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math/0002064 | REF has already been remarked. We show REF . In light of REF (applied with MATH) we can assume that MATH. Let MATH be a MATH-module and MATH an almost MATH-module; we have natural bijections MATH which proves REF . Now REF follows by inspecting the proof of REF , or by CITE (ch. III REF ). |
math/0002064 | The functor MATH is right adjoint to an exact functor, hence it preserves injectives. Now, let MATH be an injective envelope of MATH; to show that MATH is an injective envelope of MATH, it suffices to show that MATH is an essential extension of MATH. However, if MATH and MATH, then MATH, hence MATH, but MATH does not contain MATH-torsion, thus MATH. |
math/0002064 | We recall that the categories MATH and MATH are both complete and cocomplete. Now let MATH be any small indexing category and MATH be any functor. Denote by MATH the composed functor MATH. We claim that MATH. The proof is an easy application of REF . A similar argument also works for limits and for the category MATH. |
math/0002064 | REF follows easily from REF follows easily from REF . |
math/0002064 | By REF it follows that MATH is right exact. To show that it is also left exact when MATH is a flat MATH-module, it suffices to remark that MATH is left exact. Now, by REF , the functor MATH is right adjoint to an exact functor, so REF is clear. |
math/0002064 | Let MATH be a MATH-algebra, MATH a MATH-algebra and MATH a morphism of MATH-algebras. By REF we obtain a natural MATH-linear morphism MATH. Together with the structure morphism MATH this yields a map MATH which is easily seen to be a ring homomorphism. It is equally clear that the ideal MATH defined above is mapped to zero by MATH, hence the latter factors through a map of MATH-algebras MATH. Conversely, such a map induces a morhism of MATH-algebras MATH just by taking localisation. It is easy to check that the two procedures are inverse to each other, which shows the assertion. |
math/0002064 | REF : we remark only that MATH for all MATH and leave the details to the reader. We prove REF . From CITE REF it follows easily that MATH. It then suffices to show that MATH is almost zero. We have MATH and the product MATH is again a NAME product. Next let MATH. If MATH then MATH and MATH. In view of REF , MATH is almost zero, hence we reduce to showing that MATH is almost zero. But MATH for all MATH. On the other hand, by REF we get MATH, hence MATH and finally MATH which means that MATH is a surjective inverse system, so its MATH vanishes and the result follows. |
math/0002064 | REF is easy and we leave it to the reader. To prove REF , take a finitely generated ideal MATH such that MATH, pick a morphism MATH whose cokernel is annihilated by MATH, and apply the following REF . |
math/0002064 | We need the following Let MATH be a finitely generated MATH-module and suppose that we are given MATH and a (not necessarily commutative) diagram MATH such that MATH, MATH. Let MATH be an ideal such that MATH has a finitely generated submodule containing MATH. Then MATH has a finitely generated submodule containing MATH. Let MATH be the submodule of MATH given by the assumption. We have MATH and MATH. We take MATH. Clearly MATH, so MATH, hence MATH and finally MATH. Now, let MATH and MATH. By assumption there is a complex MATH with MATH, MATH. Letting MATH, MATH, MATH, MATH, one checks easily that MATH and MATH can be given such that all the assumptions of the above claim are fulfilled. So, with MATH we get that MATH lies in a finitely generated submodule of MATH. But MATH is contained in an ideal generated by finitely many such products MATH. |
math/0002064 | The ``only if" part in REF (respectively, REF ) is first checked when MATH is finitely generated (respectively, finitely presented) and then extended to the general case. We leave the details to the reader and we proceed to verify the ``if" part. For REF , choose a set MATH and an epimorphism MATH. Let MATH be the directed set of finite subsets of MATH, ordered by inclusion. For MATH, let MATH. Then MATH, so the assumption gives MATH, that is, MATH is almost zero, so, for every MATH, the image of MATH in the above colimit is MATH, that is, there exists MATH such that MATH, which proves the contention. For REF , we present MATH as a filtered colimit MATH, where each MATH is finitely presented (this can be done for example, by taking such a presentation of the MATH-module MATH and applying MATH). The assumption of REF gives that MATH is an almost isomorphism, hence, for every MATH there is MATH and MATH such that MATH, where MATH is the natural morphism to the colimit. If such a MATH exists for MATH, then it exists for every MATH. Hence, if MATH is a finitely generated subideal, say MATH, then there exist MATH and MATH such that MATH for MATH. Hence MATH is contained in MATH and contains MATH. Hence MATH has a finitely generated submodule MATH containing MATH. Choose a presentation MATH. Then one can lift MATH to a finitely generated submodule MATH of MATH. Then MATH is a finitely generated submodule of MATH containing MATH. Since we also have MATH and MATH is arbitrary, the conclusion follows from REF . |
math/0002064 | These facts can be deduced from REF , or proved directly. |
math/0002064 | Left to the reader. |
math/0002064 | REF : for given MATH, we consider any MATH-module MATH and we apply the functor MATH to REF : MATH which implies MATH for all MATH. Since MATH is arbitrary, REF follows from REF . REF : by hypothesis, for arbitrary MATH we can find MATH and a morphism MATH such that MATH. Let MATH be the image of MATH, so that MATH factors as MATH. Also MATH factors as MATH . Since by REF is almost projective, the natural morphism induced by MATH is an epimorphism. Then for arbitrary MATH the morphism MATH is in the image of MATH, in other words, there exists a MATH-linear morphism MATH such that MATH. If now we take MATH, it is clear that MATH. This proves REF , since the MATH satisfying the assertion of REF form an ideal. REF : for a given MATH-module MATH, apply the functor MATH to the sequence REF. This yields MATH. Now the claim follows from REF . |
math/0002064 | We have a spectral sequence : MATH . On the other hand we have also natural isomorphisms MATH . Hence : MATH which is the claim. |
math/0002064 | REF : let MATH be such a MATH-module. Let MATH and pick a three term complex MATH such that MATH. Set MATH; this is a finitely presented MATH-module and MATH factors through a morphism MATH. Let MATH; from REF we see that MATH is the image of some element MATH. If we define MATH and MATH, MATH by MATH and MATH, then clearly MATH. Let MATH. Then MATH and the map MATH factors through a morphism MATH. Similarly the map MATH factors through a morphism MATH. Let MATH and MATH. The reader can check that MATH. By REF the claim follows. REF : let MATH be such an almost finitely generated almost projective MATH-module. For any finitely generated ideal MATH pick a morphism MATH such that MATH. If MATH is a set of generators for MATH, a standard argument shows that, for any MATH, MATH lifts to a morphism MATH; then, since MATH is almost projective, MATH lifts to a morphism MATH. Now REF applies with MATH, MATH, MATH, MATH and MATH and shows that MATH has a finitely generated submodule MATH containing MATH. Then the span of all such MATH is a finitely generated submodule of MATH containing MATH. By REF , the claim follows. |
