paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0005100 | One has MATH . It is easy to see that MATH and one verifies directly that MATH. |
math/0005100 | Put MATH. Then one has MATH . |
math/0005100 | Since MATH it follows from REF that the number of summands of MATH is equal to the rank of MATH. We want to show that the summands of MATH are a strongly exceptional collection. To do this we have to compute the MATH between the summands of MATH. We first compute the MATH's using REF . The result is as follows. MATH . ... |
math/0005100 | MATH. We have already pointed out that MATH implies MATH CITE, and MATH implies MATH is obvious. MATH. Assume that MATH is finitely generated. If MATH for a finite dimensional hereditary MATH-algebra MATH, then MATH has a tilting object. If MATH where MATH is a sheaf of hereditary orders over MATH, it follows from REF ... |
math/0005100 | MATH . When MATH has a tilting object MATH, it follows from CITE that MATH is derived equivalent to the finite dimensional algebra End-MATH. MATH . Since the hereditary category MATH is derived equivalent to a finite dimensional algebra MATH, it follows that MATH must have finite global dimension. Hence MATH, and conse... |
math/0005100 | That REF are equivalent follows from CITE. To prove the other equivalences we first review some generalities. First of all if MATH is a MATH-module, then by CITE the NAME dimension of MATH as MATH-module is equal to the NAME dimension of MATH as MATH-module. Furthermore we claim that MATH if and only if MATH for all MA... |
math/0005102 | Since MATH is a closed subgroup of MATH and the open set MATH is MATH invariant, it suffices to show that MATH acts freely on MATH. To do this we must show that the map MATH given on MATH points by MATH is a closed embedding. First we show that the image MATH of MATH is closed in MATH. Let MATH be matrix coordinates on... |
math/0005102 | CASE: Because the action of MATH on MATH is completely reducible, we can decompose MATH as MATH-module. As MATH, we have MATH. Hence MATH is a MATH-submodule and we get the desired direct sum decomposition of MATH. CASE: Let MATH be a basis for MATH. The map MATH defines an isomorphism of the span of MATH (denoted MATH... |
math/0005102 | If a group MATH has good representations, then so does MATH, so we may assume that MATH is algebraically closed. Suppose that MATH is a representation of MATH, and let MATH be the complement of a finite set of invariant linear subspaces MATH. We will show that MATH does not act properly on MATH. The strategy of the pro... |
math/0005102 | Suppose MATH where MATH is reductive and MATH unipotent. If MATH has a good representation then so does MATH. As proved above, this implies that MATH is a torus, so MATH is solvable. |
math/0005102 | For this proof only, we will use ``good" to mean ``set-theoretically freely good". Let MATH be a good representation of MATH, with MATH a finite set of proper invariant subspaces containing the vectors with nontrivial stabilizers. Because MATH is affine CITE, the vector bundle MATH is generated by a finite dimensional ... |
math/0005104 | Our argument closely follows CITE. We can first rule out the case MATH, because for it we only need to show that MATH is not standard. But a standard polyhedron without boundary must have vertices, while MATH. So we proceed assuming that MATH is irreducible. We will now prove that if MATH is not standard then MATH, and... |
math/0005104 | There are no closed bricks of complexity zero, since MATH, MATH, MATH, and MATH can be obtained assembling respectively two copies of MATH, two copies of MATH, one copy of MATH and one of MATH, and two copies of MATH. Moreover MATH, MATH, and MATH are not non-trivial assemblings of each other, and the conclusion follow... |
math/0005104 | Let MATH be an edge of MATH, and let MATH be the triple of (possibly not distinct) faces of MATH incident to MATH. The number of MATH's that separate MATH from MATH is even; it follows that MATH is a surface away from MATH. Let MATH be a boundary component of MATH, containing the triod MATH. Since MATH is a disc, which... |
math/0005104 | Assume a face MATH of MATH contains no vertices, and let MATH be incident to the triods MATH. Then MATH is a connected component of MATH, but MATH is standard without boundary by REF , so MATH, whence MATH and MATH. A contradiction. |
