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math/9906161 | The following pairing formula was proved by CITE for short-range MATH, but the same proof also applies when MATH is long-range. Also, the proof can be easily localized, see CITE. Suppose that for MATH, MATH, MATH and MATH. Let MATH. Then MATH . We apply this result with MATH, MATH. By the construction of MATH we conclude that MATH, MATH, while for MATH we see directly from the definition of MATH and MATH that MATH, MATH. Substitution into REF proves the proposition. |
math/9906161 | Let MATH. Suppose also that there is no generalized broken geodesic of length MATH starting at some MATH and ending at MATH. That means that for any MATH we cannot have MATH, MATH, and MATH, MATH, at the same time. REF (with MATH signs instead of MATH) implies that MATH indeed, we also have MATH, so we can even replace MATH by MATH. Thus, by our assumption on MATH and MATH, and by REF, we have MATH . But the complex pairing MATH extends by continuity from MATH to MATH satisfying MATH. To see this just let MATH with MATH, MATH, and note that MATH extends as claimed. Hence, the pairing MATH defined first for MATH extends by continuity to MATH satisfying our wave front condition. In other words, MATH can be paired with every distribution whose wave front set has no elements related to MATH by the generalized broken geodesic flow at time MATH. Thus, for any MATH with MATH disjoint from the image of MATH under the generalized broken geodesic flow at time MATH, and for any MATH, MATH is defined by continuity from MATH, so MATH. But this states exactly that MATH is contained in the image of MATH under the generalized broken geodesic flow at time MATH. |
math/9906162 | Fix MATH and MATH. Let MATH be a homotopy contracting MATH to MATH. Then define MATH as follows: MATH for each MATH. Clearly, MATH contracts MATH to the point MATH. |
math/9906162 | Assume that MATH. Fix MATH. Let MATH and let MATH be such that MATH. Define the map MATH by MATH . Observe that for each MATH, we have that MATH and diam-MATH; also, MATH is homeomorphic to MATH. Thus, since MATH is a MATH (see CITE , NAME REF), we have that MATH is a MATH (see CITE , REF ). Therefore, since MATH is contractible by REF , MATH is a MATH (see CITE , REF ). Observe that for each MATH, the map MATH given by MATH is a MATH - map and the sequence MATH converges uniformly to the identity map MATH of MATH. Hence, by REF, MATH is a sequence of MATH - maps that converges uniformly to MATH. Therefore, by REF , we have that MATH. |
math/9906162 | Let MATH. By REF , MATH by a homeomorphism MATH, where MATH is a polyhedron (not necessarily compact). Note that there is a compact MATH, MATH, in MATH that is a neighborhood of the projection of MATH into MATH. Now, by REF , MATH is a NAME cube neighborhood in MATH of MATH. Therefore, MATH is a neighborhood of MATH in MATH and, by REF , MATH. |
math/9906162 | The continuity of MATH is well known. To prove the second part of the lemma, let MATH and let MATH. Since MATH, clearly, MATH. Hence, MATH. Also, MATH, which is seen as follows: Since MATH, there are MATH such that MATH and MATH; then, MATH hence, MATH. Therefore, since MATH and MATH, it is clear that MATH. |
math/9906162 | Fix MATH. Since MATH is a MATH - manifold REF , MATH is a MATH - manifold by REF . Note that MATH is a MATH (by CITE using REF , pp. REF); hence, MATH is a MATH - manifold REF . Therefore, in order to show that MATH, it suffices to obtain a MATH map, MATH, from MATH onto MATH REF . Assume that MATH, MATH and MATH are as in REF . Define MATH as follows: For each MATH, MATH . It follows easily that MATH is continuous. Also, MATH maps onto MATH since for each MATH, MATH. We prove that MATH is a MATH map by showing that the fibers of MATH are contractible ( CITE , p. REF). Fix MATH, where MATH. For any MATH and any MATH, let MATH . We show that MATH maps MATH into MATH. Let MATH and let MATH. Since MATH, MATH and MATH; hence, by REF , MATH. Therefore, MATH. Also, note that for any MATH, MATH the second equality being due to the fact that MATH. Hence, we have shown that MATH is a homotopy contracting MATH to the point MATH in MATH. Therefore, we have proved that MATH is a MATH map. |
math/9906165 | Consider MATH, MATH and MATH. Now MATH is clearly an additive functor; hence, in order to show that MATH is faithful, we just need to show that MATH implies MATH. By making use of the exact sequence REF we can see that it is enough to check it seperately for maps of semi-abelian schemes or lattices. Since torsion points are NAME dense in a semi-abelian scheme over MATH, MATH implies MATH for morphisms MATH between semi-abelian schemes. Finally MATH which is clearly faithful. If MATH induces an isomorphism MATH then by REF we have that MATH injects into MATH and MATH surjects onto MATH, therefore we have an extension of lattices MATH . Moreover by the snake lemma applied to the resulting diagram given by REF we get that MATH . Now we have that MATH is a finite group, since MATH; we can see that MATH . Thus MATH, since it injects into MATH which is torsion free. If we let MATH denote the quotient of MATH by MATH, we then get the following exact sequence of complexes MATH . Applying MATH we have that the composition of the following maps MATH is the zero map as well as an isomorphism, therefore MATH whence MATH, that is, MATH. |
math/9906165 | See REF. |
math/9906165 | Everything follows from REF combined with REF , by taking the connected components of the identity, once we know the following. MATH . Recall that MATH denotes the connected component of the identity of the kernel. In order to prove REF we consider the normalization MATH and the following commutative diagram MATH . Now, because of CITE (compare CITE) the morphism MATH is representable by an affine morphism. Then MATH, since MATH is an abelian variety. Since we obviously have MATH, we must have MATH which is the claimed equality REF . |
math/9906165 | We first claim that MATH . To see this, we consider the twisted log NAME complex MATH. It is well-known (see CITE, page REF for a quick proof) that its hypercohomology groups are the relative cohomology groups MATH, the NAME filtration on the relative cohomology is given by the subcomplexes MATH, and the corresponding hypercohomology spectral sequence degenerates at MATH. We then have MATH . The claimed isomorphism is obtained from REF for MATH. Let MATH be the sheaf on MATH given by the kernel of MATH where MATH is the inclusion. We have MATH because of REF and GAGA. We have that MATH is the kernel of MATH; therefore, by the exponential sequences on MATH and MATH, since MATH is an exact functor, we get the following induced relative exponential exact sequence of sheaves on MATH where MATH is the extension by zero functor along MATH. We then get the following exact sequence of cohomology groups MATH . The exact sequence in REF is then obtained. Since MATH is finitely generated and MATH is divisible, we get REF. From REF we then get REF. REF is well known if MATH (for example, see CITE). In order to show REF in general, we can proceed as follows. By considering relative NAME cohomology MATH we get a canonical cycle class map MATH . Moreover, MATH is an isomorphism, fitting into the following commutative diagram with exact rows MATH obtained from REF . For any closed sub-scheme MATH with MATH, we then have the following commutative diagram of cohomology groups having exact rows and columns MATH . Here MATH is the (relative) NAME cohomology of MATH with support in MATH. Let MATH be the support of a divisor MATH, that is, MATH. We then have that MATH . In particular: MATH and MATH is purely of type MATH; in fact, we have an isomorphism MATH . REF then follows from a diagram chase in the diagram above, using a general homological lemma CITE. REF then follows from the diagram as well, yielding the following isomorphism, in the category of REF over MATH, MATH where MATH denotes the extension class map determined by MATH, regarded as an extension of mixed NAME structures. |
math/9906165 | Let MATH, MATH and MATH be the corresponding open subsets; we have a diagram MATH . We let MATH, MATH denote the closed imbeddings. Let MATH be the structure morphism. Because of the canonical exact sequence MATH of sheaves on MATH we have MATH where the MATH is taken in the derived category. Thus MATH computes the singular cohomology of the pair MATH. Now we have MATH as mixed NAME structures. Similarly the complex MATH computes the cohomology of the pair MATH and we have MATH by using NAME - NAME duality, that is, MATH is left adjoint to MATH, where the dualizing complex MATH is given by MATH, and the obvious equality MATH. Now we can argue that MATH where the last equality is given by the following isomorphism MATH . The isomorphism REF can be obtained from biduality for constructible sheaves. In fact, let MATH be the dualizing sheaf; since MATH is smooth MATH, therefore, by biduality, REF is equivalent to MATH which is clear since MATH where we have used that MATH is the dualizing sheaf on MATH. Summarizing, we have obtained the following isomorphism MATH yielding the claimed duality isomorphism of groups. In order to show the compatibility of the above with the mixed NAME structures we consider the following induced pairing in the derived category MATH . This pairing is unique (up to a unique integer multiple); indeed, we have MATH where we have used REF , biduality for the constructible sheaf MATH and the standard formalism of derived categories. The same arguments apply to the constant sheaves MATH or MATH. By NAME 's theory of mixed NAME modules CITE, CITE, all of the above constructions and isomorphisms can (after MATH) be ``lifted" in a natural way to the derived category of mixed NAME modules. In particular, we see that our duality isomorphism is compatible with the mixed NAME structures as claimed. We leave to the reader the analogous proofs of the assertions about functoriality, and compatibility with NAME duality. |