math/0002064 | Pick an indexing set MATH large enough, and an epimorphism MATH. For every MATH we have the standard morphisms MATH such that MATH and MATH. For every MATH choose MATH such that MATH. It is easy to check that MATH is generated by the almost elements MATH (MATH, MATH). REF follows already. For REF , the ``only if" is clear; if MATH, then MATH for all MATH, hence MATH and therefore MATH. Next, notice that, from REF we derive MATH, that is, MATH, so REF follows directly from the definition of MATH. Since MATH is flat, to show REF we have only to verify that the functor MATH is faithful. To this purpose, it suffices to check that MATH for every proper ideal MATH of MATH. This follows easily from REF . |
math/0002064 | If MATH for some finite set MATH, then MATH and the claims are obvious. More generally, if MATH is almost projective and almost finitely generated, for any MATH there exists a finite set MATH and morphisms MATH such that MATH. We apply the natural transformation MATH to REF : an easy diagram chase allows then to conclude that the kernel and cokernel of REF are killed by MATH. As MATH is arbitrary, it follows that REF is an isomorphism in this case. An analogous argument works when MATH is almost finitely generated almost projective, so we get REF . If MATH is only almost projective, then we still have morphisms of the type REF, but now MATH is no longer necessarily finite. However, the cokernels of the induced morphisms MATH and MATH are still annihilated by MATH. Hence, to show REF (respectively, REF ) it suffices to consider the case when MATH is free and MATH is almost finitely generated (respectively, presented). By passing to almost elements, we can further reduce to the analogous question for usual rings and modules, and by the usual juggling we can even replace MATH by a finitely generated (respectively, presented) MATH-module and MATH by a free MATH-module. This case is easily dealt with, and REF follow. REF (respectively, REF ) is similar : one considers almost elements and replaces MATH by a free MATH-module (respectively, and MATH by a finitely generated MATH-module). In REF (respectively, REF ), for every finitely generated submodule MATH of MATH we can find, by REF , a finitely generated (respectively, presented) MATH-module MATH and a morphism MATH whose kernel and cokernel are annihilated by MATH. It follows easily that we can replace MATH by MATH and suppose that MATH is finitely generated (respectively, presented). Then the argument in CITE (Ch.I REF ) can be taken over verbatim to show REF (respectively, REF ). |
math/0002064 | REF is an easy consequence of REF . To prove REF , we we apply the natural transformation REF to REF : by diagram chase one sees that the kernel and cokernel of the morphism MATH are killed by MATH. |
math/0002064 | Let MATH. For any MATH let MATH. Then MATH. For MATH set MATH and let MATH be the natural morphism. An easy induction shows that MATH for all MATH. Since MATH we obtain MATH for all MATH. Therefore MATH for all MATH. Since MATH, the claim follows. |
math/0002064 | Let MATH. By the above REF it suffices to show that MATH vanishes for all MATH and all MATH-modules MATH. The maps MATH define a map MATH such that we have a short exact sequence MATH. Applying the long exact MATH sequence one obtains a short exact sequence (cp. REF ) MATH . Then REF implies that MATH for all MATH and moreover MATH is isomorphic to MATH. Let MATH be the transpose of MATH and write MATH as a composition MATH, so that MATH, the composition of the respective transposed morphims. We have monomorphisms MATH for all MATH. Hence MATH for all MATH. Since MATH is a NAME product, REF shows that MATH and the assertion follows. |
math/0002064 | Let MATH be a flat MATH-module and MATH an injective map of MATH-modules. Denote by MATH the kernel of the induced map MATH; we have MATH. We obtain an exact sequence MATH. But one sees easily that MATH and MATH, which shows that MATH is a flat MATH-module. Similarly, let MATH be two MATH-modules. Then the natural map MATH is an almost isomorphism and the assertion follows from REF . |
math/0002064 | Using the idempotent MATH we get a MATH-linear decomposition MATH where the bimodule structure on MATH is given by the zero morphisms and the bimodule structure on MATH is given by MATH and MATH. We have to prove that MATH is equivalent to a one-point category. By REF we can assume that MATH or MATH. By REF we have MATH and on MATH the bimodule actions are the zero morphisms. So it is enough to consider MATH. In this case, if MATH is any extension, MATH factors through a morphism MATH and composing with MATH we get a right inverse of MATH, which shows that MATH is the split extension. Then it is easy to see that MATH does not have any non-trivial automorphisms, which proves the assertion. |
math/0002064 | REF : the hypothesis MATH implies that MATH is a morphism of (non-unitary) MATH-monoids. We can then replace MATH by MATH and thereby assume that MATH, MATH and MATH is a (non-unitary) MATH-bimodule and the right and left actions on MATH coincide. The assertion to prove is that MATH lifts to a unique idempotent MATH. However, this follows easily from REF . To show REF , we observe that, by REF , the unit MATH of MATH lifts uniquely to an idempotent MATH. We have to show that MATH is a unit for MATH. Let us show the left unit property. Via MATH we can view the extension MATH as an exact sequence of left MATH-modules. We can then split MATH as the direct sum MATH where MATH is a sequence of unitary MATH-modules and MATH is a sequence of MATH-modules with trivial actions. But by hypothesis, on MATH and on MATH the MATH-module structure is unitary, so MATH and this is the left unit property. |
math/0002064 | Of course REF is an immediate consequence of REF . To show REF , let MATH be any object of MATH. Using REF one sees easily that the sequence MATH is right exact; MATH won't be exact in general, unless MATH (and therefore MATH) is an exact algebra. In any case, the kernel of MATH is almost zero, so we get an extension of MATH by a quotient of MATH which maps to MATH. In particular we get by pushout an extension MATH by MATH, that is, an object of MATH and in fact the assignment MATH is an essential inverse for the functor REF. |
math/0002064 | Using the adjunction REF we are reduced to showing that the natural map MATH is a bijection for all MATH-modules MATH. Given MATH we construct MATH. We extend MATH to MATH by setting it equal to zero on MATH. Then it is easy to check that the resulting map descends to MATH, hence giving a MATH-derivation MATH. This procedure yields a right inverse MATH to MATH. To show that MATH is injective, suppose that MATH is an almost zero MATH-derivation. Composing with the natural MATH-linear map MATH we obtain an almost zero map MATH. But MATH, hence MATH. This implies that in fact MATH, and the assertion follows. |
math/0002064 | To ease notation, set MATH and MATH. We have natural isomorphisms : MATH . But it is easy to see that the natural map MATH is an isomorphism. |
math/0002064 | With the notation of the proof of REF we have natural isomorphisms MATH where the last isomorphism follows directly from REF and the subsequent REF . Finally, REF shows that MATH, as required. |
