Z be the complement of U in V ; thus Z D VnmC1. Let Z1; : : : ; Zr be the irreducible components of Z. On applying the induction to the restriction of ' to the map '1.Zj / ! Zj for each j , we obtain the result. Recall that a regular map 'W W ! V of algebraic varieties is closed if, for example, W is complete (7.7). PROPOSITION 9.11. Let 'W W ! V be a regular surjective closed map of varieties, and let n 2 N. If V is irreducible and all fibres '1.P / of ' are irreducible of dimension n, then W is irreducible of dimension dim.V / C n. PROOF. Let Z be an irreducible closed subset of W , and consider the map 'jZW Z ! V ; it has fibres .'jZ/1.P / D '1.P / \ Z. There are three possibilities. (a) '.Z/ ¤ V . Then '.Z/ is a proper closed subset of V . (b) '.Z/ D V , dim.Z/ < n C dim.V /. Then (b) of (9.9) shows that there is a nonempty open subset U of V such that for P 2 U , dim.'1.P / \ Z/ D dim.Z/ dim.V / < n: Thus, for P 2 U , the fibre '1.P / is not contained in Z. (c) '.Z/ D V , dim.Z/ n C dim.V /. Then 9.9(b) shows that dim.'1.P / \ Z/ dim.Z/ dim.V / n for all P ; thus '1.P / Z for all P 2 V , and so Z D W ; moreover dim Z D dim V C n. Now let Z1; : : : ; Zr be the irreducible components of W . I claim that (c) holds for at least one of the Zi . Otherwise, there will be an open subset U of V such that for P in U , '1.P / is contained in none of the Zi ; but '1.P / is irreducible and '1.P / D S.'1.P / \ Zi /, and so this is impossible. CAUTION. It is possible for all the fibres of regular map W ! V to be reducible without 2y2 D 0 is irreducible, W being reducible. The variety in A but the fibres of the projection to the first factor (obtained by fixing the values of y1 and y2) are all reducible. Pass to the projective closure to extend this to P 2 with equation x2 1y1 x2 2 A 2 P 2. c. Flat maps and their fibres 201 c. Flat maps and their fibres Flat maps Let A be a ring, and let B be an A-algebra. If the sequence of A-modules 0 ! N 0 ˛! N ˇ! N 00 ! 0 is exact, then the sequence of B-modules B ˝A N 0 1˝˛! B ˝A N 1˝ˇ! B ˝A N 00 ! 0 is exact,1 but B˝A N 0 ! B˝A N need not be injective. For example, when we tensor the exact sequence of kŒX-modules 0 ! kŒX f 7!Xf ! kŒX f 7!f mod .X/ ! kŒX=.X / ! 0 with k, we get the sequence 0! k id! k ! 0: k DEFINITION 9.12. An A-algebra B is flat if M ! N injective H) B ˝A M ! B ˝A N injective. It is faithfully flat if, in addition, B ˝A M D 0 H) M D 0: Therefore, an A-algebra B is flat if and only if the functor M B ˝A M from A- modules to B-modules is exact. EXAMPLE 9.13. (a) Let S be a multiplicative subset of A. Then S 1A is a flat A-algebra (1.18). (b) Every open immersion is flat (obv
ious). (c) The composite of two flat maps is flat (obvious). PROPOSITION 9.14. Let A ! A0 be a homomorphism of rings. If A ! B is flat, then so also is A0 ! B ˝A A0. PROOF. For any A0-module M , .B ˝A A0/ ˝A0 M ' B ˝A .A0 ˝A0 M / ' B ˝A M: In other words, tensoring an A0-module M with B ˝A A0 is the same as tensoring M (regarded as an A-module) with B. Therefore it preserves exact sequences. 1The surjectivity of 1 ˝ ˇ is obvious. Let B ˝A N ! Q be the cokernel of 1 ˝ ˛. Because .1 ˝ ˇ/ ı .1 ˝ ˛/ D 1 ˝ .ˇ ı ˛/ D 0; there is a unique A-linear map f W Q ! B ˝A N 00 such that f ı D 1 ˝ ˇ. We shall construct an inverse g to f . Let b 2 B, and let n 2 N . If ˇ.n/ D 0, then n D ˛.n0/ for some n0 2 N 0; hence b ˝ n D b ˝ ˛.n0/, and so .b ˝ n/ D 0. It follows by linearity that .b ˝ n1/ D .b ˝ n2/ if ˇ.n1/ D ˇ.n2/, and so the A-bilinear map B N ! Q; .b; n/ 7! .b ˝ n/ factors through B N 00. It therefore defines an A-linear map gW B ˝A N 00 ! Q. To show that f and g are inverse, it suffices to check that g ı f D idQ on elements of the form .b ˝ n/ and that f ı g D idB˝AN 00 on elements of the form b ˝ ˇ.n/ — both are obvious. 202 9. REGULAR MAPS AND THEIR FIBRES PROPOSITION 9.15. A homomorphism ˛W A ! B of rings is flat if and only if, for all maximal ideals n in B, the map A˛1.n/ ! Bn is flat. PROOF. Let n be a prime ideal of B, and let m D ˛1.n/ — it is a prime ideal in A. If A ! B is flat, then so is Am ! Am ˝A B ' S 1 m B (9.14). The map Bn is flat (9.13a), and so the composite Am ! Bn is flat (9.13c). For the converse, let N 0 ! N be an injective homomorphism of A-modules, and let n be a maximal ideal of B. Then Am ˝A .N 0 ! N / is injective (9.13). Therefore, the map Bn ˝A .N 0 ! N / ' Bn ˝Am .Am ˝A .N 0 ! N // is injective, and so the kernel M of B ˝A .N 0 ! N / has the property that Mn D 0. Let x 2 M , and let a D fb 2 B j bx D 0g. For each maximal ideal n of B, x maps to zero in Mn, and so a contains an element not in n. Hence a D B, and so x D 0. PROPOSITION 9.16. A flat homomorphism 'W A ! B is faithfully flat if and only if every maximal ideal m of A is of the form '1.n/ for some maximal ideal n of B. PROOF. ): Let m be a maximal ideal of A, and let M D A=m; then B ˝A M ' B='.m/B: As B ˝A M ¤ 0, we see that '.m/B ¤ B. Therefore '.m/ is contained in a maximal ideal n of B. Now '1.n/ is a proper ideal in A containing m, and hence equals m. (: Let M be a nonzero A-module. Let x be a nonzero element of M , and let a D ann.x/ defD fa 2 A j ax D 0g. Then a is an ideal in A, and M 0 defD Ax ' A=a. Moreover, B ˝A M 0 ' B='.a/ B and, because A ! B is flat, B ˝A M 0 is a submodule of B ˝A M . Because a is proper, it is contained in a maximal ideal m of A, and therefore '.a/ '.m/ n for some maximal ideal n of A. Hence '.a/ B n ¤ B, and so B ˝A M B ˝A M 0 ¤ 0. COROLLARY 9.17. A flat local homomorphism A ! B of local rings is faithfully flat. PROOF. Let m and n be the (unique) maximal ideals of A and B. By hypothesis, nc D m, and so the statement follows from the proposition. Properties of flat maps LEMMA 9.18. Let B be an A-algebra, and let p be a prime ideal of A. The prime ideals of B contracting to p are in natural one-to-one correspondence with the prime ideals of B ˝A .p/. PROOF. Let S D A X p. Then .p/ D S 1.A=p/. Therefore we obtain B ˝A .p/ from B by first passing to B=pB and then making the elements of A not in p act invertibly. After the first step, we are left with the prime ideals q of B such that qc p, and after the second step only with those such that qc \ S D ;, i.e., such that qc D p. PROPOSITION 9.19. Let B be a faithfully flat A-algebra. Every prime ideal p of A is of the form qc for some prime ideal q of B. c. Flat maps and their fibres 203 PROOF. The ring B ˝A .p/ is nonzero, because .p/ ¤ 0 and A ! B is faithfully flat, and so it has a prime (even maximal) ideal q. For this ideal, qc D p. SUMMARY 9.20. A flat homomorphism 'W A ! B is faithfully flat if the image of spec.'/W spec.B/ ! spec.A/ includes all maximal ideals of A, in which case it includes all prime ideals of A. PROPOSITION 9.21 (GOING-DOWN THEOREM FOR FLAT MAPS). Let A ! B be a flat homomorphism. Let p p0 be prime ideals in A, and let q be a prime ideal in B such that qc D p. Then q contains a prime ideal q0 such that q0c D p0: B A q q0 p p0: PROOF. Because A ! B is flat, the homomorphism Ap ! Bq is flat, and because pAp D .qBq/c, it is faithfully flat (9.16). The ideal p0Ap is prime (1.14), and so there exists a prime ideal of Bq lying over p0Ap (by 9.19). The contraction of this ideal to B is contained in q and contracts to p0 in A. DEFINITION 9.22. A regular map 'W W ! V of algebraic varieties is flat if, for all P 2 W , the map W;P is flat, and it is faithfully flat if it is flat and surjective. OV;'.P / ! O PROPOSITION 9.23. A regular map 'W W ! V of affine algebraic varieties is flat (resp. faithfully flat) if and only if the map f 7! f ı 'W kŒV ! kŒW is flat (resp. faithfully flat). PROOF. Apply (9.15) and (9.16). PROPOSITION 9.24. Let 'W W ! V be a flat map of affine algebraic varieties. Let S S 0 be closed irreducible subsets of V , and let T be a closed irreducible subset of W such that '.T / is a dense subset of S. Then there exists a closed irreducible subset T 0 of W containing T and such that '.T 0/ is a dense subset of S 0. PROOF. Let p D I.S/, p0 D I.S 0/, and q D I.T /. Then p p0 because S S 0. Moreover '! S is dominant and so the map kŒS D kŒV =p ! kŒT =q is injective. qc D p because T According to (9.21), there exists a prime ideal q0 in kŒW contained in q and such that q0c D p0. Now V .q0/ has the required properties. THEOREM 9.25 (GENERIC FLATNESS). For every regular map 'W W ! V of irreducible '! U is algebraic varieties, there exists a nonempty open subset U of V such that '1.U / faithfully flat. PROOF. We may assume that W and V are affine, say, V D Spm.A/ and W D Spm.B/. Let F be the field of fractions of A. We regard B as a subring of F ˝A B. As F ˝A B is a finitely generated F -algebra, the Noether normalization theorem (2.45) shows that there exist elements x1; : : : ; xm of F ˝A B such that F Œx1; : : : ; xm is a polynomial ring over F and F ˝A B is a finite F Œx1; : : : ; xm-algebra. After multiplying each xi by an element of A, we may suppose that it lies in B. Let b1; : : : ; bn generate B as an Aalgebra. Each bi satisfies a monic polynomial equation with coefficients in F Œx1; : : : ; xm. 204 9. REGULAR MAPS AND THEIR FIBRES Let a 2 A be a common denominator for the coefficients of these polynomials. Then each bi is integral over Aa. As the bi generate Ba as an Aa-algebra, this shows that Ba is a finite AaŒx1; : : : ; xm-algebra (1.36). Therefore, after replacing A with Aa and B with Ba, we may suppose that B is a finite AŒx1; : : : ; xm-algebra. injective B finite F ˝A B finite E ˝AŒx1;:::;xm B finite AŒx1; : : : ; xm F Œx1; : : : ; xm E defD F .x1; : : : ; xm/ A F: Let E D F .x1; : : : ; xm/ be the field of fractions of AŒx1; : : : ; xm, and let b1; : : : ; br be elements of B that form a basis for E ˝AŒx1;:::;xm B as an E-vector space. Each element of B can be expressed as a linear combination of the bi with coefficients in E. Let q be a common denominator for the coefficients arising from a set of generators for B as an AŒx1; : : : ; xm-module. Then b1; : : : ; br generate Bq as an AŒx1; : : : ; xmq-module. In other words, the map .c1; : : : ; cr / 7! P ci bi W AŒx1; : : : ; xmr q ! Bq () is surjective. This map becomes an isomorphism when tensored with E over AŒx1; : : : ; xmq, which implies that each element of its kernel is killed by a nonzero element of AŒx1; : : : ; xmq and so is zero (because AŒx1; : : : ; xnq is an integral domain). Hence the map () is an isomorphism, and so Bq is free of finite rank over AŒx1; : : : ; xmq. Let a be some nonzero coefficient of the polynomial q, and consider the maps Aa ! AaŒx1; : : : ; xm ! AaŒx1; : : : ; xmq ! Baq: The first and third arrows realize their targets as nonzero free modules over their sources, and so are faithfully flat. The middle arrow is flat by (9.13). Let m be a maximal ideal in Aa. Then mAaŒx1; : : : ; xm does not contain the polynomial q because the coefficient a of q is invertible in Aa. Hence mAaŒx1; : : : ; xmq is a proper ideal of AaŒx1; : : : ; xmq, and so the map Aa ! AaŒx1; : : : ; xmq is faithfully flat (apply 9.16). This completes the proof. LEMMA 9.26. Let V be an algebraic variety. A constructible subset C of V is closed if it has the following property: let Z be a closed irreducible subset of V ; if Z \ C contains a dense open subset of Z, then Z C . PROOF. Let Z be an irreducible component of NC . Then Z \ C is constructible and it is dense in Z, and so it contains a nonempty open subset U of Z (9.6). Hence Z C . THEOREM 9.27. A flat map 'W W ! V of algebraic varieties is open. PROOF. Let U be an open subset of W . Then '.U / is constructible (9.7) and the goingdown theorem (9.21) implies that V X '.U / satisfies the hypotheses of the lemma. Therefore V X '.U / is closed. COROLLARY 9.28. Let 'W W ! V be a regular map of irreducible algebraic varieties. Then there exists a dense open subset U of W such that '.U / is open, U D '1.'U /, and U '! '.U / is flat. c. Flat maps and their fibres 205 '! U PROOF. According to 9.25, there exists a dense open subset U of V such that '1.U / is flat. In particular, '.'1.U // is open in V (9.27). Note that '1.'.'1.U // D '1.U /. Let U 0 D '1.U /. Then U 0 is a dense open subset of W , '.U 0/ is open, U 0 D '1.'U 0/, and U 0 '! '.U 0/ is flat. Fibres and flatness The notion of flatness allows us to sharpen our earlier results. PROPOSITION 9.29. Let 'W W ! V be a dominant map of irreducible algebraic varieties. Let P 2 '.W /. Then dim '1.P / dim.W / dim.V /; (38) and equality holds if ' is flat. PROOF. The inequality was proved in 9.9. If ' is flat, then we shall prove (more precisely) that, if Z is an irreducible component of '1.P /, then dim.Z/ D dim.W / dim.V /: After replacing V with an o
pen neighbourhood of P and W with an open subset intersecting Z, we may suppose that both V and W are affine. Let V V1 Vm D fP g be a maximal chain of distinct irreducible closed subsets of V (so m D dim.V /). Now '.Z/ D fP g, and so (see 9.24) there exists a chain of irreducible closed subsets W W1 Wm D Z such that '.Wi / is a dense subset of Vi . Let Z Z1 Zn be a maximal chain of distinct irreducible closed subsets of V (so n D dim.Z/). The existence of the chain W W1 Wm Z1 Zn shows that dim.W / m C n D dim.V / C dim.Z/: Together with (38), this implies that we have equality. PROPOSITION 9.30. Let 'W W ! V be a dominant map of irreducible algebraic varieties. Let P 2 '.W /. Then dim '1.P / dim.W / dim.V /: There exists a dense open subset U of W such that '.U / is open in V , U D '1.'.U //, and equality holds for all P 2 '.U /. PROOF. Let U be an open subset of W as in 9.28. 206 9. REGULAR MAPS AND THEIR FIBRES PROPOSITION 9.31. Let 'W W ! V be a dominant map of irreducible varieties. Let S be a closed irreducible subset of V , and let T be an irreducible component of '1.S/ such that '.T / is dense in S . Then dim.T / dim.S / C dim.W / dim.V /; and equality holds if ' is flat. PROOF. The inequality can be proved by a similar argument to that in 9.9 — see, for example, Hochschild 1981, X, Theorem 2.1.2 The equality can be deduced by the same argument as in 9.29. PROPOSITION 9.32. Let 'W W ! V be a dominant map of irreducible varieties. There exists '! '.U / a nonempty open subset U of W such that '.U / is open, U D '1.'U /, and U is flat. If S is a closed irreducible subset of V meeting '.U /, and T is an irreducible component of '1.S/ meeting U , then dim.T / D dim.S / C dim.W / dim.V /: PROOF. Let U be an open subset of W as in 9.28. FINITE MAPS PROPOSITION 9.33. Let V be an irreducible algebraic variety. A finite map 'W W ! V is flat if and only if is independent of P 2 V . Q7!P X dimk O Q=mP Q O PROOF. It suffices to prove this with V affine, in which case it follows from CA 12.6 (equivalence of (d) and (e)). The integer dimk O Q is the multiplicity of Q in its fibre. The theorem says that a finite map is flat if and only if the number of points in each fibre (counting multiplicities) is constant. Q=mP O For example, let V be the subvariety of A nC1 defined by an equation X m C a1X m1 C C am D 0; ai 2 kŒT1; : : : ; Tn and let 'W V ! A set of points .P; c/ with c a root of the polynomial n be the projection map (see p. 51). The fibre over a point P of A n is the X m C a1.P /X m1 C C am.P / D 0: The multiplicity of .P; c/ in its fibre is the multiplicity of c as a root of the polynomial. Therefore P Q D m for every P , and so the map ' is flat. Q=mP Q7!P dimk O O 2Hochschild, Gerhard P., Basic theory of algebraic groups and Lie algebras. Springer, 1981. c. Flat maps and their fibres 207 Criteria for flatness THEOREM 9.34. Let 'W A ! B be a local homomorphism of noetherian local rings, and let m be the maximal ideal of A. If A is regular, B is Cohen-Macaulay, and dim.B/ D dim.A/ C dim.B=mB/; then ' is flat. PROOF. See Matsumura 1986, 23.1.3 9.35. We don’t define the notion of being Cohen-Macaulay here (see ibid. p. 134), but merely list some of its properties. (a) A noetherian ring A is Cohen-Macaulay if and only if Am is Cohen-Macaulay for every maximal ideal m of A (this is part of the definition). (b) Zero-dimensional and reduced one-dimensional noetherian rings are Cohen-Macaulay (ibid. p. 139). (c) Regular noetherian rings are Cohen-Macaulay (ibid. p. 137). (d) Let 'W A ! B be a flat local homomorphism of noetherian local rings, and let m be the maximal ideal of A. Then B is Cohen-Macaulay if and only if both A and B=mB are Cohen-Macaulay (ibid. p. 181). PROPOSITION 9.36. Let 'W A ! B be a finite homomorphism noetherian rings with A regular. Then ' is flat if and only if B is Cohen-Macaulay. PROOF. Note that B=mB/ is zero-dimensional,4 hence Cohen-Macaulay, for every maximal ideal m of A (9.35b), and that ht.n/ D ht.nc/ for every maximal ideal n of B. If ' is flat, then B is Cohen-Macaulay by (9.35d). Conversely, if B is Cohen-Macaulay, then ' is flat by (9.34). EXAMPLE 9.37. Let A be a finite kŒX1; : : : ; Xn-algebra (cf. 2.45). The map kŒX1; : : : ; Xn ! A is flat if and only if A is Cohen-Macaulay. An algebraic variety V is said to be Cohen-Macaulay if V;P is Cohen-Macaulay for all P 2 V . An affine algebraic variety V is Cohen-Macaulay if and only if kŒV is Cohen-Macaulay (9.35a). A nonsingular variety is Cohen-Macaulay (9.35c). O THEOREM 9.38. Let V and W be algebraic varieties with V nonsingular and W CohenMacaulay. A regular map 'W W ! V is flat if and only if dim '1.P / D dim W dim V for all P 2 V . PROOF. Immediate consequence of (9.34). (39) 3Matsumura, Hideyuki, Commutative ring theory. Cambridge University Press, Cambridge, 1986. 4Note that C defD B=mB D B ˝A A=m is a finite k-algebra. Therefore it has only finitely many maximal ideals. Every prime ideal in C is an intersection of maximal ideals (2.18), but a prime ideal can equal a finite intersection of ideals only if it equals one of the ideals. 208 9. REGULAR MAPS AND THEIR FIBRES ASIDE 9.39. The theorem fails with “nonsingular” weakened to “normal”. Let Z=2Z act on W defD A 2 by .x; y/ 7! .x; y/. The quotient of W by this action is the quadric cone V A 3 defined by T V D U 2. The quotient map 'W W ! V is .x; y/ 7! .t; u; v/ D .x2; xy; y2/. The variety W is nonsingular, and V is normal because kŒV D kŒX; Y G (cf. CA 23.12). Moreover ' is finite, and so its fibres have constant dimension 0, but it is not flat because X Q7!P dimk O Q=mP Q D O 3 2 if P D .0; 0; 0/ otherwise (see 9.33). See mo117043. d. Lines on surfaces As an application of some of the above results, we consider the problem of describing the set 3. To avoid possible problems, we assume for the rest of lines on a surface of degree m in P of this chapter that k has characteristic zero. We first need a way of describing lines in P 3 as being one-dimensional subspaces in k4, and lines in P 3. Recall that we can associate with each n an affine cone over QV in knC1. This allows us to think of points projective variety V P 3 as being two-dimensional in P subspaces in k4. To such a subspace W k4, we can attach a one-dimensional subspace V2 W in V2 k4 k6, that is, to each line L in P 5. Not 3 should form a every point in P 3 corresponds to choosing a four-dimensional set. (Fix two planes in P point on each of the planes.) We shall show that there is natural one-to-one correspondence 5. Rather between the set of lines in P than using exterior algebras, I shall usually give the old-fashioned proofs. 5 should be of the form p.L/ — heuristically, the lines in P 3 and the set of points on a certain hyperspace ˘ P 3, we can attach point p.L/ in P 3; giving a line in P Let L be a line in P 3 and let x D .x0 W x1 W x2 W x3/ and y D .y0 W y1 W y2 W y3/ be distinct points on L. Then p.L/ D .p01 W p02 W p03 W p12 W p13 W p23/ 2 P 5; pij defD ˇ ˇ ˇ ˇ xi xj yi yj ˇ ˇ ˇ ˇ ; depends only on L. The pij are called the Pl¨ucker coordinates of L, after Pl¨ucker (18011868). In terms of exterior algebras, write e0, e1, e2, e3 for the canonical basis for k4, so that x, regarded as a point of k4 is P xi ei , and y D P yi ei ; then V2 k4 is a 6-dimensional vector space with basis ei ^ej , 0 i < j 3, and x^y D P pij ei ^ej with pij given by the above formula. We define pij for all i; j , 0 i; j 3 by the same formula — thus pij D pj i . LEMMA 9.40. The line L can be recovered from p.L/ as follows: L D f.P j aj p1j W P j aj p0j W P j aj p2j W P PROOF. Let QL be the cone over L in k4 — it is a two-dimensional subspace of k4 — and let x D .x0; x1; x2; x3/ and y D .y0; y1; y2; y3/ be two linearly independent vectors in QL. Then QL D ff .y/x f .x/y j f W k4 ! k linearg: j aj p3j / j .a0 W a1 W a2 W a3/ 2 P 3g: Write f D P aj Xj ; then f .y/x f .x/y D .P aj p0j ; P aj p1j ; P aj p2j ; P aj p3j /: d. Lines on surfaces 209 LEMMA 9.41. The point p.L/ lies on the quadric ˘ P 5 defined by the equation X01X23 X02X13 C X03X12 D 0: PROOF. This can be verified by direct calculation, or by using that x0 x1 x2 x3 y0 y1 y2 y3 x0 x1 x2 x3 y0 y1 y2 y3 .p01p23 p02p13 C p03p12/ (expansion in terms of 2 2 minors). LEMMA 9.42. Every point of ˘ is of the form p.L/ for a unique line L. PROOF. Assume p03 ¤ 0; then the line through the points .0 W p01 W p02 W p03/ and .p03 W p13 W p23 W 0/ has Pl¨ucker coordinates .p01p03 W p02p03 W p2 03 W p01p23 p02p13 „ … ƒ‚ p03p12 W p03p13 W p03p23/ D .p01 W p02 W p03 W p12 W p13 W p23/: A similar construction works when one of the other coordinates is nonzero, and this way we get inverse maps. Thus we have a canonical one-to-one correspondence flines in P 3g $ fpoints on ˘ gI 3 with the points of an algebraic variety. We that is, we have identified the set of lines in P may now use the methods of algebraic geometry to study the set. (This is a special case of the Grassmannians discussed in 6.) We next consider the set of homogeneous polynomials of degree m in 4 variables, F .X0; X1; X2; X3/ D X ai0i1i2i3X i0 0 : : : X i3 3 : i0Ci1Ci2Ci3Dm LEMMA 9.43. The set of homogeneous polynomials of degree m in 4 variables is a vector space of dimension 3Cm m PROOF. See the footnote p. 141. Let D 3Cm m 1 D .mC1/.mC2/.mC3/ as the projective space attached to the vector space of homogeneous polynomials of degree m in 4 variables (p. 145). Then we have a surjective map 1, and regard P 6 ! fsurfaces of degree m in P 3g; P .: : : W ai0i1i2i3 W : : :/ 7! V .F /; F D X ai0i1i2i3X i0 0 X i1 1 X i2 2 X i3 3 : The map is not quite injective — for example, X 2Y and XY 2 define the same surface — as being (possibly but nevertheless, we can (somewhat loosely) think of the points of P degenerate) surfaces of degree m in P 3. Let m ˘ P P 5 P be the set of pairs .L; F / consisting of a line L in P 3 lying on the surface F .X0; X1; X2; X3/ D 0. 210 9. REGULAR MAPS AND THEIR FIBRES THEOREM 9.44. The set m is an irreducible cl
osed subset of ˘ P projective variety. The dimension of m is m.mC1/.mC5/ C 3. 6 ; it is therefore a EXAMPLE 9.45. For m D 1; m is the set of pairs consisting of a plane in P 3 and a line on the plane. The theorem says that the dimension of 1 is 5. Since there are 13 planes in P 3, and each has 12 lines on it, this seems to be correct. PROOF. We first show that m is closed. Let p.L/ D .p01 W p02 W : : :/ F D X ai0i1i2i3X i0 0 X i3 3 : From 9.40 we see that L lies on the surface F .X0; X1; X2; X3/ D 0 if and only if F .P bj p0j W P bj p1j W P bj p2j W P bj p3j / D 0, all .b0; : : : ; b3/ 2 k4: Expand this out as a polynomial in the bj with coefficients polynomials in the ai0i1i2i3 and pij . Then F .:::/ D 0 for all b 2 k4 if and only if the coefficients of the polynomial are all zero. But each coefficient is of the form P .: : : ; ai0i1i2i3; : : : I p01; p02 W : : :/ with P homogeneous separately in the a’s and p’s, and so the set is closed in ˘ P the discussion in 6.51). (cf. It remains to compute the dimension of m. We shall apply Proposition 9.11 to the projection map .L; F / m ˘ P L ' ˘: For L 2 ˘ , '1.L/ consists of the homogeneous polynomials of degree m such that L V .F / (taken up to nonzero scalars). After a change of coordinates, we can assume that L is the line X0 D 0 X1 D 0; i.e., L D f.0; 0; ; /g. Then L lies on F .X0; X1; X2; X3/ D 0 if and only if X0 or X1 occurs in each nonzero monomial term in F , i.e., F 2 '1.L/ ” ai0i1i2i3 D 0 whenever i0 D 0 D i1: Thus '1.L/ is a linear subspace of P ; in particular, it is irreducible. We now compute its dimension. Recall that F has C 1 coefficients altogether; the number with i0 D 0 D i1 is m C 1, and so '1.L/ has dimension .m C 1/.m C 2/.m C 3/ 6 1 .m C 1/ D m.m C 1/.m C 5/ 6 1: We can now deduce from 9.11 that m is irreducible and that dim.m/ D dim.˘ / C dim.'1.L// D m.m C 1/.m C 5/ 6 C 3; as claimed. d. Lines on surfaces 211 Now consider the other projection. By definition 1.F / D fL j L lies on V .F /g: EXAMPLE 9.46. Let m D 1. Then D 3 and dim 1 D 5. The projection W 1 ! P 3 is surjective (every plane contains at least one line), and (9.9) tells us that dim 1.F / 2. In fact of course, the lines on any plane form a 2-dimensional family, and so 1.F / D 2 for all F . THEOREM 9.47. When m > 3, the surfaces of degree m containing no line correspond to an open subset of P . PROOF. We have dim m dim P D m.m C 1/.m C 5/ 6 C 3 .m C 1/.m C 2/.m C 3/ 6 C 1 D 4 .m C 1/: Therefore, if m > 3, then dim m < dim P P . This proves the claim. , and so .m/ is a proper closed subvariety of We now look at the case m D 2. Here dim m D 10, and D 9, which suggests that should be surjective and that its fibres should all have dimension 1. We shall see that this is correct. A quadric is said to be nondegenerate if it is defined by an irreducible polynomial of degree 2. After a change of variables, any nondegenerate quadric will be defined by an equation This is just the image of the Segre mapping (see 6.26) XW D Y Z: .a0 W a1/, .b0 W b1/ 7! .a0b0 W a0b1 W a1b0 W a1b1/ W P 1 P 1 ! P 3: There are two obvious families of lines on P vertical family; each is parametrized by P two families of lines on the quadric: 1 P 1, namely, the horizontal family and the 1, and so is called a pencil of lines. They map to t0X D t1Z t0Y D t1W and t0X D t1Y t0Z D t1W: Since a degenerate quadric is a surface or a union of two surfaces, we see that every quadric surface contains a line, that is, that W 2 ! P 9 is surjective. Thus (9.9) tells us that all the fibres have dimension 1, and the set where the dimension is > 1 is a proper closed subset. In fact the dimension of the fibre is > 1 exactly on the set of reducible F ’s, which we know to be closed (this was a homework problem in the original course). It follows from the above discussion that if F is nondegenerate, then 1.F / is isomorphic to the disjoint union of two lines, 1.F / P 1. Classically, one defines a regulus to be a nondegenerate quadric surface together with a choice of a pencil of lines. One can show that the set of reguli is, in a natural way, an algebraic variety R, and that, over the set of nondegenerate quadrics, factors into the composite of two regular maps: 2 1.S / D pairs, .F; L/ with L on set of reguli; D set of nondegenerate quadrics. 212 9. REGULAR MAPS AND THEIR FIBRES 1/, The fibres of the top map are connected, and of dimension 1 (they are all isomorphic to P and the second map is finite and two-to-one. Factorizations of this type occur quite generally (see the Stein factorization theorem, 8.64). We now look at the case m D 3. Here dim 3 D 19; D 19 W we have a map W 3 ! P THEOREM 9.48. The set of cubic surfaces containing exactly 27 lines corresponds to an 19; the remaining surfaces either contain an infinite number of lines or a open subset of P nonzero finite number 27. 19: EXAMPLE 9.49. (a) Consider the Fermat surface C X 3 2 Let be a primitive cube root of one. There are the following lines on the surface, 0 i; j 2: X 3 0 X0 C i X1 D 0 X2 C j X3 D 0 X0 C i X2 D 0 X1 C j X3 D 0 X0 C i X3 D 0 X1 C j X2 D 0: There are three sets, each with nine lines, for a total of 27 lines. (b) Consider the surface X1X2X3 D X 3 0 : In this case, there are exactly three lines. To see this, look first in the affine space where X0 ¤ 0 — here we can take the equation to be X1X2X3 D 1. A line in A 3 can be written in parametric form Xi D ai t C bi , but a direct inspection shows that no such line lies on the surface. Now look where X0 D 0, that is, in the plane at infinity. The intersection of the surface with this plane is given by X1X2X3 D 0 (homogeneous coordinates), which is the union of three lines, namely, X1 D 0; X2 D 0; X3 D 0: Therefore, the surface contains exactly three lines. (c) Consider the surface X 3 1 C X 3 2 D 0: Here there is a pencil of lines: t0X1 D t1X0 t0X2 D t1X0: (In the affine space where X0 ¤ 0, the equation is X 3 C Y 3 D 0, which contains the line X D t, Y D t, all t:/ We now discuss the proof of Theorem 9.48. If W 3 ! P 19 were not surjective, then 19, and the nonempty fibres would all have .3/ would be a proper closed subvariety of P dimension 1 (by 9.9), which contradicts two of the above examples. Therefore the map is 19 where the fibres have dimension 0; outside surjective, and there is an open subset U of P U , the fibres have dimension > 0. Given that every cubic surface has at least one line, it is not hard to show that there is an open subset U 0 where the cubics have exactly 27 lines (see Reid 1988, pp. 106–110).5 In fact, U 0 can be taken to be the set of nonsingular cubics. According to 8.26, the restriction of to 1.U / is finite, and so we can apply 8.40 to see that all cubics in U U 0 have fewer than 27 lines. 5Reid, Miles Undergraduate algebraic geometry. LMS Student Texts, 12, CUP, Cambridge, 1988. According to Reid, p. 126, every adult algebraic geometer knows the proof that every cubic contains a line. e. Bertini’s theorem 213 REMARK 9.50. The twenty-seven lines on a cubic surface were discovered in 1849 by Salmon and Cayley, and have been much studied — see A. Henderson, The Twenty-Seven Lines Upon the Cubic Surface, Cambridge University Press, 1911. For example, it is known that the group of permutations of the set of 27 lines preserving intersections (that is, such that L \ L0 ¤ ; ” .L/ \ .L0/ ¤ ;/ is isomorphic to the Weyl group of the root system of a simple Lie algebra of type E6, and hence has 25920 elements. It is known that there is a set of 6 skew lines on a nonsingular cubic surface V . Let L and L0 be two skew lines. Then “in general” a line joining a point on L to a point on L0 will meet the surface in exactly one further point. In this way one obtains an invertible regular 1 to an open subset of V , and hence V is birationally map from an open subset of P 2. equivalent to P 1 P e. Bertini’s theorem n be a nonsingular projective variety. The hyperplanes H in P n form a projective n_ (the “dual” projective space). The set of hyperplanes H not containing X and n_. If dim.X/ 2, then the Let X P space P such that X \ H is nonsingular, form an open subset of P intersections X \ H are connected. f. Birational classification Recall that two varieties V and W are birationally equivalent if k.V / k.W /. This means that the varieties themselves become isomorphic once a proper closed subset has been removed from each (3.36). The main problem of birational algebraic geometry is to classify algebraic varieties up to birational equivalence by finding a particularly good representative in each equivalence class. For curves this is easy: in each birational equivalence class there is exactly one nonsingular projective curve (up to isomorphism). More precisely, the functor V k.V / is a contravariant equivalence from the category of nonsingular projective algebraic curves over k and dominant maps to the category of fields finitely generated and of transcendence degree 1 over k. For surfaces, the problem is already much more difficult because many surfaces, even projective and nonsingular, will have the same function field. For example, every blow-up of a point on a surface produces a birationally equivalent surface. A nonsingular projective surface is said to be minimal if it cannot be obtained from another such surface by blowing up. The main theorem for surfaces (Enriques 1914, Kodaira 1966) says that a birational equivalence class contains either (a) a unique minimal surface, or (b) a surface of the form C P 1 for a unique nonsingular projective curve C . In higher dimensions, the problem becomes very involved, although much progress has been made — see Wikipedia: MINIMAL MODEL PROGRAM. Exercises 9-1. Let G be a connected group variety, and consider an action of G on a variety V , i.e., a regular map G V ! V such that .gg0/v D g.g0v/ for all g; g0 2 G and v 2 V . Show that 214 9. REGULAR MAPS AND THEIR FIBRES each orbit O D Gv of G is open in its closure NO, and that NO X O is a union of orbits of
strictly lower dimension. Deduce that each orbit is a nonsingular subvariety of V , and that there exists at least one closed orbit. 9-2. Let G D GL2 D V , and let G act on V by conjugation. According to the theory of Jordan canonical forms, the orbits are of three types: (a) Characteristic polynomial X 2 C aX C b; distinct roots. (b) Characteristic polynomial X 2 C aX C b; minimal polynomial the same; repeated roots. (c) Characteristic polynomial X 2 C aX C b D .X ˛/2; minimal polynomial X ˛. For each type, find the dimension of the orbit, the equations defining it (as a subvariety of V ), the closure of the orbit, and which other orbits are contained in the closure. (You may assume, if you wish, that the characteristic is zero. Also, you may assume the following (fairly difficult) result: for any closed subgroup H of an group variety G, G=H has a natural structure of an algebraic variety with the following properties: G ! G=H is regular, and a map G=H ! V is regular if the composite G ! G=H ! V is regular; dim G=H D dim G dim H .) [The enthusiasts may wish to carry out the analysis for GLn.] 9-3. Find 3d 2 lines on the Fermat projective surface ; d 3; .p; d / D 1; p the characteristic. 9-4. (a) Let 'W W ! V be a quasi-finite dominant regular map of irreducible varieties. Show that there are open subsets U 0 and U of W and V such that '.U 0/ U and 'W U 0 ! U is finite. (b) Let G be a group variety acting transitively on irreducible varieties W and V , and let 'W W ! V be G-equivariant regular map satisfying the hypotheses in (a). Then ' is finite, and hence proper. Solutions to the exercises 1-1 Use induction on n. For n D 1, use that a nonzero polynomial in one variable has only finitely many roots (which follows from unique factorization, for example). Now suppose n > 1 and write f D P gi X i n with each gi 2 kŒX1; : : : ; Xn1. If f is not the zero polynomial, then some gi is not the zero polynomial. Therefore, by induction, there exist .a1; : : : ; an1/ 2 kn1 such that f .a1; : : : ; an1; Xn/ is not the zero polynomial. Now, by the degree-one case, there exists a b such that f .a1; : : : ; an1; b/ ¤ 0. 1-2 .X C 2Y; Z/; Gaussian elimination (to reduce the matrix of coefficients to row echelon form); .1/, unless the characteristic of k is 2, in which case the ideal is .X C 1; Z C 1/. 2-1 W D Y -axis, and so I.W / D .X/. Clearly, .X 2; XY 2/ .X/ rad.X 2; XY 2/ and rad..X// D .X/. On taking radicals, we find that .X/ D rad.X 2; XY 2/. 2-2 The d d minors of a matrix are polynomials in the entries of the matrix, and the set of matrices with rank r is the set where all .r C 1/ .r C 1/ minors are zero. 2-3 Clearly V D V .Xn X n 1 ; : : : ; X2 X 2 1 /. The map Xi 7! T i W kŒX1; : : : ; Xn ! kŒT 1 ! V .] induces an isomorphism kŒV ! kŒT . [Hence t 7! .t; : : : ; t n/ is an isomorphism of affine varieties A 2-4 We use that the prime ideals are in one-to-one correspondence with the irreducible closed subsets Z of A 2. For such a set, 0 dim Z 2. Case dim Z D 2. Then Z D A Case dim Z D 1. Then Z ¤ A 2, and the corresponding ideal is .0/. 2, and so I.Z/ contains a nonzero polynomial f .X; Y /. If I.Z/ ¤ .f /, then dim Z D 0 by (2.64, 2.62). Hence I.Z/ D .f /. Case dim Z D 0. Then Z is a point .a; b/ (see 2.63), and so I.Z/ D .X a; Y b/. 2-6 The statement Homkalgebras.A ˝ Q k/ ¤ ; can be interpreted as saying that a certain set of polynomials has a zero in k.6 If the polynomials have a common zero in C, then the ideal they generate in CŒX1; : : : does not contain 1. A fortiori, the ideal they generate in QŒX1; : : : does not contain 1, and so the Nullstellensatz (2.11) implies that the polynomials have a common zero in k. Q k; B ˝ 2-7 Regard HomA.M; N / as an affine space over k; the elements not isomorphisms are the zeros of a polynomial; because M and N become isomorphic over kal, the polynomial is not identically zero; therefore it has a nonzero in k (Exercise 1-1). 6Choose bases for A and B as Q-vector spaces. Now a linear map from A to B is given by a matrix M . The condition on the coefficients of the matix for the map to be a homomorphism of algebras is polynomial. 215 216 SOLUTIONS TO THE EXERCISES 1 ! A 3-1 A map ˛W A 1 is continuous for the Zariski topology if the inverse images of finite sets are finite, whereas it is regular only if it is given by a polynomial P 2 kŒT , so it is easy to give examples, e.g., any map ˛ such that ˛1.point/ is finite but arbitrarily large. 3-3 The image omits the points on the Y -axis except for the origin. The complement of the image is not dense, and so it is not open, but any polynomial zero on it is also zero at .0; 0/, and so it not closed. 3-4 Let i be an element of k with square 1. The map .x; y/ 7! .x C iy; x iy/ from the circle to the hyperbola has inverse .x; y/ 7! ..x C y/=2; .x y/=2i/. The k-algebra kŒX; Y =.XY 1/ ' kŒX; X 1, which is not isomorphic to kŒX (too many units). 3-5 No, because both C1 and 1 map to .0; 0/. The map on rings is kŒx; y ! kŒT ; x 7! T 2 1; y 7! T .T 2 1/; which is not surjective (T is not in the image). 1. Then f jU0 D P .X/ 2 kŒX, where X is the regular function 5-1 Let f be regular on P .a0W a1/ 7! a1=a0W U0 ! k, and f jU1 D Q.Y / 2 kŒY , where Y is .a0W a1/ 7! a0=a1. On U0 \ U1, X and Y are reciprocal functions. Thus P .X / and Q.1=X/ define the same function on U0 \ U1 D A 1 X f0g. This implies that they are equal in k.X/, and must both be constant. V / D Q .Vi ; Vi / — to give a regular function on FVi is the same 5-2 Note that .V; as to give a regular function on each Vi (this is the “obvious” ringed space structure). Thus, if V is affine, it must equal Specm.Q Ai /, where Ai D .Vi ; Vi /, and so V D FSpecm.Ai / (use the description of the ideals in A B on in Section 1a). Etc.. O O O 5-5 Let H be an algebraic subgroup of G. By definition, H is locally closed, i.e., open in its Zariski closure NH . Assume first that H is connected. Then NH is a connected algebraic group, and it is a disjoint union of the cosets of H . It follows that H D NH . In the general case, H is a finite disjoint union of its connected components; as one component is closed, they all are. 4-1 (b) The singular points are the common solutions to 8 < : 4X 3 2XY 2 D 0 4Y 3 2X 2Y D 0 X 4 C Y 4 X 2Y 2 D 0: H) X D 0 or Y 2 D 2X 2 H) Y D 0 or X 2 D 2Y 2 Thus, only .0; 0/ is singular, and the variety is its own tangent cone. 4-2 Directly from the definition of the tangent space, we have that Ta.V \ H / Ta.V / \ Ta.H /. As dim Ta.V \ H / dim V \ H D dim V 1 D dim Ta.V / \ Ta.H /; we must have equalities everywhere, which proves that a is nonsingular on V \ H . (In particular, it can’t lie on more than one irreducible component.) The surface Y 2 D X 2 C Z is smooth, but its intersection with the X-Y plane is singular. No, P needn’t be singular on V \ H if H TP .V / — for example, we could have H V or H could be the tangent line to a curve. SOLUTIONS TO THE EXERCISES 217 4-4 We can assume V and W to affine, say I.V / D a kŒX1; : : : ; Xm I.W / D b kŒXmC1; : : : ; XmCn: If a D .f1; : : : ; fr / and b D .g1; : : : ; gs/, then I.V W / D .f1; : : : ; fr ; g1; : : : ; gs/. Thus, T.a;b/.V W / is defined by the equations .df1/a D 0; : : : ; .dfr /a D 0; .dg1/b D 0; : : : ; .dgs/b D 0; which can obviously be identified with Ta.V / Tb.W /. 4-5 Take C to be the union of the coordinate axes in A irreducible, then this is more difficult. . . ) n. (Of course, if you want C to be 4-6 A matrix A satisfies the equations if and only if .I C "A/tr J .I C "A/ D I Atr J C J A D 0: Such an A is of the form M N P Q with M; N; P; Q n n-matrices satisfying N tr D N; P tr D P; M tr D Q. The dimension of the space of A’s is therefore n.n C 1/ 2 (for N ) C n.n C 1/ 2 (for P ) C n2 (for M; Q) D 2n2 C n: 4-7 Let C be the curve Y 2 D X 3, and consider the map A 1 ! C , t 7! .t 2; t 3/. The corresponding map on rings kŒX; Y =.Y 2/ ! kŒT is not an isomorphism, but the map on the geometric tangent cones is an isomorphism. 4-8 The singular locus Vsing has codimension 2 in V , and this implies that V is normal. [Idea of the proof: let f 2 k.V / be integral over kŒV , f … kŒV , f D g= h, g; h 2 kŒV ; for any P 2 V .h/ X V .g/, 4-9 No! Let a D .X 2Y /. Then V .a/ is the union of the X and Y axes, and I V .a/ D .XY /. For a D .a; b/, P is not integrally closed, and so P is singular.] O .dX 2Y /a D 2ab.X a/ C a2.Y b/ .dXY /a D b.X a/ C a.Y b/. If a ¤ 0 and b D 0, then the equations .dX 2Y /a D a2Y D 0 .dXY /a D aY D 0 have the same solutions. 6-1 Let P D .a W b W c/, and assume c ¤ 0. Then the tangent line at / is @F @X P X C @F @Y P Y @F @X P C a c @F @Y P b c Z D 0: 218 SOLUTIONS TO THE EXERCISES Now use that, because F is homogeneous, F .a; b; c/ D 0 H) @F @X P a C @F @Y P C @F @Z P c D 0. (This just says that the tangent plane at .a; b; c/ to the affine cone F .X; Y; Z/ D 0 passes through the origin.) The point at 1 is .0 W 1 W 0/, and the tangent line is Z D 0, the line at 1. [The line at 1 meets the cubic curve at only one point instead of the expected 3, and so the line at 1 “touches” the curve, and the point at 1 is a point of inflexion.] 6-2 The equation defining the conic must be irreducible (otherwise the conic is singular). After a linear change of variables, the equation will be of the form X 2 C Y 2 D Z2 (this is proved in calculus courses). The equation of the line in aX C bY D cZ, and the rest is easy. [Note that this is a special case of Bezout’s theorem (6.37) because the multiplicity is 2 in case (b).] 6-3 (a) The ring kŒX; Y; Z=.Y X 2; Z X 3/ D kŒx; y; z D kŒx ' kŒX; which is an integral domain. Therefore, .Y X 2; Z X 3/ is a radical ideal. (b) The polynomial F D Z XY D .Z X 3/ X.Y X 2/ 2 I.V / and F D ZW XY . If ZW XY D .Y W X 2/f C .ZW 2 X 3/g; then, on equating terms of degree 2, we would find ZW XY D a.Y W X 2/; which is false. 6-4 Let P D .a0W : : : W an/ and Q D .b0W : : : W bn/ be two points of P that the hyperplane LcW Pci Xi D 0 pass through P and not through Q is that n, n 2. The condi
tion Pai ci D 0; Pbi ci ¤ 0: The .n C 1/-tuples .c0; : : : ; cn/ satisfying these conditions form a nonempty open subset of the hyperplane H W P ai Xi D 0 in A nC1. On applying this remark to the pairs .P0; Pi /, we find that the .n C 1/-tuples c D .c0; : : : ; cn/ such that P0 lies on the hyperplane Lc but not P1; : : : ; Pr form a nonempty open subset of H . 6-5 The subset C D f.a W b W c/ j a ¤ 0; b ¤ 0g [ f.1 W 0 W 0/g 2 is not locally closed. Let P D .1 W 0 W 0/. If the set C were locally closed, then P 2 such that U \ C is closed. When we look in of P would have an open neighbourhood U in P U0, P becomes the origin, and C \ U0 D .A 2 X fX-axisg/ [ foriging. 2 a finite number of The open neighbourhoods U of P are obtained by removing from A curves not passing through P . It is not possible to do this in such a way that U \ C is closed in U (U \ C has dimension 2, and so it can’t be a proper closed subset of U ; we can’t have U \ C D U because any curve containing all nonzero points on X-axis also contains the origin). SOLUTIONS TO THE EXERCISES 219 6-6 Let Pcij Xij D 0 be a hyperplane containing the image of the Segre map. We then have Pcij ai bj D 0 for all a D .a0; : : : ; am/ 2 kmC1 and b D .b0; : : : ; bn/ 2 knC1. In other words, aC bt D 0 for all a 2 kmC1 and b 2 knC1, where C is the matrix .cij /. This equation shows that aC D 0 for all a, and this implies that C D 0. 7-2 Define f .v/ D h.v; Q/ and g.w/ D h.P; w/, and let ' D h .f ı p C g ı q/. Then '.v; Q/ D 0 D '.P; w/, and so the rigidity theorem (7.35) implies that ' is identically zero. 8-2 For example, consider .A 1 X f1g/ ! A 1 x7!xn ! A 1 for n > 1 an integer prime to the characteristic. The map is obviously quasi-finite, but it is not finite because it corresponds to the map of k-algebras X 7! X nW kŒX ! kŒX; .X 1/1 which is not finite (the elements 1=.X 1/i , i 1, are linearly independent over kŒX, and so also over kŒX n). 8-3 Assume that V is separated, and consider two regular maps f; gW Z W . We have to show that the set on which f and g agree is closed in Z. The set where ' ı f and ' ı g agree is closed in Z, and it contains the set where f and g agree. Replace Z with the set where ' ı f and ' ı g agree. Let U be an open affine subset of V , and let Z0 D .' ı f /1.U / D .' ı g/1.U /. Then f .Z0/ and g.Z0/ are contained in '1.U /, which is an open affine subset of W , and is therefore separated. Hence, the subset of Z0 on which f and g agree is closed. This proves the result. [Note that the problem implies the following statement: if 'W W ! V is a finite regular map and V is separated, then W is separated.] 8-4 Let V D A n, and let W be the subvariety of A n A 1 defined by the polynomial Qn i D1.X Ti / D 0: The fibre over .t1; : : : ; tn/ 2 A union of the linear subspaces defined by the equations n is the set of roots of Q.X ti /. Thus, Vn D A n; Vn1 is the Ti D Tj ; 1 i; j n; i ¤ j I Vn2 is the union of the linear subspaces defined by the equations Ti D Tj D Tk; 1 i; j; k n; i; j; k distinct, and so on. 9-1 Consider an orbit O D Gv. The map g 7! gvW G ! O is regular, and so O contains an open subset U of NO (9.7). If u 2 U , then gu 2 gU , and gU is also a subset of O which is open in NO (because P 7! gP W V ! V is an isomorphism). Thus O, regarded as a topological subspace of NO, contains an open neighbourhood of each of its points, and so must be open in NO. We have shown that O is locally closed in V , and so has the structure of a subvariety. From (4.37), we know that it contains at least one nonsingular point P . But then gP is nonsingular, and every point of O is of this form. From set theory, it is clear that NO X O is a union of orbits. Since NO X O is a proper closed subset of NO, all of its subvarieties must have dimension < dim NO D dim O. Let O be an orbit of lowest dimension. The last statement implies that O D NO. 9-2 An orbit of type (a) is closed, because it is defined by the equations Tr.A/ D a; det.A/ D b; (as a subvariety of V ). It is of dimension 2, because the centralizer of 0 0 An orbit of type (b) is of dimension 2, but is not closed: it is defined by the equations , which has dimension 2. ˛ 0 0 ˇ , ˛ ¤ ˇ, is Tr.A/ D a; det.A/ D b root of X 2 C aX C b. An orbit of type (c) is closed of dimension 0: it is defined by the equation A D ˛ 0 0 ˛ . An orbit of type (b) contains an orbit of type (c) in its closure. 9-3 Let be a primitive d th root of 1. Then, for each i; j , 1 i; j d , the following equations define lines on the surface X0 C i X1 D 0 X2 C j X3 D 0 X0 C i X2 D 0 X1 C j X3 D 0 X0 C i X3 D 0 X1 C j X2 D 0: There are three sets of lines, each with d 2 lines, for a total of 3d 2 lines. 9-4 (a) Compare the proof of Theorem 9.9. (b) Use the transitivity, and apply Proposition 8.26. 220 algebra affine, 65 finite, 13 finitely generated, 13 algebraically dependent, 35 algebraically independent, 35 A analytic space, 169 axiom n, 37 separation, 101 base change, 112 basis transcendence, 35 birationally equivalent, 73, 116 boundary, 55 codimension, 55 complete intersection ideal-theoretic, 79 local, 79 set-theoretic, 79 component of a function, 50 cone, 131 affine over a set, 132 content of a polynomial, 24 convergent, 60 Cramer’s rule, 26 curve, 55 elliptic, 38, 130, 135 degree of a hypersurface, 150 of a map, 184 of a projective variety, 152 derivation, 91 desingularization, 193 differential, 87 dimension, 74, 113 Index of a topological space, 54 of an algebraic set, 54 pure, 55, 114 direct limit, 22 direct system, 22 directed set, 22 discrete valuation ring, 86 divisor, 177 effective, 177 locally principal, 176 positive, 177 prime, 176 principal, 178 support of, 177 domain factorial, 23 integrally closed, 28 normal, 28 unique factorization, 23 element integral over a ring, 26 irreducible, 23 prime, 23 F .A/, 18 faithfully flat, 201 fibre, 112 field of rational functions, 50, 113 flat, 201, 203 form leading, 83 function holomorphic, 169 rational, 63 regular, 49, 61, 100 function field, 50, 113 generate, 13 germ 221 of a function, 60 graph of a regular map, 110 group symplectic, 98 group variety, 109 homogeneous, 137 homomorphism finite, 13 of algebras, 13 hypersurface, 50, 142 hypersurface section, 142 ideal, 13 generated by a subset, 14 graded, 131, 133 homogeneous, 131 maximal, 14 prime, 14 radical, 42 immersion, 103 closed, 72, 103 open, 103 integral closure, 28 integral domain, 13 integrally closed, 28 irreducible components, 47 isolated in its fibre, 186 isomorphic locally, 97 .p/, 186 lemma Gauss’s, 24 Nakayama’s, 16 prime avoidance, 77 Zariski’s, 41 linearly equivalent, 178 local equation, 176 local ring regular, 16 local system of parameters, 120 manifold complex, 99 differentiable, 99 topological, 99 map affine, 194 bilinear, 32 birational, 116, 188 dominant, 51, 72, 115 ´etale, 116, 119 finite, 51, 74, 178, 182 Frobenius, 69 proper, 159 quasi-finite, 51, 181, 182 rational, 115 regular, 50 Segre, 144 separable, 123, 184 Veronese, 141 minimal surface, 213 morphism of affine algebraic varieties, 65 of ringed spaces, 64 mP , 42 multiplicity, 206 of a point, 84 n-fold, 55 neighbourhood ´etale, 120 nilpotent, 42 node, 84 nondegenerate quadric, 211 normalization, 175, 176 open affine, 71 open subset basic, 50 principal, 50 pencil of lines, 211 Picard group, 178 point factorial, 176 multiple, 87 nonsingular, 82, 87 normal, 173 ordinary multiple, 84 singular, 87 smooth, 82, 87 with coordinates in a ring, 123 polynomial Hilbert, 152 homogeneous, 129 primitive, 24 222 prevariety algebraic, 99 separated, 101 product fibred, 112 of algebraic varieties, 108 of objects, 105 tensor, 34 projection with centre, 144 radical of an ideal, 42 rational map, 115 real locus, 38 regular map, 100 of affine algebraic varieties, 65 of algebraic sets, 50 regulus, 211 resolution of singularities, 193 resultant, 162 ring associated graded, 93 coordinate, 48 graded, 133 local, 16 noetherian, 16 normal, 36 of dual numbers, 89 reduced, 42 ringed space, 60 section of a sheaf, 60 semisimple group, 97 Lie algebra, 97 separable degree, 185 set (projective) algebraic, 130 constructible, 197 sheaf of algebras, 59 singular locus, 83 Spm.A/, 66 spm.A/, 66 stalk, 60 subring, 13 subset algebraic, 37 analytic, 169 multiplicative, 17 subspace locally closed, 103 subvariety, 103 closed, 70 open affine, 99 surface, 55 T1 space, 46 tangent cone, 83, 93 geometric, 83, 93 tangent space, 82, 87 tensor product of modules, 33 theorem Bezout’s , 150 Chinese Remainder, 15 going-up, 31 Hilbert basis, 39 Hilbert Nullstellensatz, 40 Noether normalization, 53 Stein factorization, 191 strong Hilbert Nullstellensatz, 42 Zariski’s main, 186 topological space irreducible , 46 noetherian, 46 quasicompact, 46 topology ´etale, 121 Zariski, 40, 132 variety abelian, 166 affine algebraic, 65 algebraic, 101 Cohen-Macaulay, 207 complete, 157 factorial, 176 flag, 150 Grassmann, 147 group, 109 normal, 173 projective, 129 quasi-affine, 102 quasi-projective, 129 rational, 126 unirational, 126 zero set, 37 223
“copy” of X we work in, the rest of ˜f has to follow what f does. Proof. First we show it is open. Let y be such that ˜f1(y) = ˜f2(y). Then there is an evenly covered open neighbourhood U ⊆ X of f (y). Let ˜U be such that ˜f1(y) ∈ ˜U , p( ˜U ) = U and p\| ˜U : ˜U → U is a homeomorphism. Let V = ˜f −1 2 ( ˜U ). We will show that ˜f1 = ˜f2 on V . 1 ( ˜U ) ∩ ˜f −1 Indeed, by construction p\| ˜U ◦ ˜f1\|V = p\| ˜U ◦ ˜f2\|V . Since p\| ˜U is a homeomorphism, it follows that ˜f1\|V = ˜f2\|V . Now we show S is closed. Suppose not. Then there is some y ∈ ¯S \ S. So ˜f1(y) = ˜f2(y). Let U be an evenly covered neighbourhood of f (y). Let p−1(U ) = Uα. Let ˜f1(y) ∈ Uβ and ˜f2(y) ∈ Uγ, where β = γ. Then V = ˜f −1 2 (Uγ) is an open neighbourhood of y, and hence intersects S by deﬁnition of closure. So there is some x ∈ V such that ˜f1(x) = ˜f2(x). But ˜f1(x) ∈ Uβ and ˜f2(x) ∈ Uγ, and hence Uβ and Uγ have a non-trivial intersection. This is a contradiction. So S is closed. 1 (Uβ) ∩ ˜f −1 We just had a uniqueness statement. How about existence? Given a map, is there guarantee that we can lift it to something? Moreover, if I have ﬁxed a “copy” of X I like, can I also lift my map to that copy? We will later come up with a general criterion for when lifts exist. However, it turns out homotopies can always be lifted. Lemma (Homotopy lifting lemma). Let p : ˜X → X be a covering space, H : Y × I → X be a homotopy from f0 to f1. Let ˜f0 be a lift of f0. Then there exists a unique homotopy ˜H : Y × I → ˜X such that (i) ˜H( · , 0) = ˜f0; and (ii) ˜H is a lift of H, i.e. p ◦ ˜H = H. This lemma might be diﬃcult to comprehend at ﬁrst. We can look at the special case where Y = ∗. Then a homotopy is just a path. So the lemma specializes to Lemma (Path lifting lemma). Let p : ˜X → X be a covering space, γ : I → X a path, and ˜x0 ∈ ˜X such that p(˜x0) = x0 = γ(0). Then there exists a unique path ˜γ : I → ˜X such that (i) ˜γ(0) = ˜x0; and (ii) ˜γ is a lift of γ, i.e. p ◦ ˜γ = γ. This is exactly the picture we were drawing before. We just have to start at a point ˜x0, and then everything is determined because locally, everything upstairs in ˜X is just like X. Note that we have already proved uniqueness. So we just need to prove existence. 24 3 Covering spaces II Algebraic Topology In theory, it makes sense to prove homotopy lifting, and path lifting comes immediately as a corollary. However, the proof of homotopy lifting is big and scary. So instead, we will prove path lifting, which is something we can more easily visualize and understand, and then use that to prove homotopy lifting. Proof. Let Observe that (i) 0 ∈ S. S = {s ∈ I : ˜γ exists on [0, s] ⊆ I}. (ii) S is open. If s ∈ S and ˜γ(s) ∈ Vβ ⊆ p−1(U ), we can deﬁne ˜γ on some small neighbourhood of s by ˜γ(t) = (p\|Vβ )−1 ◦ γ(t) (iii) S is closed. If s ∈ S, then pick an evenly covered neighbourhood U of γ(s). 2 , 1] ∩ S = ∅. So S is 2 ∈ S. So (s − ε Suppose γ((s − ε, s)) ⊆ U . So s − ε closed. Since S is both open and closed, and is non-empty, we have S = I. So ˜γ exists. How can we promote this to a proof of the homotopy lifting lemma? At every point y ∈ Y , we know what to do, since we have path lifting. So ˜H(y, · ) is deﬁned. So the thing we have to do is to show that this is continuous. Steps of the proof are (i) Use compactness of I to argue that the proof of path lifting works on small neighbourhoods in Y . (ii) For each y, we pick an open neighbourhood U of y, and deﬁne a good path lifting on U × I. (iii) By uniqueness of lifts, these path liftings agree when they overlap. So we have one big continuous lifting. With the homotopy lifting lemma in our toolkit, we can start to use it to do stuﬀ. So far, we have covering spaces and fundamental groups. We are now going to build a bridge between these two, and show how covering spaces can be used to reﬂect some structures of the fundamental group. At least one payoﬀ of this work is that we are going to exhibit some non-trivial fundamental groups. We have just showed that we are allowed to lift homotopies. However, what we are really interested in is homotopy as paths. The homotopy lifting lemma does not tell us that the lifted homotopy preserves basepoints. This is what we are going to show. Corollary. Suppose γ, γ : I → X are paths x0 x1 and ˜γ, ˜γ : I → ˜X are lifts of γ and γ respectively, both starting at ˜x0 ∈ p−1(x0). If γ γ as paths, then ˜γ and ˜γ are homotopic as paths. In particular, ˜γ(1) = ˜γ(1). 25 3 Covering spaces II Algebraic Topology Note that if we cover the words “as paths” and just talk about homotopies, then this is just the homotopy lifting lemma. So we can view this as a stronger form of the homotopy lifting lemma. Proof. The homotopy lifting lemma gives us an ˜H, a lift of H with ˜H( · , 0) = ˜γ. cx0 γ H γ cx1 lift c˜x0 ˜γ ˜H ˜γ c˜x1 In this diagram, we by assumption know the bottom of the ˜H square is ˜γ. To show that this is a path homotopy from ˜γ to ˜γ, we need to show that the other edges are c˜x0, c˜x1 and ˜γ respectively. Now ˜H( · , 1) is a lift of H( · , 1) = γ, starting at ˜x0. Since lifts are unique, we must have ˜H( · , 1) = ˜γ. So this is indeed a homotopy between ˜γ and ˜γ. Now we need to check that this is a homotopy of paths. We know that ˜H(0, · ) is a lift of H(0, · ) = cx0. We are aware of one lift of cx0 , namely c˜x0 . By uniqueness of lifts, we must have ˜H(0, · ) = c˜x0 . Similarly, ˜H(1, · ) = c˜x1. So this is a homotopy of paths. So far, our picture of covering spaces is like this: x0 x1 Except. . . is it? Is it possible that we have four copies of x0 but just three copies of x1? This is obviously possible if X is not path connected — the component containing x0 and the one containing x1 are completely unrelated. But what if X is path connected? Corollary. If X is a path connected space, x0, x1 ∈ X, then there is a bijection p−1(x0) → p−1(x1). Proof. Let γ : x0 x1 be a path. We want to use this to construct a bijection between each preimage of x0 and each preimage of x1. The obvious thing to do is to use lifts of the path γ. 26 3 Covering spaces II Algebraic Topology γ x0 x1 Deﬁne a map fγ : p−1(x0) → p−1(x1) that sends ˜x0 to the end point of the unique lift of γ at ˜x0. The inverse map is obtained by replacing γ with γ−1, i.e. fγ−1 . To show this is an inverse, suppose we have some lift ˜γ : ˜x0 ˜x1, so that fγ(˜x0) = ˜x1. Now notice that ˜γ−1 is a lift of γ−1 starting at ˜x1 and ending at ˜x0. So fγ−1 (˜x1) = ˜x0. So fγ−1 is an inverse to fγ, and hence fγ is bijective. Deﬁnition (n-sheeted). A covering space p : ˜X → X of a path-connected space X is n-sheeted if \|p−1(x)\| = n for any (and hence all) x ∈ X. Each covering space has a number associated to it, namely the number of sheets. Is there any number we can assign to fundamental groups? Well, the index of a subgroup might be a good candidate. We’ll later see if this is the case. One important property of covering spaces is the following: Lemma. If p : ˜X → X is a covering map and ˜x0 ∈ ˜X, then p∗ : π1( ˜X, ˜x0) → π1(X, x0) is injective. Proof. To show that a group homomorphism p∗ is injective, we have to show that if p∗(x) is trivial, then x must be trivial. Consider a based loop ˜γ in ˜X. We let γ = p ◦ ˜γ. If γ is trivial, i.e. γ cx0 as paths, the homotopy lifting lemma then gives us a homotopy upstairs between ˜γ and c˜x0. So ˜γ is trivial. As we have originally said, our objective is to make our fundamental group act on something. We are almost there already. Let’s look again at the proof that there is a bijection between p−1(x0) and p−1(x1). What happens if γ is a loop? For any ˜x0 ∈ p−1(x0), we can look at the end point of the lift. This end point may or may not be our original ˜x0. So each loop γ “moves” our ˜x0 to another ˜x 0. However, we are not really interested in paths themselves. We are interested in equivalence classes of paths under homotopy of paths. However, this is ﬁne. If γ is homotopic to γ, then this homotopy can be lifted to get a homotopy between ˜γ and ˜γ. In particular, these have the same end points. So each (based) homotopy class gives a well-deﬁned endpoint. 27 3 Covering spaces II Algebraic Topology ˜x0 ˜x 0 x0 ˜γ γ ˜X p X Now this gives an action of π1(X, x0) on p−1(x0)! Note, however, that this will not be the sort of actions we are familiar with. We usually work with left-actions, where the group acts on the left, but now we will have right-actions, which may confuse you a lot. To see this, we have to consider what happens when we perform two operations one after another, which you shall check yourself. We write this action as ˜x0 · [γ]. When we have an action, we are interested in two things — the orbits, and the stabilizers. This is what the next lemma tells us about. Lemma. Suppose X is path connected and x0 ∈ X. (i) The action of π1(X, x0) on p−1(x0) is transitive if and only if ˜X is path connected. Alternatively, we can say that the orbits of the action correspond to the path components. (ii) The stabilizer of ˜x0 ∈ p−1(x0) is p∗(π1( ˜X, ˜x0)) ⊆ π1(X, x0). (iii) If ˜X is path connected, then there is a bijection p∗(π1( ˜X, ˜x0))\π1(X, x0) → p−1(x0). Note that p∗(π1( ˜X, ˜x0))\π1(X, x0) is not a quotient, but simply the set of cosets. We write it the “wrong way round” because we have right cosets instead of left cosets. Note that this is great! If we can ﬁnd a covering space p and a point x0 such that p−1(x0) is non-trivial, then we immediately know that π1(X, x0) is non-trivial! Proof. (i) If ˜x0, ˜x 0 ∈ p−1(x0), then since ˜X is path connected, we know that there is some ˜γ : ˜x0 ˜x 0. Then we can project this to γ = p ◦ ˜γ. Then γ is a path from x0 x0, i.e. a loop. Then by the deﬁnition of the action, ˜x0 · [γ] = ˜γ(1) = ˜x 0. (ii) Suppose [γ] ∈ stab(˜x0). Then ˜γ is a loop based at ˜x0. So ˜γ deﬁnes [˜γ] ∈ π1( ˜X, ˜x0) and γ = p ◦ ˜γ. (iii) This follows directly from the orbit-stabilizer theorem. 28 3 Covering spaces II Algebraic Topology We now want to use this to determine that the fundamental g