math/0005104 | Being normal, MATH is determined by an integer attached to each REF-component of MATH. Now we cut MATH open along MATH as explained in CITE: if a REF-component bears an integer MATH we replace the component by MATH parallel ones. We get a polyhedron MATH which contains MATH, such that MATH is the disjoint union of an o... |
math/0005104 | For MATH it was shown during the proof of REF that MATH is MATH or MATH, so we suppose MATH. By contradiction, assume MATH is not prime and let MATH be a standard minimal skeleton of MATH. Then MATH contains an essential normal sphere MATH. Such a sphere cannot be parallel to the boundary in MATH. Applying REF we get M... |
math/0005104 | Let MATH be a minimal skeleton for MATH, which is standard by REF , and let MATH be the minimal skeleton of MATH. Then MATH is a skeleton for MATH with minimal number of vertices, but MATH is not nuclear: there is a face MATH of MATH, glued to the free segment of MATH, which is incident to some vertex of MATH by REF . ... |
math/0005104 | By REF , MATH is standard. Suppose a face MATH is incident more than once to some MATH. Let MATH be an arc in MATH having endpoints MATH and MATH in two distinct edges of MATH, and let MATH be an essential closed curve in MATH with MATH. Now MATH is cut by MATH into components MATH and MATH. Since MATH is a ball, we ca... |
math/0005104 | Of course we can assume MATH. Since MATH and the inequality is easy for any non-trivial assembling of MATH, we also assume MATH. Suppose now that a face MATH is incident to two distinct triods MATH. Then there is an arc MATH, properly embedded in MATH, with endpoints MATH, and two essential loops MATH and MATH such tha... |
math/0005104 | First, MATH is prime and MATH, so MATH by the previous lemma. If MATH were not a brick then it would split as MATH with MATH. In all cases we must have MATH for some MATH, which contradicts the previous lemma. |
math/0005104 | Let MATH be a minimal skeleton of MATH. By REF we have MATH. Now MATH is super-standard, so each edge in MATH determines a different face of MATH. Then MATH, and the conclusion follows. |
math/0005104 | Suppose MATH is an edge of MATH with common endpoints; since MATH is orientable, the regular neighbourhood of MATH in MATH is an annulus, so there is a component MATH with MATH. Then MATH is a length-REF loop; this is impossible by REF , since length-REF loops are never fake. |
math/0005104 | First, MATH is a triod by REF . There are two possibilities for the regular neighbourhood MATH of MATH in MATH, which are shown in REF and lead to a sphere and a torus respectively. In the first case MATH contains three external discs MATH with MATH. By REF all the loops MATH are fake, so MATH for some MATH. In the sec... |
math/0005104 | Let MATH be an embedded face with MATH or fewer vertices. A loop in MATH very close to MATH and disjoint from MATH bounds a disc MATH parallel to MATH. Moreover MATH and MATH is not fake since MATH has only one component. Let MATH be a face incident at least twice to an edge MATH of MATH. It follows that there is a len... |
math/0005104 | The proof of REF works away from MATH. We only need to show that MATH is a surface near MATH: let MATH and MATH be the faces other than MATH incident to an edge MATH. Since MATH is orientable, MATH is adjacent to MATH on both sides and MATH is adjacent to MATH on both sides (or the converse). Therefore MATH and MATH ar... |
math/0005104 | By contradiction let MATH with MATH non-separating, and put MATH. The co-disconnecting surface MATH is by REF a closed orientable surface, which is non-empty since MATH disconnects MATH, whereas MATH does not. We assume that MATH is minimal among all mimimal skeleta of MATH for which there exists a non-separating torus... |
math/0005104 | The trace MATH is obtained from MATH as shown in REF ; it follows from the figure that if MATH then MATH and if MATH then MATH or MATH. By REF the edges of MATH have distinct ends. Using this fact one easily sees that MATH if MATH and MATH if MATH or MATH, and the conclusion follows. |
math/0005104 | The condition that MATH and MATH lie on the same of side of MATH means that MATH, during its transformation into MATH, is pushed towards MATH, and the conclusion is obvious. |
math/0005104 | It is enough to show that one of the following must hold: CASE: MATH is a non-separating torus, and MATH has a triod as a trace on some MATH; CASE: MATH bounds one of the polyhedra of type REF. So we assume REF does not hold and show REF . Our argument is long and organized in many steps. We first describe the overall ... |