math/0002064 | It follows directly from CITE |
math/0002064 | REF gives an isomorphism : MATH (and likewise for MATH). Then the claim follows from REF . |
math/0002064 | For any almost MATH-algebra MATH we let MATH denote the complex of MATH-modules MATH placed in degrees MATH; we have a distiguished triangle MATH . By the assumption, the natural map MATH is a quasi-isomorphism and MATH. On the other hand, for all MATH we have MATH . In particular MATH for all MATH. As MATH is flat over MATH, we have MATH. Then by the long exact Tor sequence associated to MATH we get the assertion for all MATH. Next we consider the natural map of distinguished triangles MATH; writing down the associated morphism of long exact Tor sequences, we obtain a diagram with exact rows : MATH . By the above, the leftmost vertical map is an isomorphism; moreover, the assumption gives MATH. Then, since MATH is injective, also MATH must be injective, which implies our assertion for the remaining case MATH. |
math/0002064 | Let us remark that the functor MATH : MATH commutes with tensor products; hence the same holds for the functor MATH (see REF ). Then, in view of REF , the theorem is reduced immediately to REF . |
math/0002064 | Since MATH for all MATH, REF applies (with MATH and MATH), giving the natural isomorphisms MATH . Since MATH, the same theorem also applies with MATH, MATH, MATH, and we notice that in this case MATH; hence we have MATH . Next we apply transitivity to the sequence MATH, to obtain (thanks to REF) MATH . Applying MATH to the second isomorphism REF we obtain MATH . Finally, composing REF we derive MATH . However, by inspection, the isomorphism REF is the sum map. Consequently MATH, as claimed. |
math/0002064 | We replace MATH by MATH and apply the functor MATH (which commutes with tensor products by REF ) thereby reducing the assertion to the above mentioned REF . |
math/0002064 | For REF use the Tor sequences. In view of REF , to show REF it suffices to know that MATH is an almost finitely presented MATH-module; but this follows from the existence of an epimorphism of MATH-modules MATH defined by MATH. Of the remaining assertions, only REF are not obvious, but the proof is just the ``almost version" of a well-known argument. Let us show REF ; the same argument applies to REF . We remark that MATH is an étale morphism, since MATH is unramified. Define MATH. By REF , MATH is étale. Define also MATH. By REF , MATH is flat (respectively, étale). The claim follows by remarking that MATH and applying REF . |
math/0002064 | Suppose that MATH is unramified. We start by showing that for every MATH there exist almost elements MATH of MATH such that MATH . Since MATH is an almost projective MATH-module, for every MATH there exists an ``approximate splitting" for the epimorphism MATH, that is, a MATH-linear morphism MATH such that MATH. Set MATH. We see that MATH. To show that MATH we use the MATH-linearity of MATH to compute MATH . Next take any almost element MATH of MATH and compute MATH . This establishes REF. Next let us take any other MATH and a corresponding almost element MATH. Both MATH and MATH are elements of MATH, hence we have MATH which implies MATH . Let us define a map MATH by the rule MATH . To show that REF does indeed determine a well defined morphism, we need to check that MATH and MATH for all MATH and all MATH. However, both identities follow easily by a repeated application of REF. It is easy to see that MATH defines an almost element with the required properties. Conversely, suppose an almost element MATH of MATH is given with the stated properties. We define MATH by MATH REF and MATH. Then REF says that MATH is a MATH-linear morphism and REF shows that MATH. Hence, by REF , MATH is unramified. |
math/0002064 | Left to the reader as an exercise. |
math/0002064 | Let MATH and MATH be given. Obviously we have MATH and MATH. Pick morphisms MATH and MATH as in REF. Using the foregoing notation, we can write : MATH from which the claim follows directly. |
math/0002064 | REF is left to the reader as an exercise. For REF we compute using REF : MATH. |
math/0002064 | Suppose first that there exists a splitting MATH for MATH, so that we can view MATH as a matrix MATH, where MATH. By additivity of the trace, we are then reduced to show that MATH. By REF , this is the same as MATH, which obviously vanishes. In general, for any MATH we consider the morphism MATH and the pull back morphism MATH : MATH . Then MATH is a split exact sequence with the endomorphism MATH, for a certain MATH. The pair of morphisms MATH induces a morphism MATH, and it is easy to check that MATH and MATH. We can therefore apply REF to deduce that MATH. By the foregoing we know that MATH, so the claim follows. |
math/0002064 | REF are left as exercises for the reader. We verify REF . For given MATH pick morphisms MATH and MATH such that MATH and MATH. If we set MATH, MATH, MATH and MATH then we have MATH. Define MATH . Using REF we can write MATH which implies immediately the claim. |
math/0002064 | Under the stated hypotheses, MATH is an almost projective MATH-module (by REF ). Let MATH and MATH the trace morphism for the morphism of almost MATH-algebras MATH. By faithful flatness, the natural morphism MATH is a monomorphism, hence it suffices to show that MATH is an epimorphism (here MATH is considered as a MATH-algebra via the second factor). However, from REF we see that MATH is a right inverse for MATH. The claim follows. |
math/0002064 | Suppose that MATH is étale. Let MATH be the idempotent almost element of MATH provided by REF . We define a morphism MATH by MATH. To start with, we remark that both MATH and MATH are MATH-linear morphisms (for the natural MATH-module structure of MATH defined in REF ). Indeed, let us pick any MATH, MATH and compute directly MATH . Next we show that MATH is a left inverse for MATH. In fact, let MATH. We have MATH . Therefore it suffices to show that MATH. However, by REF is unramified, hence REF gives a decomposition MATH such that MATH acts as the identity on the first factor and as the zero morphism on the second factor. Now, let MATH. From the above we derive a MATH-linear isomorphism MATH. We dualize and apply REF to obtain another MATH-linear isomorphism MATH . Finally, composing the isomorphism REF with the projection on the first factor, we get a split MATH-linear epimorphism MATH, hence a surjective MATH-linear morphism MATH. Such a morphism is necessarily an isomorphism, and, tracing backward, the same must hold for MATH. Conversely, suppose that the trace form is a perfect pairing. By REF the natural morphism MATH is an isomorphism and one verifies easily that MATH. In particular MATH is also an isomorphism. The multiplication gives an almost element MATH; let MATH. We derive MATH . Furthermore, we have already remarked that MATH is a MATH-linear morphism, hence MATH is a MATH-linear morphism. Consequently, for any almost element MATH of MATH we have MATH . Since by REF is an isomorphism, this implies MATH . Consider the morphism MATH defined by MATH; then MATH is MATH-linear (for the MATH-module structure defined by the second factor). Applying REF we conclude that MATH. On the other hand, REF says that this trace is equal to MATH, hence MATH . Let MATH be defined as MATH . From REF we see that both MATH and MATH are MATH-linear morphisms and from REF we know moreover that MATH. By REF we deduce that MATH is an almost projective MATH-module, that is, MATH is unramified, as claimed. |