roup of a space is non-trivial. We can be more ambitious, and try to actually ﬁnd π1(X, x0). In the best possible scenario, we would have π1( ˜X, ˜x0) trivial. Then we have a bijection between π1(X, x0) and p−1(x0). In other words, we want our covering space ˜X to be simply connected. Deﬁnition (Universal cover). A covering map p : ˜X → X is a universal cover if ˜X is simply connected. We will look into universal covers in depth later and see what they really are. Corollary. If p : ˜X → X is a universal cover, then there is a bijection : π1(X, x0) → p−1(x0). Note that the orbit-stabilizer theorem does not provide a canonical bijection between p−1(x0) and p∗π1( ˜X, ˜x0)\π1(X, x0). To obtain a bijection, we need to pick a starting point ˜x0 ∈ p−1(x0). So the above bijection depends on a choice of ˜x0. 3.2 The fundamental group of the circle and its applica- tions Finally, we can exhibit a non-trivial fundamental group. We are going to consider the space S1 and a universal covering R. R p−1(1) p 1 S1 Then our previous corollary gives Corollary. There is a bijection π1(S1, 1) → p−1(1) = Z. What’s next? We just know that π1(S1, 1) is countably inﬁnite, but can we work out the group structure? We can, in fact, prove a stronger statement: Theorem. The map : π1(S1, 1) → p−1(1) = Z is a group isomorphism. Proof. We know it is a bijection. So we need to check it is a group homomorphism. The idea is to write down representatives for what we think the elements should be. 29 3 Covering spaces II Algebraic Topology ˜u2 u2 2 1 0 −1 −2 R p S1 Let ˜un : I → R be deﬁned by t → nt, and let un = p ◦ ˜un. Since R is simply connected, there is a unique homotopy class between any two points. So for any [γ] ∈ π1(S1, 1), if ˜γ is the lift to R at 0 and ˜γ(1) = n, then ˜γ ˜un as paths. So [γ] = [un]. To show that this has the right group operation, we can easily see that um · un = ˜um+n, since we are just moving by n + m in both cases. Therefore ([um][un]) = ([um · um]) = m + n = ([um+n]). So is a group isomorphism. What have we done? In general, we might be given a horrible, crazy loop in S1. It would be rather diﬃcult to work with it directly in S1. So we pull it up to the universal covering R. Since R is nice and simply connected, we can easily produce a homotopy that “straightens out” the path. We then project this homotopy down to S1, to get a homotopy from γ to un. It is indeed possible to produce a homotopy directly inside S1 from each loop to some un, but that would be tedious work that involves messing with a lot of algebra and weird, convoluted formulas. With the fundamental group of the circle, we do many things. An immediate application is that we can properly deﬁne the “winding number” of a closed curve. Since C \ {0} is homotopy equivalent to S1, its fundamental group is Z as well. Any closed curve S1 → C \ {0} thus induces a group homomorphism Z → Z. Any such group homomorphism must be of the form t → nt, and the winding number is given by n. If we stare at it long enough, it is clear that this is exactly the number of times the curve winds around the origin. Also, we have the following classic application: Theorem (Brouwer’s ﬁxed point theorem). Let D2 = {(x, y) ∈ R2 : x2 +y2 ≤ 1} be the unit disk. If f : D2 → D2 is continuous, then there is some x ∈ D2 such that f (x) = x. Proof. Suppose not. So x = f (x) for all x ∈ D2. 30 3 Covering spaces II Algebraic Topology f (x) x g(x) We deﬁne g : D2 → S1 as in the picture above. Then we know that g is continuous and g is a retraction from D2 onto S1. In other words, the following composition is the identity: S1 ι D2 idS1 g S1 Then this induces a homomorphism of groups whose composition is the identity: Z ι∗ g∗ Z {0} idZ But this is clearly nonsense! So we must have had a ﬁxed point. But we have a problem. What about D3? Can we prove a similar theorem? Here the fundamental group is of little use, since we can show that the fundamental group of Sn for n ≥ 2 is trivial. Later in the course, we will be able to prove this theorem for higher dimensions, when we have developed more tools to deal with stuﬀ. 3.3 Universal covers We have deﬁned universal covers mysteriously as covers that are simply connected. We have just shown that p : R → S1 is a universal cover. In general, what do universal covers look like? Let’s consider a slightly more complicated example. What would be a universal cover of the torus S1 × S1? An obvious guess would be p × p : R × R → S1 × S1. How can we visualize this? First of all, how can we visualize a torus? Often, we just picture it as the surface of a doughnut. Alternatively, we can see it as a quotient of the square, where we identify the following edges: 31 3 Covering spaces II Algebraic Topology Then what does it feel like to live in the torus? If you live in a torus and look around, you don’t see a boundary. The space just extends indeﬁnitely for ever, somewhat like R2. The diﬀerence is that in the torus, you aren’t actually seeing free space out there, but just seeing copies of the same space over and over again. If you live inside the square, the universe actually looks like this: As we said, this looks somewhat likes R2, but we know that this is not R2, since we can see some symmetry in this space. Whenever we move one unit horizontally or vertically, we get back to “the same place”. In fact, we can move horizontally by n units and vertically by m units, for any n, m ∈ Z, and still get back to the same place. This space has a huge translation symmetry. What is this symmetry? It is exactly Z × Z. We see that if we live inside the torus S1 × S1, it feels like we are actually living in the universal covering space R × R, except that we have an additional symmetry given by the fundamental group Z × Z. Hopefully, you are convinced that universal covers are nice. We would like to say that universal covers always exist. However, this is not always true. Firstly, we should think — what would having a universal cover imply? Suppose X has a universal cover ˜X. Pick any point x0 ∈ X, and pick an evenly covered neighbourhood U in X. This lifts to some ˜U ⊆ ˜X. If we draw a teeny-tiny loop γ around x0 inside U , we can lift this γ to ˜γ in ˜U . But we know that ˜X is simply connected. So ˜γ is homotopic to the constant path. Hence γ is also homotopic to the constant path. So all loops (contained in U ) at x0 are homotopic to the constant path. It seems like for every x0 ∈ X, there is some neighbourhood of x0 that is simply connected. Except that’s not what we just showed above. The homotopy from ˜γ to the constant path is a homotopy in ˜X, and can pass through anything 32 3 Covering spaces II Algebraic Topology in ˜X, not just ˜U . Hence the homotopy induced in X is also a homotopy in X, not a homotopy in U . So U itself need not be simply connected. What we have is a slightly weaker notion. Deﬁnition (Locally simply connected). X is locally simply connected if for all x0 ∈ X, there is some neighbourhood U of x0 such that U is simply connected. As we mentioned, what we actually want is a weaker condition. Deﬁnition (Semi-locally simply connected). X is semi-locally simply connected if for all x0 ∈ X, there is some neighbourhood U of x0 such that any loop γ based at x0 is homotopic to cx0 as paths in X. We have just argued that if a universal cover p : ˜X → X exists, then X is semi-locally simply connected. This is really not interesting, since we don’t care if a space is semi-locally simply connected. What is important is that this is the other direction. We still need one more additional condition: Deﬁnition (Locally path connected). A space X is locally path connected if for any point x and any neighbourhood V of x, there is some open path connected U ⊆ V such that x ∈ U . It is important to note that a path connected space need not be locally path connected. It is an exercise for the reader to come up with a counterexample. Theorem. If X is path connected, locally path connected and semi-locally simply connected, then X has a universal covering. Note that we can alternatively deﬁne a universal covering as a covering space of X that is also a covering space of all other covers of X. If we use this deﬁnition, then we can prove this result easily using Zorn’s lemma. However, that proof is not too helpful since it does not tell us where the universal covering comes from. Instead, we will provide a constructive proof (sketch) that will hopefully be more indicative of what universal coverings are like. Proof. (idea) We pick a basepoint x0 ∈ X for ourselves. Suppose we have a universal covering ˜X. Then this lifts to some ˜x0 in ˜X. If we have any other point ˜x ∈ ˜X, since ˜X should be path connected, there is a path ˜α : ˜x0 ˜x. If we have another path, then since ˜X is simply connected, the paths are homotopic. Hence, we can identify each point in ˜X with a path from ˜x0, i.e. {points of ˜X} ←→ {paths ˜α from ˜x0 ∈ ˜X}/. This is not too helpful though, since we are deﬁning ˜X in terms of things in ˜X. However, by path lifting, we know that paths ˜α from ˜x0 in ˜X biject with paths α from x0 in X. Also, by homotopy lifting, homotopies of paths in X can be lifted to homotopies of paths in ˜X. So we have {points of ˜X} ←→ {paths α from x0 ∈ X}/. So we can produce our ˜X by picking a basepoint x0 ∈ X, and deﬁning ˜X = {paths α : I → X such that α(0) = x0}/. The covering map p : ˜X → X is given by [α] → α(1). One then has to work hard to deﬁne the topology, and then show this is simply connected. 33 3 Covering spaces II Algebraic Topology 3.4 The Galois correspondence Recall that at the beginning, we wanted to establish a correspondence between covering spaces and fundamental groups. We have already established the result that covering maps are injective on π1. Therefore, given a (based) covering space p : ( ˜X, ˜x0) → (X, x0), we can give a subgroup p∗π1( ˜X, ˜x0) ≤ π1(X, x0). It turns out that as long as we deﬁne carefully what we mean for based covering spaces to
be “the same”, this is a one-to-one correspondence — each subgroup corresponds to a covering space. We can have the following table of correspondences: Covering spaces Fundamental group (Based) covering spaces ←→ Subgroups of π1 Number of sheets ←→ Index Universal covers ←→ Trivial subgroup We now want to look at some of these correspondences. Recall that we have shown that π1(X, x0) acts on p−1(x0). However, this is not too interesting an action, since p−1(x0) is a discrete group with no structure. Having groups acting on a cube is fun because the cube has some structure. So we want something more “rich” for π1(X, x0) to act on. We note that we can make π1(X, x0) “act on” the universal cover. How? Recall that in the torus example, each item in the fundamental group corresponds to translating the whole universal covering by some amount. In general, a point on ˜X can be thought of as a path α on X starting from x0. Then it is easy to make a loop γ : x0 x0 act on this: use the concatenation γ · α: [γ] · [α] = [γ · α]. α ˜x0 ˜γ ˜x 0 γ α x0 ˜X p X We will use this idea and return to the initial issue of making subgroups correspond to covering spaces. We want to show that this is surjective — every subgroup arises from some cover. We want to say “For any subgroup H ≤ π1(X, x0), there is a based covering map p : ( ˜X, ˜x0) → (X, x0) such that p∗π1( ˜X, ˜x0) = H”. Except, this cannot possibly be true, since by taking the trivial subgroup, this would imply that there is a universal covering for every space. So we need some additional assumptions. Proposition. Let X be a path connected, locally path connected and semilocally simply connected space. For any subgroup H ≤ π1(X, x0), there is a based covering map p : ( ˜X, ˜x0) → (X, x0) such that p∗π1( ˜X, ˜x0) = H. 34 3 Covering spaces II Algebraic Topology Proof. Since X is a path connected, locally path connected and semi-locally simply connected space, let ¯X be a universal covering. We have an intermediate group H such that π1( ˜X, ˜x0) = 1 ≤ H ≤ π1(X, x0). How can we obtain a corresponding covering space? Note that if we have ¯X and we want to recover X, we can quotient ¯X by the action of π1(X, x0). Since π1(X, x0) acts on ¯X, so does H ≤ π1(X, x0). Now we can deﬁne our covering space by taking quotients. We deﬁne ∼H on ¯X to be the orbit relation for the action of H, i.e. ˜x ∼H ˜y if there is some h ∈ H such that ˜y = h˜x. We then let ˜X be the quotient space ¯X/∼H . We can now do the messy algebra to show that this is the covering space we want. We have just showed that every subgroup comes from some covering space, i.e. the map from the set of covering spaces to the subgroups of π1 is surjective. Now we want to prove injectivity. To do so, we need a generalization of the homotopy lifting lemma. Suppose we have path-connected spaces (Y, y0), (X, x0) and ( ˜X, ˜x0), with f : (Y, y0) → (X, x0) a continuous map, p : ( ˜X, ˜x0) → (X, x0) a covering map. When does a lift of f to ˜f : (Y, y0) → ( ˜X, ˜x0) exist? The answer is given by the lifting criterion. Lemma (Lifting criterion). Let p : ( ˜X, ˜x0) → (X, x0) be a covering map of pathconnected based spaces, and (Y, y0) a path-connected, locally path connected based space. If f : (Y, y0) → (X, x0) is a continuous map, then there is a (unique) lift ˜f : (Y, y0) → ( ˜X, ˜x0) such that the diagram below commutes (i.e. p ◦ ˜f = f ): ( ˜X, ˜x0) p (X, x0) ˜f f (Y, y0) if and only if the following condition holds: f∗π1(Y, y0) ≤ p∗π1( ˜X, ˜x0). Note that uniqueness comes from the uniqueness of lifts. So this lemma is really about existence. Also, note that the condition holds trivially when Y is simply connected, e.g. when it is an interval (path lifting) or a square (homotopy lifting). So paths and homotopies can always be lifted. Proof. One direction is easy: if ˜f exists, then f = p ◦ ˜f . So f∗ = p∗ ◦ ˜f∗. So we know that im f∗ ⊆ im p∗. So done. In the other direction, uniqueness follows from the uniqueness of lifts. So we only need to prove existence. We deﬁne ˜f as follows: Given a y ∈ Y , there is some path αy : y0 y. Then f maps this to βy : x0 f (y) in X. By path lifting, this path lifts uniquely to ˜βy in ˜X. Then we set ˜f (y) = ˜βy(1). Note that if ˜f exists, then this must be what ˜f sends y to. What we need to show is that this is well-deﬁned. 35 3 Covering spaces II Algebraic Topology Suppose we picked a diﬀerent path α y : y0 y. Then this α y would have diﬀered from αy by a loop γ in Y . Our condition that f∗π1(Y, y0) ≤ p∗π1( ˜X, ˜x0) says f ◦ γ is the image of a y also diﬀer by a loop in ˜X, and hence have the same loop in ˜X. So ˜βy and ˜β end point. So this shows that ˜f is well-deﬁned. Finally, we show that ˜f is continuous. First, observe that any open set U ⊆ ˜X can be written as a union of ˜V such that p\| ˜V : ˜V → p( ˜V ) is a homeomorphism. Thus, it suﬃces to show that if p\| ˜V : ˜V → p( ˜V ) = V is a homeomorphism, then ˜f −1( ˜V ) is open. Let y ∈ ˜f −1( ˜V ), and let x = f (y). Since f −1(V ) is open and Y is locally path-connected, we can pick an open W ⊆ f −1(V ) such that y ∈ W and W is path connected. We claim that W ⊆ ˜f −1( ˜V ). Indeed, if z ∈ W , then we can pick a path γ from y to z. Then f sends (f (γ)), this to a path from x to f (z). The lift of this path to ˜X is given by p\|−1 ˜V (f (z)) ∈ ˜V . whose end point is p\|−1 ˜V (f (z)) ∈ ˜V . So it follows that ˜f (z) = p\|−1 ˜V Now we prove that every subgroup of π1 comes from exactly one covering space. What this statement properly means is made precise in the following proposition: Proposition. Let (X, x0), ( ˜X1, ˜x1), ( ˜X2, ˜x2) be path-connected based spaced, and pi : ( ˜Xi, ˜xi) → (X, x0) be covering maps. Then we have p1∗π1( ˜X1, ˜x1) = p2∗π1( ˜X2, ˜x2) if and only if there is some homeomorphism h such that the following diagram commutes: ( ˜X1, ˜x1) h ( ˜X2, ˜x2) p1 p2 (X, x0) i.e. p1 = p2 ◦ h. Note that this is a stronger statement than just saying the two covering spaces are homeomorphic. We are saying that we can ﬁnd a nice homeomorphism that works well with the covering map p. Proof. If such a homeomorphism exists, then clearly the subgroups are equal. If the subgroups are equal, we rotate our diagram a bit: ( ˜X2, ˜x2) h= ˜p1 p2 ( ˜X1, ˜x1) (X, x0) p1 Then h = ˜p1 exists by the lifting criterion. By symmetry, we can get h−1 = ˜p2. To show ˜p2 is indeed the inverse of ˜p1, note that ˜p2 ◦ ˜p1 is a lift of p2 ◦ ˜p1 = p1. 36 3 Covering spaces II Algebraic Topology Since id ˜X1 map. Similarly, ˜p1 ◦ ˜p2 is also the identity. is also a lift, by the uniqueness of lifts, we know ˜p2 ◦ ˜p1 is the identity ( ˜X1, ˜x1) ˜p2 p1 ( ˜X1, ˜x1) ˜p1 ( ˜X2, ˜x2) (X, x0) p2 p1 Now what we would like to do is to forget about the basepoints. What happens when we change base points? Recall that the eﬀect of changing base points is that we will conjugate our group. This doesn’t actually change the group itself, but if we are talking about subgroups, conjugation can send a subgroup into a diﬀerent subgroup. Hence, if we do not specify the basepoint, we don’t get a subgroup of π1, but a conjugacy class of subgroups. Proposition. Unbased covering spaces correspond to conjugacy classes of subgroups. 37 4 Some group theory II Algebraic Topology 4 Some group theory Algebraic topology is about translating topology into group theory. Unfortunately, you don’t know group theory. Well, maybe you do, but not the right group theory. So let’s learn group theory! 4.1 Free groups and presentations Recall that in IA Groups, we deﬁned, say, the dihedral group to be D2n = r, s \| rn = s2 = e, srs = r−1. What does this expression actually mean? Can we formally assign a meaning to this expression? We start with a simple case — the free group. This is, in some sense, the “freest” group we can have. It is deﬁned in terms of an alphabet and words. Deﬁnition (Alphabet and words). We let S = {sα : α ∈ Λ} be our alphabet, and we have an extra set of symbols S−1 = {s−1 α : α ∈ Λ}. We assume that S ∩ S−1 = ∅. What do we do with alphabets? We write words with them! We deﬁne S∗ to be the set of words over S ∪ S−1, i.e. it contains n-tuples x1 · · · xn for any 0 ≤ n < ∞, where each xi ∈ S ∪ S−1. Example. Let S = {a, b}. Then words could be the empty word ∅, or a, or aba−1b−1, or aa−1aaaaabbbb, etc. We are usually lazy and write aa−1aaaaabbbb as aa−1a5b4. When we see things like aa−1, we would want to cancel them. This is called elementary reduction. Deﬁnition (Elementary reduction). An elementary reduction takes a word usαs−1 α v and gives uv, or turns us−1 α sαv into uv. Since each reduction shortens the word, and the word is ﬁnite in length, we cannot keep reducing for ever. Eventually, we reach a reduced state. Deﬁnition (Reduced word). A word is reduced if it does not admit an elementary reduction. Example. ∅, a, aba−1b−1 are reduced words, while aa−1aaaaabbbb is not. Note that there is an inclusion map S → S∗ that sends the symbol sα to the word sα. Deﬁnition (Free group). The free group on the set S, written F (S), is the set of reduced words on S∗ together with some operations: (i) Multiplication is given by concatenation followed by elementary reduction to get a reduced word. For example, (aba−1b−1) · (bab) = aba−1b−1bab = ab2 (ii) The identity is the empty word ∅. (iii) The inverse of x1 · · · xn is x−1 n · · · x−1 1 , where, of course, (s−1 α )−1 = sα. 38 4 Some group theory II Algebraic Topology The elements of S are called the generators of F (S). Note that we have not showed that multiplication is well-deﬁned — we might reduce the same word in diﬀerent ways and reach two diﬀerent reduced words. We will show that this is indeed well-deﬁned later, using topology! Some people like to deﬁne the free group in a diﬀerent way. This is a cleaner way to deﬁne the free group without messing with alphabets and words, but is (for most people) less intuitive. This deﬁnition also does not make it clear that the free group F (S) of any set S exists. We will state this deﬁnition as a l
emma. Lemma. If G is a group and φ : S → G is a set map, then there exists a unique homomorphism f : F (S) → G such that the following diagram commutes: F (S) S f φ G where the arrow not labeled is the natural inclusion map that sends sα (as a symbol from the alphabet) to sα (as a word). Proof. Clearly if f exists, then f must send each sα to φ(sα) and s−1 Then the values of f on all other elements must be determined by α to φ(sα)−1. f (x1 · · · xn) = f (x1) · · · f (xn) since f is a homomorphism. So if f exists, it must be unique. So it suﬃces to show that this f is a well-deﬁned homomorphism. This is well-deﬁned if we deﬁne F (S) to be the set of all reduced words, since each reduced word has a unique representation (since it is deﬁned to be the representation itself). To show this is a homomorphism, suppose x = x1 · · · xna1 · · · ak, y = a−1 k · · · a−1 1 y1 · · · ym, where y1 = x−1 n . Then Then we can compute xy = x1 · · · xny1 · · · ym. f (x)f (y) = φ(x1) · · · φ(xn)φ(a1) · · · φ(ak)φ(ak)−1 · · · φ(a1)−1φ(y1) · · · φ(ym) = φ(x1) · · · φ(xn) · · · φ(y1) · · · φ(ym) = f (xy). So f is a homomorphism. We call this a “universal property” of F (S). We can show that F (S) is the unique group satisfying the conditions of this lemma (up to isomorphism), by taking G = F (S) and using the uniqueness properties. 39 4 Some group theory II Algebraic Topology Deﬁnition (Presentation of a group). Let S be a set, and let R ⊆ F (S) be any subset. We denote by R the normal closure of R, i.e. the smallest normal subgroup of F (S) containing R. This can be given explicitly by R = n i=1 girig−1 i : n ∈ N, ri ∈ R, gi ∈ F (S) . Then we write S \| R = F (S)/R. This is just the usual notation we have for groups. For example, we can write D2n = r, s \| rn, s2, srsr. Again, we can deﬁne this with a universal property. Lemma. If G is a group and φ : S → G is a set map such that f (r) = 1 for all r ∈ R (i.e. if r = s±1 m , then φ(r) = φ(s1)±1φ(s2)±1 · · · φ(sm)±1 = 1), then there exists a unique homomorphism f : S \| R → G such that the following triangle commutes: 1 s±1 2 · · · s±1 S \| R f φ G S Proof is similar to the previous one. In some sense, this says that all the group S \| R does is it satisﬁes the relations in R, and nothing else. Example (The stupid canonical presentation). Let G be a group. We can view the group as a set, and hence obtain a free group F (G). There is also an obvious surjection G → G. Then by the universal property of the free group, there is a surjection f : F (G) → G. Let R = ker f = R. Then G \| R is a presentation for G, since the ﬁrst isomorphism theorem says G ∼= F (G)/ ker f. This is a really stupid example. For example, even the simplest non-trivial group Z/2 will be written as a quotient of a free group with two generators. However, this tells us that every group has a presentation. Example. a, b \| b ∼= a ∼= Z. Example. With a bit of work, we can show that a, b \| ab−3, ba−2 = Z/5. 4.2 Another view of free groups Recall that we have not yet properly deﬁned free groups, since we did not show that multiplication is well-deﬁned. We are now going to do this using topology. Again let S be a set. For the following illustration, we will just assume S = {a, b}, but what we will do works for any set S. We deﬁne X by 40 4 Some group theory II Algebraic Topology a x0 b We call this a “rose with 2 petals”. This is a cell complex, with one 0-cell and \|S\| 1-cells. For each s ∈ S, we have one 1-cell, es, and we ﬁx a path γs : [0, 1] → es that goes around the 1-cell once. We will call the 0-cells and 1-cells vertices and edges, and call the whole thing a graph. What’s the universal cover of X? Since we are just lifting a 1-complex, the result should be a 1-complex, i.e. a graph. Moreover, this graph is connected and simply connected, i.e. it’s a tree. We also know that every vertex in the universal cover is a copy of the vertex in our original graph. So it must have 4 edges attached to it. So it has to look like something this: ˜x0 In X, we know that at each vertex, there should be an edge labeled a going in; an edge labeled a going out; an edge labeled b going in; an edge labeled b going out. This should be the case in ˜X as well. So ˜X looks like this: a b b ˜x0 a a b b The projection map is then obvious — we send all the vertices in ˜X to x0 ∈ X, and then the edges according to the labels they have, in a way that respects the direction of the arrow. It is easy to show this is really a covering map. 41 4 Some group theory II Algebraic Topology We are now going to show that this tree “is” the free group. Notice that every word w ∈ S∗ denotes a unique “edge path” in ˜X starting at ˜x0, where an edge path is a sequence of oriented edges ˜e1, · · · , ˜en such that the “origin” of ˜ei+1 is equal to the “terminus” of ˜ei. For example, the following path corresponds to w = abb−1b−1ba−1b−1. ˜x0 We can note a few things: (i) ˜X is connected. So for all ˜x ∈ p−1(x0), there is an edge-path ˜γ : ˜x0 ˜x. (ii) If an edge-path ˜γ fails to be locally injective, we can simplify it. How can an edge fail to be locally injective? It is ﬁne if we just walk along a path, since we are just tracing out a line. So it fails to be locally injective if two consecutive edges in the path are the same edge with opposite orientations: ˜ei ˜ei+1 ˜ei−1 ˜ei+2 We can just remove the redundant lines and get ˜ei−1 ˜ei+2 This reminds us of two things — homotopy of paths and elementary reduction of words. (iii) Each point ˜x ∈ p−1(x0) is joined to ˜x0 by a unique locally injective edge-path. (iv) For any w ∈ S∗, from (ii), we know that ˜γ is locally injective if and only if w is reduced. We can thus conclude that there are bijections F (S) p−1(x0) π1(X, x0) 42 4 Some group theory II Algebraic Topology that send ˜x to the word w ∈ F (S) such that ˜γw is a locally injective edge-path to ˜x0 ˜x; and ˜x to [γ] ∈ π1(X, x0) such that ˜x0 · [γ] = ˜x. So there is a bijection between F (S) and π1(X, x0). It is easy to see that the operations on F (S) and π1(X, x0) are the same, since they are just concatenating words or paths. So this bijection identiﬁes the two group structures. So this induces an isomorphism F (S) ∼= π1(X, x0). 4.3 Free products with amalgamation We managed to compute the fundamental group of the circle. But we want to ﬁnd the fundamental group of more things. Recall that at the beginning, we deﬁned cell complexes, and said these are the things we want to work with. Cell complexes are formed by gluing things together. So we want to know what happens when we glue things together. Suppose a space X is constructed from two spaces A, B by gluing (i.e. X = A ∪ B). We would like to describe π1(X) in terms of π1(A) and π1(B). To understand this, we need to understand how to “glue” groups together. Deﬁnition (Free product). Suppose we have groups G1 = S1 \| R1, G2 = S2 \| R2, where we assume S1 ∩ S2 = ∅. The free product G1 ∗ G2 is deﬁned to be G1 ∗ G2 = S1 ∪ S2 \| R1 ∪ R2. This is not a really satisfactory deﬁnition. A group can have many diﬀerent presentations, and it is not clear this is well-deﬁned. However, it is clear that this group exists. Note that there are natural homomorphisms ji : Gi → G1 ∗ G2 that send generators to generators. Then we can have the following universal property of the free product: Lemma. G1 ∗ G2 is the group such that for any group K and homomorphisms φi : Gi → K, there exists a unique homomorphism f : G1 ∗ G2 → K such that the following diagram commutes: G2 j2 j1 G1 G1 ∗ G2 f φ1 φ2 K Proof. It is immediate from the universal property of the deﬁnition of presentations. Corollary. The free product is well-deﬁned. Proof. The conclusion of the universal property can be seen to characterize G1 ∗ G2 up to isomorphism. Again, we have a deﬁnition in terms of a concrete construction of the group, without making it clear this is well-deﬁned; then we have a universal property 43 4 Some group theory II Algebraic Topology that makes it clear this is well-deﬁned, but not clear that the object actually exists. Combining the two would give everything we want. However, this free product is not exactly what we want, since there is little interaction between G1 and G2. In terms of gluing spaces, this corresponds to gluing A and B when A ∩ B is trivial (i.e. simply connected). What we really need is the free product with amalgamation, as you might have guessed from the title. Deﬁnition (Free product with amalgamation). Suppose we have groups G1, G2 and H, with the following homomorphisms: i2 H G2 i1 G1 The free product with amalgamation is deﬁned to be G1 ∗ H G2 = G1 ∗ G2/{(j2 ◦ i2(h))−1(j1 ◦ i1)(h) : h ∈ H}. Here we are attempting to identify things “in H” as the same, but we need to use the maps jk and ik to map the things from H to G1 ∗ G2. So we want to say “for any h, j1 ◦ i1(h) = j2 ◦ i2(h)”, or (j2 ◦ i2(h))−1(j1 ◦ i1)(h) = e. So we quotient by the (normal closure) of these things. We have the universal property Lemma. G1 ∗ H φi : Gi → K, there exists a unique homomorphism G1 ∗ H following diagram commutes: G2 is the group such that for any group K and homomorphisms G2 → K such that the H i1 G1 i2 G2 j2 j1 G1 ∗ H G2 φ1 f φ2 K This is the language we will need to compute fundamental groups. These deﬁnitions will (hopefully) become more concrete as we see more examples. 44 5 Seifert-van Kampen theorem II Algebraic Topology 5 Seifert-van Kampen theorem 5.1 Seifert-van Kampen theorem The Seifert-van Kampen theorem is the theorem that tells us what happens when we glue spaces together. Here we let X = A ∪ B, where A, B, A ∩ B are path-connected. A x0 B We pick a basepoint x0 ∈ A ∩ B for convenience. Since we like diagrams, we can write this as a commutative diagram: A ∩ B A B X where all arrows are inclusion (i.e. injective) maps. We can consider what happens when we take the fundamental groups of each space. Then we have the induced homomorphisms π1(A ∩ B, x0) π1(B, x0) π1(A, x0) π1(X, x0) We might guess that π1(X, x0) is just the free produ
ct with amalgamation π1(X, x0) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0). The Seifert-van Kampen theorem says that, under mild hypotheses, this guess is correct. Theorem (Seifert-van Kampen theorem). Let A, B be open subspaces of X such that X = A ∪ B, and A, B, A ∩ B are path-connected. Then for any x0 ∈ A ∩ B, we have π1(X, x0) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0). Note that by the universal property of the free product with amalgamation, π1(B, x0) → we by deﬁnition know that there is a unique map π1(A, x0) ∗ π1(A∩B,x0) π1(X, x0). The theorem asserts that this map is an isomorphism. Proof is omitted because time is short. Example. Consider a higher-dimensional sphere Sn = {v ∈ Rn+1 : \|v\| = 1} for n ≥ 2. We want to ﬁnd π1(Sn). The idea is to write Sn as a union of two open sets. We let n = e1 ∈ Sn ⊆ Rn+1 be the North pole, and s = −e1 be the South pole. We let A = Sn \ {n}, 45 5 Seifert-van Kampen theorem II Algebraic Topology and B = Sn \ {s}. By stereographic projection, we know that A, B ∼= Rn. The hard part is to understand the intersection. To do so, we can draw a cylinder Sn−1 × (−1, 1), and project our A ∩ B onto the cylinder. We can similarly project the cylinder onto A ∩ B. So A ∩ B ∼= Sn−1 × (−1, 1) Sn−1, since (−1, 1) is contractible. n s We can now apply the Seifert-van Kampen theorem. Note that this works only if Sn−1 is path-connected, i.e. n ≥ 2. Then this tells us that π1(Sn) ∼= π1(Rn) ∗ π1(Sn−1) π1(Rn) ∼= 1 ∗ π1(Sn−1) 1 It is easy to see this is the trivial group. We can see this directly from the universal property of the amalgamated free product, or note that it is the quotient of 1 ∗ 1, which is 1. So for n ≥ 2, π1(Sn) ∼= 1. We have found yet another simply connected space. However, this is unlike our previous examples. Our previous spaces were simply connected because they were contractible. However, we will later see that Sn is not contractible. So this is genuinely a new, interesting example. Why did we go though all this work to prove that π1(Sn) ∼= 1? It feels like we can just prove this directly — pick a point that is not in the curve as the North pole, project stereographically to Rn, and contract it there. However, the problem is that space-ﬁlling curves exist. We cannot guarantee that we can pick a point not on the curve! It is indeed possible to prove directly that given any curve on Sn (with n ≥ 1), we can deform it slightly so that it is no longer surjective. Then the above proof strategy works. However, using Seifert-van Kampen is much neater. Example (RPn). Recall that RPn ∼= Sn/{± id}, and the quotient map Sn → RPn is a covering map. Now that we have proved that Sn is simply connected, we know that Sn is a universal cover of RPn. For any x0 ∈ RPn, we have a bijection π1(RPn, x0) ↔ p−1(x0). Since p−1(x0) has two elements by deﬁnition, we know that \|π1(RPn, x0)\| = 2. So π1(RPn, x0) = Z/2. You will prove a generalization of this in example sheet 2. 46 5 Seifert-van Kampen theorem II Algebraic Topology Example (Wedge of two circles). We are going to consider the operation of wedging. Suppose we have two topological spaces, and we want to join them together. The natural way to join them is to take the disjoint union. What if we have based spaces? If we have (X, x0) and (Y, y0), we cannot just take the disjoint union, since we will lose our base point. What we do is take the wedge sum, where we take the disjoint union and then identify the base points: X Y x0 y0 x0 ∼ y0 X ∧ Y Suppose we take the wedge sum of two circles S1 ∧ S1. We would like to pick A, B to be each of the circles, but we cannot, since A and B have to be open. So we take slightly more, and get the following: x0 A B Each of A and B now look like this: We see that both A and B retract to the circle. So π1(A) ∼= π1(B) ∼= Z, while A ∩ B is a cross, which retracts to a point. So π1(A ∩ B) = 1. Hence by the Seifert-van Kampen theorem, we get π1(S1 ∧ S1, x0) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0) ∼= Z ∗ 1 Z ∼= Z ∗ Z ∼= F2, where F2 is just F (S) for \|S\| = 2. We can see that Z ∗ Z ∼= F2 by showing that they satisfy the same universal property. Note that we had already ﬁgured this out when we studied the free group, where we realized F2 is the fundamental group of this thing. More generally, as long as x0, y0 in X and Y are “reasonable”, π1(X ∧ Y ) ∼= π1(X) ∗ π1(Y ). 47 5 Seifert-van Kampen theorem II Algebraic Topology Next, we would exhibit some nice examples of the covering spaces of S1 ∧ S1, i.e. the “rose with 2 petals”. Recall that π1(S1 ∧ S1, x0) ∼= F2 ∼= a, b. Example. Consider the map φ : F2 → Z/3 which sends a → 1, b → 1. Note that 1 is not the identity, since this is an abelian group, and the identity is 0. This exists since the universal property tells us we just have to say where the generators go, and the map exists (and is unique). Now ker φ is a subgroup of F2. So there is a based covering space of S1 ∧ S1 corresponding to it, say, ˜X. Let’s work out what it is. First, we want to know how many sheets it has, i.e. how many copies of x0 we have. There are three, since we know that the number of sheets is the index of the subgroup, and the index of ker φ is \|Z/3\| = 3 by the ﬁrst isomorphism theorem. ˜x0 Let’s try to lift the loop a at ˜x0. Since a ∈ ker φ = π1( ˜X, ˜x0), a does not lift to a loop. So it goes to another vertex. ˜x0 a Similarly, a2 ∈ ker φ = π1( ˜X, ˜x0). So we get the following a ˜x0 a Since a3 ∈ ker φ, we get a loop labelled a. a a a ˜x0 Note that ab−1 ∈ ker φ. So ab−1 gives a loop. So b goes in the same direction as a: b a ˜x0 b a a b 48 5 Seifert-van Kampen theorem II Algebraic Topology This is our covering space. This is a fun game to play at home: (i) Pick a group G (ﬁnite groups are recommended). (ii) Then pick α, β ∈ G and let φ : F2 → G send a → α, b → β. (iii) Compute the covering space corresponding to φ. 5.2 The eﬀect on π1 of attaching cells We have started the course by talking about cell complexes, but largely ignored them afterwards. Finally, we are getting back to them. The process of attaching cells is as follows: we start with a space X, get a function Sn−1 → X, and attach Dn to X by gluing along the image of f to get X ∪f Dn: x0 y0 0 Since we are attaching stuﬀ, we can use the Seifert-van Kampen theorem to analyse this. Theorem. If n ≥ 3, then π1(X ∪f Dn) ∼= π1(X). More precisely, the map π1(X, x0) → π1(X ∪f Dn, x0) induced by inclusion is an isomorphism, where x0 is a point on the image of f . This tells us that attaching a high-dimensional disk does not do anything to the fundamental group. Proof. Again, the diﬃculty of applying Seifert-van Kampen theorem is that we need to work with open sets. Let 0 ∈ Dn be any point in the interior of Dn. We let A = X ∪f (Dn \ {0}). Note that Dn \{0} deformation retracts to the boundary Sn−1. So A deformation retracts to X. Let B = ˚D, the interior of Dn. Then A ∩ B = ˚Dn \ 0 ∼= Sn−1 × (−1, 1) We cannot use y0 as our basepoint, since this point is not in A ∩ B. Instead, pick an arbitrary y1 ∈ A ∩ B. Since Dn is path connected, we have a path γ : y1 y0, and we can use this to recover the fundamental groups based at y0. Now Seifert-van Kampen theorem says π1(X ∪f Dn, y1) ∼= π1(A, y1) ∗ π1(A∩B,y1) π1(B, y1). Since B is just a disk, and A ∩ B is simply connected (n ≥ 3 implies Sn−1 is simply connected), their fundamental groups are trivial. So we get π1(X ∪f Dn, y1) ∼= π1(A, y1). We can now use γ to change base points from y1 to y0. So π1(X ∪f Dn, y0) ∼= π1(A, y0) ∼= π1(X, y0). 49 5 Seifert-van Kampen theorem II Algebraic Topology The more interesting case is when we have smaller dimensions. Theorem. If n = 2, then the natural map π1(X, x0) → π1(X ∪f Dn, x0) is surjective, and the kernel is [f ]. Note that this statement makes sense, since Sn−1 is a circle, and f : Sn−1 → X is a loop in X. This is what we would expect, since if we attach a disk onto the loop given by f , this loop just dies. Proof. As before, we get π1(X ∪f Dn, y1) ∼= π1(A, y1) ∗ π1(A∩B,y1) π1(B, y1). Again, B is contractible, and π1(B, y1) ∼= 1. However, π1(A ∩ B, y1) ∼= Z. Since π1(A ∩ B, y1) is just (homotopic to) the loop induced by f , it follows that π1(A, y1) ∗ π1(A∩B,y1) 1 = (π1(A, y1) ∗ 1)/π1(A ∩ B, y1) ∼= π1(X, x0)/f . In summary, we have π1(X ∪f Dn) = π1(X) n ≥ 3 π1(X)/f n = 2 This is a useful result, since this is how we build up cell complexes. If we want to compute the fundamental groups, we can just work up to the two-cells, and know that the higher-dimensional cells do not have any eﬀect. Moreover, whenever X is a cell complex, we should be able to write down the presentation of π1(X). Example. Let X be the 2-torus. Possibly, our favorite picture of the torus is (not a doughnut): This is already a description of the torus as a cell complex! We start with our zero complex X (0): We then add our 1-cells to get X (1): b a We now glue our square to the cell complex to get X = X (2): 50 5 Seifert-van Kampen theorem II Algebraic Topology matching up the colors and directions of arrows. So we have our torus as a cell complex. What is its fundamental group? There are many ways we can do this computation, but this time we want to do it as a cell complex. We start with X (0). This is a single point. So its fundamental group is π1(X (0)) = 1. When we add our two 1-cells, we get π1(X (1)) = F2 Finally, to get π1(X), we have to quotient out by the boundary of our square, ∼= a, b. which is just aba−1b−1. So we have π1(X (2)) = F2/aba−1b−1 = a, b \| aba−1b−1 ∼= Z2. We have the last congruence since we have two generators, and then we make them commute by quotienting the commutator out. This procedure can be reversed — given a presentation of a group, we can just add the right edges and squares to produce a cell complex with that presentation. Corollary. For any (ﬁnite) group presentation S \| R, there exists a (ﬁnite) cell complex (of dimension 2) X such that π1(X) ∼= S \| R. There really isn’t anything that requires that ﬁniteness in this proof, but ﬁniteness makes us feel more comfortabl
e. Proof. Let S = {a1, · · · , am} and R = {r1, · · · , rn}. We start with a single point, and get our X (1) by adding a loop about the point for each ai ∈ S. We then get our 2-cells e2 j for j = 1, · · · , n, and attaching them to X (1) by fi : S1 → X (1) given by a based loop representing ri ∈ F (S). Since all groups have presentations, this tells us that all groups are funda- mental groups of some spaces (at least those with ﬁnite presentations). 5.3 A reﬁnement of the Seifert-van Kampen theorem We are going to make a reﬁnement of the theorem so that we don’t have to worry about that openness problem. We ﬁrst start with a deﬁnition. Deﬁnition (Neighbourhood deformation retract). A subset A ⊆ X is a neighbourhood deformation retract if there is an open set A ⊆ U ⊆ X such that A is a strong deformation retract of U , i.e. there exists a retraction r : U → A and r idU rel A. This is something that is true most of the time, in suﬃciently sane spaces. Example. If Y is a subcomplex of a cell complex, then Y is a neighbourhood deformation retract. 51 5 Seifert-van Kampen theorem II Algebraic Topology Theorem. Let X be a space, A, B ⊆ X closed subspaces. Suppose that A, B and A ∩ B are path connected, and A ∩ B is a neighbourhood deformation retract of A and B. Then for any x0 ∈ A ∩ B. π1(X, x0) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0). A x0 B This is just like Seifert-van Kampen theorem, but usually easier to apply, since we no longer have to “fatten up” our A and B to make them open. Proof. Pick open neighbourhoods A ∩ B ⊆ U ⊆ A and A ∩ B ⊆ V ⊆ B that strongly deformation retract to A ∩ B. Let U be such that U retracts to A ∩ B. Since U retracts to A, it follows that U is path connected since path-connectedness is preserved by homotopies. Let A = A∪V and B = B ∪U . Since A = (X \B)∪V , and B = (X \A)∪U , it follows that A and B are open. Since U and V retract to A ∩ B, we know A A and B B. Also, A ∩ B = (A ∪ V ) ∩ (. In particular, it is path connected. So by Seifert van-Kampen, we get π1(A ∪ B) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0). This is basically what we’ve done all the time when we enlarge our A and B to become open. Example. Let X = S1 ∧ S1, the rose with two petals. Let A, B ∼= S1 be the circles. x0 A B Then since {x0} = A ∩ B is a neighbourhood deformation retract of A and B, we know that π1X ∼= π1S1 ∗ π1S1. 5.4 The fundamental group of all surfaces We have found that the torus has fundamental group Z2, but we already knew this, since the torus is just S1 × S1, and the fundamental group of a product is the product of the fundamental groups, as you have shown in the example sheet. So we want to look at something more interesting. We look at all surfaces. We start by deﬁning what a surface is. It is surprisingly diﬃcult to get mathematicians to agree on how we can deﬁne a surface. Here we adopt the following deﬁnition: 52 5 Seifert-van Kampen theorem II Algebraic Topology Deﬁnition (Surface). A surface is a Hausdorﬀ topological space such that every point has a neighbourhood U that is homeomorphic to R2. Some people like C more that R, and sometimes they put C2 instead of R2 in the deﬁnition, which is confusing, since that would have two complex dimensions and hence four real dimensions. Needless to say, the actual surfaces will also be diﬀerent and have diﬀerent homotopy groups. We will just focus on surfaces with two real dimensions. To ﬁnd the fundamental group of all surfaces, we rely on the following theorem that tells us what surfaces there are. Theorem (Classiﬁcation of compact surfaces). If X is a compact surface, then X is homeomorphic to a space in one of the following two families: (i) The orientable surface of genus g, Σg includes the following (please excuse my drawing skills): A more formal deﬁnition of this family is the following: we start with the 2-sphere, and remove a few discs from it to get S2 \ ∪g i=1D2. Then we take g tori with an open disc removed, and attach them to the circles. (ii) The non-orientable surface of genus n, En = {RP2, K, · · · } (where K is the Klein bottle). This has a similar construction as above: we start with the sphere S2, make a few holes, and then glue M¨obius strips to them. It would be nice to be able to compute fundamental groups of these guys. To do so, we need to write them as polygons with identiﬁcation. Example. To obtain a surface of genus two, written Σ2, we start with what we had for a torus: b a a b 53 5 Seifert-van Kampen theorem II Algebraic Topology If we just wanted a torus, we are done (after closing the loop), but now we want a surface with genus 2, so we add another torus: b1 a1 a1 b2 b1 a2 a2 b2 To visualize how this works, imagine cutting this apart along the dashed line. This would give two tori with a hole, where the boundary of the holes are just the dashed line. Then gluing back the dashed lines would give back our orientable surface with genus 2. In general, to produce Σg, we produce a polygon with 4g sides. Then we get π1Σg = a1, b1, · · · , ag, bg \| a1b1a−1 1 b−1 1 · · · agbga−1 g b−1 g . We do we care? The classiﬁcation theorem tells us that each surface is homeomorphic to some of these orientable and non-orientable surfaces, but it doesn’t tell us ∼= Σ241, via some weird homeomorphism there is no overlap. It might be that Σ6 that destroys some holes. However, this result lets us know that all these orientable surfaces are genuinely diﬀerent. While it is diﬃcult to stare at this fundamental group and say that π1Σg ∼= π1Σg for g = g, we can perform a little trick. We can take the abelianization of the group π1Σg, where we further quotient by all commutators. Then the abelianized fundamental group of Σg will simply be Z2g. These are clearly distinct for diﬀerent values of g. So all these surfaces are distinct. Moreover, they are not even homotopy equivalent. The fundamental groups of the non-orientable surfaces is left as an exercise for the reader. 54 6 Simplicial complexes II Algebraic Topology 6 Simplicial complexes So far, we have taken a space X, and assigned some things to it. The ﬁrst was easy — π0(X). It was easy to calculate and understand. We then spent a lot of time talking about π1(X). What are they good for? A question we motivated ourselves with was to prove Rm ∼= Rn implies n = m. π0 was pretty good for the case when n = 1. If Rm ∼= R, then we would have Rm \ {0} ∼= R \ {0} S0. We know that \|π0(S0)\| = 2, while \|π0(Rm \ {0})\| = 1 for m = 1. This is just a fancy way of saying that R \ {0} is disconnected while Rm \ {0} is not for m = 1. We can just add 1 to n, and add 1 to our subscript. If Rm ∼= R2, then we have Rm \ {0} ∼= R2 \ {0} S1. We know that π1(S1) ∼= Z, while π1(Rm \ {0}) ∼= π1(Sm−1) ∼= 1 unless m = 2. The obvious thing to do is to create some πn(X). But as you noticed, π1 took us quite a long time to deﬁne, and was really hard to compute. As we would expect, this only gets harder as we get to higher dimensions. This is indeed possible, but will only be done in Part III courses. The problem is that πn works with groups, and groups are hard. There are too many groups out there. We want to do some easier algebra, and a good choice is linear algebra. Linear algebra is easy. In the rest of the course, we will have things like H0(X) and H1(X), instead of πn, which are more closely related to linear algebra. Another way to motivate the abandoning of groups is as follows: recall last time we saw that if X is a ﬁnite cell, reasonably explicitly deﬁned, then we can write down a presentation S \| R for π1(X). This sounds easy, except that we don’t understand presentations. In fact there is a theorem that says there is no algorithm that decides if a particular group presentation is actually the trivial group. So even though we can compute the fundamental group in terms of presentations, this is not necessarily helpful. On the other hand, linear algebra is easy. Computers can do it. So we will move on to working with linear algebra instead. This is where homology theory comes in. It takes a while for us to deﬁne it, but after we ﬁnish developing the machinery, things are easy. 6.1 Simplicial complexes There are many ways of formulating homology theory, and these are all equivalent at least for suﬃciently sane spaces (such as cell complexes). In this course, we will be using simplicial homology, which is relatively more intuitive and can be computed directly. The drawback is that we will have to restrict to a particular kind of space, known as simplicial complexes. This is not a very serious restriction per se, since many spaces like spheres are indeed simplicial complexes. However, the deﬁnition of simplicial homology is based on exactly how we view our space as a simplicial complex, and it will take us quite a lot of work to show that the simplicial homology is indeed a property of the space itself, and not how we represent it as a simplicial complex. We now start by deﬁning simplicial complexes, and developing some general theory of simplicial complexes that will become useful later on. Deﬁnition (Aﬃne independence). A ﬁnite set of points {a1, · · · , an} ⊆ Rm is 55 6 Simplicial complexes II Algebraic Topology aﬃnely independent iﬀ n i=1 tiai = 0 with n i=1 ti = 0 ⇔ ti = 0 for all i. Example. When n = 3, the following points are aﬃnely independent: The following are not: The proper way of understanding this deﬁnition is via the following lemma: Lemma. a0, · · · , an ∈ Rm are aﬃnely independent if and only if a1−a0, · · · , an− a0 are linearly independent. Alternatively, n + 1 aﬃnely independent points span an n-dimensional thing. Proof. Suppose a0, · · · , an are aﬃnely independent. Suppose Then we can rewrite this as λi(ai − a0) = 0. n i=1 − n i=1 λi a0 + λ1a1 + · · · + λnan = 0. Now the sum of the coeﬃcients is 0. So aﬃne independence implies that all coeﬃcients are 0. So a1 − a0, · · · , an − a0 are linearly independent. On the other hand, suppose a1 − a0, · · · , an − a0 are linearly indepen
dent. Now suppose Then we can write n i=0 tiai = 0, n i=0 ti = 0. t0 = − n i=1 ti. Then the ﬁrst equation reads 0 = − n i=1 ti a0 + t1a1 + · · · + tnan = n i=1 ti(ai − a0). So linear independence implies all ti = 0. 56 6 Simplicial complexes II Algebraic Topology The relevance is that these can be used to deﬁne simplices (which are simple, as opposed to complexes). Deﬁnition (n-simplex). An n-simplex is the convex hull of (n + 1) aﬃnely independent points a0, · · · , an ∈ Rm, i.e. the set σ = a0, · · · , an = n i=0 tiai : n i=0 ti = 1, ti ≥ 0 . The points a0, · · · , an are the vertices, and are said to span σ. The (n + 1)-tuple (t0, · · · , tn) is called the barycentric coordinates for the point tiai. Example. When n = 0, then our 0-simplex is just a point: When n = 1, then we get a line: When n = 2, we get a triangle: When n = 3, we get a tetrahedron: The key motivation of this is that simplices are determined by their vertices. Unlike arbitrary subspaces of Rn, they can be speciﬁed by a ﬁnite amount of data. We can also easily extract the faces of the simplices. Deﬁnition (Face, boundary and interior). A face of a simplex is a subset (or subsimplex) spanned by a subset of the vertices. The boundary is the union of the proper faces, and the interior is the complement of the boundary. The boundary of σ is usually denoted by ∂σ, while the interior is denoted by ˚σ, and we write τ ≤ σ when τ is a face of σ. In particular, the interior of a vertex is the vertex itself. Note that these notions of interior and boundary are distinct from the topological notions of interior and boundary. Example. The standard n-simplex is spanned by the basis vectors {e0, · · · , en} in Rn+1. For example, when n = 2, we get the following: 57 6 Simplicial complexes II Algebraic Topology We will now glue simplices together to build complexes, or simplicial com- plexes. Deﬁnition. A (geometric) simplicial complex is a ﬁnite set K of simplices in Rn such that (i) If σ ∈ K and τ is a face of σ, then τ ∈ K. (ii) If σ, τ ∈ K, then σ ∩ τ is either empty or a face of both σ and τ . Deﬁnition (Vertices). The vertices of K are the zero simplices of K, denoted VK. Example. This is a simplicial complex: These are not: Technically, a simplicial complex is deﬁned to be a set of simplices, which are just collections of points. It is not a subspace of Rn. Hence we have the following deﬁnition: Deﬁnition (Polyhedron). The polyhedron deﬁned by K is the union of the simplices in K, and denoted by \|K\|. 58 6 Simplicial complexes II Algebraic Topology We make this distinction because distinct simplicial complexes may have the same polyhedron, such as the following: Just as in cell complexes, we can deﬁne things like dimensions. Deﬁnition (Dimension and skeleton). The dimension of K is the highest dimension of a simplex of K. The d-skeleton K (d) of K is the union of the n-simplices in K for n ≤ d. Note that since these are ﬁnite and live inside Rn, we know that \|K\| is always compact and Hausdorﬀ. Usually, when we are given a space, say Sn, it is not deﬁned to be a simplicial complex. We can “make” it a simplicial complex by a triangulation. Deﬁnition (Triangulation). A triangulation of a space X is a homeomorphism h : \|K\| → X, where K is some simplicial complex. Example. Let σ be the standard n-simplex. The boundary ∂σ is homeomorphic to Sn−1 (e.g. the boundary of a (solid) triangle is the boundary of the triangle, which is also a circle). This is called the simplicial (n − 1)-sphere. We can also triangulate our Sn in a diﬀerent way: Example. In Rn+1, consider the simplices ±e0, · · · , ±en for each possible combination of signs. So we have 2n+1 simplices in total. Then their union deﬁnes a simplicial complex K, and \|K\| ∼= Sn. The nice thing about this triangulation is that the simplicial complex is invariant under the antipodal map. So not only can we think of this as a triangulation of the sphere, but a triangulation of RPn as well. As always, we don’t just look at objects themselves, but also maps between them. Deﬁnition (Simplicial map). A simplicial map f : K → L is a function f : VK → VL such that if a0, · · · , an is a simplex in K, then {f (a0), · · · , f (an)} spans a simplex of L. The nice thing about simplicial maps is that we only have to specify where the vertices go, and there are only ﬁnitely many vertices. So we can completely specify a simplicial map by writing down a ﬁnite amount of information. It is important to note that we say {f (a0), · · · , f (an)} as a set span a simplex of L. In particular, they are allowed to have repeats. 59 6 Simplicial complexes II Algebraic Topology Example. Suppose we have the standard 2-simplex K as follows: a2 a0 a1 The following does not deﬁne a simplicial map because a1, a2 is a simplex in K, but {f (a1), f (a2)} does not span a simplex: f (a0), f (a1) f (a2) On the other hand, the following is a simplicial map, because now {f (a1), f (a2)} spans a simplex, and note that {f (a0), f (a1), f (a2)} also spans a 1-simplex because we are treating the collection of three vertices as a set, and not a simplex. f (a0), f (a1) f (a2) Finally, we can also do the following map: f (a0), f (a1) f (a2) The following lemma is obvious, but we will need it later on. Lemma. If K is a simplicial complex, then every point x ∈ \|K\| lies in the interior of a unique simplex. As we said, simplicial maps are nice, but they are not exactly what we want. We want to have maps between spaces. Lemma. A simplicial map f : K → L induces a continuous map \|f \| : \|K\| → \|L\|, and furthermore, we have \|f ◦ g\| = \|f \| ◦ \|g\|. There is an obvious way to deﬁne this map. We know how to map vertices, and then just extend everything linearly. Proof. For any point in a simplex σ = a0, · · · , an, we deﬁne \|f \| n i=0 tiai = n i=0 tif (ai). The result is in L because {f (ai)} spans a simplex. It is not diﬃcult to see this is well-deﬁned when the point lies on the boundary of a simplex. This is clearly continuous on σ, and is hence continuous on \|K\| by the gluing lemma. The ﬁnal property is obviously true by deﬁnition. 60 6 Simplicial complexes II Algebraic Topology 6.2 Simplicial approximation This is all very well, but we are really interested in continuous maps. So given a continuous map f , we would like to ﬁnd a related simplicial map g. In particular, we want to ﬁnd a simplicial map g that “approximates” f in some sense. The deﬁnition we will write down is slightly awkward, but it turns out this is the most useful deﬁnition. Deﬁnition (Open star and link). Let x ∈ \|K\|. The open star of x is the union of all the interiors of the simplices that contain x, i.e. StK(x) = ˚σ. x∈σ∈K The link of x, written LkK(x), is the union of all those simplices that do not contain x, but are faces of a simplex that does contain x. Deﬁnition (Simplicial approximation). Let f : \|K\| → \|L\| be a continuous map between the polyhedra. A function g : VK → VL is a simplicial approximation to f if for all v ∈ VK, f (StK(v)) ⊆ StL(g(v)). (∗) The following lemma tells us why this is a good deﬁnition: Lemma. If f : \|K\| → \|L\| is a map between polyhedra, and g : VK → VL is a simplicial approximation to f , then g is a simplicial map, and \|g\| f . Furthermore, if f is already simplicial on some subcomplex M ⊆ K, then we get g\|M = f \|M , and the homotopy can be made rel M . Proof. First we want to check g is really a simplicial map if it satisﬁes (∗). Let σ = a0, · · · , an be a simplex in K. We want to show that {g(a0), · · · , g(an)} spans a simplex in L. Pick an arbitrary x ∈ ˚σ. Since σ contains each ai, we know that x ∈ StK(ai) for all i. Hence we know that f (x) ∈ n i=0 f (StK(ai)) ⊆ n i=0 StL(g(ai)). Hence we know that there is one simplex, say, τ that contains all g(ai) whose interior contains f (x). Since each g(ai) is a vertex in L, each g(ai) must be a vertex of τ . So they span a face of τ , as required. We now want to prove that \|g\| f . We let H : \|K\| × I → \|L\| ⊆ Rm be deﬁned by (x, t) → t\|g\|(x) + (1 − t)f (x). This is clearly continuous. So we need to check that im H ⊆ \|L\|. But we know that both \|g\|(x) and f (x) live in τ and τ is convex. It thus follows that H(x × I) ⊆ τ ⊆ \|L\|. To prove the last part, it suﬃces to show that every simplicial approximation to a simplicial map must be the map itself. Then the homotopy is rel M by the construction above. This is easily seen to be true — if g is a simplicial approximation to f , then f (v) ∈ f (StK(v)) ⊆ StL(g(v)). Since f (v) is a vertex and g(v) is the only vertex in StL(g(v)), we must have f (v) = g(v). So done. 61 6 Simplicial complexes II Algebraic Topology What’s the best thing we might hope for at this point? It would be great if every map were homotopic to a simplicial map. Is this possible? Let’s take a nice example. Let’s consider the following K: How many homotopy classes of continuous maps are there K → K? Countably many, one for each winding number. However, there can only be at most 33 = 27 simplicial maps. The problem is that we don’t have enough vertices to realize all those interesting maps. The idea is to reﬁne our simplicial complexes. Suppose we have the following simplex: What we do is to add a point in the center of each simplex, and join them up: This is known as the barycentric subdivision. After we subdivide it once, we can realize more homotopy classes of maps. We will show that for any map, as long as we are willing to barycentrically subdivide the simplex many times, we can ﬁnd a simplicial approximation to it. Deﬁnition (Barycenter). The barycenter of σ = a0, · · · , an is ˆσ = n i=0 1 n + 1 ai. Deﬁnition (Barycentric subdivision). The (ﬁrst) barycentric subdivision K of K is the simplicial complex: K = {ˆσ0, · · · , ˆσn : σi ∈ K and σ0 < σ1 < · · · < σn}. If you stare at this long enough, you will realize this is exactly what we have drawn above. The rth barycentric subdivision K (r) is deﬁned inductively as the barycentric subdivision of the r − 1th barycentric subdivision, i.e. K (r) = (K (r
−1)). Proposition. \|K\| = \|K \| and K really is a simplicial complex. Proof. Too boring to be included in lectures. 62 6 Simplicial complexes II Algebraic Topology We now have a slight problem. Even though \|K \| and \|K\| are equal, the identity map from \|K \| to \|K\| is not a simplicial map. To solve this problem, we can choose any function K → VK by σ → vσ with vσ ∈ σ, i.e. a function that sends any simplex to any of its vertices. Then we can deﬁne g : K → K by sending ˆσ → vσ. Then this is a simplicial map, and indeed a simplicial approximation to the identity map \|K \| → \|K\|. We will revisit this idea later when we discuss homotopy invariance. The key theorem is that as long as we are willing to perform barycentric subdivisions, then we can always ﬁnd a simplicial approximation. Theorem (Simplicial approximation theorem). Let K and L be simplicial complexes, and f : \|K\| → \|L\| a continuous map. Then there exists an r and a simplicial map g : K (r) → L such that g is a simplicial approximation of f . Furthermore, if f is already simplicial on M ⊆ K, then we can choose g such that \|g\|\|M = f \|M . The ﬁrst thing we have to ﬁgure out is how far we are going to subdivide. To do this, we want to quantify how “ﬁne” our subdivisions are. Deﬁnition (Mesh). Let K be a simplicial complex. The mesh of K is µ(K) = max{v0 − v1 : v0, v1 ∈ K}. We have the following lemma that tells us how large our mesh is: Lemma. Let dim K = n, then µ(K (r)) ≤ n r n + 1 µ(K). The key point is that as r → ∞, the mesh goes to zero. So indeed we can make our barycentric subdivisions ﬁner and ﬁner. The proof is purely technical and omitted. Now we can prove the simplicial approximation theorem. Proof of simplicial approximation theorem. Suppose we are given the map f : \|K\| → \|L\|. We have a natural cover of \|L\|, namely the open stars of all vertices. We can use f to pull these back to \|K\| to obtain a cover of \|K\|: {f −1(StL(w)) : w ∈ VL}. The idea is to barycentrically subdivide our K such that each open star of K is contained in one of these things. By the Lebesgue number lemma, there exists some δ, the Lebesgue number of the cover, such that for each x ∈ \|K\|, Bδ(x) is contained in some element of the cover. By the previous lemma, there is an r such that µ(K (r)) < δ. Now since the mesh µ(K (r)) is the smallest distance between any two vertices, the radius of every open star StK(r)(x) is at most µ(K (r)). Hence it follows that StK(r) (x) ⊆ Bδ(x) for all vertices x ∈ VK(r) . Therefore, for all x ∈ VK(r), there is some w ∈ VL such that StK(r) (x) ⊆ Bδ(x) ⊆ f −1(StL(w)). 63 6 Simplicial complexes II Algebraic Topology Therefore deﬁning g(x) = w, we get f (StK(r) (x)) ⊆ StL(g(x)). So g is a simplicial approximation of f . The last part follows from the observation that if f is a simplicial map, then it maps vertices to vertices. So we can pick g(v) = f (v). 64 7 Simplicial homology II Algebraic Topology 7 Simplicial homology 7.1 Simplicial homology For now, we will forget about simplicial approximations and related ﬂuﬀ, but just note that it is ﬁne to assume everything can be considered to be simplicial. Instead, we are going to use this framework to move on and deﬁne some new invariants of simplicial complexes K, known as Hn(K). These are analogous to π0, π1, · · · , but only use linear algebra in the deﬁnitions, and are thus much simpler. The drawback, however, is that the deﬁnitions are slightly less intuitive at ﬁrst sight. Despite saying “linear algebra”, we won’t be working with vector spaces most of the time. Instead, we will be using abelian groups, which really should be thought of as Z-modules. At the end, we will come up with an analogous theory using Q-modules, i.e. Q-vector spaces, and most of the theory will carry over. Using Q makes some of our work easier, but as a result we would have lost some information. In the most general case, we can replace Z with any abelian group, but we will not be considering any of these in this course. Recall that we deﬁned an n-simplex as the collection of n + 1 vertices that span a simplex. As a simplex, when we permute the vertices, we still get the same simplex. What we want to do is to remember the orientation of the simplex. In particular, we want think of the simplices (a, b) and (b, a) as diﬀerent: a b a b Hence we deﬁne oriented simplices. Deﬁnition (Oriented n-simplex). An oriented n-simplex in a simplicial complex K is an (n + 1)-tuple (a0, · · · , an) of vertices ai ∈ Vk such that a0, · · · , an ∈ K, where we think of two (n + 1)-tuples (a0, · · · , an) and (aπ(0), · · · , aπ(n)) as the same oriented simplex if π ∈ Sn is an even permutation. We often denote an oriented simplex as σ, and then ¯σ denotes the same simplex with the opposite orientation. Example. As oriented 2-simplices, (v0, v1, v2) and (v1, v2, v0) are equal, but they are diﬀerent from (v2, v1, v0). We can imagine the two diﬀerent orientations of the simplices as follows: v1 v2 v0 v2 v0 v1 One and two dimensions are the dimensions where we can easily visualize the orientation. This is substantially harder in higher dimensions, and often we just work with the deﬁnition instead. Deﬁnition (Chain group Cn(K)). Let K be a simplicial complex. For each n ≥ 0, we deﬁne Cn(K) as follows: Let {σ1, · · · , σ} be the set of n-simplices of K. For each i, choose an orientation on σi. That is, choose an order for the vertices (up to an even 65 7 Simplicial homology II Algebraic Topology permutation). This choice is not important, but we need to make it. Now when we say σi, we mean the oriented simplex with this particular orientation. Now let Cn(K) be the free abelian group with basis {σ1, · · · , σ}, i.e. Cn(K) ∼= Z. So an element in Cn(K) might look like σ3 − 7σ1 + 52σ64 − 28σ1000000. In other words, an element of Cn(K) is just a formal sum of n-simplices. For convenience, we deﬁne C(K) = 0 for < 0. This will save us from making exceptions for n = 0 cases later. For each oriented simplex, we identify −σi with ¯σi, at least when n ≥ 1. In this deﬁnition, we have to choose a particular orientation for each of our simplices. If you don’t like making arbitrary choices, we could instead deﬁne Cn(K) as some quotient, but it is slightly more complicated. Note that if there are no n-simplices (e.g. when n = −1), we can still meaningfully talk about Cn(K), but it’s just 0. Example. We can think of elements in the chain group C1(X) as “paths” in X. For example, we might have the following simplex: v1 σ1 σ2 v0 σ3 v2 σ6 Then the path v0 v1 v2 v0 v1 around the left triangle is represented by the member σ1 − σ2 + σ3 + σ1 = 2σ1 − σ2 + σ3. Of course, with this setup, we can do more random things, like adding 57 copies of σ6 to it, and this is also allowed. So we could think of these as disjoint union of paths instead. When deﬁning fundamental groups, we had homotopies that allowed us to “move around”. Somehow, we would like to say that two of these paths are “equivalent” under certain conditions. To do so, we need to deﬁne the boundary homomorphisms. Deﬁnition (Boundary homomorphisms). We deﬁne boundary homomorphisms by dn : Cn(K) → Cn−1(K) (a0, · · · , an) → n i=0 (−1)i(a0, · · · , ˆai, · · · , an), where (a0, · · · , ˆai, · · · , an) = (a0, · · · , ai−1, ai+1, · · · , an) is the simplex with ai removed. This means we remove each vertex in turn and add them up. This is clear if we draw some pictures in low dimensions: 66 7 Simplicial homology II Algebraic Topology v0 v1 −v0 v1 If we take a triangle, we get v1 v1 v0 v2 v0 v2 An important property of the boundary map is that the boundary of a boundary is empty: Lemma. dn−1 ◦ dn = 0. In other words, im dn+1 ⊆ ker dn. Proof. This just involves expanding the deﬁnition and working through the mess. With this in mind, we will deﬁne the homology groups as follows: Deﬁnition (Simplicial homology group Hn(K)). The nth simplicial homology group Hn(K) is deﬁned as Hn(K) = ker dn im dn+1 . This is a nice, clean deﬁnition, but what does this mean geometrically? Somehow, Hk(K) describes all the “k-dimensional holes” in \|K\|. Since we are going to draw pictures, we are going to start with the easy case of k = 1. Our H1(K) is made from the kernel of d1 and the image of d2. First, we give these things names. Deﬁnition (Chains, cycles and boundaries). The elements of Ck(K) are called k-chains of K, those of ker dk are called k-cycles of K, and those of im dk+1 are called k-boundaries of K. Suppose we have some c ∈ ker dk. In other words, dc = 0. If we interpret c as a “path”, if it has no boundary, then it represents some sort of loop, i.e. a cycle. For example, if we have the following cycle: e1 e0 e2 We have We can then compute the boundary as c = (e0, e1) + (e1, e2) + (e2, e0). dc = (e1 − e0) + (e2 − e1) + (e0 − e2) = 0. 67 7 Simplicial homology II Algebraic Topology So this c is indeed a cycle. Now if c ∈ im d2, then c = db for some 2-chain b, i.e. c is the boundary of some two-dimensional thing. This is why we call this a 1-boundary. For example, suppose we have our cycle as above, but is itself a boundary of a 2-chain. v1 v0 v2 We see that a cycle that has a boundary has been “ﬁlled in”. Hence the “holes” are, roughly, the cycles that haven’t been ﬁlled in. Hence we deﬁne the homology group as the cycles quotiented by the boundaries, and we interpret its elements as k-dimensional “holes”. Example. Let K be the standard simplicial 1-sphere, i.e. we have the following in R3. e2 e1 e0 Our simplices are thus K = {e0, e1, e2, e0, e1, e1, e2, e2, e0}. Our chain groups are C0(K) = (e0), (e1), (e2) ∼= Z3 C1(K) = (e0, e1), (e1, e2), (e2, e0) ∼= Z3. All other chain groups are zero. Note that our notation is slightly confusing here, since the brackets · can mean the simplex spanned by the vertices, or the group generated by certain elements. However, you are probably clueful enough to distinguish the two uses. Hence, the only non-zero boundary map is d1 : C1(K) → C0(K). We can write down its matrix with respect to the given basis. 