math/0005104 | Take points MATH; we have MATH. Let MATH be a face of MATH incident to some MATH. The gluing path of MATH to MATH can be split into arcs MATH, meeting at points MATH, where MATH and MATH for all MATH, and each MATH is glued to one MATH. The map MATH is not necessarily injective, since MATH can be multiply incident to a... |
math/0005104 | The first assertion is easy and taken for granted. By construction MATH sits in MATH and it is simple, so we only need to show that MATH is an open MATH-ball. To this end we note that MATH is a ball MATH. Moreover MATH consists of two discs MATH and MATH, and MATH consists of two balls MATH and MATH, with MATH and MATH... |
math/0005106 | CASE: The left module operation on MATH is determined by MATH and the right module operation by MATH. This proves uniqueness. To prove existence, we must show that the right module operation is well-defined REF and satisfies the right module REF , furthermore that MATH is equivariant REF . NAME and axioms of the left m... |
math/0005106 | If MATH is an equivariant derivation of MATH, then MATH according to REF , so it remains to show that *[] REF MATH, if MATH is an equivariant derivation of MATH, and *[] REF MATH, if MATH is a right ideal of MATH with MATH. CASE: Let MATH be an equivariant derivation of MATH. Then MATH is an object of MATH. According t... |
math/0005110 | Let MATH be a maximal family of orthogonal projections in MATH which reduces MATH . Then MATH, where MATH, and this is an indecomposable decomposition. Suppose that MATH is another indecomposable decomposition, and let MATH . Then the projections MATH belong to MATH and reduce MATH . It follows that the projections in ... |
math/0005110 | It will be enough to establish the proposition when MATH and MATH. Let MATH be the identity embedding. Then MATH and so MATH which is to say that MATH and hence that the induced maps MATH are unitarily equivalent. From this it follows that MATH are unitarily equivalent. |
math/0005110 | Let MATH be the given presentations. Consider the MATH-module homomorphism MATH which is the composition MATH where MATH is the natural map. Suppose first that MATH and let MATH be the map MATH where MATH is a rank one projection. Then MATH, the class of MATH in MATH, has image MATH in MATH which in turn coincides with... |
math/0005110 | Let MATH be the given presentations. Consider the MATH-module homomorphism MATH which is the composition MATH where MATH is the natural map. Suppose first that MATH and let MATH, be the map MATH where MATH is a rank one projection. Then MATH, the class of MATH in MATH, has image MATH in MATH. Let MATH . Choose MATH in ... |
math/0005110 | It is clear that MATH and MATH are star-extendibly isomorphic then there is an induced isomorphism MATH of the invariants. The sufficiency direction will follow from REF once we show that MATH is naturally isomorphic to MATH with corresponding identification of scales. Let MATH be the morphism for which MATH, where MAT... |
math/0005110 | Let MATH be presentations with partial embeddings in MATH. For each MATH choose MATH large enough so that MATH . By uniform stability we assume that MATH is chosen so that we may obtain in MATH a star-extendible embedding MATH with MATH. The map MATH induces a map MATH, MATH . Furthermore, this map is graded by multipl... |
math/0005110 | We begin the proof by putting the map MATH into a standard form. First replace MATH by a unitary conjugate MATH, where the unitary MATH has the form MATH to arrange that MATH and MATH have standard orthogonal final projections of the form MATH where MATH are the matrix units of MATH and MATH is the multiplicity of the ... |
math/0005110 | Let MATH have quadruple MATH as before, let MATH have quadruple MATH and consider the composition MATH. The images of MATH and MATH give rise to the quadruple MATH for this composition. Note that MATH where MATH with a similar formula for MATH. Thus we may compute MATH . By orthogonality MATH . Thus MATH is equal to MA... |
math/0005110 | In view of REF if the sequences are asymptotically equivalent then one can construct an asymptotically commuting diagram of unital star-extendible embeddings. From this it follows that MATH and MATH are star extendibly isomorphic. On the other hand by the stability of MATH-algebras if MATH and MATH are star extendibly ... |