math/0002064 | Under the stated hypothesis, MATH is an almost projective MATH-module (by virtue of REF ). Hence, for given MATH, pick a sequence of morphisms MATH such that MATH; with the notation of REF, define MATH by MATH, so that MATH. One verifies easily that MATH for all integers MATH. Now, suppose that MATH. It follows that MATH for MATH sufficiently large, hence MATH for MATH sufficiently large. Let MATH be any prime ideal of MATH; let MATH be the natural projection and MATH the fraction field of MATH. The MATH-linear morphism MATH is nilpotent on the vector space MATH, hence MATH. This shows that MATH lies in the intersection of all prime ideals of MATH, hence it is nilpotent. Since by REF is reduced, we get MATH. Finally, this implies MATH. Now, for any MATH, the almost element MATH will be nilpotent as well, so the same conclusion applies to it. This shows that MATH. But by REF is étale over MATH, hence REF yields MATH, as required. |
math/0002064 | Let MATH be any MATH-extension of MATH by MATH. The composition MATH of the structural morphism for MATH followed by MATH coincides with the projection MATH. Therefore MATH and MATH. Hence MATH factors through MATH; the restriction of MATH to MATH defines a morphism MATH and a morphism of MATH-extensions MATH. In this way we obtain an inverse for REF. |
math/0002064 | It follows directly from the (almost version of the) local flatness criterion (see REF ). |
math/0002064 | By induction we can assume MATH. Then REF follows directly from REF . We show REF : by REF (and again REF ) a given weakly étale morphism MATH can be lifted to a unique flat morphism MATH. We need to prove that MATH is weakly étale, that is, that MATH is MATH-flat. However, it is clear that MATH is weakly étale, hence it has a flat lifting MATH. Then the composition MATH is flat and it is a lifting of MATH. We deduce that there is an isomorphism of MATH-algebras MATH lifting MATH and moreover the morphisms MATH and MATH coincide with MATH. REF follows. To show REF , suppose that MATH is étale and let MATH denote as usual the kernel of MATH. By REF there is a natural morphism of almost algebras MATH which is clearly étale. Hence MATH lifts to a weakly étale MATH-algebra MATH, and the isomorphism MATH lifts to an isomorphism MATH of MATH-algebras. It follows that MATH is an almost projective MATH-module, that is, MATH is étale, as claimed. |
math/0002064 | Under the assumptions, we can find a finitely generated MATH-module MATH such that MATH. By REF , there exists a finite filtration MATH such that each MATH is a quotient of a direct sum of copies of MATH. This implies that, for every MATH-module MATH, we have MATH . REF follows easily. Notice that if MATH and MATH, then we can take MATH in REF. For REF let MATH. We estimate MATH in two ways. By the first spectral sequence of hyperhomology we have an exact sequence MATH. By the second spectral sequence for hyperhomology we have an exact sequence MATH. Hence MATH is annihilated by the product of the three annihilators in REF and the result follows by applying REF with MATH. |
math/0002064 | REF : we have to show that MATH is almost zero for every MATH-module MATH. Let MATH; by REF is nilpotent, so by the usual devissage we may assume that MATH. If MATH is represented by an extension MATH then after tensoring by MATH and using the flatness of MATH we get an exact sequence of MATH-modules MATH. Thus MATH comes from an element of MATH which is almost zero by assumption. REF : let MATH be a finitely generated subideal of MATH. By assumption there is a map MATH such that MATH. For all MATH the morphism MATH lifts to a morphism MATH. Then MATH satisfies MATH. By REF it follows MATH for some MATH. As MATH was arbitrary, the result follows. CASE: Let MATH be as above. By REF , MATH is almost finitely generated over MATH, so we can choose a morphism MATH such that MATH. Consider MATH. By REF , there is a finitely generated submodule MATH of MATH containing MATH. Notice that MATH maps onto MATH and MATH is annihilated by MATH. Hence MATH is contained in the image of MATH and therefore we can lift a finite generating set MATH for MATH to almost elements MATH of MATH. If we quotient MATH by the span of these MATH, we get a finitely presented MATH-module MATH with a morphism MATH such that MATH is annihilated by MATH and MATH is annihilated by MATH. By REF we derive MATH for some MATH. Since MATH is arbitrary, this proves the result. |
math/0002064 | As usual we reduce to MATH. Then REF applies with MATH, MATH, MATH, MATH, MATH and MATH. We obtain a class MATH which gives the obstruction to the existence of a flat MATH-module MATH lifting MATH. Since MATH is almost projective, we know that MATH, which says that MATH for all MATH. In other words, for every MATH we can find an extension of MATH-modules MATH of MATH by MATH such that MATH. Let MATH be the class of MATH. Notice that, for any MATH, MATH is the class of an extension MATH such that MATH, hence, by REF , MATH for all MATH. Hence we can define a morphism MATH . However, one sees easily that MATH and MATH, hence we can view MATH as an element of MATH and moreover we have a spectral sequence MATH with MATH for all MATH (this spectral sequence is constructed for example, from the double complex MATH where MATH (respectively, MATH) is a projective resolution of MATH (respectively, MATH)). In particular, our MATH is an element in MATH which therefore survives in the abutment as a class of MATH. The latter can be lifted to an element MATH via the surjection MATH. Let MATH be an extension representing MATH. Checking compatibilities, we see that MATH for every MATH. Hence MATH coincides with the identity map on the submodule MATH. Since MATH, we see that MATH is actually the identity map. By the local flatness criterion, it then follows that MATH is flat over MATH, hence the MATH-module MATH is a flat lifting of MATH, so it is almost projective, by REF . Now REF follows from REF . |
math/0002064 | Since both MATH and MATH are almost projective and MATH are epimorphisms, there exist morphisms MATH and MATH such that MATH and MATH. Then we have MATH and MATH, that is, the morphism MATH (respectively, MATH) has image contained in the almost submodule MATH (respectively, MATH). Since MATH this implies MATH and MATH. Hence MATH . Define MATH. Notice that MATH . This implies the equalities MATH and MATH. Then the pair MATH satisfies PQ(MATH). Under REF , every element of MATH is a multiple of an element of the form MATH, therefore the claim follows for arbitrary MATH. |
math/0002064 | Let MATH be such a morphism. Under the assumption, we can find a finite MATH-module MATH such that MATH. One sees easily that MATH is a faithful MATH-module, so by CITE REF , MATH satisfies the following condition : MATH . Now let MATH be a MATH-module such that MATH is flat. Pick an epimorphism MATH with MATH free over MATH. Then MATH is universally exact over MATH, hence over MATH. Consider the sequence MATH. Clearly MATH. However, it is easy to check that a sequence MATH of MATH-modules is universally exact if and only if the sequence MATH is universally exact over MATH. We conclude that MATH is a universally exact sequence of MATH-modules, hence, by REF , MATH is flat over MATH, that is, MATH is flat over MATH as required. |