 −1 0 1 −1 0 1 0 1 −1   68 7 Simplicial homology II Algebraic Topology We have now everything we need to know about the homology groups, and we just need to do some linear algebra to ﬁgure out the image and kernel, and thus the homology groups. We have H0(K) = ker(d0 : C0(K) → C−1(K)) im(d1 : C1(K) → C0(K)) ∼= C0(K) im d1 ∼= Z3 im d1 . After doing some row operations with our matrix, we see that the image of d1 is a two-dimensional subspace generated by the image of two of the edges. Hence we have H0(K) = Z. What does this H0(K) represent? We initially said that Hk(K) should represent the k-dimensional holes, but when k = 0, this is simpler. As for π0, H0 just represents the path components of K. We interpret this to mean K has one path component. In general, if K has r path components, then we expect H0(K) to be Zr. Similarly, we have H1(K) = ker d1 im d2 ∼= ker d1. It is easy to see that in fact we have ker d1 = (e0, e1) + (e1, e2) + (e2, e0) ∼= Z. So we also have H1(K) ∼= Z. We see that this H1(K) is generated by precisely the single loop in the triangle. The fact that H1(K) is non-trivial means that we do indeed have a hole in the middle of the circle. Example. Let L be the standard 2-simplex (and all its faces) in R3. e2 e1 e0 Now our chain groups are C0(L) = C0(K) ∼= Z3 ∼= (e0), (e1), (e2) C1(L) = C1(K) ∼= Z3 ∼= (e0, e1), (e1, e2), (e2, e0) C2(L) ∼= Z = (e0, e1, e2). 69 7 Simplicial homology II Algebraic Topology Since d1 is the same as before, the only new interesting boundary map is d2. We compute d2((e0, e1, e2)) = (e0, e1) + (e1, e2) + (e2, e0). We know that H0(L) depends only on d0 and d1, which are the same as for K. So H0(L) ∼= Z. Again, the interpretation of this is that L is path-connected. The ﬁrst homology group is H1(L) = ker d1 im d2 = (e0, e1) + (e1, e2) + (e2, e0) (e0, e1) + (e1, e2) + (e2, e0) ∼= 0. This precisely illustrates the fact that the “hole” is now ﬁlled in L. Finally, we have H2(L) = ker d2 im d3 = ker d2 ∼= 0. This is zero since there aren’t any two-dimensional holes in L. We have hopefully gained some intuition on what these homology groups mean. We are now going to spend a lot of time developing formalism. 7.2 Some homological algebra We will develop some formalism to help us compute homology groups Hk(K) in lots of examples. Firstly, we axiomatize the setup we had above. Deﬁnition (Chain complex and diﬀerentials). A chain complex C· is a sequence of abelian groups C0, C1, C2, · · · equipped with maps dn : Cn → Cn−1 such that dn−1 ◦ dn = 0 for all n. We call these maps the diﬀerentials of C·. d0 0 C0 d1 C1 d2 C2 d3 · · · Whenever we have some of these things, we can deﬁne homology groups in exactly the same way as we deﬁned them for simplicial complexes. Deﬁnition (Cycles and boundaries). The space of n-cycles is The space of n-boundaries is Zn(C) = ker dn. Bn(C) = im dn+1. Deﬁnition (Homology group). The n-th homology group of C· is deﬁned to be ker dn im dn+1 Zn(C) Bn(C) Hn(C) = = . In mathematics, when we have objects, we don’t just talk about the objects themselves, but also functions between them. Suppose we have two chain complexes. For the sake of simplicity, we write the chain maps of both as dn. In general, we want to have maps between these two sequences. This would correspond to having a map fi : Ci → Di for each i, satisfying some commutativity relations. 70 7 Simplicial homology II Algebraic Topology Deﬁnition (Chain map). A chain map f· : C· → D· is a sequence of homomorphisms fn : Cn → Dn such that for all n. In other words, the following diagram commutes: fn−1 ◦ dn = dn ◦ fn 0 0 d0 C0 d1 C1 d2 C2 d3 · · · f0 f1 f2 D0 d0 d1 D1 d2 D2 d3 · · · We want to have homotopies. So far, chain complexes seem to be rather rigid, but homotopies themselves are rather ﬂoppy. How can we deﬁne homotopies for chain complexes? It turns out we can have a completely algebraic deﬁnition for chain homotopies. Deﬁnition (Chain homotopy). A chain homotopy between chain maps f·, g· : C· → D· is a sequence of homomorphisms hn : Cn → Dn+1 such that gn − fn = dn+1 ◦ hn + hn−1 ◦ dn. We write f· g· if there is a chain homotopy between f· and g·. The arrows can be put in the following diagram, which is not commutative: Cn−1 dn Cn gn fn hn hn−1 Dn dn+1 Dn+1 The intuition behind this deﬁnition is as follows: suppose C· = C·(K) and D· = C·(L) for K, L simplicial complexes, and f· and g· are “induced” by simplicial maps f, g : K → L (if f maps an n-simplex σ to a lower-dimensional simplex, then f•σ = 0). How can we detect if f and g are homotopic via the homotopy groups? Suppose H : \|K\| × I → \|L\| is a homotopy from f to g. We further suppose that H actually comes from a simplicial map K × I → L (we’ll skim over the technical issue of how we can make K × I a simplicial complex. Instead, you are supposed to ﬁgure this out yourself in example sheet 3). Let σ be an n-simplex of K, and here is a picture of H(σ × I): 71 7 Simplicial homology II Algebraic Topology Let hn(σ) = H(σ × I). We think of this as an (n + 1)-chain. What is its boundary? We’ve got the vertical sides plus the top and bottom. The bottom should just be f (σ), and the top is just g(σ), since H is a homotopy from f to g. How about the sides? They are what we get when we pull the boundary ∂σ up with the homotopy, i.e. H(∂σ × I) = hn−1 ◦ dn(σ). Now note that f (σ) and g(σ) have opposite orientations, so we get the result dn+1 ◦ hn(σ) = hn−1 ◦ dn(σ) + gn(σ) − fn(σ). Rearranging and dropping the σs, we get dn+1 ◦ hn − hn−1 ◦ dn = gn − fn. This looks remarkably like our deﬁnition for chain homotopies of maps, with the signs a bit wrong. So in reality, we will have to ﬁddle with the sign of hn a bit to get it right, but you get the idea. Lemma. A chain map f· : C· → D· induces a homomorphism: f∗ : Hn(C) → Hn(D) [c] → [f (c)] Furthermore, if f· and g· are chain homotopic, then f∗ = g∗. Proof. Since the homology groups are deﬁned as the cycles quotiented by the boundaries, to show that f∗ deﬁnes a homomorphism, we need to show f sends cycles to cycles and boundaries to boundaries. This is an easy check. If dn(σ) = 0, then dn(fn(σ)) = fn(dn(σ)) = fn(0) = 0. So fn(σ) ∈ Zn(D). Similarly, if σ is a boundary, say σ = dn(τ ), then fn(σ) = fn(dn(τ )) = dn(fn(τ )). So fn(σ) is a boundary. It thus follows that f∗ is well-deﬁned. Now suppose hn is a chain homotopy between f and g. For any c ∈ Zn(C), we have gn(c) − fn(c) = dn+1 ◦ hn(c) + hn−1 ◦ dn(c). Since c ∈ Zn(C), we know that dn(c) = 0. So gn(c) − fn(c) = dn+1 ◦ hn(c) ∈ Bn(D). Hence gn(c) and fn(c) diﬀer by a boundary. So [gn(c)] − [fn(c)] = 0 in Hn(D), i.e. f∗(c) = g∗(c). The following statements are easy to check: Proposition. (i) Being chain-homotopic is an equivalence relation of chain maps. 72 7 Simplicial homology II Algebraic Topology (ii) If a· : A· → C· is a chain map and f· g·, then f· ◦ a· g· ◦ a·. (iii) If f : C· → D· and g : D· → A· are chain maps, then g∗ ◦ f∗ = (g· ◦ f·)∗. (iv) (idC· )∗ = idH∗(C). The last two statements can be summarized in fancy language by saying that Hn is a functor. Deﬁnition (Chain homotopy equivalence). Chain complexes C· and D· are chain homotopy equivalent if there exist f· : C· → D· and g· : D· → C· such that f· ◦ g· idD·, g· ◦ f· idC· . We should think of this in exactly the same way as we think of homotopy equivalences of spaces. The chain complexes themselves are not necessarily the same, but the induced homology groups will be. Lemma. Let f· : C· → D· be a chain homotopy equivalence, then f∗ : Hn(C) → Hn(D) is an isomorphism for all n. Proof. Let g· be the homotopy inverse. Since f· ◦ g· idD·, we know f∗ ◦ g∗ = idH∗(D). Similarly, g∗ ◦ f∗ = idH∗(C). So we get isomorphisms between Hn(C) and Hn(D). 7.3 Homology calculations We’ll get back to topology and compute some homologies. Here, K is always a simplicial complex, and C· = C·(K). Lemma. Let f : K → L be a simplicial map. Then f induces a chain map f· : C·(K) → C·(L). Hence it also induces f∗ : Hn(K) → Hn(L). Proof. This is fairly obvious, except that simplicial maps are allowed to “squash” simplices, so f might send an n-simplex to an (n − 1)-simplex, which is not in Dn(L). We solve this problem by just killing these troublesome simplices. Let σ be an oriented n-simplex in K, corresponding to a basis element of Cn(K). Then we deﬁne fn(σ) = f (σ) 0 f (σ) is an n-simplex f (σ) is a k-simplex for k < n . More precisely, if σ = (a0, · · · , an), then fn(σ) = (f (a0), · · · , f (an)) 0 f (a0), · · · , f (an) spans an n-simplex otherwise . We then extend fn linearly to obtain fn : Cn(K) → Cn(L). It is immediate from this that this satisﬁes the chain map condition, i.e. f· commutes with the boundary operators. 73 7 Simplicial homology II Algebraic Topology Deﬁnition (Cone). A simplicial complex is a cone if, for some v0 ∈ Vk, \|K\| = StK(v0) ∪ LkK(v0). v0 We see that a cone ought to be contractible — we can just squash it to the point v0. This is what the next lemma tells us. Lemma. If K is a cone with cone point v0, then inclusion i : {v0} → \|K\| induces a chain homotopy equivalence i· : Cn({v0}) → Cn(K). Therefore Hn(K The homotopy inverse to i· is given by r·, where r : k → v0 is the only map. It is clear that r· ◦ i· = id, and we need to show that i· ◦ r· id. This chain homotopy can be deﬁned by hn : Cn(K) → Cn+1(K), where hn associates to any simplex σ in LkK(v0) the simplex spanned by σ and v0. Details are left to the reader. Corollary. If ∆n is the standard n-simplex, and L consists of ∆n and all its faces, then Hk(L) = Proof. K is a cone (on any vertex). Z k = 0 k > 0 0 What we would really like is an example of non-trivial homology groups. Moreover, we want them in higher dimensions, and not just the examples we got for fundamental groups. An obvious candidate is the standard n-sphere. Corollary. Let K be the standard (n − 1)-sphere (i.e. the proper faces of L from above). Then for n ≥ 2, we have Hk(K) =    . Proof. We write down the chain groups for K and L. 0 0 C0(L) = C0(K) dL 1 dK 1 C1(L)
= C1(K) dL n−1 Cn−1(L) dL n Cn(L) = Cn−1(K) Cn(K) = 0 dK n−1 · · · · · · 74 7 Simplicial homology II Algebraic Topology For k < n − 1, we have Ck(K) = Ck(L) and Ck+1(K) = Ck+1(L). Also, the boundary maps are equal. So Hk(K) = Hk(L) = 0. We now need to compute Hn−1(K) = ker dK n−1 = ker dL n−1 = im dL n . We get the last equality since n−1 ker dL im dL n = Hn−1(L) = 0. We also know that Cn(L) is generated by just one simplex (e0, · · · , en). So Cn(L) ∼= Z. Also dL n is injective since it does not kill the generator (e0, · · · , en). So Hn−1(K) ∼= im dL n ∼= Z. This is very exciting, because at last, we have a suggestion for a non-trivial invariant of Sn−1 for n > 2. We say this is just a “suggestion”, since the simplicial homology is deﬁned for simplicial complexes, and we are not guaranteed that if we put a diﬀerent simplicial complex on Sn−1, we will get the same homology groups. So the major technical obstacle we still need to overcome is to see that Hk are invariants of the polyhedron \|K\|, not just K, and similarly for maps. But this will take some time. We will quickly say something we’ve alluded to already: Lemma (Interpretation of H0). H0(K) ∼= Zd, where d is the number of path components of K. Proof. Let K be our simplicial complex and v, w ∈ Vk. We note that by deﬁnition, v, w represent the same homology class in H0(K) if and only if there is some c such that d1c = w − v. The requirement that d1c = w − v is equivalent to saying c is a path from v to w. So [v] = [w] if and only if v and w are in the same path component of K. Most of the time, we only care about path-connected spaces, so H0(K) ∼= Z. 7.4 Mayer-Vietoris sequence The Mayer-Vietoris theorem is exactly like the Seifert-van Kampen theorem for fundamental groups, which tells us what happens when we glue two spaces together. Suppose we have a space K = M ∪ N . M N 75 7 Simplicial homology II Algebraic Topology We will learn how to compute the homology of the union M ∪ N in terms of those of M, N and M ∩ N . Recall that to state the Seifert-van Kampen theorem, we needed to learn some new group-theoretic notions, such as free products with amalgamation. The situation is somewhat similar here. We will need to learn some algebra in order to state the Mayer-Vietoris theorem. The objects we need are known as exact sequences. Deﬁnition (Exact sequence). A pair of homomorphisms of abelian groups is exact (at B) if f A B g C im f = ker g. A collection of homomorphisms fi−1 · · · Ai fi Ai+1 fi+1 Ai+2 fi+2 · · · is exact at Ai if We say it is exact if it is exact at every Ai. ker fi = im fi−1. Recall that we have seen something similar before. When we deﬁned the chain complexes, we had d2 = 0, i.e. im d ⊆ ker d. Here we are requiring exact equivalence, which is something even better. Algebraically, we can think of an exact sequence as chain complexes with trivial homology groups. Alternatively, we see the homology groups as measuring the failure of a sequence to be exact. There is a particular type of exact sequences that is important. Deﬁnition (Short exact sequence). A short exact sequence is an exact sequence of the form 0 f A B g C 0 What does this mean? – The kernel of f is equal to the image of the zero map, i.e. {0}. So f is injective. – The image of g is the kernel of the zero map, which is everything. So g is surjective. – im f = ker g. Since we like chain complexes, we can produce short exact sequences of chain complexes. Deﬁnition (Short exact sequence of chain complexes). A short exact sequence of chain complexes is a pair of chain maps i· and j· 0 i· A· j· C· 0 B· 76 7 Simplicial homology II Algebraic Topology such that for each k, 0 Ak ik Bk jk Ck 0 is exact. Note that by requiring the maps to be chain maps, we imply that ik and jk commute with the boundary maps of the chain complexes. The reason why we care about short exact sequences of chain complexes (despite them having such a long name) is the following result: Theorem (Snake lemma). If we have a short exact sequence of complexes 0 A· i· B· j· C· 0 then a miracle happens to their homology groups. In particular, there is a long exact sequence (i.e. an exact sequence that is not short) · · · Hn(A) Hn−1(A) i∗ i∗ Hn(B) ∂∗ Hn−1(B) j∗ j∗ Hn(C) Hn−1(C) · · · where i∗ and j∗ are induced by i· and j·, and ∂∗ is a map we will deﬁne in the proof. Having an exact sequence is good, since if we know most of the terms in an exact sequence, we can ﬁgure out the remaining ones. To do this, we also need to understand the maps i∗, j∗ and ∂∗, but we don’t need to understand all, since we can deduce some of them with exactness. Yet, we still need to know some of them, and since they are deﬁned in the proof, you need to remember the proof. Note, however, if we replace Z in the deﬁnition of chain groups by a ﬁeld (e.g. Q), then all the groups become vector spaces. Then everything boils down to the rank-nullity theorem. Of course, this does not get us the right answer in exams, since we want to have homology groups over Z, and not Q, but this helps us to understand the exact sequences somewhat. If, at any point, homology groups confuse you, then you can try to work with homology groups over Q and get a feel for what homology groups are like, since this is easier. We are ﬁrst going to apply this result to obtain the Mayer-Vietoris theorem. Theorem (Mayer-Vietoris theorem). Let K, L, M, N be simplicial complexes with K = M ∪ N and L = M ∩ N . We have the following inclusion maps: i L j N M k K. Then there exists some natural homomorphism ∂∗ : Hn(K) → Hn−1(L) that 77 7 Simplicial homology II Algebraic Topology gives the following long exact sequence: ∂∗ · · · Hn(L) i∗+j∗ Hn(M ) ⊕ Hn(N ) k∗−∗ Hn(K) ∂∗ Hn−1(L) i∗+j∗ Hn−1(M ) ⊕ Hn−1(N ) k∗−∗ Hn−1(K) · · · H0(M ) ⊕ H0(N ) k∗−∗ H0(K) · · · 0 Here A ⊕ B is the direct sum of the two (abelian) groups, which may also be known as the Cartesian product. Note that unlike the Seifert-van Kampen theorem, this does not require the intersection L = M ∩ N to be (path) connected. This restriction was needed for Seifert-van Kampen since the fundamental group is unable to see things outside the path component of the basepoint, and hence it does not like non-path connected spaces well. However, homology groups don’t have these problems. Proof. All we have to do is to produce a short exact sequence of complexes. We have 0 Cn(L) in+jn Cn(M ) ⊕ Cn(N ) kn−n Cn(K) 0 Here in + jn : Cn(L) → Cn(M ) ⊕ Cn(N ) is the map x → (x, x), while kn − n : Cn(M ) ⊕ Cn(N ) → Cn(K) is the map (a, b) → a − b (after applying the appropriate inclusion maps). It is easy to see that this is a short exact sequence of chain complexes. The image of in + jn is the set of all elements of the form (x, x), and the kernel of kn − n is also these. It is also easy to see that in + jn is injective and kn − n is surjective. At ﬁrst sight, the Mayer-Vietoris theorem might look a bit scary to use, since it involves all homology groups of all orders at once. However, this is actually often a good thing, since we can often use this to deduce the higher homology groups from the lower homology groups. Yet, to properly apply the Mayer-Vietoris sequence, we need to understand the map ∂∗. To do so, we need to prove the snake lemma. Theorem (Snake lemma). If we have a short exact sequence of complexes 0 A· i· B· j· C· 0 then there is a long exact sequence · · · Hn(A) Hn−1(A) i∗ i∗ Hn(B) ∂∗ Hn−1(B) j∗ j∗ Hn(C) Hn−1(C) · · · where i∗ and j∗ are induced by i· and j·, and ∂∗ is a map we will deﬁne in the proof. 78 7 Simplicial homology II Algebraic Topology The method of proving this is sometimes known as “diagram chasing”, where we just “chase” around commutative diagrams to ﬁnd the elements we need. The idea of the proof is as follows — in the short exact sequence, we can think of A as a subgroup of B, and C as the quotient B/A, by the ﬁrst isomorphism theorem. So any element of C can be represented by an element of B. We apply the boundary map to this representative, and then exactness shows that this must come from some element of A. We then check carefully that this is well-deﬁned, i.e. does not depend on the representatives chosen. Proof. The proof of this is in general not hard. It just involves a lot of checking of the details, such as making sure the homomorphisms are well-deﬁned, are actually homomorphisms, are exact at all the places etc. The only important and non-trivial part is just the construction of the map ∂∗. First we look at the following commutative diagram: 0 0 in in−1 An dn An−1 Bn dn Bn−1 jn jn−1 Cn dn Cn−1 0 0 To construct ∂∗ : Hn(C) → Hn−1(A), let [x] ∈ Hn(C) be a class represented by x ∈ Zn(C). We need to ﬁnd a cycle z ∈ An−1. By exactness, we know the map jn : Bn → Cn is surjective. So there is a y ∈ Bn such that jn(y) = x. Since our target is An−1, we want to move down to the next level. So consider dn(y) ∈ Bn−1. We would be done if dn(y) is in the image of in−1. By exactness, this is equivalent to saying dn(y) is in the kernel of jn−1. Since the diagram is commutative, we know jn−1 ◦ dn(y) = dn ◦ jn(y) = dn(x) = 0, using the fact that x is a cycle. So dn(y) ∈ ker jn−1 = im in−1. Moreover, by exactness again, in−1 is injective. So there is a unique z ∈ An−1 such that in−1(z) = dn(y). We have now produced our z. We are not done. We have ∂∗[x] = [z] as our candidate deﬁnition, but we need to check many things: (i) We need to make sure ∂∗ is indeed a homomorphism. (ii) We need dn−1(z) = 0 so that [z] ∈ Hn−1(A); (iii) We need to check [z] is well-deﬁned, i.e. it does not depend on our choice of y and x for the homology class [x]. (iv) We need to check the exactness of the resulting sequence. We now check them one by one: (i) Since everything involved in deﬁning ∂∗ are homomorphisms, it follows that ∂∗ is also a homomorphism. 79 7 Simplicial homology II Algebraic Topology (ii) We check dn−1(z) = 0. To do so, we need to add an additional layer. 0 0 0 in in−1 An dn An−1 Bn dn Bn−1 jn jn−1 Cn dn Cn−1 dn−1 dn−1 d
n−1 An−2 in−2 Bn−2 jn−2 Cn−2 0 0 0 We want to check that dn−1(z) = 0. We will use the commutativity of the diagram. In particular, we know in−2 ◦ dn−1(z) = dn−1 ◦ in−1(z) = dn−1 ◦ dn(y) = 0. By exactness at An−2, we know in−2 is injective. So we must have dn−1(z) = 0. (iii) (a) First, in the proof, suppose we picked a diﬀerent y such that jn(y) = jn(y) = x. Then jn(y − y) = 0. So y − y ∈ ker jn = im in. Let a ∈ An be such that in(a) = y − y. Then dn(y) = dn(y − y) + dn(y) = dn ◦ in(a) + dn(y) = in−1 ◦ dn(a) + dn(y). Hence when we pull back dn(y) and dn(y) to An−1, the results diﬀer by the boundary dn(a), and hence produce the same homology class. (b) Suppose [x] = [x]. We want to show that ∂∗[x] = ∂∗[x]. This time, we add a layer above. 0 0 0 An+1 in+1 Bn+1 jn+1 Cn+1 dn+1 dn+1 dn+1 in in−1 An dn An−1 Bn dn Bn−1 jn jn−1 Cn dn Cn−1 0 0 0 By deﬁnition, since [x] = [x], there is some c ∈ Cn+1 such that x = x + dn+1(c). By surjectivity of jn+1, we can write c = jn+1(b) for some b ∈ Bn+1. By commutativity of the squares, we know x = x + jn ◦ dn+1(b). The next step of the proof is to ﬁnd some y such that jn(y) = x. Then jn(y + dn+1(b)) = x. So the corresponding y is y = y + dn+1(b). So dn(y) = dn(y), and hence ∂∗[x] = ∂∗[x]. 80 7 Simplicial homology II Algebraic Topology (iv) This is yet another standard diagram chasing argument. When reading this, it is helpful to look at a diagram and see how the elements are chased along. It is even more beneﬁcial to attempt to prove this yourself. (a) im i∗ ⊆ ker j∗: This follows from the assumption that in ◦ jn = 0. (b) ker j∗ ⊆ im i∗: Let [b] ∈ Hn(B). Suppose j∗([b]) = 0. Then there is some c ∈ Cn+1 such that jn(b) = dn+1(c). By surjectivity of jn+1, there is some b ∈ Bn+1 such that jn+1(b) = c. By commutativity, we know jn(b) = jn ◦ dn+1(b), i.e. jn(b − dn+1(b)) = 0. By exactness of the sequence, we know there is some a ∈ An such that in(a) = b − dn+1(b). Moreover, in−1 ◦ dn(a) = dn ◦ in(a) = dn(b − dn+1(b)) = 0, using the fact that b is a cycle. Since in−1 is injective, it follows that dn(a) = 0. So [a] ∈ Hn(A). Then i∗([a]) = [b] − [dn+1(b)] = [b]. So [b] ∈ im i∗. (c) im j∗ ⊆ ker ∂∗: Let [b] ∈ Hn(B). To compute ∂∗(j∗([b])), we ﬁrst pull back jn(b) to b ∈ Bn. Then we compute dn(b) and then pull it back to An+1. However, we know dn(b) = 0 since b is a cycle. So ∂∗(j∗([b])) = 0, i.e. ∂∗ ◦ j∗ = 0. (d) ker ∂∗ ⊆ im j∗: Let [c] ∈ Hn(C) and suppose ∂∗([c]) = 0. Let b ∈ Bn be such that jn(b) = c, and a ∈ An−1 such that in−1(a) = dn(b). By assumption, ∂∗([c]) = [a] = 0. So we know a is a boundary, say a = dn(a) for some a ∈ An. Then by commutativity we know dn(b) = dn ◦ in(a). In other words, dn(b − in(a)) = 0. So [b − in(a)] ∈ Hn(B). Moreover, j∗([b − in(a)]) = [jn(b) − jn ◦ in(a)] = [c]. So [c] ∈ im j∗. (e) im ∂∗ ⊆ ker i∗: Let [c] ∈ Hn(C). Let b ∈ Bn be such that jn(b) = c, and a ∈ An−1 be such that in(a) = dn(b). Then ∂∗([c]) = [a]. Then i∗([a]) = [in(a)] = [dn(b)] = 0. So i∗ ◦ ∂∗ = 0. (f) ker i∗ ⊆ im ∂∗: Let [a] ∈ Hn(A) and suppose i∗([a]) = 0. So we can ﬁnd some b ∈ Bn+1 such that in(a) = dn+1(b). Let c = jn+1(b). Then dn+1(c) = dn+1 ◦ jn+1(b) = jn ◦ dn+1(b) = jn ◦ in(a) = 0. So [c] ∈ Hn(C). Then [a] = ∂∗([c]) by deﬁnition of ∂∗. So [a] ∈ im ∂∗. 81 7 Simplicial homology II Algebraic Topology 7.5 Continuous maps and homotopy invariance This is the most technical part of the homology section of the course. The goal is to see that the homology groups H∗(K) depend only on the polyhedron, and not the simplicial structure on it. Moreover, we will show that they are homotopy invariants of the space, and that homotopic maps f g : \|K\| → \|L\| induce equal maps H∗(K) → H∗(L). Note that this is a lot to prove. At this point, we don’t even know arbitrary continuous maps induce any map on the homology. We only know simplicial maps do. We start with a funny deﬁnition. Deﬁnition (Contiguous maps). Simplicial maps f, g : K → L are contiguous if for each σ ∈ K, the simplices f (σ) and g(σ) (i.e. the simplices spanned by the image of the vertices of σ) are faces of some some simplex τ ∈ L. σ τ f g The signiﬁcance of this deﬁnition comes in two parts: simplicial approximations of the same map are contiguous, and contiguous maps induce the same maps on homology. Lemma. If f, g : K → L are simplicial approximations to the same map F , then f and g are contiguous. Proof. Let σ ∈ K, and pick some s ∈ ˚σ. Then F (s) ∈ ˚τ for some τ ∈ L. Then the deﬁnition of simplicial approximation implies that for any simplicial approximation f to F , f (σ) spans a face of τ . Lemma. If f, g : K → L are continguous simplicial maps, then f∗ = g∗ : Hn(K) → Hn(L) for all n. Geometrically, the homotopy is obvious. If f (σ) and g(σ) are both faces of τ , then we just pick the homotopy as τ . The algebraic deﬁnition is less inspiring, and it takes some staring to see it is indeed what we think it should be. Proof. We will write down a chain homotopy between f and g: hn((a0, · · · , an)) = n i=0 (−1)i[f (a0), · · · , f (ai), g(ai), · · · , g(an)], where the square brackets means the corresponding oriented simplex if there are no repeats, 0 otherwise. We can now check by direct computation that this is indeed a chain homotopy. 82 7 Simplicial homology II Algebraic Topology Now we know that if f, g : K → L are both simplicial approximations to the same F , then they induce equal maps on homology. However, we do not yet know that continuous maps induce well-deﬁned maps on homologies, since to produce simplicial approximations, we needed to perform barycentric subdivision, and we need to make sure this does not change the homology. K K Given a barycentric subdivision, we need to choose a simplicial approximation to the identity map a : K → K. It turns out this is easy to do, and we can do it almost arbitrarily. Lemma. Each vertex ˆσ ∈ K is a barycenter of some σ ∈ K. Then we choose a(ˆσ) to be an arbitrary vertex of σ. This deﬁnes a function a : VK → VK. This a is a simplicial approximation to the identity. Moreover, every simplicial approximation to the identity is of this form. Proof. Omitted. Next, we need to show this gives an isomorphism on the homologies. Proposition. Let K be the barycentric subdivision of K, and a : K → K a simplicial approximation to the identity map. Then the induced map a∗ : Hn(K ) → Hn(K) is an isomorphism for all n. Proof. We ﬁrst deal with K being a simplex σ and its faces. Now that K is just a cone (with any vertex as the cone vertex), and K is also a cone (with the barycenter as the cone vertex). Therefore Hn(K) ∼= Hn(K ) ∼= Z n = 0 0 n > 0 So only n = 0 is (a little) interesting, but it is easy to check that a∗ is an isomorphism in this case, since it just maps a vertex to a vertex, and all vertices in each simplex are in the same homology class. To ﬁnish the proof, note that K is built up by gluing up simplices, and K is built by gluing up subdivided simplices. So to understand their homology groups, we use the Mayer-Vietoris sequence. Given a complicated simplicial complex K, let σ be a maximal dimensional simplex of K. We let L = K \ {σ} (note that L includes the boundary of σ). We let S = {σ and all its faces} ⊆ K and T = L ∩ S. We can do similarly for K , and let L, S, T be the corresponding barycentric subdivisions. We have K = L ∪ S and K = L ∪ S (and L ∩ S = T ). By the previous lemma, we see our construction of a gives a(L) ⊆ L, a(S) ⊆ S and a(T ) ⊆ T . So these induce maps of the corresponding homology groups Hn(T ) Hn(S) ⊕ Hn(L) Hn(K ) Hn−1(T ) Hn−1(S) ⊕ Hn−1(L) a∗ a∗⊕a∗ a∗ a∗ a∗⊕a∗ Hn(T ) Hn(S) ⊕ Hn(L) Hn(K) Hn−1(T ) Hn−1(S) ⊕ Hn−1(L) 83 7 Simplicial homology II Algebraic Topology By induction, we can assume all but the middle maps are isomorphisms. By the ﬁve lemma, this implies the middle map is an isomorphism, where the ﬁve lemma is as follows: Lemma (Five lemma). Consider the following commutative diagram: A1 a A2 B1 b B2 C1 c C2 D1 d D2 E1 e E2 If the top and bottom rows are exact, and a, b, d, e are isomorphisms, then c is also an isomorphism. Proof. Exercise in example sheet. We now have everything we need to move from simplical maps to continuous maps. Putting everything we have proved so far together, we obtain the following result: Proposition. To each continuous map f : \|K\| → \|L\|, there is an associated map f∗ : Hn(K) → Hn(L) (for all n) given by f∗ = s∗ ◦ ν−1 K,r, where s : K (r) → L is a simplicial approximation to f , and νK,r : Hn(K (r)) → Hn(K) is the isomorphism given by composing maps Hn(K (i)) → Hn(K (i−1)) induced by simplical approximations to the identity. Furthermore: (i) f∗ does not depend on the choice of r or s. (ii) If g : \|M \| → \|K\| is another continuous map, then (f ◦ g)∗ = f∗ ◦ g∗. Proof. Omitted. Corollary. If f : \|K\| → \|L\| is a homeomorphism, then f∗ : Hn(K) → Hn(L) is an isomorphism for all n. Proof. Immediate from (ii) of previous proposition. This is good. We know now that homology groups is a property of the space itself, not simplicial complexes. For example, we have computed the homology groups of a particular triangulation of the n-sphere, and we know it is true for all triangulations of the n-sphere. We’re not exactly there yet. We have just seen that homology groups are invariant under homeomorphisms. What we would like to know is that they are invariant under homotopy. In other words, we want to know homotopy equivalent maps induce equal maps on the homology groups. The strategy is: (i) Show that “small” homotopies don’t change the maps on Hn 84 7 Simplicial homology II Algebraic Topology (ii) Note that all homotopies can be decomposed into “small” homotopies. Lemma. Let L be a simplicial complex (with \|L\| ⊆ Rn). Then there is an ε = ε(L) > 0 such that if f, g : \|K\| → \|L\| satisfy f (x) − g(x) < ε, then f∗ = g∗ : Hn(K) → Hn(L) for all n. The idea of the proof is that if f (x) − g(x) is small enough, we can barycentrically subdivide K such that we get a simplicial approximation to both f and g. Proof. By the Lebesgue number lemma, there is an ε
> 0 such that each ball of radius 2ε in \|L\| lies in some star StL(w). Now apply the Lebesgue number lemma again to {f −1(Bε(y))}y∈\|L\|, an open cover of \|K\|, and get δ > 0 such that for all x ∈ \|K\|, we have f (Bδ(x)) ⊆ Bε(y) ⊆ B2ε(y) ⊆ StL(w) for some y ∈ \|L\| and StL(w). Now since g and f diﬀer by at most ε, we know g(Bδ(x)) ⊆ B2ε(y) ⊆ StL(w). Now subdivide r times so that µ(K (r)) < 1 2 δ. So for all v ∈ VK(r) , we know StK(r)(v) ⊆ Bδ(v). This gets mapped by both f and g to StL(w) for the same w ∈ VL. We deﬁne s : VK(r) → VL sending v → w. Theorem. Let f g : \|K\| → \|L\|. Then f∗ = g∗. Proof. Let H : \|K\| × I → \|L\|. Since \|K\| × I is compact, we know H is uniformly continuous. Pick ε = ε(L) as in the previous lemma. Then there is some δ such that \|s − t\| < δ implies \|H(x, s) − H(x, t)\| < ε for all x ∈ \|K\|. Now choose 0 = t0 < t1 < · · · < tn = 1 such that ti − ti−1 < δ for all i. Deﬁne fi : \|K\| → \|L\| by fi(x) = H(x, ti). Then we know fi − fi−1 < ε for all i. Hence (fi)∗ = (fi−1)∗. Therefore (f0)∗ = (fn)∗, i.e. f∗ = g∗. This is good, since we know we can not only deal with spaces that are homeomorphic to complexes, but spaces that are homotopic to complexes. This is important, since all complexes are compact, but we would like to talk about non-compact spaces such as open balls as well. Deﬁnition (h-triangulation and homology groups). An h-triangulation of a space X is a simplicial complex K and a homotopy equivalence h : \|K\| → X. We deﬁne Hn(X) = Hn(K) for all n. Lemma. Hn(X) is well-deﬁned, i.e. it does not depend on the choice of K. Proof. Clear from previous theorem. We have spent a lot of time and eﬀort developing all the technical results and machinery of homology groups. We will now use them to do stuﬀ. 85 7 Simplicial homology II Algebraic Topology Digression — historical motivation At ﬁrst, we motivated the study of algebraic topology by saying we wanted to show that Rn and Rm are not homeomorphic. However, historically, this is not why people studied algebraic topology, since there are other ways to prove this is true. Initially, people were interested in studying knots. These can be thought of as embeddings S → S3. We really should just think of S3 as R3 ∪ {∞}, where the point at inﬁnity is just there for convenience. The most interesting knot we know is the unknot U : A less interesting but still interesting knot is the trefoil T . One question people at the time asked was whether the trefoil knot is just the unknot in disguise. It obviously isn’t, but can we prove it? In general, given two knots, is there any way we can distinguish if they are the same? The idea is to study the fundamental groups of the knots. It would obviously be silly to compute the fundamental groups of U and T themselves, since they are both isomorphic to S1 and the groups are just Z. Instead, we look at the fundamental groups of the complements. It is not diﬃcult to see that π1(S3 \ U ) ∼= Z, and with some suitable machinery, we ﬁnd π1(S3 \ T ) ∼= a, b \| a3b−2. Staring at it hard enough, we can construct a surjection π1(S3 \ T ) → S3. This tells us π1(S3 \ T ) is non-abelian, and is certainly not Z. So we know U and T are genuinely diﬀerent knots. 7.6 Homology of spheres and applications Lemma. The sphere Sn−1 is triangulable, and Hk(Sn−1) ∼= Z k = 0, n − 1 otherwise 0 86 7 Simplicial homology II Algebraic Topology Proof. We already did this computation for the standard (n − 1)-sphere ∂∆n, where ∆n is the standard n-simplex. We immediately have a few applications Proposition. Rn ∼= Rm for m = n. Proof. See example sheet 4. Theorem (Brouwer’s ﬁxed point theorem (in all dimensions)). There is no retraction Dn onto ∂Dn ∼= Sn−1. So every continuous map f : Dn → Dn has a ﬁxed point. Proof. The proof is exactly the same as the two-dimensional case, with homology groups instead of the fundamental group. We ﬁrst show the second part from the ﬁrst. Suppose f : Dn → Dn has no ﬁxed point. Then the following g : Dn → ∂Dn is a continuous retraction. f (x) x g(x) So we now show no such continuous retraction can exist. Suppose r : Dn → ∂Dn is a retraction, i.e. r ◦ i id : ∂Dn → ∂Dn. Sn−1 i Dn r Sn−1 We now take the (n − 1)th homology groups to obtain Hn−1(Sn−1) i∗ Hn−1(Dn) r∗ Hn−1(Sn−1). Since r ◦ i is homotopic to the identity, this composition must also be the identity, but this is clearly nonsense, since Hn−1(Sn−1) ∼= Z while Hn−1(Dn) ∼= 0. So such a continuous retraction cannot exist. Note it is important that we can work with continuous maps directly, and not just their simplicial approximations. Otherwise, here we can only show that every simplicial approximation of f has a ﬁxed point, but this does not automatically entail that f itself has a ﬁxed point. For the next application, recall from the ﬁrst example sheet that if n is odd, then the antipodal map a : Sn → Sn is homotopic to the identity. What if n is even? The idea is to consider the eﬀect on the homology group: a∗ : Hn(Sn) → Hn(Sn). 87 7 Simplicial homology II Algebraic Topology By our previous calculation, we know a∗ is a map a∗ : Z → Z. If a is homotopic to the identity, then a∗ should be homotopic to the identity map. We will now compute a∗ and show that it is multiplication by −1 when n is even. To do this, we want to use a triangulation that is compatible with the antipodal map. The standard triangulation clearly doesn’t work. Instead, we use the following triangulation h : \|K\| → Sn: The vertices of K are given by VK = {±e0, ±e1, · · · , ±en}. This triangulation works nicely with the antipodal map, since this maps a vertex to a vertex. To understand the homology group better, we need the following lemma: Lemma. In the triangulation of Sn given by vertices VK = {±e0, ±e1, · · · , ±en}, the element x = ε0 · · · εn(ε0e0, · · · , εnen) is a cycle and generates Hn(Sn). ε∈{±1}n+1 Proof. By direct computation, we see that dx = 0. So x is a cycle. To show it generates Hn(Sn), we note that everything in Hn(Sn) ∼= Z is a multiple of the generator, and since x has coeﬃcients ±1, it cannot be a multiple of anything else (apart from −x). So x is indeed a generator. Now we can prove our original proposition. Proposition. If n is even, then the antipodal map a id. Proof. We can directly compute that a∗x = (−1)n+1x. If n is even, then a∗ = −1, but id∗ = 1. So a id. 7.7 Homology of surfaces We want to study compact surfaces and their homology groups. To work with the simplicial homology, we need to assume they are triangulable. We will not prove this fact, and just assume it to be true (it really is). Recall we have classiﬁed compact surfaces, and have found the following orientable surfaces Σg. 88 7 Simplicial homology II Algebraic Topology We also have non-compact versions of these, known as Fg, where we take the above ones and cut out a hole: We want to compute the homology groups of these Σg. One way to do so is to come up with speciﬁc triangulations of the space, and then compute the homology groups directly. However, there is a better way to do it. Given a Σg, we can slice it apart along a circle and write it as Σg = Fg−1 ∪ F1. Then, we need to compute H∗(Fg). Fortunately, this is not too hard, since it turns out Fg is homotopic to a relatively simple space. Recall that we produced Σg by starting with a 4g-gon and gluing edges: b1 a1 a1 b2 b1 a2 a2 b2 Now what is Fg? Fg is Σg with a hole cut out of it. We’ll cut the hole out at the center, and be left with b1 a1 a1 b2 b1 a2 a2 b2 We can now expand the hole to the boundary, and get a deformation retraction from Fg to its boundary: 89 7 Simplicial homology II Algebraic Topology b1 a1 a1 b2 b1 a2 a2 b2 Gluing the edges together, we obtain the rose with 2g petals, which we shall call X2g: Using the Mayer-Vietoris sequence (exercise), we see that Hn(Fg) ∼= Hn(X2g) ∼=   Z n = 0 Z2g n = 1  n > 1 0 We now compute the homology groups of Σg. The Mayer-Vietoris sequence gives the following 0 H2(S1) H2(Fg−1) ⊕ H2(F1) H2(Σg) H1(S1) H1(Fg−1) ⊕ H1(F1) H1(Σg) H0(S1) H0(Fg−1) ⊕ H0(F1) H0(Σg) We can put in the terms we already know, and get 0 H2(Σg) Z Z2g H1(Σg) Z Z2 Z 0 0 By exactness, we know H2(Σg) = ker{Z → Z2g}. We now note that this map is indeed the zero map, by direct computation — this map sends S1 to the If we look at the picture, after the boundary of the hole of Fg−1 and F1. deformation retraction, this loop passes through each one-cell twice, once with each orientation. So these cancel, and give 0. So H2(Σg) = Z. To compute H1(Σg), for convenience, we break it up into a short exact sequence, noting that the function Z → Z2g is zero: 0 Z2g H1(Σg) ker(Z → Z2) 0. 90 7 Simplicial homology II Algebraic Topology We now claim that Z → Z2 is injective — this is obvious, since it sends 1 → (1, 1). So the kernel is zero. So H1(Σg) ∼= Z2g. This is a typical application of the Mayer-Vietoris sequence. We write down the long exact sequence, and put in the terms we already know. This does not immediately give us the answer we want — we will need to understand one or two of the maps, and then we can ﬁgure out all the groups we want. 7.8 Rational homology, Euler and Lefschetz numbers So far, we have been working with chains with integral coeﬃcients. It turns out we can use rational coeﬃcients instead. In the past, Cn(K) was an abelian group, or a Z-module. If we use rational coeﬃcients, since Q is a ﬁeld, this becomes a vector space, and we can use a lot of nice theorems about vector spaces, such as the rank-nullity theorem. Moreover, we can reasonably talk about things like the dimensions of these homology groups. Deﬁnition (Rational homology group). For a simplicial complex K, we can deﬁne the rational n-chain group Cn(K, Q) in the same way as Cn(K) = Cn(K, Z). That is, Cn(K, Q) is the vector space over Q with basis the n-simplices of K (with a choice of orientation). We can deﬁne dn, Zn, Bn as before, and the rational nth homology group is Hn(K; Q) ∼= Zn(K; Q) Bn(K; Q) . Now our homology group is a vector space, and it is much easier to work with.
However, the consequence is that we will lose some information. Fortunately, the way in which we lose information is very well-understood and rather simple. Lemma. If Hn(K) ∼= Zk ⊕ F for F a ﬁnite group, then Hn(K; Q) ∼= Qk. Proof. Exercise. Hence, when passing to rational homology groups, we lose all information about the torsion part. In some cases this information loss is not very signiﬁcant, but in certain cases, it can be. For example, in RP2, we have Hn(RP2) ∼=    Z n = 0 Z/2 n = 1 n > 1 0 If we pass on to rational coeﬃcients, then we have lost everything in H1(RP2), and RP2 looks just like a point: Example. We have Hn(RP2; Q) ∼= Hn(∗; Q). Hn(Sn; Q) ∼= Q k = 0, n otherwise 0 . 91 7 Simplicial homology II Algebraic Topology We also have Hn(Σg, Q) ∼=   Q Q2g  0 k = 0, 2 k = 1 otherwise . In this case, we have not lost any information because there was no torsion part of the homology groups. However, for the non-orientable surfaces, one can show that so Hk(En) =    Z k = 0 Zn−1 × Z/2 k = 1 0 otherwise , Hk(En; Q) =   Q Qn−1  0 k = 0 k = 1 otherwise . This time, this is diﬀerent from the integral coeﬃcient case, where we have an extra Z2 term in H1. The advantage of this is that the homology “groups” are in fact vector spaces, and we can talk about things like dimensions. Also, maps on homology groups are simply linear maps, i.e. matrices, and we can study their properties with proper linear algebra. Recall that the Euler characteristic of a polyhedron is deﬁned as “faces − edges + vertices”. This works for two-dimensional surfaces, and we would like to extend this to higher dimensions. There is an obvious way to do so, by counting higher-dimensional surfaces and putting the right signs. However, if we deﬁne it this way, it is not clear that this is a property of the space itself, and not just a property of the triangulation. Hence, we deﬁne it as follows: Deﬁnition (Euler characteristic). The Euler characteristic of a triangulable space X is χ(X) = (−1)i dimQ Hi(X; Q). i≥0 This clearly depends only on the homotopy type of X, and not the triangula- tion. We will later show this is equivalent to what we used to have. More generally, we can deﬁne the Lefschetz number. Deﬁnition (Lefschetz number). Given any map f : X → X, we deﬁne the Lefschetz number of f as L(f ) = i≥0 (−1)i tr(f∗ : Hi(X; Q) → Hi(X; Q)). Why is this a generalization of the Euler characteristic? Just note that the trace of the identity map is the number of dimensions. So we have χ(X) = L(id). 92 7 Simplicial homology II Algebraic Topology Example. We have χ(Sn) = 2 n even 0 n odd We also have χ(Σg) = 2 − 2g, χ(En) = 2 − n. Example. If α : Sn → Sn is the antipodal map, we saw that α∗ : Hn(Sn) → Hn(Sn) is multiplication by (−1)n+1. So L(α) = 1 + (−1)n(−1)n+1 = 1 − 1 = 0. We see that even though the antipodal map has diﬀerent behaviour for diﬀerent dimensions, the Lefschetz number ends up being zero all the time. We will soon see why this is the case. Why do we want to care about the Lefschetz number? The important thing is that we will use this to prove a really powerful generalization of Brouwer’s ﬁxed point theorem that allows us to talk about things that are not balls. Before that, we want to understand the Lefschetz number ﬁrst. To deﬁne the Lefschetz number, we take the trace of f∗, and this is a map of the homology groups. However, we would like to understand the Lefschetz number in terms of the chain groups, since these are easier to comprehend. Recall that the homology groups are deﬁned as quotients of the chain groups, so we would like to know what happens to the trace when we take quotients. Lemma. Let V be a ﬁnite-dimensional vector space and W ≤ V a subspace. Let A : V → V be a linear map such that A(W ) ⊆ W . Let B = A\|W : W → W and C : V /W → V /W the induced map on the quotient. Then Proof. In the right basis, tr(A) = tr(B) + tr(C). A = B A 0 C . What this allows us to do is to not look at the induced maps on homology, but just the maps on chain complexes. This makes our life much easier when it comes to computation. Corollary. Let f· : C·(K; Q) → C·(K; Q) be a chain map. Then (−1)i tr(fi : Ci(K) → Ci(K)) = i≥0 i≥0 (−1)i tr(f∗ : Hi(K) → Hi(K)), with homology groups understood to be over Q. This is a great corollary. The thing on the right is the conceptually right thing to have — homology groups are nice and are properties of the space itself, not the triangulation. However, to actually do computations, we want to work with the chain groups and actually calculate with chain groups. 93 7 Simplicial homology II Algebraic Topology Proof. There is an exact sequence 0 Bi(K; Q) Zi(K; Q) Hi(K; Q) 0 This is since Hi(K, Q) is deﬁned as the quotient of Zi over Bi. We also have the exact sequence 0 Zi(K; Q) Ci(K; Q) di Bi−1(K; Q) 0 This is true by deﬁnition of Bi−1 and Zi. Let f H i , f C i maps induced by f on the corresponding groups. Then we have i , f Z , f B i be the various L(\|f \|) = (−1)i tr(f H i ) = = i≥0 i≥0 i≥0 (−1)i(tr(f Z i ) − tr(f B i )) (−1)i(tr(f C i ) − tr(f B i−1) − tr(f B i )). Because of the alternating signs in dimension, each f B with opposite signs. So all f B i cancel out, and we are left with i appears twice in the sum L(\|f \|) = i≥0 (−1)i tr(f C i ). Since tr(id \|Ci(K;Q)) = dimQ Ci(K; Q), which is just the number of i-simplices, this tells us the Euler characteristic we just deﬁned is the usual Euler characteristic, i.e. χ(X) = (−1)i number of i-simplices. Finally, we get to the important theorem of the section. i≥0 Theorem (Lefschetz ﬁxed point theorem). Let f : X → X be a continuous map from a triangulable space to itself. If L(f ) = 0, then f has a ﬁxed point. Proof. We prove the contrapositive. Suppose f has no ﬁxed point. We will show that L(f ) = 0. Let δ = inf{\|x − f (x)\| : x ∈ X} thinking of X as a subset of Rn. We know this is non-zero, since f has no ﬁxed point, and X is compact (and hence the inﬁmum point is achieved by some x). Choose a triangulation L : \|K\| → X such that µ(K) < δ 2 . We now let g : K (r) → K be a simplicial approximation to f . Since we picked our triangulation to be so ﬁne, for x ∈ σ ∈ K, we have \|f (x) − g(x)\| < δ 2 94 7 Simplicial homology II Algebraic Topology since the mesh is already less than δ 2 . Also, we know So we have \|f (x) − x\| ≥ δ. \|g(x) − x\| > δ 2 . So we must have g(x) ∈ σ. The conclusion is that for any σ ∈ K, we must have g(σ) ∩ σ = ∅. Now we compute L(f ) = L(\|g\|). The only complication here is that g is a map from K (r) to K, and the domains and codomains are diﬀerent. So we need to compose it with si : Ci(K; Q) → Ci(K (r); Q) induced by inverses of simplicial approximations to the identity map. Then we have L(\|g\|) = = i≥0 i≥0 (−1)i tr(g∗ : Hi(X; Q) → Hi(X; Q)) (−1)i tr(gi ◦ si : Ci(K; Q) → Ci(K; Q)) Now note that si takes simplices of σ to sums of subsimplices of σ. So gi ◦ si takes every simplex oﬀ itself. So each diagonal terms of the matrix of gi ◦ si is 0! Hence the trace is L(\|g\|) = 0. Example. If X is any contractible polyhedron (e.g. a disk), then L(f ) = 1 for any f : X → X, which is obvious once we contract the space to a point. So f has a ﬁxed point. Example. Suppose G is a path-connected topological group, i.e. G is a group and a topological space, and inverse and multiplication are continuous maps. If g = 1, then the map rg : G → G γ → γg has no ﬁxed point. This implies L(rg) = 0. However, since the space is path connected, rg id, where the homotopy is obtained by multiplying elements along the path from e to rg. So χ(G) = L(idG) = L(rg) = 0. So if G = 1, then in fact χ(G) = 0. This is quite fun. We have worked with many surfaces. Which can be topological groups? The torus is, since it is just S1 × S1, and S1 is a topological group. The Klein bottle? Maybe. However, other surfaces cannot since they don’t have Euler characteristic 0. 95
exact sequence of complexes 0 A· i· B· q· C· 0 . Then there are maps such that there is a long exact sequence ∂ : Hn(C·) → Hn−1(A·) · · · Hn(A) Hn−1(A) i∗ i∗ Hn(B) ∂∗ Hn−1(B) q∗ q∗ Hn(C) . Hn−1(C) · · · The method of proving this is sometimes known as “diagram chasing”, where we just “chase” around commutative diagrams to ﬁnd the elements we need. The idea of the proof is as follows — in the short exact sequence, we can think of A as a subgroup of B, and C as the quotient B/A, by the ﬁrst isomorphism theorem. So any element of C can be represented by an element of B. We apply the boundary map to this representative, and then exactness shows that this must come from some element of A. We then check carefully that these is well-deﬁned, i.e. does not depend on the representatives chosen. Proof. The proof of this is in general not hard. It just involves a lot of checking of the details, such as making sure the homomorphisms are well-deﬁned, are actually homomorphisms, are exact at all the places etc. The only important and non-trivial part is just the construction of the map ∂∗. First we look at the following commutative diagram: 0 0 An dn An−1 in in−1 Bn dn Bn−1 qn qn−1 Cn dn Cn−1 0 0 To construct ∂∗ : Hn(C) → Hn−1(A), let [x] ∈ Hn(C) be a class represented by x ∈ Zn(C). We need to ﬁnd a cycle z ∈ An−1. By exactness, we know the map qn : Bn → Cn is surjective. So there is a y ∈ Bn such that qn(y) = x. Since our target is An−1, we want to move down to the next level. So consider dn(y) ∈ Bn−1. We would be done if dn(y) is in the image of in−1. By exactness, this is equivalent saying dn(y) is in the kernel of qn−1. Since the diagram is commutative, we know qn−1 ◦ dn(y) = dn ◦ qn(y) = dn(x) = 0, 25 3 Four major tools of (co)homology III Algebraic Topology using the fact that x is a cycle. So dn(y) ∈ ker qn−1 = im in−1. Moreover, by exactness again, in−1 is injective. So there is a unique z ∈ An−1 such that in−1(z) = dn(y). We have now produced our z. We are not done. We have ∂∗[x] = [z] as our candidate deﬁnition, but we need to check many things: (i) We need to make sure ∂∗ is indeed a homomorphism. (ii) We need dn−1(z) = 0 so that [z] ∈ Hn−1(A); (iii) We need to check [z] is well-deﬁned, i.e. it does not depend on our choice of y and x for the homology class [x]. (iv) We need to check the exactness of the resulting sequence. We now check them one by one: (i) Since everything involved in deﬁning ∂∗ are homomorphisms, it follows that ∂∗ is also a homomorphism. (ii) We check dn−1(z) = 0. To do so, we need to add an additional layer. 0 0 0 An dn An−1 in in−1 Bn dn Bn−1 qn qn−1 Cn dn Cn−1 dn−1 dn−1 dn−1 An−2 in−2 Bn−2 qn−2 Cn−2 0 0 0 We want to check that dn−1(z) = 0. We will use the commutativity of the diagram. In particular, we know in−2 ◦ dn−1(z) = dn−1 ◦ in−1(z) = dn−1 ◦ dn(y) = 0. By exactness at An−2, we know in−2 is injective. So we must have dn−1(z) = 0. (iii) (a) First, in the proof, suppose we picked a diﬀerent y such that qn(y) = qn(y) = x. Then qn(y − y) = 0. So y − y ∈ ker qn = im in. Let a ∈ An be such that in(a) = y − y. Then dn(y) = dn(y − y) + dn(y) = dn ◦ in(a) + dn(y) = in−1 ◦ dn(a) + dn(y). Hence when we pull back dn(y) and dn(y) to An−1, the results diﬀer by the boundary dn(a), and hence produce the same homology class. 26 3 Four major tools of (co)homology III Algebraic Topology (b) Suppose [x] = [x]. We want to show that ∂∗[x] = ∂∗[x]. This time, we add a layer above. 0 0 0 An+1 in+1 Bn+1 qn+1 Cn+1 dn+1 dn+1 dn+1 An dn An−1 in in−1 Bn dn Bn−1 qn qn−1 Cn dn Cn−1 0 0 0 By deﬁnition, since [x] = [x], there is some c ∈ Cn+1 such that x = x + dn+1(c). By surjectivity of qn+1, we can write c = qn+1(b) for some b ∈ Bn+1. By commutativity of the squares, we know x = x + qn ◦ dn+1(b). The next step of the proof is to ﬁnd some y such that qn(y) = x. Then qn(y + dn+1(b)) = x. So the corresponding y is y = y + dn+1(b). So dn(y) = dn(y), and hence ∂∗[x] = ∂∗[x]. (iv) This is yet another standard diagram chasing argument. When reading this, it is helpful to look at a diagram and see how the elements are chased along. It is even more beneﬁcial to attempt to prove this yourself. (a) im i∗ ⊆ ker q∗: This follows from the assumption that in ◦ qn = 0. (b) ker q∗ ⊆ im i∗: Let [b] ∈ Hn(B). Suppose q∗([b]) = 0. Then there is some c ∈ Cn+1 such that qn(b) = dn+1(c). By surjectivity of qn+1, there is some b ∈ Bn+1 such that qn+1(b) = c. By commutativity, we know qn(b) = qn ◦ dn+1(b), i.e. qn(b − dn+1(b)) = 0. By exactness of the sequence, we know there is some a ∈ An such that in(a) = b − dn+1(b). Moreover, in−1 ◦ dn(a) = dn ◦ in(a) = dn(b − dn+1(b)) = 0, using the fact that b is a cycle. Since in−1 is injective, it follows that dn(a) = 0. So [a] ∈ Hn(A). Then i∗([a]) = [b] − [dn+1(b)] = [b]. So [b] ∈ im i∗. 27 3 Four major tools of (co)homology III Algebraic Topology (c) im q∗ ⊆ ker ∂∗: Let [b] ∈ Hn(B). To compute ∂∗(q∗([b])), we ﬁrst pull back qn(b) to b ∈ Bn. Then we compute dn(b) and then pull it back to An+1. However, we know dn(b) = 0 since b is a cycle. So ∂∗(q∗([b])) = 0, i.e. ∂∗ ◦ q∗ = 0. (d) ker ∂∗ ⊆ im q∗: Let [c] ∈ Hn(C) and suppose ∂∗([c]) = 0. Let b ∈ Bn be such that qn(b) = c, and a ∈ An−1 such that in−1(a) = dn(b). By assumption, ∂∗([c]) = [a] = 0. So we know a is a boundary, say a = dn(a) for some a ∈ An. Then by commutativity we know dn(b) = dn ◦ in(a). In other words, dn(b − in(a)) = 0. So [b − in(a)] ∈ Hn(B). Moreover, q∗([b − in(a)]) = [qn(b) − qn ◦ in(a)] = [c]. So [c] ∈ im q∗. (e) im ∂∗ ⊆ ker i∗: Let [c] ∈ Hn(C). Let b ∈ Bn be such that qn(b) = c, and a ∈ An−1 be such that in(a) = dn(b). Then ∂∗([c]) = [a]. Then i∗([a]) = [in(a)] = [dn(b)] = 0. So i∗ ◦ ∂∗ = 0. (f) ker i∗ ⊆ im ∂∗: Let [a] ∈ Hn(A) and suppose i∗([a]) = 0. So we can ﬁnd some b ∈ Bn+1 such that in(a) = dn+1(b). Let c = qn+1(b). Then dn+1(c) = dn+1 ◦ qn+1(b) = qn ◦ dn+1(b) = qn ◦ in(a) = 0. So [c] ∈ Hn(C). Then [a] = ∂∗([c]) by deﬁnition of ∂∗. So [a] ∈ im ∂∗. Another piece of useful algebra is known as the 5-lemma: Lemma (Five lemma). Consider the following commutative diagram If the two rows are exact, m and p are isomorphisms, q is injective and is surjective, then n is also an isomorphism. Proof. The philosophy is exactly the same as last time. We ﬁrst show that n is surjective. Let c ∈ C . Then we obtain d = t(c) ∈ D. Since p is an isomorphism, we can ﬁnd d ∈ D such that p(d) = d. Then we have q(j(d)) = u(p(d)) = u(f (c)) = 0. Since q is injective, we know j(d) = 0. Since the sequence is exact, there is some c ∈ C such that h(c) = d. We are not yet done. We do not know that n(c) = c. All we know is that d(n(c)) = d(c). So d(c − n(c)) = 0. By exactness at C , we can ﬁnd some b 28 3 Four major tools of (co)homology III Algebraic Topology such that s(b) = n(c) − c. Since m was surjective, we can ﬁnd b ∈ B such that m(b) = b. Then we have So we have n(g(b)) = n(c) − c. n(c − g(b)) = c. So n is surjective. Showing that n is injective is similar. Corollary. Let f : (X, A) → (Y, B) be a map of pairs, and that any two of f∗ : H∗(X, A) → H∗(Y, B), H∗(X) → H∗(Y ) and H∗(A) → H∗(B) are isomorphisms. Then the third is also an isomorphism. Proof. Follows from the long exact sequence and the ﬁve lemma. That wasn’t too bad, as it is just pure algebra. Proof of homotopy invariance The next goal is want to show that homotopy of continuous maps does not aﬀect the induced map on the homology groups. We will do this by showing that homotopies of maps induce homotopies of chain complexes, and chain homotopic maps induce the same map on homology groups. To make sense of this, we need to know what it means to be a homotopy of chain complexes. Deﬁnition (Chain homotopy). A chain homotopy between chain maps f·, g· : C· → D· is a collection of homomorphisms Fn : Cn → Dn+1 such that gn − fn = dD n+1 ◦ Fn + Fn−1 ◦ dC n : Cn → Dn for all n. The idea of the chain homotopy is that Fn(σ) gives us an n + 1 simplex whose boundary is gn − fn, plus some terms arising from the boundary of c itself: gn(σ) gn(σ) Fn(σ) : dFn(σ) = + Fn−1(dσ) fn(σ) fn(σ) We will not attempt to justify the signs appearing in the deﬁnition; they are what are needed for it to work. The relevance of this deﬁnition is the following result: Lemma. If f· and g· are chain homotopic, then f∗ = g∗ : H∗(C·) → H∗(D·). Proof. Let [c] ∈ Hn(C·). Then we have gn(c) − fn(c) = dD n+1Fn(c) + Fn−1(dC n (c)) = dD n+1Fn(c), where the second term dies because c is a cycle. So we have [gn(c)] = [fn(c)]. 29 3 Four major tools of (co)homology III Algebraic Topology That was the easy part. What we need to do now is to show that homotopy of maps between spaces gives a chain homotopy between the corresponding chain maps. We will change notation a bit. Notation. From now on, we will just write d for dC n . For f : X → Y , we will write f# : Cn(X) → Cn(Y ) for the map σ → f ◦ σ, i.e. what we used to call fn. Now if H : [0, 1] × X → Y is a homotopy from f to g, and σ : ∆n → X is an n-chain, then we get a homotopy [0, 1] × ∆n [0,1]×σ [0, 1] × X H Y from f#(σ) to g#(σ). Note that we write [0, 1] for the identity map [0, 1] → [0, 1]. The idea is that we are going to cut up [0, 1] × ∆n into n + 1-simplices. Suppose we can ﬁnd a collection of chains Pn ∈ Cn+1([0, 1] × ∆n) for n ≥ 0 such that d(Pn) = i1 − i0 − (−1)j([0, 1] × δj)#(Pn−1), n where j=0 i0 : δn = {0} × ∆n → [0, 1] × ∆n i1 : δn = {0} × ∆n → [0, 1] × ∆n and δj : ∆n−1 → ∆n is the inclusion of the jth face. These are “prisms” connecting the top and bottom face. Intuitively, the prism P2 looks like this: and the formula tells us its boundary is the top and bottom triangles, plus the side faces given by the prisms of the edges. Suppose we managed to ﬁnd such prisms. We can then deﬁne Fn : Cn(X) → Cn+1(Y ) by sending (σ : ∆n → X) → (H ◦ ([0, 1] × σ))#(Pn). 30 3 Four major tools of (co)homology III Algebraic Topology We now calculate. dFn(σ) = d((H ◦ (1 × ∆n))#(Pn)) = (H ◦ ([0, 1] × σ))#(d(Pn))  = (H ◦ ([0, 1] × σ))# i1 − i0 − (−1)j([0, 1] × δj)#(Pn−1)   n j=0 = H ◦
([0, 1] × σ) ◦ i1 − H ◦ ([0, 1] × σ) ◦ i0 − n j=0 (−1)jH# ◦ (([0, 1] × σ) ◦ δj)#(Pn−1 (−1)jH# ◦ (([0, 1] × σ) ◦ δj)#(Pn−1) j=0 = g ◦ σ − f ◦ σ − Fn−1(dσ) = g#(σ) − f#(σ) − Fn−1(dσ). So we just have to show that Pn exists. We already had a picture of what it looks like, so we just need to ﬁnd a formula that represents it. We view [0, 1] × ∆n ∈ R × Rn+1. Write {v0, v1, · · · , vn} for the vertices of {0} × ∆n ⊆ [1, 0] × ∆n, and {w0, · · · , wn} for the corresponding vertices of {1} × ∆n. Now if {x0, x1, · · · , xn+1} ⊆ {v0, · · · , vn} ∪ {w0, · · · , wn}, we let by [x0, · · · , xn] : ∆n → [0, 1] × ∆n (t0, · · · , tn+1) = tixi. This is still in the space by convexity. We let Pn = n i=0 (−1)i[v0, v1, · · · , vi, wi, wi+1, · · · , wn] ∈ Cn+1([0, 1] × ∆n). It is a boring check that this actually works, and we shall not bore the reader with the details. Proof of excision and Mayer-Vietoris Finally, we prove excision and Mayer-Vietoris together. It turns out both follow easily from what we call the “small simplices theorem”. Deﬁnition (C U subspaces of X such that their interiors cover X, i.e. n (X) and H U n (X)). We let U = {Uα}α∈I be a collection of ˚Uα = X. α∈I Let C U n (X) ⊆ Cn(X) be the subgroup generated by those singular n-simplices σ : ∆n → X such that σ(∆n) ⊆ Uα for some α. It is clear that if σ lies in Uα, then so do its faces. So C U n (X) is a sub-chain complex of C·(X). We write H U n (X) = Hn(C U· (X)). 31 3 Four major tools of (co)homology III Algebraic Topology It would be annoying if each choice of open cover gives a diﬀerent homology theory, because this would be too many homology theories to think about. The small simplices theorem says that the natural map H U ∗ (X) → H∗(X) is an isomorphism. Theorem (Small simplices theorem). The natural map H U isomorphism. ∗ (X) → H∗(X) is an The idea is that we can cut up each simplex into smaller parts by barycentric subdivision, and if we do it enough, it will eventually lie in on of the open covers. We then go on to prove that cutting it up does not change homology. Proving it is not hard, but technically annoying. So we ﬁrst use this to deduce our theorems. Proof of Mayer-Vietoris. Let X = A ∪ B, with A, B open in X. We let U = {A, B}, and write C·(A + B) = C U· (X). Then we have a natural chain map C·(A) ⊕ C·(B) jA−jB C·(A + B) that is surjective. The kernel consists of (x, y) such that jA(x) − jB(y) = 0, i.e. jA(x) = jB(y). But j doesn’t really do anything. It just forgets that the simplices lie in A or B. So this means y = x is a chain in A ∩ B. We thus deduce that we have a short exact sequence of chain complexes C·(A ∩ B) (iA,iB ) C·(A) ⊕ C·(B) jA−jB C·(A + B). Then the snake lemma shows that we obtain a long exact sequence of homology groups. So we get a long exact sequence of homology groups · · · Hn(A ∩ B) (iA,iB ) Hn(A) ⊕ Hn(B) jA−jB H U n (X) · · · . By the small simplices theorem, we can replace H U obtain Mayer-Vietoris. n (X) with Hn(X). So we Now what does the boundary map ∂ : Hn(X) → Hn−1(A ∩ B) do? Suppose we have c ∈ Hn(X) represented by a cycle a + b ∈ C U n (X), with a supported in A and b supported in B. By the small simplices theorem, such a representative always exists. Then the proof of the snake lemma says that ∂([a + b]) is given by tracing through Cn(A) ⊕ Cn(B) jA−jB Cn(A + B) Cn−1(A ∩ B) (iA,iB ) Cn−1(A) ⊕ Cn−1(B) d We now pull back a + b along jA − jB to obtain (a, −b), then apply d to obtain (da, −db). Then the required object is [da] = [−db]. We now move on to prove excision. 32 3 Four major tools of (co)homology III Algebraic Topology Proof of excision. Let X ⊇ A ⊇ Z be such that Z ⊇ ˚A. Let B = X \ Z. Then again take U = {A, B}. By assumption, their interiors cover X. We consider the short exact sequences 0 0 C·(A) C·(A + B) C·(A + B)/C·(A) C·(A) C·(X) C·(X, A) 0 0 Looking at the induced map between long exact sequences on homology, the middle and left terms induce isomorphisms, so the right term does too by the 5-lemma. On the other hand, the map C·(B)/C·(A ∩ B) C·(A + B)/C·(A) is an isomorphism of chain complexes. Since their homologies are H·(B, A ∩ B) and H·(X, A), we infer they the two are isomorphic. Recalling that B = X \ ¯Z, we have shown that H∗(X \ Z, A \ Z) ∼= H∗(X, A). We now provide a sketch proof of the small simplices theorem. As mentioned, the idea is to cut our simplices up, and one method to do so is barycentric subdivision. Given a 0-simplex {v0}, its barycentric subdivision is itself. If x = {x0, · · · , xn} ⊆ Rn spans an n-simplex σ, we let bx = 1 n + 1 n i=0 xi be its barycenter . If we have a 1-simplex Then the barycentric subdivision is obtained as We can degenerately describe this as ﬁrst barycentrically subdividing the boundary (which does nothing in this case), and then add the barycenter. In the case of a 2-simplex: we ﬁrst barycentrically subdivide the boundary: 33 3 Four major tools of (co)homology III Algebraic Topology Then add the barycenter bx, and for each standard simplex in the boundary, we “cone it oﬀ” towards bx: More formally, in the standard n-simplex ∆n ⊆ Rn+1, we let Bn be its barycenter. For each singular i-simplex σ : ∆i → ∆n, we deﬁne Cone∆n i (σ) : ∆i+1 → ∆n by (t0, t1, · · · , ti+1) → t0bn + (1 − t0) · σ (t1, · · · , ti+1) 1 − t0 . We can then extend linearly to get a map Cone∆n i : Ci(∆n) → Ci+1(∆n). Example. In the 2-simplex the cone of the bottom edge is the simplex in orange: Since this increases the dimension, we might think this is a chain map. Then for i > 0, we have dCone∆n i (σ) = i+1 j=0 Cone∆n i (σ) ◦ δj i+1 = σ + (−1)jCone∆n i−1(σ ◦ δj−1) j=1 = σ − Cone∆n i−1(dσ). For i = 0, we get dCone∆n i (σ) = σ − ε(σ) · bn, 34 3 Four major tools of (co)homology III Algebraic Topology In total, we have i + Cone∆n where ci = 0 for i > 0, and c0(σ) = ε(σ)bn is a map C·(∆n) → C·(∆n). i−1d = id − c·, dCone∆n We now use this cone map to construct a barycentric subdivision map ρX n : Cn(X) → Cn(X), and insist that it is natural that if f : X → Y is a map, then f# ◦ ρX n ◦ f#, i.e. the diagram n = ρY Cn(X) f# Cn(Y ) ρX n ρY n Cn(X) f# . Cn(Y ) So if σ : ∆n → X, we let ιn : ∆n → ∆n ∈ Cn(∆n) be the identity map. Then we must have n (σ) = ρX ρX n (σ#ιn) = σ#ρ∆n n (ιn). So if know how to do barycentric subdivision for ιn itself, then by naturality, we have deﬁned it for all spaces! Naturality makes life easier for us, not harder! So we deﬁne ρX n recursively on n, for all spaces X at once, by (i) ρX 0 = idC0(X) (ii) For n > 0, we deﬁne the barycentric subdivision of ιn by n (ιn) = Cone∆n ρ∆n n−1(ρ∆n n−1(dιn)), and then extend by naturality. This has all the expected properties: Lemma. ρX· is a natural chain map. Lemma. ρX· is chain homotopic to the identity. Proof. No one cares. Lemma. The diameter of each subdivided simplex in (ρ∆n n n+1 diam(∆n). k n )k(ιn) is bounded by Proof. Basic geometry. Proposition. If c ∈ C U n (X), then pX (c) ∈ C U Moreover, if c ∈ Cn(X), then there is some k such that (ρX n (X). n )k(c) ∈ C U n (X). Proof. The ﬁrst part is clear. For the second part, note that every chain is a ﬁnite sum of simplices. So we only have to check it for single simplices. We let σ be a simplex, and let V = {σ−1˚Uα} be an open cover of ∆n. By the Lebesgue number lemma, there is some ε > 0 such that any set of diameter < ε is contained in some σ−1˚Uα. So we can choose 35 3 Four major tools of (co)homology III Algebraic Topology k > 0 such that (ρ∆n So each lies in some σ−1˚Uα. So n )(ιn) is a sum of simplices which each has diameter < ε. (ρ∆n n )k(ιn) = C V n (∆n). So applying σ tells us (ρ∆n n )k(σ) ∈ C U n (X). Finally, we get to the theorem. Theorem (Small simplices theorem). The natural map U : H U is an isomorphism. ∗ (X) → H∗(X) Proof. Let [c] ∈ Hn(X). By the proposition, there is some k > 0 such that (ρX n is chain homotopic to the identity. Thus so is (ρX n (x). We know that ρX n )k(c)] = [c]. So the map H U n (X) → Hn(X) is surjective. n )k. So [(ρX n )k(c) ∈ C U To show that it is injective, we suppose U ([c]) = 0. Then we can ﬁnd some z ∈ Hn+1(X) such that dz = c. We can then similarly subdivide z enough such that it lies in C U n+1(X). So this shows that [c] = 0 ∈ H U n (X). That’s it. We move on to (slightly) more interesting stuﬀ. The next few sections will all be slightly short, as we touch on various diﬀerent ideas. 36 4 Reduced homology III Algebraic Topology 4 Reduced homology Deﬁnition (Reduced homology). Let X be a space, and x0 ∈ X a basepoint. We deﬁne the reduced homology to be ˜H∗(X) = H∗(X, {x0}). Note that by the long exact sequence of relative homology, we know that ˜Hn(X) ∼= Hn(X) for n ≥ 1. So what is the point of deﬁning a new homology theory that only diﬀers when n = 0, which we often don’t care about? It turns out there is an isomorphism between H∗(X, A) and ˜H∗(X/A) for suitably “good” pairs (X, A). Deﬁnition (Good pair). We say a pair (X, A) is good if there is an open set U containing ¯A such that the inclusion A → U is a deformation retract, i.e. there exists a homotopy H : [0, 1] × U → U such that H(0, x) = x H(1, x) ∈ A H(t, a) = a for all a ∈ A, t ∈ [0, 1]. Theorem. If (X, A) is good, then the natural map H∗(X, A) H∗(X/A, A/A) = ˜H∗(X/A) is an isomorphism. Proof. As i : A → U is in particular a homotopy equivalence, the map H∗(A) H∗(U ) is an isomorphism. So by the ﬁve lemma, the map on relative homology H∗(X, A) H∗(X, U ) is an isomorphism as well. As i : A → U is a deformation retraction with homotopy H, the inclusion is also a deformation retraction. So again by the ﬁve lemma, the map {∗} = A/A → U/A H∗(X/A, A/A) H∗(X/A, U/A) is also an isomorphism. Now we have Hn(X, A) Hn(X/A, A/A) ∼ ∼ Hn(X, U ) excise A Hn(X \ A, U \ A) Hn(X/A, U/A) excise A/A Hn We now notice that X \ A = X vertical map is actually an isomorphism. So the result follows. A and . So the right-hand 37 5 Cell complexes III Algebraic Topology 5 Cell complexes So far, everything we’ve said is true for arbitrary spaces. This includes, for example, the topological space with thre
e points a, b, c, whose topology is {∅, {a}, {a, b, c}}. However, these spaces are horrible. We want to restrict our attention to nice spaces. Spaces that do feel like actual, genuine spaces. The best kinds of space we can imagine would be manifolds, but that is a bit too strong a condition. For example, the union of the two axes in R2 is not a manifold, but it is still a sensible space to talk about. Perhaps we can just impose conditions like Hausdorﬀness and maybe second countability, but we can still produce nasty spaces that satisfy these properties. So the idea is to provide a method to build spaces, and then say we only consider spaces built this way. These are known as cell complexes, or CW complexes Deﬁnition (Cell complex). A cell complex is any space built out of the following procedure: (i) Start with a discrete space X 0. The set of points in X 0 are called I0. (ii) If X n−1 has been constructed, then we may choose a family of maps {ϕα : Sn−1 → X n−1}α∈In , and set X n = X n−1 α∈In Dn α /{x ∈ ∂Dn α ∼ ϕα(x) ∈ X n−1}. We call X n the n-skeleton of X We call the image of Dn open cell eα. α \ ∂Dn α in X n the (iii) Finally, we deﬁne X = X n n≥0 with the weak topology, namely that A ⊆ X is open if A ∩ X n is open in X n for all n. We write Φα : Dn α → X n for the obvious inclusion map. This is called the characteristic map for the cell eα. Deﬁnition (Finite-dimensional cell complex). If X = X n for some n, we say X is ﬁnite-dimensional. Deﬁnition (Finite cell complex). If X is ﬁnite-dimensional and In are all ﬁnite, then we say X is ﬁnite. Deﬁnition (Subcomplex). A subcomplex A of X is a simplex obtained by using a subset I n ⊆ In. Note that we cannot simply throw away some cells to get a subcomplex, as the higher cells might want to map into the cells you have thrown away, and you need to remove them as well. We note the following technical result without proof: Lemma. If A ⊆ X is a subcomplex, then the pair (X, A) is good. 38 5 Cell complexes III Algebraic Topology Proof. See Hatcher 0.16. Corollary. If A ⊆ X is a subcomplex, then Hn(X, A) ∼ ˜Hn(X/A) is an isomorphism. We are next going to show that we can directly compute the cohomology of a cell complex by looking at the cell structures, instead of going through all the previous rather ad-hoc mess we’ve been through. We start with the following lemma: Lemma. Let X be a cell complex. Then (i) Hi(X n, X n−1∈In (ii) Hi(X n) = 0 for all i > n. (iii) Hi(X n) → Hi(X) is an isomorphism for i < n. Proof. (i) As (X n, X n−1) is good, we have an isomorphism Hi(X n, X n−1) ∼ ˜Hi(X n/X n−1) . But we have X n/X n−1 ∼= Sn α, α∈In the space obtained from Y = α by collapsing down the subspace Z = {xα : α ∈ In}, where each xα is the south pole of the sphere. To compute the homology of the wedge X n/X n−1, we then note that (Y, Z) is good, and so we have a long exact sequence α∈In Sn Hi(Z) Hi(Y ) ˜Hi(Y /Z) Hi−1(Z) Hi−1(Y ) . Since Hi(Z) vanishes for i ≥ 1, the result then follows from the homology of the spheres plus the fact that Hi( Xα) = Hi(Xα). (ii) This follows by induction on n. We have (part of) a long exact sequence Hi(X n−1) Hi(X n) Hi(X n, X n−1) We know the ﬁrst term vanishes by induction, and the third term vanishes for i > n. So it follows that Hi(X n) vanishes. 39 5 Cell complexes III Algebraic Topology (iii) To avoid doing too much point-set topology, we suppose X is ﬁnitedimensional, so X = X m for some m. Then we have a long exact sequence Hi+1(X n+1, X n) Hi(X n) Hi(X n+1) Hi(X n+1, X n) Now if i < n, we know the ﬁrst and last groups vanish. So we have Hi(X n) ∼= Hi(X n+1). By continuing, we know that Hi(X n) ∼= Hi(X n+1) ∼= Hi(X n+2) ∼= · · · ∼= Hi(X m) = Hi(X). To prove this for the general case, we need to use the fact that any map from a compact space to a cell complex hits only ﬁnitely many cells, and then the result would follow from this special case. For a cell complex X, let n (X) = Hn(X n, X n−1) ∼= C cell Z. α∈In We deﬁne dcell n : C cell n (X) → C cell n−1(X) by the composition Hn(X n, X n−1) ∂ Hn−1(X n−1) q Hn−1(X n−1, X n−2) . We consider 0 0 Hn(X n+1) Hn(X n) ∂ qn Hn+1(X n+1, X n) dcell n+1 Hn(X n, X n−1) dcell n Hn−1(X n−1, X n−2) ∂ qn−1 Hn−1(X n−1) 0 Hn−1(X n) Referring to the above diagram, we see that n ◦ dcell dcell n+1 = qn−1 ◦ ∂ ◦ qn ◦ ∂ = 0, (X), dcell· ) is a since the middle ∂ ◦ qn is part of an exact sequence. So (C cell· chain complex, and the corresponding homology groups are known as the cellular homology of X, written H cell n (X). 40 5 Cell complexes III Algebraic Topology Theorem. Proof. We have Hn(X) ∼= Hn(X n+1) H cell n (X) ∼= Hn(X). = Hn(X n)/ im(∂ : Hn+1(X n+1, X n) → Hn(X n)) Since qn is injective, we apply it to and bottom to get = qn(Hn(X n))/ im(dcell n+1 : Hn+1(X n+1, X n) → Hn(X n, X n−1)) By exactness, the image of qn is the kernel of ∂. So we have = ker(∂ : Hn(X n, X n−1) → Hn−1(X n−1))/ im(dcell = ker(dcell = H cell n )/ im(dcell n+1) n (X). n+1) Corollary. If X is a ﬁnite cell complex, then Hn(X) is a ﬁnitely-generated abelian group for all n, generated by at most \|In\| elements. In particular, if there are no n-cells, then Hn(X) vanishes. If X has a cell-structure with cells in even-dimensional cells only, then H∗(X) are all free. We can similarly deﬁne cellular cohomology. Deﬁnition (Cellular cohomology). We deﬁne cellular cohomology by C n cell(X) = H n(X n, X n−1) and let dn cell be the composition H n(X n, X n−1) q∗ H n(X n) ∂ H n+1(X n+1, X n). This deﬁnes a cochain complex C· cell(X) with cohomology H ∗ cell(X) ∼= H ∗(X). H ∗ cell(X), and we have One can directly check that C· cell(X) ∼= Hom(C cell· (X), Z). This is all very good, because cellular homology is very simple and concrete. However, to actually use it, we need to understand what the map dcell n : C cell n (X) = α∈In Z{eα} → C cell n−1(X) = Z{eβ} β∈In−1 is. In particular, we want to ﬁnd the coeﬃcients dαβ such that dcell n (eα) = dαβeβ. It turn out this is pretty easy 41 5 Cell complexes III Algebraic Topology Lemma. The coeﬃcients dαβ are given by the degree of the map α = ∂Dn Sn−1 α ϕα X n−1 X n−1/X n−2 = γ∈In−1 Sn−1 γ Sn−1 β , fαβ where the ﬁnal map is obtained by collapsing the other spheres in the wedge. In the case of cohomology, the maps are given by the transposes of these. This is easier in practice that in sounds. In practice, the map is given by “the obvious one”. Proof. Consider the diagram Hn(Dn α, ∂Dn α) (Φα)∗ Hn(X n, X n−1) ∂ ∼ ∂ Hn−1(∂Dn α) ˜Hn−1(Sn−1 β ) (ϕα)∗ collapse Hn−1(X n−1) ˜Hn−1 Sn−1 γ dcell n q Hn−1(X n−1, X n−2) excision ∼ ˜Hn−1(X n−1/X n−2) By the long exact sequence, the top left horizontal map is an isomorphism. Now let’s try to trace through the diagram. We can ﬁnd isomorphism 1 fαβ dαβ 1 eα dαγeγ dαγeγ So the degree of fαβ is indeed dαβ. Example. Let K be the Klein bottle. b π a v We give it a cell complex structure by – K 0 = {v}. Note that all four vertices in the diagram are identiﬁed. v 42 5 Cell complexes III Algebraic Topology – K 1 = {a, b}. a x0 b – K 2 is the unique 2-cell π we see in the picture, where ϕπ : S1 → K 1 given by aba−1b. The cellular chain complex is given by 0 C cell 2 (K) dcell 2 C cell 1 (K) dcell 1 C cell 0 (K) Zπ Za ⊕ Zb Zv We can now compute the maps dcell i . The d1 map is easy. We have d1(a) = d1(b) = v − v = 0. In the d2 map, we can ﬁgure it out by using local degrees. Locally, the attaching map is just like the identity map, up to an orientation ﬂip, so the local degrees are ±1. Moreover, the inverse image of each point has two elements. If we think hard enough, we realize that for the attaching map of a, the two elements have opposite degree and cancel each other out; while in the case of b they have the same sign and give a degree of 2. So we have So we have d2(π) = 0a + 2b. H0(K) = Z H1(K) = Z ⊕ Z 2b H2(K) = 0 = Z ⊕ Z/2Z We can similarly compute the cohomology. By dualizing, we have C 2 cell(K)(K) C 1 cell(K) (0) C 0 cell(K) (0 2) Z Z ⊕ Z Z So we have H0(K) = Z H1(K) = Z H2(K) = Z/2Z. Note that the second cohomology is not the dual of the second homology! However, if we forget where each factor is, and just add all the homology groups together, we get Z⊕Z⊕Z/2Z. Also, if we forget all the torsion components Z/2Z, then they are the same! 43 5 Cell complexes III Algebraic Topology This is a general phenomenon. For a cell complex, if we know what all the homology groups are, we can ﬁnd the cohomologies by keeping the Z’s unchanged and moving the torsion components up. The general statement will be given by the universal coeﬃcient theorem. Example. Consider RPn = Sn/(x ∼ −x). We notice that for any point in RPn, if it is not in the equator, then it is represented by a unique element in the northern hemisphere. Otherwise, it is represented by two points. So we have RPn ∼= Dn/(x ∼ −x for x ∈ ∂Dn). This is a nice description, since if we throw out the interior of the disk, then we are left with an Sn−1 with antipodal points identiﬁed, i.e. an RPn−1! So we can immediately see that RPn = RPn−1 ∪f Dn, for f : Sn−1 → RPn−1 given by f (x) = [x]. So RPn has a cell structure with one cell in every degree up to n. What are the boundary maps? We write ei for the i-th degree cell. We know that ei is attached along the map f described above. More concretely, we have f : Si−1 s ϕi RPi−1 RPi−1/RPi−2 = Si−1 t . The open upper hemisphere and lower hemisphere of Si−1 morphically to Si−1 \ {∗}. Furthermore, s t are mapped homeo- f \|upper = f \|lower ◦ a, where a is the antipodal map. But we know that deg(a) = (−1)i. So we have a zero map if i is odd, and a 2 map if i is even. Then we have · · · 2 Ze3 0 Ze2 2 Ze1 0 Ze0 . What happens on the left end depends on whether n is even or odd. So we have Hi(RPn) =    Z i = 0 Z/2Z i odd, i < n 0 Z i even, 0 < i < n i = n is odd otherwise 0 . We can immediately work out the cohomology too. We will just write out the answer: H i(RPn) =    Z i = 0 i odd, i < n 0 Z/2Z i even is odd otherwise 0 44 6 (Co)homology with coeﬃcients III