math/9906165 | We first make a reduction to the case when MATH is equidimensional. Let MATH be the union of the MATH-dimensional irreducible components of MATH. Then by definition, MATH. On the other hand, the natural map MATH is an isomorphism of mixed NAME structures. Now for equidimensional MATH, let MATH be a resolution, with a good normal crossing compactification MATH with boundary MATH. As before, let MATH be the singular locus of MATH, MATH, and MATH the NAME closure of MATH. Associated to the cartesian square MATH there is a NAME - NAME long exact sequence of mixed NAME structures on singular homology yielding the following extension MATH where MATH . Now we claim: CASE: MATH, CASE: MATH is the proper push-forward of algebraic cycles and CASE: MATH as mixed NAME structures. In fact MATH is the free abelian group generated by the compact irreducible MATH-dimensional components of MATH, and MATH has a similar description. Thus REF - REF are clear and REF follows from REF because MATH. Moreover we have that the mapping MATH induced by the inclusion MATH, is just the cycle map relative to MATH, that is, the following diagram MATH commutes. Since, by definition, the kernel of MATH is MATH (compare REF ), the lattice MATH is canonically isomorphic to MATH. Moreover the exact sequence REF modulo torsion is canonically isomorphic to the following exact sequence MATH in the category of torsion free mixed NAME structures. But MATH is torsion-free, by the universal coefficient theorem in topology; hence so is MATH. The NAME structure on MATH is pure of weight zero and type MATH; we then have MATH . We also have the following extension of mixed NAME structures MATH . Thus the weight filtration of MATH admits the following description. Let MATH then MATH and MATH . Since MATH is pure of weight MATH, we have MATH whence MATH and MATH . Thus REF-motive associated (by NAME) to MATH is given by the following MATH (MATH was defined in REF ; MATH, MATH are similarly defined). Since MATH by NAME 's construction, we are reduced to showing that MATH in the category of REF over MATH. By REF (compare REF , where MATH and MATH is the above abelian variety) we have that MATH . According to our definition of MATH we are left to check that the following MATH commutes. We will deduce this from REF The NAME exact sequence yielding REF is given by the following commutative diagram of mixed NAME structures MATH which yields the following diagram of mixed NAME structures MATH where MATH . Let MATH; then MATH is a closed subset of MATH such that MATH, and MATH is homologically equivalent to zero relative to MATH; we let MATH . We have the following diagram of torsion-free mixed NAME structures MATH where the middle vertical mapping is obtained as follows. By REF we have MATH . Since MATH and MATH we have the following canonical map of mixed NAME structures MATH . The claimed map is obtained by composition of the duality isomorphism and the latter inclusion. Thus REF commutes by the functoriality assertion in REF . By diagram chase on REF one can then see that the image of MATH under the mapping MATH is the image of MATH under the extension class map MATH determined by the top row of REF . Thus REF commutes by REF . REF is proved. |
math/9906165 | From the NAME spectral sequence along MATH for the sheaf MATH we get a functorial map MATH . We can then consider the long exact sequence REF and compare with the corresponding sequence of étale cohomology groups. Since MATH and MATH for MATH we then get the result. |
math/9906165 | The description above can be easily obtained by modifying the original argument for absolute MATH (compare CITE). |
math/9906165 | The same proof of REF applies here to the étale sheaf MATH. |
math/9906165 | As in the proof of REF , we reduce immediately to the case when MATH is equidimensional. Now we fix a choice of resolution MATH, good compactification MATH, etc. By definition, MATH is given by REF-motive MATH. We have the following commutative diagram MATH where the bottom row is given by the NAME sequence for étale homology and the duality REF (MATH is a certain non-negative integer), and the top exact sequence is given by REF. We get the mapping MATH above by taking limit of MATH, and MATH is the induced map. Note that MATH may be viewed as the analogue of MATH for the variety MATH. It is also easy to see from the definitions that MATH as well, such that MATH is an isomorphism. Granting this, we are left to show our claim holds true for smooth schemes, that is, that MATH is an isomorphism. The latter follows from the fact that the relative NAME group of MATH is finitely generated, whence MATH, and, by REF and the NAME sequence in REF , we have MATH, since MATH is divisible. |
math/9906165 | In order to show REF we consider the canonical mapping which associates to any line bundle MATH with an integrable connection MATH, trivialized along MATH (in the appropriate sense), the cohomology class of a NAME cocycle given by the transition functions defining MATH and the induced forms. Since the following sequence (defined by the obvious maps) MATH is exact, we get the claimed isomorphism: note that MATH consists of closed REF-forms, since char. MATH. The exact sequence in REF is obtained by the exact sequence of complexes given by the columns in REF : in fact, the following equation holds MATH by REF . - From the above discussion we get the following diagram with exact rows and columns MATH . Therefore we see that MATH is the group of MATH-points of the pull-back of the group scheme MATH . The latter is the universal extension of the abelian variety MATH therefore MATH is the universal extension of the semi-abelian variety MATH and REF is proved. REF is standard (for example, can be obtained in a manner similar to the corresponding result for the usual NAME functors, by computing MATH-points as in CITE). |
math/9906165 | The proofs of REF are very similar to those in REF . In fact we have relative residue sequences given by the first row of the following commutative diagram (compare CITE) MATH where the isomorphisms are because MATH. Here recall that MATH are the irreducible components of MATH. For the latter REF we proceed as follows. Let MATH . From REF above we get the following push-out diagram MATH where ``res" is the ordinary residue of forms and MATH is the residue of connections. Therefore we are left to show that the canonical induced map MATH is an isomorphism: in fact, granting REF follows from the above push-out diagram, REF and the construction of the universal extension as being given in our REF. In order to show the isomorphism in REF we consider the following commutative diagram MATH where: by REF we know that MATH and MATH is the restriction of the canonical isomorphism MATH. Thus the relative residue sequence REF yields REF as well. The NAME algebra computation yielding REF is then straightforward. |
math/9906165 | We can consider the following pairing MATH . It will suffices to show that such a pairing yields non degenerate pairings on hypercohomology MATH . Since we are in characteristic zero we are left to show it for MATH for which it is clear from the proof of REF . Alternately, one can deduce the duality isomorphism, as in the proof of NAME duality for algebraic NAME cohomology, by reducing to NAME duality (compare CITE). |
math/9906165 | As usual, we can reduce to the case when MATH is equidimensional. Fix a resolution MATH with good normal crossing compactification MATH and boundary MATH. As above, let MATH be the union of all compact components of MATH. We clearly have the following (see REF ) relative residue sequence MATH where MATH are the smooth irreducible components of MATH. Moreover MATH by excision and duality, that is, REF , and we have the following pull-back diagram (compare with REF ) MATH by duality and the NAME sequence for NAME homology. Consider the following pull-back diagram of REF MATH . We then get the following commutative diagram (whose middle column implies the theorem) MATH where: REF top and bottom rows are obtained by applying MATH to the earlier diagram of REF, and are exact by construction compare REF the second row is exact according to REF the vertical isomorphisms are then obtained by applying REF , yielding the top row of REF as the top row of NAME realizations of REF . |
math/9906165 | The normalisation MATH of MATH clearly factors through the semi-normal curve MATH and the morphism MATH is bijective on points, and so induces an isomorphism on the groups of NAME divisors. We therefore have that MATH. On the other hand, MATH as well, from NAME 's definition. We then can assume MATH itself to be semi-normal; let MATH be the normalisation. First consider the compact case, that is, MATH. We then have a canonical quasi isomorphism MATH where MATH is the imbedding of the finite set MATH of singular points and MATH the imbedding of the inverse image of MATH: therefore, we get an isomorphism MATH . From the exact sequences REF we get the following diagram MATH showing that MATH is NAME dual of MATH (compare CITE). If MATH is not compact, let MATH be a smooth compactification of the normalization MATH, and set MATH; then MATH dualizes to MATH . One can then see that the symmetric avatars of MATH and MATH are the same, for example, by making use of the ``classical" REF . |