math/0002064 | In the weakly étale case, we have to show that the multiplication morphism MATH is flat. As MATH is nilpotent, the local flatness criterion reduces the question to the situation over MATH. So we may assume that MATH is a monomorphism. Then MATH is a monomorphism, but MATH is the multiplication morphism of MATH, which is flat by assumption. Therefore, by REF , MATH is flat. For the étale case, we have to show that MATH is almost finitely presented as a MATH-module. By REF is almost finitely presented as a MATH-module and we know already that MATH is flat as a MATH-module, so by REF (applied to the finite morphism MATH) the claim follows. |
math/0002064 | General nonsense. |
math/0002064 | Indeed, MATH is the natural morphism. So, the assertions follow by applying MATH to the short exact sequence of MATH-modules MATH where MATH and MATH. |
math/0002064 | Indeed, in this case, REF is split exact as a sequence of MATH-modules, and it remains such after tensoring by MATH. |
math/0002064 | To fix ideas, suppose that MATH is an epimorphism. Consider any object MATH of MATH. Let MATH; we deduce a natural morphism MATH such that MATH. It follows that MATH is injective, hence it is an isomorphism, by REF . We derive a commutative diagram with exact rows : MATH . From the snake lemma we deduce MATH . Since MATH we have MATH. But by REF is an epimorphism, so also MATH is an epimorphism. Then MATH implies that MATH. But MATH, thus MATH as well. We look at the exact sequence MATH : applying MATH we obtain an epimorphism MATH. From MATH it follows that MATH. In conclusion, MATH is an isomorphism. Hence the same is true for MATH, and again MATH, MATH show that MATH is an isomorphism as well, which implies the claim. |
math/0002064 | Suppose that MATH is an epimorphism and let MATH be its kernel. Let MATH; it suffices to show that MATH is a flat MATH-module. However, in view of REF , the assumption implies that MATH is a flat MATH-module. MATH is the kernel of the epimorphism MATH. Moreover, MATH identifies naturally with an ideal of MATH and MATH. Then the desired conclusion follows from CITE (lemma in CITE). |
math/0002064 | The assertion for flat almost modules follows directly from REF . Set MATH. To establish the second equivalence, it suffices to show that, if MATH is a MATH-module such that MATH is almost projective over MATH, then MATH is almost projective over MATH, or which is the same, that MATH for all MATH and any MATH-module MATH. We know already that MATH is flat. Let MATH be any MATH-module and MATH any MATH-module. The standard isomorphism MATH yields a natural isomorphism MATH, whenever MATH for every MATH. In particular, we have MATH whenever MATH comes from either a MATH-module, or a MATH-module. For a general MATH-module MATH there is a REF-step filtration such that MATH, MATH, MATH and MATH. By an easy devissage, we reduce to verify that MATH for every MATH and MATH. However, MATH is a MATH-module and MATH is a MATH-module, so the required vanishing follows for MATH. Moreover, applying MATH to REF, we derive a short exact sequence : MATH . Here again, the leftmost term of REF is a MATH-module, and the middle term is a MATH-module, so the same devissage yields the sought vanishing also for MATH. |
math/0002064 | For any MATH, denote by MATH (respectively, MATH) the MATH-fold tensor product of MATH (respectively, MATH) with itself over MATH (respectively, MATH), and by MATH the natural morphism. First of all we claim that, for every MATH, the natural diagram of almost algebras MATH is cartesian (where MATH and MATH are MATH-fold multiplication morphisms). For this, we need to verify that, for every MATH, the induced morphism MATH (defined by multiplication of the first two factors) is an isomorphism. It then suffices to check that the natural morphism MATH is an isomorphism for all MATH. Indeed, consider the commutative diagram MATH . From MATH, it follows that MATH is an epimorphism. Hence also MATH is an epimorphism. Since MATH is a monomorphism, it follows that MATH is also a monomorphism, hence MATH is an isomorphism and the claim follows easily. We consider first the case MATH; we see that REF is a diagram of the kind considered in REF, hence, for every MATH, we have the associated functor MATH and also its right adjoint MATH. Denote by MATH REF the usual morphisms, and similarly define MATH. Suppose there is given a descent datum MATH for MATH, relative to MATH. The cocycle REF implies easily that MATH is the identity on MATH. It follows that the pair MATH defines an isomorphism MATH in the category MATH. Hence MATH is an isomorphism. However, we remark that either morphism MATH yields a section for MATH, hence we are in the situation contemplated in REF , and we derive an isomorphism MATH. We claim that MATH is an object of MATH, that is, that MATH verifies the cocycle REF . Indeed, we can compute: MATH and by construction we have MATH and MATH. Therefore, the cocycle identity for MATH implies the equality MATH. If we now apply the functor MATH to this equality, and then invoke again REF , the required cocycle identity for MATH will ensue. Clearly MATH is the only descent datum on MATH lifting MATH. This proves that REF is essentially surjective. The same sort of argument also shows that the functor REF induces bijections on morphisms, so the lemma follows in this case. Next, the case MATH can be deduced formally from the previous case, by applying repeatedly natural isomorphisms of the kind MATH (MATH). Finally, the ``étaleness" of an object of MATH can be checked on its projection onto MATH, hence also the cases MATH and MATH follow directly. |
math/0002064 | Let us say that MATH is an epimorphism with kernel MATH. Then MATH is also an ideal of MATH and we have MATH and MATH. We intend to apply REF to the morphism MATH. However, the induced morphism MATH in MATH has a section, and hence it is of universal effective descent for every fibred category. Thus, we can replace in REF the category MATH by MATH, and thereby, identify (up to equivalence) the target of REF with REF-fibred product MATH. The latter is equivalent to the category MATH and the resulting functor MATH is canonically isomorphic to REF, which gives the claim. |