Algebraic Topology 6 (Co)homology with coeﬃcients Recall that when we deﬁned (co)homology, we constructed these free Z-modules from our spaces. However, we did not actually use the fact that it was Z, we might as well replace it with any abelian group A. Deﬁnition ((Co)homology with coeﬃcients). Let A be an abelian group, and X be a topological space. We let C·(X; A) = C·(X) ⊗ A with diﬀerentials d ⊗ idA. In other words C·(X; A) is the abelian group obtained by taking the direct sum of many copies of A, one for each singular simplex. We let We can also deﬁne Hn(X; A) = Hn(C·(X; A), d ⊗ idA). n (X; A) = Hn(C cell· H cell (X) ⊗ A), and the same proof shows that H cell n (X; A) = Hn(X; A). Similarly, we let C·(X; A) = Hom(C·(X), A), with the usual (dual) diﬀerential. We again set H n(X; A) = H n(C·(X; A)). We similarly deﬁne cellular cohomology. If A is in fact a commutative ring, then these are in fact R-modules. We call A the “coeﬃcients”, since a general member of C·(X; A) looks like nσs, where nσ ∈ A, σ : ∆n → X. We will usually take A = Z, Z/nZ or Q. Everything we’ve proved for homology holds for these with exactly the same proof. Example. In the case of C cell· C cell· (RPn, Z/2), all the diﬀerentials are 0. So we have (RPn), the diﬀerentials are all 0 or 2. So in Hi(RPn, Z/2) = Z/ Similarly, the cohomology groups are the same. On the other hand, if we take the coeﬃcients to be Q, then multiplication by 2 is now an isomorphism. Then we get for n not too large. C cell· (RPn, Q) Q n odd n even 0 45 7 Euler characteristic III Algebraic Topology 7 Euler characteristic There are many ways to deﬁne the Euler characteristic, and they are all equivalent. So to deﬁne it, we pick a deﬁnition that makes it obvious it is a number. Deﬁnition (Euler characteristic). Let X be a cell complex. We let χ(X) = n (−1)n number of n-cells of X ∈ Z. From this deﬁnition, it is not clear that this is a property of X itself, rather than something about its cell decomposition. We similarly deﬁne χZ(X) = n (−1)n rank Hn(X; Z). For any ﬁeld F, we deﬁne χF(X) = n (−1)n dimF Hn(X; F). Theorem. We have χ = χZ = χF. Proof. First note that the number of n cells of X is the rank of C cell we will just write as Cn. Let n (X), which Zn = ker(dn : Cn → Cn−1) Bn = im(dn+1 : Cn+1 → Cn). We are now going to write down two short exact sequences. By deﬁnition of homology, we have 0 Bn Zn Hn(X; Z) 0 . Also, the deﬁnition of Zn and Bn give us 0 Zn Cn Bn−1 0 . We will now use the ﬁrst isomorphism theorem to know that the rank of the middle term is the sum of ranks of the outer terms. So we have χZ(X) = (−1)n rank Hn(X) = (−1)n(rank Zn − rank Bn). We also have So we have rank Bn = rank Cn+1 − rank Zn+1. χZ(X) = = n (−1)n(rank Zn − rank Cn+1 + rank Zn+1) (−1)n+1 rank Cn+1 n = χ(X). For χF, we use the fact that rank Cn = dimF Cn ⊗ F. 46 8 Cup product III Algebraic Topology 8 Cup product So far, homology and cohomology are somewhat similar. We computed them, saw they are not the same, but they seem to contain the same information nevertheless. However, cohomology is 10 times better, because we can deﬁne a ring structure on them, and rings are better than groups. Just like the case of homology and cohomology, we will be able to write down the deﬁnition easily, but will struggle to compute it. Deﬁnition (Cup product). Let R be a commutative ring, and φ ∈ C k(X; R), ψ ∈ C (X; R). Then φ ψ ∈ C k+(X; R) is given by (φ ψ)(σ : ∆k+ → X) = φ(σ\|[v0,...,vk]) · ψ(σ\|[vk,...,vk+]). Here the multiplication is multiplication in R, and v0, · · · , v are the vertices of ∆k+ ⊆ Rk++1, and the restriction is given by σ\|[x0,...,xi](t0, . . . , ti) = σ . tjxj This is a bilinear map. Notation. We write H ∗(X; R) = H n(X; R). n≥0 This is the deﬁnition. We can try to establish some of its basic properties. We want to know how this interacts with the diﬀerential d with the cochains. The obvious answer d(φ ψ) = (dφ) (dψ) doesn’t work, because the degrees are wrong. What we have is: Lemma. If φ ∈ C k(X; R) and ψ ∈ C (X; R), then d(φ ψ) = (dφ) ψ + (−1)kφ (dψ). This is like the product rule with a sign. Proof. This is a straightforward computation. Let σ : ∆k++1 → X be a simplex. Then we have ((dφ) ψ)(σ) = (dφ)(σ\|[v0,...,vk+1]) · ψ(σ\|[vk+1,...,vk++1]) = φ k+1 i=0 (−1)iσ\|[v0,...,ˆvi,...,vk+1] · ψ(σ\|[vk+1,...,vk++1]) (φ (dψ))(σ) = φ(σ\|[v0,...,vk]) · (dψ)(σ\|vk,...,vk++1]) = φ(σ\|[v0,...,vk]) · ψ k++1 (−1)i−kσ\|[vk,...,ˆvi,...,vk++1] i=k = (−1)kφ(σ\|[v0,...,vk]) · ψ k++1 (−1)iσ\|[vk,...,ˆvi,...,vk++1] . i=k We notice that the last term of the ﬁrst expression, and the ﬁrst term of the second expression are exactly the same, except the signs diﬀer by −1. Then the remaining terms overlap in exactly 1 vertex, so we have ((dφ) ψ)(σ) + (−1)kφ (dψ)(σ) = (φ ψ)(dσ) = (d(φ ψ))(σ) as required. 47 8 Cup product III Algebraic Topology This is the most interesting thing about these things, because it tells us this gives a well-deﬁned map on cohomology. Corollary. The cup product induces a well-deﬁned map : H k(X; R) × H (X; R) H k+(X; R) ([φ], [ψ]) [φ ψ] Proof. To see this is deﬁned at all, as dφ = 0 = dψ, we have d(φ ψ) = (dφ) ψ ± φ (dψ) = 0. So φ ψ is a cocycle, and represents the cohomology class. To see this is well-deﬁned, if φ = φ + dτ , then φ ψ = φ ψ + dτ ψ = φ ψ + d(τ ψ) ± τ (dψ). Using the fact that dψ = 0, we know that φ ψ and φ ψ diﬀer by a boundary, so [φ ψ] = [φ ψ]. The case where we change ψ is similar. Note that the operation is associative on cochains, so associative on H ∗ too. Also, there is a map 1 : C0(X) → R sending σ → 1 for all σ. Then we have [1] [φ] = [φ]. So we have Proposition. (H ∗(X; R), , [1]) is a unital ring. Note that this is not necessarily commutative! Instead, we have the following graded commutative condition. Proposition. Let R be a commutative ring. If α ∈ H k(X; R) and β ∈ H (X; R), then we have α β = (−1)kβ α Note that this is only true for the cohomology classes. It is not true in general for the cochains. So we would expect that this is rather annoying to prove. The proof relies on the following observation: Proposition. The cup product is natural, i.e. if f : X → Y is a map, and α, β ∈ H ∗(Y ; R), then f ∗(α β) = f ∗(α) f ∗(β). So f ∗ is a homomorphism of unital rings. Proof of previous proposition. Let ρn : Cn(X) → Cn(x) be given by σ → (−1)n(n+1)/2σ\|[vn,vn−1,...,v0] The σ\|[vn,vn−1,...,v0] tells us that we reverse the order of the vertices, and the factor of (−1)n(n+1)/2 is the sign of the permutation that reverses 0, · · · , n. For convenience, we write εn = (−1)n(n+1)/2. 48 8 Cup product III Algebraic Topology Claim. We claim that ρ· is a chain map, and is chain homotopic to the identity. We will prove this later. Suppose the claim holds. We let φ ∈ C k(X; R) represent α and ψ ∈ C (X; R) represent β. Then we have (ρ∗φ ρ∗ψ)(σ) = (ρ∗φ)(σ\|[v0,...,vk](ρ∗ψ)(σ\|[vk,...,vk+]) = φ(εk · σ\|[vk,...,v0])ψ(εσ\|[vk+,...,vk]). Thus, we can compute ρ∗(ψ φ)(σ) = (ψ φ)(εk+σ\|[vk+,...,v0]) = εk+ψ(σ\|[vk+,...,vk ])φ(σ\|[vk,...,v0]) = εk+εkε(ρ∗φ ρ∗ψ)(σ). By checking it directly, we can see that εn+εkε = (−1)k. So we have α β = [φ ψ] = [ρ∗φ ρ∗ψ] = (−1)k[ρ∗(ψ φ)] = (−1)k[ψ φ] = (−1)klβ α. Now it remains to prove the claim. We have dρ(σ) = εn n (−1)jσ\|[vn,...,ˆvn−i,...,v0] i=0 n ρ(dσ) = ρ (−1)iσ\|[v0,...,ˆvi,....,vn] i=0 n = εn−1 j=0 (−1)jσ\|[vn,...,ˆvj ,v0]. We now notice that εn−1(−1)n−i = εn(−1)i. So this is a chain map! We now deﬁne a chain homotopy. This time, we need a “twisted prism”. We let Pn = i (−1)iεn−i[v0, · · · , vi, wn, · · · , wi] ∈ Cn+1([0, 1] × ∆n), where v0, · · · , vn are the vertices of {0} × ∆n and w0, · · · , wn are the vertices of {1} × ∆n. We let π : [0, 1] × ∆n → ∆n be the projection, and let F X n : Cn(X) → Cn+1(X) be given by σ → (σ ◦ π)#(Pn). 49 8 Cup product III Algebraic Topology We calculate dF X n (σ) = (σ ◦ π)#(dPn) = (σ ◦ π#)     i j≤i (−1)j(−1)iεn−i[v0, · · · , ˆvj, · · · , vi, w0, · · · , wi]    +  j≥i (−1)n+i+1−j(−1)iεn−i[v0, · · · , vi, wn, · · · , ˆwj, · · · , vi]   .   The terms with j = i give (σ ◦ π)# i εn−i[v0, · · · , vi−1, wn, · · · , wi] (−1)n+1(−1)iεn−i[v0, · · · , vi, wn, · · · , wi+1] + i = (σ ◦ π)#(εn[wn, · · · , w0] − [v0, · · · , vn]) = ρ(σ) − σ The terms with j = i are precisely −F X n−1(dσ) as required. It is easy to see that the terms are indeed the right terms, and we just have to check that the signs are right. I’m not doing that. There are some other products we can deﬁne. One example is the cross product: Deﬁnition (Cross product). Let πX : X × Y → X, πY : X × Y → Y be the projection maps. Then we have a cross product × : H k(X; R) ⊗R H (Y ; R) a ⊗ b H k+(X × Y ; R) X a) (π∗ Y b) (π∗ . Note that the diagonal map ∆ : X → X × X given by ∆(x) = (x, x) satisﬁes ∆∗(a × b) = a b for all a, b ∈ H ∗(X; R). So these two products determine each other. There is also a relative cup product : H k(X, A; R) ⊗ H k(X; R) → H k+(X, A; R) given by the same formula. Indeed, to see this is properly deﬁned, note that if φ ∈ C k(X, A; R), then φ is a map φ : Ck(X, A) = Ck(X) Ck(A) → R. In other words, it is a map Ck(X) → R that vanishes on Ck(A). Then if σ ∈ Ck+(A) and ψ ∈ C (X; R), then (φ ψ)(σ) = φ(σ\|[v0,...,vk]) · ψ(σ\|[vk,...,vk+]). 50 8 Cup product III Algebraic Topology We now notice that [v0, · · · , vk] ∈ Ck(A). So φ kills it, and this vanishes. So this is a term in H k+(X, A; R). You might ﬁnd it weird that the two factors of the cup product are in diﬀerent things, but note that a relative cohomology class is in particular a cohomology class. So this restricts to a map : H k(X, A; R) ⊗ H k(X, A; R) → H k+(X, A; R), but the result we gave is more general. Example. Suppose X is a space such that the cohomology classes are given by k 1 H k(X, Z What can x x be? By the graded commutativity property, we have So we know 2(x x) = 0 ∈ H 6(X, Z) ∼= Z. So we must have x x = 0. x x = −x x. 51 9 K¨unneth theorem and universal coeﬃcients theorem III Algebraic Topology 9 K¨unneth theorem and universal coeﬃcients theorem We are going to prove two th
eorems of similar ﬂavour — K¨unneth’s theorem and the universal coeﬃcients theorem. They are both fairly algebraic results that relate diﬀerent homology and cohomology groups. They will be very useful when we prove things about (co)homologies in general. In both cases, we will not prove the “full” theorem, as they require knowledge of certain objects known as Tor and Ext. Instead, we will focus on a particular case where the Tor and Ext vanish, so that we can avoid mentioning them at all. We start with K¨unneth’s theorem. Theorem (K¨unneth’s theorem). Let R be a commutative ring, and suppose that H n(Y ; R) is a free R-module for each n. Then the cross product map H k(X; R) ⊗ H (Y ; R) × H n(X × Y ; R) k+=n is an isomorphism for every n, for every ﬁnite cell complex X. It follows from the ﬁve lemma that the same holds if we have a relative complex (Y, A) instead of just Y . For convenience, we will write H ∗(X; R)⊗H ∗(Y ; R) for the graded R-module which in grade n is given by k+=n H k(X; R) ⊗ H (Y ; R). Then K¨unneth says the map given by H ∗(X; R) ⊗ H ∗(Y ; R) × H ∗(X × Y ; R) is an isomorphism of graded rings. Proof. Let F n(−) = k+=n H k(−; R) ⊗ H (Y ; R). We similarly deﬁne Gn(−) = H n(− × Y ; R). We observe that for each X, the cross product gives a map × : F n(X) → Gn(X), and, crucially, we know ×∗ : F n(∗) → Gn(∗) is an isomorphism, since F n(∗) ∼= Gn(∗) ∼= H n(Y ; R). The strategy is to show that F n(−) and Gn(−) have the same formal structure as cohomology and agree on a point, and so must agree on all ﬁnite cell complexes. It is clear that both F n and Gn are homotopy invariant, because they are built out of homotopy invariant things. 52 9 K¨unneth theorem and universal coeﬃcients theorem III Algebraic Topology We now want to deﬁne the cohomology of pairs. This is easy. We deﬁne F n(X, A) = i+j=n H i(X, A; R) ⊗ H j(Y ; R) Gn(X, A) = H n(X × Y, A × Y ; R). Again, the relative cup product gives us a relative cross product, which gives us a map F n(X, A) → Gn(X, A). It is immediate Gn has a long exact sequence associated to (X, A) given by the usual long exact sequence of (X × Y, A × Y ). We would like to say F has a long exact sequence as well, and this is where our hypothesis comes in. If H ∗(Y ; R) is a free R-module, then we can take the long exact sequence of (X, A) · · · H n(A; R) ∂ H n(X, A; R) H n(X; R) H n(A; R) · · · , and then tensor with H j(Y ; R). This preserves exactness, since H j(Y ; R) ∼= Rk for some k, so tensoring with H j(Y ; R) just takes k copies of this long exact sequence. By adding the diﬀerent long exact sequences for diﬀerent j (with appropriate translations), we get a long exact sequence for F . We now want to prove K¨unneth by induction on the number of cells and the dimension at the same time. We are going to prove that if X = X ∪f Dn for some Sn−1 → X , and × : F (X ) → G(X ) is an isomorphism, then × : F (X) → G(X) is also an isomorphism. In doing so, we will assume that the result is true for attaching any cells of dimension less than n. Suppose X = X ∪f Dn for some f : Sn−1 → X . We get long exact sequences F ∗−1(X ) F ∗(X, X ) F ∗(X) F ∗(X ) F ∗+1(X, X ) ×∼ × × ×∼ × G∗−1(X ) G∗(X, X ) G∗(X) G∗(X ) G∗+1(X, X ) Note that we need to manually check that the boundary maps ∂ commute with the cross product, since this is not induced by maps of spaces, but we will not do it here. Now by the ﬁve lemma, it suﬃces to show that the maps on the relative cohomology × : F n(X, X ) → Gn(X, X ) is an isomorphism. We now notice that F ∗(−) and G∗(−) have excision. Since (X, X ) is a good pair, we have a commutative square F ∗(Dn, ∂Dn) × G∗(Dn, ∂Dn) F ∗(X, X ) × G∗(X, X ) ∼ ∼ So we now only need the left-hand map to be an isomorphism. We look at the long exact sequence for (Dn, ∂Dn)! F ∗−1(∂Dn) F ∗(Dn, ∂Dn) F ∗(Dn) F ∗(∂Dn) F ∗+1(Dn, ∂Dn) ×∼ ×∼ × ×∼ ×× G∗−1(∂Dn) G∗(Dn, ∂Dn) G∗(Dn) G∗(∂Dn) G∗+1(Dn, ∂Dn) 53 9 K¨unneth theorem and universal coeﬃcients theorem III Algebraic Topology But now we know the vertical maps for Dn and ∂Dn are isomorphisms — the ones for Dn are because they are contractible, and we have seen the result of ∗ already; whereas the result for ∂Dn follows by induction. So we are done. The conditions of the theorem require that H n(Y ; R) is free. When will this hold? One important example is when R is actually a ﬁeld, in which case all modules are free. Example. Consider H ∗(S1, Z). We know it is Z in ∗ = 0, 1, and 0 elsewhere. Let’s call the generator of H 0(S1, Z) “1”, and the generator of H 1(S1, Z) as x. Then we know that x x = 0 since there isn’t anything in degree 2. So we know H ∗(S1, Z) = Z[x]/(x2). Then K¨unneth’s theorem tells us that H ∗(T n, Z) ∼= H ∗(S1; Z)⊗n, where T n = (S1)n is the n-torus, and this is an isomorphism of rings. So this is H ∗(T n, Z) ∼= Z[x1, · · · , xn]/(x2 i , xixj + xjxi), using the fact that xi, xj have degree 1 so anti-commute. Note that this has an interesting cup product! This ring is known as the exterior algebra in n generators. Example. Let f : Sn → T n be a map for n > 1. We claim that this induces the zero map on the nth cohomology. We look at the induced map on cohomology: f ∗ : H n(T n; Z) H n(Sn, Z) . Looking at the presentation above, we know that H n(T n, Z) is generated by x1 · · · xn, and f ∗ sends it to (f ∗x1) · · · (f ∗xn). But f ∗xi ∈ H 1(Sn, Z) = 0 for all n > 1. So f ∗(x1 · · · xn) = 0. Note that the statement does not involve cup products at all, but it would be much more diﬃcult to prove this without using cup products! We are now going to prove another useful result. Theorem (Universal coeﬃcients theorem for (co)homology). Let R be a PID and M an R-module. Then there is a natural map H∗(X; R) ⊗ M → H∗(X; M ). If H∗(X; R) is a free module for each n, then this is an isomorphism. Similarly, there is a natural map H ∗(X; M ) → HomR(H∗(X; R), M ), , which is an isomorphism again if H ∗(X; R) is free. 54 9 K¨unneth theorem and universal coeﬃcients theorem III Algebraic Topology In particular, when R = Z, then an R-module is just an abelian group, and this tells us how homology and cohomology with coeﬃcients in an abelian group relate to the usual homology and cohomology theory. Proof. Let Cn be Cn(X; R) and Zn ⊆ Cn be the cycles and Bn ⊆ Zn the boundaries. Then there is a short exact sequence 0 Zn i Cn g Bn−1 0 , and Bn−1 ≤ Cn−1 is a submodule of a free R-module, and is free, since R is a PID. So by picking a basis, we can ﬁnd a map s : Bn−1 → Cn such that g ◦ s = idBn−1. This induces an isomorphism i ⊕ s : Zn ⊕ Bn−1 ∼ Cn. Now tensoring with M , we obtain 0 Zn ⊗ M Cn ⊗ M Bn−1 ⊗ M 0 , which is exact because we have Cn ⊗ M ∼= (Zn ⊕ Bn−1) ⊗ M ∼= (Zn ⊗ M ) ⊕ (Bn−1 ⊗ M ). So we obtain a short exact sequence of chain complexes 0 (Zn ⊗ M, 0) (Cn ⊗ M, d ⊗ id) (Bn−1 ⊗ M, 0) 0 , which gives a long exact sequence in homology: · · · Bn ⊗ M Zn ⊗ M Hn(X; M ) Bn−1 ⊗ M · · · We’ll leave this for a while, and look at another short exact sequence. By deﬁnition of homology, we have a long exact sequence 0 Bn Zn Hn(X; R) 0 . As Hn(X; R) is free, we have a splitting t : Hn(X; R) → Zn, so as above, tensoring with M preserves exactness, so we have 0 Bn ⊗ M Zn ⊗ M Hn(X; R) ⊗ M 0 . Hence we know that Bn ⊗ M → Zn ⊗ M is injective. So our previous long exact sequence breaks up to 0 Bn ⊗ M Zn ⊗ M Hn(X; M ) 0. Since we have two short exact sequence with ﬁrst two terms equal, the last terms have to be equal as well. The cohomology version is similar. 55 10 Vector bundles III Algebraic Topology 10 Vector bundles 10.1 Vector bundles We now end the series of random topics, and work on a more focused topic. We are going to look at vector bundles. Intuitively, a vector bundle over a space X is a continuous assignment of a vector space to each point in X. In the ﬁrst section, we are just going to look at vector bundles as topological spaces. In the next section, we are going to look at homological properties of vector bundles. Deﬁnition (Vector bundle). Let X be a space. A (real) vector bundle of dimension d over X is a map π : E → X, with a (real) vector space structure on each ﬁber Ex = π−1({x}), subject to the local triviality condition: for each x ∈ X, there is a neighbourhood U of x and a homeomorphism ϕ : E\|U = π−1(U ) → U × Rd such that the following diagram commutes E\|U ϕ π U U × Rd , π1 and for each y ∈ U , the restriction ϕ\|Ey : Ey → {y} × Rd is a linear isomorphism for each y ∈ U . This maps is known as a local trivialization. We have an analogous deﬁnition for complex vector bundles. Deﬁnition (Section). A section of a vector bundle π : E → X is a map s : X → E such that π ◦ s = id. In other words, s(x) ∈ Ex for each x. Deﬁnition (Zero section). The zero section of a vector bundle is s0 : X → E given by s0(x) = 0 ∈ Ex. Note that the composition E π X s0 E is homotopic to the identity map on idE, since each Ex is contractible. One important operation we can do on vector bundles is pullback : Deﬁnition (Pullback of vector bundles). Let π : E → X be a vector bundle, and f : Y → X a map. We deﬁne the pullback f ∗E = {(y, e) ∈ Y × E : f (y) = π(e)}. This has a map f ∗π : f ∗E → Y given by projecting to the ﬁrst coordinate. The vector space structure on each ﬁber is given by the identiﬁcation (f ∗E)y = Ef (y). It is a little exercise in topology to show that the local trivializations of π : E → X induce local trivializations of f ∗π : f ∗E → Y . Everything we can do on vector spaces can be done on vector bundles, by doing it on each ﬁber. 56 10 Vector bundles III Algebraic Topology Deﬁnition (Whitney sum of vector bundles). Let π : E → F and ρ : F → X be vector bundles. The Whitney sum is given by E ⊕ F = {(e, f ) ∈ E × F : π(e) = ρ(f )}. This has a natural map π ⊕ ρ : E ⊕ F → X given by (π ⊕ ρ)(e, f ) = π(e) = ρ(f ). This is again a vector bundle, with (E ⊕ F )x = Ex ⊕ Fx and again local trivializations of E and F induce one for E ⊕ F . Tensor products can be deﬁned similarly. Similarly, we have the notion of subbundles. Deﬁnition (Vector subb
undle). Let π : E → X be a vector bundle, and F ⊆ E a subspace such that for each x ∈ X there is a local trivialization (U, ϕ) E\|U ϕ π U U × Rd , π1 such that ϕ takes F \|U to U × Rk, where Rk ⊆ Rd. Then we say F is a vector sub-bundle. Deﬁnition (Quotient bundle). Let F be a sub-bundle of E. Then E/F , given by the ﬁberwise quotient, is a vector bundle and is given by the quotient bundle. We now look at one of the most important example of a vector bundle. In some sense, this is the “universal” vector bundle, as we will see later. Example (Grassmannian manifold). We let X = Grk(Rn) = {k-dimensional linear subgroups of Rn}. To topologize this space, we pick a ﬁxed V ∈ Grk(Rn). Then any k-dimensional subspace can be obtained by applying some linear map to V . So we obtain a surjection GLn(R) → Grk(Rn) M → M (V ). So we can given Grk(Rn) the quotient (ﬁnal) topology. For example, Gr1(Rn+1) = RPn. Now to construct a vector bundle, we need to assign a vector space to each point in X. But a point in Grk(Rn) is a vector space, so we have an obvious deﬁnition E = {(V, v) ∈ Grk(Rn) × Rn : v ∈ V }. This has the evident projection π : E → X given by the ﬁrst projection. We then have EV = V. 57 10 Vector bundles III Algebraic Topology To see that this is a vector bundle, we have to check local triviality. We ﬁx a V ∈ Grk(Rn), and let U = {W ∈ Grk(Rn) : W ∩ V ⊥ = {0}}. We now construct a map ϕ : E\|U → U × V ∼= U × Rk by mapping (W, w) to (W, prV (w)), where prV : Rn → V is the orthogonal projection. Now if w ∈ U , then prV (w) = 0 since W ∩ V ⊥ = {0}. So ϕ is a homeomor- phism. We call this bundle γR k,n → Grk(Rn). In the same way, we can get a canonical complex vector bundle γC k,n → Grk(Cn). Example. Let M be a smooth d-dimensional manifold, then it naturally has a d-dimensional tangent bundle π : T M → M with (T M )\|x = TxM . If M ⊆ N is a smooth submanifold, with i the inclusion map, then T M is a subbundle of i∗T N . Note that we cannot say T M is a smooth subbundle of T N , since they have diﬀerent base space, and thus cannot be compared without pulling back. The normal bundle of M in N is νM ⊆N = i∗T N T M . Here is a theorem we have to take on faith, because proving it will require some diﬀerential geometry. Theorem (Tubular neighbourhood theorem). Let M ⊆ N be a smooth submanifold. Then there is an open neighbourhood U of M and a homeomorphism νM ⊆N → U , and moreover, this homeomorphism is the identity on M (where we view M as a submanifold of νM ⊆N by the image of the zero section). This tells us that locally, the neighbourhood of M in N looks like νM ⊆N . We will also need some results from point set topology: Deﬁnition (Partition of unity). Let X be a compact Hausdorﬀ space, and {Uα}α∈I be an open cover. A partition of unity subordinate to {Uα} is a collection of functions λα : X → [0, ∞) such that (i) supp(λα) = {x ∈ X : λα(x) > 0} ⊆ Uα. (ii) Each x ∈ X lies in ﬁnitely many of these supp(λα). (iii) For each x, we have α∈I λα(x) = 1. Proposition. Partitions of unity exist for any open cover. You might have seen this in diﬀerential geometry, but this is easier, since we do not require the partitions of unity to be smooth. Using this, we can prove the following: Lemma. Let π : E → X be a vector bundle over a compact Hausdorﬀ space. Then there is a continuous family of inner products on E. In other words, there is a map E ⊗ E → R which restricts to an inner product on each Ex. 58 10 Vector bundles III Algebraic Topology Proof. We notice that every trivial bundle has an inner product, and since every bundle is locally trivial, we can patch these up using partitions of unity. Let {Uα}α∈I be an open cover of X with local trivializations ϕα : E\|Uα → Uα × Rd. The inner product on Rd then gives us an inner product on E\|Uα , say · , · α. We let λα be a partition of unity associated to {Uα}. Then for u ⊗ v ∈ E ⊗ E, we deﬁne u, v = λα(π(u))u, vα. α∈I Now if π(u) = π(v) is not in Uα, then we don’t know what we mean by u, vα, but it doesn’t matter, because λα(π(u)) = 0. Also, since the partition of unity is locally ﬁnite, we know this is a ﬁnite sum. It is then straightforward to see that this is indeed an inner product, since a positive linear combination of inner products is an inner product. Similarly, we have Lemma. Let π : E → X be a vector bundle over a compact Hausdorﬀ space. Then there is some N such that E is a vector subbundle of X × RN . Proof. Let {Uα} be a trivializing cover of X. Since X is compact, we may wlog assume the cover is ﬁnite. Call them U1, · · · , Un. We let ϕi : E\|Ui → Ui × Rd. We note that on each patch, E\|Ui embeds into a trivial bundle, because it is a trivial bundle. So we can add all of these together. The trick is to use a partition of unity, again. We deﬁne fi to be the composition E\|Ui ϕi Ui × Rd π2 Rd . Then given a partition of unity λi, we deﬁne f : E → X × (Rd)n v → (π(v), λ1(π(v))f1(v), λ2(π(v))f2(v), · · · , λn(π(v))fn(v)). We see that this is injective. If v, w belong to diﬀerent ﬁbers, then the ﬁrst coordinate distinguishes them. If they are in the same ﬁber, then there is some Ui with λi(π(u)) = 0. Then looking at the ith coordinate gives us distinguishes them. This then exhibits E as a subbundle of X × Rn. Corollary. Let π : E → X be a vector bundle over a compact Hausdorﬀ space. Then there is some p : F → X such that E ⊕ F ∼= X × Rn. In particular, E embeds as a subbundle of a trivial bundle. Proof. By above, we can assume E is a subbundle of a trivial bundle. We can then take the orthogonal complement of E. 59 10 Vector bundles III Algebraic Topology Now suppose again we have a vector bundle π : E → X over a compact Hausdorﬀ X. We can then choose an embedding E ⊆ X × RN , and then we get a map fπ : X → Grd(RN ) sending x to Ex ⊆ RN . Moreover, if we pull back the tautological bundle along fπ, then we have πγR f ∗ k,N ∼= E. So every vector bundle is the pullback of the canonical bundle γR k,N over a Grassmannian. However, there is a slight problem. Diﬀerent vector bundles will require diﬀerent N ’s. So we will have to overcome this problem if we want to make a statement of the sort “a vector bundle is the same as a map to a Grassmannian”. The solution is to construct some sort of Grd(R∞). But inﬁnite-dimensional vector spaces are weird. So instead we take the union of all Grd(RN ). We note that for each N , there is an inclusion Grd(RN ) → Grd(Rn+1), which induces an inclusion of the canonical bundle. We can then take the union to obtain Grd(R∞) with a canonical bundle γR d . Then the above result shows that each vector bundle over X is the pullback of the canoncial bundle γR d along some map f : X → Grd(R∞). Note that a vector bundle over X does not uniquely specify a map X → Grd(R∞), as there can be multiple embeddings of X into the trivial bundle X × RN . Indeed, if we wiggle the embedding a bit, then we can get a new bundle. So we don’t have a coorespondence between vector bundles π : E → X and maps fπ : X → Grd(R∞). The next best thing we can hope for is that the “wiggle the embedding a bit” is all we can do. More precisely, two maps f, g : X → Grd(R∞) pull back isomorphic vector bundles if and only if they are homotopic. This is indeed true: Theorem. There is a correspondence    homotopy classess of maps f : X → Grd(R∞)    d-dimensional vector bundles π : E → X [f ] [fπ] f ∗γR d π The proof is mostly technical, and is left as an exercise on the example sheet. 10.2 Vector bundle orientations We are now going to do something that actually involves algebraic topology. Unfortunately, we have to stop working with arbitrary bundles, and focus on orientable bundles instead. So the ﬁrst thing to do is to deﬁne orientability. What we are going to do is to come up with a rather reﬁned notion of orientability. For each commutative ring R, we will have the notion of Rorientability. The strength of this condition will depend on what R is — any vector bundle is F2 orientable, while Z-orientability is the strongest — if a vector bundle is Z-orientable, then it is R-orientable for all R. 60 10 Vector bundles III Algebraic Topology While there isn’t really a good geometric way to think about general Rorientability, the reason for this more reﬁned notion is that whenever we want things to be true for (co)homology with coeﬃcients in R, then we need the bundle to be R-orientable. Let’s begin. Let π : E → X be a vector bundle of dimension d. We write E# = E \ s0(X). We now look at the relative homology groups H i(Ex, E# x ; R), where E# x = Ex \ {0}. We know Ex is a d-dimensional vector space. So we can choose an isomorphism Ex → Rd. So after making this choice, we know that H i(Ex, E# x ; R) ∼= H i(Rd, Rd \ {0}; R) = R i = d 0 otherwise However, there is no canonical generator of H d(Ex, E# we had to pick an isomorphism Ex ∼= Rd. x ; R) as an R-module, as Deﬁnition (R-orientation). A local R-orientation of E at x ∈ X is a choice of R-module generator εx ∈ H d(Ex, E# x ; R). An R-orientation is a choice of local R-orientation {εx}x∈X which are compatible in the following way: if U ⊆ X is open on which E is trivial, and x, y ∈ U , then under the homeomorphisms (and in fact linear isomorphisms): Ex Ey hx E\|U ϕα ∼= U × Rd π2 Rd hy the map y ◦ (h−1 h∗ x )∗ : H d(Ex, E# x ; R) → H d(Ey, E# y ; R) sends εx to εy. Note that this deﬁnition does not depend on the choice of ϕU , because we used it twice, and they cancel out. It seems pretty horriﬁc to construct an orientation. However, it isn’t really that bad. For example, we have Lemma. Every vector bundle is F2-orientable. Proof. There is only one possible choice of generator. In the interesting cases, we are usually going to use the following result to construct orientations: 61 10 Vector bundles III Algebraic Topology Lemma. If {Uα}α∈I is a family of covers such that for each α, β ∈ I, the homeomorphism (Uα ∩ Uβ) × Rd ϕα ∼= E\|Uα∩Uβ ∼= ϕβ (Uα ∩ Uβ) × Rd gives an orientation preserving map from (Uα ∩ Uβ) × Rd to itself, i.e. has a positive det
erminant on each ﬁber, then E is orientable for any R. Note that we don’t have to check the determinant at each point on Uα ∩ Uβ. By continuity, we only have to check it for one point. Proof. Choose a generator u ∈ H d(Rd, Rd \ {0}; R). Then for x ∈ Uα, we deﬁne εx by pulling back u along Ex E\|Uα ϕα Uα × Rd π2 Rd . (†α) If x ∈ Uβ as well, then the analogous linear isomorphism †α diﬀers from †β by post-composition with a linear map L : Rd → Rd of positive determinant. We now use the fact that any linear map of positive determinant is homotopic to the identity. Indeed, both L and id lies in GL+ d (R), a connected group, and a path between them will give a homotopy between the maps they represent. So we know (†α) is homotopic to (†β). So they induce the same maps on cohomology classes. Now if we don’t know that the maps have positive determinant, then (†α) and (†β) might diﬀer by a sign. So in any ring R where 2 = 0, we know every vector bundle is R-orientable. This is a generalization of the previous result for F2 we had. 10.3 The Thom isomorphism theorem We now get to the main theorem about vector bundles. Theorem (Thom isomorphism theorem). Let π : E → X be a d-dimensional vector bundle, and {εx}x∈X be an R-orientation of E. Then (i) H i(E, E#; R) = 0 for i < d. (ii) There is a unique class uE ∈ H d(E, E#; R) which restricts to εx on each ﬁber. This is known as the Thom class. (iii) The map Φ given by the composition H i(X; R) π∗ H i(E; R) −uE H i+d(E, E#; R) is an isomorphism. Note that (i) follows from (iii), since H i(X; R) = 0 for i < 0. Before we go on and prove this, we talk about why it is useful. Deﬁnition (Euler class). Let π : E → X be a vector bundle. We deﬁne the Euler class e(E) ∈ H d(X; R) by the image of uE under the composition H d(E, E#; R) H d(E; R) s∗ 0 H d(X; R) . 62 10 Vector bundles III Algebraic Topology This is an example of a characteristic class, which is a cohomology class related to an oriented vector bundle that behaves nicely under pullback. More precisely, given a vector bundle π : E → X and a map f : Y → X, we can form a pullback ˆf f f ∗E f ∗π Y E . π X ∼= Ef (y), an R-orientation for E Since we have a ﬁberwise isomorphism (f ∗E)y induces one for f ∗E, and we know f ∗(uE) = uf ∗E by uniqueness of the Thom class. So we know e(f ∗(E)) = f ∗e(E) ∈ H d(Y ; R). Now the Euler class is a cohomology class of X itself, so we can use the Euler class to compare and contrast diﬀerent vector bundles. How can we think of the Euler class? It turns out the Euler class gives us an obstruction to the vector bundle having a non-zero section. Theorem. If there is a section s : X → E which is nowhere zero, then e(E) = 0 ∈ H d(X; R). Proof. Notice that any two sections of E → X are homotopic. So we have e ≡ s∗ 0uE = s∗uE. But since uE ∈ H d(E, E#; R), and s maps into E#, we have s∗uE. Perhaps more precisely, we look at the long exact sequence for the pair (E, E#), giving the diagram H d(E, E#; R) H d(E; R) H d(E#; R) s∗ 0 s∗ H d(X; R) Since s and s0 are homotopic, the diagram commutes. Also, the top row is exact. So uE ∈ H d(E, E#; R) gets sent along the top row to 0 ∈ H d(E#; R), and thus s∗ sends it to 0 ∈ H d(X; R). But the image in H d(X; R) is exactly the Euler class. So the Euler class vanishes. Now cohomology is not only a bunch of groups, but also a ring. So we can ask what happens when we cup uE with itself. Theorem. We have uE uE = Φ(e(E)) = π∗(e(E)) uE ∈ H ∗(E, E#; R). This is just tracing through the deﬁnitions. Proof. By construction, we know the following maps commute: H d(E, E#; R) ⊗ H d(E, E#; R) H 2d(E, E#; R) q∗⊗id H d(E; R) ⊗ H d(E, E#; R) 63 10 Vector bundles III Algebraic Topology We claim that the Thom class uE ⊗ uE ∈ H d(E, E#; R) ⊗ H d(E, E#; R) is sent to π∗(e(E)) ⊗ uE ∈ H d(E; R) ⊗ H d(E, E#; R). By deﬁnition, this means we need q∗uE = π∗(e(E)), and this is true because π∗ is homotopy inverse to s∗ 0 and e(E) = s∗ 0q∗uE. So if we have two elements Φ(c), Φ(d) ∈ H ∗(E, E#; R), then we have Φ(c) Φ(d) = π∗c uE π∗d uE = ±π∗c π∗d uE uE = ±π∗(c d e(E)) uE = ±Φ(c d e(E)). So e(E) is precisely the information necessary to recover the cohomology ring H ∗(E, E#; R) from H ∗(X; R). Lemma. If π : E → X is a d-dimensional R-module vector bundle with d odd, then 2e(E) = 0 ∈ H d(X; R). Proof. Consider the map α : E → E given by negation on each ﬁber. This then gives an isomorphism a∗ : H d(E, E#; R) ∼= H d(E, E#; R). This acts by negation on the Thom class, i.e. a∗(uE) = −uE, as on the ﬁber Ex, we know a is given by an odd number of reﬂections, each x ; R) by −1 (by the analogous result on Sn). So we of which acts on H d(Ex, E# change εx by a sign. We then lift this to a statement about uE by the fact that uE is the unique thing that restricts to εx for each x. But we also know which implies a ◦ s0 = s0, 0(a∗(uE)) = s∗ s∗ 0(uE). Combining this with the result that a∗(uE) = −uE, we get that 2e(E) = 2s∗ 0(uE) = 0. This is a disappointing result, because if we already know that H d(X; R) has no 2-torsion, then e(E) = 0. After all that fun, we prove the Thom isomorphism theorem. Proof of Thom isomorphism theorem. We will drop the “R” in all our diagrams for readability (and also so that it ﬁts in the page). We ﬁrst consider the case where the bundle is trivial, so E = X × Rd. Then we note that H ∗(Rd, Rd \ {0}) = R ∗ = d ∗ = d 0 . 64 10 Vector bundles III Algebraic Topology In particular, the modules are free, and (a relative version of) K¨unneth’s theorem tells us the map × : H ∗(X) ⊗ H ∗(Rd, Rd \ {0}) ∼= H ∗(X × Rd, X × (Rd \ {0})) is an isomorphism. Then the claims of the Thom isomorphism theorem follow immediately. (i) For i < d, all the summands corresponding to H i(X × Rd, X × (Rd \ {0})) vanish since the H ∗(Rd, Rd \ {0}) term vanishes. (ii) The only non-vanishing summand for H d(X × Rd, X × (Rd \ {0}) is H 0(X) ⊗ H d(Rd, Rd \ {0}). Then the Thom class must be 1 ⊗ u, where u is the object corresponding to εx ∈ H d(Ex, E# x ) = H d(Rd, Rd \ {0}), and this is unique. (iii) We notice that Φ is just given by Φ(x) = π∗(x) uE = x × uE, which is an isomorphism by K¨unneth. We now patch the result up for a general bundle. Suppose π : E → X is a bundle. Then it has an open cover of trivializations, and moreover if we assume our X is compact, there are ﬁnitely many of them. So it suﬃces to show that if U, V ⊆ X are open sets such that the Thom isomorphism the holds for E restricted to U, V, U ∩ V , then it also holds on U ∪ V . The relative Mayer-Vietoris sequence gives us H d−1(E\|U ∩V , E#\|U ∩V ) ∂M V H d(E\|U ∪V , E#\|U ∪V ) H d(E\|U , E#\|U ) ⊕ H d(E\|V , E#\|V ) H d(E\|U ∩V , E#\|U ∩V ). We ﬁrst construct the Thom class. We have uE\|V ∈ H d(E\|V , E#), uE\|U ∈ H d(E\|U , E#). We claim that (uE\|U , uE\|V ) ∈ H d(E\|U , E#\|U ) ⊕ H d(E\|V , E#\|V ) gets sent to 0 by i∗ V . Indeed, both the restriction of uE\|U and uE\|V to U ∩ V are Thom classes, so they are equal by uniqueness, so the diﬀerence vanishes. U − i∗ Then by exactness, there must be some uE\|U ∪V ∈ H d(E\|U ∪V , E#\|U ∪V ) that restricts to uE\|U and uE\|V in U and V respectively. Then this must be a Thom class, since the property of being a Thom class is checked on each ﬁber. Moreover, we get uniqueness because H d−1(E\|U ∩V , E#\|U ∩V ) = 0, so uE\|U and uE\|V must be the restriction of a unique thing. The last part in the Thom isomorphism theorem come from a routine application of the ﬁve lemma, and the ﬁrst part follows from the last as previously mentioned. 65 10 Vector bundles III Algebraic Topology 10.4 Gysin sequence Now we do something interesting with vector bundles. We will come up with a long exact sequence associated to a vector bundle, known as the Gysin sequence. We will then use the Gysin sequence to deduce something about the base space. Suppose we have a d-dimensional vector bundle π : E → X that is R-oriented. We want to talk about the unit sphere in every ﬁber of E. But to do so, we need to have a notion of length, and to do that, we want an inner product. But luckily, we do have one, and we know that any two norms on a ﬁnite-dimensional vector space are equivalent. So we might as well arbitrarily choose one. Deﬁnition (Sphere bundle). Let π : E → X be a vector bundle, and let · , · : E ⊗ E → R be an inner product, and let S(E) = {v ∈ E; v, v = 1} ⊆ E. This is the sphere bundle associated to E. Since the unit sphere is homotopy equivalent to Rd \ {0}, we know the inclusion j : S(E) E# is a homotopy equivalence, with inverse given by normalization. The long exact sequence for the pair (E, E#) gives (as before, we do not write the R): H i+d(E, E#) H i+d(E) H i+d(E#) H i+d+1(E, E#) Φ π∗ s∗ 0 j∗ Φ H i(X) · e(E) H i+d(X) p∗ H i+d(S(E)) p! H i+1(X) where p : S(E) → E is the projection, and p! is whatever makes the diagram commutes (since j∗ and Φ are isomorphisms). The bottom sequence is the Gysin sequence, and it is exact because the top row is exact. This is in fact a long exact sequence of H ∗(X; R)-modules, i.e. the maps commute with cup products. Example. Let L = γC 1,n+1 → CPn = Gr1(Cn+1) be the tautological 1 complex dimensional vector bundle on Gr1(Cn+1). This is Z-oriented as any complex vector bundle is, because if we consider the inclusion GL1(C) → GL2(R) obtained by pretending C is R2, we know GL1(C) is connected, so lands in the component of the identity, so has positive determinant. The sphere bundle consists of S(L) = {(V, v) ∈ CPn × Cn+1 : v ∈ V, \|v\| = 1} ∼= {v ∈ Cn+1 : \|v\| = 1} ∼= S2n+1, where the middle isomorphism is given by (V, v) (v, v) v v 66 10 Vector bundles III Algebraic Topology The Gysin sequence is H i+1(S2n+1) p! H i(CPn) e(L) H i+2(CPn) p∗ H i+2(S2n+1) Now if i ≤ 2n − 2, then both outer terms are 0. So the maps in the middle are isomorphisms. Thus we get isomorphisms H 0(CPn) e(L) H 2(CPn) e(L) H 4(CPn(L) Z · e(L)2 Similarly, we know that the terms in the odd degree vanish. Checking what happens at the end points carefully, the conclusion is that H ∗(CPn) = Z[e(L)]/(e(L)n+1) as a ring. Example. We do the real case
of the above computation. We have K = γR 1,n+1 → RPn = Gr1(Rn+1). The previous trick doesn’t work, and indeed this isn’t Z-orientable. However, it is F2-oriented as every vector bundle is, and by exactly the same argument, we know S(L) ∼= Sn. So by the same argument as above, we will ﬁnd that H ∗(RPn, F2) ∼= F2[e(L)]/(e(L)n+1). Note that this is diﬀerent from the one we had before, because here deg e(L) = 1, while the complex case had degree 2. 67 11 Manifolds and Poincar´e duality III Algebraic Topology 11 Manifolds and Poincar´e duality We are going to prove Poincar´e duality, and then use it to prove a lot of things about manifolds. Poincar´e duality tells us that for a compact oriented manifold M of dimension n, we have Hd(M ) ∼= H n−d(M ). To prove this, we will want to induct over covers of M . However, given a compact manifold, the open covers are in general not compact. We get compactness only when we join all of them up. So we need to come up with a version of Poincar´e duality that works for non-compact manifolds, which is less pretty and requires the notion of compactly supported cohomology. 11.1 Compactly supported cohomology Deﬁnition (Support of cochain). Let ϕ ∈ C n(X) be a cochain. We say ϕ has support in S ⊆ X if whenever σ : ∆n → X \ S ⊆ X, then ϕ(σ) = 0. In this case, dϕ also has support in S. Note that this has a slight subtlety. The deﬁnition only requires that if σ lies completely outside S, then ϕ(σ) vanishes. However, we can have simplices that extends very far and only touches S slightly, and the support does not tell us anything about the value of σ. Later, we will get around this problem by doing suﬃcient barycentric subdivision. Deﬁnition (Compactly-supported cochain). Let C·c (X) ⊆ C·(X) be the subchain complex consisting of these ϕ which has support in some compact K ⊆ X. Note that this makes sense — we have seen that if ϕ has support in K, then dϕ has support in K. To see it is indeed a sub-chain complex, we need to show that C·c (X) is a subgroup! Fortunately, if ϕ has support on K, and ψ has support in L, then ϕ + ψ has support in K ∪ L, which is compact. Deﬁnition (Compactly-supported cohomology). The compactly supported cohomology of X is c (X) = H ∗(C· H ∗ c (X)). Note that we can write C· c (X) = C·(X, X \ K) ⊆ C·(X). K compact We would like to say that the compactly supported cohomology is also “built out of” those relative cohomology, but we cannot just take the union, because the relative cohomology is not a subgroup of H ∗(X). To do that, we need something more fancy. Deﬁnition (Directed set). A directed set is a partial order (I, ≤) such that for all i, j ∈ I, there is some k ∈ I such that i ≤ k and j ≤ k. Example. Any total order is a directed system. Example. N with divisibility \| as the partial order is a directed system. 68 11 Manifolds and Poincar´e duality III Algebraic Topology Deﬁnition (Direct limit). Let I be a directed set. An direct system of abelian groups indexed by I is a collection of abelian groups Gi for each i ∈ I and homomorphisms for all i, j ∈ I such that i ≤ j, such that ρij : Gi → Gj and ρii = idGi ρik = ρjk ◦ ρij whenever i ≤ j ≤ k. We deﬁne the direct limit on the system (Gi, ρij) to be Gi = lim −→ i∈I i∈I Gi /x − ρij(x) : x ∈ Gi. The underlying set of it is Gi /{x ∼ ρij(x) : x ∈ Gi}. i∈I In terms of the second description, the group operation is given as follows: given x ∈ Gi and y ∈ Gj, we ﬁnd some k such that i, j ≤ k. Then we can view x, y as elements as Gk and do the operation there. It is an exercise to show that these two descriptions are indeed the same. Now observe that if J ⊆ I is a sub-directed set such that for all a ∈ I, there is some b ∈ J such that a ≤ b. Then we have Gi lim −→ i∈J ∼= lim −→ i∈I Gi. So our claim is now Theorem. For any space X, we let K(X) = {K ⊆ X : K is compact}. This is a directed set under inclusion, and the map K → H n(X, X \ K) gives a direct system of abelian groups indexed by K(X), where the maps ρ are given by restriction. Then we have H ∗ c (X) ∼= lim −→ K(X) H n(X, X \ K). Proof. We have c (X) ∼= lim C n −→ K(X) C n(X, X \ K), 69 11 Manifolds and Poincar´e duality III Algebraic Topology where we have a map C n(X, X \ K) → C n c (X) lim −→ K(α) given in each component of the direct limit by inclusion, and it is easy to see that this is well-deﬁned and bijective. It is then a general algebraic fact that H ∗ commutes with inverse limits, and we will not prove it. Lemma. We have H i c(Rd; R) ∼= R i = d 0 otherwise . Proof. Let B ∈ K(Rd) be the balls, namely B = {nDd, n = 0, 1, 2, · · · }. Then since every compact set is contained in one of them, we have H n c (X) ∼= lim −→ K∈K(Rd) H n(Rd, Rd \ K; R) ∼= lim −→ nDd∈B H n(Rd, Rd \ nDd; R) We can compute that directly. Since Rd is contractible, the connecting map H i(Rd, Rd \ nDd; R) → H i−1(Rd \ nDd; R) in the long exact sequence is an isomorphism. Moreover, the following diagram commutes: H i(Rd, Rd \ nDn; R) ρn,n+1 H i(Rd, Rd \ (n + 1)Dd; R) ∂ ∂ H i−1(Rd \ nDd; R) H i−1(Rd \ (n + 1)Dd; R) But all maps here are isomorphisms because the horizontal maps are homotopy equivalences. So we know H i(Rd, Rd \ nDd; R) ∼= H i(Rd, Rd \ {0}; R) ∼= H i−1(Rd \ {0}; R). lim −→ So it follows that H i(Rd, Rd \ {0}; R) = R i = d 0 otherwise . In general, this is how we always compute compactly-supported cohomology — we pick a suitable subset of K(X) and compute the limit of that instead. Note that compactly-supported cohomology is not homotopy-invariant! It knows about the dimension of Rd, since the notion of compactness is not homotopy invariant. Even worse, in general, a map f : X → Y does not induce a map f ∗ : H ∗ c (X). Indeed, the usual map does not work because the preimage of a compact set of a compact set is not necessarily compact. c (Y ) → H ∗ 70 11 Manifolds and Poincar´e duality III Algebraic Topology Deﬁnition (Proper map). A map f : X → Y of spaces is proper if the preimage of a compact space is compact. Now if f is proper, then it does induce a map H ∗ c ( · ) by the usual construction. From now on, we will assume all spaces are Hausdorﬀ, so that all compact subsets are closed. This isn’t too bad a restriction since we are ultimately interested in manifolds, which are by deﬁnition Hausdorﬀ. Let i : U → X be the inclusion of an open subspace. We let K ⊆ U be compact. Then by excision, we have an isomorphism H ∗(U, U \ K) ∼= H ∗(X, X \ K), since the thing we cut out, namely X \ U , is already closed, and U \ K is open, since K is closed. So a compactly supported cohomology class on U gives one on X. So we get a map i∗ : H ∗ c (U ) → H ∗ We call this “extension by zero”. Indeed, this is how the cohomology class works — if you have a cohomology class φ on U supported on K ⊆ U , then given any simplex in X, if it lies inside U , we know how to evaluate it. If it lies outside K, then we just send it to zero. Then by barycentric subdivision, we can assume every simplex is either inside U or outside K, so we are done. c (X). Example. If i : U → Rd is an open ball, then the map iRd) is an isomorphism. So each cohomology class is equivalent to something with a support as small as we like. Since it is annoying to write H n(X, X \ K) all the time, we write H n(X \| K; R) = H n(X, X \ K; R). By excision, this depends only on a neighbourhood of K in X. In the case where K is a point, this is local cohomology at a point. So it makes sense to call this local cohomology near K ⊆ X. Our end goal is to produce a Mayer-Vietoris for compactly-supported coho- mology. But before we do that, we ﬁrst do it for local cohomology. Proposition. Let K, L ⊆ X be compact. Then there is a long exact sequence H n(X \| K ∩ L) H n(X \| K) ⊕ H n(X \| L) H n(X \| K ∪ L) H n+1(X \| K ∩ L) ∂ H n+1(X \| K) ⊕ H n+1(X \| L) · · · , where the unlabelled maps are those induced by inclusion. We are going to prove this by relating it to a Mayer-Vietoris sequence of some sort. 71 11 Manifolds and Poincar´e duality III Algebraic Topology Proof. We cover X \ K ∩ L by U = {X \ K, X \ L}. We then draw a huge diagram (here ∗ denotes the dual, i.e. X ∗ = Hom(X; R), and C·(X \| K) = C·(X, X \ K)): 0 0 0 0 0 0 ∗ C·(X) CU· (X\K∩L) C·(X \| K) ⊕ C·(X \| L) C·(X \| K ∪ L) C·(X) (id,− id) C·(X) ⊕ C·(X) id + id C·(X) C· (j∗ U (X \ K ∩ L) C·(X \ K) ⊕ C·(X \ L) 1 ,−j∗ 2 ) 1 +i∗ i∗ 2 C·(X \ K ∪ L) 0 0 0 0 0 0 This is a diagram. Certainly. The bottom two rows and all columns are exact. By a diagram chase (the nine lemma), we know the top row is exact. Taking the long exact sequence almost gives what we want, except the ﬁrst term is a funny thing. We now analyze that object. We look at the left vertical column: 0 Hom C·(x) CU· (X\K∩L) , R C·(X) Hom(C U· (X \ K ∩ L), R) 0 Now by the small simplices theorem, the right hand object gives the same (co)homology as C·(X \ K ∩ L; R). So we can produce another short exact sequence: 0 0 Hom C·(x) CU· (X\(K∩L)) , R C·(X) Hom(C U· (X \ K ∩ L), R) C·(X, X \ K ∩ L) C·(X) Hom(C·(X \ K ∩ L), R) 0 0 Now the two right vertical arrows induce isomorphisms when we pass on to homology. So by taking the long exact sequence, the ﬁve lemma tells us the left hand map is an isomorphism on homology. So we know H∗ Hom C·(x) C U· (X \ (K ∩ L)) , R ∼= H ∗(X \| K ∩ L). So the long exact of the top row gives what we want. 72 11 Manifolds and Poincar´e duality III Algebraic Topology Corollary. Let X be a manifold, and X = A ∪ B, where A, B are open sets. Then there is a long exact sequence H n c (A ∩ B) H n c (A) ⊕ H n c (B) H n c (X) H n+1 c (A ∩ B) H n+1 c (A) ⊕ H n+1 c (B) · · · ∂ Note that the arrows go in funny directions, which is diﬀerent from both homology and cohomology versions! Proof. Let K ⊆ A and L ⊆ B be compact sets. Then by excision, we have isomorphisms H n(X \| K) ∼= H n(A \| K) H n(X \| L) ∼= H n(B \| L) H n(X \| K ∩ L) ∼= H n(A ∩ B \| K ∩ L). So the long exact sequence from the previous proposition gives us H n(A ∩ B \| K ∩ L) H n(A \| K) ⊕ H n(B \| L) H n(X \| K ∪ L) H n+1(A ∩ B \| K ∩ L) H n+1(A \| K) ⊕ H n+1(
B \| L) · · · ∂ The next step is to take the direct limit over K ∈ K(A) and L ∈ K(B). We need to make sure that these do indeed give the right compactly supported cohomology. The terms H n(A \| K) ⊕ H n(B \| L) are exactly right, and the one for H n(A ∩ B \| K ∩ L) also works because every compact set in A ∩ B is a compact set in A intersect a compact set in B (take those to be both the original compact set). So we get a long exact sequence H n c (A ∩ B) H n c (A) ⊕ H n c (B) lim −→ K∈K(A) L∈K(B) H n(X \| K ∪ L) H n+1 ∂ c (A ∩ B) To show that that funny direct limit is really what we want, we have to show that every compact set C ∈ X lies inside some K ∪ L, where K ⊆ A and L ⊆ B are compact. Indeed, as X is a manifold, and C is compact, we can ﬁnd a ﬁnite set of closed balls in X, each in A or in B, such that their interiors cover C. So done. (In general, this will work for any locally compact space) This is all we have got to say about compactly supported cohomology. We now start to talk about manifolds. 11.2 Orientation of manifolds Similar to the case of vector bundles, we will only work with manifolds with orientation. The deﬁnition of orientation of a manifold is somewhat similar 73 11 Manifolds and Poincar´e duality III Algebraic Topology to the deﬁnition of orientation of a vector bundle. Indeed, it is true that an orientation of a manifold is just an orientation of the tangent bundle, but we will not go down that route, because the description we use for orientation here will be useful later on. After deﬁning orientation, we will prove a result similar to (the ﬁrst two parts of) the Thom isomorphism theorem. For a d-manifold M and x ∈ M , we know that Hi(M \| x; R) ∼= R i = d i = d 0 . We can then deﬁne a local orientation of a manifold to be a generator of this group. Deﬁnition (Local R-orientation of manifold). Fr a d-manifold M , a local Rorientation of M at x is an R-module generator µx = Hd(M \| x; R). Deﬁnition (R-orientation). An R-orientation of M is a collection {µx}x∈M of local R-orientations such that if is a chart of M , and p, q ∈ Rd, then the composition of isomorphisms ϕ : Rd → U ⊆ M Hd(M \| ϕ(p)) Hd(M \| ϕ(q)) ∼ ∼ Hd(U \| ϕ(p)) Hd(U \| ϕ(q)) ∼ ϕ∗ ∼ ϕ∗ Hd(Rd \| p) ∼ Hd(Rd \| q) sends µx to µy, where the vertical isomorphism is induced by a translation of Rd. Deﬁnition (Orientation-preserving homeomorphism). For a homomorphism f : U → V with U, V ∈ Rd open, we say f is R-orientation-preserving if for each x ∈ U , and y = f (x), the composition Hd(Rd \| 0; R) translation Hd(Rd \| x; R) excision Hd(U \| x; R) Hd(Rd \| 0; R) translation Hd(Rd \| y; R) excision Hd(V \| y; R) f∗ is the identity Hd(Rd \| 0; R) → Hd(Rd \| 0; R). As before, we have the following lemma: Lemma. (i) If R = F2, then every manifold is R-orientable. (ii) If {ϕα : Rd → Uα ⊆ M } is an open cover of M by Euclidean space such that each homeomorphism Rd ⊇ ϕ−1 α (Uα ∩ Uβ) ϕ−1 α Uα ∩ Uβ ϕ−1 β β (Uα ∩ Uβ) ⊆ Rd ϕ−1 is orientation-preserving, then M is R-orientable. 74 11 Manifolds and Poincar´e duality III Algebraic Topology Proof. (i) F2 has a unique F2-module generator. (ii) For x ∈ Uα, we deﬁne µx to be the image of the standard orientation of Rd via Hd(M \| x) ∼= Hα(Ud \| x) (ϕα)∗ Hd(Rd \| ϕ−1 α (x)) Rd(Rd \| 0) trans. If this is well-deﬁned, then it is obvious that this is compatible. However, we have to check it is well-deﬁned, because to deﬁne this, we need to pick a chart. If x ∈ Uβ as well, we need to look at the corresponding µ instead. But they have to agree by deﬁnition of orientation-preserving. x deﬁned using Uβ Finally, we get to the theorem: Theorem. Let M be an R-oriented manifold and A ⊆ M be compact. Then (i) There is a unique class µA ∈ Hd(M \| A; R) which restricts to µx ∈ Hd(M \| x; R) for all x ∈ A. (ii) Hi(M \| A; R) = 0 for i > d. Proof. Call a compact set A “good” if it satisﬁes the conclusion of the theorem. Claim. We ﬁrst show that if K, L and K ∩ L is good, then K ∪ L is good. This is analogous to the proof of the Thom isomorphism theorem, and we will omit this. Now our strategy is to prove the following in order: (i) If A ⊆ Rd is convex, then A is good. (ii) If A ⊆ Rd, then A is good. (iii) If A ⊆ M , then A is good. Claim. If A ⊆ Rd is convex, then A is good. Let x ∈ A. Then we have an inclusion Rd \ A → Rd \ {x}. This is in fact a homotopy equivalence by scaling away from x. Thus the map Hi(Rd \| A) → Hi(Rd \| x) is an isomorphism by the ﬁve lemma for all i. Then in degree d, there is some µA corresponding to µx. This µA is then has the required property by deﬁnition of orientability. The second part of the theorem also follows by what we know about Hi(Rd \| x). Claim. If A ⊆ Rd, then A is good. 75 11 Manifolds and Poincar´e duality III Algebraic Topology For A ⊆ Rd compact, we can ﬁnd a ﬁnite collection of closed balls Bi such that A ⊆ n i=1 ˚Bi = B. Moreover, if U ⊇ A for any open U , then we can in fact take Bi ⊆ U . By induction on the number of balls n, the ﬁrst claim tells us that any B of this form is good. We now let G = {B ⊆ Rd : A ⊆ ˚B, B compact and good}. We claim that this is a directed set under inverse inclusion. To see this, for B, B ∈ G, we need to ﬁnd a B ∈ G such that B ⊆ B, B and B is good and compact. But the above argument tells us we can ﬁnd one contained in ˚B ∪ ˚B. So we are safe. Now consider the directed system of groups given by and there is an induced map B → Hi(Rd \| B), Hi(Rd \| B) → Hi(Rd \| A), lim −→ B∈G since each Hi(Rd \| B) maps to Hi(Rd \| A) by inclusion, and these maps are compatible. We claim that this is an isomorphism. We ﬁrst show that this is surjective. Let [c] ∈ Hi(Rd \| A). Then the boundary of c ∈ Ci(Rd) is a ﬁnite sum of simplices in Rd \ A. So it is a sum of simplices in some compact C ⊆ Rd \ A. But then A ⊆ Rd \ C, and Rd \ C is an open neighbourhood of A. So we can ﬁnd a good B such that A ⊆ B ⊆ Rd \ C. Then c ∈ Ci(Rd \| B) is a cycle. So we know [c] ∈ Hi(Rd \| B). So the map is surjective. Injectivity is obvious. An immediate consequence of this is that for i > d, we have Hi(Rd \| A) = 0. Also, if i = d, we know that µA is given uniquely by the collection {µB}B∈G (uniqueness follows from injectivity). Claim. If A ⊆ M , then A is good. This follows from the fact that any compact A ⊆ M can be written as a ∼= Rd. So Aα and their intersections ﬁnite union of compact Aα with Aα ⊆ Uα are good. So done. Corollary. If M is compact, then we get a unique class [M ] = µM ∈ Hn(M ; R) such that it restricts to µx for each x ∈ M . Moreover, H i(M ; R) = 0 for i > d. This is not too surprising, actually. If we have a triangulation of the manifold, then this [M ] is just the sum of all the triangles. Deﬁnition (Fundamental class). The fundamental class of an R-oriented manifold is the unique class [M ] that restricts to µx for each x ∈ M . 76 11 Manifolds and Poincar´e duality III Algebraic Topology 11.3 Poincar´e duality We now get to the main theorem — Poincar´e duality: Theorem (Poincar´e duality). Let M be a d-dimensional R-oriented manifold. Then there is a map DM : H k c (M ; R) → Hd−k(M ; R) that is an isomorphism. The majority of the work is in deﬁning the map. Afterwards, proving it is an isomorphism is a routine exercise with Mayer-Vietoris and the ﬁve lemma. What does this tell us? We know that M has no homology or cohomology in negative dimensions. So by Poincar´e duality, there is also no homology or cohomology in dimensions > d. Moreover, if M itself is compact, then we know H 0 c (M ; R) has a special element 1. So we also get a canonical element of Hd(M ; R). But we know there is a special element of Hd(M ; R), namely the fundamental class. They are in fact the same, and this is no coincidence. This is in fact how we are going to produce the map. To deﬁne the map DM , we need the notion of the cap product Deﬁnition (Cap product). The cap product is deﬁned by · · : Ck(X; R) × C (X; R) → Ck−(X; R) (σ, ϕ) → ϕ(σ\|[v0,...,v])σ\|[v,...,vk]. We want this to induce a map on homology. To do so, we need to know how it interacts with diﬀerentials. Lemma. We have d(σ ϕ) = (−1)d((dσ) ϕ − σ (dϕ)). Proof. Write both sides out. As with the analogous formula for the cup product, this implies we get a well-deﬁned map on homology and cohomology, i.e. We obtain a map Hk(X; R) × H (X; R) → Hk−(X; R). As with the cup product, there are also relative versions : Hk(X, A; R) × H (X; R) → Hk−(X, A; R) and : Hk(X, A; R) × H (X, A; R) → Hk−(X; R). We would like to say that the cap product is natural, but since maps of spaces induce maps of homology and cohomology in opposite directions, this is rather tricky. What we manage to get is the following: 77 11 Manifolds and Poincar´e duality III Algebraic Topology Lemma. If f : X → Y is a map, and x ∈ Hk(X; R) And y ∈ H (Y ; R), then we have f∗(x) y = f∗(x f ∗(y)) ∈ Hk−(Y ; R). In other words, the following diagram commutes: Hk(Y ; R) × H (Y ; R) Hk−(Y ; R) f∗×id Hk(X; R) × H (Y ; R) id ×f ∗ f∗ Hk(X; R) × H (X; R) Hk−(X; R) Proof. We check this on the cochain level. We let x = σ : ∆k → X. Then we have f#(σ f #y) = f# (f #y)(σ\|[v0,...,v])σ\|[v,...,vk] = y(f#(σ\|[v0,...,v]))f#(σ\|[v,...,vk]) = y((f#σ)\|[v0,...,v])(f#σ)\|[v,...,vk] = (f#σ) y. So done. Now if M is compact, then we simply deﬁne the duality map as DM = [M ] · : H d(M ; R) → Hd−(M ; R). If not, we note that H d c (M ; R) is the directed limit of H d(M, K; R) with K compact, and each H d(M, L; R) has a fundamental class. So we can deﬁne the required map for each K, and then put them together. More precisely, if K ⊆ L ⊆ M are such that K, L are compact, then we have an inclusion map (id, inc) : (M, M \ L) → (M, M \ K). Then we have an induced map (id, inc) : Hd(M \| L; R) → Hd(M \| K; R) that sends the fundamental class µL to µK, by uniqueness of the fundamental class. Then the relative version of our lemma tells us that the following map commutes: H (M \| K; R) (id,inc)∗ H (M \| L; R) µK · µL · Hd−(M ; R) id Hd−(M ; R) Indeed, this is just saying that (id)∗(µL (id, inc)∗(ϕ)) = µK ϕ. 78 11 Manifolds and Poincar´e dual
ity III Algebraic Topology So we get a duality map DM = lim −→ (µK · ) : lim −→ H (M \| K; R) → Hd−(M ; R). Now we can prove Poincar´e duality. Proof. We say M is “good” if the Poincar´e duality theorem holds for M . We now do the most important step in the proof: Claim 0. Rd is good. The only non-trivial degrees to check are = 0, d, and = 0 is straightforward. For = d, we have shown that the maps H d c (Rd; R) ∼ H d(Rd \| 0; R) UCT HomR(Hd(Rd \| 0; R), R) are isomorphisms, where the last map is given by the universal coeﬃcients theorem. Under these isomorphisms, the map H d c (Rd; R) DRd H0(Rd; R) ε R corresponds to the map HomK(Hd(Rd \| 0; R), R) → R is given by evaluating a function at the fundamental class µ0. But as µ0 ∈ Hd(Rd \| 0; R) is an R-module generator, this map is an isomorphism. Claim 1. If M = U ∪ V and U, V, U ∩ V are good, then M is good. Again, this is an application of the ﬁve lemma with the Mayer-Vietoris sequence. We have H c (U V ) H c (U ) ⊕ H d c (V ) DUV DU ⊕DV H c (M ) DM H +1 c (U V ) DUV Hd−(U V ) Hd−(U ) ⊕ Hd−(V ) Hd−(M ) Hd−−1(U V ) We are done by the ﬁve lemma if this commutes. But unfortunately, it doesn’t. It only commutes up to a sign, but it is suﬃcient for the ﬁve lemma to apply if we trace through the proof of the ﬁve lemma. Claim 2. If U1 ⊆ U2 ⊆ · · · with M = good. n Un, and Ui are all good, then M is Any compact set in M lies in some Un, so the map lim −→ H c (Un) → H c (Un) is an isomorphism. Similarly, since simplices are compact, we also have Hd−k(M ) = lim −→ Hd−k(Un). Since the direct limit of open sets is open, we are done. Claim 3. Any open subset of Rd is good. 79 11 Manifolds and Poincar´e duality III Algebraic Topology Any U is a countable union of open balls (something something rational points something something). For ﬁnite unions, we can use Claims 0 and 1 and induction. For countable unions, we use Claim 2. Claim 4. If M has a countable cover by Rd’s it is good. Same argument as above, where we instead use Claim 3 instead of Claim 0 for the base case. Claim 5. Any manifold M is good. Any manifold is second-countable by deﬁnition, so has a countable open cover by Rd. Corollary. For any compact d-dimensional R-oriented manifold M , the map [M ] · : H (M ; R) → Hd−(M ; R) is an isomorphism. Corollary. Let M be an odd-dimensional compact manifold. Then the Euler characteristic χ(M ) = 0. Proof. Pick R = F2. Then M is F2-oriented. Since we can compute Euler characteristics using coeﬃcients in F2. We then have χ(M ) = 2n+1 (−1)i dimF2 Hi(M, F2). r=0 But we know Hi(M, F2) ∼= H 2n+1−i(M, F2) ∼= (H2n+1−i(M, F2))∗ ∼= H2n+1−i(M, F2) by Poincar´e duality and the universal coeﬃcients theorem. But the dimensions of these show up in the sum above with opposite signs. So they cancel, and χ(M ) = 0. What is the relation between the cap product and the cup product? If σ ∈ Ck+(X; R), ϕ ∈ C k(X; R) and ψ ∈ C (X; R), then ψ(σ ϕ) = ψ(ϕ(σ\|[v0,...,vk])σ\|[vk,...,vk+]) = ϕ(σ\|[v0,...,vk])ψ(σ\|[vk,...,vk+]) = (ϕ ψ)(σ), Since we like diagrams, we can express this equality in terms of some diagram commuting. The map h : H k(X; R) → HomR(Hk(X; R), R) in the universal coeﬃcients theorem is given by [ϕ] → ([σ] → ϕ(σ)). This map exists all the time, even if the hypothesis of the universal coeﬃcients theorem does not hold. It’s just that it need not be an isomorphism. The formula ψ(σ ϕ) = (ϕ ψ)(σ) 80 11 Manifolds and Poincar´e duality III Algebraic Topology then translates to the following diagram commuting: H (X; R) ϕ · H k+(X; R) h h HomR(H(X; R), R) ( · ϕ)∗ HomR(H+k(X; R), R) Now when the universal coeﬃcient theorem applies, then the maps h are isomorphisms. So the cup product and cap product determine each other, and contain the same information. Now since Poincar´e duality is expressed in terms of cap products, this correspondence gives us some information about cupping as well. Theorem. Let M be a d-dimensional compact R-oriented manifold, and consider the following pairing: · , · : H k(M ; R) ⊗ H d−k(M, R) R . [ϕ] ⊗ [ψ] (ϕ ψ)[M ] If H∗(M ; R) is free, then · , · is non-singular, i.e. both adjoints are isomorphisms, i.e. both H k(M ; R) Hom(H d−k(M ; R), R) [ϕ] ([ψ] → ϕ, ψ) and the other way round are isomorphisms. Proof. We have as we know ϕ, ψ = (−1)\|ϕ\|\|ψ\|ψ, ϕ, ϕ ψ = (−1)\|ϕ\|\|ψ\|ψ ϕ. So if one adjoint is an isomorphism, then so is the other. To see that they are isomorphsims, we notice that we have an isomorphism H k(M ; R) UCT HomR(Hk(M ; R), R) D∗ m HomR(H d−k(M ; R), R) . [ϕ] ([σ] → ϕ(σ)) ([ψ] → ϕ([M ] ψ)) But we know ϕ([M ] ψ) = (ψ ϕ)([M ]) = ψ, ϕ. So this is just the adjoint. So the adjoint is an isomorphism. This is a very useful result. We have already seen examples where we can ﬁgure out if a cup product vanishes. But this result tells us that certain cup products are not zero. This is something we haven’t seen before. 81 11 Manifolds and Poincar´e duality III Algebraic Topology Example. Consider CPn. We have previously computed H∗(CPn, Z) = Z ∗ = 2i, 0 otherwise 0 ≤ i ≤ n Also, we know CPn is Z-oriented, because it is a complex manifold. Let’s forget that we have computed the cohomology ring structure, and compute it again. Suppose for induction, that we have found H ∗(CPn−1, Z) = Z[x]/(xn). As CPn is obtained from CPn−1 by attaching a 2n cell, the map given by inclusion i∗ : H 2(CPn, Z) → H 2(CPn−1, Z) is an isomorphism. Then the generator x ∈ H 2(CPn−1, Z) gives us a generator y ∈ H 2(CPn, Z). Now if we can show that yk ∈ H 2k(CPn, Z) ∼= Z is a generator for all k, then H ∗(CPn, Z) ∼= Z[y]/(yn+1). But we know that yn−1 generates H 2n−2(CPn, Z), since it pulls back under i∗ to xn−1, which is a generator. Finally, consider H 2(CP2, Z) ⊗ H 2n−2(CP2, Z) Z y ⊗ yn−1 yn[CPn]. Since this is non-singular, we know yn ∈ H 2n(CPn, Z) must be a generator. Of course, we can get H ∗(RPn, F2) similarly. 11.4 Applications We go through two rather straightforward applications, before we move on to bigger things like the intersection product. Signature We focus on the case where d = 2n is even. Then we have, in particular, a non-degenerate bilinear form · , · : H n(M ; R) ⊗ H n(M ; R) → R. Also, we know a, b = (−1)n2 b, a = (−1)nb, a. So this is a symmetric form if n is even, and skew-symmetric form if n is odd. These are very diﬀerent scenario. For example, we know a symmetric matrix is diagonalizable with real eigenvalues (if R = R), but a skew-symmetric form does not have these properties. So if M is 4k-dimensional, and Z-oriented, then in particular M is R-oriented. Then the map · , · : H 2k(M ; R) ⊗ H 2k(M ; R) → R can be represented by a symmetric real matrix, which can be diagonalized. This has some real eigenvalues. The eigenvalues can be positive or negative. 82 11 Manifolds and Poincar´e duality III Algebraic Topology Deﬁnition (Signature of manifold). Let M be a 4k-dimensional Z-oriented manifold. Then the signature is the number of positive eigenvalues of · , · : H 2k(M ; R) ⊗ H 2k(M ; R) → R minus the number of negative eigenvalues. We write this as sgn(M ). By Sylvester’s law of inertia, this is well-deﬁned. Fact. If M = ∂W for some compact 4k + 1-dimensional manifold W with boundary, then sgn(M ) = 0. Example. CP2 has H 2(CP2; R) ∼= R, and the bilinear form is represented by the matrix (1). So the signature is 1. So CP2 is not the boundary of a manifold. Degree Recall we deﬁned the degree of a map from a sphere to itself. But if we have a Z-oriented space, we can have the fundamental class [M ], and then there is an obvious way to deﬁne the degree. Deﬁnition (Degree of map). If M, N are d-dimensional compact connected Z-oriented manifolds, and f : M → N , then f∗([M ]) ∈ Hd(N, Z) ∼= Z[N ]. So f∗([M ]) = k[N ] for some k. This k is called the degree of f , written deg(f ). If N = M = Sn and we pick the same orientation for them, then this recovers our previous deﬁnition. By exactly the same proof, we can compute this degree using local degrees, just as in the case of a sphere. Corollary. Let f : M → N be a map between manifolds. If F is a ﬁeld and deg(f ) = 0 ∈ F, then then the induced map f ∗ : H ∗(N, F) → H ∗(M, F) is injective. This seems just like an amusement, but this is powerful. We know this is in fact a map of rings. So if we know how to compute cup products in H ∗(M ; F), then we can do it for H ∗(N ; F) as well. Proof. Suppose not. Let α ∈ H k(N, F) be non-zero but f ∗(α) = 0. As · , · : H k(N, F) ⊗ H d−k(N, F) → F is non-singular, we know there is some β ∈ H d−k(N, F) such that Then we have α, β = (α β)[N ] = 1. deg(f ) = deg(f ) · 1 = (α β)(deg(f )[N ]) = (α β)(f∗[M ]) = (f ∗(α) f ∗(β))([M ]) = 0. This is a contradiction. 83 11 Manifolds and Poincar´e duality III Algebraic Topology 11.5 Intersection product Recall that cohomology comes with a cup product. Thus, Poincar´e duality gives us a product on homology. Our goal in this section is to understand this product. We restrict our attention to smooth manifolds, so that we can talk about the tangent bundle. Recall (from the example sheet) that an orientation of the manifold is the same as an orientation on the tangent bundle. We will consider homology classes that come from submanifolds. For concreteness, let M be a compact smooth R-oriented manifold, and N ⊆ M be an n-dimensional R-oriented submanifold. Let i : N → M be the inclusion. Suppose dim M = d and dim N = n. Then we obtain a canonical homology class i∗[N ] ∈ Hn(M ; R). We will abuse notation and write [N ] for i∗[N ]. This may or may not be zero. Our objective is to show that under suitable conditions, the product of [N1] and [N2] is [N1 ∩ N2]. To do so, we will have to understand what is the cohomology class Poinacr´e dual to [N ]. We claim that, suitably interpreted, it is the Thom class of the normal bundle. Write νN ⊆M for the normal bundle of N in M . Picking a metric on T M , we can decompose i∗T M ∼= T N ⊕ νN ⊆M , Since T M is oriented, we obtain an orientation on the pullback i∗T M . Similarly, T N is also oriented by assumption. In general, we have the