math/9906165 | We first note that the homomorphism MATH is well-defined; in fact if MATH is another such pair of points, then we easily see that MATH lies in the image of MATH (first we consider the case when the pair of points MATH, as well as MATH, each lie in an irreducible component of MATH; then we can deduce the general case). By REF it is clear that the character group of the torus MATH is given by the lattice MATH. The following pull-back homomorphism between abelian varieties MATH is dual to the following push-forward homomorphism MATH . Thus MATH as claimed. In order to check that the map MATH is NAME dual to MATH, it suffices to show that MATH coincides, on each generator MATH of MATH, with the analogous homomorphism for the NAME dual REF-motive. Choosing points MATH, MATH which are smooth on MATH, one can reduce (by considering the normalization of an irreducible curve passing through the pair of points, and standard functoriality for NAME and NAME varieties) to checking the duality assertion when MATH is a smooth connected projective curve, and MATH consists of REF points, for which it is ``classical" (see CITE for a more general statement; see also CITE, and CITE). |
math/9906165 | First consider the case when MATH is a proper, normal surface. The proposition is true in this case because the intersection matrix of the exceptional divisor of a desingularization of a normal surface singularity is known CITE to be negative definite: the group MATH is zero since any non-zero linear combination of compact components of MATH cannot be numerically equivalent to zero. For higher dimensional proper MATH, we take MATH to be smooth and projective; now by choosing successive hyperplane sections, we can find a complete intersection smooth surface MATH in MATH and a commutative square MATH where MATH is a reduced normal crossing divisor. Since MATH is general MATH injects into MATH. If MATH is the normalization of the image of MATH in MATH, then MATH is a resolution of singularities of a normal proper surface, with exceptional divisor MATH; hence MATH by the case of surfaces considered above, and so MATH as well. If MATH is open we just notice that MATH is contained in MATH; however, the latter group can be assumed to vanish, since MATH can be chosen to be a projective resolution of a normal compactification of MATH. |
math/9906165 | We have the formula MATH in the category of mixed NAME structures. NAME duality for REF-motives and REF then yield the result. |
math/9906165 | This follows from the NAME - NAME sequence of mixed NAME structures MATH where MATH is pure of weight MATH and MATH is pure of weight MATH. In fact, the NAME REF-motive canonically associated to MATH is exactly the claimed semi-abelian variety but, by REF , the NAME realization of MATH is MATH and the NAME realization functor is fully faithful. |
math/9906165 | This follows from the formula MATH and REF . |
math/9906165 | By REF , as above, we get the result. |
math/9906165 | The identification of MATH with isomorphism classes of pairs MATH is easy, and left to the reader. For a proof of the cohomological description, see REF. |
math/9906165 | See REF. |
math/9906165 | From REF , by taking connected components of the identity of the group schemes in REF , where MATH is connected, we claim that MATH surjects onto the abelian variety MATH: by the spectral sequence REF , the image of MATH is the kernel of the following edge homomorphism MATH which vanishes on the connected component of the identity of the domain. |
math/9906165 | From the simplicial exponential sequence, since the complex MATH is quasi-isomorphic to MATH on MATH, we have that MATH because MATH . Since we have a spectral sequence MATH such that MATH for MATH, we obtain MATH, and moreover MATH whence MATH. The following diagram MATH commutes, showing the claimed description of MATH (note that MATH is computed using the NAME topology). To show that the cycle class coincides with the extension class for the mixed NAME structure on MATH, we consider the following commutative diagram of cohomology groups having exact rows and columns MATH . The result then follows from a diagram chase (compare the proof of REF and CITE). |
math/9906165 | We have an exact sequence of mixed NAME structures MATH where MATH by universal cohomological descent: the claim then follows from REF . |
math/9906165 | Consider the NAME spectral sequence along MATH. Since MATH we then have a canonical functorial map MATH . Consider the canonical spectral sequence MATH A similar spectral sequence is clearly available for NAME cohomology groups, and MATH is compatible with a morphism between the respective spectral sequences. Since we have that MATH for all MATH and MATH, via MATH, we then get that MATH is an isomorphism. |
math/9906165 | Taking into account REF , the proof is an easy modification of CITE. |
math/9906165 | If we let MATH be given by REF-motive MATH for a choosen hypercovering and compactification with normal crossing boundary, we get the following commutative diagram MATH where REF the bottom row is just the exact sequence of cohomology with supports, REF we have MATH since MATH is a universal cohomological descent morphism, and REF the top exact sequence is given by REF. We get the mapping MATH above by taking the inverse limit of MATH; MATH is the induced map, which can also be regarded as the analogue of MATH for the case when MATH is proper (that is, MATH is proper and smooth over MATH). From the above description the mapping MATH is an isomorphism: in fact, is easy to see that we have an isomorphism MATH such that the following diagram MATH commutes (here MATH and MATH are the profinite completions of MATH and MATH, respectively). Granting this, we are left to show our claim is true for proper smooth simplicial MATH-schemes, that is, that MATH is an isomorphism. The latter follows from the fact that the NAME group of such a scheme (that is, the group of connected components of MATH) is finitely generated, whence MATH and, by the simplicial variants of NAME 's REF and NAME theory (see REF ), we have MATH . |
math/9906165 | It follows from REF and a simplicial version of CITE according to the general hint given by NAME in CITE. |
math/9906165 | The universal MATH-extension of any semi-abelian scheme is obtained as a pullback from the universal extension of its abelian quotient. The abelian quotient of MATH is MATH . By CITE it is easy to see that the universal MATH-extension of MATH is given by the group scheme (compare REF) MATH and we then have that MATH . Everything then follows from the following diagram with exact rows and columns, MATH where MATH is the toric part of MATH and the middle row and column are exact by REF . Therefore, by taking associated sheaves, we see that MATH is representable by the pull-back of the universal extension of MATH. Finally, since the NAME algebra of MATH is MATH, from REF , we get the last claim by taking NAME algebras (compare CITE). |
math/9906165 | A variant of the proof of REF (compare REF ). |
math/9906165 | Let MATH be given by REF-motive MATH for a choosen hypercovering and compactification with normal crossing boundary MATH. We have the following exact sequence of complexes MATH where MATH is just the quotient MATH. We therefore have the following push-out diagram MATH where the top row is REF and the bottom row is obtained from the dlog map as well. From the latter we are then left to show that MATH . In fact, granting REF , we have that, by the push-out diagram and REF , the universal MATH-extension of MATH is given by MATH where the lifting MATH above of MATH can be described as in REF via REF . Therefore MATH . Moreover, this isomorphism is clearly compatible with the NAME filtrations, provided we shift the index of the filtration on the right by REF. In order to show REF we consider the following commutative diagram with exact columns MATH where the horizontal maps are the canonical maps induced by universality; from the previous REF we know that the horizontal map on top is an isomorphism, so that we are left to show that the horizontal map at the bottom is an isomorphism. If MATH is smooth this last claim is clear since we have a simplicial surjective NAME residue map MATH and therefore MATH. In general, since the subschemes MATH are normal crossing divisors, for each MATH, we have exact sequences (compare CITE) MATH where the index MATH (MATH fixed) runs over the smooth components of MATH, that is, MATH. These sequences are compatible via the face and degeneracy maps of the simplicial scheme MATH. Because of this construction, and the definition of global sections of a simplicial sheaf, we clearly get a canonical identification MATH . We finally then get MATH as claimed. |
math/9906165 | In fact, by NAME theory and cohomological descent we get the following commutative square of isomorphisms MATH . Therefore, since the NAME groups are finitely generated, the NAME module of MATH is isomorphic to MATH. To conclude we remark that MATH is the group of MATH-points of a semi-abelian variety, in which torsion points are NAME dense (compare REF ). |
math/9906165 | To check that MATH is well defined is left as an exercise. We recall that MATH is an extension of the abelian variety MATH by the torus MATH (see REF for a description of the torus). Now, MATH is the character group of the torus MATH, MATH is the dual abelian variety; the claimed map between them is obtained from NAME duality - as in the proof of the corresponding assertion of REF , using standard functoriality properties of NAME and NAME varieties, one can reduce to the case of the standard smooth proper hypercovering of an irreducible projective curve with REF node; now we further reduce to determining the NAME dual of MATH of this singular curve, which is treated in CITE. If MATH is normal, then MATH, and so MATH is injective, by the NAME spectral sequence for the sheaf MATH along MATH; therefore, from REF we get MATH . Since MATH is normal, MATH is an abelian variety CITE, therefore MATH. |