math/0002064 | REF : suppose that MATH for a finite MATH-algebra MATH; then MATH is a finite MATH-algebra and MATH. It remains to show that MATH is nilpotent. Suppose that MATH is generated by MATH elements as a MATH-module and let MATH (respectively, MATH) be the NAME ideal of MATH (respectively,of MATH); we have MATH (see CITE (Chap. XIX REF )); on the other hand MATH, so the claim is clear. REF : we shall consider the fibred category MATH; the same argument applies also to étale almost algebras. We begin by establishing a very special case : REF holds when MATH, where MATH and MATH are ideals in MATH such that MATH is nilpotent. First of all we remark that the situation considered in the claim is stable under arbitrary base change, therefore it suffices to show that MATH is of MATH-REF-descent in this case. Then we factor MATH as a composition MATH and we remark that MATH is of MATH-REF-descent by REF ; since a composition of morphisms of MATH-REF-descent is again of MATH-REF-descent, we are reduced to show that MATH is of MATH-REF-descent, that is, we can assume that MATH. However, in this case the claim follows easily from REF . More generally, REF holds when MATH, where MATH are ideals of MATH, such that MATH is nilpotent. We prove this by induction on MATH, the case MATH being covered by REF . Therefore, suppose that MATH, and set MATH. By induction, the morphism MATH is of universal MATH-REF-descent. However, according to CITE (Chap. II REF ), the sieves of universal MATH-REF-descent form a topology on MATH; for this topology, MATH is a covering family of MATH and MATH is a covering morphism, hence MATH is a covering family of MATH, and then, by composition of covering families, MATH is a covering family of MATH, which is equivalent to the claim. Now, let MATH be a general strictly finite morphism, so that MATH for some finite MATH-algebra MATH. Pick generators MATH of the MATH-module MATH, and monic polynomials MATH such that MATH for MATH. There exists a finite and faithfully flat extension MATH of MATH such that the images in MATH of MATH,.,MATH split as products of monic linear factors. This extension MATH can be obtained as follows. It suffices to find, for each MATH, an extension MATH that splits MATH, because then MATH will split them all, so we can assume that MATH and MATH; moreover, by induction on the degree of MATH, it suffices to find an extension MATH such that MATH factors in MATH as a product of the form MATH, where MATH is a monic polynomial of degree MATH. Clearly we can take MATH. Given a MATH as in REF , we remark that the morphism MATH is of universal MATH-REF-descent. Considering again the topology of universal MATH-REF-descent, it follows that MATH is of universal MATH-REF-descent if and only if the same holds for the induced morphism MATH. Therefore, in proving REF we can replace MATH by MATH and assume from start that the polynomials MATH factor in MATH as product of linear factors. Now, let MATH and MATH (for MATH). We get a surjective homomorphism of MATH-algebras MATH by the rule MATH (MATH). Moreover, any sequence MATH yields a homomorphism MATH, determined by the assignment MATH. A simple combinatorial argument shows that MATH, where MATH runs over all the sequences as above. Hence the product map MATH has nilpotent kernel. We notice that the MATH-algebra MATH is a quotient of MATH, hence it can be written as a product of rings of the form MATH, for various ideals MATH. By REF , the kernel of the induced homomorphism MATH is nilpotent, hence the same holds for the kernel of the composition MATH, which is therefore of the kind considered in REF . Hence MATH is of universal MATH-REF-descent. Since such morphisms form a topology, it follows that also MATH is of universal MATH-REF-descent, which concludes the proof of REF . Finally, let MATH be as in REF and pick again MATH as in the proof of REF . By REF , MATH is almost projective over MATH if and only if MATH is almost projective over MATH; hence we can replace MATH by MATH, and by arguing as in the proof of REF , we can assume from start that MATH for ideals MATH, MATH such that MATH is nilpotent. By an easy induction, we can furthermore reduce to the case MATH. We factor MATH as MATH; by REF it follows that MATH is almost projective over MATH, and then REF says that MATH itself is almost projective. |
math/0002064 | REF : if MATH is an inverse of MATH up to MATH and MATH is an inverse of MATH up to MATH, then MATH is an inverse of MATH up to MATH. REF : given an inverse MATH of MATH up to MATH and an inverse MATH of MATH up to MATH, let MATH. We compute : MATH . REF is similar and REF is an easy diagram chasing left to the reader. REF follows from REF and the snake lemma. |
math/0002064 | REF : given MATH, construct a morphism MATH using the morphism MATH coming from MATH and MATH. REF : the assertion for MATH is clear, and the assertion for MATH follows from REF . |
math/0002064 | REF : we first show that MATH is faithfully flat. Since MATH is flat, it remains to show that if MATH is a MATH-module such that MATH, then MATH. It suffice to do this for MATH, for an arbitrary ideal MATH of MATH. After base change by MATH, we reduce to show that MATH implies MATH. However, MATH is invertible up to MATH, so MATH which means MATH. In particular, MATH is a monomorphism, hence the proof is complete in case that MATH is an epimorphism. In general, consider the composition MATH. From REF it follows that MATH is invertible up to MATH; then REF says that MATH is invertible up to MATH. The latter is also weakly étale; by the foregoing we derive that it is an isomorphism. Consequently MATH is an isomorphism, and finally, by faithful flatness, MATH itself is an isomorphism. REF : the morphisms MATH and MATH are invertible up to MATH. By REF it follows that MATH is invertible up to MATH; hence, by REF , the morphism MATH induced by MATH and MATH is invertible up to MATH (in fact one verifies that it is invertible up to MATH). But MATH is a morphism of weakly étale MATH-algebras, so it is weakly étale, so it is an isomorphism by REF . |
math/0002064 | We first consider REF for the special case where MATH. The functor MATH induces a functor MATH, and by restriction (see REF ) we obtain a functor MATH; by REF , the latter is isomorphic to the functor MATH of the lemma. Furthermore, from REF we derive a natural ring isomorphism MATH, hence an essentially commutative diagram MATH where MATH and MATH are the equivalences of REF . Clearly MATH and MATH restrict to equivalences on the corresponding categories of étale algebras, hence the lemma follows in this case. For the general case of REF , let MATH be a morphism as in REF induces an essentially commutative diagram of the corresponding categories of algebras, so by the previous case, the functor MATH has both a left essential inverse and a right essential inverse; these essential inverses must be isomorphic, so MATH has an essential inverse as desired. Finally, we remark that the map in REF is the same as the map MATH, and the latter is a bijection in view of REF . |
math/0002064 | In light of REF , it suffices to show that MATH is invertible up to a power of MATH. For this, factor the identity morphism of MATH as MATH and argue as in the proof of REF . |
math/0002064 | Suppose first that MATH is an isomorphism; in this case we claim that MATH is an epimorphism and MATH for any MATH-algebra MATH. Indeed, since by REF, MATH acts on MATH, hence MATH annihilates MATH, hence annihilates its image MATH, whence the claim. If, moreover, MATH is a MATH-algebra, REF implies that MATH is invertible up to a power of MATH. In the general case, consider the intermediate almost MATH-algebra MATH equipped with the ideal MATH. In REF is an epimorphism with nilpotent kernel, hence it remains such after any base change MATH. To prove REF , it suffices then to consider the morphism MATH, hence we can assume from start that MATH is an isomorphism, which is the case already dealt with. To prove REF , it suffices to consider the cases of MATH and MATH. The second case is treated above. In the first case, we do not necessarily have MATH and the assertion to be checked is that, for every MATH-algebra MATH, the morphism MATH is invertible up to a power of MATH. We apply REF to the commutative diagram with exact rows: MATH to deduce that MATH is invertible up to some power