math/0005110 | In the proof we will indicate the set of elementary compression type maps for MATH and MATH by MATH and MATH respectively. Suppose first that MATH is the elementary compression type homomorphism MATH where MATH where MATH is a compression type embedding determined by an interval projection of MATH of rank MATH and wher... |
math/0005110 | The equivalence of REF is elementary and we have already noted the equivalnece of REF . Plainly REF implies REF . Assume REF . Let MATH have MATH block decomposition MATH-with MATH. By assumption each MATH is a partial isometry. Consider a product MATH. The MATH block entry is given by the sum MATH . Since MATH is regu... |
math/0005110 | The first part of the theorem follows from the arguments above which show that the indecomposable maps MATH are labelled by the partially defined order preserving functions MATH whose domains are intervals. The second assertion follows from the combinatorial discussion. |
math/0005110 | The sufficiency direction follows from REF . The necessity of the condition, that is, the fact that MATH is an invariant, will follow from REF once we show the stability of MATH. However this follows readily form the perturbational stability of MATH. Let MATH be star extendible and suppose that MATH . Then consider the... |
math/0005110 | We may assume that MATH. First note that if MATH is irregular then for at least one of the marix units MATH of the triple MATH the partial isometry MATH has the block form MATH where the operator MATH is not a partial isometry. To see this we argue by contradiction and assume otherwise. Since MATH and MATH are block di... |
math/0005110 | We first note the immediate consequence of REF that if MATH and MATH are MATH-algebra systems for which none of the embeddings MATH, MATH and their system compositions are of MATH-character then a commuting diagram isomorphism between MATH and MATH is necessarily regular. In general consider the commuting diagram MATH ... |
math/0005110 | Note that MATH and that MATH . Thus we have the following calculation. MATH where MATH is a unimodular multiple of the rank one partial isometry MATH . We now want to show that the composition MATH is not merely locally regular, which is what the above calculation shows, but that it is regular, that is, REF-decomposabl... |
math/0005110 | Let MATH, be the maps MATH . Then for all MATH the compositions MATH are regular by the last lemma. Also, since MATH is also a primitive root of unity the lemma shows that the compositions MATH are regular. Thus the maps MATH provide a commuting diagram between two regular systems in MATH consisting of irregular maps. ... |
math/0005113 | The proper subgroup MATH maps onto MATH (for instance under MATH), so MATH. Then the proper subgroup MATH of MATH maps onto MATH (under MATH), so MATH; etc. |
math/0005113 | MATH is a subgroup of MATH, which clearly is residually finite: it is approximated by its finite quotients given by the action on MATH, for any MATH. |
math/0005113 | First, we will prove that if MATH is central then MATH must be in MATH. Let MATH where MATH and MATH. If MATH and MATH for some MATH then MATH does not commute with the elements MATH such that MATH. If MATH then choose MATH with MATH and consider MATH . It is clear that MATH since MATH, which, along with the fact that ... |
math/0005113 | Take any MATH, and consider the coordinates of MATH. They all belong to MATH, whence MATH is a finite group of order at most MATH, and MATH is finite, since MATH is of finite index in MATH. |
math/0005113 | The proof is by induction on the length of MATH and it will be done for all MATH simultaneously. The statement is clear for the empty word. Next, no word of length MATH represents the identity in any group MATH. Now assume that the claim is true for all words of length less than MATH, with MATH and let MATH be a word o... |
math/0005113 | First, note that MATH is generated as a subgroup by the set MATH . Indeed, conjugation of any generator in MATH by an element from MATH gives another generator, conjugation by MATH is unimportant since MATH and, for MATH, MATH . The subgroup generated by MATH is thus normal. On the other hand MATH and MATH so that MATH... |
math/0005113 | Define MATH to be the number of MATH-letters from MATH appearing in the words at the level MATH and MATH to be the number of simple reductions performed to get the words MATH on the level MATH from their unreduced versions MATH. A reduced word MATH of length MATH has at most MATH-letters. Every MATH-letter in MATH that... |