math/9906165 | It follows from the explicit construction by NAME in CITE that MATH is equal to its ``NAME variety" in the sense of CITE, and the morphism MATH is then universal by CITE, that is, any torus bundle on MATH which is trivial on MATH is a push-out of MATH. If moreover MATH is normal and proper, and MATH is a projective resolution of singularities, then MATH, as seen in the proof of REF ; therefore the character group of the torus vanishes. |
math/9906165 | Since MATH, the claimed torus bundle is obtained as the NAME dual of the following injective map of REF MATH . Since MATH the description of MATH is clear. |
math/9906165 | It follows from NAME duality and the isomorphism of mixed NAME structures MATH because of REF . |
math/9906165 | We will assume, for simplicity of exposition, that MATH and MATH are irreducible; we leave the necessary modifications (mainly notational) for the general case to the reader. If the morphism MATH is not dominant we define MATH to be the zero homomorphism. If MATH is dominant, we choose resolutions MATH, MATH and good compactifications MATH and MATH with normal crossing boundaries MATH, MATH, such that there is a morphism MATH compatible with MATH, and hence satisfying MATH. Let MATH. The push-forward MATH, as a NAME divisor, clearly belongs to MATH. We therefore just need to show that there is an induced push-forward of relative line bundles which is compatible with the push-forward of NAME divisors. This is the content of the following lemma. |
math/9906165 | By considering the obvious map MATH, we reduce immediately to the case when MATH. Now we can construct a NAME factorization diagram MATH where MATH is a normal, proper variety of dimension MATH, MATH is a finite, surjective morphism, and MATH is birational and proper with connected fibres. Further, MATH, and MATH. Define MATH, so that MATH. Let MATH denote the union of the components of MATH which are not contained in MATH. Let MATH, MATH, MATH. Then MATH, MATH and MATH are each closed subsets of MATH, MATH and MATH, respectively, which have codimension MATH. Let MATH, MATH, MATH, so that we have an induced commutative triangle of proper morphisms MATH which is the NAME factorization of MATH. Also define MATH, MATH, MATH. We now make the following claims. CASE: There is a homomorphism MATH, which fits into a commutative triangle MATH CASE: There is a norm map MATH, such that REF the composition MATH is multiplication by MATH, and REF MATH for the class of any NAME (MATH . NAME) divisor MATH on MATH with support disjoint from MATH. CASE: The natural restriction map MATH is an isomorphism. Granting these claims, the desired map MATH is the composition MATH . This obviously factors through the subgroup MATH, which is the maximal divisible subgroup of MATH. We now proceed to prove the claims, in the order stated. First, we consider the map MATH. We have that MATH where for a scheme MATH and a closed subscheme MATH, we let MATH. By the NAME spectral sequence for MATH, we obtain an exact sequence MATH and MATH is the composition of MATH with the natural map MATH . In fact MATH since MATH, and the natural map MATH is injective. This means we have an exact sequence MATH . So to construct the map MATH and the commutative triangle in REF , it suffices to prove that the natural map MATH vanishes. Thus it suffices to prove that for each closed point MATH, the map to the stalk at MATH vanishes. We may identify MATH with MATH, where MATH . So we want to show that the maps MATH vanish, for all MATH. If MATH, then MATH. It suffices to see that the natural map MATH vanishes. Now MATH is a non-singular point of MATH. Thus we can find a non-singular proper variety MATH, containing MATH as a dense open subset; we can find a non-singular proper variety MATH containing MATH as a dense open set, and dominating MATH. Then MATH, and evidently the map MATH vanishes, as it factors through MATH. So we may take MATH. Now the fiber MATH is contained in MATH. Let MATH be the completion of MATH, and let MATH . Then we have a natural homomorphism MATH . Since MATH is faithfully flat, we see easily that this homomorphism is injective. So we are reduced to proving that MATH vanishes. For each MATH, let MATH be the closed subscheme defined by the MATH-th power of the ideal sheaf of the reduced fiber MATH. Let MATH denote the scheme theoretic intersection MATH. Then MATH, since MATH. There is a natural homomorphism MATH . We claim that it is an isomorphism. This follows, using the five lemma, from the NAME Existence Theorem CITE, which gives isomorphisms MATH and analogous isomorphisms on unit groups. Hence we are reduced to proving that for each MATH, the natural restriction maps MATH are zero. This is clear for MATH since MATH, so that MATH. For MATH, one has that MATH is an affine algebraic group which is purely of additive type (that is, is a vector group) CITE. Hence any homomorphism from a semi-abelian variety to MATH must vanish. This completes the proof of REF . Now we construct the norm map MATH of REF . First note that MATH, since the relative NAME group of a semi-local pair vanishes. Hence we have an identification MATH . Since MATH are integral and normal, and MATH is finite surjective, the norm map on functions induces a homomorphism MATH. We claim this induces a map on subsheaves MATH, or equivalently, that the composition MATH vanishes. Since MATH injects into the direct sum of constant sheaves associated to its stalks at the generic points of MATH, it suffices to show that for any such generic point MATH, the map on stalks MATH vanishes. Now MATH is the function field of an irreducible component of MATH, and is the residue field of the discrete valuation ring MATH. The stalk MATH is the unit group of the (semi-local) integral closure of MATH in the function field of MATH; denote this semi-local ring by MATH. The stalk MATH is the subgroup of MATH of units congruent to MATH modulo the NAME radical (which is the ideal defining MATH in MATH). Now MATH is a free module over MATH of rank equal to the degree of MATH, and for any MATH, the norm of MATH equals the determinant of the endomorphism of the free MATH-module MATH given by multiplication by MATH. So it suffices to observe that if MATH is congruent to MATH modulo the NAME radical, then this endomorphism is of the form MATH, where the matrix entries of MATH lie in the maximal ideal of MATH; hence the determinant of this matrix maps to MATH in the residue field of MATH. This proves that MATH vanishes. Now we define the map MATH to be the map MATH induced by the sheaf map MATH . This evidently has the property that MATH is multiplication by MATH, since this is true at the sheaf level. To see the compatibility with the push-forward for divisors MATH with support MATH disjoint from MATH, we compare the above map MATH with the analogous map MATH . This completes the proof of REF . To prove REF , it suffices to note that MATH and that MATH . All of these follow from the choice of the open set MATH, such that MATH has codimension MATH in MATH, and MATH has codimension MATH in MATH (recall that MATH is integral, non-singular and complete, and MATH is a reduced, normal crossing divisor in MATH, and is hence a complete, equidimensional and NAME scheme; thus MATH is locally connected in codimension REF). |
math/9906165 | Let MATH be the pullback along a choosen resolution of singularities MATH. We can choose a ``NAME compactification" MATH of the resolution MATH, that is, we can also get a locally free sheaf MATH on MATH which extends MATH (to construct a NAME compactification, first choose an arbitrary one, and a coherent extension MATH of MATH; then resolve singularities of the NAME blow-up associated to MATH, on which the pull-back of MATH, modulo torsion, is a locally free sheaf). We can then assume that MATH extends to MATH on MATH, and the boundary MATH is a normal crossing divisor in MATH which is a projective bundle over the normal crossing boundary MATH of MATH. Since the NAME varieties of MATH and MATH are also isomorphic, the exact sequence REF (compare REF ) yields an isomorphism of semi-abelian varieties MATH. NAME of divisors from MATH to MATH yields a compatible isomorphism between lattices, giving rise to the claimed isomorphism for MATH; that for MATH follows from NAME duality. For MATH and MATH, we argue as follows. Consider a NAME compactification MATH of MATH, that is, such that MATH extends to a locally free sheaf MATH on MATH. We can find a smooth proper hypercovering MATH of MATH such that the induced reduced hypercovering of MATH is a normal crossing divisor MATH in MATH. Then MATH yields a smooth proper hypercovering of MATH, and MATH is a smooth compactification with normal crossing boundary. Now MATH is a smooth proper hypercovering of MATH, and we can get an induced compactification of MATH which has normal crossing boundary. We then see easily that MATH because of the exact sequence REF (compare REF ); similarly the lattices are isomorphic. |
math/9906165 | Consider a NAME compactification MATH of MATH, so that MATH extends to a locally free sheaf MATH on MATH, and let MATH. We let MATH be a smooth proper hypercovering of MATH such that the reduced inverse image of MATH is a normal crossing divisor, and let MATH be the simplicial vector bundle on MATH obtained by the pull-back of MATH along the hypercovering. We take MATH to be the compactification of MATH with normal crossing boundary. We then have to show that MATH where MATH is the normal crossing boundary of MATH, considered as a compactification of MATH. We have MATH . Thus it is clear that the groups of divisors supported on MATH and on MATH, which are algebraically equivalent to zero (that is, have classes in MATH) on the respective proper simplicial schemes, are naturally isomorphic; hence the lattices of our two REF-motives are naturally isomorphic. From the short exact sequence MATH we conclude that MATH, and we are done. |