following result: Lemma. Let X be a space and V a vector bundle over X. If V = U ⊕ W , then orientations for any two of U, W, V give an orientation for the third. Proof. Say dim V = d, dim U = n, dim W = m. Then at each point x ∈ X, by K¨unneth’s theorem, we have an isomorphism H d(Vx, V # x ; R) ∼= H n(Ux, U # x ; R) ⊗ H m(Wx, W # x ; R) ∼= R. So any local R-orientation on any two induces one on the third, and it is straightforward to check the local compatibility condition. Can we ﬁnd a more concrete description of this orientation on νN ⊆M ? By c(Rd), we know the same argument as when we showed that H i(Rn \| {0}) ∼= H i H i(νN ⊆M , ν# N ⊆M ; R) ∼= H i c(νN ⊆M ; R). Also, by the tubular neighbourhood theorem, we know νN ⊆M is homeomorphic to an open neighbourhood U of N in M . So we get isomorphisms H i c(νN ⊆M ; R) ∼= H i c(U ; R) ∼= Hd−i(U ; R) ∼= Hd−i(N ; R), where the last isomorphism comes from the fact that N is homotopy-equivalent to U . In total, we have an isomorphism H i(νN ⊆M , ν# N ⊆M ; R) ∼= Hd−i(N ; R). 84 11 Manifolds and Poincar´e duality III Algebraic Topology Under this isomorphism, the fundamental class [N ] ∈ Hn(N ; R) corresponds to some EN ⊆M ∈ H d−n(νN ⊆M , ν# N ⊆M ; R) But we know νN ⊆M has dimension d − n. So EN ⊆M is in the right dimension to be a Thom class, and is a generator for H d−n(νN ⊆M , ν# N ⊆M ; R), because it is a generator for Hd−n(N ; R). One can check that this is indeed the Thom class. How is this related to the other things we’ve had? We can draw the commu- tative diagram Hn(N ; R) ∼ Hn(U ; R) i∗ Hn(M ; R) ∼ ∼ H d−n c (U ; R) extension by 0 H d−n(M ; R) The commutativity of the square is a straightforward naturality property of Poincar´e duality. Under the identiﬁcation H d−n c (U ; R) ∼= H d−n(νN ⊆M , ν# N ⊆M ; R), the above (U ; R) is the Thom class of the says that the image of [N ] ∈ Hn(N ; R) in H d−n normal bundle νN ⊆M . c On the other hand, if we look at this composition via the bottom map, then [N ] gets sent to D−1 M ([N ]). So we know that Theorem. The Poincar´e dual of a submanifold is (the extension by zero of) the normal Thom class. Now if we had two submanifolds N, W ⊆ M . Then the normal Thom classes give us two cohomology classes of M . As promised, When the two intersect nicely, the cup product of their Thom classes is the Thom class of [N ∩ W ]. The niceness condition we need is the following: Deﬁnition (Transverse intersection). We say two submanifolds N, W ⊆ M intersect transversely if for all x ∈ N ∩ W , we have Example. We allow intersections like TxN + TxW = TxM. but not this: 85 11 Manifolds and Poincar´e duality III Algebraic Topology It is a fact that we can always “wiggle” the submanifolds a bit so that the intersection is transverse, so this is not too much of a restriction. We will neither make this precise nor prove it. Whenever the intersection is transverse, the intersection N ∩ W will be a submanifold of M , and of N and W as well. Moreover, (νN ∩W ⊆M )x = (νN ⊆M )x ⊕ (νW ⊆M )x. Now consider the inclusions iN : N ∩ W → N iW : N ∩ W → W. Then we have νN ∩W ⊆M = i∗ N (νN ⊆M ) ⊕ i∗ W (νW ⊆M ). So with some abuse of notation, we can write N EN ⊆M i∗ i∗ W EW ⊆M ∈ H ∗(νN ∩W ⊆M , ν# N ∩W ⊆M ; R), and we can check this gives the Thom class. So we have D−1 M ([N ]) D−1 M ([W ]) = D−1 M ([N ∩ W ]). The slogan is “cup product is Poincar´e dual to intersection”. One might be slightly worried about the graded commutativity of the cup product, because N ∩ W = W ∩ N as manifolds, but in general, D−1 M ([N ]) D−1 M ([W ]) = D−1 M ([W ]) D−1 M ([N ]). The ﬁx is to say that N ∩ W are W ∩ N are not the same as oriented manifolds in general, and sometimes they diﬀer by a sign, but we will not go into details. More generally, we can deﬁne Deﬁnition (Intersection product). The intersection product on the homology of a compact manifold is given by Hn−k(M ) ⊗ Hn−(M ) Hn−k−(M ) (a, b) Example. We know that a · b = DM (D−1 M (a) D−1 M (b)) H2k(CPn, Z) ∼= Z 0 ≤ k ≤ n otherwise 0 . Moreover, we know the generator for these when we computed it using cellular homology, namely [CPk] ≡ yk ∈ H2k(CPn, Z). To compute [CP] · [CP], if we picked the canonical copies of CP, CPk ⊆ CPn, then one would be contained in the other, and this is exactly the opposite of intersecting transversely. 86 11 Manifolds and Poincar´e duality III Algebraic Topology Instead, we pick CPk = {[z0 : z1 · · · zk : 0 : · · · : 0]}, CP = {[0 : · · · : 0 : w0 : · · · : w]}. It is a fact that any two embeddings of CPk → CPn are homotopic, so we can choose these. Now these two manifolds intersect transversely, and the intersection is So this says that CPk ∩ CP = CPk+−n. yk · y = ±yl+−n, where there is some annoying sign we do not bother to ﬁgure out. So if xk is Poincar´e dual to yn−k, then xk x = xk+, which is what we have previously found out. Example. Consider the manifold with three holes b1 a1 b2 a2 b3 a3 Then we have ai · bi = {pt}, ai · bj = 0 for i = j. So we ﬁnd the ring structure of H∗(Eg, Z), hence the ring structure of H ∗(Eg, Z). This is so much easier than everything else we’ve been doing. Here we know that the intersection product is well-deﬁned, so we are free to pick our own nice representatives of the loop to perform the calculation. Of course, this method is not completely general. For example, it would be diﬃcult to visualize this when we work with manifolds with higher dimensions, and more severely, not all homology classes of a manifold have to come from submanifolds (e.g. twice the fundamental class)! 11.6 The diagonal Again, let M be a compact Q-oriented d-dimensional manifold. Then M × M is a 2d-dimensional manifold. We can try to orient it as follows — K¨unneth gives us an isomorphism H d+d(M × M ; Q) ∼= H d(M ; Q) ⊗ H d(M ; Q), as H k(M ; Q) = 0 for k > d. By the universal coeﬃcients theorem, plus the fact that duals commute with tensor products, we have an isomorphism Hd+d(M × M ; Q) ∼= Hd(M ; Q) ⊗ Hd(M ; Q). 87 11 Manifolds and Poincar´e duality III Algebraic Topology Thus the fundamental class [M ] gives us a fundamental class [M × M ] = [M ] ⊗ [M ]. We are going to do some magic that involves the diagonal map ∆ : M M × M x (x, x). This gives us a cohomology class δ = D−1 M ×M (∆∗[M ]) ∈ H d(M × M, Q) ∼= i+j=d H i(M, Q) ⊗ H j(M, Q). It turns out a lot of things we want to do with this δ can be helped a lot by doing the despicable thing called picking a basis. We let {ai} be a basis for H ∗(M, Q). On this vector space, we have a non-singular form · , · : H ∗(M, Q) ⊗ H ∗(M ; Q) → Q given by ϕ, ψ → (ϕ ψ)([M ]). Last time we were careful about the degrees of the cochains, but here we just say that if ϕ ψ does not have dimension d, then the result is 0. Now let {bi} be the dual basis to the ai using this form, i.e. It turns out δ has a really nice form expressed using this basis: ai, bj = δij. Theorem. We have δ = (−1)\|ai\|ai ⊗ bi. i Proof. We can certainly write δ = i, Cijai ⊗ bj for some Cij. So we need to ﬁgure out what the coeﬃcients Cij are. We try to compute ((bk ⊗ a) δ)[M × M ] = = = Cij(bk ⊗ a) (ai ⊗ bj)[M × M ] Cij(−1)\|a\|\|ai\|(bk ai) ⊗ (a bi)[M ] ⊗ [M ] Cij(−1)\|a\|\|ai\|(δik(−1)\|ai\|\|bk\|)δj = (−1)\|ak\|\|a\|+\|ak\|\|bk\|Ck. But we can also compute this a diﬀerent way, using the deﬁnition of δ: (bk ⊗ a δ)[M × M ] = (bk ⊗ a)(∆∗[M ]) = (bk a)[M ] = (−1)\|a\|\|bk\|δk. So we see that Ck = δk(−1)\|a\|. 88 11 Manifolds and Poincar´e duality III Algebraic Topology Corollary. We have the Euler characteristic. ∆∗(δ)[M ] = χ(M ), Proof. We note that for a ⊗ b ∈ H n(M × M ), we have ∆∗(a ⊗ b) = ∆∗(π∗ 1a π∗ 2b) = a b because πi ◦ ∆ = id. So we have ∆∗(δ) = (−1)\|ai\|ai bi. Thus ∆∗(δ)[M ] = (−1)\|ai\| = i k (−1)k dimQ H k(M ; Q) = χ(M ). So far everything works for an arbitrary manifold. Now we suppose further that M is smooth. Then δ is the Thom class of the normal bundle νM ⊆M ×M of M → M × M. By deﬁnition, pulling this back along ∆ to M gives the Euler class of the normal ∼= T M , because the ﬁber at x is the cokernel of bundle. But we know νM ⊆M ×M the map TxM ∆ TxM ⊕ TxM and the inverse given by gives us an isomorphism Corollary. We have TxM ⊕ TxM → TxM (v, w) → v − w TxM ⊕ TxM ∆TxM ∼= TxM. e(T M )[M ] = χ(M ). Corollary. If M has a nowhere-zero vector ﬁeld, then χ(M ) = 0. More generally, this tells us that χ(M ) is the number of zeroes of a vector ﬁeld M → T M (transverse to the zero section) counted with sign. Lemma. Suppose we have R-oriented vector bundles E → X and F → X with Thom classes uE, uF . Then the Thom class for E ⊕ F → X is uE uF . Thus e(E ⊕ F ) = e(E) e(F ). Proof. More precisely, we have projection maps πE E E ⊕ F 89 πF . F 11 Manifolds and Poincar´e duality III Algebraic Topology We let U = π−1 E (E#) and V = π−1 F (F #). Now observe that U ∪ V = (E ⊕ F )#. So if dim E = e, dim F = f , then we have a map H e(E, E#) ⊗ H f (F, F #) π∗ E ⊗π∗ F H e(E ⊕ F, U ) ⊗ H f (E ⊕ F, V ) , H e+f (E ⊕ F, (E ⊕ F )#) and it is easy to see that the image of uE ⊗ uF is the Thom class of E ⊕ F by checking the ﬁbers. Corollary. T S2n has no proper subbundles. Proof. We know e(T S2n) = 0 as e(T S2n)[S2] = χ(S2n) = 2. But it cannot be a proper cup product of two classes, since there is nothing in lower cohomology groups. So T S2n is not the sum of two subbundles. Hence T S2n cannot have a proper subbundle E or else T S2n = E ⊕E⊥ (for any choice of inner product). 11.7 Lefschetz ﬁxed point theorem Finally, we are going to prove the Lefschetz ﬁxed point theorem. This is going to be better than the version you get in Part II, because this time we will know how many ﬁxed points there are. So let M be a compact d-dimensional manifold that is Z-oriented, and f : M → M be a map. Now if we want to count the number of ﬁxed points, then we want to make sure the map is “transverse” in some sense, so that there aren’t inﬁnitely many ﬁxed points. It turns out the right condition is that the graph Γf = {(x, f (x)) ∈ M × M } ⊆ M × M has to be transverse to the diagonal. Since Γf ∩ ∆ is exactly the ﬁxed points of f ,
this is equivalent to requiring that for each ﬁxed point x, the map Dx∆ ⊕ DxF : TxM ⊕ TxM → TxM ⊕ TxM is an isomorphism. We can write the matrix of this map, which is I Dxf . I I Doing a bunch of row and column operations, this is equivalent to requiring that I Dxf 0 I − Dxf is invertible. Thus the condition is equivalent to requiring that 1 is not an eigenvalue of Dxf . The claim is now the following: 90 11 Manifolds and Poincar´e duality III Algebraic Topology Theorem (Lefschetz ﬁxed point theorem). Let M be a compact d-dimensional Z-oriented manifold, and let f : M → M be a map such that the graph Γf and diagonal ∆ intersect transversely. Then Then we have x∈ﬁx(f ) sgn det(I − Dxf ) = (−1)k tr(f ∗ : H i(M ; Q) → H k(M ; Q)). k Proof. We have [Γf ] · [∆(M )] ∈ H0(M × M ; Q). We now want to calculate ε of this. By Poincar´e duality, this is equal to (D−1 M ×M [Γf ] D−1 M ×M [∆(M )])[M × M ] ∈ Q. This is the same as (D−1 M ×M [∆(M )])([Γf ]) = δ(F∗[M ]) = (F ∗δ)[M ], where F : M → M × M is given by F (x) = (x, f (x)). δ = (−1)\|ai\|ai ⊗ bi. F ∗δ = (−1)\|ai\|ai ⊗ f ∗bi. f ∗bi = Cijbj. We now use the fact that So we have We write Then we have (F ∗δ)[M ] = i,j (−1)\|ai\|Cij(ai ⊗ bj)[M ] = (−1)\|ai\|Cii, i and Cii is just the trace of f ∗. We now compute this product in a diﬀerent way. As Γf and ∆(M ) are transverse, we know Γf ∩ ∆(M ) is a 0-manifold, and the orientation of Γf and ∆(M ) induces an orientation of it. So we have [Γf ] · [∆(m)] = [Γf ∩ ∆(M )] ∈ H0(M × M ; Q). We know this Γf ∩ ∆(M ) has \| ﬁx(f )\| many points, so [Γf ∩ ∆(M )] is the sum of \| ﬁx(f )\| many things, which is what we’ve got on the left above. We have to ﬁgure out the sign of each term is actually sgn det(I − Dxf ), and this is left as an exercise on the example sheet. Example. Any map f : CP2n → CP2n has a ﬁxed point. We can’t prove this using the normal ﬁxed point theorem, but we can exploit the ring structure of cohomology to do this. We must have f ∗(x) = λx ∈ H 2(CP2; Q) = Qx. 91 11 Manifolds and Poincar´e duality III Algebraic Topology for some λ ∈ Q. So we must have f ∗(xi) = λixi. We can now very easily compute the right hand side of the ﬁxed point theorem (−1)k tr(f ∗ : H k → H k) = 1 + λ + λ2 + · · · + λ2n. k and this cannot be zero. 92 Index Index n (X), 31 n (X), 31 C U H U R-orientation, 61 manifold, 74 Grk(Rn), 57 γR k,n, 58 , 47 n-skeleton, 38 barycenter, 33 barycentric subdivision, 33 boundary, 6 Brouwer’s ﬁxed point theorem, 20 cap product, 77 cell complex, 38 ﬁnite, 38 ﬁnite-dimensional, 38 subcomplex, 38 cellular cohomology, 41 cellular homology, 40 chain complex, 5 chain homotopy, 29 chain map, 6 characteristic class, 63 characteristic map, 38 coboundary, 6 cochain complex, 5 cochain maps, 6 cocycle, 6 cohomology, 5 with coeﬃcients, 45 cohomology class, 5 compactly-supported cochain, 68 compactly-supported cohomology, 68 cross product, 50 cup product, 47 CW complexes, 38 cycle, 6 degree of map, 19, 83 diﬀerentials, 5 direct limit, 69 direct system, 69 directed set, 68 III Algebraic Topology Euler class, 62 exact sequence, 14 exact sequence for relative homology, 16 excision theorem, 17 face of standard simplex, 7 ﬁber, 56 ﬁnite cell complex, 38 ﬁnite-dimensional cell complex, 38 ﬁxed point, 20 ﬁxed point theorem Lefschetz, 91 fundamental class, 76 good pair, 37 graded commutative, 48 Grassmannian manifold, 57 Gysin sequence, 66 Hairy ball theorem, 21 homeomorphism orientation preserving, 74 homology, 5 with coeﬃcients, 45 homology class, 5 homotopy, 3 homotopy equivalence, 3 homotopy invariance theorem, 14 homotopy inverse, 3 intersect transversely, 85 intersection product, 86 K¨unneth’s theorem, 52 Lefschetz ﬁxed point theorem, 91 local R-orientation, 61 manifold, 74 local cohomology, 71 local degree, 24 local homology, 23 local orientation, 61 local trivialization, 56 manifold R-orientation, 74 local R-orientation, 74 signature, 83 Euler characteristic, 46 map of pairs, 17 93 Index III Algebraic Topology Mayer-Vietoris theorem, 15 nine lemma, 72 normal bundle, 58 open cell, 38 orientation, 61 manifold, 74 orientation preserving homeomorphism, 74 partition of unity, 58 proper map, 71 pullback vector bundle, 56 quotient bundle, 57 reduced homology, 37 relative homology, 16 section, 56 zero, 56 short exact sequence, 15 signature of manifold, 83 singular n-simplex, 8 singular cohomology, 9 singular homology, 9 small simplices theorem, 32, 36 snake lemma, 25 sphere bundle, 66 standard n-simplex, 7 subcomplex, 38 support, 68 cochain, 68 tangent bundle, 58 Thom class, 62 transverse intersection, 85 tubular neighbourhood theorem, 58 universal coeﬃcient theorem for (co)homology, 54 vector bundle, 56 pullback, 56 quotient, 57 subbundle, 57 Whitney sum, 57 vector sub-bundle, 57 weak topology, 38 Whitney sum of vector bundles, 57 zero section, 56 94
be invertible, i.e. a unit, as they cannot all lie in the Jacobson radical. We may wlog assume α1 ◦ β1 is a ∼= M 1. unit. If this is the case, then both α1 and β1 have to be invertible. So M1 Consider the map id −θ = φ, where θ : M α−1 1 M1 M 1 M m i=2 Mi M. Then φ(M 1) = M1. So φ\|M 1 looks like α1. Also m φ i=2 Mi = m i=2 Mi, So φ\|m surjective. However, if φ(x) = 0, this says x = θ(x), So i=2 Mi looks like the identity map. So in particular, we see that φ is x ∈ m i=2 Mi. 22 1 Artinian algebras III Algebras But then θ(x) = 0. Thus x = 0. Thus φ is an automorphism of m with φ(M 1) = φ(M1). So this gives an isomorphism between m i=2 Mi = M M1 ∼= M M1 ∼= n i=2 M i , and so we are done by induction. Now it remains to prove that the endomorphism rings are local. Recall the following result from linear algebra. Lemma (Fitting). Suppose M is a module with both the ACC and DCC on submodules, and let f ∈ EndA(M ). Then for large enough n, we have M = im f n ⊕ ker f n. Proof. By ACC and DCC, we may choose n large enough so that f n : f n(M ) → f 2n(M ) is an isomorphism, as if we keep iterating f , the image is a descending chain and the kernel is an ascending chain, and these have to terminate. If m ∈ M , then we can write for some m1. Then f n(m) = f 2n(m1) m = f n(m1) + (m − f n(m1)) ∈ im f n + ker f n, im f n ∩ ker f n = ker(f n : f n(M ) → f 2n(M )) = 0. and also So done. Lemma. Suppose M is an indecomposable module satisfying ACC and DCC on submodules. Then B = EndA(M ) is local. Proof. Choose a maximal left ideal of B, say I. It’s enough to show that if x ∈ I, then x is left invertible. By maximality of I, we know B = Bx + I. We write 1 = λx + y, for some λ ∈ B and y ∈ I. Since y ∈ I, it has no left inverse. So it is not an isomorphism. By Fitting’s lemma and the indecomposability of M , we see that ym = 0 for some m. Thus (1 + y + y2 + · · · + ym−1)λx = (1 + y + · · · + ym−1)(1 − y) = 1. So x is left invertible. Corollary. Let A be a left Artinian algebra. Then A has the unique decomposition property. Proof. We know A satisﬁes the ACC and DCC condition. So AA is a ﬁnite direct sum of indecomposables. 23 1 Artinian algebras III Algebras So if A is an Artinian algebra, we know A can be uniquely decomposed as a direct sum of indecomposable projectives, A = Pj. For convenience, we will work with right Artinian algebras and right modules instead of left ones. It turns out that instead of studying projectives in A, we can study idempotent elements instead. Recall that End(AA) ∼= A. The projection onto Pj is achieved by left multiplication by an idempotent ej, Pj = ejA. The fact that the A decomposes as a direct sum of the Pj translates to the condition ej = 1, eiej = 0 for i = j. Deﬁnition (Orthogonal idempotent). A collection of idempotents {ei} is orthogonal if eiej = 0 for i = j. The indecomposability of Pj is equivalent to ej being primitive: Deﬁnition (Primitive idempotent). An idempotent is primitive if it cannot be expressed as a sum e = e + e, where e, e are orthogonal idempotents, both non-zero. We see that giving a direct sum decomposition of A is equivalent to ﬁnding an orthogonal collection of primitive idempotents that sum to 1. This is rather useful, because idempotents are easier to move around that projectives. Our current plan is as follows — given an Artinian algebra A, we can quotient out by J(A), and obtain a semi-simple algebra A/J(A). By Artin–Wedderburn, we know how we can decompose A/J(A), and we hope to be able to lift this decomposition to one of A. The point of talking about idempotents instead is that we know what it means to lift elements. Proposition. Let N be a nilpotent ideal in A, and let f be an idempotent of A/N ≡ ¯A. Then there is an idempotent e ∈ A with f = ¯e. In particular, we know J(A) is nilpotent, and this proposition applies. The proof involves a bit of magic. Proof. We consider the quotients A/N i for i ≥ 1. We will lift the idempotents successively as we increase i, and since N is nilpotent, repeating this process will eventually land us in A. Suppose we have found an idempotent fi−1 ∈ A/N i−1 with ¯fi−1 = f . We want to ﬁnd fi ∈ A/N i such that ¯fi = f . For i > 1, we let x be an element of A/N i with image fi−1 in A/N i−1. Then since x2 − x vansishes in A/N i−1, we know x2 − x ∈ N i−1/N i. Then in particular, (x2 − x)2 = 0 ∈ A/N i. (†) 24 1 Artinian algebras III Algebras We let fi = 3x2 − 2x3. Then by a direct computation using (†), we ﬁnd f 2 i = fi, and fi has image 3fi−1 − 2fi−1 = fi−1 in A/N i−1 (alternatively, in characteristic p, we can use fi = xp). Since N k = 0 for some k, this process gives us what we want. Just being able to lift idempotents is not good enough. We want to lift decompositions as projective indecomposables. So we need to do better. Corollary. Let N be a nilpotent ideal of A. Let ¯1 = f1 + · · · + fr with {fi} orthogonal primitive idempotents in A/N . Then we can write 1 = e1 + · · · + er, with {ei} orthogonal primitive idempotents in A, and ¯ei = fi. Proof. We deﬁne a sequence e i ∈ A inductively. We set e 1 = 1. Then for each i > 1, we pick e inductive hypothesis we know that fi + · · · + ft ∈ e i a lift of fi + · · · + ft ∈ e i−1Ae i−1Ae i−1/N . Then i−1, since by e ie i+1 = e i+1 = e i+1e i. We let Then Also, if j > i, then and so Similarly ejei = 0. ei = e i − e i+1. ¯ei = fi. ej = e i+1eje i+1, eiej = (e i − e i+1)e i+1eje i+1 = 0. We now apply this lifting of idempotents to N = J(A), which we know is nilpotent. We know A/N is the direct sum of simple modules, and thus the decomposition corresponds to ¯1 = f1 + · · · + ft ∈ A/J(A), and these fi are orthogonal primitive idempotents. Idempotent lifting then gives 1 = e1 + · · · + et ∈ A, and these are orthogonal primitive idempotents. So we can write A = ejA = Pi, 25 1 Artinian algebras III Algebras where Pi = eiA are indecomposable projectives, and Pi/PiJ(A) = Si is simple. By Krull–Schmidt, any indecomposable projective isomorphic to one of these Pj. The ﬁnal piece of the picture is to ﬁgure out when two indecomposable projectives lie in the same block. Recall that if M is a right A-module and e is idempotent, then M e ∼= HomA(eA, M ). In particular, if M = f A for some idempotent f , then Hom(eA, f A) ∼= f Ae. However, if e and f are in diﬀerent blocks, say B1 and B2, then f Ae ∈ B1 ∩ B2 = 0, since B1 and B2 are (two-sided!) ideals. So we know Hom(eA, f A) = 0. So if Hom(eA, f A) = 0, then they are in the same block. The existence of a homomorphism can alternatively be expressed in terms of composition factors. We have seen that each indecomposable projective P has a simple “top” P/P J(A) ∼= S. Deﬁnition (Composition factor). A simple module S is a composition factor of a module M if there are submodules M1 ≤ M2 with M2/M1 ∼= S. Suppose S is a composition factor of a module M . Then we have a diagram P S 0 M2 So by deﬁnition of projectivity, we obtain a non-zero diagonal map P → M2 ≤ M as shown. Lemma. Let P be an indecomposable projective, and M an A-module. Then Hom(P, M ) = 0 iﬀ P/P J(A) is a composition factor of M . Proof. We have proven ⇒. Conversely, suppose there is a non-zero map f : P → M . Then it factors as S = P P J(A) → im f (im f )J(A) . Now we cannot have im f = (im f )J(A), or else we have im f = (im f )J(A)n = 0 for suﬃciently large n since J(A) is nilpotent. So this map must be injective, hence an isomorphism. So this exhibits S as a composition factor of M . We deﬁne a (directed) graph whose vertices are labelled by indecomposable projectives, and there is an edge P1 → P2 if the top S1 of P1 is a composition factor of P2. 26 1 Artinian algebras III Algebras Theorem. Indecomposable projectives P1 and P2 are in the same block if and only if they lie in the same connected component of the graph. Proof. It is clear that P1 and P2 are in the same connected component, then they are in the same block. Conversely, consider a connected component X, and consider I = P. P ∈X We show that this is in fact a left ideal, hence an ideal. Consider any x ∈ A. Then for each P ∈ X, left-multiplication gives a map P → A, and if we decompose A = Pi, then this can be expressed as a sum of maps fi : P → Pi. Now such a map can be non-zero only if P is a composition factor of Pi. So if fi = 0, then Pi ∈ X. So left-multiplication by x maps I to itself, and it follows that I is an ideal. 1.5 K0 We now brieﬂy talk about the notion of K0. Deﬁnition (K0). For any associative k-algebra A, consider the free abelian group with basis labelled by the isomorphism classes [P ] of ﬁnitely-generated projective A-modules. Then introduce relations [P1] + [P2] = [P1 ⊕ P2], This yields an abelian group which is the quotient of the free abelian group by the subgroup generated by [P1] + [P2] − [P1 ⊕ P2]. The abelian group is K0(A). Example. If A is an Artinian algebra, then we know that any ﬁnitely-generated projective is a direct sum of indecomposable projectives, and this decomposition is unique by Krull-Schmidt. So K0(A) = abelian group generated by the isomorphism classes of indecomposable projectives . So K0(A) ∼= Zr, where r is the number of isomorphism classes of indecomposable projectives, which is the number of isomorphism classes of simple modules. Here we’re using the fact that two indecomposable projectives are isomorphic iﬀ their simple tops are isomorphic. It turns out there is a canonical map K0(A) → A/[A, A]. Recall we have met A/[A, A] when we were talking about the number of simple modules. We remarked that it was the 0th Hochschild homology group, and when A = kG, there is a k-basis of A/[A, A] given by gi + [A, A], where gi are conjugacy class representatives. 27 1 Artinian algebras III Algebras To construct this canonical map, we ﬁrst look at the trace map Mn(A) → A/[A, A]. This is a k-linear map, invariant under conjugation. We also note that the canonical inclusion Mn(A) → Mn+1(A) X → X 0 0 0 is compatible with the trace map. We observe that the trace induces an isomorphism Mn(A) [Mn(A), Mn
(A)] → A [A, A] , by linear algebra. Now if P is ﬁnitely generated projective. It is a direct summand of some An. Thus we can write An = P ⊕ Q, for P, Q projective. Moreover, projection onto P corresponds to an idempotent e in Mn(A) = EndA(An), and that and we have P = e(An). EndA(P ) = eMn(A)e. Any other choice of idempotent yields an idempotent e1 conjugate to e in M2n(A). Therefore the trace of an endomorphism of P is well-deﬁned in A/[A, A], independent of the choice of e. Thus we have a trace map EndA(P ) → A/[A, A]. In particular, the trace of the identity map on P is the trace of e. We call this the trace of P . Note that if we have ﬁnitely generated projectives P1 and P2, then we have P1 ⊕ Q1 = An P2 ⊕ Q2 = Am Then we have So we deduce that (P1 ⊕ P2) ⊕ (Q1 ⊕ Q2) = Am+n. tr(P1 ⊕ P2) = tr P1 + tr P2. Deﬁnition (Hattori-Stallings trace map). The map K0(A) → A/[A, A] induced by the trace is the Hattori–Stallings trace map. 28 1 Artinian algebras III Algebras Example. Let A = kG, and G be ﬁnite. Then A/[A, A] is a k-vector space with basis labelled by a set of conjugacy class representatives {gi}. Then we know, for a ﬁnitely generated projective P , we can write tr P = rP (gi)gi, where the rp(gi) may be regarded as class functions. However, P may be regarded as a k-vector space. So the there is a trace map EndK(P ) → k, and also the “character” χp : G → k, where χP (g) = tr g. Hattori proved that if CG(g) is the centralizer of g ∈ G, then χp(g) = \|CG(G)\|rp(g−1). (∗) If char k = 0 and k is is algebraically closed, then we know kG is semi-simple. So every ﬁnitely generated projective is a direct sum of simples, and with r the number of simples, and (∗) implies that the trace map K0(kG) ∼= Zr Zr ∼= K0(kG) → kG [kG, kG] ∼= kr is the natural inclusion. This is the start of the theory of algebraic K-theory, which is a homology theory telling us about the endomorphisms of free A-modules. We can deﬁne K1(A) to be the abelianization of GL(A) = lim n→∞ GLn(A). K2(A) tells us something about the relations required if you express GL(A) in terms of generators and relations. We’re being deliberately vague. These groups are very hard to compute. Just as we saw in the i = 0 case, there are canonical maps Ki(A) → HHi(A), where HH∗ is the Hochschild homology. The i = 1 case is called the Dennis trace map. These are analogous to the Chern maps in topology. 29 2 Noetherian algebras III Algebras 2 Noetherian algebras 2.1 Noetherian algebras In the introduction, we met the deﬁnition of Noetherian algebras. Deﬁnition (Noetherian algebra). An algebra is left Noetherian if it satisﬁes the ascending chain condition (ACC ) on left ideals, i.e. if I1 ≤ I2 ≤ I3 ≤ · · · is an ascending chain of left ideals, then there is some N such that IN +m = IN for all m ≥ 0. Similarly, we say an algebra is Noetherian if it is both left and right Noethe- rian. We’ve also met a few examples. Here we are going to meet lots more. In fact, most of this ﬁrst section is about establishing tools to show that certain algebras are Noetherian. One source of Noetherian algebras is via constructing polynomial and power series rings. Recall that in IB Groups, Rings and Modules, we proved the Hilbert basis theorem: Theorem (Hilbert basis theorem). If A is Noetherian, then A[X] is Noetherian. Note that our proof did not depend on A being commutative. The same proof works for non-commutative rings. In particular, this tells us k[X1, · · · , Xn] is Noetherian. It is also true that power series rings of Noetherian algebras are also Noetherian. The proof is very similar, but for completeness, we will spell it out completely. Theorem. Let A be left Noetherian. Then A[[X]] is Noetherian. Proof. Let I be a left ideal of A[[X]]. We’ll show that if A is left Noetherian, then I is ﬁnitely generated. Let Jr = {a : there exists an element of I of the form aX r + higher degree terms}. We note that Jr is a left ideal of A, and also note that J0 ≤ J1 ≤ J2 ≤ J3 ≤ · · · , as we can always multiply by X. Since A is left Noetherian, this chain terminates at JN for some N . Also, J0, J1, J2, · · · , JN are all ﬁnitely generated left ideals. We suppose ai1, · · · , aisi generates Ji for i = 1, · · · , N . These correspond to elements fij(X) = aijX j + higher odder terms ∈ I. We show that this ﬁnite collection generates I as a left ideal. Take f (X) ∈ I, and suppose it looks like with bn = 0. bnX n + higher terms, 30 2 Noetherian algebras III Algebras Suppose n < N . Then bn ∈ J n, and so we can write So bn = cnjanj. f (X) − cnjfnj(X) ∈ I has zero coeﬃcient for X n, and all other terms are of higher degree. Repeating the process, we may thus wlog n ≥ N . We get f (X) of the form dN X N + higher degree terms. The same process gives f (X) − cN jfN j(X) with terms of degree N + 1 or higher. We can repeat this yet again, using the fact JN = JN +1, so we obtain f (X) − cN jfN j(x) − dN +1,jXfN j(X) + · · · . So we ﬁnd f (X) = ej(X)fN j(X) for some ej(X). So f is in the left ideal generated by our list, and hence so is f . Example. It is straightforward to see that quotients of Noetherian algebras are Noetherian. Thus, algebra images of the algebras A[x] and A[[x]] would also be Noetherian. For example, ﬁnitely-generated commutative k-algebras are always Noetherian. Indeed, if we have a generating set xi of A as a k-algebra, then there is an algebra homomorphism k[X1, · · · , Xn] Xi A xi We also saw previously that Example. Any Artinian algebra is Noetherian. The next two examples we are going to see are less obviously Noetherian, and proving that they are Noetherian takes some work. Deﬁnition (nth Weyl algebra). The nth Weyl algebra An(k) is the algebra generated by X1, · · · , Xn, Y1, · · · , Yn with relations for all i, and everything else commutes. YiXi − XiYi = 1, This algebra acts on the polynomial algebra k[X1, · · · , Xn] with Xi acting by left multiplication and Yi = ∂ . Thus k[X1, · · · , Xn] is a left An(k) module. ∂Xi This is the prototype for thinking about diﬀerential algebras, and D-modules in general (which we will not talk about). The other example we have is the universal enveloping algebra of a Lie algebra. 31 2 Noetherian algebras III Algebras Deﬁnition (Universal enveloping algebra). Let g be a Lie algebra over k, and take a k-vector space basis x1, · · · , xn. We form an associative algebra with generators x1, · · · , xn with relations xixj − xjxi = [xi, xj], and this is the universal enveloping algebra U(g). Example. If g is abelian, i.e. [xi, xj] = 0 in g, then the enveloping algebra is the polynomial algebra in x1, · · · , xn. Example. If g = sl2(k), then we have a basis 1 . They satisfy [e, f ] = h, [h, e] = 2e, [h, f ] = −2f, To prove that An(k) and U(g) are Noetherian, we need some machinery, that involves some “deformation theory”. The main strategy is to make use of a natural ﬁltration of the algebra. Deﬁnition (Filtered algebra). A (Z-)ﬁltered algebra A is a collection of k-vector spaces · · · ≤ A−1 ≤ A0 ≤ A1 ≤ A2 ≤ · · · such that Ai · Aj ⊆ Ai+j for all i, j ∈ Z, and 1 ∈ A0. For example a polynomial ring is naturally ﬁltered by the degree of the polynomial. The deﬁnition above was rather general, and often, we prefer to talk about more well-behaved ﬁltrations. Deﬁnition (Exhaustive ﬁltration). A ﬁltration is exhaustive if Ai = A. Deﬁnition (Separated ﬁltration). A ﬁltration is separated if Ai = {0}. Unless otherwise speciﬁed, our ﬁltrations are exhaustive and separated. For the moment, we will mostly be interested in positive ﬁltrations. Deﬁnition (Positive ﬁltration). A ﬁltration is positive if Ai = 0 for i < 0. Our canonical source of ﬁltrations is the following construction: Example. If A is an algebra generated by x1, · · · , xn, say, we can set – A0 is the k-span of 1 – A1 is the k-span of 1, x1, · · · , xn – A1 is the k-span of 1, x1, · · · , xn, xixj for i, j ∈ {1, · · · , n}. In general, Ar is elements that are of the form of a (non-commutative) polynomial expression of degree ≤ r. 32 2 Noetherian algebras III Algebras Of course, the ﬁltration depends on the choice of the generating set. Often, to understand a ﬁltered algebra, we consider a nicer object, known as the associated graded algebra. Deﬁnition (Associated graded algebra). Given a ﬁltration of A, the associated graded algebra is the vector space direct sum gr A = Ai Ai−1 . This is given the structure of an algebra by deﬁning multiplication by (a + Ai−1)(b + Aj−1) = ab + Ai+j−1 ∈ Ai+j Ai+j−1 . In our example of a ﬁnitely-generated algebra, the graded algebra is generated by x1 + A0, · · · , xn + A0 ∈ A1/A0. The associated graded algebra has the natural structure of a graded algebra: Deﬁnition (Graded algebra). A (Z-)graded algebra is an algebra B that is of the form B = Bi, i∈Z where Bi are k-subspaces, and BiBj ⊆ Bi+j. The Bi’s are called the homogeneous components. A graded ideal is an ideal of the form Ji, where Ji is a subspace of Bi, and similarly for left and right ideals. There is an intermediate object between a ﬁltered algebra and its associated graded algebra, known as the Rees algebra. Deﬁnition (Rees algebra). Let A be a ﬁltered algebra with ﬁltration {Ai}. Then the Rees algebra Rees(A) is the subalgebra AiT i of the Laurent polynomial algebra A[T, T −1] (where T commutes with A). Since 1 ∈ A0 ⊆ A1, we know T ∈ Rees(A). The key observation is that – Rees(A)/(T ) ∼= gr A. – Rees(A)/(1 − T ) ∼= A. Since An(k) and U(g) are ﬁnitely-generated algebras, they come with a natural ﬁltering induced by the generating set. It turns out, in both cases, the associated graded algebras are pretty simple. Example. Let A = An(k), with generating set X1, · · · , Xn and Y1, · · · , Yn. We take the ﬁltration as for a ﬁnitely-generated algebra. Now observe that if ai ∈ Ai, and aj ∈ Aj, then So we see that gr A is commutative, and in fact aiaj − ajai ∈ Ai+j−2. gr An(k) ∼= k[ ¯X1, · · · , ¯Xn, ¯Y1, · · · , ¯Yn], where ¯Xi, ¯Yi are the images of Xi and Yi in A1/A0 respectively. This is not hard to prove, b
ut is rather messy. It requires a careful induction. 33 2 Noetherian algebras III Algebras Example. Let g be a Lie algebra, and consider A = U(g). This has generating set x1, · · · , xn, which is a vector space basis for g. Again using the ﬁltration for ﬁnitely-generated algebras, we get that if ai ∈ Ai and aj ∈ Aj, then aiaj − ajai ∈ Ai+j−1. So again gr A is commutative. In fact, we have gr A ∼= k[¯x1, · · · , ¯xn]. The fact that this is a polynomial algebra amounts to the same as the Poincar´eBirkhoﬀ-Witt theorem. This gives a k-vector space basis for U(g). In both cases, we ﬁnd that gr A are ﬁnitely-generated and commutative, and therefore Noetherian. We want to use this fact to deduce something about A itself. Lemma. Let A be a positively ﬁltered algebra. If gr A is Noetherian, then A is left Noetherian. By duality, we know that A is also right Noetherian. Proof. Given a left ideal I of A, we can form gr I = I ∩ Ai I ∩ Ai−1 , where I is ﬁltered by {I ∩ Ai}. By the isomorphism theorem, we know I ∩ Ai I ∩ Ai−1 ∼= I ∩ Ai + Ai−1 Ai−1 ⊆ Ai Ai−1 . Then gr I is a left graded ideal of gr A. Now suppose we have a strictly ascending chain I1 < I2 < · · · of left ideals. Since we have a positive ﬁltration, for some Ai, we have I1 ∩ Ai I2 ∩ Ai and I1 ∩ Ai−1 = I2 ∩ Ai−1. Thus gr I1 gr I2 gr I3 · · · . This is a contradiction since gr A is Noetherian. So A must be Noetherian. Where we need positivity is the existence of that transition from equality to non-equality. If we have a Z-ﬁltered algebra instead, then we need to impose some completeness assumption, but we will not go into that. Corollary. An(k) and U(g) are left/right Noetherian. Proof. gr An(k) and gr U(g) are commutative and ﬁnitely generated algebras. 34 2 Noetherian algebras III Algebras Note that there is an alternative ﬁltration for An(k) yielding a commutative associated graded algebra, by setting A0 = k[X1, · · · , Xn] and A1 = k[X1, · · · , Xn] + n j=1 k[X1, · · · , Xn]Yj, i.e. linear terms in the Y , and then keep on going. Essentially, we are ﬁltering on the degrees of the Yi only. This also gives a polynomial algebra as an associative graded algebra. The main diﬀerence is that when we take the commutator, we don’t go down by two degrees, but one only. Later, we will see this is advantageous when we want to get a Poisson bracket on the associated graded algebra. We can look at further examples of Noetherian algebras. Example. The quantum plane kq[X, Y ] has generators X and Y , with relation XY = qY X for some q ∈ k×. This thing behaves diﬀerently depending on whether q is a root of unity or not. This quantum plane ﬁrst appeared in mathematical physics. Example. The quantum torus kq[X, X −1, Y, Y −1] has generators X, X −1, Y , Y −1 with relations XX −1 = Y Y −1 = 1, XY = qY X. The word “quantum” in this context is usually thrown around a lot, and doesn’t really mean much apart from non-commutativity, and there is very little connection with actual physics. These algebras are both left and right Noetherian. We cannot prove these by ﬁltering, as we just did. We will need a version of Hilbert’s basis theorem which allows twisting of the coeﬃcients. This is left as an exercise on the second example sheet. In the examples of An(k) and U(g), the associated graded algebras are commutative. However, it turns out we can still capture the non-commutativity of the original algebra by some extra structure on the associated graded algebra. So suppose A is a (positively) ﬁltered algebra whose associated graded algebra gr A is commutative. Recall that the ﬁltration has a corresponding Rees algebra, and we saw that Rees A/(T ) ∼= gr A. Since gr A is commutative, this means [Rees A, Rees A] ⊆ (T ). This induces a map Rees(A) × Rees(A) → (T )/(T 2), sending (r, s) → T 2 + [r, s]. Quotienting out by (T ) on the left, this gives a map gr A × gr A → (T ) (T 2) . We can in fact identify the right hand side with gr A as well. Indeed, the map gr A ∼= Rees(A) (T ) mult. by T (T ) (T 2) , 35 2 Noetherian algebras III Algebras is an isomorphism of gr A ∼= Rees A/(T )-modules. We then have a bracket { · , · } : gr A × gr A gr A . (¯r, ¯s) {r, s} Note that in our original ﬁltration of the Weyl algebra An(k), since the commutator brings us down by two degrees, this bracket vanishes identically, but using the alternative ﬁltration does not give a non-zero { · , · }. This { · , · } is an example of a Poisson bracket. Deﬁnition (Poisson algebra). An associative algebra B is a Poisson algebra if there is a k-bilinear bracket { · , · } : B × B → B such that – B is a Lie algebra under { · , · }, i.e. {r, s} = −{s, r} and {{r, s}, t} + {{s, t}, r} + {{t, r}, s} = 0. – We have the Leibnitz rule {r, st} = s{r, t} + {r, s}t. The second condition says {r, · } : B → B is a derivation. 2.2 More on An(k) and U(g) Our goal now is to study modules of An(k). The ﬁrst result tells us we must focus on inﬁnite dimensional modules. Lemma. Suppose char k = 0. Then An(k) has no non-zero modules that are ﬁnite-dimensional k-vector spaces. Proof. Suppose M is a ﬁnite-dimensional module. Then we’ve got an algebra homomorphism θ : An(k) → Endk(M ) ∼= Mm(k), where m = dimk M . In An(k), we have Y1X1 − X1Y1 = 1. Applying the trace map, we know tr(θ(Y1)θ(X1) − θ(X1)θ(Y1)) = tr I = m. But since the trace is cyclic, the left hand side vanishes. So m = 0. So M is trivial. A similar argument works for the quantum torus, but using determinants instead. We’re going to make use of our associated graded algebras from last time, which are isomorphic to polynomial algebras. Given a ﬁltration {Ai} of A, we may ﬁlter a module with generating set S by setting Mi = AiS. 36 2 Noetherian algebras III Algebras Note that which allows us to form an associated graded module AjMi ⊆ Mi+j, gr M = Mi Mi+1 . This is a graded gr A-module, which is ﬁnitely-generated if M is. So we’ve got a ﬁnitely-generated graded module over a graded commutative algebra. To understand this further, we prove some results about graded modules over commutative algebras, which is going to apply to our gr A and gr M . Deﬁnition (Poincar´e series). Let V be a graded module over a graded algebra S, say ∞ V = Vi. Then the Poincar´e series is i=0 P (V, t) = ∞ (dim Vi)ti. i=0 Theorem (Hilbert-Serre theorem). The Poincar´e series P (V, t) of a ﬁnitelygenerated graded module ∞ Vi V = over a ﬁnitely-generated generated commutative algebra i=0 S = ∞ i=0 Si with homogeneous generating set x1, · · · , xm is a rational function of the form f (t) (1 − tki) , where f (t) ∈ Z[t] and ki is the degree of the generator xi. Proof. We induct on the number m of generators. If m = 0, then S = S0 = k, and V is therefore a ﬁnite-dimensional k-vector space. So P (V, t) is a polynomial. Now suppose m > 0. We assume the theorem is true for < m generators. Consider multiplication by xm. This gives a map xm Vi Vi+km , and we have an exact sequence 0 Ki xm Vi Vi+km Li+km → 0, (∗) where K = Ki = ker(xm : V → V ) 37 2 Noetherian algebras III Algebras and L = Li+km = coker(xm : V → V ). Then K is a graded submodule of V and hence is a ﬁnitely-generated S-module, using the fact that S is Noetherian. Also, L = V /xmV is a quotient of V , and it is thus also ﬁnitely-generated. Now both K and L are annihilated by xm. So they may be regarded as S0[x1, · · · , xm−1]-modules. Applying dimk to (∗), we know dimk(Ki) − dimk(Vi) + dim(Vi+km) − dim(Li+km ) = 0. We multiply by ti+km , and sum over i to get tkmP (K, t) − tkm P (V, t) + P (V, t) − P (L, t) = g(t), where g(t) is a polynomial with integral coeﬃcients arising from consideration of the ﬁrst few terms. We now apply the induction hypothesis to K and L, and we are done. Corollary. If each k1, · · · , km = 1, i.e. S is generated by S0 = k and homogeneous elements x1, · · · , xm of degree 1, then for large enough i, then dim Vi = φ(i) for some polynomial φ(t) ∈ Q[t] of d − 1, where d is the order of the pole of P (V, t) at t = 1. Moreover, i j=0 dim Vj = χ(i), where χ(t) ∈ Q[t] of degree d. Proof. From the theorem, we know that P (V, t) = f (t) (1 − t)d , for some d with f (1) = 0, f ∈ Z[t]. But (1 − t)−1 = 1 + t + t2 + · · · By diﬀerentiating, we get an expression (1 − t)− ti. f (t) = a0 + a1t + · · · + asts, If then we get dim Vi = a0 d + i − 1 d − 1 + a1 d + i − 2 d − 1 + · · · + as , where we set r give φ(i) for a polynomial φ(t) ∈ Q[t], valid for i − s > 0. In fact, we have = 0 if r < d − 1, and this expression can be rearranged to d−1 φ(t) = f (1) (d − 1)! td−1 + lower degree term. 38 2 Noetherian algebras III Algebras Since f (1) = 0, this has degree d − 1. This implies that i j=0 dim Vj is a polynomial in Q[t] of degree d. This φ(t) is the Hilbert polynomial , and χ(t) the Samuel polynomial . Some people call χ(t) the Hilbert polynomial instead, though. We now want to apply this to our cases of gr A, where A = An(k) or U(g), ﬁltered as before. Then we deduce that i 0 dim Mj Mj−1 = χ(i), for a polynomial χ(t) ∈ Q[t]. But we also know i j=0 dim Mj Mj−1 = dim Mi. We are now in a position to make a deﬁnition. Deﬁnition (Gelfand-Kirillov dimension). Let A = An(k) or U(g) and M a ﬁnitely-generated A-module, ﬁltered as before. Then the Gelfand-Kirillov dimension d(M ) of M is the degree of the Samuel polynomial of gr M as a gr A-module. This makes sense because gr A is a commutative algebra in this case. A priori, it seems like this depends on our choice of ﬁltering on M , but actually, it doesn’t. For a more general algebra, we can deﬁne the dimension as below: Deﬁnition (Gelfand-Kirillov dimension). Let A be a ﬁnitely-generated k-algebra, which is ﬁltered as before, and a ﬁnitely-generated A-module M , ﬁltered as before. Then the GK-dimension of M is d(M ) = lim sup n→∞ log(dim Mn) log n . In the case of A = An(k) or U(g), this matches the previous deﬁnition. Again, this does not actually depend on the choice of generating sets. Recall we showed that no non-zero An(k)-module M can have ﬁnite dimension as a k-vector space. So we know d(M ) >
0. Also, we know that d(M ) is an integer for cases A = An or U(g), since it is the degree of a polynomial. However, for general M = A, we can get non-integral values. In fact, the values we can get are 0, 1, 2, and then any real number ≥ 2. We can also have ∞ if the lim sup doesn’t exist. Example. If A = kG, then we have GK-dim(kG) < ∞ iﬀ G has a subgroup H of ﬁnite index with H embedding into the strictly upper triangular integral matrices, i.e. matrices of the form  1 0   ...   0 ∗ 1 ... 0 · · · · · · . . . · · ·  ∗ ∗   ...   1 . This is a theorem of Gromoll, and is quite hard to prove. 39 2 Noetherian algebras III Algebras Example. We have GK-dim(A) = 0 iﬀ A is ﬁnite-dimensional as a k-vector space. We have and in general Indeed, we have GK-dim(k[X]) = 1, GK-dim(k[X1, · · · , Xn]) = n. dimk(mth homogeneous component) = m + n n . So we have χ(t) = t + n n This is of degree n, with leading coeﬃcient 1 n! . We can make the following deﬁnition, which we will not use again: Deﬁnition (Multiplicity). Let A be a commutative algebra, and M an A-module. The multiplicity of M with d(M ) = d is d! × leading coeﬃcient of χ(t). On the second example sheet, we will see that the multiplicity is integral. We continue looking at more examples. Example. We have d(An(k)) = 2n, and d(U(g)) = dimk g. Here we are using the fact that the associated graded algebras are polynomial algebras. Example. We met k[X1, · · · , Xn] as the “canonical” An(k)-module. The ﬁltration of the module matches the one we used when thinking about the polynomial algebra as a module over itself. So we get d(k[X1, · · · , Xn]) = n. Lemma. Let M be a ﬁnitely-generated An-module. Then d(M ) ≤ 2n. Proof. Take generators m1, · · · , ms of M . Then there is a surjective ﬁltered module homomorphism An ⊕ · · · ⊕ An M (a1, · · · , as) aimi It is easy to see that quotients can only reduce dimension, so GK-dim(M ) ≤ d(An ⊕ · · · ⊕ An). χAn⊕···⊕An = sχAn But has degree 2n. More interestingly, we have the following result: 40 2 Noetherian algebras III Algebras Theorem (Bernstein’s inequality). Let M be a non-zero ﬁnitely-generated An(k)-module, and char k = 0. Then d(M ) ≥ n. Deﬁnition (Holonomic module). An An(k) module M is holonomic iﬀ d(M ) = n. If we have a holonomic module, then we can quotient by a maximal submodule, and get a simple holonomic module. For a long time, people thought all simple modules are holonomic, until someone discovered a simple module that is not holonomic. In fact, most simple modules are not holonomic, but we something managed to believe otherwise. Proof. Take a generating set and form the canonical ﬁltrations {Ai} of An(k) and {Mi} of M . We let χ(t) be the Samuel polynomial. Then for large enough i, we have χ(i) = dim Mi. We claim that dim Ai ≤ dim Homk(Mi, M2i) = dim Mi × dim M2i. Assuming this, for large enough i, we have But we know dim Ai ≤ χ(i)χ(2i). dim Ai = i + 2 2n , which is a polynomial of degree 2n. But χ(t)χ(2t) is a polynomial of degree 2d(M ). So we get that n ≤ d(M ). So it remains to prove the claim. It suﬃces to prove that the natural map Ai → Homk(Mi, M2i), given by multiplication is injective. So we want to show that if a ∈ Ai = 0, then aMi = 0. We prove this by induction on i. When i = 0, then A0 = k, and M0 is a ﬁnite-dimensional k-vector space. Then the result is obvious. If i > 0, we suppose the result is true for smaller i. We let a ∈ Ai is non-zero. If aMi = 0, then certainly a ∈ k. We express a = cαβX α1 1 X α2 2 · · · X αn n Y β1 1 · · · Y βn n , where α = (α1, · · · , αn), β = (β1, · · · , βn), and cα,β ∈ k. If possible, pick a j such that cα,α = 0 for some α with αj = 0 (this happens when there is an X involved). Then [Yj, a] = αjcα,βX α1 1 · · · X αj −1 j 41 · · · X αn n Y β1 1 · · · Y βn n , 2 Noetherian algebras III Algebras and this is non-zero, and lives in Ai−1. If aMi = 0, then certainly aMi−1 = 0. Hence [Yj, a]Mi−1 = (Yja − aYj)Mi−1 = 0, using the fact that YjMi−1 ⊆ Mi. This is a contradiction. If a only has Y ’s involved, then we do something similar using [Xj, a]. There is also a geometric way of doing this. We take k = C. We know gr An is a polynomial algebra gr An = k[ ¯X1, · · · , ¯Xn, ¯Y1, · · · , ¯Yn], which may be viewed as the coordinate algebra of the cotangent bundle on aﬃne n-space Cn. The points of this correspond to the maximal ideals of gr An. If I is a left ideal of An(C), then we can form gr I and we can consider the set of maximal ideals containing it. This gives us the characteristic variety Ch(An/I). We saw that there was a Poisson bracket on gr An, and this may be used to deﬁne a skew-symmetric form on the tangent space at any point of the cotangent bundle. In this case, this is non-degenerate skew-symmetric form. We can consider the tangent space U of Ch(An/I) at a non-singular point, and there’s a theorem of Gabber (1981) which says that U ⊇ U ⊥, where ⊥ is with respect to the skew-symmetric form. By non-degeneracy, we must have dim U ≥ n, and we also know that dim Ch(An/I) = d(An/I). So we ﬁnd that d(An/I) ≥ n. In the case of A = U (g), we can think of gr A as the coordinate algebra on g∗, the vector space dual of g. The Poisson bracket leads to a skew-symmetric form on tangent spaces at points of g∗. In this case, we don’t necessarily get non-degeneracy. However, on g, we have the adjoint action of the corresponding Lie group G, and this induces a co-adjoint action on g∗. Thus g∗ is a disjoint union of orbits. If we consider the induced skew-symmetric form on tangent spaces of orbits (at non-singular points), then it is non-degenerate. 2.3 Injective modules and Goldie’s theorem The goal of this section is to prove Goldie’s theorem. Theorem (Goldie’s theorem). Let A be a right Noetherian algebra with no non-zero ideals all of whose elements are nilpotent. Then A embeds in a ﬁnite direct sum of matrix algebras over division algebras. The outline of the proof is as follows — given any A, we embed A in an “injective hull” E(A). We will then ﬁnd that similar to what we did in Artin– Wedderburn, we can decompose End(E(A)) into a direct sum of matrix algebras over division algebras. But we actually cannot. We will have to ﬁrst quotient End(E(A)) by some ideal I. On the other hand, we do not actually have an embedding of A ∼= EndA(A) into End(E(A)). Instead, what we have is only a homomorphism EndA(A) → End(E(A))/I, where we quotient out by the same ideal I. So actually the two of our problems happen to cancel each other out. 42 2 Noetherian algebras III Algebras We will then prove that the kernel of this map contains only nilpotent elements, and then our hypothesis implies this homomorphism is indeed an embedding. We begin by ﬁrst constructing the injective hull. This is going to involve talking about injective modules, which are dual to the notion of projective modules. Deﬁnition (Injective module). An A-module E is injective if for every diagram of A-module maps 0 θ N , ψ M φ E such that θ is injective, there exists a map ψ that makes the diagram commute. Equivalently, Hom( · , E) is an exact functor. Example. Take A = k. Then all k-vector spaces are injective k-modules. Example. Take A = k[X]. Then k(X) is an injective k[X]-module. Lemma. Every direct summand of an injective module is injective, and direct products of injectives is injective. Proof. Same as proof for projective modules. Lemma. Every A-module may be embedded in an injective module. We say the category of A-modules has enough injectives. The dual result for projectives was immediate, as free modules are projective. Proof. Let M be a right A-module. Then Homk(A, M ) is a right A-module via We claim that Homk(A, M ) is an injective module. Suppose we have (f a)(x) = f (ax). 0 θ N1 M1 φ Homk(A, M ) We consider the k-module diagram 0 N1 θ β M1 α M where α(m1) = φ(m1)(1). Since M is injective as a k-module, we can ﬁnd the β such that α = βθ. We deﬁne ψ : N1 → Homk(A, M ) by ψ(n1)(x) = β(n1x). It is straightforward to check that this does the trick. Also, we have an embedding M → Homk(A, M ) by m → (φn : x → mx). 43 2 Noetherian algebras III Algebras The category theorist will write the proof in a line as HomA( · , Homk(A, M )) ∼= Homk( · ⊗A A, M ) ∼= Homk( · , M ), which is exact since M is injective as a k-module. Note that neither the construction of Homk(A, M ), nor the proof that it is injective requires the right A-modules structure of M . All we need is that M is an injective k-module. Lemma. An A-module is injective iﬀ it is a direct summand of every extension of itself. Proof. Suppose E is injective and E is an extension of E. Then we can form the diagram 0 E , ψ E id E and then by injectivity, we can ﬁnd ψ. So E = E ⊕ ker ψ. Conversely, suppose E is a direct summand of every extension. But by the previous lemma, we can embed E in an injective E. This implies that E is a direct summand of E, and hence injective. There is some sort of “smallest” injective a module embeds into, and this is called the injective hull, or injective envelope. This is why our injectives are called E. The “smallness” will be captured by the fact that it is an essential extension. Deﬁnition (Essential submodule). An essential submodule M of an A-module N is one where M ∩ V = {0} for every non-zero submodule V of N . We say N is an essential extension of M . Lemma. An essential extension of an essential extension is essential. Proof. Suppose M < E < F are essential extensions. Then given N ≤ F , we know N ∩ E = {0}, and this is a submodule of E. So (N ∩ E) ∩ M = N ∩ M = 0. So F is an essential extension of M . Lemma. A maximal essential extension is an injective module. Such maximal things exist by Zorn’s lemma. Proof. Let E be a maximal essential extension of M , and consider any embedding E → F . We shall show that E is a direct summand of F . Let S be the set of all non-zero submodules V of F with V ∩ E = {0}. We apply Zorn’s lemma to get a maximal such module, say V1. Then E embeds into F/V1 as an essential submodule.
By transitivity of essential extensions, F/V1 is an essential extension of M , but E is maximal. So E ∼= F/V1. In other words, F = E ⊕ V1. 44 2 Noetherian algebras III Algebras We can now make the following deﬁnition: Deﬁnition (Injective hull). A maximal essential extension of M is the injective hull (or injective envelope) of M , written E(M ). Proposition. Let M be an A-module, with an inclusion M → I into an injective module. Then this extends to an inclusion E(M ) → I. Proof. By injectivity, we can ﬁll in the diagram 0 E(M ) . ψ M I We know ψ restricts to the identity on M . So ker ψ ∩ M = {0}. By Since E(M ) is essential, we must have ker ψ = 0. So E(M ) embeds into I. Proposition. Suppose E is an injective essential extension of M . Then E ∼= E(M ). In particular, any two injective hulls are isomorphic. Proof. By the previous lemma, E(M ) embeds into E. But E(M ) is a maximal essential extension. So this forces E = E(M ). Using what we have got, it is not hard to see that Proposition. E(M1 ⊕ M2) = E(M1) ⊕ E(M2). Proof. We know that E(M1) ⊕ E(M2) is also injective (since ﬁnite direct sums are the same as direct products), and also M1 ⊕ M2 embeds in E(M1) ⊕ E(M2). So it suﬃces to prove this extension is essential. Let V ≤ E(M1) ⊕ E(M2). Then either V /E(M1) = 0 or V /E(M2) = 0. We wlog it is the latter. Note that we can naturally view V E(M2) ≤ E(M1) ⊕ E(M2) E(M2) ∼= E(M1). Since M1 ⊆ E(M1) is essential, we know M1 ∩ (V /E(M2)) = 0. So there is some m1 + m2 ∈ V such that m2 ∈ E(M2) and m1 ∈ M1. Now consider {m ∈ E(M2) : am1 + m ∈ V for some a ∈ A}. This is a non-empty submodule of E(M2), and so contains an element of M2, say n. Then we know am1 + n ∈ V ∩ (M1 ⊕ M2), and we are done. The next two examples of injective hulls will be stated without proof: Example. Take A = k[X], and M = k[X]. Then E(M ) = k(X). 45 2 Noetherian algebras III Algebras Example. Let A = k[X] and V = k be the trivial module, where X acts by 0. Then E(V ) = k[X, X −1] Xk[X] , which is a quotient of A-modules. We note V embeds in this as V ∼= k[X] Xk[X] → k[X, X −1] Xk[X] . Deﬁnition (Uniform module). A non-zero module V is uniform if given non-zero submodules V1, V2, then V1 ∩ V2 = {0}. Lemma. V is uniform iﬀ E(V ) is indecomposable. Proof. Suppose E(V ) = A ⊕ B, with A, B non-zero. Then V ∩ A = {0} and V ∩B = {0} since the extension is essential. So we have two non-zero submodules of V that intersect trivially. Conversely, suppose V is not uniform, and let V1, V2 be non-zero submodules that intersect trivially. By Zorn’s lemma, we suppose these are maximal submodules that intersect trivially. We claim E(V1) ⊕ E(V2) = E(V1 ⊕ V2) = E(V ) To prove this, it suﬃces to show that V is an essential extension of V1 ⊕ V2, so that E(V ) is an injective hull of V1 ⊕ V2. Let W ≤ V be non-zero. If W ∩ (V1 ⊕ V2) = 0, then V1 ⊕ (V2 ⊕ W ) is a larger pair of submodules with trivial intersection, which is not possible. So we are done. Deﬁnition (Domain). An algebra is a domain if xy = 0 implies x = 0 or y = 0. This is just the deﬁnition of an integral domain, but when we have non- commutative algebras, we usually leave out the word “integral”. To show that the algebras we know and love are indeed domains, we again do some deformation. Lemma. Let A be a ﬁltered algebra, which is exhaustive and separated. Then if gr A is a domain, then so is A. Proof. Let x ∈ Ai \ Ai−1, and y ∈ Aj \ Aj−1. We can ﬁnd such i, j for any elements x, y ∈ A because the ﬁltration is exhaustive and separated. Then we have ¯x = x + Ai−1 = 0 ∈ Ai/Ai−1 ¯y = y + Aj−1 = 0 ∈ Aj/Aj−1. If gr A is a domain, then we deduce ¯x¯y = 0. So we deduce that xy ∈ Ai+j−1. In particular, xy = 0. Corollary. An(k) and U(g) are domains. Lemma. Let A be a right Noetherian domain. Then AA is uniform, i.e. E(AA) is indecomposable. 46 2 Noetherian algebras III Algebras Proof. Suppose not, and so there are xA and yA non-zero such that xA ∩ yA = {0}. So xA ⊕ yA is a direct sum. But A is a domain and so yA ∼= A as a right A-module. Thus yxA ⊕ yyA is a direct sum living inside yA. Further decomposing yyA, we ﬁnd that xA ⊕ yxA ⊕ y2xA ⊕ · · · ⊕ ynxA is a direct sum of non-zero submodules. But this is an inﬁnite strictly ascending chain as n → ∞, which is a contradiction. Recall that when we proved Artin–Wedderburn, we needed to use Krull– Schmidt, which told us the decomposition is unique up to re-ordering. That relied on the endomorphism algebra being local. We need something similar here. Lemma. Let E be an indecomposable injective right module. Then EndA(E) is a local algebra, with the unique maximal ideal given by I = {f ∈ End(E) : ker f is essential}. Note that since E is indecomposable injective, given any non-zero V ≤ E, we know E(V ) embeds into, and hence is a direct summand of E. Hence E(V ) = E. So ker f being essential is the same as saying ker f being non-zero. However, this description of the ideal will be useful later on. Proof. Let f : E → E and ker f = {0}. Then f (E) is an injective module, and so is a direct summand of E. But E is indecomposable. So f is surjective. So it is an isomorphism, and hence invertible. So it remains to show that I = {f ∈ End(E) : ker f is essential} is an ideal. If ker f and ker g are essential, then ker(f + g) ≥ ker f ∩ ker g, and the intersection of essential submodules is essential. So ker(f + g) is also essential. Also, if ker g is essential, and f is arbitrary, then ker(f ◦ g) ≥ ker g, and is hence also essential. So I is a maximal left ideal. The point of this lemma is to allow us to use Krull–Schmidt. Lemma. Let M be a non-zero Noetherian module. Then M is an essential extension of a direct sum of uniform submodules N1, · · · , Nr. Thus E(M ) ∼= E(N1) ⊕ · · · E(Nr) is a direct sum of ﬁnitely many indecomposables. This decomposition is unique up to re-ordering (and isomorphism). Proof. We ﬁrst show any non-zero Noetherian module contains a uniform one. Suppose not, and M is in particular not uniform. So it contains non-zero V1, V 2 with V1 ∩ V 2 is not uniform by assumption. So it contains non-zero V2 and V 3 with zero intersection. We keep on repeating. Then we get 2 = 0. But V V1 ⊕ V2 ⊕ · · · ⊕ Vn 47 2 Noetherian algebras III Algebras is a strictly ascending chain of submodules of M , which is a contradiction. Now for non-zero Noetherian M , pick N1 uniform in M . Either N1 is essential 2 = 0. We pick 2 non-zero with N1 ∩ N in M , and we’re done, or there is some N N2 uniform in N 2. Then either N1 ⊕ N2 is essential, or. . . And we are done since M is Noetherian. Taking injective hulls, we get E(M ) = E(N1) ⊕ · · · ⊕ E(Nr), and we are done by Krull–Schmidt and the previous lemma. This is the crucial lemma, which isn’t really hard. This allows us to deﬁne yet another dimension for Noetherian algebras. Deﬁnition (Uniform dimension). The uniform dimension, or Goldie rank of M is the number of indecomposable direct summands of E(M ). This is analogous to vector space dimensions in some ways. Example. The Goldie rank of domains is 1, as we showed AA is uniform. This is true for An(k) and U(g). Lemma. Let E1, · · · , Er be indecomposable injectives. Put E = E1 ⊕ · · · ⊕ Er. Let I = {f ∈ EndA(E) : ker f is essential}. This is an ideal, and then EndA(E)/I ∼= Mn1(D1) ⊕ · · · ⊕ Mns (Ds) for some division algebras Di. Proof. We write the decomposition instead as E = En1 1 ⊕ · · · ⊕ Enr r . Then as in basic linear algebra, we know elements of End(E) can be written as an r × r matrix whose (i, j)th entry is an element of Hom(Eni i Now note that if Ei ∼= Ej, then the kernel of a map Ei → Ej is essential in , Enj j ). Ei. So quotienting out by I kills all of these “oﬀ-diagonal” entries. Also Hom(Eni i , Eni i ) = Mni(End(Ei)), and so quotienting out by I gives Mni(End(Ei)/{essential kernel}) ∼= Mni(Di), where Di ∼= End(Ei) essential kernel , which we know is a division algebra since I is a maximal ideal. The ﬁnal piece to proving Goldie’s theorem is the following piece Lemma. If A is a right Noetherian algebra, then any f : AA → AA with ker f essential in AA is nilpotent. Proof. Consider 0 < ker f ≤ ker f 2 ≤ · · · . Suppose f is not nilpotent. We claim that this is a strictly increasing chain. Indeed, for all n, we have f n(AA) = 0. Since ker f is essential, we know This forces ker f n+1 > ker f n, which is a contradiction. f n(AA) ∩ ker f = {0}. 48 2 Noetherian algebras III Algebras We can now prove Goldie’s theorem. Theorem (Goldie’s theorem). Let A be a right Noetherian algebra with no non-zero ideals all of whose elements are nilpotent. Then A embeds in a ﬁnite direct sum of matrix algebras over division algebras. Proof. As usual, we have a map A x EndA(AA) left multiplication by x For a map AA → AA, it lifts to a map E(AA) → E(AA) by injectivity: 0 AA f AA θ θ E(AA) f E(AA) We can complete the diagram to give a map f : E(AA) → E(AA), which restricts to f on AA. This is not necessarily unique. However, if we have two lifts f and f , then the diﬀerence f − f has AA in the kernel, and hence has an essential kernel. So it lies in I. Thus, if we compose maps AA EndA(AA) End(E(AA))/I . The kernel of this consists of A which when multiplying on the left has essential kernel. This is an ideal all of whose elements is nilpotent. By assumption, any such ideal vanishes. So we have an embedding of A in End(E(AA))/I, which we know to be a direct sum of matrix algebras over division rings. Goldie didn’t present it like this. This work in injective modules is due to Matlis. We saw that (right Noetherian) domains had Goldie rank 1. So we get that End(E(A))/I ∼= D for some division algebra D. So by Goldie’s theorem, a right Noetherian algebra embeds in a division algebra. In particular, this is true for An(k) and U(g). 49 3 Hochschild homology and cohomology III Algebras 3 Hochschild homology and cohomology 3.1 Introduction We now move on to talk about Hochschild (co)homology. We will mostly talk about Hochschild cohomology, as that is the one that is interesting. Roughly speakin
g, given a k-algebra A and an A-A-bimodule M , Hochschild cohomology is an inﬁnite sequence of k-vector spaces HH n(A, M ) indexed by n ∈ N associated to the data. While there is in theory an inﬁnite number of such vector spaces, we are mostly going to focus on the cases of n = 0, 1, 2, and we will see that these groups can be interpreted as things we are already familiar with. The construction of these Hochschild cohomology groups might seem a bit arbitrary. It is possible to justify these a priori using the general theory of homological algebra and/or model categories. On the other hand, Hochschild cohomology is sometimes used as motivation for the general theory of homological algebra and/or model categories. Either way, we are not going to develop these general frameworks, but are going to justify Hochschild cohomology in terms of its practical utility. Unsurprisingly, Hochschild (co)homology was ﬁrst developed by Hochschild in 1945, albeit only working with algebras of ﬁnite (vector space) dimension. It was introduced to give a cohomological interpretation and generalization of some results of Wedderburn. Later in 1962/1963, Gerstenhaber saw how Hochschild cohomology was relevant to the deformations of algebras. More recently, it’s been realized that that the Hochschild cochain complex has additional algebraic structure, which allows yet more information about deformation. As mentioned, we will work with A-A-bimodules over an algebra A. If our algebra has an augmentation, i.e. a ring map to k, then we can have a decent theory that works with left or right modules. However, for the sake of simplicity, we shall just work with bimodules to make our lives easier. Recall that a A-A-bimodule is an algebra with both left and right A actions that are compatible. For example, A is an A-A-bimodule, and we sometimes write it as AAA to emphasize this. More generally, we can view A⊗(n+2) = A⊗k · · ·⊗k A as an A-A-bimodule by x(a0 ⊗ a1 ⊗ · · · ⊗ an+1)y = (xa0) ⊗ a1 ⊗ · · · ⊗ (an+1y). The crucial property of this is that for any n ≥ 0, the bimodule A⊗(n+2) is a free A-A-bimodule. For example, A ⊗k A is free on a single generator 1 ⊗k 1, whereas if {xi} is a k-basis of A, then A ⊗k A ⊗k A is free on {1 ⊗k xi ⊗k 1}. The general theory of homological algebra says we should be interested in such free things. Deﬁnition (Free resolution). Let A be an algebra and M an A-A-bimodule. A free resolution of M is an exact sequence of the form d2 · · · F2 d1 F1 d0 F0 M , where each Fn is a free A-A-bimodule. More generally, we can consider a projective resolution instead, where we allow the bimodules to be projective. In this course, we are only interested in one particular free resolution: 50 3 Hochschild homology and cohomology III Algebras Deﬁnition (Hochschild chain complex). Let A be a k-algebra with multiplication map µ : A ⊗ A. The Hochschild chain complex is d1 · · · A ⊗k A ⊗k A d0 A ⊗k A µ A 0. We refer to A⊗k(n+2) as the degree n term. The diﬀerential d : A⊗k(n+3) → A⊗k(n+2) is given by d(a0 ⊗k · · · ⊗k an+1) = n+1 (−1)ia0 ⊗k · · · ⊗k aiai+1 ⊗k · · · ⊗k an+2. i=0 This is a free resolution of AAA (the exactness is merely a computation, and we shall leave that as an exercise to the reader). In a nutshell, given an A-Abimodule M , its Hochschild homology and cohomology is obtained by applying · ⊗A-A M and HomA-A( · , M ) to the Hochschild chain complex, and then taking the homology and cohomology of the resulting chain complex. We shall explore in more detail what this means. It is a general theorem that we could have applied the functors · ⊗A-A M and HomA-A( · , M ) to any projective resolution of AAA and take the (co)homology, and the resulting vector spaces will be the same. However, we will not prove that, and will just always stick to this standard free resolution all the time. 3.2 Cohomology As mentioned, the construction of Hochschild cohomology involves applying HomA-A( · , M ) to the Hochschild chain complex, and looking at the terms HomA-A(A⊗(n+2), M ). This is usually not very convenient to manipulate, as it involves talking about bimodule homomorphisms. However, we note that A⊗(n+2) is a free A-A-bimodule generated by a basis of A⊗n. Thus, there is a canonical isomorphism HomA-A(A⊗(n+2), M ) ∼= Homk(A⊗n, M ), and k-linear maps are much simpler to work with. Deﬁnition (Hochschild cochain complex). The Hochschild cochain complex of an A-A-bimodule M is what we obtain by applying HomA-A( · , M ) to the Hochschild chain complex of A. Explicitly, we can write it as Homk(k, M ) δ0 Homk(A, M ) δ1 Homk(A ⊗ A, M ) · · · , where (δ0f )(a) = af (1) − f (1)a (δ1f )(a1 ⊗ a2) = a1f (a2) − f (a1a2) + f (a1)a2 (δ2f )(a1 ⊗ a2 ⊗ a3) = a1f (a2 ⊗ a3) − f (a1a2 ⊗ a3) + f (a1 ⊗ a2a3) − f (a1 ⊗ a2)a3 (δn−1f )(a1 ⊗ · · · ⊗ an) = a1f (a2 ⊗ · · · ⊗ an) + n i=1 (−1)if (a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an) + (−1)n+1f (a1 ⊗ · · · ⊗ an−1)an 51 3 Hochschild homology and cohomology III Algebras The reason the end ones look diﬀerent is that we replaced HomA-A(A⊗(n+2), M ) with Homk(A⊗n, M ). The crucial observation is that the exactness of the Hochschild chain complex, and in particular the fact that d2 = 0, implies im δn−1 ⊆ ker δn. Deﬁnition (Cocycles). The cocycles are the elements in ker δn. Deﬁnition (Coboundaries). The coboundaries are the elements in im δn. These names came from algebraic topology. Deﬁnition (Hochschild cohomology groups). We deﬁne HH 0(A, M ) = ker δ0 ker δn im δn−1 HH n(A, N ) = These are k-vector spaces. If we do not want to single out HH 0, we can extend the Hochschild cochain complex to the left with 0, and setting δn = 0 for n < 0 (or equivalently extending the Hochschild chain complex to the right similarly), Then HH 0(A, M ) = ker δ0 im δ−1 = ker δ0. The ﬁrst thing we should ask ourselves is when the cohomology groups vanish. There are two scenarios where we can immediately tell that the (higher) cohomology groups vanish. Lemma. Let M be an injective bimodule. Then HH n(A, M ) = 0 for all n ≥ 1. Proof. HomA-A( · , M ) is exact. Lemma. If AAA is a projective bimodule, then HH n(A, M ) = 0 for all M and all n ≥ 1. If we believed the previous remark that we could compute Hochschild cohomology with any projective resolution, then this result is immediate — indeed, we can use · · · → 0 → 0 → A → A → 0 as the projective resolution. However, since we don’t want to prove such general results, we shall provide an explicit computation. Proof. If AAA is projective, then all A⊗n are projective. At each degree n, we can split up the Hochschild chain complex as the short exact sequence 0 A⊗(n+3) ker dn dn A⊗(n+2) dn−1 im dn−1 0 The im d is a submodule of A⊗(n+1), and is hence projective. So we have A⊗(n+2) ∼= A⊗(n+3) ker dn ⊕ im dn−1, 52 3 Hochschild homology and cohomology III Algebras and we can write the Hochschild chain complex at n as ker dn ⊕ A⊗(n+3) ker dn dn A⊗(n+3) ker dn ⊕ im dn−1 dn−1 A⊗(n+1) im dn−1 ⊕ im dn−1 (a, b) (b, 0) (c, d) (0, d) Now Hom( · , M ) certainly preserves the exactness of this, and so the Hochschild cochain complex is also exact. So we have HH n(A, M ) = 0 for n ≥ 1. This is a rather strong result. By knowing something about A itself, we deduce that the Hochschild cohomology of any bimodule whatsoever must vanish. Of course, it is not true that HH n(A, M ) vanishes in general for n ≥ 1, or else we would have a rather boring theory. In general, we deﬁne Deﬁnition (Dimension). The dimension of an algebra A is Dim A = sup{n : HH n(A, M ) = 0 for some A-A-bimodule M }. This can be inﬁnite if such a sup does not exist. Thus, we showed that AAA embeds as a direct summand in A ⊗ A, then Dim A = 0. Deﬁnition (k-separable). An algebra A is k-separable if AAA embeds as a direct summand of A ⊗ A. Since A ⊗ A is a free A-A-bimodule, this condition is equivalent to A being projective. However, there is some point in writing the deﬁnition like this. Note that an A-A-bimodule is equivalently a left A ⊗ Aop-module. Then AAA is a direct summand of A ⊗ A if and only if there is a separating idempotent e ∈ A ⊗ Aop so that AAA viewed as A ⊗ Aop-bimodule is (A ⊗ Aop)e. This is technically convenient, because it is often easier to write down a separating idempotent than to prove directly A is projective. Note that whether we write A⊗Aop or A⊗A is merely a matter of convention. They have the same underlying set. The notation A ⊗ A is more convenient when we take higher powers, but we can think of A ⊗ Aop as taking A as a left-A right-k module and Aop as a left-k right-A, and tensoring them gives a A-A-bimodule. We just proved that separable algebras have dimension 0. Conversely, we have Lemma. If Dim A = 0, then A is separable. Proof. Note that there is a short exact sequence 0 ker µ A ⊗ A µ A 0 If we can show this splits, then A is a direct summand of A ⊗ A. To do so, we need to ﬁnd a map A ⊗ A → ker µ that restricts to the identity on ker µ. 53 3 Hochschild homology and cohomology III Algebras To do so, we look at the ﬁrst few terms of the Hochschild chain complex · · · d im d ⊕ ker µ A ⊗ A µ A 0 . By assumption, for any M , applying HomA-A( · , M ) to the chain complex gives an exact sequence. Omitting the annoying A-A subscript, this sequence looks like 0 −→ Hom(A, M ) µ∗ −→ Hom(A ⊗ A, M ) (∗) −→ Hom(ker µ, M ) ⊕ Hom(im d, M ) d∗ −→ · · · Now d∗ sends Hom(ker µ, M ) to zero. So Hom(ker µ, M ) must be in the image of (∗). So the map Hom(A ⊗ A, M ) −→ Hom(ker µ, M ) must be surjective. This is true for any M . In particular, we can pick M = ker µ. Then the identity map idker µ lifts to a map A ⊗ A → ker µ whose restriction to ker µ is the identity. So we are done. Example. Mn(k) is separable. It suﬃces to write down the separating idempotent. We let Eij be the elementary matrix with 1 in the (i, j)th slot and 0 otherwise. We ﬁx j, and then Eij ⊗ Eji ∈ A ⊗ Aop i is a separating idempotent. Example. Let A = kG with char k \|G\|. Then A ⊗ Aop = kG ⊗ (kG)op ∼= kG ⊗ kG. But this is just isomorphic to k(G × G), which is again semi