math/0005113 | Let MATH denote the growth of MATH with respect to word length. We will obtain MATH for some positive constants MATH. Then, MATH applications of REF yield, neglecting the (unimportant) constant MATH, MATH, from which the theorem's claim follows. See CITE for a similar proof. We now prove REF . Choose some MATH with MAT... |
math/0005113 | Let a minimal form of MATH be MATH . Then MATH can be written in the form MATH and rewritten in the form MATH where MATH. Clearly, MATH, which yields MATH . Now, observe that if the MATH-generator MATH is of weight MATH with MATH then MATH has as components one MATH-generator of weight MATH and one MATH-generator (of w... |
math/0005113 | Let MATH. Choose MATH so that MATH is big enough in order that MATH be satisfied for all MATH and all MATH. Such a choice is possible because MATH for MATH and the two expressions are equal for MATH. Define a function MATH on MATH by MATH . We prove, by induction on MATH, that MATH. If the weight of MATH is MATH, the p... |
math/0005113 | Let MATH be a MATH-factorable sequence, factored in complete words of lengths MATH, with all MATH. We can define a modification of the portrait of an element by requiring that whenever we ``blow up" a leaf on the level MATH because its size is too big, we expand it MATH levels down (that is, the original word is expand... |
math/0005113 | For MATH, denote by MATH, MATH and MATH the number of MATH-letters, MATH-letters and MATH-letters from MATH, respectively, in the words on the level MATH of the decomposition of depth MATH of MATH. Clearly, MATH. Also, MATH since every MATH-letter from the level MATH contributes at most one MATH-letter to the next leve... |
math/0005113 | We will prove that the order of any element MATH in MATH divides some power of MATH. The proof is by induction on the length MATH of MATH and it will be done for all MATH simultaneously. The statement is clear for MATH and MATH. Assume that it is true for all words of length less than MATH, where MATH, and consider an ... |
math/0005113 | MATH, yielding MATH. Using an argument similar to that in REF and the observations on the structure of the words MATH given above, we conclude that MATH . |
math/0005113 | As usual, we use induction on MATH and we prove the statement simultaneously for all MATH-homogeneous MATH. We will prove that MATH where MATH. The statement is obvious for MATH. Consider an element MATH of length MATH, MATH and let MATH be its period sequence. We know that MATH because the word MATH is MATH-homogeneou... |
math/0005113 | Let MATH be the abelianization map. Recall that MATH is the elementary MATH-group of rank MATH. We recast the construction of the period sequence as follows: in the graph below, nodes correspond to images of elements MATH under MATH; arrows indicate taking a projection, MATH or MATH. Double arrows indicate a squaring w... |
math/0005113 | Assume that MATH lies in MATH and that MATH contains an element MATH satisfying the following conditions: CASE: MATH is of order MATH; CASE: MATH has a representation of length MATH with MATH odd; CASE: this representation is of the form MATH, with all the MATH in MATH except for one, which is in MATH. We shall constru... |
math/0005119 | The proof is standard. Let us consider the following exact sequence of functors from MATH to the category of MATH-linear spaces. MATH . Here MATH, MATH, and MATH, MATH, MATH are natural transformations given as follows MATH where MATH. The two middle terms in the exact sequence REF are exact functors (both with respect... |
math/0005119 | All the statements except the functoriality of MATH are obvious. The functoriality of MATH follows from properties of the NAME characteristic (namely, additivity with respect to algebraic stratifications, and multiplicativity for fiber bundles). |
math/0005119 | Associativity follows from functorial REF of the maps MATH. |
math/0005119 | REF is a simple calculation using the definition of the MATH-product, all the rest is obvious. |
math/0005119 | REF have been proven by Ch. CITE. REF follows from REF follows from REF and from the fact that MATH for any MATH. |