math/9906165 | We may assume without loss of generality that MATH is equidimensional. Let MATH be the normalization, MATH a resolution of singularities, and MATH the induced resolution of singularities. Since MATH is ``good", the scheme MATH is a curve which is smooth at MATH. Denote by MATH the pull-back curve MATH. Let MATH be the restriction of MATH. Then MATH, and the normalisation MATH of the curve MATH clearly factors through MATH. Let MATH be the induced map. Then there is a natural pull-back map on NAME varieties MATH. Thus, in order to get the claimed map MATH on REF-motives, it is enough to show that any divisor MATH pulls back to a divisor MATH. Since MATH is smooth at the finite set of points MATH it will suffices to show that MATH where MATH is the canonical induced imbedding. Now let MATH denote the support of MATH. Then MATH is mapped to MATH, and therefore MATH, which is the support of MATH, is mapped to MATH. We thus have REF 's homological NAME groups MATH by NAME 's compatibility result CITE between pull-back and NAME maps for locally complete intersection morphisms. Since the push-forward of MATH vanishes as a cycle on MATH, the pull-back of MATH to MATH pushes forward to zero in MATH. Since MATH is a reduced REF-dimensional scheme, the latter push-forward to MATH is in fact zero as a cycle on MATH. |
math/9906165 | We recall that MATH and MATH. Let MATH be the semi-normalization of MATH; the canonical identification MATH (see REF ) together with the pull-back map MATH, which is just the semi-abelian quotient of MATH, yields the result. |
math/9906165 | This follows easily from the dual statement, that is, that the following diagram MATH commutes. Moreover, MATH injects into MATH. |
math/9906165 | Let MATH be the induced map on smooth compactifications MATH and MATH, compatibly with the normal crossing boundaries MATH and MATH. We then have the following diagram of REF MATH yielding the claimed map, where since MATH is normal we have that MATH, and we have a pull-back map on relative line bundles MATH. |
math/9906165 | Since MATH by REF we are left to prove our claim for regular imbeddings. We then have the following diagram MATH where MATH for cycles exists trivially, and therefore, by REF , the composite of MATH and MATH factors through MATH. In order to show that the construction is independent of the factorisation, we observe that it is so on the étale realizations (where it coincides with the NAME map obtained via NAME duality), and therefore, by REF , we are done. |
math/9906165 | The following sequence MATH is exact and natural in MATH. To show pro-representability we use NAME 's criterion in CITE, saying that MATH needs to be left exact on the subcategory of artinian algebras. From REF one can see that MATH is pro-representable. Then there is a topological algebra MATH such that MATH. We show that the local components are noetherian by using NAME 's criterion CITE. In fact, the local component at a point MATH is pro-represented by the localization MATH, where MATH, and in order to show that MATH is noetherian it will suffices to show that MATH is finite dimensional. For any MATH-scheme MATH the MATH-point MATH of MATH induces an element MATH by pulling back along the structural morphism; we then get an automorphism MATH by adding MATH in MATH. Thus we can assume MATH, therefore we have MATH . By REF we conclude that PREF is satisfied. REF - REF follow from a diagram chase, because of REF and the assumptions REF - iii. To show PREF we proceed as follows. Let MATH be a point, that is, MATH. We have to show that MATH is a closed subscheme of MATH. Let MATH be the induced point of MATH and let MATH be the closed embedding. Then MATH actually belongs to MATH since it yields zero in MATH. Then MATH is a closed subscheme of MATH hence of MATH. We can see that MATH. In fact, if MATH is a point such that MATH then MATH belongs to MATH. Conversely, if MATH is such that MATH then MATH belongs to MATH as well which means that MATH where MATH whence MATH. |
math/9906165 | Let MATH be a non-singular projective MATH-curve, MATH a finite set of closed points. For any divisor MATH on MATH, let MATH be the map given by MATH. Clearly MATH is a homomorphism from divisors on MATH to MATH. If MATH is an effective divisor of degree MATH, then MATH determines a point MATH in an obvious way, and hence a homomorphism MATH . There is a homomorphism MATH where MATH is the MATH projection. Clearly the image is contained in the subgroup of invariants for the action of the permutation group MATH. Hence there exists a map of sets (not necessarily unique, or even a homomorphism) MATH such that MATH is a pre-image in MATH of MATH, for any MATH, that is, the diagram MATH commutes. We claim that for any effective divisor MATH of degree MATH on MATH, we have MATH . Indeed, let MATH, where MATH is repeated MATH times as a coordinate. Then MATH is a preimage in MATH of MATH. Hence MATH . On the other hand, from the definitions, it is clear that MATH . Now suppose MATH such that MATH. Let MATH, and consider MATH as a morphism MATH. There is an induced morphism MATH where MATH. The map MATH has the property that if MATH as a divisor, then MATH. Let MATH, MATH be the divisors of poles and zeroes of MATH. Then the lemma asserts that MATH . To prove this, by REF , it suffices to show that MATH . Since MATH, it follows that MATH and so we are reduced to proving that MATH. If MATH is the structure morphism, then we are given that MATH while clearly MATH and MATH both equal the identity on MATH. Hence MATH as desired. |
math/9906170 | It is easy to see that any common Lagrangian complement at MATH extends to a common Lagrangian complement in a neighborhood MATH. The existence at MATH of such a complement is standard if the residue field is infinite. The following lemma deals with the remaining case: |
math/9906170 | Let MATH. Then MATH, and similarly MATH. On dimensional grounds, we must indeed have MATH. As a result MATH and MATH are complementary Lagrangian subspaces of MATH. Moreover, by hypothesis they are even-dimensional. Let MATH be a system of vectors in MATH mapping onto a basis of MATH. Since MATH and MATH are complementary Lagrangian subspaces of MATH, the symmetric bilinear form MATH associated to MATH induces a perfect pairing between them. So there exists a system of vectors MATH in MATH such that MATH for all MATH. Let MATH be the subspace spanned by the MATH and MATH. Then MATH is nondegenerate, so there is an orthogonal direct sum decomposition MATH such that MATH and MATH are both nondegenerate. Moreover, MATH is a Lagrangian subspace, for which there exists a complementary Lagrangian subspace MATH by our previous remarks. Let MATH be a basis of MATH. One may now check that MATH form a basis for a Lagrangian subspace MATH complementary to both MATH and MATH. |
math/9906170 | The equivalence of REF follows from the fact that of MATH. The implication MATH is standard. To prove REF , use REF to cover MATH by open subsets MATH over each of which MATH have a common Lagrangian complement MATH. Then by REF there exist alternating maps MATH such that MATH. Let MATH be the isomorphism induced by MATH, and let MATH and MATH be the two projections from MATH. MATH . Then MATH is an isomorphism such that MATH is alternating. |
math/9906170 | REF follows from REF . CASE: Since MATH and MATH are complementary, their images MATH are also complementary. Thus the map MATH is an isomorphism. This is equivalent to the composition MATH being invertible, or to MATH being invertible. It follows that MATH is also invertible. CASE: We now have isometries MATH . The left-to-right composition is the identity on the first summand MATH, and sends MATH (corresponding to MATH on the left) onto the second summand MATH (corresponding to MATH on the right), compatibly with the hyperbolic quadratic form on MATH. Consequently, the left-to-right composition is MATH. To find the image of MATH on the right, one first goes to the left (where its image is MATH), and then applies the left-to-right composition. Therefore the image of MATH on the right is the composite image The above composite map is MATH, so the image of MATH on the right is MATH. But MATH and this completes the proof. |
math/9906170 | It is enough to prove the lemma in the case where MATH and where MATH is universal. So let MATH, and let MATH be the polynomial ring in the independent variables MATH (MATH). Let MATH and MATH be the MATH matrices with coefficients in MATH given by MATH let MATH, let MATH, and set MATH. Thus MATH. We have to show that the two ideals MATH in MATH coincide. However, we may notice the following three facts: The ideal MATH is prime. This is because the ideal of MATH generated by the MATH . NAME of MATH is prime (compare NAME Fra REF), so its extensions to MATH (which is a polynomial algebra over over MATH) and to MATH are also prime. There is an involution of MATH exchanging MATH and MATH. Since MATH is a polynomial algebra over MATH in variables which are the entries of MATH and MATH, one can specify a morphism MATH by specifying alternating matrices MATH and MATH. Thus we may define MATH by MATH . One computes that MATH, so MATH is the invertible element MATH. Hence MATH extends uniquely to a morphism MATH. One checks that MATH and that MATH, so MATH is an involution. Since MATH exchanges MATH and MATH, it exchanges MATH and MATH. The ideals MATH and MATH define the same algebraic subset of MATH. This is equivalent to showing that a morphism MATH with MATH a field factors through MATH if and only if it factors through MATH. But giving such a MATH is equivalent to giving alternating matrices MATH and MATH with coefficients in MATH such that MATH is invertible. Such a MATH factors through MATH if and only if MATH, and it factors through MATH if and only if MATH . Since MATH is invertible, the two conditions are equivalent. These three facts show that (in the generic case) MATH and MATH are prime ideals defining the same algebraic subsets. This proves that MATH are MATH are equal in the generic case, and therefore equal in all cases. This proves the lemma. |