-simple. Thus, as a bimodule, A ⊗ A is completely reducible. So the quotient of AAA is a direct summand (of bimodules). So we know that whenever char k \|G\|, then kG is k-separable. The obvious question is — is this notion actually a generalization of separable ﬁeld extensions? This is indeed the case. Fact. Let L/K be a ﬁnite ﬁeld extension. Then L is separable as a K-algebra iﬀ L/K is a separable ﬁeld extension. However k-separable algebras must be ﬁnite-dimensional k-vector spaces. So this doesn’t pass on to the inﬁnite case. In the remaining of the chapter, we will study what Hochschild cohomology in the low dimensions mean. We start with HH 0. The next two propositions follow from just unwrapping the deﬁnitions: Proposition. We have HH 0(A, M ) = {m ∈ M : am − ma = 0 for all a ∈ A}. In particular, HH 0(A, A) is the center of A. 54 3 Hochschild homology and cohomology III Algebras Proposition. ker δ1 = {f ∈ Homk(A, M ) : f (a1a2) = a1f (a2) + f (a1)a2}. These are the derivations from A to M . We write this as Der(A, M ). On the other hand, im δ0 = {f ∈ Homk(A, M ) : f (a) = am − ma for some m ∈ M }. These are called the inner derivations from A to M . So HH 1(A, M ) = Der(A, M ) InnDer(A, M ) . Setting A = M , we get the derivations and inner derivations of A. Example. If A = k[x], and char k = 0, then DerA = p(X) d dx : p(x) ∈ k[X] , and there are no (non-trivial) inner derivations because A is commutative. So we ﬁnd HH 1(k[X], k[X]) ∼= k[X]. In general, Der(A) form a Lie algebra, since D1D2 − D2D1 ∈ Endk(A) is in fact a derivation if D1 and D2 are. There is another way we can think about derivations, which is via semi-direct products. Deﬁnition (Semi-direct product). Let M be an A-A-bimodule. We can form the semi-direct product of A and M , written A M , which is an algebra with elements (a, m) ∈ A × M , and multiplication given by (a1, m1) · (a2, m2) = (a1a2, a1m2 + m1a2). Addition is given by the obvious one. Alternatively, we can write A M ∼= A + M ε, where ε commutes with everything and ε2 = 0. Then M ε forms an ideal with (M ε)2 = 0. In particular, we can look at A A ∼= A + Aε. This is often written as A[ε]. Previously, we saw ﬁrst cohomology can be understood in terms of derivations. We can formulate derivations in terms of this semi-direct product. Lemma. We have Derk(A, M ) ∼= {algebra complements to M in A M isomorphic to A}. 55 3 Hochschild homology and cohomology III Algebras Proof. A complement to M is an embedded copy of A in A M , A a A M (a, Da) The function A → M given by a → Da is a derivation, since under the embedding, we have ab → (ab, aDb + Dab). Conversely, a derivation f : A → M gives an embedding of A in A M given by a → (a, f (a)). We can further rewrite this characterization in terms of automorphisms of the semi-direct product. This allows us to identify inner derivations as well. Lemma. We have Der(A, M ) ∼= automorphisms of A M of the form a → a + f (a)ε, mε → mε , where we view A M ∼= A + M ε. Moreover, the inner derivations correspond to automorphisms achieved by conjugation by 1 + mε, which is a unit with inverse 1 − mε. The proof is a direct check. This applies in particular when we pick M = AAA, and the Lie algebra of derivations of A may be thought of as the set of inﬁnitesimal automorphisms. Let’s now consider HH 2(A, M ). This is to be understood in terms of exten- sions, of which the “trivial” example is the semi-direct product. Deﬁnition (Extension). Let A be an algebra and M and A-A-bimodule. An extension of A by M . An extension of A by M is a k-algebra B containing a 2-sided ideal I such that – I 2 = 0; – B/I ∼= A; and – M ∼= I as an A-A-bimodule. Note that since I 2 = 0, the left and right multiplication in B induces an A-A-bimodule structure on I, rather than just a B-B-bimodule. We let π : B → A be the canonical quotient map. Then we have a short exact sequence 0 I B A 0 . Then two extensions B1 and B2 are isomorphic if there is a k-algebra isomorphism θ : B1 → B2 such that the following diagram commutes: 0 I B1 θ B2 A . 0 Note that the semi-direct product is such an extension, called the split extension. 56 3 Hochschild homology and cohomology III Algebras Proposition. There is a bijection between HH 2(A, M ) with the isomorphism classes of extensions of A by M . This is something that should be familiar if we know about group cohomology. Proof. Let B be an extension with, as usual, π : B → A, I = M = ker π, I 2 = 0. We now try to produce a cocycle from this. Let ρ be any k-linear map A → B such that π(ρ(a)) = a. This is possible since π is surjective. Equivalently, ρ(π(b)) = b mod I. We deﬁne a k-linear map fρ : A ⊗ A → I ∼= M by Note that the image lies in I since a1 ⊗ a2 → ρ(a1)ρ(a2) − ρ(a1a2). ρ(a1)ρ(a2) ≡ ρ(a1a2) (mod I). It is a routine check that fρ is a 2-cocycle, i.e. it lies in ker δ2. If we replace ρ by any other ρ, we get fρ, and we have fρ(a1 ⊗ a2) − fρ(a1 ⊗ a2) = ρ(a1)(ρ(a2) − ρ(a2)) − (ρ(a1a2) − ρ(a1a2)) + (ρ(a1) − ρ(a1))ρ(a2) = a1 · (ρ(a2) − ρ(a2)) − (ρ(a1a2) − ρ(a1a2)) + (ρ(a1) − ρ(a1)) · a2, where · denotes the A-A-bimodule action in I. Thus, we ﬁnd fρ − fρ = δ1(ρ − ρ), noting that ρ − ρ actually maps to I. So we obtain a map from the isomorphism classes of extensions to the second cohomology group. Conversely, given an A-A-bimodule M and a 2-cocycle f : A ⊗ A → M , we let as k-vector spaces. We deﬁne the multiplication map Bf = A ⊕ M (a1, m1)(a2, m2) = (a1a2, a1m2 + m1a2 + f (a1 ⊗ a2)). This is associative precisely because of the 2-cocycle condition. The map (a, m) → a yields a homomorphism π : B → A, with kernel I being a two-sided ideal of B which has I 2 = 0. Moreover, I ∼= M as an A-A-bimodule. Taking ρ : A → B by ρ(a) = (a, 0) yields the 2-cocycle we started with. Finally, let f be another 2 co-cycle cohomologous to f . Then there is a linear map τ : A → M with f − f = δ1τ. That is, f (a1 ⊗ A2) = f (a1 ⊗ a2) + a1 · τ (a2) − τ (a1a2) + τ (a1) · a2. Then consider the map Bf → B f given by (a, m) → (a, m + τ (a)). One then checks this is an isomorphism of extensions. And then we are done. 57 3 Hochschild homology and cohomology III Algebras In the proof, we see 0 corresponds to the semi-direct product. Corollary. If HH 2(A, M ) = 0, then all extensions are split. We now prove some theorems that appear to be trivialities. However, they are trivial only because we now have the machinery of Hochschild cohomology. When they were ﬁrst proved, such machinery did not exist, and they were written in a form that seemed less trivial. Theorem (Wedderburn, Malcev). Let B be a k-algebra satisfying – Dim(B/J(B)) ≤ 1. – J(B)2 = 0 Then there is an subalgebra A ∼= B/J(B) of B such that B = A J(B). Furthermore, if Dim(B/J(B)) = 0, then any two such subalgebras A, A are conjugate, i.e. there is some x ∈ J(B) such that Notice that 1 + x is a unit in B. A = (1 + x)A(1 + x)−1. In fact, the same result holds if we only require J(B) to be nilpotent. This follows from an induction argument using this as a base case, which is messy and not really interesting. Proof. We have J(B)2 = 0. Since we know Dim(B/J(B)) ≤ 1, we must have where HH 2(A, J(B)) = 0 A ∼= B J(B) . Note that we regard J(B) as an A-A-bimodule here. So we know that all extension of A by J(B) are semi-direct, as required. Furthermore, if Dim(B/J(B)) = 0, then we know HH 1(A, J(A)) = 0. So by our older lemmas, we see that complements are all conjugate, as required. Corollary. If k is algebraically closed and dimk B < ∞, then there is a subalgebra A of B such that A ∼= B J(B) , and B = A J(B). Moreover, A is unique up to conjugation by units of the form 1+x with x ∈ J(B). Proof. We need to show that Dim(A) = 0. But we know B/J(B) is a semisimple k-algebra of ﬁnite dimension, and in particular is Artinian. So by Artin–Wedderburn, we know B/J(B) is a direct sum of matrix algebras over k (since k is algebraically closed and dimk(B/J(B))). We have previously observed that Mn(k) is k-separable. Since k-separability behaves well under direct sums, we know B/J(B) is k-separable, hence has dimension zero. It is a general fact that J(B) is nilpotent. 58 3 Hochschild homology and cohomology III Algebras 3.3 Star products We are now going to do study some deformation theory. Suppose A is a k-algebra. We write V for the underlying vector space of A. Then there is a natural algebra structure on V ⊗k k[[t]] = V [[t]], which we may write as A[[t]]. However, we might we to consider more interesting algebra structures on this vector space. Of course, we don’t want to completely forget the algebra structure on A. So we make the following deﬁnition: Deﬁnition (Star product). Let A be a k-algebra, and let V be the underlying vector space. A star product is an associative k[[t]]-bilinear product on V [[t]] that reduces to the multiplication on A when we set t = 0. Can we produce non-trivial star products? It seems diﬃcult, because when we write down an attempt, we need to make sure it is in fact associative, and that might take quite a bit of work. One example we have already seen is the following: Example. Given a ﬁltered k-algebra A, we formed the Rees algebra associated with the ﬁltration, and it embeds as a vector space in (gr A)[[t]]. Thus we get a product on (gr A)[[t]]. There are two cases where we are most interested in — when A = An(k) or A = U(g). We saw that gr A was actually a (commutative) polynomial algebra. However, the product on the Rees algebra is non-commutative. So the ∗-product will be non-commutative. In general, the availability of star products is largely controlled by the Hochschild cohomology of A. To understand this, let’s see what we actually need to specify to get a star product. Since we required the product to be a k[[t]]-bilinear map f : V [[t]] × V [[t]] → V [[t]], all we need to do is to specify what elements of V = A are sent to. Let a, b ∈ V = A. We write f (a, b) = ab + tF1(a, b) + t2F2(a, b) + · · · . Because of bilinearity, we know Fi are k-bilinear maps, and so correspond to k-linear
maps V ⊗ V → V . For convenience, we will write F0(a, b) = ab. The only non-trivial requirement f has to satisfy is associativity: f (f (a, b), c) = f (a, f (b, c)). What condition does this force on our Fi? By looking at coeﬃcients of t, this implies that for all λ = 0, 1, 2, · · · , we have Fm(Fn(a, b), c) − Fm(a, Fn(b, c)) = 0. (∗) m+n=λ m,n≥0 For λ = 0, we are just getting the associativity of the original multiplication on A. When λ = 1, then this says aF1(b, c) − F1(ab, c) + F1(a, bc) − F1(a, b)c = 0. 59 3 Hochschild homology and cohomology III Algebras All this says is that F1 is a 2-cocycle! This is not surprising. Indeed, we’ve just seen (a while ago) that working mod t2, the extensions of A by AAA are governed by HH 2. Thus, we will refer to 2-cocycles as inﬁnitesimal deformations in this context. Note that given an arbitrary 2 co-cycle A ⊗ A → A, it may not be possible to produce a star product with the given 2-cocycle as F1. Deﬁnition (Integrable 2-cocycle). Let f : A ⊗ A → A be a 2-cocycle. Then it is integrable if it is the F1 of a star product. We would like to know when a 2-cocycle is integrable. Let’s rewrite (∗) as (†λ): m+n=λ m,n>0 Fm(Fn(a, b), c) − Fm(a, Fn(b, c)) = (δ2Fλ)(a, b, c). (†λ) Here we are identifying Fλ with the corresponding k-linear map A ⊗ A → A. For λ = 2, this says F1(F1(ab), c) − F1(a, F1(b, c)) = (δ2F2)(a, b, c). If F1 is a 2-cocycle, then one can check that the LHS gives a 3-cocycle. If F1 is integrable, then the LHS has to be equal to the RHS, and so must be a coboundary, and thus has cohomology class zero in HH 3(A, A). In fact, if F1, · · · , Fλ−1 satisfy (†1), · · · , (†λ−1), then the LHS of (†λ) is also a 3-cocycle. If it is a coboundary, and we have deﬁned F1, · · · , Fλ1, then we can deﬁne Fλ such that (†λ) holds. However, if it is not a coboundary, then we get stuck, and we see that our choice of F1, · · · , Fλ−1 does not lead to a ∗-product. The 3-cocycle appearing on the LHS of (†λ) is an obstruction to integrability. If, however, they are always coboundaries, then we can inductively deﬁne F1, F2, · · · to give a ∗-product. Thus we have proved Theorem (Gerstenhaber). If HH 3(A, A) = 0, then all inﬁnitesimal deformations are integrable. Of course, even if HH 3(A, A) = 0, we can still get ∗-products, but we need to pick our F1 more carefully. Now after producing star products, we want to know if they are equivalent. Deﬁnition (Equivalence of star proeducts). Two star products f and g are equivalent on V ⊗ k[[t]] if there is a k[[t]]-linear automorphism Φ of V [[t]] of the form Φ(a) = a + tφ1(a) + t2φ2(a) + · · · sch that Equivalently, the following diagram has to commute: f (a, b) = Φ−1g(Φ(a), Φ(b)). V [[t]] ⊗ V [[t]] Φ⊗Φ V [[t]] ⊗ V [[t]] f g V [[t]] Φ V [[t]] Star products equivalent to the usual product on A ⊗ k[[t]] are called trivial . 60 3 Hochschild homology and cohomology III Algebras Theorem (Gerstenhaber). Any non-trivial star product f is equivalent to one of the form g(a, b) = ab + tnGn(a, b) + tn+1Gn+1(a, b) + · · · , where Gn is a 2-cocycle and not a coboundary. In particular, if HH 2(A, A) = 0, then any star product is trivial. Proof. Suppose as usual f (a, b) = ab + tF1(a, b) + t2F2(a, b) + · · · , and suppose F1, · · · , Fn−1 = 0. Then it follows from (†) that If Fn is a coboundary, then we can write δ2Fn = 0. Fn = −δφn for some φn : A → A. We set Then we can compute that Φn(a) = a + tnφn(a). Φ−1 n (f (Φn(a), Φn(b))) is of the form ab + tn+1Gn+1(a, b) + · · · . So we have managed to get rid of a further term, and we can keep going until we get the ﬁrst non-zero term not a coboundary. Suppose this never stops. Then f is trivial — we are using that · · · ◦ Φn+2 ◦ Φn+1 ◦ Φn converges in the automorphism ring, since we are adding terms of higher and higher degree. We saw that derivations can be thought of as inﬁnitesimal automorphisms. One can similarly consider k[[t]]-linear maps of the form Φ(a) = a + tφ1(a) + t2φ2(a) + · · · and consider whether they deﬁne automorphisms of A ⊗ k[[t]]. Working modulo t2, we have already done this problem — we are just considering automorphisms of A[ε], and we saw that these automorphisms correspond to derivations. Deﬁnition (Integrable derivation). We say a derivation is integrable if there is an automorphism of A ⊗ k[[t]] that gives the derivation when we mod t2. In this case, the obstructions are 2-cocycles which are not coboundaries. Theorem (Gerstenhaber). Suppose HH 2(A, A) = 0. Then all derivations are integrable. The proof is an exercise on the third example sheet. We haven’t had many examples so far, because Hochschild cohomology is diﬃcult to compute. But we can indeed do some examples. 61 3 Hochschild homology and cohomology III Algebras Example. Let A = k[x]. Since A is commutative, we have HH 0(A, A) = A. Since A is commutative, A has no inner derivations. So we have HH 1(A, A) = DerA = f (x) d dx : f (x) ∈ k[x] . For any i > 1, we have So we have HH i(A, A) = 0. Dim(A) = 1. We can do this by explicit calculation. complex, we had a short exact sequence If we look at our Hochschild chain 0 ker µ A ⊗ A A 0 (∗) and thus we have a map A ⊗ A ⊗ A d A ⊗ A whose image is ker µ . The point is that ker µ is a projective A-A-bimodule. This will mean that HH i(A, M ) = 0 for i ≥ 2 in the same way we used to show that when AAA is a projective A-A-bimodule for i ≥ 1. In particular, HH i(A, A) = 0 for i ≥ 2. To show that ker µ is projective, we notice that A ⊗ A = k[X] ⊗k k[X] ∼= k[X, Y ]. So the short exact sequence (∗) becomes 0 (X − Y )k[X, Y ] k[X, Y ] k[X] 0 . So (X − Y )k[X, Y ] is a free k[X, Y ] module, and hence projective. We can therefore use our theorems to see that any extension of k[X] by k[X] is split, and any ∗-product is trivial. We also get that any derivation is integrable. Example. If we take A = k[X1, X2], then again this is commutative, and HH 0(A, A) = A HH 1(A, A) = DerA. We will talk about HH 2 later, and similarly HH i(A, A) = 0 for i ≥ 3. From this, we see that we may have star products other than the trivial ones, and in fact we know we have, because we have one arising from the Rees algebra of A1(k). But we know that any inﬁnitesimal deformation yields a star product. So there are much more. 62 3 Hochschild homology and cohomology III Algebras 3.4 Gerstenhaber algebra We now want to understand the equations (†) better. To do so, we consider the graded vector space HH·(A, A) = ∞ HH n(A, A), A=0 as a whole. It turns out this has the structure of a Gerstenhaber algebra The ﬁrst structure to introduce is the cup product. They are a standard tool in cohomology theories. We will write Sn(A, A) = Homk(A⊗n, A) = HomA-A(A⊗(n+2), AAA). The Hochschild chain complex is then the graded chain complex S·(A, A). Deﬁnition (Cup product). The cup product : Sm(A, A) ⊗ Sn(A, A) → Sm+n(A, A) is deﬁned by (f g)(a1 ⊗ · · · ⊗ am ⊗ b1 ⊗ · · · ⊗ bn) = f (a1 ⊗ · · · ⊗ am) · g(b1 ⊗ · · · ⊗ bn), where ai, bj ∈ A. Under this product, S·(A, A) becomes an associative graded algebra. Observe that δ(f g) = δf g + (−1)mnf δg. So we say δ is a (left) graded derivation of the graded algebra S·(A, A). In homological (graded) algebra, we often use the same terminology but with suitable sign changes which depends on the degree. Note that the cocycles are closed under . So cup product induces a product on HH·(A, A). If f ∈ Sm(A, A) and g ∈ Sn(A, A), and both are cocycles, then (−1)m(g f − (−1)mn(f g)) = δ(f ◦ g), where f ◦ g is deﬁned as follows: we set f ◦i g(a1 ⊗ · · · ⊗ ai−1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ ai+1 · · · ⊗ am) = f (a1 ⊗ · · · ⊗ ai−1 ⊗ g(b1 ⊗ · · · ⊗ bn) ⊗ ai+1 ⊗ · · · ⊗ am). Then we deﬁne f ◦ g = m i=1 (−1)(n−1)(i−1)f ◦i g. This product ◦ is not an associative product, but is giving a pre-Lie structure to S·(A, A). Deﬁnition (Gerstenhaber bracket). The Gerstenhaber bracket is [f, g] = f ◦ g − (−1)(n+1)(m+1)g ◦ f 63 3 Hochschild homology and cohomology III Algebras This deﬁnes a graded Lie algebra structure on the Hochschild chain complex, It is a grade Lie algebra on but notice that we have a degree shift by 1. S·+1(A, A). Of course, we really should deﬁne what a graded Lie algebra is. Deﬁnition (Graded Lie algebra). A graded Lie algebra is a vector space L = Li with a bilinear bracket [ · , · ] : L × L → L such that – [Li, Lj] ⊆ Li+j; – [f, g] − (−1)mn[g, f ]; and – The graded Jacobi identity holds: (−1)mp[[f, g], h] + (−1)mn[[g, h], f ] + (−1)np[[h, f ], g] = 0 where f ∈ Lm, g ∈ Ln, h ∈ Lp. In fact, S·+1(A, A) is a diﬀerential graded Lie algebra under the Gerstenhaber bracket. Lemma. The cup product on HH·(A, A) is graded commutative, i.e. f g = (−1)mn(g f ). when f ∈ HH m(A, A) and g ∈ HH n(A, A). Proof. We previously “noticed” that (−1)m(g f − (−1)mn(f g)) = δ(f ◦ g), Deﬁnition (Gerstenhaber algebra). A Gerstenhaber algebra is a graded vector space H = H i with H·+1 a graded Lie algebra with respect to a bracket [ · , · ] : H m × H n → H m+n−1, and an associative product : H m × H n → H m+n which is graded commutative, such that if f ∈ H m, then [f, · ] acts as a degree m − 1 graded derivation of : [f, g h] = [f, g] h + (−1)(m−1)ng [f, h] if g ∈ H n. This is analogous to the deﬁnition of a Poisson algebra. We’ve seen that HH·(A, A) is an example of a Gerstenhaber algebra. We can look at what happens in low degrees. We know that H 0 is a commutative k-algebra, and : H 0 × H 1 → H 1 is a module action. Also, H 1 is a Lie algebra, and [ · , · ] : H 1 × H 0 → H 0 is a Lie module action, i.e. H 0 gives us a Lie algebra representation of H 1. In other words, 64 3 Hochschild homology and cohomology III Algebras the corresponding map [ · , · ] : H 1 → Endk(H 0) gives us a map of Lie algebras H 1 → Der(H 0). The prototype Gerstenhaber algebra is the exterior algebra DerA for a commutative algebra A (with A in degree 0). Explicitly, to deﬁne the exterior product over A, we ﬁrst consider the tensor product over A of two A-modules V and W , deﬁned by V ⊗A W = V ⊗k W av ⊗ w − v ⊗ aw The exterior product is then V ∧A V = V ⊗A V v ⊗ v
: v ∈ V . The product is given by the wedge, and the Schouten bracket is given by [λ1 ∧ · · · ∧ λm] = (−1)(m−1)(n−1) (−1)i+j[λi, λj] λ1 ∧ · · · λm ith missing ∧ λ 1 ∧ · · · ∧ λ n jth missing . i,j For any Gerstenhaber algebra H = H i, there is a canonical homomorphism of Gerstenhaber algebras H 1 → H. Theorem (Hochschild–Kostant–Ronsenberg (HKR) theorem). If A is a “smooth” commutative k-algebra, and char k = 0, then the canonical map H 0 (DerA) → HH ∗(A, A) A is an isomorphism of Gerstenhaber algebras. We will not say what “smooth” means, but this applies if A = k[X1, · · · , Xn], or if k = C or R and A is appropriate functions on smooth manifolds or algebraic varieties. In the 1960’s, this was stated just for the algebra structure, and didn’t think about the Lie algebra. Example. Let A = k[X, Y ], with char k = 0. Then HH 0(A, A) = A and HH 1(A, A) = DerA ∼= p(X, Y ) ∂ ∂y + q(X, Y ) : p, q ∈ A . ∂ ∂Y So we have which is generated as an A-modules by ∂ HH 2(A, A) = DerA ∧A DerA, ∂Y . Then ∂X ∧ ∂ HH i(A, A) = 0 for all i ≥ 3 We can now go back to talk about star products. Recall when we considered possible star products on V ⊗k k[[t]], where V is the underlying vector space of the algebra A. We found that associativity of the star product was encapsulated by some equations (†λ). Collectively, these are equivalent to the statement 65 3 Hochschild homology and cohomology III Algebras Deﬁnition (Maurer–Cartan equation). The Maurer–Cartan equation is δf + 1 2 [f, f ]Gerst = 0 f = tλFλ, for the element where F0(a, b) = ab. When we write [ · , · ]Gerst, we really mean the k[[t]]-linear extension of the Gerstenhaber bracket. If we want to think of things in cohomology instead, then we are looking at things modulo coboundaries. For the graded Lie algebra ·+1(DerA), the Maurer–Cartan elements, i.e. solutions of the Maurer–Cartan equation, are the formal Poisson structures. They are formal power series of the form for πi ∈ DerA ∧ DerA, satisfying Π = tiπi [Π, Π] = 0. There is a deep theorem of Kontsevich from the early 2000’s which implies Theorem (Kontsevich). There is a bijection equivalence classes of star products ←→ classes of formal Poisson structures This applies for smooth algebras in char 0, and in particular for polynomial algebras A = k[X1, · · · , Xn]. This is a diﬃcult theorem, and the ﬁrst proof appeared in 2002. An unnamed lecturer once tried to give a Part III course with this theorem as the punchline, but the course ended up lasting 2 terms, and they never reached the punchline. 3.5 Hochschild homology We don’t really have much to say about Hochschild homology, but we are morally obliged to at least write down the deﬁnition. To do Hochschild homology, we apply · ⊗A-A M for an A-A-bimodule M to the Hochschild chain complex. · · · d A ⊗k A ⊗k A d A ⊗k A µ A 0. , We will ignore the → A → 0 bit. We need to consider what · ⊗A-A · means. If we have bimodules V and W , we can regard V as a right A ⊗ Aop-module. We can also think of W as a left A ⊗ Aop module. We let B = A ⊗ Aop, 66 3 Hochschild homology and cohomology III Algebras and then we just consider V ⊗B W = V ⊗k W vx ⊗ w − v ⊗ xw : w ∈ B = V ⊗k W ava ⊗ w − v ⊗ awa . Thus we have b1 · · · (A ⊗k A ⊗k A) ⊗A-A M b0 (A ⊗k A) ⊗A-A M ∼= M , Deﬁnition (Hochschild homology). The Hochschild homology groups are HH0(A, M ) = HHi(A, M ) = M im b0 ker bi−1 im bi for i > 0. A long time ago, we counted the number of simple kG-modules for k alge[A,A] , braically closed of characteristic p when G is ﬁnite. In the proof, we used A and we pointed out this is HH0(A, A). Lemma. In particular, Proof. Exercise. HH0(A, M ) = M xm − mx : m ∈ M, x ∈ A . HH0(A, A) = A [A, A] . 67 4 Coalgebras, bialgebras and Hopf algebras III Algebras 4 Coalgebras, bialgebras and Hopf algebras We are almost at the end of the course. So let’s deﬁne an algebra. Deﬁnition (Algebra). A k-algebra is a k-vector space A and k-linear maps → xy u : k → A λ → λI called the multiplication/product and unit such that the following two diagrams commute: A ⊗ A ⊗ A µ⊗id A ⊗ A k ⊗ A u⊗id A ⊗ A id ⊗µ A ⊗ A µ µ A ∼= µ A A ⊗ k id ⊗u ∼= These encode associativity and identity respectively. Of course, the point wasn’t to actually deﬁne an algebra. The point is to deﬁne a coalgebra, whose deﬁnition is entirely dual. Deﬁnition (Coalgebra). A coalgebra is a k-vector space C and k-linear maps ∆ : called comultiplication/coproduct and counit respectively, such that the following diagrams commute: C ⊗ C ⊗ C id ⊗∆ C ⊗ C k ⊗ C ε⊗id C ⊗ C id ⊗ε C ⊗ k ∆⊗id C ⊗ C ∆ C ∆ ∼= µ C ∼= These encode coassociativity and coidentity A morphism of coalgebras f : C → D is a k-linear map such that the following diagrams commute subspace I of C is a co-ideal if ∆(I) ≤ C ⊗ I + I ⊗ C, and ε(I) = 0. In this case, C/I inherits a coproduct and counit. A cocommutative coalgebra is one for which τ ◦ ∆ = ∆, where τ : V ⊗ W → W ⊗ V given by the v ⊗ w → w ⊗ v is the “twist map”. It might be slightly diﬃcult to get one’s head around what a coalgebra actually is. It, of course, helps to look at some examples, and we will shortly do so. It also helps to know that for our purposes, we don’t really care about coalgebras per se, but things that are both algebras and coalgebras, in a compatible way. 68 4 Coalgebras, bialgebras and Hopf algebras III Algebras There is a very natural reason to be interested in such things. Recall that when doing representation theory of groups, we can take the tensor product of two representations and get a new representation. Similarly, we can take the dual of a representation and get a new representation. If we try to do this for representations (ie. modules) of general algebras, we see that this is not possible. What is missing is that in fact, the algebras kG and U(g) also have the structure of coalgebras. In fact, they are Hopf algebras, which we will deﬁne soon. We shall now write down some coalgebra structures on kG and U(g). Example. If G is a group, then kG is a co-algebra, with ∆(g) = g ⊗ g ε λg (g) = λg. We should think of the speciﬁcation ∆(g) = g ⊗ g as saying that our groups act diagonally on the tensor products of representations. More precisely, if V, W are representations and v ∈ V, w ∈ W , then g acts on v ⊗ w by ∆(g) · (v ⊗ w) = (g ⊗ g) · (v ⊗ w) = (gv) ⊗ (gw). Example. For a Lie algebra g over k, the universal enveloping algebra U(g) is a co-algebra with ∆(x) = x ⊗ 1 + 1 ⊗ x for x ∈ g, and we extend this by making it an algebra homomorphism. To deﬁne ε, we note that elements of U(g) are uniquely of the form λ + λi1,...,in xi1 1 · · · xin n , where {xi} is a basis of g (the PBW theorem). Then we deﬁne λ + ε λi1,...,in xi1 1 · · · xin n = λ. This time, the speciﬁcation of ∆ is telling us that if X ∈ g and v, w are elements of a representation of g, then X acts on the tensor product by ∆(X) · (v ⊗ w) = Xv ⊗ w + v ⊗ Xw. Example. Consider O(Mn(k)) = k[Xij : 1 ≤ i, j ≤ n], the polynomial functions on n × n matrices, where Xij denotes the ijth entry. Then we deﬁne n ∆(Xij) = Xi ⊗ Xj, and These are again algebra maps. i=1 ε(Xij) = δij. 69 4 Coalgebras, bialgebras and Hopf algebras III Algebras We can also talk about O(GLn(k)) and O(SLn(k)). The formula of the determinant gives an element D ∈ O(Mn(k)). Then O(GLn(k)) is given by adding a formal inverse to D in O(GLn(k)), and O(SLn(k)) is obtained by quotienting out O(GLn(k)) by the bi-ideal D − 1. From an algebraic geometry point of view, these are the coordinate algebra of the varieties Mn(k), GLn(k) and SLn(k). This is dual to matrix multiplication. We have seen that we like things that are both algebras and coalgebras, compatibly. These are known as bialgebras. Deﬁnition (Bialgebra). A bialgebra is a k-vector space B and maps µ, υ, ∆, ε such that (i) (B, µ, u) is an algebra. (ii) (B, ∆, ε) is a coalgebra. (iii) ∆ and ε are algebra morphisms. (iv) µ and u are coalgebra morphisms. Being a bialgebra means we can take tensor products of modules and still get modules. If we want to take duals as well, then it turns out the right notion is that of a Hopf algebra: Deﬁnition (Hopf algebra). A bialgebra (H, µ, u, ∆, ε) is a Hopf algebra if there is an antipode S : H → H that is a k-linear map such that µ ◦ (S ⊗ id) ◦ ∆ = µ ◦ (id ⊗S) ◦ ∆ = u ◦ ε. Example. kG is a Hopf algebra with S(g) = g−1. Example. U(g) is a Hopf algebra with S(x) = −x for x ∈ U(g). Note that our examples are all commutative or co-commutative. The term quantum groups usually refers to a non-commutative non-co-commutative Hopf algebras. These are neither quantum nor groups. As usual, we write V ∗ for Homk(V, k), and we note that if we have α : V → W , then this induces a dual map α∗ : W ∗ → V ∗. Lemma. If C is a coalgebra, then C ∗ is an algebra with multiplication ∆∗ (that is, ∆∗\|C∗⊗C∗ ) and unit ε∗. If C is co-commutative, then C ∗ is commutative. However, if an algebra A is inﬁnite dimensional as a k-vector space, then A∗ may not be a coalgebra. The problem is that (A∗ ⊗ A∗) is a proper subspace of (A ⊗ A)∗, and µ∗ of an inﬁnite dimensional A need not take values in A∗ ⊗ A∗. However, all is ﬁne for ﬁnite dimensional A, or if A is graded with ﬁnite dimensional components, where we can form a graded dual. In general, for a Hopf algebra H, one can deﬁne the Hopf dual , H 0 = {f ∈ H ∗ : ker f contains an ideal of ﬁnite codimension}. 70 4 Coalgebras, bialgebras and Hopf algebras III Algebras Example. Let G be a ﬁnite group. Then (kG)∗ is a commutative non-cocommutative Hopf algebra if G is non-abelian. Let {g} be the canonical basis for kG, and {φg} be the dual basis of (kG)∗. Then ∆(φg) = h1h2=g φh1 ⊗ φh2 . There is an easy way of producing non-commutative non-co-commutative Hopf algebras — we take a non-commutative Hopf algebra and a non-co-commutative Hopf algebra, and take the tensor product of them, but this is silly. The easiest non-trivial example of a non-commutative non-co-commutative Hopf algebra is the Drinfeld double, or quantum double, which is a general construction from a ﬁnite dimensional hopf
algebra. Deﬁnition (Drinfeld double). Let G be a ﬁnite group. We deﬁne D(G) = (kG)∗ ⊗k kG as a vector space, and the algebra structure is given by the crossed product (kG)∗ G, where G acts on (kG)∗ by Then the product is given by f g(x) = f (gxg−1). (f1 ⊗ g1)(f2 ⊗ g2) = f1f g−1 1 2 ⊗ g1g2. The coalgebra structure is the tensor of the two coalgebras (kG)∗ and kG, with ∆(φg ⊗ h) = g1g2=g φg1 ⊗ h ⊗ φg2 ⊗ h. D(G) is quasitriangular , i.e. there is an invertible element R of D(G) ⊗ D(G) such that R∆(x)R−1 = τ (∆(x)), where τ is the twist map. This is given by R = R1 = g g (φg ⊗ 1) ⊗ (1 ⊗ g) (φg ⊗ 1) ⊗ (1 ⊗ g−1). The equation R∆R−1 = τ ∆ results in an isomorphism between U ⊗ V and V ⊗ U for D(G)-bimodules U and V , given by ﬂip follows by the action of R. If G is non-abelian, then this is non-commutative and non-co-commutative. The point of deﬁning this is that the representations of D(G) correspond to the G-equivariant k-vector bundles on G. As we said, this is a general construction. Theorem (Mastnak, Witherspoon (2008)). The bialgebra cohomology H· bi(H, H) for a ﬁnite-dimensional Hopf algebra is equal to HH·(D(H), k), where k is the trivial module, and D(H) is the Drinfeld double. 71 4 Coalgebras, bialgebras and Hopf algebras III Algebras In 1990, Gerstenhaber and Schack deﬁned bialgebra cohomology, and proved results about deformations of bialgebras analogous to our results from the previous chapter for algebras. In particular, one can consider inﬁnitesimal deformations, and up to equivalence, these correspond to elements of the 2nd cohomology group. There is also the question as to whether an inﬁnitesimal deformation is integrable to give a bialgebra structure on V ⊗ k[[t]], where V is the underlying vector space of the bialgebra. Theorem (Gerstenhaber–Schack). Every deformation is equivalent to one where the unit and counit are unchnaged. Also, deformation preserves the existence of an antipode, though it might change. Theorem (Gerstenhaber–Schack). All deformations of O(Mn(k)) or O(SLn(k)) are equivalent to one in which the comultiplication is unchanged. We nwo try to deform O(M2(k)). By the previous theorems, we only have to change the multiplication. Consider Oq(M2(k)) deﬁned by X12X11 = qX11X12 X22X12 = qX12X22 X21X11 = qX11X21 X22X21 = qX21X22 X21X12 = X12X21 X11X22 − X22X11 = (q−1 − q)X12X21. We deﬁne the quantum determinant = X11X22 − q−1X12X21 = X22X11 − qX12X21. det q Then Then we deﬁne ∆(det q ) = det q , ⊗ det q ε(det q ) = 1. O(SL2(k)) = O(M2(k)) (detq −1) , where we are quotienting by the 2-sided ideal. It is possible to deﬁne an antipode, given by S(X11) = X22 S(X12) = −qX12 S(X21) = −q−1X21 S(X22) = X11, and this gives a non-commutative and non-co-commutative Hopf algebra. This is an example that we pulled out of a hat. But there is a general construction due to Faddeev–Reshetikhin–Takhtajan (1988) via R-matrices, which are a way of producing a k-linear map V ⊗ V → V ⊗ V, 72 4 Coalgebras, bialgebras and Hopf algebras III Algebras where V is a ﬁntie-dimesnional vector space. We take a basis e1, · · · , en of V , and thus a basis e1 ⊗ ej of V ⊗ V . We write Rm ij for the matrix of R, deﬁned by R(ei ⊗ ej) = Rm ij e ⊗ em. ,m The rows are indexed by pairs (, m), and the columns by pairs (i, j), which are put in lexicographic order. The action of R on V ⊗ V induces 3 diﬀerent actions on V ⊗ V ⊗ V . For s, t ∈ {1, 2, 3}, we let Rst be the invertible map which acts like R on the sth and tth components, and identity on the other. So for example, R12(e1 ⊗ e2 ⊗ v) = m i,j e ⊗ em ⊗ v. Deﬁnition (Yang–Baxter equation). R satisﬁes the quantum Yang–Baxter equation (QYBE ) if and the braided form of QYBE (braid equation) if R12R13R23 = R23R13R12 R12R23R12 = R23R12R23. Note that R satisﬁes QYBE iﬀ Rτ satisﬁes the braid equation. Solutions to either case are R-matrices. Example. The identity map and the twist map τ satisﬁes both. Take V to be 2-dimensional, and R to be the map Rm ij =  − , where q = 0 ∈ K. Thus, we have R(e1 ⊗ e1) = qe1 ⊗ e2 R(e2 ⊗ e1) = e2 ⊗ e1 R(e1 ⊗ e2) = e1 ⊗ e2 + (q − q−1)e2 ⊗ e1 R(e2 ⊗ e2) = qe2 ⊗ e2, and this satisﬁes QYBE. Similarly, (Rτ )m ij =  − satisﬁes the braid equation. We now deﬁne the general construction. 73 4 Coalgebras, bialgebras and Hopf algebras III Algebras Deﬁnition (R-symmetric algebra). Given the tensor algebra T (V ) = ∞ n=0 V ⊗n, we form the R-symmetric algebra SR(V ) = T (V ) z − R(z) : z ∈ V ⊗ V . Example. If R is the identity, then SR(V ) = T (V ). Example. If R = τ , then SR(V ) is the usual symmetric algebra. Example. The quantum plane Oq(k2) can be written as SR(V ) with R(e1 ⊗ e2) = qe2 ⊗ e1 R(e1 ⊗ e1) = e1 ⊗ e1 R(e2 ⊗ e1) = q−1e1 ⊗ e2 R(e2 ⊗ e2) = e2 ⊗ e2. Generally, given a V which is ﬁnite-dimensional as a vector space, we can identify (V ⊗ V )∗ with V ∗ ⊗ V ∗. We set E = V ⊗ V ∗ ∼= Endk(V ) ∼= Mn(k). We deﬁne R13 and R∗ 24 : E ⊗ E → E ⊗ E, where R13 acts like R on terms 1 and 3 in ∗, and 24 acts like R∗ on terms 2 and 4. identity on the rest; R∗ Deﬁnition (Coordinate algebra of quantum matrices). The coordinate algebra of quantum matrices associated with R is R13(z) − R∗ T (E) 24(z) : z ∈ E ⊗ E = SR(E), where T = R∗ 24R−1 13 . The coalgebra structure remains the same as O(Mn(k)), and for the antipode, we write E1 for the image of e1 in SR(V ), and similarly Fj for fj. Then we map E1 → Fj → n j=1 n i=1 Xij ⊗ Ej Fi ⊗ Xij. This is the general construction we are up for. Example. We have for Oq(M2(k)) = ARτ (V ) Rm ij =  − , 74 Index Index 2-cocycle integrable, 60 2-cocycle condition, 19 A M , 55 A[ε], 55 Aop, 7 An, 31 An(k), 31 Fi, 59 G-graded algebra, 18 J(A), 8 K0, 27 R-symmetric algebra, 74 GK-dim(M ), 39 HH n(A, M ), 52 Z-ﬁltered algebra, 32 Z-graded algebra, 33 gr M , 37 Der(A, M ), 55 d(M ), 39 f ◦ g, 63 f ◦i g, 63 k-algebra, 3, 68 k-separable algebra, 53 kG, 15 kq[X, X −1, Y, Y −1], 35 kq[X, Y ], 35 p-regular elements, 16 ACC, 4, 30 algebra, 3, 68 k-separable, 53 Artinian, 3 central simple, 19 cocommutative, 68 ﬁltered, 32 graded, 33 homogeneous components, 33 local, 21 Noetherian, 4, 30 opposite, 7 Rees algebra, 33 semisimple, 9 simple, 9 annihilator, 7 antipode, 70 Artin–Wedderburn theorem, 4, 13 III Algebras Artinian algebra, 3 ascending chain condition, 4, 30 associated graded algebra, 33 associated graded module, 37 associativity, 68 Bernstein’s inequality, 41 bialgebra, 70 bimodule, 6 block, 21 braid equation, 73 central simple algebra, 19 chain complex Hochschild, 51 characteristic variety, 42 circle product, 63 co-ideal, 68 coalgebra, 68 coassociativity, 68 coboundaries, 52 cochain complex Hochschild, 51 cocommutative coalgebra, 68 cocycle integrable, 60 cocycles, 52 coidentity, 68 complement, 10 completely reducible, 9 composition factor, 26 comultiplication, 68 coordinate algebra associated with quantum matrices, 74 coproduct, 68 counit, 68 crossed product, 18 cup product, 63 DCC, 3 deformation inﬁnitesimal, 60 degree n, 51 Dennis trace map, 29 derivation graded, 63 integrable, 61 75 Index III Algebras derivations, 55 descending chain condition, 3 diﬀerential graded Lie algebra, 64 dimension, 53 holonomic module, 41 homogeneous components, 33 Hopf algebra, 70 Hopf dual, 70 uniform, 48 division algebra, 3 domain, 46 Drinfeld double, 71 enough injectives, 43 equivalence of star products, 60 essential extension, 44 essential submodule, 44 exhaustive ﬁltration, 32 extension, 56 essential, 44 ﬁltered algebra, 32 Rees algebra, 33 ﬁltration exhaustive, 32 positive, 32 separated, 32 formal Poisson structures, 66 free resolution, 50 Gelfand-Kirillov dimension, 39 Gerstenhaber algebra, 64 Gerstenhaber bracket, 63 Goldie rank, 48 Goldie’s theorem, 5, 49 graded algebra, 33 homogeneous components, 33 graded derivation, 63 graded ideal, 33 graded Jacobi identity, 64 graded Lie algebra, 64 group algebra, 15 Hattori–Stallings trace map, 28 Hilbert basis theorem, 30 Hilbert polynomial, 39 Hilbert-Serre theorem, 37 HKR theorem, 65 Hochschild chain complex, 51 Hochschild cochain complex, 51 Hochschild cohomology groups, 52 Hochschild homology, 67 Hochschild–Kostant–Ronsenberg ideal, 3 graded, 33 prime, 7 idempotent orthogonal, 24 primitive, 24 identity, 68 indecomposable, 21 inﬁnitesimal deformations, 60 injective envelope, 45 injective hull, 45 injective module, 43 inner derivations, 55 integrable derivation, 61 integrable 2-cocycle, 60 irreducible module, 7 Jacobi identity graded, 64 Jacobson radical, 8 Krull–Schmidt theorem, 22 left regular representation, 6 Leibnitz rule, 36 Lie algebra graded, 64 universal enveloping algebra, 32 local algebra, 21 Maschke’s theorem, 15 Maurer–Cartan equation, 66 module, 6 injective, 43 irreducible, 7 projective, 19 simple, 7 uniform, 46 morphism, 68 coalgebras, 68 multiplication, 68 multiplicity, 40 theorem, 65 Nakayama lemma, 8 76 Index III Algebras Noetherian algebra, 4, 30 obstruction, 60 opposite algebra, 7 orthogonal idempotents, 24 Poincar´e series, 37 Poincar´e-Birkhoﬀ-Witt theorem, 34 Poisson algebra, 36 positive ﬁltration, 32 pre-Lie structure, 63 prime ideal, 7 primitive idempotent, 24 product, 68 projective module, 19 projective resolution, 50 quantum double, 71 quantum groups, 70 quantum plane, 35, 74 quantum torus, 35 quantum Yang–Baxter equation, 73 quasitriangular, 71 QYBE, 73 radical, 11 Rees algebra, 33 resolution free, 50 projective, 50 Samuel polynomial, 39 Schouten bracket, 65 Schur’s lemma, 13 semi-direct product, 55 semisimple algebra, 9 separable algebra, 53 separated ﬁltration, 32 separating idempotent, 53 simple algebra, 9 simple module, 7 skew ﬁelds, 3 split extension, 56 star product, 59 equivalence, 60 trivial, 60 submodule essential, 44 trace of projective, 28 trivial star product, 60 uniform dimension, 48 uniform module, 46 unique decomposition property, 22 unit, 68 universal enveloping algebra, 4, 32 Weyl algebra, 4, 31 77
> 0)(∀y ∈ A) \|y − a\| < δ ⇒ \|f (y) − f (a)\| < ε. Intuitively, f is continuous at a if we can obtain f (a) as accurately as we wish by using more accurate values of a (the deﬁnition says that if we want to approximate f (a) by f (y) to within accuracy ε, we just have to get our y to within δ of a for some δ). For example, suppose we have the function f (x . Suppose that we don’t know what the function actually is, but we have a computer program that computes this function. We want to know what f (π) is. Since we cannot input π (it has inﬁnitely many digits), we can try 3, and it gives 0. Then we try 3.14, and it gives 0 again. If we try 3.1416, it gives 1 (since π = 3.1415926 · · · < 3.1416). We keep giving more and more digits of π, but the result keeps oscillating between 0 and 1. We have no hope of what f (π) might be, even approximately. So this f is discontinuous at π. However, if we have the function g(x) = x2, then we can ﬁnd the (approximate) value of g(π). We can ﬁrst try g(3) and obtain 9. Then we can try g(3.14) = 9.8596, g(3.1416) = 9.86965056 etc. We can keep trying and obtain more and more accurate values of g(π). So g is continuous at π. Example. – Constant functions are continuous. – The function f (x) = x is continuous (take δ = ε). The deﬁnition of continuity of a function looks rather like the deﬁnition of convergence. In fact, they are related by the following lemma: Lemma. The following two statements are equivalent for a function f : A → R. – f is continuous – If (an) is a sequence in A with an → a, then f (an) → f (a). Proof. (i)⇒(ii) Let ε > 0. Since f is continuous at a, (∃δ > 0)(∀y ∈ A) \|y − a\| < δ ⇒ \|f (y) − f (a)\| < ε. 23 4 Continuous functions IA Analysis I We want N such that ∀n ≥ N , \|f (an) − f (a)\| < ε. By continuity, it is enough to ﬁnd N such that ∀n ≥ N , \|an − a\| < δ. Since an → a, such an N exists. (ii)⇒(i) We prove the contrapositive: Suppose f is not continuous at a. Then (∃ε > 0)(∀δ > 0)(∃y ∈ A) \|y − a\| < δ and \|f (y) − f (a)\| ≥ ε. For each n, we can therefore pick an ∈ A such that \|an − a\| < 1 f (a)\| ≥ ε. But then an → a (by Archimedean property), but f (an) → f (a). n and \|f (an) − Example. (i) Let f (x) = −1 = f (0). f (− 1 . Then f is not continuous because − 1 n → 0 but (ii) Let f : Q → R with f (x) = 1 x2 > 2 0 x2 < 2 Then f is continuous. For every a ∈ Q, we can ﬁnd an interval about a on which f is constant. So f is continuous at a. (iii) Let f (x) = sin 1 0 x x = 0 x = 0 Then f (a) is discontinuous. For example, let an = 1/[(2n + 0.5)π]. Then an → 0 and f (an) → 1 = f (0). We can use this sequence deﬁnition as the deﬁnition for continuous functions. This has the advantage of being cleaner to write and easier to work with. In particular, we can reuse a lot of our sequence theorems to prove the analogous results for continuous functions. Lemma. Let A ⊆ R and f, g : A → R be continuous functions. Then (i) f + g is continuous (ii) f g is continuous (iii) if g never vanishes, then f /g is continuous. Proof. (i) Let a ∈ A and let (an) be a sequence in A with an → a. Then (f + g)(an) = f (an) + g(an). But f (an) → f (a) and g(an) → g(a). So f (an) + g(an) → f (a) + g(a) = (f + g)(a). (ii) and (iii) are proved in exactly the same way. 24 4 Continuous functions IA Analysis I With this lemma, from the fact that constant functions and f (x) = x are continuous, we know that all polynomials are continuous. Similarly, rational functions P (x)/Q(x) are continuous except when Q(x) = 0. Lemma. Let A, B ⊆ R and f : A → B, g : B → R. Then if f and g are continuous, g ◦ f : A → R is continuous. Proof. We oﬀer two proofs: (i) Let (an) be a sequence in A with an → a ∈ A. Then f (an) → f (a) since f is continuous. Then g(f (an)) → g(f (a)) since g is continuous. So g ◦ f is continuous. (ii) Let a ∈ A and ε > 0. Since g is continuous at f (a), there exists η > 0 such that ∀z ∈ B, \|z − f (a)\| < η ⇒ \|g(z) − g(f (a))\| < ε. Since f is continuous at a, ∃δ > 0 such that ∀y ∈ A, \|y − a\| < δ ⇒ \|f (y) − f (a)\| < η. Therefore \|y − a\| < δ ⇒ \|g(f (y)) − g(f (a))\| < ε. There are two important theorems regarding continuous functions — the maximum value theorem and the intermediate value theorem. Theorem (Maximum value theorem). Let [a, b] be a closed interval in R and let f : [a, b] → R be continuous. Then f is bounded and attains its bounds, i.e. f (x) = sup f for some x, and f (y) = inf f for some y. Proof. If f is not bounded above, then for each n, we can ﬁnd xn ∈ [a, b] such that f (xn) ≥ n for all n. By Bolzano-Weierstrass, since xn ∈ [a, b] and is bounded, the sequence (xn) has a convergent subsequence (xnk ). Let x be its limit. Then since f is continuous, f (xnk ) → f (x). But f (xnk ) ≥ nk → ∞. So this is a contradiction. Now let C = sup{f (x) : x ∈ [a, b]}. Then for every n, we can ﬁnd xn n . So by Bolzano-Weierstrass, (xn) has a convergent ≤ f (xnk ) ≤ C, f (xnk ) → C. Therefore if such that f (xn) ≥ C − 1 subsequence (xnk ). Since C − 1 nk x = lim xnk , then f (x) = C. A similar argument applies if f is unbounded below. Theorem (Intermediate value theorem). Let a < b ∈ R and let f : [a, b] → R be continuous. Suppose that f (a) < 0 < f (b). Then there exists an x ∈ (a, b) such that f (x) = 0. Proof. We have several proofs: (i) Let A = {x : f (x) < 0} and let s = sup A. We shall show that f (s) = 0 (this is similar to the proof that If f (s) < 0, then setting ε = \|f (s)\| in the deﬁnition of continuity, we can ﬁnd δ > 0 such that ∀y, \|y − s\| < δ ⇒ f (y) < 0. Then s + δ/2 ∈ A, so s is not an upper bound. Contradiction. 2 exists in Numbers and Sets). √ If f (s) > 0, by the same argument, we can ﬁnd δ > 0 such that ∀y, \|y − s\| < δ ⇒ f (y) > 0. So s − δ/2 is a smaller upper bound. (ii) Let a0 = a, b0 = b. By repeated bisection, construct nested intervals and f (an) < 0 ≤ f (bn). Then by the [an, bn] such that bn − an = b0−a0 2n 25 4 Continuous functions IA Analysis I nested intervals property, we can ﬁnd x ∈ ∩∞ an, bn → x. Since f (an) < 0 for every n, f (x) ≤ 0. Similarly, since f (bn) ≥ 0 for every n, f (x) ≥ 0. So f (x) = 0. n=0[an, bn]. Since bn − an → 0, It is easy to generalize this to get that, if f (a) < c < f (b), then ∃x ∈ (a, b) such that f (x) = c, by applying the result to f (x) − c. Also, we can assume instead that f (b) < c < f (a) and obtain the same result by looking at −f (x). Corollary. Let f : [a, b] → [c, d] be a continuous strictly increasing function with f (a) = c, f (b) = d. Then f is invertible and its inverse is continuous. Proof. Since f is strictly increasing, it is an injection (suppose x = y. wlog, x < y. Then f (x) < f (y) and so f (x) = f (y)). Now let y ∈ (c, d). By the intermediate value theorem, there exists x ∈ (a, b) such that f (x) = y. So f is a surjection. So it is a bijection and hence invertible. Let g be the inverse. Let y ∈ [c, d] and let ε > 0. Let x = g(y). So f (x) = y. Let u = f (x − ε) and v = f (x + ε) (if y = c or d, make the obvious adjustments). Then u < y < v. So we can ﬁnd δ > 0 such that (y − δ, y + δ) ⊆ (u, v). Then \|z − y\| < δ ⇒ g(z) ∈ (x − ε, x + ε) ⇒ \|g(z) − g(y)\| < ε. With this corollary, we can create more continuous functions, e.g. √ x. 4.2 Continuous induction* Continuous induction is a generalization of induction on natural numbers. It provides an alternative mechanism to prove certain results we have shown. Proposition (Continuous induction v1). Let a < b and let A ⊆ [a, b] have the following properties: (i) a ∈ A (ii) If x ∈ A and x = b, then ∃y ∈ A with y > x. (iii) If ∀ε > 0, ∃y ∈ A : y ∈ (x − ε, x], then x ∈ A. Then b ∈ A. Proof. Since a ∈ A, A = ∅. A is also bounded above by b. So let s = sup A. Then ∀ε > 0, ∃y ∈ A such that y > s − ε. Therefore, by (iii), s ∈ A. If s = b, then by (ii), we can ﬁnd y ∈ A such that y > s. It can also be formulated as follows: Proposition (Continuous induction v2). Let A ⊆ [a, b] and suppose that (i) a ∈ A (ii) If [a, x] ⊆ A and x = b, then there exists y > x such that [a, y] ⊆ A. (iii) If [a, x) ⊆ A, then [a, x] ⊆ A. Then A = [a, b] 26 4 Continuous functions IA Analysis I Proof. We prove that version 1 ⇒ version 2. Suppose A satisﬁes the conditions of v2. Let A = {x ∈ [a, b] : [a, x] ⊆ A}. Then a ∈ A. If x ∈ A with x = b, then [a, x] ⊆ A. So ∃y > x such that [a, y] ⊆ A. So ∃y > x such that y ∈ A. If ∀ε > 0, ∃y ∈ (x − ε, x] such that [a, y] ⊆ A, then [a, x) ⊆ A. So by (iii), [a, x] ⊆ A, so x ∈ A. So A satisﬁes properties (i) to (iii) of version 1. Therefore b ∈ A. So [a, b] ⊆ A. So A = [a, b]. We reprove intermediate value theorem here: Theorem (Intermediate value theorem). Let a < b ∈ R and let f : [a, b] → R be continuous. Suppose that f (a) < 0 < f (b). Then there exists an x ∈ (a, b) such that f (x) = 0. Proof. Assume that f is continuous. Suppose f (a) < 0 < f (b). Assume that (∀x) f (x) = 0, and derive a contradiction. Let A = {x : f (x) < 0} Then a ∈ A. If x ∈ A, then f (x) < 0, and by continuity, we can ﬁnd δ > 0 such that \|y − x\| < δ ⇒ f (y) < 0. So if x = b, then we can ﬁnd y ∈ A such that y > x. We prove the contrapositive of the last condition, i.e. x ∈ A ⇒ (∃δ > 0)(∀y ∈ A) y ∈ (x − δ, x]. If x ∈ A, then f (x) > 0 (we assume that f is never zero. If not, we’re done). Then by continuity, ∃δ > 0 such that \|y − x\| < δ ⇒ f (y) > 0. So y ∈ A. Hence by continuous induction, b ∈ A. Contradiction. Now we prove that continuous functions in closed intervals are bounded. Theorem. Let [a, b] be a closed interval in R and let f : [a, b] → R be continuous. Then f is bounded. Proof. Let f : [a, b] be continuous. Let A = {x : f is bounded on [a, x]}. Then a ∈ A. If x ∈ A, x = b, then ∃δ > 0 such that \|y − x\| < δ ⇒ \|f (y) − f (x)\| < 1. So ∃y > x (e.g. take min{x + δ/2, b}) such that f is bounded on [a, y], which implies that y ∈ A. Now suppose that ∀ε > 0, ∃y ∈ (x, −ε, x] such that y ∈ A. Again, we can ﬁnd δ > 0 such that f is bounded on (x − δ, x + δ), and in particular on (x − δ, x]. Pick y such that f is bounded on [a, y] and y > x − δ. Then f is bounded on [a, x]. So x ∈ A. So we are done by c
ontinuous induction. Finally, we can prove a theorem that we have not yet proven. Deﬁnition (Cover of a set). Let A ⊆ R. A cover of A by open intervals is a set {Iγ : γ ∈ Γ} where each Iγ is an open interval and A ⊆ γ∈Γ Iγ. A ﬁnite subcover is a ﬁnite subset {Iγ1 , · · · , Iγn } of the cover that is still a cover. Not every cover has a ﬁnite subcover. For example, the cover {( 1 n , 1) : n ∈ N} of (0, 1) has no ﬁnite subcover. Theorem (Heine-Borel*). Every cover of a closed, bounded interval [a, b] by open intervals has a ﬁnite subcover. We say closed intervals are compact (cf. Metric and Topological Spaces). 27 4 Continuous functions IA Analysis I Proof. Let {Iγ : γ ∈ Γ} be a cover of [a, b] by open intervals. Let A = {x : [a, x] can be covered by ﬁnitely many of the Iγ}. Then a ∈ A since a must belong to some Iγ. If x ∈ A, then pick γ such that x ∈ Iγ. Then if x = b, since Iγ is an open interval, it contains [x, y] for some y > x. Then [a, y] can be covered by ﬁnitely many Iγ, by taking a ﬁnite cover for [a, x] and the Iγ that contains x. Now suppose that ∀ε > 0, ∃y ∈ A such that y ∈ (x − ε, x]. Let Iγ be an open interval containing x. Then it contains (x − ε, x] for some ε > 0. Pick y ∈ A such that y ∈ (x−ε, x]. Now combine Iγ with a ﬁnite subcover of [a, y] to get a ﬁnite subcover of [a, x]. So x ∈ A. Then done by continuous induction. We can use Heine-Borel to prove that continuous functions on [a, b] are bounded. Theorem. Let [a, b] be a closed interval in R and let f : [a, b] → R be continuous. Then f is bounded and attains it bounds, i.e. f (x) = sup f for some x, and f (y) = inf f for some y. Proof. Let f : [a, b] → R be continuous. Then by continuity, (∀x ∈ [a, b])(∃δx > 0)(∀y) \|y − x\| < δx ⇒ \|f (y) − f (x)\| < 1. Let γ = [a, b] and for each x ∈ γ, let Ix = (x − δx, x + δx). So by Heine-Borel, we can ﬁnd x1, · · · , xn such that [a, b] ⊆ n 1 (xi − δxi, xi + δxi ). But f is bounded in each interval (xi − δxi, xi + δxi) by \|f (xi)\| + 1. So it is bounded on [a, b] by max \|f (xi)\| + 1. 28 5 Diﬀerentiability IA Analysis I 5 Diﬀerentiability In the remainder of the course, we will properly develop calculus, and put diﬀerentiation and integration on a rigorous foundation. Every notion will be given a proper deﬁnition which we will use to prove results like the product and quotient rule. 5.1 Limits First of all, we need the notion of limits. Recall that we’ve previously had limits for sequences. Now, we will deﬁne limits for functions. Deﬁnition (Limit of functions). Let A ⊆ R and let f : A → R. We say or “f (x) → as x → a”, if lim x→a f (x) = , (∀ε > 0)(∃δ > 0)(∀x ∈ A) 0 < \|x − a\| < δ ⇒ \|f (x) − \| < ε. We couldn’t care less what happens when x = a, hence the strict inequality 0 < \|x − a\|. In fact, f doesn’t even have to be deﬁned at x = a. Example. Let f (x Then lim x→2 = 2, even though f (2) = 3. Example. Let f (x) = sin x x . Then f (0) is not deﬁned but lim x→0 f (x) = 1. We will see a proof later after we deﬁne what sin means. We notice that the deﬁnition of the limit is suspiciously similar to that of continuity. In fact, if we deﬁne g(x) = f (x) x = a x = a Then f (x) → as x → a iﬀ g is continuous at a. Alternatively, f is continuous at a if f (x) → f (a) as x → a. It follows also that f (x) → as x → a iﬀ f (xn) → for every sequence (xn) in A with xn → a. The previous limit theorems of sequences apply here as well Proposition. If f (x) → and g(x) → m as x → a, then f (x) + g(x) → + m, f (x)g(x) → m, and f (x) m if g and m don’t vanish. g(x) → 5.2 Diﬀerentiation Similar to what we did in IA Diﬀerential Equations, we deﬁne the derivative as a limit. 29 5 Diﬀerentiability IA Analysis I Deﬁnition (Diﬀerentiable function). f is diﬀerentiable at a with derivative λ if lim x→a f (x) − f (a) x − a = λ. lim h→0 f (a + h) − f (a) h = λ. Equivalently, if We write λ = f (a). Here we see why, in the deﬁnition of the limit, we say that we don’t care what happens when x = a. In our deﬁnition here, our function is 0/0 when x = a, and we can’t make any sense out of what happens when x = a. Alternatively, we write the deﬁnition of diﬀerentiation as f (x + h) − f (x) h = f (x) + ε(h), where ε(h) → 0 as h → 0. Rearranging, we can deduce that f (x + h) = f (x) + hf (x) + hε(h), Note that by the deﬁnition of the limit, we don’t have to care what value ε takes when h = 0. It can be 0, π or 101010 . However, we usually take ε(0) = 0 so that ε is continuous. Using the small-o notation, we usually write o(h) for a function that satisﬁes o(h) h → 0 as h → 0. Hence we have Proposition. f (x + h) = f (x) + hf (x) + o(h). We can interpret this as an approximation of f (x + h): f (x + h) = f (x) + hf (x) linear approximation + o(h) error term . And diﬀerentiability shows that this is a very good approximation with small o(h) error. Conversely, we have Proposition. If f (x + h) = f (x) + hf (x) + o(h), then f is diﬀerentiable at x with derivative f (x). Proof. f (x + h) − f (x) h = f (x) + o(h) h → f (x). We can take derivatives multiple times, and get multiple derivatives. Deﬁnition (Multiple derivatives). This is deﬁned recursively: f is (n + 1)times diﬀerentiable if it is n-times diﬀerentiable and its nth derivative f (n) is diﬀerentiable. We write f (n+1) for the derivative of f (n), i.e. the (n + 1)th derivative of f . Informally, we will say f is n-times diﬀerentiable if we can diﬀerentiate it n times, and the nth derivative is f (n). 30 5 Diﬀerentiability IA Analysis I We can prove the usual rules of diﬀerentiation using the small o-notation. It can also be proven by considering limits directly, but the notation will become a bit more daunting. Lemma (Sum and product rule). Let f, g be diﬀerentiable at x. Then f + g and f g are diﬀerentiable at x, with (f + g)(x) = f (x) + g(x) (f g)(x) = f (x)g(x) + f (x)g(x) Proof. (f + g)(x + h) = f (x + h) + g(x + h) = f (x) + hf (x) + o(h) + g(x) + hg(x) + o(h) = (f + g)(x) + h(f (x) + g(x)) + o(h) f g(x + h) = f (x + h)g(x + h) = [f (x) + hf (x) + o(h)][g(x) + hg(x) + o(h)] = f (x)g(x) + h[f (x)g(x) + f (x)g(x)] + o(h)[g(x) + f (x) + hf (x) + hg(x) + o(h)] + h2f (x)g(x) error term By limit theorems, the error term is o(h). So we can write this as = f g(x) + h(f (x)g(x) + f (x)g(x)) + o(h). Lemma (Chain rule). If f is diﬀerentiable at x and g is diﬀerentiable at f (x), then g ◦ f is diﬀerentiable at x with derivative g(f (x))f (x). Proof. If one is suﬃciently familiar with the small-o notation, then we can proceed as g(f (x + h)) = g(f (x) + hf (x) + o(h)) = g(f (x)) + hf (x)g(f (x)) + o(h). If not, we can be a bit more explicit about the computations, and use hε(h) instead of o(h): (g ◦ f )(x + h) = g(f (x + h)) = g[f (x) + hf (x) + hε1(h) ] the “h” term = g(f (x)) + f g(x) + hε1(h)g(f (x)) + hf (x) + hε1(h)ε2(hf (x) + hε1(h)) = g ◦ f (x) + hg(f (x))f (x) ε1(h)g(f (x)) + f (x) + ε1(h)ε2 + h error term hf (x) + hε1(h) . We want to show that the error term is o(h), i.e. it divided by h tends to 0 as h → 0. But ε1(h)g(f (x)) → 0, f (x)+ε1(h) is bounded, and ε2(hf (x)+hε1(h)) → 0 because hf (x) + hε1(h) → 0 and ε2(0) = 0. So our error term is o(h). 31 5 Diﬀerentiability IA Analysis I We usually don’t write out the error terms so explicitly, and just use heuristics like f (x + o(h)) = f (x) + o(h); o(h) + o(h) = o(h); and g(x) · o(h) = o(h) for any (bounded) function g. Example. (i) Constant functions are diﬀerentiable with derivative 0. (ii) f (x) = λx is diﬀerentiable with derivative λ. (iii) Using the product rule, we can show that xn is diﬀerentiable with derivative nxn−1 by induction. (iv) Hence all polynomials are diﬀerentiable. Example. Let f (x) = 1/x. If x = 0, then f (x + h) − f (x) h = by limit theorems. 1 x+h − 1 h x −h x(x+h) h = = −1 x(x + h) → −1 x2 Lemma (Quotient rule). If f and g are diﬀerentiable at x, and g(x) = 0, then f /g is diﬀerentiable at x with derivative f g (x) = f (x)g(x) − g(x)f (x) g(x)2 . Proof. First note that 1/g(x) = h(g(x)) where h(y) = 1/y. So 1/g(x) is diﬀeren−1 g(x)2 g(x) by the chain rule. tiable at x with derivative By the product rule, f /g is diﬀerentiable at x with derivative f (x) g(x) − f (x) g(x) g(x)2 = f (x)g(x) − f (x)g(x) g(x)2 . Lemma. If f is diﬀerentiable at x, then it is continuous at x. Proof. As y → x, f (y) − f (x) y − x → f (x). Since, y − x → 0, f (y) − f (x) → 0 by product theorem of limits. So f (y) → f (x). So f is continuous at x. Theorem. Let f : [a, b] → [c, d] be diﬀerentiable on (a, b), continuous on [a, b], and strictly increasing. Suppose that f (x) never vanishes. Suppose further that f (a) = c and f (b) = d. Then f has an inverse g and for each y ∈ (c, d), g is diﬀerentiable at y with derivative 1/f (g(y)). In human language, this states that if f is invertible, then the derivative of f −1 is 1/f . Note that the conditions will (almost) always require f to be diﬀerentiable on open interval (a, b), continuous on closed interval [a, b]. This is because it doesn’t make sense to talk about diﬀerentiability at a or b since the deﬁnition of f (a) requires f to be deﬁned on both sides of a. 32 5 Diﬀerentiability IA Analysis I Proof. g exists by an earlier theorem about inverses of continuous functions. Let y, y + k ∈ (c, d). Let x = g(y), x + h = g(y + k). Since g(y + k) = x + h, we have y + k = f (x + h). So k = f (x + h) − y = f (x + h) − f (x). So g(y + k) − g(y) k = (x + h) − x f (x + h) − f (x) = f (x + h) − f (x) h −1 . As k → 0, since g is continuous, g(y + k) → g(y). So h → 0. So g(y + k) − g(y) k → [f (x)]−1 = [f (g(y)]−1. Example. Let f (x) = x1/2 for x > 0. Then f is the inverse of g(x) = x2. So f (x) = 1 g(f (x)) = 1 2x1/2 = 1 2 x−1/2. Similarly, we can show that the derivative of x1/q is 1 q x1/q−1. Then let’s take xp/q = (x1/q)p. By the chain rule, its derivative is p(x1/q)p−1 · 1 q x1/q−1 = p q x 5.3 Diﬀerentiation theorems p−1 q + 1 q −1 = p q −1. x p q Everything we’ve had so far is something we already know. It’s just that now we can prove them rigorously. In this section, we will come up with genuinely new t
heorems, including but not limited to Taylor’s theorem, which gives us Taylor’s series. Theorem (Rolle’s theorem). Let f be continuous on a closed interval [a, b] (with a < b) and diﬀerentiable on (a, b). Suppose that f (a) = f (b). Then there exists x ∈ (a, b) such that f (x) = 0. It is intuitively obvious: if you move up and down, and ﬁnally return to the same point, then you must have changed direction some time. Then f (x) = 0 at that time. Proof. If f is constant, then we’re done. Otherwise, there exists u such that f (u) = f (a). wlog, f (u) > f (a). Since f is continuous, it has a maximum, and since f (u) > f (a) = f (b), the maximum is not attained at a or b. Suppose maximum is attained at x ∈ (a, b). Then for any h = 0, we have f (x + h) − f (x since f (x + h) − f (x) ≤ 0 by maximality of f (x). By considering both sides as we take the limit h → 0, we know that f (x) ≤ 0 and f (x) ≥ 0. So f (x) = 0. 33 5 Diﬀerentiability IA Analysis I Corollary (Mean value theorem). Let f be continuous on [a, b] (a < b), and diﬀerentiable on (a, b). Then there exists x ∈ (a, b) such that f (x) = f (b) − f (a) b − a . Note that f (b)−f (a) b−a is the slope of the line joining f (a) and f (b). f (x) f (b) f (a) The mean value theorem is sometimes described as “rotate your head and apply Rolle’s”. However, if we actually rotate it, we might end up with a non-function. What we actually want is a shear. Proof. Let Then g(x) = f (x) − f (b) − f (a) b − a x. g(b) − g(a) = f (b) − f (a) − f (b) − f (a) b − a (b − a) = 0. So by Rolle’s theorem, we can ﬁnd x ∈ (a, b) such that g(x) = 0. So f (x) = f (b) − f (a) b − a , as required. We’ve always assumed that if a function has a positive derivative everywhere, then the function is increasing. However, it turns out that this is really hard to prove directly. It does, however, follow quite immediately from the mean value theorem. Example. Suppose f (x) > 0 for every x ∈ (a, b). Then for u, v in [a, b], we can ﬁnd w ∈ (u, v) such that f (v) − f (u) v − u = f (w) > 0. It follows that f (v) > f (u). So f is strictly increasing. Similarly, if f (x) ≥ 2 for every x and f (0) = 0, then f (1) ≥ 2, or else we can ﬁnd x ∈ (0, 1) such that 2 ≤ f (x) = f (1) − f (0) 1 − 0 = f (1). 34 5 Diﬀerentiability IA Analysis I Theorem (Local version of inverse function theorem). Let f be a function with continuous derivative on (a, b). Let x ∈ (a, b) and suppose that f (x) = 0. Then there is an open interval (u, v) containing x on which f is invertible (as a function from (u, v) to f ((u, v))). Moreover, if g is the inverse, then g(f (z)) = 1 f (z) for every z ∈ (u, v). This says that if f has a non-zero derivative, then it has an inverse locally and the derivative of the inverse is 1/f . Note that this not only requires f to be diﬀerentiable, but the derivative itself also has to be continuous. Proof. wlog, f (x) > 0. By the continuity, of f , we can ﬁnd δ > 0 such that f (z) > 0 for every z ∈ (x − δ, x + δ). By the mean value theorem, f is strictly increasing on (x − δ, x + δ), hence injective. Also, f is continuous on (x − δ, x + δ) by diﬀerentiability. Then done by the inverse function theorem. Finally, we are going to prove Taylor’s theorem. To do so, we will ﬁrst need some lemmas. Theorem (Higher-order Rolle’s theorem). Let f be continuous on [a, b] (a < b) and n-times diﬀerentiable on an open interval containing [a, b]. Suppose that f (a) = f (a) = f (2)(a) = · · · = f (n−1)(a) = f (b) = 0. Then ∃x ∈ (a, b) such that f (n)(x) = 0. Proof. Induct on n. The n = 0 base case is just Rolle’s theorem. Suppose we have k < n and xk ∈ (a, b) such that f (k)(xk) = 0. Since f (k)(a) = 0, we can ﬁnd xk+1 ∈ (a, xk) such that f (k+1)(xk+1) = 0 by Rolle’s theorem. So the result follows by induction. Corollary. Suppose that f and g are both diﬀerentiable on an open interval containing [a, b] and that f (k)(a) = g(k)(a) for k = 0, 1, · · · , n − 1, and also f (b) = g(b). Then there exists x ∈ (a, b) such that f (n)(x) = g(n)(x). Proof. Apply generalised Rolle’s to f − g. Now we shall show that for any f , we can ﬁnd a polynomial p of degree at most n that satisﬁes the conditions for g, i.e. a p such that p(k)(a) = f (k)(a) for k = 0, 1, · · · , n − 1 and p(b) = f (b). A useful ingredient is the observation that if then Therefore, if Qk(x) = (x − a)k k! , Q(j) k (a(x) = n−1 k=0 f (k)(a)Qk(x), 35 5 Diﬀerentiability IA Analysis I then Q(j)(a) = f (j)(a) for j = 0, 1, · · · , n − 1. To get p(b) = f (b), we use our nth degree polynomial term: p(x) = Q(x) + (x − a)n (b − a)n f (b) − Q(b). Then our ﬁnal term does not mess up our ﬁrst n − 1 derivatives, and gives p(b) = f (b). By the previous corollary, we can ﬁnd x ∈ (a, b) such that f (n)(x) = p(n)(x). f (n)(x) = n! (b − a)n f (b) − Q(b). f (b) = Q(b) + (b − a)n n! f (n)(x). That is, Therefore Alternatively, f (b) = f (a) + (b − a)f (a) + · · · + (b − a)n−1 (n − 1)! f (n−1)(a) + (b − a)n n! f (n)(x). Setting b = a + h, we can rewrite this as Theorem (Taylor’s theorem with the Lagrange form of remainder). f (a + h) = f (a) + hf (a) + · · · + hn−1 (n − 1)! f (n−1)(a) + (n−1)-degree approximation to f near a hn f (n)(x) n! error term . for some x ∈ (a, a + h). Strictly speaking, we only proved it for the case when h > 0, but we can easily show it holds for h < 0 too by considering g(x) = f (−x). Note that the remainder term is not necessarily small, but this often gives us the best (n − 1)-degree approximation to f near a. For example, if f (n) is bounded by C near a, then hn n! f (n)(x) ≤ C n! \|h\|n = o(hn−1). Example. Let f : R → R be a diﬀerentiable function such that f (0) = 1 and f (x) = f (x) for every x (intuitively, we know it is ex , but that thing doesn’t exist!). Then for every x, we have f (x) = 1 + x + x2 2! + x3 3! + · · · = ∞ n=0 xn n! . While it seems like we can prove this works by diﬀerentiating it and see that f (x) = f (x), the sum rule only applies for ﬁnite sums. We don’t know we can diﬀerentiate a sum term by term. So we have to use Taylor’s theorem. 36 5 Diﬀerentiability IA Analysis I Since f (x) = f (x), it follows that all derivatives exist. By Taylor’s theorem, f (x) = f (0) + f (0)x + f (2)(0) 2! x2 + · · · + f (n−1)(0) (n − 1)! xn−1 + f (n)(u) n! xn. for some u between 0 and x. This equals to f (x) = n−1 k=0 xk k! + f (n)(u) n! xn. We must show that the remainder term f (n)(u) that x is ﬁxed, but u can depend on n. n! xn → 0 as n → ∞. Note here But we know that f (n)(u) = f (u), but since f is diﬀerentiable, it is continuous, and is bounded on [0, x]. Suppose \|f (u)\| ≤ C on [0, x]. Then f (n)(u) n! xn ≤ C n! \|x\|n → 0 from limit theorems. So it follows that f (x) = 1 + x + x2 2! + x3 3! + · · · = ∞ n=0 xn n! . 5.4 Complex diﬀerentiation Deﬁnition (Complex diﬀerentiability). Let f : C → C. Then f is diﬀerentiable at z with derivative f (z) if lim h→0 f (z + h) − f (z) h exists and equals f (z). Equivalently, f (z + h) = f (z) + hf (z) + o(h). This is exactly the same deﬁnition with real diﬀerentiation, but has very diﬀerent properties! All the usual rules — chain rule, product rule etc. also apply (with the same proofs). Also the derivatives of polynomials are what you expect. However, there are some more interesting cases. Example. f (z) = ¯z is not diﬀerentiable. z + h − z h = ¯h h = 1 −1 h is purely imaginary h is real If this seems weird, this is because we often think of C as R2, but they are not the same. For example, reﬂection is a linear map in R2, but not in C. A linear map in C is something in the form x → bx, which can only be a dilation or rotation, not reﬂections or other weird things. Example. f (z) = \|z\| is also not diﬀerentiable. If it were, then \|z\|2 would be as well (by the product rule). So would \|z\|2 z = ¯z when z = 0 by the quotient rule. At z = 0, it is certainly not diﬀerentiable, since it is not even diﬀerentiable on R. 37 6 Complex power series IA Analysis I 6 Complex power series Before we move on to integration, we ﬁrst have a look at complex power series. This will allow us to deﬁne the familiar exponential and trigonometric functions. Deﬁnition (Complex power series). A complex power series is a series of the form ∞ anzn. n=0 when z ∈ C and an ∈ C for all n. When it converges, it is a function of z. When considering complex power series, a very important concept is the radius of convergence. To make sense of this concept, we ﬁrst need the following lemma: Lemma. Suppose that anzn converges and \|w\| < \|z\|, then anwn converges (absolutely). Proof. We know that \|anwn\| = \|anzn\| · Since anzn converges, the terms anzn are bounded. So pick C such that . n w z for every n. Then \|anzn\| ≤ C 0 ≤ ∞ n=0 \|anwn\| ≤ ∞ n=0 C n w z , which converges (geometric series). So by the comparison test, anwn converges absolutely. It follows that if anzn does not converge and \|w\| > \|z\|, then anwn does not converge. Now let R = sup{\|z\| : anzn converges } (R may be inﬁnite). If \|z\| < R, then we can ﬁnd z0 with \|z0\| ∈ (\|z\|, R] such that ∞ 0 converges. So by lemma above, anzn converges. If \|z\| > R, then anzn diverges by deﬁnition of R. n anzn Deﬁnition (Radius of convergence). The radius of convergence of a power series anzn is R = sup \|z\| : anzn converges . {z : \|z\| < R} is called the circle of convergence.1. If \|z\| < R, then anzn converges. If \|z\| > R, then anzn diverges. When \|z\| = R, the series can converge at some points and not the others. Example. ∞ n=0 zn has radius of convergence of 1. When \|z\| = 1, it diverges (since the terms do not tend to 0). 1Note to pedants: yes it is a disc, not a circle 38 6 Complex power series IA Analysis I ∞ n=0 zn n Example. to nth is has radius of convergence 1, since the ratio of (n + 1)th term zn+1/(n + 1) zn/. So if \|z\| < 1, then the series converges by the ratio test. If \|z\| > 1, then eventually the terms are increasing in modulus. If z = 1, then it diverges (harmonic series). If \|z\| = 1 and z = 1, it converges by Abel’s test. Example. The series ∞ n=1 zn n2 converges for \|z\| ≤ 1 and diverges for \|z\| > 1. As