of MATH, hence so is MATH, which implies the assertion. As for REF , we remark that the ``only if" part is trivial; and we assume therefore that MATH is MATH-covering (respectively, MATH-bicovering). Consider first the assertion for ``covering". We need to show that MATH is of universal effective descent for MATH, where MATH is either one of our two fibered categories. In light of REF , this is reduced to the assertion that MATH is of effective descent for MATH. We notice that MATH is bicovering for MATH (in REF by REF , in REF by REF ). As MATH is an isomorphism, the assertion is reduced to the case where MATH is an isomorphism. In this case, by REF , there is a natural equivalence: MATH. Then the assertion follows easily from REF . Finally suppose that MATH is bicovering. The foregoing already says that MATH is covering, so it remains to show that MATH is also covering. The above argument again reduces to the case where MATH is an isomorphism. Then, as in the proof of REF , the induced morphism MATH is an isomorphism as well. Thus the assertion for ``bicovering" is reduced to the assertion for ``covering". |
math/0002064 | REF : let MATH be a quasi-isomorphism of MATH-algebras. Clearly the induced morphism MATH is still a quasi-isomorphism of MATH-algebras. But by REF , MATH and MATH are exact simplicial almost MATH-algebras; moreover, it follows from REF that MATH is a quasi-isomorphism of MATH-modules. Then the claim follows easily from the exactness of the sequence REF. Now REF is clear. |
math/0002064 | We show by induction on MATH that VAN REF MATH . For MATH the claim follows immediately from REF . Therefore suppose that MATH and that VAN REF is known for all almost isomorphisms of MATH-algebras MATH and all MATH. Let MATH. Then by transitivity REF we have a distinguished triangle in MATH . We deduce that VAN REF and VAN REF imply VAN(MATH), thus we can assume that MATH is either injective or surjective. Let MATH be the simplicial MATH-algebra augmented over MATH defined by MATH. It is a simplicial resolution of MATH by free MATH-algebras, in particular the augmentation is a quasi-isomorphism of simplicial MATH-algebras. Set MATH. This is a simplicial MATH-algebra augmented over MATH via a quasi-isomorphism. Moreover, the induced morphisms MATH are isomorphisms. By REF there is a quasi-isomorphism MATH. On the other hand we have a spectral sequence MATH . It follows easily that MATH for all MATH implies MATH. Therefore we are reduced to the case where MATH is a free MATH-algebra and MATH is either injective or surjective. We examine separately these two cases. If MATH is surjective, then we can find a right inverse MATH for MATH. By applying transitivity to the sequence MATH we get a distinguished triangle MATH . Since MATH there follows an isomorphism : MATH. Furthermore, since MATH is an isomorphism, MATH is an isomorphism as well, hence by induction (and by a spectral sequence of the type REF ) MATH. The claim follows in this case. Finally suppose that MATH is injective. Write MATH, for a free MATH-module MATH and set MATH; since MATH is an isomorphism, MATH. We apply transitivity to the sequence MATH. By arguing as above we are reduced to showing that MATH . We know that MATH and we will show that MATH for MATH. To this purpose we apply transitivity to the sequence MATH. As MATH and MATH are flat MATH-modules, REF yields MATH for MATH and MATH is a flat MATH-module. In particular MATH for all MATH. Consequently MATH for all MATH and MATH. The latter module is easily seen to be almost zero. |
math/0002064 | Apply the base change theorem (CITE II. REFEF) to the (flat) projections of MATH onto MATH and respectively MATH to deduce that the natural map MATH is a quasi-isomorphism in MATH. By REF the induced morphism MATH is still a quasi-isomorphism. There are spectral sequences MATH . On the other hand, by REF we have MATH for all MATH. Then the theorem follows directly from REF and transitivity. |
math/0002064 | By transitivity we may assume MATH. Let MATH be the standard resolution of MATH (see CITE II. REFEF). Each MATH contains MATH as a direct summand, hence it is exact, so that we have an exact sequence of simplicial MATH-modules MATH. The augmentation MATH is a quasi-isomorphism and we deduce that MATH is a quasi-isomorphism; hence MATH is a quasi-isomorphism as well. We have MATH for a free MATH-module MATH and the map MATH is identified with MATH, whence MATH is identified with MATH. By REF the map MATH is a quasi-isomorphism. In view of REF we derive that MATH is a quasi-isomorphism, that is, MATH is a quasi-isomorphism. Since MATH is flat and MATH is a quasi-isomorphism, we get the desired conclusion. |
math/0002064 | Let MATH be the map corresponding to MATH under the bijection REF. By inspection, the compositions MATH and MATH are induced by scalar multiplication. Pick any MATH and lift it to an element MATH; define MATH by MATH for all MATH. Then MATH and MATH. This easily implies that MATH annihilates MATH and MATH. In light of REF , the claim follows. |
math/0002064 | Fix an element MATH. For each MATH-algebra MATH and each element MATH we get a map MATH by MATH, hence a map MATH. For varying MATH we obtain a map of sets MATH : MATH. According to the terminology of CITE, the system of maps MATH for MATH ranging over all MATH-algebras forms a homogeneous polynomial law of degree MATH from MATH to MATH, so it factors through the universal homogeneous degree MATH polynomial law MATH . The resulting MATH-linear map MATH depends MATH-linearly on MATH, hence we derive a MATH-linear map MATH. Next notice that by REF acts trivially on MATH so MATH acts trivially on MATH and we get an isomorphism MATH. We deduce a natural map MATH. Now, in order to prove the proposition for the case MATH, it suffices to show that this last map is just the natural isomorphism that ``reorders the factors". Indeed, let MATH and MATH such that MATH; then MATH sends the generator MATH to MATH. On the other hand, pick any MATH and let MATH be the polynomial MATH-algebra in MATH variables; write MATH with MATH. Then MATH (see CITE pp. REF) and the claim follows easily. Next notice that MATH is flat, so that tensoring with MATH commutes with taking coinvariants (respectively, invariants) under the action of the symmetric group; this implies the assertion for MATH (respectively, MATH). To deal with MATH recall that for any MATH-module MATH and MATH we have the antisymmetrization operator MATH and a surjection MATH which is an isomorphism for MATH free, hence for MATH flat. The result for MATH (and again the flatness of MATH) then gives MATH, hence the assertion for MATH and MATH flat. For general MATH let MATH be a presentation with MATH free. Define MATH by MATH and MATH. By functoriality we derive an exact sequence MATH which reduces the assertion to the flat case. For MATH the same reduction argument works as well (compare CITE p. REF) and for flat modules the assertion for MATH follows from the corresponding assertion for MATH. |
math/0002064 | It is deduced directly from REF by applying MATH. |
math/0002064 | This is deduced from REF applied to MATH which is a quasi-isomorphism of chain complexes of flat MATH-modules. We note that CITE deals with a more general mixed simplicial construction of MATH which applies to bounded above complexes, but one can check that it reduces to the simplicial definition for complexes in MATH. |