math/9906170 | Let MATH and MATH be the inclusions, and let MATH be the kernel in the natural sequence MATH . If MATH is the isomorphism induced by the quadratic form MATH, and MATH, then we get a commutative diagram MATH . Since the diagonals are short exact sequences, the kernels of MATH and of MATH are both equal to MATH. In addition, MATH. We claim that MATH is a line bundle, and that the complexes MATH are exact and are locally free resolutions of MATH for the subscheme MATH of grade MATH, with MATH. We will prove these claims locally by making MATH locally alternating and applying the NAME structure theorem CITE. Now the vector bundles MATH and MATH may be of even or odd rank. If the rank is even, we use the same trick as in REF and replace MATH by MATH without changing the kernel and cokernel of MATH. Thus we may assume that MATH and MATH are of odd rank. By REF is also odd for all MATH. Therefore, by REF , MATH is locally alternating, that is, MATH is covered by open subsets MATH over which there are isomorphisms MATH such that MATH is alternating. Thus we see that our complexes REF are locally isomorphic to complexes MATH such that MATH is alternating with kernel MATH in the notation of REF . Now MATH is of odd rank, MATH is alternating, and the ideal MATH generated by its submaximal minors is of grade MATH. So the NAME structure theorem CITE applies. Therefore the kernel MATH is a line bundle, the map MATH is given by the submaximal NAME of MATH, and the complex REF is exact and is a resolution of MATH. We can also identify the ideal sheaf MATH generated by the submaximal minors of MATH with MATH. This works because on MATH the sheaf MATH is generated by the submaximal NAME MATH of the alternating map MATH, while MATH is generated by the submaximal minors. Since the MATH-th submaximal minor is MATH (CITE Appendix), we do indeed get MATH. This verifies our claims. Now we set MATH and twist. We get two dual resolutions MATH . The alternating product of the determinant line bundles in each resolution is trivial, and therefore MATH. Let us now verify that MATH satisfies REF - REF . The duality between the two resolutions of MATH shows that MATH is relatively NAME of codimension MATH. The duality also induces an isomorphism MATH, making MATH subcanonical. Clearly MATH is of finite local projective dimension. Moreover, MATH is the NAME extension class of REF and is thus the image of the class in MATH of MATH . So MATH is strongly subcanonical with respect to MATH. |
math/9906170 | The only things we need to show for REF which do not already follow from REF are that REF commutes and that MATH is alternating. But MATH is a totally isotropic subbundle, so any local section MATH satisfies MATH . Thus MATH is alternating, and the central part of REF commutes. The rest is easy and left to the reader. For REF the fact that REF is a quasi-isomorphism implies that MATH is a subbundle of MATH. It is totally isotropic by the same calculation REF. |
math/9906170 | We begin by choosing bases for MATH and MATH and the dual basis of MATH, so that we can treat MATH and MATH as matrices. Since MATH is a direct summand of MATH, the columns of the total matrix MATH are linearly independent even modulo MATH. Moreover, by REF above MATH is an alternating matrix because MATH is a totally isotropic submodule. We now begin a series of row and column operations on MATH and MATH which will put them into the required form. The column operations (respectively, row operations) correspond to changes of basis of MATH (respectively, of MATH and MATH) and to the action of invertible matrices MATH (respectively, MATH) on MATH and MATH via MATH and MATH. Choose a maximal invertible minor of MATH. After row and column operations, we can assume that the corresponding submatrix is an identity block lying in the upper left corner of MATH and that the blocks below and to the right of it are MATH. Thus we can assume that MATH where the blocks of the two matrices are of the same size, the on-diagonal blocks are square, and the coefficients of MATH lie in MATH. Since MATH is alternating, we see that all the coefficients of MATH also lie in MATH. Hence all the coefficients in the last block of columns of MATH lie in MATH except those in MATH. Since these columns must be linearly independent modulo MATH, it follows that MATH must be invertible. Applying a new set of column operations to MATH and MATH, we may assume that MATH and that MATH. Moreover, MATH remains alternating, which actually means that MATH and MATH are alternating, and MATH. A final set of row and column operations using the matrices MATH and MATH puts MATH and MATH into the form required by the lemma. |
math/9906170 | CASE: We put the matrices of MATH and of MATH in the special form of REF , and we set MATH. The determinant of the matrix of MATH is then MATH. Consequently the determinant of the matrix of MATH with respect to any bases of MATH and MATH is of the form MATH, with MATH an invertible element of MATH coming from the determinants of the change-of-basis matrices. CASE: Using the special forms for MATH and MATH given in REF , we find that MATH is generated by the submaximal NAME MATH of MATH, while the ideal MATH is generated by MATH. Since we suppose MATH and therefore MATH are not zero-divisors, this gives REF . |
math/9906170 | CASE: Extending MATH to a coherent sheaf on the closure MATH and then applying NAME 's Theorem A, we see that there exists a surjection of the form MATH. Pulling back gives us a class MATH. Let MATH be the homogeneous coordinate ring of MATH, and let MATH be the homogeneous ideal of strictly positive degree elements vanishing on the closed subset MATH. Extend MATH to a coherent sheaf on MATH, and let MATH be a finitely generated graded MATH-module whose associated sheaf is this extension of MATH. Then MATH. Consequently, MATH, as a member of a local cohomology module, is annihilated by some power MATH of MATH. A finite set of homogeneous generators of MATH gives surjections MATH and MATH, such that the pullback of MATH along the induced map MATH vanishes. CASE: For the same reasons as in REF , MATH is killed by some power of MATH. For convenience we assume that the same MATH as in REF kills MATH. Let MATH be the surjection used in REF . Then the surjection MATH kills MATH because the exterior square factors as MATH. |
math/9906170 | By hypothesis, MATH lifts to a class in MATH. By REF there exists a vector bundle MATH and a surjection and kernel MATH such that MATH lifts further to a class MATH. This defines an extension MATH. Attaching these extensions gives an acyclic complex MATH . We claim that MATH is locally free. Our reasoning is as follows. Since the local projective dimension of MATH is at most MATH, the local projective dimension of MATH is at most MATH. By REF , MATH will be locally free if MATH generates the sheaf MATH. Moreover the sheaves MATH, MATH, and MATH are all isomorphic, and their respective global sections MATH, MATH, and MATH correspond under these isomorphisms. Since MATH generates MATH, the section MATH generates MATH. Thus MATH is locally free. The complex MATH is now a locally free resolution of MATH. As in CITE and CITE, we try to make this into a commutative associative differential graded algebra resolution of MATH by constructing a map MATH from the divided square covering the identity in degree MATH: MATH . Now MATH maps into the kernel MATH of MATH. Hence the first problem in filling in the dotted arrows above is to carry out a lifting The obstruction to carrying out the lifting is a class MATH. There is no reason for this class to vanish. So the liftings sought in REF need not exist. But there is a way around this. By REF there is a surjection from another vector bundle MATH such that the pullback of MATH to MATH vanishes. We now redo the construction of the complex and get commutative diagrams with exact rows and columns MATH . This allows us to construct a new complex MATH . One sees easily that MATH and therefore MATH are also vector bundles. But this time, the composite map MATH, lifts to MATH since the obstruction is the class in MATH which we got to vanish using REF . Since the square marked with the MATH is cartesian, we get a lifting MATH. The other liftings MATH now occur automatically. We therefore get a chain map MATH which makes MATH into a commutative associative differential graded algebra with divided powers. We now claim that having this differential graded algebra structure gives us all the properties we want and puts us into the situation of REF . Indeed, as in CITE, the multiplication gives pairings MATH, and therefore maps MATH. These maps are compatible with the differential, and as a result, the following diagram commutes: The top row is exact by construction, and the bottom row is exact because it is the dual of the top row which is a resolution of a sheaf of grade MATH. Since MATH is an isomorphism, one sees that MATH is exact. Thus MATH embeds in MATH as a subbundle which is totally isotropic for the hyperbolic symmetric bilinear form on MATH. The subbundle MATH is even totally isotropic for the hyperbolic quadratic form, since the restriction of this form to local sections of MATH is the function MATH, and this function vanishes because the composite map from REF factors through MATH and hence vanishes identically. Thus MATH is a Lagrangian subbundle of MATH. This completes the proof. |