evidenced by the above examples, the ratio test can be used to ﬁnd the radius of convergence. We also have an alternative test based on the nth root. Lemma. The radius of convergence of a power series anzn is R = 1 lim sup n\|an\| . Often n\|an\| converges, so we only have to ﬁnd the limit. Proof. Suppose \|z\| < 1/ lim sup n\|an\|. Then \|z\| lim sup n\|an\| < 1. Therefore there exists N and ε > 0 such that \|z\| n\|an\| ≤ 1 − ε sup n≥N by the deﬁnition of lim sup. Therefore \|anzn\| ≤ (1 − ε)n for every n ≥ N , which implies (by comparison with geometric series) that anzn converges absolutely. On the other hand, if \|z\| lim sup n\|an\| > 1, it follows that \|z\| n\|an\| ≥ 1 for inﬁnitely many n. Therefore \|anzn\| ≥ 1 for inﬁnitely many n. So anzn does not converge. Example. The radius of convergence of So lim sup n\|an\| = 1 2 . So 1/ lim sup n\|an\| = 2. zn 2n is 2 because n\|an\| = 1 2 for every n. But often it is easier to ﬁnd the radius convergence from elementary methods such as the ratio test, e.g. for n2zn. 6.1 Exponential and trigonometric functions Deﬁnition (Exponential function). The exponential function is ez = ∞ n=0 zn n! . By the ratio test, this converges on all of C. 39 6 Complex power series IA Analysis I A fundamental property of this function is that ez+w = ezew. Once we have this property, we can say that Proposition. The derivative of ez is ez. Proof. But So ez+h − ez h = ez eh − 1 h 1 + = ez h 2! + h2 3! + · · · h 2! + h2 3! + · · · ≤ \|h\| 2 + \|h\|2 4 + \|h\|3 8 + · · · = \|h\|/2 1 − \|h\|/2 → 0. ez+h − ez h → ez. But we must still prove that ez+w = ezew. Consider two sequences (an), (bn). Their convolution is the sequence (cn) deﬁned by cn = a0bn + a1bn−1 + a2bn−2 + · · · + anb0. The relevance of this is that if you take N n=0 anzn N n=0 bnzn and N n=0 cnzn, and equate coeﬃcients of zn, you get cn = a0bn + a1bn−1 + a2bn−2 + · · · + anb0. n=0 an and ∞ Theorem. Let ∞ let (cn) be the convolution of the sequences (an) and (bn). Then ∞ converges (absolutely), and n=0 bn be two absolutely convergent series, and n=0 cn ∞ n=0 ∞ ∞ bn . an cn = n=0 n=0 Proof. We ﬁrst show that a rearrangement of cn converges absolutely. Hence it converges unconditionally, and we can rearrange it back to cn. Consider the series (a0b0) + (a0b1 + a1b1 + a1b0) + (a0b2 + a1b2 + a2b2 + a2b1 + a2b0) + · · · (∗) Let SN = N n=0 an, TN = N n=0 bn, UN = N n=0 \|an\|, VN = N n=0 \|bn\|. 40 6 Complex power series IA Analysis I Also let SN → S, TN → T, UN → U, VN → V (these exist since an and bn converge absolutely). If we take the modulus of the terms of (∗), and consider the ﬁrst (N + 1)2 terms (i.e. the ﬁrst N + 1 brackets), the sum is UN VN . Hence the series converges absolutely to U V . Hence (∗) converges. The partial sum up to (N + 1)2 of the series (∗) itself is SN TN , which converges to ST . So the whole series converges to ST . Since it converges absolutely, it converges unconditionally. Now consider a rearrangement: a0b0 + (a0b1 + a1b0) + (a0b2 + a1b1 + a2b0) + · · · Then this converges to ST as well. But the partial sum of the ﬁrst 1 + 2 + · · · + N terms is c0 + c1 + · · · + cN . So N n=0 cn → ST = ∞ ∞ bn . an n=0 n=0 Corollary. ezew = ez+w. Proof. By theorem above (and deﬁnition of ez), ∞ n=0 ∞ ezew = ezew = 1 · wn n! + 1 n! wn + z 1! n 1 wn−1 (n − 1)! + zwn−1 + z2 2! n 2 wn−2 (n − 2)! + · · · + · 1 z2wn−2 + · · · + zn zn n! n n n=0 ∞ = (z + w)n by the binomial theorem n=0 = ez+w. Note that if (cn) is the convolution of (an) and (bn), then the convolution of (anzn) and (bnzn) is (cnzn). Therefore if both anzn and bnzn converge absolutely, then their product is cnzn. Note that we have now completed the proof that the derivative of ez is ez. Now we deﬁne sin z and cos z: Deﬁnition (Sine and cosine). sin z = cos z = eiz − e−iz 2i eiz + e−iz 2 = z − = 1 − z3 3! z2 2! + + z5 5! z4 4! − − z7 7! z6 6! + · · · + · · · . We now prove certain basic properties of sin and cos, using known properties of ez. 41 6 Complex power series IA Analysis I Proposition. d dz d dz sin z = cos z = ieiz + ie−iz 2i ieiz − ie−iz 2 = cos z = − sin z sin2 z + cos2 z = e2iz + 2 + e−2iz 4 + e2iz − 2 + e−2iz −4 = 1. It follows that if x is real, then \| cos x\| and \| sin x\| are at most 1. Proposition. Proof. cos(z + w) = cos z cos w − sin z sin w sin(z + w) = sin z cos w + cos z sin w cos z cos w − sin z sin w = (eiz + e−iz)(eiw + e−iw) 4 + (eiz − e−iz)(eiw − e−iw) 4 = ei(z+w) + e−i(z+w) 2 = cos(z + w). Diﬀerentiating both sides wrt z gives − sin z cos w − cos z sin w = − sin(z + w). So sin(z + w) = sin z cos w + cos z sin w. When x is real, we know that cos x ≤ 1. Also sin 0 = 0, and d dx sin x = cos x ≤ 1. So for x ≥ 0, sin x ≤ x, “by the mean value theorem”. Also, cos 0 = 1, and d dx cos x = − sin x, which, for x ≥ 0, is greater than −x. From this, it follows that when x ≥ 0, cos x ≥ 1 − x2 2 comes from “integrating” −x, (or ﬁnding a thing whose derivative is −x)). 2 (the 1 − x2 Continuing in this way, we get that for x ≥ 0, if you take truncate the power series for sin x or cos x, it will be ≥ sin x, cos x if you stop at a positive term, and ≤ if you stop at a negative term. For example, sin x ≥ x − x3 3! + x5 5! − x7 7! + x9 9! − x11 11! . In particular, cos 2 ≤ 1 − 24 4! Since cos 0 = 1, it follows by the intermediate value theorem that there exists some x ∈ (0, 2) such that cos x = 0. Since cos x ≥ 1 − x2 2 , we can further deduce that x > 1. = 1 − 2 + 22 2! < 0. 2 3 + Deﬁnition (Pi). Deﬁne the smallest x such that cos x = 0 to be π 2 . 42 6 Complex power series IA Analysis I sin π Since sin2 z + cos2 z = 1, it follows that sin π 2 ≥ 0 by the mean value theorem. So sin π Thus 2 = ±1. Since cos x > 0 on [0, π 2 ], 2 = 1. Proposition. Proof. z + z + cos sin π 2 π 2 = − sin z = cos z cos(z + π) = − cos z sin(z + π) = − sin z cos(z + 2π) = cos z sin(z + 2π) = sin z z + cos π 2 = cos z cos π 2 = − sin z sin = − sin z π 2 − sin z sin π 2 and similarly for others. 6.2 Diﬀerentiating power series We shall show that inside the circle of convergence, the derivative of ∞ given by the obvious formula ∞ n=1 nanzn−1. n=0 an z is We ﬁrst prove some (seemingly arbitrary and random) lemmas to build up the proof of the above statement. They are done so that the ﬁnal proof will not be full of tedious algebra. Lemma. Let a and b be complex numbers. Then bn − an − n(b − a)an−1 = (b − a)2(bn−2 + 2abn−3 + 3a2bn−4 + · · · + (n − 1)an−2). Proof. If b = a, we are done. Otherwise, bn − an b − a = bn−1 + abn−2 + a2bn−3 + · · · + an−1. Diﬀerentiate both sides with respect to a. Then −nan−1(b − a) + bn − an (b − a)2 = bn−2 + 2abn−3 + · · · + (n − 1)an−2. Rearranging gives the result. Alternatively, we can do bn − an = (b − a)(bn−1 + abn−2 + · · · + an−1). Subtract n(b − a)an−1 to obtain (b − a)[bn−1 − an−1 + a(bn−2 − an−2) + a2(bn−3 − an−3) + · · · ] and simplify. 43 6 Complex power series IA Analysis I This implies that (z + h)n − zn − nhzn−1 = h2((z + h)n−2 + 2z(z + h)n−3 + · · · + (n − 1)zn−2), which is actually the form we need. Lemma. Let anzn have radius of convergence R, and let \|z\| < R. Then nanzn−1 converges (absolutely). Proof. Pick r such that \|z\| < r < R. Then \|an\|rn converges, so the terms \|an\|rn are bounded above by, say, C. Now n\|anzn−1\| = n\|an\|rn−1 n−1 \|z\| r ≤ C r n \|z\| r n−1 The series n by the comparison test. \|z\| r n−1 converges, by the ratio test. So n\|anzn−1\| converges, Corollary. Under the same conditions, ∞ n=2 n 2 anzn−2 converges absolutely. Proof. Apply Lemma above again and divide by 2. Theorem. Let anzn be a power series with radius of convergence R. For \|z\| < R, let ∞ ∞ f (z) = anzn and g(z) = nanzn−1. Then f is diﬀerentiable with derivative g. n=0 n=1 Proof. We want f (z + h) − f (z) − hg(z) to be o(h). We have f (z + h) − f (z) − hg(z) = ∞ n=2 an((z + h)n − zn − hnzn). We started summing from n = 2 since the n = 0 and n = 1 terms are 0. Using our ﬁrst lemma, we are left with h2 ∞ n=2 (z + h)n−2 + 2z(z + h)n−3 + · · · + (n − 1)zn−2 an We want the huge inﬁnite series to be bounded, and then the whole thing is a bounded thing times h2, which is deﬁnitely o(h). Pick r such that \|z\| < r < R. If h is small enough that \|z + h\| ≤ r, then the last inﬁnite series is bounded above (in modulus) by ∞ n=2 \|an\|(rn−2 + 2rn−2 + · · · + (n − 1)rn−2) = ∞ n=2 \|an\| n 2 rn−2, which is bounded. So done. 44 6 Complex power series IA Analysis I In IB Analysis II, we will prove the same result using the idea of uniform convergence, which gives a much nicer proof. Example. The derivative of is ez = 1 + z + z2 2! + z3 3! + · · · 1 + z + z2 2! + · · · = ez. So we have another proof that of this fact. Similarly, the derivatives of sin z and cos z work out as cos z and − sin z. 6.3 Hyperbolic trigonometric functions Deﬁnition (Hyperbolic sine and cosine). We deﬁne cosh z = sinh z = ez + e−z 2 ez − e−z 2 = 1 + = z + z2 2! z3 3! + + z4 4! z5 5! + + z6 6! z7 7! + · · · + · · · Either from the deﬁnition or from diﬀerentiating the power series, we get that Proposition. d dz d dz cosh z = sinh z sinh z = cosh z Also, by deﬁnition, we have Proposition. Also, Proposition. cosh iz = cos z sinh iz = i sin z cosh2 z − sinh2 z = 1, 45 7 The Riemann Integral IA Analysis I 7 The Riemann Integral Finally, we can get to integrals. There are many ways to deﬁne an integral, which can have some subtle diﬀerences. The deﬁnition we will use here is the Riemann integral, which is the simplest deﬁnition, but is also the weakest one, in the sense that many functions are not Riemann integrable but integrable under other deﬁnitions. Still, the deﬁnition of the Riemann integral is not too straightforward, and requires a lot of preliminary deﬁnitions. 7.1 Riemann Integral Deﬁnition (Dissections). Let [a, b] be a closed interval. A dissection of [a, b] is a sequence a = x0 < x1 < x2 < · · · < xn = b. Deﬁnition (Upper and lower sums). Given a dissection D, the upper sum and lower sum are deﬁned by the formulae UD(f ) = LD(f ) = n i=1 n i=1 (xi − xi−1) sup x∈[xi−1,xi] f (x) (xi − xi−1) inf x∈[xi−1,xi] f (x) Sometimes we use the shorthand M
i = sup x∈[xi−1,xi] f (x), mi = inf x∈[xi−1−xi] f (x). The upper sum is the total area of the red rectangles, while the lower sum is the total area of the black rectangles: y · · · · · · a x1 x2 x3 xi xi+1 · · · b x Deﬁnition (Reﬁning dissections). If D1 and D2 are dissections of [a, b], we say that D2 reﬁnes D1 if every point of D1 is a point of D2. Lemma. If D2 reﬁnes D1, then UD2f ≤ UD1f and LD2f ≥ LD1 f. 46 7 The Riemann Integral IA Analysis I Using the picture above, this is because if we cut up the dissections into smaller pieces, the red rectangles can only get chopped into shorter pieces and the black rectangles can only get chopped into taller pieces. y y x0 x1 x x0 x1 x2 x3 x Proof. Let D be x0 < x1 < · · · < xn. Let D2 be obtained from D1 by the addition of one point z. If z ∈ (xi−1, xi), then UD2f − UD1f = (z − xi−1) sup x∈[xi−1,z] f (x) + (xi − z) sup x∈[z,xi] f (x) − (xi − xi−1)Mi. But supx∈[xi−1,z] f (x) and supx∈[z,xi] f (x) are both at most Mi. So this is at most Mi(z − xi−1 + xi − z − (xi − xi−1)) = 0. So UD2 f ≤ UD1 f. By induction, the result is true whenever D2 reﬁnes D1. A very similar argument shows that LD2f ≥ LD1 f . Deﬁnition (Least common reﬁnement). If D1 and D2 be dissections of [a, b]. Then the least common reﬁnement of D1 and D2 is the dissection made out of the points of D1 and D2. Corollary. Let D1 and D2 be two dissections of [a, b]. Then UD1f ≥ LD2 f. Proof. Let D be the least common reﬁnement (or indeed any common reﬁnement). Then by lemma above (and by deﬁnition), UD1f ≥ UDf ≥ LDf ≥ LD2 f. Finally, we can deﬁne the integral. Deﬁnition (Upper, lower, and Riemann integral). The upper integral is The lower integral is b a f (x) dx = inf D UDf. b a f (x) dx = sup D LDf. 47 7 The Riemann Integral IA Analysis I If these are equal, then we call their common value the Riemann integral of f , and is denoted b a f (x) dx. If this exists, we say f is Riemann integrable. We will later prove the fundamental theorem of calculus, which says that integration is the reverse of diﬀerentiation. But why don’t we simply deﬁne integration as anti-diﬀerentiation, and prove that it is the area of the curve? There are things that we cannot ﬁnd (a closed form of) the anti-derivative of, like e−x2 . In these cases, we wouldn’t want to say the integral doesn’t exist — it surely does according to this deﬁnition! There is an immediate necessary condition for Riemann integrability — boundedness. If f is unbounded above in [a, b], then for any dissection D, there must be some i such that f is unbounded on [xi−1, xi]. So Mi = ∞. So UDf = ∞. Similarly, if f is unbounded below, then LDf = −∞. So unbounded functions are not Riemann integrable. Example. Let f (x) = x on [a, b]. Intuitively, we know that the integral is (b2 − a2)/2, and we will show this using the deﬁnition above. Let D = x0 < x1 < · · · < xn be a dissection. Then UDf = n i=1 (xi − xi−1)xi We know that the integral is b2−a2 2 . So we put each term of the sum into the form i −x2 x2 2 i−1 plus some error terms. = = = n i=1 x2 i 2 − x2 i−1 2 + x2 i 2 − xi−1xi + x2 i−1 2 1 2 1 2 n i=1 (x2 i − x2 i−1 + (xi − xi−1)2) (b2 − a2) + 1 2 n (xi − xi−1)2. i=1 Deﬁnition (Mesh). The mesh of a dissection D is maxi(xi+1 − xi). Then if the mesh is < δ, then 1 2 n (xi − xi−1)2 ≤ i=1 δ 2 n (xi − xi−1) = i=1 δ 2 (b − a). So by making δ small enough, we can show that for any ε > 0, Similarly, So b a b a x dx < x dx > 1 2 1 2 (b2 − a2) + ε. (b2 − a2) − ε. b a x dx = 1 2 (b2 − a2). 48 7 The Riemann Integral IA Analysis I Example. Deﬁne f : [0, 1] → R by f (x . Let x0 < x1 < · · · < xn be a dissection. Then for every i, we have mi = 0 (since there is an irrational in every interval), and Mi = 1 (since there is a rational in every interval). So UDf = n i=1 Mi(xi − xi−1) = n (xi − xi−1) = 1. i=1 Similarly, LDf = 0. Since D was arbitrary, we have 1 0 f (x) dx = 1, 1 0 f (x) dx = 0. So f is not Riemann integrable. Of course, this function is not interesting at all. The whole point of its existence is to show undergraduates that there are some functions that are not integrable! Note that it is important to say that the function is not Riemann integrable. There are other notions for integration in which this function is integrable. For example, this function is Lebesgue-integrable. Using the deﬁnition to show integrability is often cumbersome. Most of the time, we use the Riemann’s integrability criterion, which follows rather immediately from the deﬁnition, but is much nicer to work with. Proposition (Riemann’s integrability criterion). This is sometimes known as Cauchy’s integrability criterion. Let f : [a, b] → R. Then f is Riemann integrable if and only if for every ε > 0, there exists a dissection D such that UD − LD < ε. Proof. (⇒) Suppose that f is integrable. Then (by deﬁnition of Riemann integrability), there exist D1 and D2 such that and UD1 < LD2 > b a b a f (x) dx + f (x) dx − ε 2 , ε 2 . Let D be a common reﬁnement of D1 and D2. Then UDf − LDf ≤ UD1 f − LD2 f < ε. (⇐) Conversely, if there exists D such that UDf − LDf < ε, 49 7 The Riemann Integral IA Analysis I then which is, by deﬁnition, that inf UDf − sup LDf < ε, b a f (x) dx − b a f (x) dx < ε. Since ε > 0 is arbitrary, this gives us that b a f (x) dx = b a f (x) dx. So f is integrable. The next big result we want to prove is that integration is linear, ie b a (λf (x) + µg(x)) dx = λ b a f (x) dx + µ b a g(x) dx. We do this step by step: Proposition. Let f : [a, b] → R be integrable, and λ ≥ 0. Then λf is integrable, and b b λf (x) dx = λ f (x) dx. Proof. Let D be a dissection of [a, b]. Since a a sup x∈[xi−1,xi] λf (x) = λ sup x∈[xi−1,xi] f (x), and similarly for inf, we have UD(λf ) = λUDf LD(λf ) = λLDf. So if we choose D such that UDf − LDf < ε/λ, then UD(λf ) − LD(λf ) < ε. So the result follows from Riemann’s integrability criterion. Proposition. Let f : [a, b] → R be integrable. Then −f is integrable, and b a −f (x) dx = − b a f (x) dx. Proof. Let D be a dissection. Then sup x∈[xi−1,xi] inf x∈[xi−1,xi] −f (x) = − inf x∈[xi−1,xi] f (x) −f (x) = − sup f (x). x∈[xi−1,xi] Therefore UD(−f ) = n (xi − xi−1)(−mi) = −LD(f ). i=1 50 7 The Riemann Integral IA Analysis I Similarly, So LD(−f ) = −UDf. UD(−f ) − LD(−f ) = UDf − LDf. Hence if f is integrable, then −f is integrable by the Riemann integrability criterion. Proposition. Let f, g : [a, b] → R be integrable. Then f + g is integrable, and b a (f (x) + g(x)) dx = b a f (x) dx + b a g(x) dx. Proof. Let D be a dissection. Then UD(f + g) = n (xi − xi−1) i=1 n (xi − xi−1) ≤ i=1 = UDf + UDg Therefore, (f (x) + g(x)) sup x∈[xi−1,xi] sup u∈[xi−1,xi] f (u) + sup g(v) v∈[xi−1,xi] b a (f (x) + g(x)) dx ≤ b a f (x) dx + Similarly, b a (f (x) + g(x)) dx ≥ b a b a g(x) dx = f (x) dx + b a b a f (x) dx + b a g(x) dx. g(x) dx. So the upper and lower integrals are equal, and the result follows. So we now have that b a (λf (x) + µg(x)) dx = λ b a f (x) dx + µ b a g(x) dx. We will prove more “obvious” results. Proposition. Let f, g : [a, b] → R be integrable, and suppose that f (x) ≤ g(x) for every x. Then b b f (x) dx ≤ g(x) dx. Proof. Follows immediately from the deﬁnition. a a Proposition. Let f : [a, b] → R be integrable. Then \|f \| is integrable. Proof. Note that we can write sup x∈[xi−1,xi] f (x) − inf x∈[xi−1,xi] f (x) = sup u,v∈[xi−1,xi] \|f (u) − f (v)\|. 51 7 The Riemann Integral IA Analysis I Similarly, sup x∈[xi−1,xi] \|f (x)\| − inf x∈[xi−1,xi] \|f (x)\| = sup u,v∈[xi−1,xi] \|\|f (u)\| − \|f (v)\|\|. For any pair of real numbers, x, y, we have that \|\|x\| − \|y\|\| ≤ \|x − y\| by the triangle inequality. Then for any interval u, v ∈ [xi−1, xi], we have \|\|f (u)\| − \|f (v)\|\| ≤ \|f (u) − f (v)\|. Hence we have sup x∈[xi−1,xi] \|f (x)\| − inf x∈[xi−1,xi] \|f (x)\| ≤ sup x∈[xi−1,xi] f (x) − inf x∈[xi−1,xi] f (x). So for any dissection D, we have UD(\|f \|) − LD(\|f \|) ≤ UD(f ) − LD(f ). So the result follows from Riemann’s integrability criterion. Combining these two propositions, we get that if \|f (x) − g(x)\| ≤ C, for every x ∈ [a, b], then b a f (x) dx − b a g(x) dx ≤ C(b − a). Proposition (Additivity property). Let f : [a, c] → R be integrable, and let b ∈ (a, c). Then the restrictions of f to [a, b] and [b, c] are Riemann integrable, and b c c f (x) dx + f (x) dx = f (x) dx a b a Similarly, if f is integrable on [a, b] and [b, c], then it is integrable on [a, c] and the above equation also holds. Proof. Let ε > 0, and let a = x0 < x1 < · · · < xn = c be a dissection of D of [a, c] such that and UD(f ) ≤ LD(f ) ≥ f (x) dx + ε, f (x) dx − ε. c a c a Let D be the dissection made of D plus the point b. Let D1 be the dissection of [a, b] made of points of D from a to b, and D2 be the dissection of [b, c] made of points of D from b to c. Then and UD1 (f ) + UD2(f ) = UD(f ) ≤ UD(f ), LD1 (f ) + LD2(f ) = LD(f ) ≥ LD(f ). 52 7 The Riemann Integral IA Analysis I Since UD(f ) − LD(f ) < 2ε, and both UD2 (f ) − LD2(f ) and UD1 (f ) − LD1(f ) are non-negative, we have UD1(f ) − LD1(f ) and UD2(f ) − LD2(f ) are less than 2ε. Since ε is arbitrary, it follows that the restrictions of f to [a, b] and [b, c] are both Riemann integrable. Furthermore, b a f (x) dx + Similarly, b a f (x) dx + c b c b f (x) dx ≤ UD1(f ) + UD2(f ) = UD(f ) ≤ UD(f ) ≤ c a f (x) dx + ε. f (x) dx ≥ LD1(f ) + LD2 (f ) = LD(f ) ≥ LD(f ) ≥ c a f (x) dx − ε. Since ε is arbitrary, it follows that b a f (x) dx + c b f (x) dx = c a f (x) dx. The other direction is left as an (easy) exercise. Proposition. Let f, g : [a, b] → R be integrable. Then f g is integrable. Proof. Let C be such that \|f (x)\|, \|g(x)\| ≤ C for every x ∈ [a, b]. Write Li and i for the sup and inf of g in [xi−1, xi]. Now let D be a dissection, and for each i, let ui and vi be two points in [xi−1, xi]. We will pretend that ui and vi are the minimum and maximum when we write the proof, but we cannot assert that they are, since f g need not have maxima and minima. We will then note that since our results hold for arbitrary ui and vi, it must hold when f g is at its supre
mum and inﬁmum. We ﬁnd what we pretend is the diﬀerence between the upper and lower sum: xi − xi−1)(f (vi)g(vi) − f (ui)g(ui) (xi − xi−1)f (vi)(g(vi) − g(ui)) + (f (vi) − f (ui))g(ui) n i=1 n i=1 n = ≤ C(Li − i) + (Mi − mi)C i=1 = C(UDg − LDg + UDf − LDf ). Since ui and vi are arbitrary, it follows that UD(f g) − LD(f g) ≤ C(UDf − LDf + UDg − LDg). Since C is ﬁxed, and we can get UDf − LDf and UDg − LDg arbitrary small (since f and g are integrable), we can get UD(f g) − LD(f g) arbitrarily small. So the result follows. 53 7 The Riemann Integral IA Analysis I Theorem. Every continuous function f on a closed bounded interval [a, b] is Riemann integrable. Proof. wlog assume [a, b] = [0, 1]. n , 2 Suppose the contrary. Let f be non-integrable. This means that there exists some ε such that for every dissection D, UD − LD > ε. In particular, for every n , · · · , n n, let Dn be the dissection 0, 1 n . Since UDn − LDn > ε, there exists some interval k in which sup f − inf f > ε. Suppose the supremum and inﬁmum are attained at xn and yn respectively. Then we have \|xn − yn\| < 1 n and f (xn) − f (yn) > ε. n , k+1 By Bolzano Weierstrass, (xn) has a convergent subsequence, say (xni). Say xni → x. Since \|xn − yn\| < 1 n → 0, we must have yni → x. By continuity, we must have f (xni) → f (x) and f (yni) → f (x), but f (xni ) and f (yni ) are always apart by ε. Contradiction. n With this result, we know that a lot of things are integrable, e.g. e−x2 To prove this, we secretly used the property of uniform continuity: . Deﬁnition (Uniform continuity*). Let A ⊆ R and let f : A → R. Then f is uniformly continuous if (∀ε)(∃δ > 0)(∀x)(∀y) \|x − y\| < δ ⇒ \|f (x) − f (y)\| ≤ ε. This is diﬀerent from regular continuity. Regular continuity says that at any point x, we can ﬁnd a δ that works for this point. Uniform continuity says that we can ﬁnd a δ that works for any x. It is easy to show that a uniformly continuous function is integrable, since by uniformly continuity, as long as the mesh of a dissection is suﬃciently small, the diﬀerence between the upper sum and the lower sum can be arbitrarily small by uniform continuity. Thus to prove the above theorem, we just have to show that continuous functions on a closed bounded interval are uniformly continuous. Theorem (non-examinable). Let a < b and let f : [a, b] → R be continuous. Then f is uniformly continuous. Proof. Suppose that f is not uniformly continuous. Then (∃ε)(∀δ > 0)(∃x)(∃y) \|x − y\| < δ and \|f (x) − f (y)\| ≥ ε. Therefore, we can ﬁnd sequences (xn), (yn) such that for every n, we have \|xn − yn\| ≤ 1 n and \|f (xn) − f (yn)\| ≥ ε. Then by Bolzano-Weierstrass theorem, we can ﬁnd a subsequence (xnk ) converging to some x. Since \|xnk −ynk \| ≤ 1 , ynk → x as well. But \|f (xnk )−f (ynk )\| ≥ ε nk for every k. So f (xnk ) and f (ynk ) cannot both converge to the same limit. So f is not continuous at x. This proof is very similar to the proof that continuous functions are integrable. In fact, the proof that continuous functions are integrable is just a fuse of this proof and the (simple) proof that uniformly continuously functions are integrable. Theorem. Let f : [a, b] → R be monotone. Then f is Riemann integrable. 54 7 The Riemann Integral IA Analysis I Note that monotone functions need not be “nice”. It can even have inﬁnitely many discontinuities. For example, if f : [0, 1] → R maps x to the 1/(ﬁrst non-zero digit in the binary expansion of x), with f (0) = 0. Proof. let ε > 0. Let D be a dissection of mesh less than ε f (b)−f (a) . Then UDf − LDf = n i=1 (xi − xi−1)(f (xi) − f (xi−1)) ≤ ε f (b) − f (a) n i=1 = ε. (f (xi) − f (xi−1)) Pictorially, we see that the diﬀerence between the upper and lower sums is total the area of the red rectangles. y x To calculate the total area, we can stack the red areas together to get something f (b)−f (a) and height f (b) − f (a). So the total area is just ε. of width ε Lemma. Let a < b and let f be a bounded function from [a, b] → R that is continuous on (a, b). Then f is integrable. An example where this would apply is 1 x . It gets nasty near x = 0, but its “nastiness” is conﬁned to x = 0 only. So as long as its nastiness is suﬃciently contained, it would still be integrable. 0 sin 1 The idea of the proof is to integrate from a point x1 very near a up to a point xn−1 very close to b. Since f is bounded, the regions [a, x1] and [xn−1, b] are small enough to not cause trouble. Proof. Let ε > 0. Suppose that \|f (x)\| ≤ C for every x ∈ [a, b]. Let x0 = a and pick x1 such that x1 − x0 < ε 8C . Also choose z between x1 and b such that b − z < ε Then f is continuous [x1, z]. Therefore it is integrable on [x1, z]. So we can ﬁnd a dissection D with points x1 < x2 < · · · < xn−1 = z such that 8C . UDf − LDf < ε 2 . 55 7 The Riemann Integral IA Analysis I Let D be the dissection a = x0 < x1 < · · · < xn = b. Then UDf − LDf < ε 8C · 2C + ε 2 + ε 8C · 2C = ε. So done by Riemann integrability criterion. Example. – f (x) = sin 1 0 x x = 0 x = 0 deﬁned on [−1, 1] is integrable. – g(x) = x x ≤ 1 x2 + 1 x > 1 deﬁned on [0, 1] is integrable. Corollary. Every piecewise continuous and bounded function on [a, b] is integrable. Proof. Partition [a, b] into intervals I1, · · · , Ik, on each of which f is (bounded and) continuous. Hence for every Ij with end points xj−1, xj, f is integrable on [xj−1, xj] (which may not equal Ij, e.g. Ij could be [xj−1, xj)). But then by the additivity property of integration, we get that f is integrable on [a, b] We deﬁned Riemann integration in a very general way — we allowed arbitrary dissections, and took the extrema over all possible dissection. Is it possible to just consider some particular nice dissections instead? Perhaps unsurprisingly, yes! It’s just that we opt to deﬁne it the general way so that we can easily talk about things like least common reﬁnements. Lemma. Let f : [a, b] → R be Riemann integrable, and for each n, let Dn be the dissection a = x0 < x1 < · · · < xn = b, where xi = a + i(b−a) for each i. Then n and UDn f → LDn f → b a b a f (x) dx f (x) dx. Proof. Let ε > 0. We need to ﬁnd an N . The only thing we know is that f is Riemann integrable, so we use it: Since f is integrable, there is a dissection D, say u0 < u1 < · · · < um, such that UDf − b a f (x) dx < ε 2 . We also know that f is bounded. Let C be such that \|f (x)\| ≤ C. For any n, let D be the least common reﬁnement of Dn and D. Then UDf ≤ UDf. Also, the sums UDn f and UDf are the same, except that at most m of the subintervals [xi−1, xi] are subdivided in D. 56 7 The Riemann Integral IA Analysis I For each interval that gets chopped up, the upper sum decreases by at most b−a n · 2C. Therefore UDn f − UDf ≤ b − a n 2C · m. Pick n such that 2Cm(b − a)/n < ε 2 . Then So UDnf − UDf < ε 2 . UDn f − b a f (x) dx < ε. This is true whenever n > 4C(b−a)m therefore ε . Since we also have UDn f ≥ b a f (x) dx, UDn f → b a f (x) dx. The proof for lower sums is similar. For convenience, we deﬁne the following: Notation. If b > a, we deﬁne a b f (x) dx = − b a f (x) dx. We now prove that the fundamental theorem of calculus, which says that integration is the reverse of diﬀerentiation. Theorem (Fundamental theorem of calculus, part 1). Let f : [a, b] → R be continuous, and for x ∈ [a, b], deﬁne F (x) = x a f (t) dt. Then F is diﬀerentiable and F (x) = f (x) for every x. Proof. F (x + h) − F (x) h = 1 h x+h x f (t) dt Let ε > 0. Since f is continuous, at x, then there exists δ such that \|y − x\| < δ implies \|f (y) − f (x)\| < ε. If \|h\| < δ, then 1 h x+h x f (t) dt − f (x) = ≤ ≤ x+h (f (t) − f (x)) dt x x+h x \|f (t) − f (x)\| dt 1 h 1 \|h\| ε\|h\| \|h\| = ε. 57 7 The Riemann Integral IA Analysis I Corollary. If f is continuously diﬀerentiable on [a, b], then Proof. Let b a f (t) dt = f (b) − f (a). g(x) = x a f (t) dt. Then d dx Since g(x) − f (x) = 0, g(x) − f (x) must be a constant function by the mean value theorem. We also know that g(x) = f (x) = (f (x) − f (a)). g(a) = 0 = f (a) − f (a) So we must have g(x) = f (x) − f (a) for every x, and in particular, for x = b. Theorem (Fundamental theorem of calculus, part 2). Let f : [a, b] → R be a diﬀerentiable function, and suppose that f is integrable. Then b a f (t) dt = f (b) − f (a). Note that this is a stronger result than the corollary above, since it does not require that f is continuous. Proof. Let D be a dissection x0 < x1 < · · · < xn. We want to make use of this dissection. So write f (b) − f (a) = n i=1 (f (xi) − f (xi−1)). For each i, there exists ui ∈ (xi−1, xi) such that f (xi) − f (xi−1j) = (xi − xi−1)f (ui) by the mean value theorem. So f (b) − f (a) = n (xi − xi−1)f (ui). i=1 We know that f (ui) is somewhere between sup x∈[xi,xi−1] f (x) and inf x∈[xi,xi−1] f (x) by deﬁnition. Therefore LDf ≤ f (b) − f (a) ≤ UDf . Since f is integrable and D was arbitrary, LDf and UDf can both get arbitrarily close to b a f (t) dt. So f (b) − f (a) = b a f (t) dt. 58 7 The Riemann Integral IA Analysis I Note that the condition that f is integrable is essential. It is possible to ﬁnd a diﬀerentiable function whose derivative is not integrable! You will be asked to ﬁnd it in the example sheet. Using the fundamental theorem of calculus, we can easily prove integration by parts: Theorem (Integration by parts). Let f, g : [a, b] → R be integrable such that everything below exists. Then b a f (x)g(x) dx = f (b)g(b) − f (a)g(a) − b a f (x)g(x) dx. Proof. By the fundamental theorem of calculus, b a (f (x)g(x) + f (x)g(x)) dx = b a The result follows after rearrangement. (f g)(x) dx = f (b)g(b) − f (a)g(a). Recall that when we ﬁrst had Taylor’s theorem, we said it had the Lagrange form of the remainder. There are many other forms of the remainder term. Here we will look at the integral form: Theorem (Taylor’s theorem with the integral form of the remainder). Let f be n + 1 times diﬀerentiable on [a, b] with with f (n+1) continuous. Then f (b) = f (a) + (b − a)f (a) + + (b − a)n n! f (n)(a) + b a (b − a)2 2! (b − t)n n! f (2)(a) + · · · f (
n+1)(t) dt. Proof. Induction on n. When n = 0, the theorem says f (b) − f (a) = b a f (t) dt. which is true by the fundamental theorem of calculus. Now observe that b a (b − t)n n! f (n+1)(t) dt = = b f (n+1)(t) a f (n+1)(t) dt −(b − t)n+1 (n + 1)! b + a (b − a)n+1 (n + 1)! (b − t)n+1 (n + 1)! f (n+1)(a) + b a (b − t)n+1 (n + 1)! f (n+2)(t) dt. So the result follows by induction. Note that the form of the integral remainder is rather weird and unexpected. How could we have come up with it? We might start with the fundamental 59 7 The Riemann Integral IA Analysis I theorem of algebra and integrate by parts. The ﬁrst attempt would be to integrate 1 to t and diﬀerentiate f (t) to f (2)(t). So we have f (b) = f (a) + b a f (t) dt = f (a) + [tf (t)]b a − b a tf (2)(t) dt = f (a) + bf (b) − af (a) − b a tf (2)(t) dt We want something in the form (b − a)f (a), so we take that out and see what we are left with. = f (a) + (b − a)f (a) + b(f (b) − f (a)) − b a tf (2)(t) dt Then we note that f (b) − f (a) = b a f (2)(t) dt. So we have = f (a) + (b − a)f (a) + b a (b − t)f (2)(t) dt. Then we can see that the right thing to integrate is (b − t) and continue to obtain the result. Theorem (Integration by substitution). Let f : [a, b] → R be continuous. Let g : [u, v] → R be continuously diﬀerentiable, and suppose that g(u) = a, g(v) = b, and f is deﬁned everywhere on g([u, v]) (and still continuous). Then b a f (x) dx = v u f (g(t))g(t) dt. Proof. By the fundamental theorem of calculus, f has an anti-derivative F deﬁned on g([u, v]). Then v u f (g(t))g(t) dt = = v u v u F (g(t))g(t) dt (F ◦ g)(t) dt = F ◦ g(v) − F ◦ g(u) = F (b) − F (a) = b a f (x) dx. We can think of “integration by parts” as what you get by integrating the product rule, and “integration by substitution” as what you get by integrating the chain rule. 7.2 Improper integrals It is sometimes sensible to talk about integrals of unbounded functions or integrating to inﬁnity. But we have to be careful and write things down nicely. 60 7 The Riemann Integral IA Analysis I Deﬁnition (Improper integral). Suppose that we have a function f : [a, b] → R b a+ε f (x) dx such that, for every ε > 0, f is integrable on [a + ε, b] and lim ε→0 exists. Then we deﬁne the improper integral b a f (x) dx to be lim ε→0 b a+ε f (x) dx. even if the Riemann integral does not exist. We can do similarly for [a, b − ε], or integral to inﬁnity: ∞ a f (x) dx = lim b→∞ b a f (x) dx. when it exists. Example. So 1 ε x−1/2 dx = 2x−1/21 ε = 2 − 2ε1/2 → 2. 1 0 x−1/2 dx = 2, even though x−1/2 is unbounded on [0, 1]. Note that oﬃcially we are required to make f (x) = x−1/2 a function with domain [0, 1]. So we can assign f (0) = π, or any number, since it doesn’t matter. Example. x 1 1 t2 dt = − as x → ∞ by the fundamental theorem of calculus. So 1 x2 dx = 1. Finally, we can prove the integral test, whose proof we omitted when we ﬁrst 1 ∞ began. Theorem (Integral test). Let f : [1, ∞] → R be a decreasing non-negative function. Then ∞ n=1 f (n) converges iﬀ ∞ 1 f (x) dx < ∞. Proof. We have n+1 n f (x) dx ≤ f (n) ≤ n n−1 f (x) dx, since f is decreasing (the right hand inequality is valid only for n ≥ 2). It follows that N +1 N N f (x) dx ≤ f (n) ≤ f (x) dx + f (1) 1 n=1 1 So if the integral exists, then f (n) is increasing and bounded above by ∞ 1 f (x) dx, so converges. 1 f (x) dx is unbounded. Then N n=1 f (n) If the integral does not exist, then N is unbounded, hence does not converge. Example. Since x 1 1 t2 dt < ∞, it follows that ∞ n=1 1 n2 converges. 61
emann integral*). Let f : [a, b] → R be a bounded function, and let Df be the set of points of discontinuities of f . Then f is Riemann integrable if and only if Df has measure zero. Using this result, a lot of our theorems follow easily of these. Apart from the easy ones like the sum and product of integrable functions is integrable, we can also easily show that the composition of a continuous function with an integrable function is integrable, since composing with a continuous function will not introduce more discontinuities. Similarly, we can show that the uniform limit of integrable functions is integrable, since the points of discontinuities of the uniform limit is at most the (countable) union of all discontinuities of the functions in the sequence. Proof is left as an exercise for the reader, in the example sheet. 24 4 Rn as a normed space IB Analysis II 4 Rn as a normed space 4.1 Normed spaces Our objective is to extend most of the notions we had about functions of a single variable f : R → R to functions of multiple variables f : Rn → R. More generally, we want to study functions f : Ω → Rm, where Ω ⊆ Rn. We wish to define analytic notions such as continuity, differentiability and even integrability (even though we are not doing integrability in this course). In order to do this, we need more structure on Rn. We already know that Rn is a vector space, which means that we can add, subtract and multiply by scalars. But to do analysis, we need something to replace our notion of \|x − y\| in R. This is known as a norm. It is useful to define and study this structure in an abstract setting, as opposed to thinking about Rn specifically. This leads to the general notion of normed spaces. Definition (Normed space). Let V be a real vector space. A norm on V is a function ∥ · ∥ : V → R satisfying (i) ∥x∥ ≥ 0 with equality iff x = 0 (non-negativity) (ii) ∥λx∥ = \|λ\|∥x∥ (linearity in scalar multiplication) (iii) ∥x + y∥ ≤ ∥x∥ + ∥y∥ (triangle inequality) A normed space is a pair (V, ∥ · ∥). If the norm is understood, we just say V is a normed space. We do have to be slightly careful since there can be multiple norms on a vector space. Intuitively, ∥x∥ is the length or magnitude of x. Example. We will first look at finite-dimensional spaces. This is typically Rn with different norms. – Consider Rn, with the Euclidean norm ∥x∥2 = 2 . x2 i This is also known as the usual norm. It is easy to check that this is a norm, apart from the triangle inequality. So we’ll just do this. We have ∥x + y∥2 = n (xi + yi)2 i=1 = ∥x∥2 + ∥y∥2 + 2 xiyi ≤ ∥x∥2 + ∥y∥2 + 2∥x∥y∥ = (∥x∥2 + ∥y∥2), where we used the Cauchy-Schwarz inequality. So done. – We can have the following norm on Rn: ∥x∥1 = \|xi\|. It is easy to check that this is a norm. 25 4 Rn as a normed space IB Analysis II – We can also have the following norm on Rn: ∥x∥∞ = max{\|xi\| : 1 ≤ i ≤ n}. It is also easy to check that this is a norm. – In general, we can define the p norm (for p ≥ 1) by ∥x∥p = \|xi\|p1/p . It is, however, not trivial to check the triangle inequality, and we will not do this. We can show that as p → ∞, ∥x∥p → ∥x∥∞, which justifies our notation above. We also have some infinite dimensional examples. Often, we can just extend our notions on Rn to infinite sequences with some care. We write RN for the set of all infinite real sequences (xk). This is a vector space with termwise addition and scalar multiplication. – Define ℓ1 = (xk) ∈ RN : \|xk\| < ∞ . This is a linear subspace of RN. We define the norm by ∥(xk)∥1 = ∥(xk)∥ℓ1 = \|xk\|. – Similarly, we can define ℓ2 by ℓ2 = (xk) ∈ RN : x2 k < ∞ . The norm is defined by ∥(xk)∥2 = ∥(xk)∥ℓ2 = 1/2 . x2 k We can also write this as ∥(xk)∥ℓ2 = lim n→∞ ∥(x1, · · · , xn)∥2. So the triangle inequality for the Euclidean norm implies the triangle inequality for ℓ2. – In general, for p ≥ 1, we can define ℓp = (xk) ∈ RN : \|xk\|p < ∞ with the norm ∥(xk)∥p = ∥(xk)∥ℓp = \|xk\|p1/p . 26 4 Rn as a normed space IB Analysis II – Finally, we have ℓ∞, where ℓ∞ = {(xk) ∈ RN : sup \|xk\| < ∞}, with the norm ∥(xk)∥∞ = ∥(xk)∥ℓ∞ = sup \|xk\|. Finally, we can have examples where we look at function spaces, usually C([a, b]), the set of continuous real functions on [a, b]. – We can define the L1 norm by ∥f ∥L1 = ∥f ∥1 = b a \|f \| dx. – We can define L2 similarly by ∥f ∥L2 = ∥f ∥2 = 1 2 f 2 dx b a – In general, we can define Lp for p ≥ 1 by ∥f ∥Lp = ∥f ∥p = 1 p f p dx b a – Finally, we have L∞ by ∥f ∥L∞ = ∥f ∥∞ = sup \|f \|. . . This is also called the uniform norm, or the supremum norm. Later, when we define convergence for general normed space, we will show that convergence under the uniform norm is equivalent to uniform convergence. To show that L2 is actually a norm, we can use the Cauchy-Schwarz inequality for integrals. Lemma (Cauchy-Schwarz inequality (for integrals)). If f, g ∈ C([a, b]), f, g ≥ 0, then b a f g dx ≤ b 1/2 b f 2 dx) 1/2 g2 dx . a a a f 2 dx = 0, then f = 0 (since f is continuous). So the inequality Proof. If b holds trivially. Otherwise, let A2 = b a f 2 dx ̸= 0, B2 = b b ϕ(t) = (g − tf )2 dt ≥ 0. a g2 dx. Consider the function for every t. We can expand this as a ϕ(t) = t2A2 − 2t b a gf dx + B2. 27 4 Rn as a normed space IB Analysis II The conditions for a quadratic in t to be non-negative is exactly 2 gf dx − A2B2 ≤ 0. b a So done. Note that the way we defined Lp is rather unsatisfactory. To define the ℓp spaces, we first have the norm defined as a sum, and then ℓp to be the set of all sequences for which the sum converges. However, to define the Lp space, we restrict ourselves to C([0, 1]), and then define the norm. Can we just define, say, L1 to be the set of all functions such that 1 0 \|f \| dx exists? We could, but then the norm would no longer be the norm, since if we have the function f (x) = 1 x = 0.5 0 x ̸= 0.5 , then f is integrable with integral 0, but is not identically zero. So we cannot expand our vector space to be too large. To define Lp properly, we need some more sophisticated notions such as Lebesgue integrability and other fancy stuff, which will be done in the IID Probability and Measure course. We have just defined many norms on the same space Rn. These norms are clearly not the same, in the sense that for many x, ∥x∥1 and ∥x∥2 have different values. However, it turns out the norms are all “equivalent” in some sense. This intuitively means the norms are “not too different” from each other, and give rise to the same notions of, say, convergence and completeness. A precise definition of equivalence is as follows: Definition (Lipschitz equivalence of norms). Let V be a (real) vector space. Two norms ∥ · ∥, ∥ · ∥′ on V are Lipschitz equivalent if there are real constants 0 < a < b such that a∥x∥ ≤ ∥x∥′ ≤ b∥x∥ for all x ∈ V . It is easy to show this is indeed an equivalence relation on the set of all norms on V . We will show that if two norms are equivalent, the “topological” properties of the space do not depend on which norm we choose. For example, the norms will agree on which sequences are convergent and which functions are continuous. It is possible to reformulate the notion of equivalence in a more geometric way. To do so, we need some notation: Definition (Open ball). Let (V, ∥ · ∥) be a normed space, a ∈ V , r > 0. The open ball centered at a with radius r is Br(a) = {x ∈ V : ∥x − a∥ < r}. Then the requirement that a∥x∥ ≤ ∥x∥′ ≤ b∥x∥ for all x ∈ V is equivalent to saying B1/b(0) ⊆ B′ 1(0) ⊆ B1/a(0), where B′ is the ball with respect to ∥ · ∥′, while B is the ball with respect to ∥ · ∥. Actual proof of equivalence is on the second example sheet. 28 4 Rn as a normed space IB Analysis II Example. Consider R2. Then the norms ∥ · ∥∞ and ∥ · ∥2 are equivalent. This is easy to see using the ball picture: where the blue ones are the balls with respect to ∥ · ∥∞ and the red one is the ball with respect to ∥ · ∥2. In general, we can consider Rn, again with ∥ · ∥2 and ∥ · ∥∞. We have ∥x∥∞ ≤ ∥x∥2 ≤ √ n∥x∥∞. These are easy to check manually. However, later we will show that in fact, any two norms on a finite-dimensional vector space are Lipschitz equivalent. Hence it is more interesting to look at infinite dimensional cases. Example. Let V = C([0, 1]) with the norms ∥f ∥1 = 1 0 \|f \| dx, ∥f ∥∞ = sup [0,1] \|f \|. We clearly have the bound ∥f ∥1 ≤ ∥f ∥∞. However, there is no constant b such that ∥f ∥∞ ≤ b∥f ∥1 for all f . This is easy to show by constructing a sequence of functions fn by y 1 1 n x n and the height is 1. Then ∥fn∥∞ = 1 but ∥fn∥1 = 1 where the width is 2 Example. Similarly, consider the space ℓ2 = (xn) : x2 regular ℓ2 norm and the ℓ∞ norm. We have n → 0. n < ∞ under the ∥(xk)∥∞ ≤ ∥(xk)∥ℓ2 , 29 4 Rn as a normed space IB Analysis II but there is no b such that ∥(xk)∥ℓ2 ≤ b∥(xk)∥∞. For example, we can consider the sequence xn = (1, 1, · · · , 1, 0, 0, · · · ), where the first n terms are 1. So far in all our examples, out of the two inequalities, one holds and one does not. Is it possible for both inequalities to not hold? The answer is yes. This is an exercise on the second example sheet as well. This is all we are going to say about Lipschitz equivalence. We are now going to define convergence, and study the consequences of Lipschitz equivalence to convergence. Definition (Bounded subset). Let (V, ∥ · ∥) be a normed space. A subset E ⊆ V is bounded if there is some R > 0 such that E ⊆ BR(0). Definition (Convergence of sequence). Let (V, ∥ · ∥) be a normed space. A sequence (xk) in V converges to x ∈ V if ∥xk − x∥ → 0 (as a sequence in R), i.e. (∀ε > 0)(∃N )(∀k ≥ N ) ∥xk − x∥ < ε. These two definitions, obviously, depends on the chosen norm, not just the vector space V . However, if two norms are equivalent, then they agree on what is bounded and what converges. Proposition. If ∥ · ∥ and ∥ · ∥′ are Lipschitz equivalent norms on a vector space V , then (i) A subset E ⊆ V is bounded with respect to ∥ · ∥ if and only if it is bounded with respect to ∥ · ∥′. (ii) A sequence xk converges to x with respect to ∥ · ∥ if and o
nly if it converges to x with respect to ∥ · ∥′. Proof. (i) This is direct from definition of equivalence. (ii) Say we have a, b such that a∥y∥ ≤ ∥y∥′ ≤ b∥y∥ for all y. So a∥xk − x∥ ≤ ∥xk − x∥′ ≤ b∥xk − x∥. So ∥xk − x∥ → 0 if and only if ∥xk − x∥′ → 0. So done. What if the norms are not equivalent? It is not surprising that there are some sequences that converge with respect to one norm but not another. More surprisingly, it is possible that a sequence converges to different limits under different norms. This is, again, on the second example sheet. We have some easy facts about convergence: Proposition. Let (V, ∥ · ∥) be a normed space. Then (i) If xk → x and xk → y, then x = y. 30 4 Rn as a normed space IB Analysis II (ii) If xk → x, then axk → ax. (iii) If xk → x, yk → y, then xk + yk → x + y. Proof. (i) ∥x − y∥ ≤ ∥x − xk∥ + ∥xk − y∥ → 0. So ∥x − y∥ = 0. So x = y. (ii) ∥axk − ax∥ = \|a\|∥xk − x∥ → 0. (iii) ∥(xk + yk) − (x + y)∥ ≤ ∥xk − x∥ + ∥yk − y∥ → 0. Proposition. Convergence in Rn (with respect to, say, the Euclidean norm) is equivalent to coordinate-wise convergence, i.e. x(k) → x if and only if x(k) j → xj for all j. Proof. Fix ε > 0. Suppose x(k) → x. Then there is some N such that for any k ≥ N such that ∥x(k) − x∥2 2 = (x(k) j − xj)2 < ε. n j=1 Hence \|x(k) j − xj\| < ε for all k ≤ N . On the other hand, for any fixed j, there is some Nj such that k ≥ Nj implies j − xj\| < ε√ n . So if k ≥ max{Nj : j = 1, · · · , n}, then \|x(k) ∥x(k) − x∥2 =   n  1 2 (x(k) j − xj)2  < ε. So done j=1 Another space we would like to understand is the space of continuous functions. It should be clear that uniform convergence is the same as convergence under the uniform norm, hence the name. However, there is no norm such that convergence under the norm is equivalent to pointwise convergence, i.e. pointwise convergence is not normable. In fact, it is not even metrizable. However, we will not prove this. We’ll now generalize the Bolzano-Weierstrass theorem to Rn. Theorem (Bolzano-Weierstrass theorem in Rn). Any bounded sequence in Rn (with, say, the Euclidean norm) has a convergent subsequence. Proof. We induct on n. The n = 1 case is the usual Bolzano-Weierstrass on the real line, which was proved in IA Analysis I. Assume the theorem holds in Rn−1, and let x(k) = (x(k) bounded sequence in Rn. Then let y(k) = (x(k) know that 1 , · · · , x(k) 1 , · · · , x(k) n ) be a n−1). Since for any k, we ∥y(k)∥2 + \|x(k) n \|2 = ∥x(k)∥2, it follows that both (y(k)) and (x(k) n ) are bounded. So by the induction hypothesis, there is a subsequence (kj) of (k) and some y ∈ Rn−1 such that y(kj ) → y. Also, 31 4 Rn as a normed space IB Analysis II by Bolzano-Weierstrass in R, there is a further subsequence (x that converges to, say, yn ∈ R. Then we know that (kjℓ ) n ) of (x(kj ) n ) So done. x(kjℓ ) → (y, yn). Note that this is generally not true for normed spaces. Finite-dimensionality is important for both of these results. Example. Consider (ℓ∞, ∥ · ∥∞). We let e(k) j = δjk be the sequence with 1 in the kth component and 0 in other components. Then e(k) j → 0 for all fixed j, and hence e(k) converges componentwise to the zero element 0 = (0, 0, · · · ). However, e(k) does not converge to the zero element since ∥e(k) − 0∥∞ = 1 for all k. Also, this is bounded but does not have a convergent subsequence for the same reasons. We know that all finite dimensional vector spaces are isomorphic to Rn as vector spaces for some n, and we will later show that all norms on finite dimensional spaces are equivalent. This means every finite-dimensional normed space satisfies the Bolzano-Weierstrass property. Is the converse true? If a normed vector space satisfies the Bolzano-Weierstrass property, must it be finite dimensional? The answer is yes, and the proof is in the example sheet. Example. Let C([0, 1]) have the ∥ · ∥L2 norm. Consider fn(x) = sin 2nπx. We know that 1 ∥fn∥2 L2 = \|fn\|2 = 1 2 . 0 So it is bounded. However, it doesn’t have a convergent subsequence. If it did, say fnj → f in L2, then we must have ∥fnj − fnj+1∥2 → 0. However, by direct calculation, we know that ∥fnj − fnj+1∥2 = 1 0 (sin 2njπx − sin 2nj+1πx)2 = 1. Note that the same argument shows also that the sequence (sin 2nπx) has no subsequence that converges pointwise on [0, 1]. To see this, we need the result that if (fj) is a sequence in C([0, 1]) that is uniformly bounded with fj → f pointwise, then fj converges to f under the L2 norm. However, we will not be able to prove this (in a nice way) without Lebesgue integration from IID Probability and Measure. 4.2 Cauchy sequences and completeness Definition (Cauchy sequence). Let (V, ∥ · ∥) be a normed space. A sequence (x(k)) in V is a Cauchy sequence if (∀ε)(∃N )(∀n, m ≥ N ) ∥x(n) − x(m)∥ < ε. 32 4 Rn as a normed space IB Analysis II Definition (Complete normed space). A normed space (V, ∥ · ∥) is complete if every Cauchy sequence converges to an element in V . We’ll start with some easy facts about Cauchy sequences and complete spaces. Proposition. Any convergent sequence is Cauchy. Proof. If xk → x, then ∥xk − xℓ∥ ≤ ∥xk − x∥ + ∥xℓ − x∥ → 0 as k, ℓ → ∞. Proposition. A Cauchy sequence is bounded. Proof. By definition, there is some N such that for all n ≥ N , we have ∥xN − xn∥ < 1. So ∥xn∥ < 1 + ∥xN ∥ for n ≥ N . So, for all n, ∥xn∥ ≤ max{∥x1∥, · · · , ∥xN −1∥, 1 + ∥xN ∥}. Proposition. If a Cauchy sequence has a subsequence converging to an element x, then the whole sequence converges to x. Proof. Suppose xkj → x. Since (xk) is Cauchy, given ε > 0, we can choose an N such that ∥xn − xm∥ < ε 2 for all n, m ≥ N . We can also choose j0 such that − x∥ < ε 2 . Then for any n ≥ N , we have kj0 ≥ n and ∥xkj0 ∥xn − x∥ ≤ ∥xn − xkj0 ∥ + ∥x − xkj0 ∥ < ε. Proposition. If ∥ · ∥′ is Lipschitz equivalent to ∥ · ∥ on V , then (xk) is Cauchy with respect to ∥ · ∥ if and only if (xk) is Cauchy with respect to ∥ · ∥′. Also, (V, ∥ · ∥) is complete if and only if (V, ∥ · ∥′) is complete. Proof. This follows directly from definition. Theorem. Rn (with the Euclidean norm, say) is complete. Proof. The important thing is to know this is true for n = 1, which we have proved from Analysis I. If (xk) is Cauchy in Rn, then (x(k) ) is a Cauchy sequence of real numbers for j j → xj ∈ R each j ∈ {1, · · · , n}. By the completeness of the reals, we know that xk for some x. So xk → x = (x1, · · · , xn) since convergence in Rn is equivalent to componentwise convergence. Note that the spaces ℓ1, ℓ2, ℓ∞ are all complete with respect to the standard norms. Also, C([0, 1]) is complete with respect to ∥ · ∥∞, since uniform Cauchy convergence implies uniform convergence, and the uniform limit of continuous functions is continuous. However, C([0, 1]) with the L1 or L2 norms are not complete (see example sheet). The incompleteness of L1 tells us that C([0, 1]) is not large enough to to be complete under the L1 or L2 norm. In fact, the space of Riemann integrable functions, say R([0, 1]), is the natural space for the L1 norm, and of course contains C([0, 1]). As we have previously mentioned, this time R([0, 1]) is too large for ∥ · ∥ to be a norm, since 1 0 \|f \| dx = 0 does not imply f = 0. This is a problem we can solve. We just have to take the equivalence classes of Riemann integrable functions, where f and g are equivalent if 1 0 \|f − g\| dx = 0. But still, 33 4 Rn as a normed space IB Analysis II L1 is not complete on R([0, 1])/∼. This is a serious problem in the Riemann integral. This eventually lead to the Lebesgue integral, which generalizes the Riemann integral, and gives a complete normed space. Note that when we quotient our R([0, 1]) by the equivalence relation f ∼ g if 1 0 \|f − g\| dx = 0, we are not losing too much information about our functions. We know that for the integral to be zero, f − g cannot be non-zero at a point of continuity. Hence they agree on all points of continuities. We also know that by Lebesgue’s theorem, the set of points of discontinuity has Lebesgue measure zero. So they disagree on at most a set of Lebesgue measure zero. Example. Let V = {(xn) ∈ RN : xj = 0 for all but finitely many j}. Take the supremum norm ∥ · ∥∞ on V . This is a subspace of ℓ∞ (and is sometimes denoted ℓ0). Then (V, ∥ · ∥∞) is not complete. We define x(k) = (1, 1 k , 0, 0, · · · ) for k = 1, 2, 3, · · · . Then this is Cauchy, since x(k) − x(ℓ)∥ = 1 min{ℓ, k} + 1 → 0, but it is not convergent in V . If it actually converged to some x, then x(k) So we must have xj = 1 j , but this sequence not in V . j → xj. We will later show that this is because V is not closed, after we define what it means to be closed. Definition (Open set). Let (V, ∥ · ∥) be a normed space. A subspace E ⊆ V is open in V if for any y ∈ E, there is some r > 0 such that Br(y) = {x ∈ V : ∥x − y∥ < r} ⊆ E. We first check that the open ball is open. Proposition. Br(y) ⊆ V is an open subset for all r > 0, y ∈ V . Proof. Let x ∈ Br(y). Let ρ = r − ∥x − y∥ > 0. Then Bρ(x) ⊆ Br(y). x y Definition (Limit point). Let (V, ∥ · ∥) be a normed space, E ⊆ V . A point y ∈ V is a limit point of E if there is a sequence (xk) in E with xk ̸= y for all k and xk → y. (Some people allow xk = y, but we will use this definition in this course) Example. Let V = R, E = (0, 1). Then 0, 1 are limit points of E. The set of all limit points is [0, 1]. If E′ = (0, 1) ∪ {2}. Then the set of limit points of E′ is still [0, 1]. 34 4 Rn as a normed space IB Analysis II There is a nice result characterizing whether a set contains all its limit points. Proposition. Let E ⊆ V . Then E contains all of its limit points if and only if V \ E is open in V . Using this proposition, we define the following: Definition (Closed set). Let (V, ∥ · ∥) be a normed space. Then E ⊆ V is closed if V \ E is open, i.e. E contains all its limit points. Note that sets can be both closed or open; or neither closed nor open. Before we prove the proposition, we first have a lemma: Lemma. Let (V, ∥ · ∥) be a normed space, E any subset of V . Then a point y ∈ V
is a limit point of E if and only if (Br(y) \ {y}) ∩ E ̸= ∅ for every r. Proof. (⇒) If y is a limit point of E, then there exists a sequence (xk) ∈ E with xk ̸= y for all k and xk → y. Then for every r, for sufficiently large k, xk ∈ Br(y). Since xk ̸= {y} and xk ∈ E, the result follows. (⇐) For each k, let r = 1 k . By assumption, we have some xk ∈ (B 1 {y}) ∩ E. Then xk → y, xk ̸= y and xk ∈ E. So y is a limit point of E. k (y) \ Now we can prove our proposition. Proposition. Let E ⊆ V . Then E contains all of its limit points if and only if V \ E is open in V . Proof. (⇒) Suppose E contains all its limit points. To show V \ E is open, we let y ∈ V \ E. So y is not a limit point of E. So for some r, we have (Br(y) \ {y}) ∩ E = ∅. Hence it follows that Br(y) ⊆ V \ E (since y ̸∈ E). (⇐) Suppose V \ E is open. Let y ∈ V \ E. Since V \ E is open, there is some r such that Br(y) ⊆ V \ E. By the lemma, y is not a limit point of E. So all limit points of E are in E. 4.3 Sequential compactness In general, there are two different notions of compactness — “sequential compactness” and just “compactness”. However, in normed spaces (and metric spaces, as we will later encounter), these two notions are equivalent. So we will be lazy and just say “compactness” as opposed to “sequential compactness”. Definition ((Sequentially) compact set). Let V be a normed vector space. A subset K ⊆ V is said to be compact (or sequentially compact) if every sequence in K has a subsequence that converges to a point in K. There are things we can immediately know about the spaces: Theorem. Let (V, ∥ · ∥) be a normed vector space, K ⊆ V a subset. Then (i) If K is compact, then K is closed and bounded. 35 4 Rn as a normed space IB Analysis II (ii) If V is Rn (with, say, the Euclidean norm), then if K is closed and bounded, then K is compact. Proof. (i) Let K be compact. Boundedness is easy: if K is unbounded, then we can generate a sequence xk such that ∥xk∥ → ∞. Then this cannot have a convergent subsequence, since any subsequence will also be unbounded, and convergent sequences are bounded. So K must be bounded. To show K is closed, let y be a limit point of K. Then there is some yk ∈ K such that yk → y. Then by compactness, there is a subsequence of yk converging to some point in K. But any subsequence must converge to y. So y ∈ K. (ii) Let K be closed and bounded. Let xk be a sequence in K. Since V = Rn and K is bounded, (xk) is a bounded sequence in Rn. So by BolzanoWeierstrass, this has a convergent subsequence xkj . By closedness of K, we know that the limit is in K. So K is compact. 4.4 Mappings between normed spaces We are now going to look at functions between normed spaces, and see if they are continuous. Let (V, ∥ · ∥), (V ′, ∥ · ∥′) be normed spaces, and let E ⊆ K be a subset, and f : E → V ′ a mapping (which is just a function, although we reserve the terminology “function” or “functional” for when V ′ = R). Definition (Continuity of mapping). Let y ∈ E. We say f : E → V ′ is continuous at y if for all ε > 0, there is δ > 0 such that the following holds: (∀x ∈ E) ∥x − y∥V < δ ⇒ ∥f (x) − f (y)∥V ′ < ε. Note that x ∈ E and ∥x − y∥ < δ is equivalent to saying x ∈ Bδ(y) ∩ E. Similarly, ∥f (x) − f (y)∥ < ε is equivalent to f (x) ∈ Bε(f (y)). In other words, x ∈ f −1(Bε(f (y))). So we can rewrite this statement as there is some δ > 0 such that E ∩ Bδ(y) ⊆ f −1(Bε(f (y))). We can use this to provide an alternative characterization of continuity. Theorem. Let (V, ∥ · ∥), (V ′, ∥ · ∥′) be normed spaces′. Then f is continuous at y ∈ E if and only if for any sequence yk → y in E, we have f (yk) → f (y). Proof. (⇒) Suppose f is continuous at y ∈ E, and that yk → y. Given ε > 0, by continuity, there is some δ > 0 such that Bδ(y) ∩ E ⊆ f −1(Bε(f (y))). For sufficiently large k, yk ∈ Bδ(y) ∩ E. So f (yk) ∈ Bε(f (y)), or equivalently, So done. \|f (yk) − f (y)\| < ε. 36 4 Rn as a normed space IB Analysis II (⇐) If f is not continuous at y, then there is some ε > 0 such that for any k, we have B 1 k (y) ̸⊆ f −1(Bε(f (y))). Choose yk ∈ B 1 f (y)∥ ≥ ε, contrary to the hypothesis. k (y) \ f −1(Bε(f (y))). Then yk → y, yk ∈ E, but ∥f (yk) − Definition (Continuous function). f : E → V ′ is continuous if f is continuous at every point y ∈ E. Theorem. Let (V, ∥ · ∥) and (V ′, ∥ · ∥′) be normed spaces, and K a compact subset of V , and f : V → V ′ a continuous function. Then (i) f (K) is compact in V ′ (ii) f (K) is closed and bounded (iii) If V ′ = R, then the function attains its supremum and infimum, i.e. there is some y1, y2 ∈ K such that f (y1) = sup{f (y) : y ∈ K}, f (y2) = inf{f (y) : y ∈ K}. Proof. (i) Let (xk) be a sequence in f (K) with xk = f (yk) for some yk ∈ K. By compactness of K, there is a subsequence (ykj ) such that ykj → y. By the previous theorem, we know that f (yjk ) → f (y). So xkj → f (y) ∈ f (K). So f (K) is compact. (ii) This follows directly from (i), since every compact space is closed and bounded. (iii) If F is any bounded subset of R, then either sup F ∈ F or sup F is a limit point of F (or both), by definition of the supremum. If F is closed and bounded, then any limit point must be in F . So sup F ∈ F . Applying this fact to F = f (K) gives the desired result, and similarly for infimum. Finally, we will end the chapter by proving that any two norms on a finite dimensional space are Lipschitz equivalent. The key lemma is the following: Lemma. Let V be an n-dimensional vector space with a basis {v1, · · · , vn}. Then for any x ∈ V , write x = n j=1 xjvj, with xj ∈ R. We define the Euclidean norm by ∥x∥2 = 1 2 . x2 j Then this is a norm, and S = {x ∈ V : ∥x∥2 = 1} is compact in (V, ∥ · ∥2). After we show this, we can easily show that every other norm is equivalent to this norm. This is not hard to prove, since we know that the unit sphere in Rn is compact, and we can just pass our things on to Rn. 37 4 Rn as a normed space IB Analysis II Proof. ∥ · ∥2 is well-defined since x1, · · · , xn are uniquely determined by x (by (a certain) definition of basis). It is easy to check that ∥ · ∥2 is a norm. Given a sequence x(k) in S, if we write x(k) = n j=1 x(k) j vj. We define the following sequence in Rn: ˜x(k) = (x(k) 1 , · · · , x(k) n ) ∈ ˜S = {˜x ∈ Rn : ∥˜x∥Euclid = 1}. As ˜S is closed and bounded in Rn under the Euclidean norm, it is compact. Hence there exists a subsequence ˜x(kj ) and ˜x ∈ ˜S such that ∥˜x(kj ) − ˜x∥Euclid → 0. This says that x = n j=1 xjvj ∈ S, and ∥xkj − x∥2 → 0. So done. Theorem. Any two norms on a finite dimensional vector space are Lipschitz equivalent. The idea is to pick a basis, and prove that any norm is equivalent to ∥ · ∥2. To show that an arbitrary norm ∥ · ∥ is equivalent to ∥ · ∥2, we have to show that for any ∥x∥, we have We can divide by ∥x∥2 and obtain an equivalent requirement: a∥x∥2 ≤ ∥x∥ ≤ b∥x∥2. a ≤ x ∥x∥2 ≤ b. We know that any x/∥x∥2 lies in the unit sphere S = {x ∈ V : ∥x∥2 = 1}. So we want to show that the image of ∥ · ∥ is bounded. But we know that S is compact. So it suffices to show that ∥ · ∥ is continuous. Proof. Fix a basis {v1, · · · , vn} for V , and define ∥ · ∥2 as in the lemma above. Then ∥ · ∥2 is a norm on V , and S = {x ∈ V : ∥x∥2 = 1}, the unit sphere, is compact by above. To show that any two norms are equivalent, it suffices to show that if ∥ · ∥ is any other norm, then it is equivalent to ∥ · ∥2, since equivalence is transitive. For any we have x = n j=1 xjvj, xjvj n j=1 \|xj\|∥vj∥ ∥xvj∥2  ≤ ∥x∥2 j=1 by the Cauchy-Schwarz inequality. So ∥x∥ ≤ b∥x∥2 for b = ∥vj∥2 1 2 . To find a such that ∥x∥ ≥ a∥x∥2, consider ∥ · ∥ : (S, ∥ · ∥2) → R. By above, we know that ∥x − y∥ ≤ b∥x − y∥2 38 4 Rn as a normed space IB Analysis II By the triangle inequality, we know that ≤ ∥x − y∥. So when x is close to y under ∥ · ∥2, then ∥x∥ and ∥y∥ are close. So ∥ · ∥ : (S, ∥ · ∥2) → R is continuous. So there is some x0 ∈ S such that ∥x0∥ = inf x∈S ∥x∥ = a, say. Since ∥x∥ > 0, we know that ∥x0∥ > 0. So ∥x∥ ≥ a∥x∥2 for all x ∈ V . ∥x∥ − ∥y∥ The key to the proof is the compactness of the unit sphere of (V, ∥ · ∥). On the other hand, compactness of the unit sphere also characterizes finite dimensionality. As you will show in the example sheets, if the unit sphere of a space is compact, then the space must be finite-dimensional. Corollary. Let (V, ∥ · ∥) be a finite-dimensional normed space. (i) The Bolzano-Weierstrass theorem holds for V , i.e. any bounded sequence sequence in V has a convergent subsequence. (ii) A subset of V is compact if and only if it is closed and bounded. Proof. If a subset is bounded in one norm, then it is bounded in any Lipschitz equivalent norm. Similarly, if it converges to x in one norm, then it converges to x in any Lipschitz equivalent norm. Since these results hold for the Euclidean norm ∥ · ∥2, it follows that they hold for arbitrary finite-dimensional vector spaces. Corollary. Any finite-dimensional normed vector space (V, ∥ · ∥) is complete. Proof. This is true since if a space is complete in one norm, then it is complete in any Lipschitz equivalent norm, and we know that Rn under the Euclidean norm is complete. 39 5 Metric spaces IB Analysis II 5 Metric spaces We would like to extend our notions such as convergence, open and closed subsets, compact subsets and continuity from normed spaces to more general sets. Recall that when we defined these notions, we didn’t really use the vector space structure of a normed vector space much. Moreover, we mostly defined these things in terms of convergence of sequences. For example, a space is closed if it contains all its limits, and a space is open if its complement is closed. So what do we actually need in order to define convergence, and hence all the notions we’ve been using? Recall we define xk → x to mean ∥xk − x∥ → 0 as a sequence in R. What is ∥xk − x∥ really about? It is measuring the distance between xk and x. So what we really need is a measure of distance. To do so, we can define a distance function d : V ×V → R
by d(x, y) = ∥x−y∥. Then we can define xk → x to mean d(xk, x) → 0. Hence, given any function d : V × V → R, we can define a notion of “convergence” as above. However, we want this to be well-behaved. In particular, we would want the limits of sequences to be unique, and any constant sequence xk = x should converge to x. We will come up with some restrictions on what d can be based on these requirements. We can look at our proof of uniqueness of limits (for normed spaces), and see what properties of d we used. Recall that to prove the uniqueness of limits, we first assume that xk → x and xk → y. Then we noticed ∥x − y∥ ≤ ∥x − xk∥ + ∥xk − y∥ → 0, and hence ∥x − y∥ = 0. So x = y. We can reformulate this argument in terms of d. We first start with d(x, y) ≤ d(x, xk) + d(xk, y). To obtain this equation, we are relying on the triangle inequality. So we would want d to satisfy the triangle inequality. After obtaining this, we know that d(xk, y) → 0, since this is just the definition of convergence. However, we do not immediately know d(x, xk) → 0, since we are given a fact about d(xk, x), not d(x, xk). Hence we need the property that d(xk, x) = d(x, xk). This is symmetry. Combining this, we know that d(x, y) ≤ 0. From this, we want to say that in fact, d(x, y) = 0, and thus x = y. Hence we need the property that d(x, y) ≥ 0 for all x, y, and that d(x, y) = 0 implies x = y. Finally, to show that a constant sequence has a limit, suppose xk = x for all k ∈ N. Then we know that d(x, xk) = d(x, x) should tend to 0. So we must have d(x, x) = 0 for all x. We will use these properties to define metric spaces. 5.1 Preliminary definitions Definition (Metric space). Let X be any set. A metric on X is a function d : X × X → R that satisfies 40 5 Metric spaces IB Analysis II – d(x, y) ≥ 0 with equality iff x = y – d(x, y) = d(y, x) – d(x, y) ≤ d(x, z) + d(z, y) The pair (X, d) is called a metric space. (non-negativity) (symmetry) (triangle inequality) We have seen that we can define convergence in terms of a metric. Hence, we can also define open subsets, closed subsets, compact spaces, continuous functions etc. for metric spaces, in a manner consistent with what we had for normed spaces. Moreover, we will show that many of our theorems for normed spaces are also valid in metric spaces. Example. (i) Rn with the Euclidean metric is a metric space, where the metric is defined by d(x, y) = ∥x − y∥ = (xj − yj)2. (ii) More generally, if (V, ∥ · ∥) is a normed space, then d(x, y) = ∥x − y∥ defines a metric on V . (iii) Discrete metric: let X be any set, and define d(x, y) = 0 x = y 1 x ̸= y . (iv) Given a metric space (X, d), we define g(x, y) = min{1, d(x, y)}. Then this is a metric on X. Similarly, if we define h(x, y) = d(x, y) 1 + d(x, y) is also a metric on X. In both cases, we obtain a bounded metric. The axioms are easily shown to be satisfied, apart from the triangle inequality. So let’s check the triangle inequality for h. We’ll use a general fact that for numbers a, c ≥ 0, b, d > 0 we have ⇔ ≤ ≤ . Based on this fact, we can start with d(x, y) ≤ d(x, z) + d(z, y). Then we obtain d(x, y) 1 + d(x, y) So done. ≤ = ≤ d(x, z) + d(z, y) 1 + d(x, z) + d(z, y) d(x, z) 1 + d(x, z) + d(z, y) + d(z, y) 1 + d(x, z) + d(z, y) d(x, z) 1 + d(x, z) + d(z, y) 1 + d(z, y) . 41 5 Metric spaces IB Analysis II We can also extend the notion of Lipschitz equivalence to metric spaces. Definition (Lipschitz equivalent metrics). Metrics d, d′ on a set X are said to be Lipschitz equivalent if there are (positive) constants A, B such that Ad(x, y) ≤ d′(x, y) ≤ Bd(x, y) for all x, y ∈ X. Clearly, any Lipschitz equivalent norms give Lipschitz equivalent metrics. Any metric coming from a norm in Rn is thus Lipschitz equivalent to the Euclidean metric. We will later show that two equivalent norms induce the same topology. In some sense, Lipschitz equivalent norms are indistinguishable. Definition (Metric subspace). Given a metric space (X, d) and a subset Y ⊆ X, the restriction d\|Y ×Y → R is a metric on Y . This is called the induced metric or subspace metric. Note that unlike vector subspaces, we do not require our subsets to have any structure. We can take any subset of X and get a metric subspace. Example. Any subspace of Rn is a metric space with the Euclidean metric. Definition (Convergence). Let (X, d) be a metric space. A sequence xn ∈ X is said to converge to x if d(xn, x) → 0 as a real sequence. In other words, (∀ε)(∃K)(∀k > K) d(xk, x) < ε. Alternatively, this says that given any ε, for sufficiently large k, we get xk ∈ Bε(x). Again, Br(a) is the open ball centered at a with radius r, defined as Br(a) = {x ∈ X : d(x, a) < r}. Proposition. The limit of a convergent sequence is unique. Proof. Same as that of normed spaces. Note that notions such as convergence, open and closed subsets and continuity of mappings all make sense in an even more general setting called topological spaces. However, in this setting, limits of convergent sequences can fail to be unique. We will not worry ourselves about these since we will just focus on metric spaces. 5.2 Topology of metric spaces We will define open subsets of a metric space in exactly the same way as we did for normed spaces. Definition (Open subset). Let (X, d) be a metric space. A subset U ⊆ X is open if for every y ∈ U , there is some r > 0 such that Br(y) ⊆ U . 42 5 Metric spaces IB Analysis II This means we can write any open U as a union of open balls: U = y∈U Br(y)(y) for appropriate choices of r(y) for every y. It is easy to check that every open ball Br(y) is an open set. The proof is exactly the same as what we had for normed spaces. Note that two different metrics d, d′ on the same set X may give rise to the same collection of open subsets. Example. Lipschitz equivalent metrics give rise to the same collection of open sets, i.e. if d, d′ are Lipschitz equivalent, then a subset U ⊆ X is open with respect to d if and only if it is open with respect to d′. Proof is left as an easy exercise. The converse, however, is not necessarily true. Example. Let X = R, d(x, y) = \|x − y\| and d′(x, y) = min{1, \|x − y\|}. It is easy to check that these are not Lipschitz equivalent, but they induce the same set collection of open subsets. Definition (Topology). Let (X, d) be a metric space. The topology on (X, d) is the collection of open subsets of X. We say it is the topology induced by the metric. Definition (Topological notion). A notion or property is said to be a topological notion or property if it only depends on the topology, and not the metric. We will introduce a useful terminology before we go on: Definition (Neighbourhood). Given a metric space X and a point x ∈ X, a neighbourhood of x is an open set containing x. Some people do not require the set to be open. Instead, it requires a neighbourhood to be a set that contains an open subset that contains x, but this is too complicated, and we could as well work with open subsets directly. Clearly, being a neighbourhood is a topological property. Proposition. Let (X, d) be a metric space. Then xk → x if and only if for every neighbourhood V of x, there exists some K such that xk ∈ V for all k ≥ K. Hence convergence is a topological notion. Proof. (⇒) Suppose xk → X, and let V be any neighbourhood of x. Since V is open, by definition, there exists some ε such that Bε(x) ⊆ V . By definition of convergence, there is some K such that xk ∈ Bε(x) for k ≥ K. So xk ∈ V whenever k ≥ K. (⇒) Since every open ball is a neighbourhood, this direction follows directly from definition. Theorem. Let (X, d) be a metric space. Then (i) The union of any collection of open sets is open (ii) The intersection of finitely many open sets is open. 43 5 Metric spaces IB Analysis II (iii) ∅ and X are open. Proof. (i) Let U = α Vα, where each Vα is open. If x ∈ U , then x ∈ Vα for some α. Since Vα is open, there exists δ > 0 such that Bδ(x) ⊆ Vα. So Bδ(x) ⊆ (ii) Let U = n i=1 Vα, where each Vα is open. If x ∈ V , then x ∈ Vi for all i = 1, · · · , n. So ∃δi > 0 with Bδi(x) ⊆ Vi. Take δ = min{δ1, · · · , δn}. So Bδ(x) ⊆ Vi for all i. So Bδ(x) ⊆ V . So V is open. α Vα = U . So U is open. (iii) ∅ satisfies the definition of an open subset vacuously. X is open since for any x, B1(x) ⊆ X. This theorem is not important in this course. However, this will be a key defining property we will use when we define topological spaces in IB Metric and Topological Spaces. We can now define closed subsets and characterize them using open subsets, in exactly the same way as for normed spaces. Definition (Limit point). Let (X, d) be a metric space and E ⊆ X. A point y ∈ X is a limit point of E if there exists a sequence xk ∈ E, xk ̸= y such that xk → y. Definition (Closed subset). A subset E ⊆ X is closed if E contains all its limit points. Proposition. A subset is closed if and only if its complement is open. Proof. Exactly the same as that of normed spaces. It is useful to observe that y ∈ X is a limit point of E if and only if (Br(y) \ {y}) ∩ E ̸= ∅ for all r > 0. We can write down an analogous theorem for closed sets: Theorem. Let (X, d) be a metric space. Then (i) The intersection of any collection of closed sets is closed (ii) The union of finitely many closed sets is closed. (iii) ∅ and X are closed. Proof. By taking complements of the result for open subsets. Proposition. Let (X, d) be a metric space and x ∈ X. Then the singleton {x} is a closed subset, and hence any finite subset is closed. Proof. Let y ∈ X \ {x}. So d(x, y) > 0. Then Bd(y,x)(x) ⊆ X \ {x}. So X \ {x} is open. So {x} is closed. Alternatively, since {x} has no limit points, it contains all its limit points. So it is closed. 44 5 Metric spaces IB Analysis II 5.3 Cauchy sequences and completeness Definition (Cauchy sequence). Let (X, d) be a metric space. A sequence (xn) in X is Cauchy if (∀ε)(∃N )(∀n, m ≥ N ) d(xn, xm) < ε. Proposition. Let (X, d) be a metric space. Then (i) Any convergent sequence is Cauchy. (ii) If a Cauchy sequ