math/0002072 | This is an immediate consequence of REF , since the second term on the right-hand side is positive semi-definite, and zero iff MATH. |
math/0002072 | Suppose MATH is a NAME critical point; that is, a point at which MATH is stationary. (There is no loss of generality in omitting a possible gauge transformation on one side of this equation.) By considering just MATH of the form MATH, for MATH, it is clear from REF that MATH is a necessary condition for MATH to be a critical point. It then follows that MATH must be a critical point of the energy. The converse is immediate. |
math/0002072 | A simple calculus of variations computation shows that MATH extremizes MATH iff MATH . On the other hand, this condition also ensures that the covariantly constant curvatures MATH, related by the non-Abelian analog of NAME Theorem mentioned previously, are continuous at MATH. This was the only place MATH might have failed to be smooth. |
math/0002075 | Write MATH, where the prime denotes the imaginary part. Then MATH . All these products, except for the cross-product, are commutative, and REF follows. From the same formula with MATH we obtain MATH. This vanishes if and only if MATH is real or purely imaginary. Together with REF we obtain REF. |
math/0002075 | CASE: If MATH and if MATH is a unit vector orthogonal to REF, then MATH. Hence MATH works for MATH, and the uniqueness, up to sign, follows easily from MATH and MATH. If MATH is arbitrary, and MATH then put MATH. Clearly, MATH. Moreover, MATH works for MATH if and only if MATH works for MATH. CASE: If MATH, and MATH is an orthonormal basis of MATH, then MATH satifies the requirements: Use the geometric properties of the cross product. CASE: The above argument shows that MATH has MATH-eigenspaces of real dimension REF. Since MATH is orthogonal, so are its eigenspaces. |
math/0002075 | The tangent vector MATH is identified with the homomorphism MATH, that maps MATH into MATH. |
math/0002075 | We consider MATH as a (right) complex vector space with imaginary unit MATH. Then MATH is MATH-linear and has a (complex) eigenvalue MATH. If MATH, then MATH . Hence MATH. We choose a basis MATH of MATH such that MATH, that is, MATH for some MATH, and MATH. Then MATH implies MATH . For the affine parametrization MATH we get: MATH . This is a real-linear equation for MATH, with associated homogeneuos equation MATH . By REF this is of real dimension REF, and any real REF-plane can be realized this way. |
math/0002075 | Choose non-zero basis vectors MATH. Then elements in MATH and endomorphisms of MATH or of MATH are represented by quaternionic MATH-matrices, and therefore the assertion reduces to that of REF . |
math/0002075 | The vector bundle MATH with fibre MATH has a total space of real dimension MATH. Therefore there exists MATH such that MATH has no zero. REF yields a unique holomorphic structure MATH such that MATH. Now any MATH is of the form MATH for some MATH. Then, by REF, for any section MATH we have MATH . Note that MATH implies MATH, and this allows us to replace MATH by MATH in the last term as well: MATH . This contains no reference to MATH, hence MATH is independent of the choice of MATH such that MATH has no zero. But the last equality shows MATH for any MATH with MATH. |
math/0002075 | Recall from REF MATH . Therefore, using MATH, MATH . But MATH by the following type argument: Using that MATH we have MATH . Similarly MATH, because MATH is left MATH and MATH is right MATH. |
math/0002075 | We first need a formula for the derivative of MATH-forms MATH which stabilize MATH, that is, MATH. If MATH, then for MATH where we wedge over composition. Note that the composition MATH makes sense, because MATH, and MATH is annihilated by MATH. We apply this to MATH and MATH. Since MATH we have on MATH, by REF , MATH . By a type argument similar to REF, we get MATH. Further, MATH and similarly for the remaining term. We obtain MATH or MATH . Since MATH and MATH for MATH is an isomorphism, we get MATH. |
math/0002075 | We have MATH, and therefore MATH . Then, from REF, MATH . Similarly, MATH . But, by a property of the real trace, MATH . |
math/0002075 | Let MATH be a variation of MATH in MATH with variational vector field MATH. Then MATH and MATH . Using the wedge REF and MATH, we get MATH . Thus MATH . Therefore MATH is harmonic if and only if MATH is normal. For the other equivalences, first note MATH . Now, together with MATH and MATH, this implies MATH . |
math/0002075 | Recall from REF MATH . Further MATH by type. Therefore MATH and similarly for MATH. |
math/0002075 | We may assume MATH. Then MATH, and MATH implies MATH. Let MATH, and MATH . Then MATH, and MATH implies MATH . Therefore MATH is imaginary, too. It follows MATH, and MATH . |
math/0002075 | MATH . Because MATH we similarly have MATH see REF. Because MATH, we have MATH . This proves REF. The positivity follows from REF . |
math/0002075 | CASE: We consider REF-form on MATH defined by MATH . Then MATH is a linear combination of terms of the form MATH . But if MATH, we get MATH hence MATH. Therefore, if MATH is the inclusion, MATH . CASE: We have MATH and REF yields the formula. The topological invariance under deformations of MATH follows from NAME theorem: If MATH deforms MATH into MATH, then MATH . |
math/0002075 | Let MATH be a variation, and MATH its mean curvature sphere. Note that for MATH to stay conformal the complex structure, that is, the operator MATH, varies, too. The variation has a variational vector field MATH given by MATH . As usual, we abbreviate MATH by a dot. Note that for MATH . We now compute the variation of the energy, which is as good as the NAME functional as long as we vary MATH. By contrast, in the proof of REF the conformal structure on MATH was fixed, and no MATH was involved. MATH . In general MATH, because MATH. Hence MATH . Next we claim MATH . On MATH let MATH, that is, MATH. Then we have MATH, and MATH . But MATH is conformal, see REF , therefore MATH . Differentiation with respect to MATH yields MATH for all MATH. Using this with MATH we get REF. Now, we compute the integral MATH. MATH . We shall show in the following lemma that MATH . Therefore we can consider MATH as a REF-form MATH and continue MATH . Now MATH is tangential by REF, and hence anti-commutes with MATH. Thus MATH . We therefore showed MATH . Since MATH, this vanishes for all variational vector fields MATH if and only if MATH . |
math/0002075 | For MATH because MATH. But MATH is right MATH, and MATH is left MATH. Hence, by type, MATH . This shows the right hand inclusion. Also, MATH . |
math/0002075 | We know from REF that MATH is an involution with the tangent space as its fixed point set: MATH . Its MATH-eigenspace is the normal space, so we need to compute MATH . But differentiation of REF yields MATH or MATH . |
math/0002075 | By definition of the trace, MATH but MATH . If follows that MATH and MATH . Similarly for MATH. |
math/0002075 | MATH . Therefore MATH . This proves the formula for MATH. Using REF and the NAME equation MATH we find, after a similar computation, MATH . On this we use REF to obtain REF. |
math/0002075 | REF give MATH . |
math/0002075 | We only have to consider the reformulation of MATH. But MATH . |
math/0002075 | MATH . Now see REF and, for the second equality, REF. |