math/9906170 | The map MATH may be identified with an element of MATH. The subscheme MATH satisfies REF for MATH if and only if MATH is in the image of MATH. Local duality and NAME duality give identifications MATH and MATH which are compatible with the inclusions. So MATH satisfies REF if and only if MATH. |
math/9906170 | We show how to start the proof off. But we will stop when we reach the point where it becomes identical to the proof of the main result of CITE. Suppose that MATH satisfy REF , and REF. REF implies that the map MATH is an isomorphism. So MATH and NAME duality induce a symmetric perfect pairing MATH which pairs MATH with MATH for all MATH. REF implies that MATH contains a Lagrangian submodule MATH for this symmetric perfect pairing. Indeed if MATH is odd, one can pick MATH. If MATH is even, then there exists a Lagrangian subspace MATH, and one can pick MATH. The two submodules MATH and MATH are orthogonal complements of each other under the NAME duality pairing; see for example CITE. Hence REF , that MATH, implies that MATH and therefore that MATH. Now the orthogonal complement of MATH under the NAME duality pairing corresponds to the orthogonal complement of MATH under our pairing REF. So REF implies that MATH. In other words MATH is sub-Lagrangian. It now follows that there exists a Lagrangian submodule MATH such that MATH. For instance, pick MATH (compare REF ). One easily checks that MATH for MATH, and that MATH for MATH. Consequently MATH is of finite length. It has an induced nondegenerate symmetric bilinear form, and it has a Lagrangian submodule MATH. We now claim that we can construct a locally free resolution MATH with MATH alternating and such that MATH, and MATH. Moreover, MATH induces a surjection MATH. Different pieces of the resolution contribute different pieces of the cohomology module MATH. The submodule MATH is contributed by MATH; the piece MATH by MATH; the piece MATH by MATH; and the piece MATH is contributed by MATH. The construction of this resolution and the verification of its properties can be done using the NAME correspondence by the same method as in CITE. It is quite long and we omit the details. |
math/9906170 | The only delicate part is REF . Because MATH is locally NAME of codimension MATH, it has a locally free resolution MATH. The symmetric isomorphism MATH corresponds to a morphism in the derived category This MATH is a member of the hyperext MATH, which in turn is the abutment of the hyperext spectral sequence MATH . The differentials MATH define complexes (indexed by MATH) MATH whose cohomology groups are the MATH. In particular, MATH is the space of homotopy classes of chain maps MATH. Hence MATH will be the class of an honest chain map if and only if it comes from MATH. However, according to the MATH-term exact sequence MATH the obstruction lies in MATH. As in the proof of REF , this obstruction may be nonzero, but it can be killed by pulling back along a suitable epimorphism MATH (compare REF ). The proof may now be completed with arguments taken from the proof of REF . |
math/9906171 | We apply REF with MATH and MATH the universal Lagrangian subbundle. The degeneracy locus is MATH and it has resolutions Moreover, MATH since MATH. It remains to show that MATH is not Pfaffian. Let MATH be a general totally isotropic subspace of dimension MATH, and let MATH be the natural embedding. If MATH were Pfaffian, its symmetric resolution would be of the form MATH (compare REF in the introduction). It would thus pull back to a symmetric resolution of MATH on MATH of the same form. Now MATH parametrizes one family of MATH's contained in a smooth hyperquadric in MATH. Thus MATH (see CITE or CITE), and MATH is a codimension MATH . NAME subvariety of MATH and indeed a point MATH. If we pull the resolution REF back to MATH and twist, we get a resolution MATH . This gives MATH, which is a contradiction. This completes the proof. |
math/9906171 | The existence of MATH and the resolutions follow from REF , using the fact that MATH. The subcanonical subvariety MATH is not Pfaffian because MATH and MATH (as one sees from REF). Using the resolutions and a computer algebra package, one computes the NAME polynomial of MATH as: MATH . So the degree of MATH is MATH. In characteristic MATH the smoothness of the general MATH follows from NAME 's transversality theorem as explained after REF . In characteristic MATH the smoothness follows from REF below. |
math/9906171 | Let MATH . We will show that MATH is nonempty without invoking NAME 's theorem on the transversality of a general translate. We will do this by showing that the complement has dimension at most MATH. Let MATH, and let MATH be the NAME parametrizing three-dimensional subspaces of MATH. Let MATH . To choose a point of MATH, one first chooses MATH, then MATH in a fiber of a NAME bundle with fiber MATH , then MATH in a fiber of an orthogonal NAME bundle with fiber MATH. We see that MATH is nonsingular with MATH. A similar computation shows that MATH is of dimension equal to MATH. So the image of MATH under the projection MATH is not dominant. Over a point MATH, one has MATH. Hence it is enough to show that the dimension of MATH is at most MATH. The essential question is: if MATH, so that MATH, when is the dimension of MATH equal to MATH, and when is it more? Now MATH. In order to identify this space we must identify MATH and the two subspaces MATH and MATH. There is a well known isomorphism MATH. Essentially, given a first-order deformation of MATH as a Lagrangian subspace of MATH, any vector MATH deforms within the deforming subspace as a MATH with MATH well-defined modulo MATH. This gives a map MATH. The vector MATH remains isotropic if and only if MATH is alternating. To describe MATH as an alternating bilinear form on MATH, one looks at MATH . If MATH is a space MATH of dimension MATH, then MATH according to REF . That is, MATH is the kernel of the natural map MATH. Now let MATH. Then there is a natural isomorphism MATH. There is also a natural isomorphism MATH. A vector MATH corresponds to a first-order infinitesimal deformation MATH of MATH, and hence to the first-order infinitesimal deformation MATH of MATH. The corresponding alternating bilinear form on MATH is computed by MATH . If we use the isomorphism MATH, then the corresponding alternating bilinear form on MATH is therefore MATH . Hence if we think of MATH as a three-dimensional subspace of MATH, then MATH is the orthogonal complement of the image of MATH in MATH. As a result MATH if and only if the map MATH is not injective. Every vector in MATH is of the form MATH. We therefore look at the set MATH . We stratify this locus according to the rank of the map MATH defined by multiplication by MATH. Generically this map is injective, and the dual multiplication map MATH is surjective. For MATH to be in MATH, MATH must be in the codimension-MATH subspace MATH. Hence MATH contains a stratum of dimension MATH where MATH is injective. The locus MATH contains two other strata, but they are of smaller dimension. In one of them the MATH have the form MATH with MATH and MATH indecomposable. Since MATH is really only well-defined up to a scalar multiple, and MATH is really well-defined only in MATH, these form a locus of dimension MATH in MATH. For these MATH the rank of MATH is MATH, so this stratum in MATH has dimension MATH. In the final stratum the MATH are of the form MATH, a locus of dimension MATH. For these MATH the rank of MATH is MATH, so this stratum of MATH has dimension MATH. Thus MATH. As a result the locus MATH is of dimension MATH. Consequently the locus MATH is of dimension at most MATH. For each MATH, to give a MATH such that MATH is equivalent to giving a member of one of the families of Lagrangian subspaces of MATH. Consequently, such MATH form a family of dimension MATH. As a result, the locus MATH is of dimension at most MATH. Finally MATH is of dimension at most MATH. But the part of MATH where MATH is exactly the part of MATH lying in MATH. So MATH. As a result MATH does not dominate MATH, and MATH is a nonempty open subset. As before, MATH must then contain a rational point because the base field is infinite, and we conclude. |
math/9906171 | If MATH is a basis of MATH and MATH, then the quadratic form on MATH given by MATH is nondegenerate and has the property that if MATH is a Lagrangian subspace of MATH, then MATH is a Lagrangian subspace of MATH. Consequently MATH is a Lagrangian subbundle of MATH. Also, if we make a standard identification MATH, then the symmetric bilinear form associated to MATH is MATH. We now choose a general Lagrangian subspace MATH and apply the construction of REF . This produces a subcanonical fourfold MATH with MATH. The degree of MATH can be computed from the formula after REF . If MATH, then MATH is nonsingular by REF . If MATH is infinite of positive characteristic, MATH is nonsingular by an argument similar to REF whose details we leave to the reader. Let MATH. The isomorphism MATH induces an isomorphism MATH which corresponds to the cup product pairing of NAME duality REF. Our problem is to identify this map. Now twisting REF produces a commutative diagram with exact rows (where we write MATH to keep things within the margins): The rows of the diagram produce natural isomorphisms MATH and MATH. Moreover, the NAME duality pairing MATH corresponds to the canonical pairing MATH since the bottom resolution is dual to the top resolution, so the hypercohomology of the bottom resolution is naturally NAME dual to the hypercohomology of the top resolution. It remains to identify the map MATH with the map MATH. We do this by using the following commutative diagram with exact rows. The maps MATH and MATH in this diagram come from the identification MATH, which yields a short exact sequence MATH . The exactness of REF implies that the top middle square of REF commutes, and that MATH induces an isomorphism MATH, while MATH induces an isomorphism MATH. As a result the second row of the diagram is exact, and the squares in the top row are all commutative. To see that the squares in the bottom row are all commutative, recall how the maps MATH, MATH, MATH and MATH are defined. The inclusions of MATH and MATH in MATH, together with the bilinear form and the NAME complex, give rise to a diagram with a commutative square such that MATH . Since the square in REF is commutative, we get MATH and MATH. The first equality means that the bottom set of squares in REF commutes. The second equality means that the composite of the two chain maps of REF is the chain map of REF . Now the first and second rows of REF produce compatible identifications MATH. And the second and third rows of REF show that the map MATH induced by MATH may be identified with MATH. This completes the proof of the theorem. |