math/0005119 | Given MATH let MATH, where MATH . We fix an object of the form MATH (respectively, MATH, MATH) in the isomorphism class of MATH for each MATH (respectively, MATH for each MATH). Using MATH and MATH we construct a functor MATH. Namely the action of MATH on objects is given by REF, and the action on morphisms is the natu... |
math/0005119 | Follows from definitions. Note that being a NAME functor MATH induces algebraic maps between varieties MATH, MATH, MATH (used in REF) for the quiver MATH and the corresponding varieties for the quiver MATH. |
math/0005119 | The proposition follows from REF and from the fact that MATH is full, faithful, exact, and has épaisse image. |
math/0005119 | An easy calculation using the definition of MATH. See also REF . |
math/0005119 | Follows from the definition. |
math/0005119 | REF follow from the definition of MATH, REF Let MATH be a sink. Then there is the following split exact sequence MATH for any MATH. Here MATH, where MATH . If MATH then MATH. The proof for MATH being a source is analogous. REF follows from REF follows from the Snake Lemma. |
math/0005119 | One has to check that the bracket is skew-symmetric and satisfies the NAME identity. Both statements follow from the definition of MATH. |
math/0005119 | Induction using the fact that for any MATH there exists MATH such that MATH CITE. |
math/0005119 | Note that MATH because MATH is simply laced. Now the proposition follows from REF, and REF. The action of MATH on the generators of MATH is given by MATH where the notation is as in REF. |
math/0005119 | Follows from the fact that MATH and from REF. |
math/0005119 | One has to check that the bracket is skew-symmetric and satisfies the NAME identity. Both statements follow (after lengthy, but straightforward calculations) from the definition of MATH. A more conceptual proof is to consider MATH as embedded into an algebra of vertex operators, associated with the quiver MATH (see CIT... |
math/0005119 | Induction on MATH, where MATH for a root MATH. |
math/0005119 | An element MATH of MATH is, by definition, a MATH matrix such that MATH. It follows that MATH for some set MATH. |
math/0005119 | REF follow from the fact that MATH is projective. REF follow from the fact that MATH is injective. Let us prove REF. We need to find the value of the function MATH on the isomorphism class of the representation MATH. We give below the calculation for MATH. The case of arbitrary MATH is completely analogous, and, moreov... |
math/0005119 | We prove the proposition by induction. It follows from the definitions that MATH . Suppose that MATH . Then using REF we have MATH which gives MATH, and MATH which provides the induction step. One can avoid some of these calculations using an argument similar to the proof of REF to get MATH for MATH (note that the inde... |
math/0005119 | Follows from REF . |
math/0005119 | An exercise in (graded) linear algebra (see, for example, CITE). |
math/0005119 | Follows from the fact that if MATH is a subobject of MATH then MATH is isomorphic to MATH for some MATH and MATH, and MATH is isomorphic to MATH. |
math/0005119 | Let us temporarily denote the map given by the formulas in the formulation of the proposition by MATH. It follows from REF that MATH is a homomorphism of NAME algebras, whose value on generators MATH of MATH coincides with the value of MATH. Thus MATH, which proves the proposition. Another possible proof is an inductio... |
math/0005119 | Follows from REF . |
math/0005119 | The proposition follows from REF , and REF . |
math/0005119 | REF follow from REF follow from REF follows from the definition of root, REF, and REF follows from REF, and REF. Let us prove REF. Let MATH be an extending vertex (that is, MATH). It follows from the following equality MATH that either MATH or there exists MATH such that MATH, which proves REF. |
math/0005119 | The theorem follows from REF. |
math/0005119 | Given MATH the statement is true for one MATH, say MATH, because of REF . Any other MATH can be obtained from MATH by repeated applications of the NAME element (see REF). Then one can use the sequence of reflection functors corresponding to the sequence of reflections in the NAME element, and reason similarly to the pr... |
math/0005119 | The proposition follows from REF, and REF. |
math/0005119 | Note that MATH is linearly independent with MATH because MATH whereas MATH. Now the proposition follows from REF. |
math/0005119 | The map MATH is surjective by construction. REF implies that MATH. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.