ence has a convergent subsequence, then the original sequence converges to the same limit. Proof. (i) If xk → x, then as m, n → ∞. d(xm, xn) ≤ d(xm, x) + d(xn, x) → 0 (ii) Suppose xkj → x. Since (xk) is Cauchy, given ε > 0, we can choose an 2 for all n, m ≥ N . We can also choose j0 such N such that d(xn, xm) < ε that kj0 ≥ n and d(xkj0 , x) < ε 2 . Then for any n ≥ N , we have d(xn, x) ≤ d(xn, xkj0 ) + d(x, xkj0 ) < ε. Definition (Complete metric space). A metric space (X, d) is complete if all Cauchy sequences converge to a point in X. Example. Let X = Rn with the Euclidean metric. Then X is complete. It is easy to produce incomplete metric spaces. Since arbitrary subsets of metric spaces are subspaces, we can just remove some random elements to make it incomplete. Example. Let X = (0, 1) ⊆ R with the Euclidean metric. Then this is incomplete, since 1 is Cauchy but has no limit in X. Similarly, X = R \ {0} is incomplete. Note, however, that it is possible to construct a metric d′ on X = R \ {0} such that d′ induces the same topology on X, but makes X complete. This shows that completeness is not a topological property. The actual construction is left as an exercise on the example sheet. k Example. We can create an easy example of an incomplete metric on Rn. We start by defining h : Rn → Rn by h(x) = x 1 + ∥x∥ , where ∥ · ∥ is the Euclidean norm. We can check that this is injective: h(x) = h(y), taking the norm gives if ∥x∥ 1 + ∥x∥ = ∥y∥ 1 + ∥y∥ . 45 5 Metric spaces IB Analysis II So we must have ∥x∥ = ∥y∥, i.e. x = y. So h(x) = h(y) implies x = y. Now we define d(x, y) = ∥h(x) − h(y)∥. It is an easy check that this is a metric on Rn. In fact, we can show that h : Rn → B1(0), and h is a homeomorphism (i.e. continuous bijection with continuous inverse) between Rn and the unit ball B1(0), both with the Euclidean metric. To show that this metric is incomplete, we can consider the sequence xk = (k − 1)e1, where e1 = (1, 0, 0, · · · , 0) is the usual basis vector. Then (xk) is Cauchy in (Rn, d). To show this, first note that h(xk) = 1 − e1. 1 k Hence we have d(xn, xm) = ∥h(xn) − h(xm)∥ = 1 n − 1 m → 0. So it is Cauchy. To show it does not converge in (Rn, d), suppose d(xk, x) → 0 for some x. Then since d(xk, x) = ∥h(xk) − h(x)∥ ≥ ∥h(xk)∥ − ∥h(x)∥ , We must have However, there is no element with ∥h(x)∥ = 1. ∥h(x)∥ = lim k→∞ ∥h(xk)∥ = 1. What is happening in this example, is that we are pulling in the whole Rn in to the unit ball. Then under this norm, a sequence that “goes to infinity” in the usual norm will be Cauchy in this norm, but we have nothing at infinity for it to converge to. Suppose we have a complete metric space (X, d). We know that we can form arbitrary subspaces by taking subsets of X. When will this be complete? Clearly it has to be closed, since it has to include all its limit points. It turns it closedness is a sufficient condition. Theorem. Let (X, d) be a metric space, Y ⊆ X any subset. Then (i) If (Y, d\|Y ×Y ) is complete, then Y is closed in X. (ii) If (X, d) is complete, then (Y, d\|Y ×Y ) is complete if and only if it is closed. Proof. (i) Let x ∈ X be a limit point of Y . Then there is some sequence xk → x, where each xk ∈ Y . Since (xk) is convergent, it is a Cauchy sequence. Hence it is Cauchy in Y . By completeness of Y , (xk) has to converge to some point in Y . By uniqueness of limits, this limit must be x. So x ∈ Y . So Y contains all its limit points. (ii) We have just showed that if Y is complete, then it is closed. Now suppose Y is closed. Let (xk) be a Cauchy sequence in Y . Then (xk) is Cauchy in X. Since X is complete, xk → x for some x ∈ X. Since x is a limit point of Y , we must have x ∈ Y . So xk converges in Y . 46 5 Metric spaces IB Analysis II 5.4 Compactness Definition ((Sequential) compactness). A metric space (X, d) is (sequentially) compact if every sequence in X has a convergent subsequence. A subset K ⊆ X is said to be compact if (K, d\|K×K) is compact. In other words, K is compact if every sequence in K has a subsequence that converges to some point in K. Note that when we say every sequence has a convergent subsequence, we do not require it to be bounded. This is unlike the statement of the BolzanoWeierstrass theorem. In particular, R is not compact. It follows from definition that compactness is a topological property, since it is defined in terms of convergence, and convergence is defined in terms of open sets. The following theorem relates completeness with compactness. Theorem. All compact spaces are complete and bounded. Note that X is bounded iff X ⊆ Br(x0) for some r ∈ R, x0 ∈ X (or X is empty). Proof. Let (X, d) be a compact metric space. Let (xk) be Cauchy in X. By compactness, it has some convergent subsequence, say xkj → x. So xk → x. So it is complete. If (X, d) is not bounded, by definition, for any x0, there is a sequence (xk) such that d(xk, x0) > k for every k. But then (xk) cannot have a convergent subsequence. Otherwise, if xkj → x, then d(xkj , x0) ≤ d(xkj , x) + d(x, x0) and is bounded, which is a contradiction. This implies that if (X, d) is a metric space and E ⊆ X, and E is compact, then E is bounded, i.e. E ⊆ BR(x0) for some x0 ∈ X, R > 0, and E with the subspace metric is complete. Hence E is closed as a subset of X. The converse is not true. For example, recall if we have an infinite-dimensional normed vector space, then the closed unit sphere is complete and bounded, but not compact. Alternatively, we can take X = R with the metric d(x, y) = min{1, \|x − y\|}. This is clearly bounded (by 1), and it is easy to check that this is complete. However, this is not compact since the sequence xk = k has no convergent subsequence. However, we can strengthen the condition of boundedness to total boundedness, and get the equivalence between “completeness and total boundedness” and compactness. Definition (Totally bounded*). A metric space (X, d) is said to be totally bounded if for all ε > 0, there is an integer N ∈ N and points x1, · · · , xN ∈ X such that N X = Bε(xi). It is easy to check that being totally bounded implies being bounded. We then have the following strengthening of the previous theorem. i=1 47 5 Metric spaces IB Analysis II Theorem. (non-examinable) Let (X, d) be a metric space. Then X is compact if and only if X is complete and totally bounded. Proof. (⇐) Let X be complete and totally bounded, (yi) ∈ X. For every j ∈ N, there exists a finite set of points Ej such that every point is within 1 j of one of these points. Now since E1 is finite, there is some x1 ∈ E1 such that there are infinitely many yi’s in B(x1, 1). Pick the first yi in B(x1, 1) and call it yi1. Now there is some x2 ∈ E2 such that there are infinitely many yi’s in 2 ). Pick the one with smallest value of i > i1, and call this yi2. B(x1, 1) ∩ B(x2, 1 Continue till infinity. This procedure gives a sequence xi ∈ Ei and subsequence (yik ), and also yin ∈ n j=1 B xj, . 1 j It is easy to see that (yin) is Cauchy since if m > n, then d(yim , yin ) < 2 completeness of X, this subsequence converges. n . By (⇒) Compactness implying completeness is proved above. Suppose X is not totally bounded. We show it is not compact by constructing a sequence with no Cauchy subsequence. Suppose ε is such that there is no finite set of points x1, · · · , xN with X = N i=1 Bε(xi). We will construct our sequence iteratively. Start by picking an arbitrary y1. Pick y2 such that d(y1, y2) ≥ ε. This exists or else Bε(y1) covers all of X. Now given y1, · · · , yn such that d(yi, yj) ≥ ε for all i, j = 1, · · · , n, i ̸= j, we pick yn+1 such that d(yn+1, yj) ≥ ε for all j = 1, · · · , n. Again, this exists, or else n i=1 Bε(yi) covers X. Then clearly the sequence (yn) is not Cauchy. So done. In IID Linear Analysis, we will prove the Arzel`a-Ascoli theorem that characterizes the compact subsets of the space C([a.b]) in a very concrete way, which is in some sense a strengthening of this result. 5.5 Continuous functions We are going to look at continuous mappings between metric spaces. Definition (Continuity). Let (X, d) and (X ′, d′) be metric spaces. A function f : X → X ′ is continuous at y ∈ X if (∀ε > 0)(∃δ > 0)(∀x) d(x, y) < δ ⇒ d′(f (x), f (y)) < ε. This is true if and only if for every ε > 0, there is some δ > 0 such that Bδ(y) ⊆ f −1Bε(f (x)). f is continuous if f is continuous at each y ∈ X. 48 5 Metric spaces IB Analysis II Definition (Uniform continuity). f is uniformly continuous on X if (∀ε > 0)(∃δ > 0)(∀x, y ∈ X) d(x, y) < δ ⇒ d(f (x), f (y)) < ε. This is true if and only if for all ε, there is some δ such that for all y, we have Bδ(y) ⊆ f −1(Bε(f (y))). Definition (Lipschitz function and Lipschitz constant). f is said to be Lipschitz on X if there is some K ∈ [0, ∞) such that for all x, y ∈ X, d′(f (x), f (y)) ≤ Kd(x, y) Any such K is called a Lipschitz constant. It is easy to show Lipschitz ⇒ uniform continuity ⇒ continuity. We have seen many examples that continuity does not imply uniform continuity. To show that uniform continuity does not imply Lipschitz, take X = X ′ = R. We define the metrics as d(x, y) = min{1, \|x − y\|}, d′(x, y) = \|x − y\|. Now consider the function f : (X, d) → (X ′, d′) defined by f (x) = x. We can then check that this is uniformly continuous but not Lipschitz. Note that the statement that metrics d and d′ are Lipschitz equivalent is equivalent to saying the two identity maps i : (X, d) → (X, d′) and i′ : (X, d′) → (X, d) are Lipschitz, hence the name. Note also that the metric itself is also a Lipschitz map for any metric. Here we are viewing the metric as a function d : X × X → R, with the metric on X × X defined as ˜d((x1, y1), (x2, y2)) = d(x1, x2) + d(y1, y2). This is a consequence of the triangle inequality, since d(x1, y1) ≤ d(x1, x2) + d(x2, y2) + d(y1, y2). Moving the middle term to the left gives d(x1, y1) − d(x2, y2) ≤ ˜d((x1, y1), (x2, y2)) Swapping the theorems around, we can put in the absolute value to obtain \|d(x1, y1) − d(x2, y2)\| ≤ ˜d((x1, y1)
, (x2, y2)) Recall that at the very beginning, we proved that a continuous map from a closed, bounded interval is automatically uniformly continuous. This is true whenever the domain is compact. Theorem. Let (X, d) be a compact metric space, and (X ′, d′) is any metric space. If f : X → X ′ be continuous, then f is uniformly continuous. 49 5 Metric spaces IB Analysis II This is exactly the same proof as what we had for the [0, 1] case. Proof. We are going to prove by contradiction. Suppose f : X → X ′ is not uniformly continuous. Since f is not uniformly continuous, there is some ε > 0 such that for all δ = 1 n but d′(f (xn), f (yn)) > ε. n , there is some xn, yn such that d(xn, yn) < 1 By compactness of X, (xn) has a convergent subsequence (xni) → x. Then we also have yni → x. So by continuity, we must have f (xni) → f (x) and f (yni) → f (x). But d′(f (xni), f (yni)) > ε for all ni. This is a contradiction. In the proof, we have secretly used (part of) the following characterization of continuity: Theorem. Let (X, d) and (X ′, d′) be metric spaces, and f : X → X ′. Then the following are equivalent: (i) f is continuous at y. (ii) f (xk) → f (y) for every sequence (xk) in X with xk → y. (iii) For every neighbourhood V of f (y), there is a neighbourhood U of y such that U ⊆ f −1(V ). Note that the definition of continuity says something like (iii), but with open balls instead of open sets. So this should not be surprising. Proof. – (i) ⇔ (ii): The argument for this is the same as for normed spaces. – (i) ⇒ (iii): Let V be a neighbourhood of f (y). Then by definition there is ε > 0 such that Bε(f (y)) ⊆ V . By continuity of f , there is some δ such that Bδ(y) ⊆ f −1(Bε(f (y))) ⊆ f −1(V ). Set U = Bε(y) and done. – (iii) ⇒ (i): for any ε, use the hypothesis with V = Bε(f (y)) to get a neighbourhood U of y such that U ⊆ f −1(V ) = f −1(Bε(f (y))). Since U is open, there is some δ such that Bδ(y) ⊆ U . So we get Bδ(y) ⊆ f −1(Bε(f (y))). So we get continuity. Corollary. A function f : (X, d) → (X ′, d′) is continuous if f −1(V ) is open in X whenever V is open in X ′. Proof. Follows directly from the equivalence of (i) and (iii) in the theorem above. 50 5 Metric spaces IB Analysis II 5.6 The contraction mapping theorem If you have already taken IB Metric and Topological Spaces, then you were probably bored by the above sections, since you’ve already met them all. Finally, we get to something new. This section is comprised of just two theorems. The first is the contraction mapping theorem, and we will use it to prove PicardLindel¨of existence theorem. Later, we will prove the inverse function theorem using the contraction mapping theorem. All of these are really powerful and important theorems in analysis. They have many more applications and useful corollaries, but we do not have time to get into those. Definition (Contraction mapping). Let (X, d) be metric space. A mapping f : X → X is a contraction if there exists some λ with 0 ≤ λ < 1 such that d(f (x), f (y)) ≤ λd(x, y). Note that a contraction mapping is by definition Lipschitz and hence (uni- formly) continuous. Theorem (Contraction mapping theorem). Let X be a (non-empty) complete metric space, and if f : X → X is a contraction, then f has a unique fixed point, i.e. there is a unique x such that f (x) = x. Moreover, if f : X → X is a function such that f (m) : X → X (i.e. f composed with itself m times) is a contraction for some m, then f has a unique fixed point. We can see finding fixed points as the process of solving equations. One important application we will have is to use this to solve differential equations. Note that the theorem is false if we drop the completeness assumption. For example, f : (0, 1) → (0, 1) defined by x 2 is clearly a contraction with no fixed point. The theorem is also false if we drop the assumption λ < 1. In fact, it is not enough to assume d(f (x), f (y)) < d(x, y) for all x, y. A counterexample is to be found on example sheet 3. Proof. We first focus on the case where f itself is a contraction. Uniqueness is straightforward. By assumption, there is some 0 ≤ λ < 1 such that d(f (x), f (y)) ≤ λd(x, y) for all x, y ∈ X. If x and y are both fixed points, then this says d(x, y) = d(f (x), f (y)) ≤ λd(x, y). This is possible only if d(x, y) = 0, i.e. x = y. To prove existence, the idea is to pick a point x0 and keep applying f . Let x0 ∈ X. We define the sequence (xn) inductively by xn+1 = f (xn). We first show that this is Cauchy. For any n ≥ 1, we can compute d(xn+1, xn) = d(f (xn), f (xn−1)) ≤ λd(xn, xn−1) ≤ λnd(x1, x0). 51 5 Metric spaces IB Analysis II Since this is true for any n, for m > n, we have d(xm, xn) ≤ d(xm, xm−1) + d(xm−1, xm−2) + · · · + d(xn+1, xn) = = m−1 j=n m−1 j=n d(xj+1, xj) λjd(x1, x0) ≤ d(x1, x0) ∞ j=n λj = λn 1 − λ d(x1, x0). Note that we have again used the property that λ < 1. This implies d(xm, xn) → 0 as m, n → ∞. So this sequence is Cauchy. By the completeness of X, there exists some x ∈ X such that xn → x. Since f is a contraction, it is continuous. So f (xn) → f (x). However, by definition f (xn) = xn+1. So taking the limit on both sides, we get f (x) = x. So x is a fixed point. Now suppose that f (m) is a contraction for some m. Hence by the first part, there is a unique x ∈ X such that f (m)(x) = x. But then f (m)(f (x)) = f (m+1)(x) = f (f (m)(x)) = f (x). So f (x) is also a fixed point of f (n)(x). By uniqueness of fixed points, we must have f (x) = x. Since any fixed point of f is clearly a fixed point of f (n) as well, it follows that x is the unique fixed point of f . Based on the proof of the theorem, we have the following error estimate in the contraction mapping theorem: for x0 ∈ X and xn = f (xn−1), we showed that for m > n, we have d(xm, xn) ≤ λn 1 − λ d(x1, x0). If xn → x, taking the limit of the above bound as m → ∞ gives d(x, xn) ≤ λn 1 − λ d(x1, x0). This is valid for all n. We are now going to use this to obtain the Picard-Lindel¨of existence theorem for ordinary differential equations. The objective is as follows. Suppose we are given a function F = (F1, F2, · · · , Fn) : R × Rn → Rn. We interpret the R as time and the Rn as space. Given t0 ∈ R and x0 ∈ Rn, we want to know when can we find a solution to the ODE df dt = F(t, f (t)) 52 5 Metric spaces IB Analysis II subject to f (t0) = x0. We would like this solution to be valid (at least) for all t in some interval I containing t0. More explicitly, we want to understand when will there be some ε > 0 and a differentiable function f = (f1, · · · , fn) : (t0 − ε, t0 + ε) → Rn (i.e. fj : (t0 − ε, t0 + ε) → R is differentiable for all j) satisfying dfj dt = Fj(t, f1(t), · · · , fn(t)) such that fj(t0) = x(j) 0 for all j = 1, . . . , n and t ∈ (t0 − ε, t0 + ε). We can imagine this scenario as a particle moving in Rn, passing through x0 at time t0. We then ask if there is a trajectory f (t) such that the velocity of the particle at any time t is given by F(t, f (t)). This is a complicated system, since it is a coupled system of many variables. Explicit solutions are usually impossible, but in certain cases, we can prove the existence of a solution. Of course, solutions need not exist for arbitrary F . For example, there will be no solution if F is everywhere discontinuous, since any derivative is continuous in a dense set of points. The Picard-Lindel¨of existence theorem gives us sufficient conditions for a unique solution to exists. We will need the following notation Notation. For x0 ∈ Rn, R > 0, we let BR(x0) = {x ∈ Rn : ∥x − x0∥2 ≤ R}. Then the theorem says Theorem (Picard-Lindel¨of existence theorem). Let x0 ∈ Rn, R > 0, a < b, t0 ∈ [a, b]. Let F : [a, b] × BR(x0) → Rn be a continuous function satisfying ∥F(t, x) − F(t, y)∥2 ≤ κ∥x − y∥2 for some fixed κ > 0 and all t ∈ [a, b], x ∈ BR(x0). In other words, F (t, · ) : Rn → Rn is Lipschitz on BR(x0) with the same Lipschitz constant for every t. Then (i) There exists an ε > 0 and a unique differentiable function f : [t0 − ε, t0 + ε] ∩ [a, b] → Rn such that df dt = F(t, f (t)) (∗) and f (t0) = x0. (ii) If sup [a,b]×BR(x0) ∥F∥2 ≤ R b − a , then there exists a unique differential function f : [a, b] → Rn that satisfies the differential equation and boundary conditions above. Even n = 1 is an important, special, non-trivial case. Even if we have only one dimension, explicit solutions may be very difficult to find, if not impossible. For example, df dt = f 2 + sin f + ef 53 5 Metric spaces IB Analysis II would be almost impossible to solve. However, the theorem tells us there will be a solution, at least locally. Note that any differentiable f satisfying the differential equation is automatically continuously differentiable, since the derivative is F(t, f (t)), which is continuous. Before we prove the theorem, we first show the requirements are indeed necessary. We first look at that ε in (i). Without the addition requirement in (ii), there might not exist a solution globally on [a, b]. For example, we can consider the n = 1 case, where we want to solve df dt = f 2, with boundary condition f (0) = 1. Our F (t, f ) = f 2 is a nice, uniformly Lipschitz function on any [0, b] × BR(1) = [0, b] × [1 − R, 1 + R]. However, we will shortly see that there is no global solution. If we assume f ̸= 0, then for all t ∈ [0, b], the equation is equivalent to d dt (t + f −1) = 0. So we need t + f −1 to be constant. The initial conditions tells us this constant is 1. So we have f (t) = 1 1 − t . Hence the solution on [0, 1) is on [0, 1). So if b ≥ 1, then there is no solution in [0, b]. 1 1−t . Any solution on [0, b] must agree with this The Lipschitz condition is also necessary to guarantee uniqueness. Without this condition, existence of a solution is still guaranteed (but is another theorem, the Cauchy-Peano theorem), but we could have many different solutions. For example, we can consider the differential equation df dt = \|f \| with f (0) = 0. Here F (t, x) = \|x\| is not Lipschitz near x = 0. It is easy to
see that both f = 0 and f (t) = 1 4 t2 are both solutions. In fact, for any α ∈ [0, b], the function fα(tt − α)2 α ≤ t ≤ b is also a solution. So we have an infinite number of solutions. We are now going to use the contraction mapping theorem to prove this. In general, this is a very useful idea. It is in fact possible to use other fixed point theorems to show the existence of solutions to partial differential equations. This is much more difficult, but has many far-reaching important applications to theoretical physics and geometry, say. For these, see Part III courses. Proof. First, note that (ii) implies (i). We know that sup [a,b]×BR(x) ∥F∥ 54 5 Metric spaces IB Analysis II is bounded since it is a continuous function on a compact domain. So we can pick an ε such that 2ε ≤ R sup[a,b]×BR(x) ∥F∥ . Then writing [t0 − ε, t0 + ε] ∩ [a, b] = [a1, b1], we have sup [a1,b1]×BR(x) ∥F∥ ≤ sup [a,b]×BR(x) ∥F∥ ≤ R 2ε ≤ R b1 − a1 . So (ii) implies there is a solution on [t0 − ε, t0 + ε] ∩ [a, b]. Hence it suffices to prove (ii). To apply the contraction mapping theorem, we need to convert this into a fixed point problem. The key is to reformulate the problem as an integral equation. We know that a differentiable f : [a, b] → Rn satisfies the differential equation (∗) if and only if f : [a, b] → BR(x0) is continuous and satisfies f (t) = x0 + t t0 F(s, f (s)) ds by the fundamental theorem of calculus. Note that we don’t require f is differentiable, since if a continuous f satisfies this equation, it is automatically differentiable by the fundamental theorem of calculus. This is very helpful, since we can work over the much larger vector space of continuous functions, and it would be easier to find a solution. We let X = C([a, b], BR(x0)). We equip X with the supremum metric ∥g − h∥ = sup t∈[a,b] ∥g(t) − h(t)∥2. We see that X is a closed subset of the complete metric space C([a, b], Rn) (again taken with the supremum metric). So X is complete. For every g ∈ X, we define a function T g : [a, b] → Rn by (T g)(t) = x0 + t t0 F(s, g(s)) ds. Our differential equation is thus f = T f . So we first want to show that T is actually mapping X → X, i.e. T g ∈ X whenever g ∈ X, and then prove it is a contraction map. We have ∥T g(t) − x0∥2 = ≤ ≤ t t0 t t0 F(s, g(s)) ds ∥F(s, g(s))∥2 ds sup [a,b]×BR(x0) ∥F∥ · \|b − a\| ≤ R 55 5 Metric spaces IB Analysis II Hence we know that T g(t) ∈ BR(x0). So T g ∈ X. Next, we need to show this is a contraction. However, it turns out T need not be a contraction. Instead, what we have is that for g1, g2 ∈ X, we have ∥T g1(t) − T g2(t)∥2 = ≤ t t0 t t0 F(s, g1(s)) − F(s, g2(s)) ds ∥F(s, g1(s)) − F(s, g2(s))∥2 ds 2 by the Lipschitz condition on F . If we indeed have ≤ κ(b − a)∥g1 − g2∥∞ then the contraction mapping theorem gives an f ∈ X such that κ(b − a) < 1, (†) i.e. T f = f , f = x0 + t t0 F(s, f (s)) ds. However, we do not necessarily have (†). There are many ways we can solve this problem. Here, we can solve it by finding an m such that T (m is a contraction map. We will in fact show that this map satisfies the bound ∥T (m)g1(t) − T (m)g2(t)∥ ≤ sup t∈[a,b] (b − a)mκm m! sup t∈[a,b] ∥g1(t) − g2(t)∥. (‡) The key is the m!, since this grows much faster than any exponential. Given this bound, we know that for sufficiently large m, we have (b − a)mκm m! < 1, i.e. T (m) is a contraction. So by the contraction mapping theorem, the result holds. So it only remains to prove the bound. To prove this, we prove instead the pointwise bound: for any t ∈ [a, b], we have ∥T (m)g1(t) − T (m)g2(t)∥2 ≤ (\|t − t0\|)mκm m! sup s∈[t0,t] ∥g1(s) − g2(s)∥. From this, taking the supremum on the left, we obtain the bound (‡). To prove this pointwise bound, we induct on m. We wlog assume t > t0. We know that for every m, the difference is given by ∥T (m)g1(t) − T (m)g2(t)∥2 = t t0 t t0 ≤ κ F (s, T (m−1)g1(s)) − F (s, T (m−1)g2(s)) ds 2 . ∥T (m−1)g1(s) − T (m−1)g2(s)∥2 ds. 56 5 Metric spaces IB Analysis II This is true for all m. If m = 1, then this gives ∥T g1(t) − T g2(t)∥ ≤ κ(t − t0) sup [t0,t] ∥g1 − g2∥2. So the base case is done. For m ≥ 2, assume by induction the bound holds with m − 1 in place of m. Then the bounds give ∥T (m)g1(t) − T (m)g2(t)∥ ≤ κ t km−1(s − t0)m−1 (m − 1)! t0 κm sup (m − 1)! [t0,t] κm(t − t0)m m! ≤ = ∥g1 − g2∥2 t0 ∥g1 − g2∥2. sup [t0,t] sup [t0,s] t ∥g1 − g2∥2 ds (s − t0)m−1 ds So done. Note that to get the factor of m!, we had to actually perform the integral, instead of just bounding (s − t0)m−1 by (t − t0). In general, this is a good strategy if we want tight bounds. Instead of bounding b a f (x) dx ≤ (b − a) sup \|f (x)\|, we write f (x) = g(x)h(x), where h(x) is something easily integrable. Then we can have a bound b a f (x) dx ≤ sup \|g(x)\| b a \|h(x)\| dx. 57 6 Differentiation from Rm to Rn IB Analysis II 6 Differentiation from Rm to Rn 6.1 Differentiation from Rm to Rn We are now going to investigate differentiation of functions f : Rn → Rm. The hard part is to first come up with a sensible definition of what this means. There is no obvious way to generalize what we had for real functions. After defining it, we will need to do some hard work to come up with easy ways to check if functions are differentiable. Then we can use it to prove some useful results like the mean value inequality. We will always use the usual Euclidean norm. To define differentiation in Rn, we first we need a definition of the limit. Definition (Limit of function). Let E ⊆ Rn and f : E → Rm. Let a ∈ Rn be a limit point of E, and let b ∈ Rm. We say if for every ε > 0, there is some δ > 0 such that lim x→a f (x) = b (∀x ∈ E) 0 < ∥x − a∥ < δ ⇒ ∥f (x) − b∥ < ε. As in the case of R in IA Analysis I, we do not impose any requirements on F when x = a. In particular, we don’t assume that a is in the domain E. We would like a definition of differentiation for functions f : Rn → R (or more generally f : Rn → Rm) that directly extends the familiar definition on the real line. Recall that if f : (b, c) → R and a ∈ (b, c), we say f is differentiable if the limit Df (a) = f ′(a) = lim h→0 exists (as a real number). This cannot be extended to higher dimensions directly, since h would become a vector in Rn, and it is not clear what we mean by dividing by a vector. We might try dividing by ∥h∥ instead, i.e. require that (∗) f (a + h) − f (a) h lim h→0 f (a + h) − f (a) ∥h∥ exists. However, this is clearly wrong, since in the case of n = 1, this reduces to the existence of the limit f (a + h) − f (a) \|h\| , which almost never exists, e.g. when f (x) = x. It is also possible that this exists while the genuine derivative does not, e.g. when f (x) = \|x\|, at x = 0. So this is clearly wrong. Now we are a bit stuck. We need to divide by something, and that thing better be a scalar. ∥h∥ is not exactly what we want. What should we do? The idea is move f ′(a) to the other side of the equation, and (∗) becomes f (a + h) − f (a) − f ′(a)h h lim h→0 = 0. Now if we replace h by \|h\|, nothing changes. So this is equivalent to f (a + h) − f (a) − f ′(a)h \|h\| lim h→0 = 0. 58 6 Differentiation from Rm to Rn IB Analysis II In other words, the function f is differentiable if there is some A such that lim h→0 f (a + h) − f (a) − Ah \|h\| = 0, and we call A the derivative. We are now in a good shape to generalize. Note that if f : Rn → R is a real-valued function, then f (a + h) − f (a) is a scalar, but h is a vector. So A is not just a number, but a (row) vector. In general, if our function f : Rn → Rm is vector-valued, then our A should be an m × n matrix. Alternatively, A is a linear map from Rn to Rm. Definition (Differentiation in Rn). Let U ⊆ Rn be open, f : Rn → Rm. We say f is differentiable at a point a ∈ U if there exists a linear map A : Rn → Rm such that lim h→0 f (a + h) − f (a) − Ah ∥h∥ = 0. We call A the derivative of f at a. We write the derivative as Df (a). This is equivalent to saying lim x→a f (x) − f (a) − A(x − a) ∥x − a∥ = 0. Note that this is completely consistent with our usual definition the case where n = m = 1, as we have discussed above, since a linear transformation α : R → R is just given by α(h) = Ah for some real A ∈ R. One might instead attempt to define differentiability as follows: for any f : Rm → R, we say f is differentiable at x if f is differentiable when restricted to any line passing through x. However, this is a weaker notion, and we will later see that if we define differentiability this way, then differentiability will no longer imply continuity, which is bad. Having defined differentiation, we want to show that the derivative is unique. Proposition (Uniqueness of derivative). Derivatives are unique. Proof. Suppose A, B : Rn → Rm both satisfy the condition lim h→0 lim h→0 f (a + h) − f (a) − Ah ∥h∥ f (a + h) − f (a) − Bh ∥h∥ = 0 = 0. By the triangle inequality, we get ∥(B − A)h∥ ≤ ∥f (a + h) − f (a) − Ah∥ + ∥f (a + h) − f (a) − Bh∥. So ∥(B − A)h∥ ∥h∥ → 0 as h → 0. We set h = tu in this proof to get ∥(B − A)tu∥ ∥tu∥ → 0 59 6 Differentiation from Rm to Rn IB Analysis II as t → 0. Since (B − A) is linear, we know ∥(B − A)tu∥ ∥tu∥ = ∥(B − A)u∥ ∥u∥ . So (B − A)u = 0 for all u ∈ Rn. So B = A. Notation. We write L(Rn; Rm) for the space of linear maps A : Rn → Rm. So Df (a) ∈ L(Rn; Rm). To avoid having to write limits and divisions all over the place, we have the following convenient notation: Notation (Little o notation). For any function α : Br(0) ⊆ Rn → Rm, write if α(h) = o(h) α(h) ∥h∥ → 0 as h → 0. In other words, α → 0 faster than ∥h∥ as h → 0. Note that officially, α(h) = o(h) as a whole is a piece of notation, and does not represent equality. Then the condition for differentiability can be written as: f : U → Rm is differentiable at a ∈ U if there is some A with Alternatively, f (a + h) − f (a) − Ah = o(h). f (a + h) = f (a) + Ah + o(h). Note that we require the domain U of f to be open, so that for each a ∈ U , there is a small ball around a on which f is defined, so f (a + h) is defined for for sufficiently small h. We co
uld relax this condition and consider “one-sided” derivatives instead, but we will not look into these in this course. We can interpret the definition of differentiability as saying we can find a “good” linear approximation (technically, it is affine, not linear) to the function f near a. While the definition of the derivative is good, it is purely existential. This is unlike the definition of differentiability of real functions, where we are asked to compute an explicit limit — if the limit exists, that’s the derivative. If not, it is not differentiable. In the higher-dimensional world, this is not the case. We have completely no idea where to find the derivative, even if we know it exists. So we would like an explicit formula for it. The idea is to look at specific “directions” instead of finding the general derivative. As always, let f : U → Rm be differentiable at a ∈ U . Fix some u ∈ Rn, take h = tu (with t ∈ R). Assuming u ̸= 0, differentiability tells lim t→0 f (a + tu) − f (a) − Df (a)(tu) ∥tu∥ = 0. 60 6 Differentiation from Rm to Rn IB Analysis II This is equivalent to saying lim t→0 f (a + tu) − f (a) − tDf (a)u \|t\|∥u∥ = 0. Since ∥u∥ is fixed, This in turn is equivalent to lim t→0 f (a + tu) − f (a) − tDf (a)u t = 0. This, finally, is equal to Df (a)u = lim t→0 f (a + tu) − f (a) t . We derived this assuming u ̸= 0, but this is trivially true for u = 0. So this valid for all u. This is of the same form as the usual derivative, and it is usually not too difficult to compute this limit. Note, however, that this says if the derivative exists, then the limit above is related to the derivative as above. However, even if the limit exists for all u, we still cannot conclude that the derivative exists. Regardless, even if the derivative does not exist, this limit is still often a useful notion. Definition (Directional derivative). We write Duf (a) = lim t→0 f (a + tu) − f (a) t whenever this limit exists. We call Duf (a) the directional derivative of f at a ∈ U in the direction of u ∈ Rn. By definition, we have Duf (a) = d dt t=0 f (a + tu). Often, it is convenient to focus on the special cases where u = ej, a member of the standard basis for Rn. This is known as the partial derivative. By convention, this is defined for real-valued functions only, but the same definition works for any Rm-valued function. Definition (Partial derivative). The jth partial derivative of f : U → R at a ∈ U is Dej f (a) = lim t→∞ f (a + tej) − f (a) t , when the limit exists. We often write this as Dej f (a) = Djf (a) = ∂f ∂xj . Note that these definitions do not require differentiability of f at a. We will see some examples shortly. Before that, we first establish some elementary properties of differentiable functions. Proposition. Let U ⊆ Rn be open, a ∈ U . 61 6 Differentiation from Rm to Rn IB Analysis II (i) If f : U → Rm is differentiable at a, then f is continuous at a. (ii) If we write f = (f1, f2, · · · , fm) : U → Rm, where each fi : U → R, then f is differentiable at a if and only if each fj is differentiable at a for each j. (iii) If f, g : U → Rm are both differentiable at a, then λf + µg is differentiable at a with D(λf + µg)(a) = λDf (a) + µDg(a). (iv) If A : Rn → Rm is a linear map, then A is differentiable for any a ∈ Rn with DA(a) = A. (v) If f is differentiable at a, then the directional derivative Duf (a) exists for all u ∈ Rn, and in fact Duf (a) = Df (a)u. (vi) If f is differentiable at a, then all partial derivatives Djfi(a) exist for j = 1, · · · , n; i = 1, · · · , m, and are given by Djfi(a) = Dfi(a)ej. (vii) If A = (Aij) be the matrix representing Df (a) with respect to the standard basis for Rn and Rm, i.e. for any h ∈ Rn, Then A is given by Df (a)h = Ah. Aij = ⟨Df (a)ej, bi⟩ = Djfi(a). where {e1, · · · , en} is the standard basis for Rn, and {b1, · · · , bm} is the standard basis for Rm. The second property is useful, since instead of considering arbitrary Rm- valued functions, we can just look at real-valued functions. Proof. (i) By definition, if f is differentiable, then as h → 0, we know f (a + h) − f (a) − Df (a)h → 0. Since Df (a)h → 0 as well, we must have f (a + h) → f (h). (ii) Exercise on example sheet 4. (iii) We just have to check this directly. We have (λf + µg)(a + h) − (λf + µg)(a) − (λDf (a) + µDg(a)) ∥h∥ f (a + h) − f (a) − Df (a)h ∥h∥ g(a + h) − g(a) − Dg(a)h ∥h∥ = λ + µ . which tends to 0 as h → 0. So done. 62 6 Differentiation from Rm to Rn IB Analysis II (iv) Since A is linear, we always have A(a + h) − A(a) − Ah = 0 for all h. (v) We’ve proved this in the previous discussion. (vi) We’ve proved this in the previous discussion. (vii) This follows from the general result for linear maps: for any linear map represented by (Aij)m×n, we have Aij = ⟨Aej, bi⟩. Applying this with A = Df (a) and note that for any h ∈ Rn, Df (a)h = (Df1(a)h, · · · , Dfm(a)h). So done. The above says differentiability at a point implies the existence of all directional derivatives, which in turn implies the existence of all partial derivatives. The converse implication does not hold in either of these. Example. Let f 2 : R2 → R be defined by f (x, y) = 0 xy = 0 1 xy ̸= 0 Then the partial derivatives are df dx (0, 0) = df dy (0, 0) = 0, In other directions, say u = (1, 1), we have f (0 + tu) − f (0) t = 1 t which diverges as t → 0. So the directional derivative does not exist. Example. Let f : R2 → R be defined by x3 y 0 f (x, y) = y ̸= 0 y = 0 Then for u = (u1, u2) ̸= 0 and t ̸= 0, we can compute So f (0 + tu) − f (0) t = tu3 1 u2 0 u2 ̸= 0 u2 = 0 Duf (0) = lim t→0 f (0 + tu) − f (0) t = 0, and the directional derivative exists. However, the function is not differentiable at 0, since it is not even continuous at 0, as diverges as δ → 0. f (δ, δ4) = 1 δ 63 6 Differentiation from Rm to Rn IB Analysis II Example. Let f : R2 → R be defined by f (x, y) = x3 x2+y2 0 (x, y) ̸= (0, 0) (x, y) = (0, 0) . It is clear that f continuous at points other than 0, and f is also continuous at 0 since \|f (x, y)\| ≤ \|x\|. We can compute the partial derivatives as ∂f ∂x (0, 0) = 1, ∂f ∂y (0, 0) = 0. In fact, we can compute the difference quotient in the direction u = (u1, u2) ̸= 0 to be f (0 + tu) − f (0) t = u3 1 1 + u2 u2 2 . So we have Duf (0) = u3 1 1 + u2 u2 2 . We can now immediately conclude that f is not differentiable at 0, since if it were, then we would have Duf (0) = Df (0)u, which should be a linear expression in u, but this is not. Alternatively, if f were differentiable, then we have Df (0)h = 1 0 h1 h2 = h1. However, we have f (0 + h) − f (0) − Df (0)h ∥h∥ = h3 1 h2 1+h2 2 h2 − h1 1 + h2 2 = − h1h2 2 1 + h2 2 h2 3 , which does not tend to 0 as h → 0. For example, if h = (t, t), this quotient is − 1 23/2 for t ̸= 0. To decide if a function is differentiable, the first step would be to compute the partial derivatives. If they don’t exist, then we can immediately know the function is not differentiable. However, if they do, then we have a candidate for what the derivative is, and we plug it into the definition to check if it actually is the derivative. This is a cumbersome thing to do. It turns out that while existence of partial derivatives does not imply differentiability in general, it turns out we can get differentiability if we add some more slight conditions. Theorem. Let U ⊆ Rn be open, f : U → Rm. Let a ∈ U . Suppose there exists some open ball Br(a) ⊆ U such that 64 6 Differentiation from Rm to Rn IB Analysis II (i) Djfi(x) exists for every x ∈ Br(a) and 1 ≤ i ≤ m, 1 ≤ j ≤ n (ii) Djfi are continuous at a for all 1 ≤ i ≤ m, 1 ≤ j ≤ n. Then f is differentiable at a. Proof. It suffices to prove for m = 1, by the long proposition. For each h = (h1, · · · , hn) ∈ Rn, we have f (a + h) − f (a) = n j=1 f (a + h1e1 + · · · + hjej) − f (a + h1e1 + · · · + hj−1ej−1). Now for convenience, we can write h(j) = h1e1 + · · · + hjej = (h1, · · · , hj, 0, · · · , 0). Then we have f (a + h) − f (a) = = n j=1 n j=1 f (a + h(j)) − f (a + h(j−1)) f (a + h(j−1) + hjej) − f (a + h(j−1)). Note that in each term, we are just moving along the coordinate axes. Since the partial derivatives exist, the mean value theorem of single-variable calculus applied to g(t) = f (a + h(j−1) + tej) on the interval t ∈ [0, hj] allows us to write this as f (a + h) − f (a) = = n j=1 n j=1 hjDjf (a + h(j−1) + θjhjej) hjDjf (a) + hj n j=1 Djf (a + h(j−1) + θjhjej) − Djf (a) for some θj ∈ (0, 1). Note that Djf (a + h(j−1) + θjhjej) − Djf (a) → 0 as h → 0 since the partial derivatives are continuous at a. So the second term is o(h). So f is differentiable at a with n Df (a)h = Djf (a)hj. This is a very useful result. For example, we can now immediately conclude j=1 that the function   → 3x2 + 4 sin y + e6z xyze14x   x y z is differentiable everywhere, since it has continuous partial derivatives. This is much better than messing with the definition itself. 65 6 Differentiation from Rm to Rn IB Analysis II 6.2 The operator norm So far, we have only looked at derivatives at a single point. We haven’t discussed much about the derivative at, say, a neighbourhood or the whole space. We might want to ask if the derivative is continuous or bounded. However, this is not straightforward, since the derivative is a linear map, and we need to define these notions for functions whose values are linear maps. In particular, we want to understand the map Df : Br(a) → L(Rn; Rm) given by x → Df (x). To do so, we need a metric on the space L(Rn; Rm). In fact, we will use a norm. Let L = L(Rn; Rm). This is a vector space over R defined with addition and scalar multiplication defined pointwise. In fact, L is a subspace of C(Rn, Rm). To prove this, we have to prove that all linear maps are continuous. Let {e1, · · · , en} be the standard basis for Rn, and for x = n j=1 xjej, A(x) = n j=1 xjAej. and A ∈ L, we have By Cauchy-Schwarz, we know ∥A(x)∥ ≤ n j=1 \|xj\|∥A(ej)∥ ≤ ∥x∥ n j=1 ∥A(ej)∥2. So we see A is Lipschitz, and is hence continuous. Alternatively, this follows from the
fact that linear maps are differentiable and hence continuous. We can use this fact to define the norm of linear maps. Since L is finitedimensional (it is isomorphic to the space of real m × n matrices, as vector spaces, and hence have dimension mn), it really doesn’t matter which norm we pick as they are all Lipschitz equivalent, but a convenient choice is the sup norm, or the operator norm. Definition (Operator norm). The operator norm on L = L(Rn; Rm) is defined by ∥A∥ = sup x∈Rn:∥x∥=1 ∥Ax∥. Proposition. (i) ∥A∥ < ∞ for all A ∈ L. (ii) ∥ · ∥ is indeed a norm on L. (iii) ∥A∥ = sup Rn\{0} ∥Ax∥ ∥x∥ . (iv) ∥Ax∥ ≤ ∥A∥∥x∥ for all x ∈ Rn. 66 6 Differentiation from Rm to Rn IB Analysis II (v) Let A ∈ L(Rn; Rm) and B ∈ L(Rm; Rp). Then BA = B ◦ A ∈ L(Rn; Rp) and Proof. ∥BA∥ ≤ ∥B∥∥A∥. (i) This is since A is continuous and {x ∈ Rn : ∥x∥ = 1} is compact. (ii) The only non-trivial part is the triangle inequality. We have ∥A + B∥ = sup ∥x∥=1 ∥Ax + Bx∥ ≤ sup ∥x∥=1 (∥Ax∥ + ∥Bx∥) ≤ sup ∥x∥=1 ∥Ax∥ + sup ∥x∥=1 ∥Bx∥ = ∥A∥ + ∥B∥ (iii) This follows from linearity of A, and for any x ∈ Rn, we have x ∥x∥ = 1. (iv) Immediate from above. (v) ∥BA∥ = sup Rn\{0} ∥BAx∥ ∥x∥ ≤ sup Rn\{0} ∥B∥∥Ax∥ ∥x∥ = ∥B∥∥A∥. For certain easy cases, we have a straightforward expression for the operator norm. Proposition. (i) If A ∈ L(R, Rm), then A can be written as Ax = xa for some a ∈ Rm. Moreover, ∥A∥ = ∥a∥, where the second norm is the Euclidean norm in Rn (ii) If A ∈ L(Rn, R), then Ax = x · a for some fixed a ∈ Rn. Again, ∥A∥ = ∥a∥. Proof. (i) Set A(1) = a. Then by linearity, we get Ax = xA(1) = xa. Then we have So we have ∥Ax∥ = \|x\|∥a∥. ∥Ax∥ \|x\| = ∥a∥. (ii) Exercise on example sheet 4. Theorem (Chain rule). Let U ⊆ Rn be open, a ∈ U , f : U → Rm differentiable at a. Moreover, V ⊆ Rm is open with f (U ) ⊆ V and g : V → Rp is differentiable at f (a). Then g ◦ f : U → Rp is differentiable at a, with derivative D(g ◦ f )(a) = Dg(f (a)) Df (a). 67 6 Differentiation from Rm to Rn IB Analysis II Proof. The proof is very easy if we use the little o notation. Let A = Df (a) and B = Dg(f (a)). By differentiability of f , we know f (a + h) = f (a) + Ah + o(h) g(f (a) + k) = g(f (a)) + Bk + o(k) Now we have ) g ◦ f (a + h) = g(f (a) + Ah + o(h) k = g(f (a)) + B(Ah + o(h)) + o(Ah + o(h)) = g ◦ f (a) + BAh + B(o(h)) + o(Ah + o(h)). We just have to show the last term is o(h), but this is true since B and A are bounded. By boundedness, ∥B(o(h))∥ ≤ ∥B∥∥o(h)∥. So B(o(h)) = o(h). Similarly, ∥Ah + o(h)∥ ≤ ∥A∥∥h∥ + ∥o(h)∥ ≤ (∥A∥ + 1)∥h∥ for sufficiently small ∥h∥. So o(Ah + o(h)) is in fact o(h) as well. Hence g ◦ f (a + h) = g ◦ f (a) + BAh + o(h). 6.3 Mean value inequalities So far, we have just looked at cases where we assume the function is differentiable at a point. We are now going to assume the function is differentiable in a region, and see what happens to the derivative. Recall the mean value theorem from single-variable calculus: if f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), then f (b) − f (a) = f ′(c)(b − a) for some c ∈ (a, b). This is our favorite theorem, and we have used it many times in IA Analysis. Here we have an exact equality. However, in general, for vector-valued functions, i.e. if we are mapping to Rm, this is no longer true. Instead, we only have an inequality. We first prove it for the case when the domain is a subset of R, and then reduce the general case to this special case. Theorem. Let f : [a, b] → Rm be continuous on [a, b] and differentiable on (a, b). Suppose we can find some M such that for all t ∈ (a, b), we have ∥Df (t)∥ ≤ M . Then ∥f (b) − f (a)∥ ≤ M (b − a). Proof. Let v = f (b) − f (a). We define g(t) = v · f (t) = m i=1 vifi(t). 68 6 Differentiation from Rm to Rn IB Analysis II Since each fi is differentiable, g is continuous on [a, b] and differentiable on (a, b) with Hence, we know g′(t) = vif ′ i (t). \|g′(t)\| ≤ m i=1 vif ′ i (t) ≤ ∥v∥ 1/2 f ′2 i (t) n i=1 = ∥v∥∥Df (t)∥ ≤ M ∥v∥. We now apply the mean value theorem to g to get g(b) − g(a) = g′(t)(b − a) for some t ∈ (a, b). By definition of g, we get v · (f (b) − f (a)) = g′(t)(b − a). By definition of v, we have ∥f (b) − f (a)∥2 = \|g′(t)(b − a)\| ≤ (b − a)M ∥f (b) − f (a)∥. If f (b) = f (a), then there is nothing to prove. Otherwise, divide by ∥f (b) − f (a)∥ and done. We now apply this to prove the general version. Theorem (Mean value inequality). Let a ∈ Rn and f : Br(a) → Rm be differentiable on Br(a) with ∥Df (x)∥ ≤ M for all x ∈ Br(a). Then ∥f (b1) − f (b2)∥ ≤ M ∥b1 − b2∥ for any b1, b2 ∈ Br(a). Proof. We will reduce this to the previous theorem. Fix b1, b2 ∈ Br(a). Note that tb1 + (1 − t)b2 ∈ Br(a) for all t ∈ [0, 1]. Now consider g : [0, 1] → Rm. g(t) = f (tb1 + (1 − t)b2). By the chain rule, g is differentiable and g′(t) = Dg(t) = (Df (tb1 + (1 − t)b2))(b1 − b2) Therefore ∥Dg(t)∥ ≤ ∥Df (tb1 + (1 − t)b2)∥∥b1 − b2∥ ≤ M ∥b1 − b2∥. Now we can apply the previous theorem, and get ∥f (b1) − f (b2)∥ = ∥g(1) − g(0)∥ ≤ M ∥b1 − b2∥. 69 6 Differentiation from Rm to Rn IB Analysis II Note that here we worked in a ball. In general, we could have worked in a convex set, since all we need is for tb1 + (1 − t)b2 to be inside the domain. But with this, we have the following easy corollary. Corollary. Let f : Br(a) ⊆ Rn → Rm have Df (x) = 0 for all x ∈ Br(a). Then f is constant. Proof. Apply the mean value inequality with M = 0. We would like to extend this corollary. Does this corollary extend to differen- tiable maps f with Df = 0 defined on any open set U ⊆ Rn? The answer is clearly no. Even for functions f : R → R, this is not true, since we can have two disjoint intervals [1, 2] ∪ [3, 4], and define f (t) to be 1 on [1, 2] and 2 on [3, 4]. Then Df = 0 but f is not constant. f is just locally constant on each interval. The problem with this is that the sets are disconnected. We cannot connect points in [1, 2] and points in [3, 4] with a line. If we can do so, then we would be able to show that f is constant. Definition (Path-connected subset). A subset E ⊆ Rn is path-connected if for any a, b ∈ E, there is a continuous map γ : [0, 1] → E such that γ(0) = a, γ(1) = b. Theorem. Let U ⊆ Rn be open and path-connected. Then for any differentiable f : U → Rm, if Df (x) = 0 for all x ∈ U , then f is constant on U . A naive attempt would be to replace tb1 − (1 − t)b2 in the proof of the mean value theorem with a path γ(t). However, this is not a correct proof, since this has to assume γ is differentiable. So this doesn’t work. We have to think some more. Proof. We are going to use the fact that f is locally constant. wlog, assume m = 1. Given any a, b ∈ U , we show that f (a) = f (b). Let γ : [0, 1] → U be a (continuous) path from a to b. For any s ∈ (0, 1), there exists some ε such that Bε(γ(s)) ⊆ U since U is open. By continuity of γ, there is a δ such that (s − δ, s + δ) ⊆ [0, 1] with γ((s − δ, s + δ)) ⊆ Bε(γ(s)) ⊆ U . Since f is constant on Bε(γ(s)) by the previous corollary, we know that g(t) = f ◦ γ(t) is constant on (s − δ, s + δ). In particular, g is differentiable at s with derivative 0. This is true for all s. So the map g : [0, 1] → R has zero derivative on (0, 1) and is continuous on (0, 1). So g is constant. So g(0) = g(1), i.e. f (a) = f (b). If γ were differentiable, then this is much easier, since we can show g′ = 0 by the chain rule: g′(t) = Df (γ(t))γ′(t). 6.4 Inverse function theorem Now, we get to the inverse function theorem. This is one of the most important theorems of the course. This has many interesting and important consequences, but we will not have time to get to these. Before we can state the inverse function theorem, we need a definition. 70 6 Differentiation from Rm to Rn IB Analysis II Definition (C 1 function). Let U ⊆ Rn be open. We say f : U → Rm is C 1 on U if f is differentiable at each x ∈ U and Df : U → L(Rn, Rm) is continuous. We write C 1(U ) or C 1(U ; Rm) for the set of all C 1 maps from U to Rm. First we get a convenient alternative characterization of C 1. Proposition. Let U ⊆ Rn be open. Then f = (f1, · · · , fn) : U → Rn is C 1 on U if and only if the partial derivatives Djfi(x) exists for all x ∈ U , 1 ≤ i ≤ n, 1 ≤ j ≤ n, and Djfi : U → R are continuous. Proof. (⇒) Differentiability of f at x implies Djfi(x) exists and is given by Djfi(x) = ⟨Df (x)ej, bi⟩, where {e1, · · · , en} and {b1, · · · , bm} are the standard basis for Rn and Rm. So we know \|Djfi(x) − Djfi(y)\| = \|⟨(Df (x) − Df (y))ej, bi⟩\| ≤ ∥Df (x) − Df (y)∥ since ej and bi are unit vectors. Hence if Df is continuous, so is Djfi. (⇐) Since the partials exist and are continuous, by our previous theorem, we know that the derivative Df exists. To show Df : U → L(Rm; Rn) is continuous, note the following general fact: For any linear map A ∈ L(Rn; Rm) represented by (aij) so that Ah = aijhj, then for x = (x1, · · · , xn), we have ∥Ax∥2 = m   n i=1 j=1  2 Aijxj  By Cauchy-Schwarz, we have m ≤   n  a2 ij    n  x2 j  Dividing by ∥x∥2, we know i=1 j=1 j=1 = ∥x∥2 m n i=1 j=1 a2 ij. ∥A∥ ≤ a2 ij. Applying this to A = Df (x) − Df (y), we get ∥Df (x) − Df (y)∥ ≤ (Djfi(x) − Djfi(y))2. So if all Djfi are continuous, then so is Df . 71 6 Differentiation from Rm to Rn IB Analysis II If we do not wish to go through all that algebra to show the inequality ∥A∥ ≤ a2 ij, a2 ij is a norm on L(Rn, Rm), since it is just the we can instead note that Euclidean norm if we treat the matrix as a vector written in a funny way. So by the equivalence of norms on finite-dimensional vector spaces, there is some C such that ∥A∥ ≤ C a2 ij, and then the result follows. Finally, we can get to the inverse function theorem. Theorem (Inverse function theorem). Let U ⊆ Rn be open, and f : U → Rm be a C 1 map. Let a ∈ U , and suppose that Df (a) is invertible as a linear map Rn → Rn. Then there exists open sets V, W ⊆ Rn with a ∈ V , f (a) ∈ W , V ⊆ U such that f \|V : V → W is a bijection. Moreover, the inverse map f \|−1 V : W → V is also C 1. We have a fancy name for these functions. Definitio
n (Diffeomorphism). Let U, U ′ ⊆ Rn are open, then a map g : U → U ′ is a diffeomorphism if it is C 1 with a C 1 inverse. Note that different people have different definitions for the word “diffeomorphism”. Some require it to be merely differentiable, while others require it to be infinitely differentiable. We will stick with this definition. Then the inverse function theorem says: if f is C 1 and Df (a) is invertible, then f is a local diffeomorphism at a. Before we prove this, we look at the simple case where n = 1. Suppose f ′(a) ̸= 0. Then there exists a δ such that f ′(t) > 0 or f ′(t) < 0 in t ∈ (a−δ, a+δ). So f \|(a−δ,a+δ) is monotone and hence is invertible. This is a triviality. However, this is not a triviality even for n = 2. Proof. By replacing f with (Df (a))−1f (or by rotating our heads and stretching it a bit), we can assume Df (a) = I, the identity map. By continuity of Df , there exists some r > 0 such that ∥Df (x) − I∥ < 1 2 for all x ∈ Br(a). By shrinking r sufficiently, we can assume Br(a) ⊆ U . Let W = Br/2(f (a)), and let V = f −1(W ) ∩ Br(a). That was just our setup. There are three steps to actually proving the theorem. Claim. V is open, and f \|V : V → W is a bijection. Since f is continuous, f −1(W ) is open. So V is open. To show f \|V : V → W is bijection, we have to show that for each y ∈ W , then there is a unique x ∈ V such that f (x) = y. We are going to use the contraction mapping theorem to 72 6 Differentiation from Rm to Rn IB Analysis II prove this. This statement is equivalent to proving that for each y ∈ W , the map T (x) = x − f (x) + y has a unique fixed point x ∈ V . Let h(x) = x − f (x). Then note that So by our choice of r, for every x ∈ Br(a), we must have Dh(x) = I − Df (x). ∥Dh(x)∥ ≤ 1 2 . Then for any x1, x2 ∈ Br(a), we can use the mean value inequality to estimate ∥h(x1) − h(x2)∥ ≤ 1 2 ∥x1 − x2∥. Hence we know ∥T (x1) − T (x2)∥ = ∥h(x1) − h(x2)∥ ≤ 1 2 ∥x1 − x2∥. Finally, to apply the contraction mapping theorem, we need to pick the right domain for T , namely Br(a). For any x ∈ Br(a), we have ∥T (x) − a∥ = ∥x − f (x) + y − a∥ = ∥x − f (x) − (a − f (a)) + y − f (a)∥ ≤ ∥h(x) − h(a)∥ + ∥y − f (a)∥ ≤ 1 2 r 2 = r. < ∥x − a∥ + ∥y − f (a)∥ + r 2 So T : Br(a) → Br(a) ⊆ Br(a). Since Br(a) is complete, T has a unique fixed point x ∈ Br(a), i.e. T (x) = x. Finally, we need to show x ∈ Br(a), since this is where we want to find our fixed point. But this is true, since T (x) ∈ Br(a) by above. So we must have x ∈ Br(a). Also, since f (x) = y, we know x ∈ f −1(W ). So x ∈ V . So we have shown that for each y ∈ W , there is a unique x ∈ V such that f (x) = y. So f \|V : V → W is a bijection. We have done the hard work now. It remains to show that f \|V is invertible with C 1 inverse. Claim. The inverse map g = f \|−1 In fact, we have V : W → V is Lipschitz (and hence continuous). ∥g(y1) − g(y2)∥ ≤ 2∥y1 − y2∥. For any x1, x2 ∈ V , by the triangle inequality, know ∥x1 − x2∥ − ∥f (x1) − f (x2)∥ ≤ ∥(x1 − f (x1)) − (x2 − f (x2))∥ = ∥h(x1) − h(x0)∥ ∥x1 − x2∥. ≤ 1 2 73 6 Differentiation from Rm to Rn IB Analysis II Hence, we get ∥x1 − x2∥ ≤ 2∥f (x1) − f (x2)∥. Apply this to x1 = g(y1) and x2 = g(y2), and note that f (g(yj)) = yj to get the desired result. Claim. g is in fact C 1, and moreover, for all y ∈ W , Dg(y) = Df (g(y))−1. (∗) Note that if g were differentiable, then its derivative must be given by (∗), since by definition, we know and hence the chain rule gives f (g(y)) = y, Df (g(y)) · Dg(y) = I. Also, we immediately know Dg is continuous, since it is the composition of continuous functions (the inverse of a matrix is given by polynomial expressions of the components). So we only need to check that Df (g(y))−1 satisfies the definition of the derivative. First we check that Df (x) is indeed invertible for every x ∈ Br(a). We use the fact that If Df (x)v = 0, then we have ∥Df (x) − I∥ ≤ 1 2 . ∥v∥ = ∥Df (x)v − v∥ ≤ ∥Df (x) − I∥∥v∥ ≤ 1 2 ∥v∥. So we must have ∥v∥ = 0, i.e. v = 0. So ker Df (x) = {0}. So Df (g(y))−1 exists. Let x ∈ V be fixed, and y = f (x). Let k be small and In other words, h = g(y + k) − g(y). f (x + h) − f (x) = k. Since g is invertible, whenever k ̸= 0, h ̸= 0. Since g is continuous, as k → 0, h → 0 as well. We have g(y + k) − g(y) − Df (g(y))−1k ∥k∥ = = = h − Df (g(y))−1k ∥k∥ Df (x)−1(Df (x)h − k) ∥k∥ −Df (x)−1(f (x + h) − f (x) − Df (x)h) ∥k∥ = −Df (x)−1 = −Df (x)−1 f (x + h) − f (x) − Df (x)h ∥h∥ f (x + h) − f (x) − Df (x)h ∥h∥ · · ∥h∥ ∥k∥ ∥g(y + k) − g(y)∥ ∥(y + k) − y∥ . 74 6 Differentiation from Rm to Rn IB Analysis II As k → 0, h → 0. The first factor −Df (x)−1 is fixed; the second factor tends to 0 as h → 0; the third factor is bounded by 2. So the whole thing tends to 0. So done. Note that in the case where n = 1, if f : (a, b) → R is C 1 with f ′(x) ̸= 0 for every x, then f is monotone on the whole domain (a, b), and hence f : (a, b) → f ((a, b)) is a bijection. In higher dimensions, this is not true. Even if we know that Df (x) is invertible for all x ∈ U , we cannot say f \|U is a bijection. We still only know there is a local inverse. Example. Let U = R2, and f : R2 → R2 be given by f (x, y) = ex cos y ex sin y . Then we can directly compute Df (x, y) = ex cos y −ex sin y ex cos y. ex sin y Then we have det(Df (x, y)) = ex ̸= 0 for all (x, y) ∈ R2. However, by periodicity, we have f (x, y + 2nπ) = f (x, y) for all n. So f is not injective on R2. One major application of the inverse function theorem is to prove the implicit function theorem. We will not go into details here, but an example of the theorem can be found on example sheet 4. 6.5 2nd order derivatives We’ve done so much work to understand first derivatives. For real functions, we can immediately know a lot about higher derivatives, since the derivative is just a normal real function again. Here, it slightly more complicated, since the derivative is a linear operator. However, this is not really a problem, since the space of linear operators is just yet another vector space, so we can essentially use the same definition. Definition (2nd derivative). Let U ⊆ Rn be open, f : U → Rm be differentiable. Then Df : U → L(Rn; Rm). We say Df is differentiable at a ∈ U if there exists A ∈ L(Rn; L(Rn; Rm)) such that lim h→0 1 ∥h∥ (Df (a + h) − Df (a) − Ah) = 0. For this to make sense, we would need to put a norm on L(Rn; Rm) (e.g. the operator norm), but A, if it exists, is independent of the choice of the norm, since all norms are equivalent for a finite-dimensional space. This is, in fact, the same definition as our usual differentiability, since L(Rn; Rm) is just a finite-dimensional space, and is isomorphic to Rnm. So Df is 75 6 Differentiation from Rm to Rn IB Analysis II differentiable if and only if Df : U → Rnm is differentiable with A ∈ L(Rn; Rnm). This allows use to recycle our previous theorems about differentiability. In particular, we know Df is differentiable is implied by the existence of partial derivatives Di(Djfk) in a neighbourhood of a, and their continuity at a, for all k = 1, · · · , m and i, j = 1, · · · , n. Notation. Write Dijf (a) = Di(Djf )(a) = ∂2 ∂xi∂xj f (a). Let’s now go back to the initial definition, and try to interpret it. By linear algebra, in general, a linear map ϕ : Rℓ → L(Rn; Rm) induces a bilinear map Φ : Rℓ × Rn → Rm by Φ(u, v) = ϕ(u)(v) ∈ Rm. In particular, we know Φ(au + bv, w) = aΦ(u, w) + bΦ(v, w) Φ(u, av + bw) = aΦ(u, v) + bΦ(u, w). Conversely, if Φ : Rℓ × Rn → Rm is bilinear, then ϕ : Rℓ → L(Rn; Rm) defined by ϕ(u) = (v → Φ(u, v)) is linear. These are clearly inverse operations to each other. So there is a one-to-one correspondence between bilinear maps ϕ : Rℓ × Rn → Rm and linear maps Φ : Rℓ → L(Rn; Rm). In other words, instead of treating our second derivative as a weird linear map in L(Rn; L(Rn; Rm)), we can view it as a bilinear map Rn × Rn → Rm. Notation. We define D2f (a) : Rn × Rn → Rm by D2f (a)(u, v) = D(Df )(a)(u)(v). We know D2f (a) is a bilinear map. In coordinates, if u = n j=1 ujej, v = n j=1 vjej, where {e1, · · · , en} are the standard basis for Rn, then using bilinearity, we have D2f (a)(u, v) = n n i=1 j=1 D2f (a)(ei, ej)uivj. This is very similar to the case of first derivatives, where the derivative can be completely specified by the values it takes on the basis vectors. In the definition of the second derivative, we can again take h = tei. Then we have lim t→0 Df (a + tei) − Df (a) − tD(Df )(a)(ei) t = 0. 76 6 Differentiation from Rm to Rn IB Analysis II Note that the whole thing at the top is a linear map in L(Rn; Rm). We can let the whole thing act on ej, and obtain lim t→0 Df (a + tei)(ej) − Df (a)(ej) − tD(Df )(a)(ei)(ej) t = 0. for all i, j = 1, · · · , n. Taking the D2f (a)(ei, ej) to the other side, we know D2f (a)(ei, ej) = lim t→0 Df (a + tei)(ej) − Df (a)(ej) t = lim t→0 Dej f (a + tei) − Dej f (a) t = DeiDej f (a). In other words, we have D2f (ei, ej) = m k=1 Dijfk(a)bk, where {b1, · · · , bm} is the standard basis for Rm. So we have D2f (u, v) = n m i,j=1 k=1 Dijfk(a)uivjbk We have been very careful to keep the right order of the partial derivatives. However, in most cases we care about, it doesn’t matter. Theorem (Symmetry of mixed partials). Let U ⊆ Rn be open, f : U → Rm, a ∈ U , and ρ > 0 such that Bρ(a) ⊆ U . Let i, j ∈ {1, · · · , n} be fixed and suppose that DiDjf (x) and DjDif (x) exist for all x ∈ Bρ(a) and are continuous at a. Then in fact DiDjf (a) = DjDif (a). The proof is quite short, when we know what to do. Proof. wlog, assume m = 1. If i = j, then there is nothing to prove. So assume i ̸= j. Let gij(t) = f (a + tei + tej) − f (a + tei) − f (a + tej) + f (a). Then for each fixed t, define ϕ : [0, 1] → R by ϕ(s) = f (a + stei + tej) − f (a + stei). Then we get gij(t) = ϕ(1) − ϕ(0). By the mean value theorem and the chain rule, there is some θ ∈ (0, 1) such that gij(t) = ϕ′(θ) = t Dif (a + θtei + tej) − Dif (a + θtei) . 77 6 Differentiation from Rm to Rn IB Analysis II Now apply mean value theorem to the function
s → Dif (a + θtei + stej), there is some η ∈ (0, 1) such that gij(t) = t2DjDif (a + θtei + ηtej). We can do the same for gji, and find some ˜θ, ˜η such that gji(t) = t2DiDjf (a + ˜θtei + ˜ηtej). Since gij = gji, we get t2DjDif (a + θtei + ηtej) = t2DiDjf (a + ˜θtei + ˜ηtej). Divide by t2, and take the limit as t → 0. By continuity of the partial derivatives, we get DjDif (a) = DiDjf (a). This is nice. Whenever the second derivatives are continuous, the order does not matter. We can alternatively state this result as follows: Proposition. If f : U → Rm is differentiable in U such that DiDjf (x) exists in a neighbourhood of a ∈ U and are continuous at a, then Df is differentiable at a and D2f (a)(u, v) = DiDjf (a)uivj. is a symmetric bilinear form. j i Proof. This follows from the fact that continuity of second partials implies differentiability, and the symmetry of mixed partials. Finally, we conclude with a version of Taylor’s theorem for multivariable functions. Theorem (Second-order Taylor’s theorem). Let f : U → R be C 2, i.e. DiDjf (x) are continuous for all x ∈ U . Let a ∈ U and Br(a) ⊆ U . Then f (a + h) = f (a) + Df (a)h + 1 2 D2f (h, h) + E(h), where E(h) = o(∥h∥2). Proof. Consider the function g(t) = f (a + th). Then the assumptions tell us g is twice differentiable. By the 1D Taylor’s theorem, we know g(1) = g(0) + g′(0) + for some s ∈ [0, 1]. 78 1 2 g′′(s) 6 Differentiation from Rm to Rn IB Analysis II In other words, f (a + h) = f (a) + Df (a)h + = f (a) + Df (a)h + 1 2 1 2 D2f (a + sh)(h, h) D2f (a)(h, h) + E(h), where E(h) = 1 2 D2f (a + sh)(h, h) − D2f (a)(h, h) . By definition of the operator norm, we get \|E(h)\| ≤ 1 2 ∥D2f (a + sh) − D2f (a)∥∥h∥2. By continuity of the second derivative, as h → 0, we get ∥D2f (a + sh) − D2f (a)∥ → 0. So E(h) = o(∥h∥2). So done. 79
deﬁnition of a weak = U ηε(x − y)Dαu(y) dy = ηε ∗ Dαu. It is an exercise to verify that we can indeed move the derivative past the integral. Thus, if we ﬁx V U . Then by the previous parts, we see that Dαuε → Dαu in Lp(V ) as ε → 0 for \|α\| ≤ k. So uε − up W k.p(V ) = \|α\|≤k Dαuε − Dαup Lp(V ) → 0 as ε → 0. Theorem (Global approximation). Let 1 ≤ p < ∞, and U ⊆ Rn be open and bounded. Then C∞(U ) ∩ W k,p(U ) is dense in W k,p(U ). Our main obstacle to overcome is the fact that the molliﬁcations are only deﬁned on Uε, and not U . Proof. For i ≥ 1, deﬁne Ui = x ∈ U \| dist(x, ∂U ) > 1 Vi = Ui+3 − ¯Ui+1 Wi = Ui+4 − ¯Ui. i=1 Ui, and we can choose V0 U such that U = ∞ i We clearly have U = ∞ Let {ζi}∞ 0 ≤ ζi ≤ 1, ζi ∈ C∞ i=0 Vi. i=0 be a partition of unity subordinate to {Vi}. Thus, we have c (Vi) and ∞ i=0 ζi = 1 on U . Fix δ > 0. Then for each i, we can choose εi suﬃciently small such that ui = ηεi ∗ ζiu satisﬁes supp ui ⊆ Wi and ui − ζiuW k.p(U ) = ui − ζiuW k.p(Wi) ≤ δ 2i+1 . Now set v = ∞ i=0 ui ∈ C∞(U ). Note that we do not know (yet) that v ∈ W k.p(U ). But it certainly is when we restrict to some V U . In any such subset, the sum is ﬁnite, and since u = ∞ i=0 ζiu, we have v − uW k,p(V ) ≤ ∞ i=0 ui − ζiuW k.p(V ) ≤ δ ∞ i=0 2−(i+1) = δ. Since the bound δ does not depend on V , by taking the supremum over all V , we have v − uW k.p(U ) ≤ δ. So we are done. 25 3 Function spaces III Analysis of PDEs It would be nice for C∞( ¯U ) to be dense, instead of just C∞(U ). It turns out this is possible, as long as we have a sensible boundary. Deﬁnition (C k,δ boundary). Let U ⊆ Rn be open and bounded. We say ∂U is C k,δ if for any point in the boundary p ∈ ∂U , there exists r > 0 and a function γ ∈ C k,δ(Rn−1) such that (possibly after relabelling and rotating axes) we have U ∩ Br(p) = {(x, xn) ∈ Br(p) : xn > γ(x)}. Thus, this says our boundary is locally the graph of a C k,δ function. Theorem (Smooth approximation up to boundary). Let 1 ≤ p < ∞, and U ⊆ Rn be open and bounded. Suppose ∂U is C 0,1. Then C∞( ¯U ) ∩ W k,p(U ) is dense in W k,p(U ). Proof. Previously, the reason we didn’t get something in C∞( ¯U ) was that we had to glue together inﬁnitely many molliﬁcations whose domain collectively exhaust U , and there is no hope that the resulting function is in C∞( ¯U ). In the current scenario, we know that U locally looks like x0 The idea is that given a u deﬁned on U , we can shift it downwards by some ε. It is a known result that translation is continuous, so this only changes u by a tiny bit. We can then mollify with a ¯ε < ε, which would then give a function deﬁned on U (at least locally near x0). So ﬁx some x0 ∈ ∂U . Since ∂U is C 0,1, there exists r > 0 such that γ ∈ C 0,1(Rn−1) such that U ∩ Br(x0) = {(x, xn) ∈ Br(x) \| xn > γ(x)}. Set V = U ∩ Br/2(x0). Deﬁne the shifted function uε to be uε(x) = u(x + εen). Now pick ¯ε suﬃciently small such that vε,¯ε = η¯ε ∗ uε is well-deﬁned. Note that here we need to use the fact that ∂U is C 0,1. Indeed, we can see that if the slope of ∂U is very steep near a point x: ε 26 3 Function spaces III Analysis of PDEs then we need to choose a ¯ε much smaller than ε. By requiring that γ is 1-H¨older continuous, we can ensure there is a single choice of ¯ε that works throughout V . As long as ¯ε is small enough, we know that vε,¯ε ∈ C∞( ¯V ). Fix δ > 0. We can now estimate vε,˜ε − uW k.p(V ) = vε,˜ε − uε + uε − uW k,p(V ) ≤ vε,˜ε − uεW k,p(V ) + uε − uW k.p(V ). Since translation is continuous in the Lp norm for p < ∞, we can pick ε > 0 such that uε − uW k.p(V ) < δ 2 . Having ﬁxed such an ε, we can pick ˜ε so small that we also have vε,˜ε − uεW k.p(V ) < δ 2 . The conclusion of this is that for any x0 ∈ ∂U , we can ﬁnd a neighbourhood V ⊆ U of x0 in U such that for any u ∈ W k,p(U ) and δ > 0, there exists v ∈ C∞( ¯V ) such that u − vW k,p(V ) ≤ δ. It remains to patch all of these together using a partition of unity. By the compactness of ∂U , we can cover ∂U by ﬁnitely many of these V , say V1, . . . , VN . We further pick a V0 such that V0 U and N Vi. U = i=0 We can pick approximations vi ∈ C∞( ¯Vi) for i = 0, . . . , N (the i = 0 case is given by the previous global approximation theorem), satisfying vi − uW k,p(Vi) ≤ δ. Pick a partition of unity {ζi}N i=0 of ¯U subordinate to {Vi}. Deﬁne v = N i=0 ζivi. Clearly v ∈ C∞( ¯U ), and we can bound Dαv − DαuLp(U ) = Dα N i=0 ζivi − Dα N i=0 ζiu Lp(U ) N ≤ Ck vi − uW k.p(Vi) i=0 ≤ Ck(1 + N )δ, where Ck is a constant that solely depends on the derivatives of the partition of unity, which are ﬁxed. So we are done. 3.4 Extensions and traces If U ⊆ Rn is open and bounded, then there is of course a restriction map W 1,p(Rn) → W 1,p(U ). It turns out under mild conditions, there is an extension map going in the other direction as well. Theorem (Extension of W 1.p functions). Suppose U is open, bounded and ∂U is C 1. Pick a bounded V such that U V . Then there exists a bounded linear operator for 1 ≤ p < ∞ such that for any u ∈ W 1,p(U ), E : W 1,p(U ) → W 1.p(Rn) 27 3 Function spaces III Analysis of PDEs (i) Eu = u almost everywhere in U (ii) Eu has support in V (iii) EuW 1,p(Rn) ≤ CuW 1,p(U ), where the constant C depends on U, V, p but not u. Proof. First note that C 1( ¯U ) is dense in W 1,p(U ). So it suﬃces to show that the above theorem holds with W 1,p(U ) replaced with C 1( ¯U ), and then extend by continuity. We ﬁrst show that we can do this locally, and then glue them together using partitions of unity. Suppose x0 ∈ ∂U is such that ∂U near x0 lies in the plane {xn = 0}. In other words, there exists r > 0 such that B+ = Br(x0) ∩ {xn ≥ 0} ⊆ ¯U B− = Br(x0) ∩ {xn ≤ 0} ⊆ Rn \ U. The idea is that we want to reﬂect u\|B+ across the xn = 0 boundary to get a function on B−, but the derivative will not be continuous if we do this. So we deﬁne a “higher order reﬂection” by ¯u(x) = u(x) −3u(x, −xn) + 4 ux, − xn x ∈ B+ x ∈ B− 2 u −x − x 2 x xn We see that this is a continuous function. Moreover, by explicitly computing the partial derivatives, we see that they are continuous across the boundary. So we know ¯u ∈ C 1(Br(x0)). We can then easily check that we have ¯uW 1,p(Br(x0)) ≤ CuW 1,p(B+) for some constant C. 28 3 Function spaces III Analysis of PDEs If ∂U is not necessarily ﬂat near x0 ∈ ∂U , then we can use a C 1 diﬀeomorphism to straighten it out. Indeed, we can pick r > 0 and γ ∈ C 1(Rn−1) such that U ∩ Br(p) = {(x, xn) ∈ Br(p) \| xn > γ(x)}. We can then use the C 1-diﬀeomorphism Φ : Rn → Rn given by Φ(x)i = xi Φ(x)n = xn − γ(x1, . . . , xn) i = 1, . . . , n − 1 Then since C 1 diﬀeomorphisms induce bounded isomorphisms between W 1,p, this gives a local extension. Since ∂U is compact, we can take a ﬁnite number of points x0 i ∈ ∂W , sets Wi and extensions ui ∈ C 1(Wi) extending u such that ∂U ⊆ N i=1 Wi. Further pick W0 U so that U ⊆ N subordinate to {Wi}. Write i=0 Wi. Let {ζi}N i=0 be a partition of unity ¯u = N i=0 ζi ¯ui where ¯u0 = u. Then ¯u ∈ C 1(Rn), ¯u = u on U , and we have ¯uW 1,p(Rn) ≤ CuW 1,p(U ). By multiplying ¯u by a cut-oﬀ, we may assume supp ¯u ⊆ V for some V U . Now notice that the whole construction is linear in u. So we have constructed a bounded linear operator from a dense subset of W 1,p(U ) to W 1,p(V ), and there is a unique extension to the whole of W 1,p(U ) by the completeness of W 1,p(V ). We can see that the desired properties are preserved by this extension. Trace theorems A lot of the PDE problems we are interested in are boundary value problems, namely we want to solve a PDE subject to the function taking some prescribed values on the boundary. However, a function u ∈ Lp(U ) is only deﬁned up to sets of measure zero, and ∂U is typically a set of measure zero. So naively, we can’t naively deﬁne u\|∂U . We would hope that if we require u to have more regularity, then perhaps it now makes sense to deﬁne the value at the boundary. This is true, and is given by the trace theorem Theorem (Trace theorem). Assume U is bounded and has C 1 boundary. Then there exists a bounded linear operator T : W 1,p(U ) → Lp(∂U ) for 1 ≤ p < ∞ such that T u = u\|∂U if u ∈ W 1,p(U ) ∩ C( ¯U ). We say T u is the trace of u. Proof. It suﬃces to show that the restriction map deﬁned on C∞ functions is a bounded linear operator, and then we have a unique extension to W 1,p(U ). The gist of the argument is that Stokes’ theorem allows us to express the integral of 29 3 Function spaces III Analysis of PDEs a function over the boundary as an integral over the whole of U . In fact, the proof is indeed just the proof of Stokes’ theorem. By a general partition of unity argument, it suﬃces to show this in the case where U = {xn > 0} and u ∈ C∞ ¯U with supp u ⊆ BR(0) ∩ ¯U . Then Rn−1 \|u(x, 0)\|p dx = ∞ Rn−1 0 ∂ ∂xn \|u(x, xn)\|p dxn dx = U p\|u\|p−1uxn sgn u dxn dx. We estimate this using Young’s inequality to get Rn−1 \|u(x, 0)\|p dx ≤ Cp U \|u\|p + \|uxn \|p dU ≤ Cpup W 1,p(U ). So we are done. We can apply this to each derivative to deﬁne trace maps W k,p(U ) → W k−1,p(U ). In general, this trace map is not surjective. So in some sense, we don’t actually need to use up a whole unit of diﬀerentiability. In the example sheet, we see that in the case p = 2, we only lose “half” a derivative. c (U ) is dense in W 1,p c (U ). (U ). In fact, the converse is true — if T u = 0, then (U ), and the trace vanishes on C∞ 0 Note that C∞ So T vanishes on W 1,p u ∈ W 1,p (U ). 0 0 3.5 Sobolev inequalities Before we can move on to PDE’s, we have to prove some Sobolev inequalities. These are inequalities that compare diﬀerent norms, and allows us to “trade” diﬀerent desirable properties. One particularly important thing we can do is to trade diﬀerentiability for continuity. So we will know that if u ∈ W k,p(U ) for some large k, then in fact u ∈ C m(U ) for some (small) m. The utility of these results is that we would like to construct our solutions in W k,p spaces, since these are easier to work with, but ultimately, we want an actual, smooth solution to our equation. Sob

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