math/9906171 | The subvariety MATH is a locally NAME surface with MATH but MATH. Thus it is subcanonical but not Pfaffian by REF . |
math/9906171 | It is enough to prove the lemma for MATH a field, since one can reduce to the case where the coefficients MATH are independent indeterminates, and MATH is the field of rational functions in these indeterminates. We have to show that REF is invariant under changes of basis in MATH. But since MATH is now assumed to be a field, MATH is generated by three special kinds of changes of basis: permutations of the basis elements, operations of the form MATH, and operations of the form MATH. REF is almost trivially invariant under the first two kinds of operations, and it is invariant under the third kind of operation because the portions MATH do not change. |
math/9906171 | Suppose first that MATH is a multiple of the divided-square quadratic form, and that MATH and MATH. Choosing a basis of MATH of the form MATH and applying REF , one sees easily that MATH. Conversely, suppose that MATH for all MATH and MATH. Write MATH. The hypothesis implies that MATH for all MATH. It also implies that each MATH. Hence MATH if the multi-indices MATH and MATH are not disjoint. Finally MATH. In other words MATH if MATH and MATH are disjoint, and MATH and MATH are disjoint, and MATH and MATH contain exactly MATH common indices. One then easily deduces that all the MATH with MATH and MATH disjoint must be equal. But then MATH is the same as the quadratic form of REF up to a constant. |
math/9906171 | The general MATH is a subcanonical surface of degree MATH with MATH and with the resolutions REF according to REF . It is smooth by an argument similar to REF . Since MATH and MATH for MATH, it is a nonclassical NAME surface by the NAME classification. Since the degree is MATH, it is a NAME model. Moreover, according to the resolution, it is not contained in a hyperquadric. Hence it is unnodal by REF . |
math/9906171 | The NAME polynomial of MATH is MATH by NAME. The reader may easily verify that the sheaves MATH have cohomology as given in the following table: Now according to the table MATH is MATH-regular; the only group obstructing its MATH-regularity is MATH; and the comultiplication MATH is dual to the multiplication MATH and thus is an isomorphism. So we may apply REF to see that all the generators of MATH are in degrees MATH, and all the relations are in degrees MATH. But for all MATH the restriction maps MATH are bijections. So the generators and relations of MATH and of MATH in degrees MATH are the same - one generator in degree MATH and no relations. Hence MATH has only a single generator in degree MATH, and its relations are all in degree MATH. In other words MATH, and the homogeneous ideal MATH is generated by a MATH-dimensional vector space MATH of cubic equations. We now go through the argument in REF which constructs a pair of symmetrically quasi-isomorphic locally free resolutions of a subcanonical subscheme of codimension MATH. We have an exact sequence MATH . There is a natural isomorphism MATH and a nonzero member of this group induces an extension which we can attach to give a locally free resolution MATH . The intermediate cohomology of MATH comes from the intermediate cohomology of MATH and is given by MATH for MATH and MATH. According to NAME 's Theorem CITE, this means that MATH. Since MATH and MATH are both of rank MATH, they are isomorphic. This gives the top locally free resolution of REF. The rest of REF is now constructed as in the proof of REF . It remains to identify the quadratic spaces MATH and MATH, and to show that MATH is uniquely determined by MATH. However, REF gives us an exact sequence MATH . This is the unique nontrivial extension of MATH by MATH, and thus coincides with the truncated NAME complex REF. There is therefore a natural identification of MATH with MATH which identifies the hyperbolic quadratic form on MATH with a quadratic form MATH on MATH for which MATH is a Lagrangian subbundle. By REF , MATH is necessarily the divided square quadratic form. The identification of the quadratic spaces MATH and MATH is unique up to homothety, once the morphisms MATH are chosen. However, REF is unique up to an alternating homotopy MATH. One may check that this alternating homotopy does not change the Lagrangian subspace MATH but only the choice of the Lagrangian complement MATH (compare REF). Thus MATH is uniquely determined by MATH. This completes the proof. |
math/9906174 | Suppose first that MATH. Then MATH and, by REF , the elements of MATH may be characterized as those MATH such that MATH and MATH are both eigenvectors for each MATH and MATH. Since MATH and MATH, the first claim follows. If MATH then calculation gives MATH where MATH. With this observation, the first claim follows in this case from REF . Finally, MATH and the second claim is established. |
math/9906174 | If MATH then, by REF , MATH may be characterized as the set of MATH in MATH such that MATH and MATH are eigenvectors of MATH and MATH. Since MATH and MATH, the claim follows. If MATH then MATH and a similar argument works on setting MATH in the notation of REF . This leaves the case where MATH. We use the notation of REF . If we set MATH for some MATH then, using the equation MATH, we have MATH and so MATH is MATH. Thus MATH. Finally, we have MATH and the last claim follows from this. |
math/9906174 | We have MATH for some MATH. Since MATH and MATH both satisfy REF , MATH and if we put MATH then MATH and so MATH. From REF, MATH and the conclusion follows. |
math/9906174 | Let MATH be chosen as above and put MATH. Then MATH, MATH and if MATH then MATH, so that MATH satisfies REF in this case. It follows that MATH because MATH and MATH. This establishes the first claim and the second then follows from the observation that MATH is the identity map. |
math/9906174 | Since the group MATH is unimodular MATH. So MATH . Therefore MATH. |
math/9906174 | Identifying MATH with MATH and MATH with MATH, we define MATH (respectively, MATH) to be the set of ideles of MATH (respectively, MATH) with absolute value one. Let MATH and MATH be the measures on MATH and MATH, such that MATH, MATH for MATH . Note that if MATH then the absolute value of MATH as an idele of MATH is MATH. Therefore, MATH for MATH. Since MATH this implies that MATH. So MATH . Since MATH this proves the proposition. |
math/9906174 | For each MATH let MATH be such that MATH. Then, from REF , MATH where we have used REF in passing from the first line to the second. Applying REF to MATH we find that MATH. Since MATH, MATH and so the NAME product formula implies that MATH. Now taking the product over all MATH on both sides of REF proves the identity. |
math/9906174 | From REF we have MATH . |
math/9906174 | The set MATH is open and MATH is a continuous function on it. We may therefore find an open set MATH containing MATH, having compact closure MATH and such that MATH for MATH. We can then choose MATH in such a way that MATH and MATH. Now REF implies that MATH doesn't vanish on MATH and hence it is bounded both above and below by positive constants on this compactum. Thus MATH is entire. REF also implies that MATH and hence MATH. |
math/9906174 | For each MATH we choose MATH such that MATH. Let MATH. For MATH we have MATH unless MATH and hence MATH by REF . Using REF , this formula implies the first statement. Now choose MATH for MATH so that MATH. It follows directly from the definition that MATH for all MATH and so the residue of MATH at MATH is MATH . We have MATH for MATH and hence MATH . Combining the last two equations shows that the residue of MATH at MATH is MATH and using the definition of MATH and the values of MATH and MATH (see the end of REF) gives the second claim. |
math/9906174 | We have MATH where the sum is over all MATH-tuples MATH which extend the MATH-tuple MATH. The claim follows immediately. |
math/9906174 | In the following, sums over MATH will be understood to include the conditions MATH and MATH as well as any further conditions which may be explicitly imposed. We have MATH where MATH . Applying the NAME theorem (CITE, p. REF) to MATH we obtain, in light of REF , MATH . We shall denote the right-hand side of this equation by MATH. Note that MATH is the right-hand side of the equation in the statement. Since MATH for all MATH and MATH we obtain MATH for all MATH and so MATH. It follows that there is a constant MATH such that MATH for all MATH (note that if MATH then the sum is MATH). Furthermore, MATH . It follows that, for all MATH, MATH and letting MATH we obtain MATH since MATH. The theorem follows. |
math/9906174 | This identity is perhaps the simplest instance of what is known as a NAME relation (see CITE, p. REF, for instance). For the reader's convenience we sketch the proof from the theory of the NAME zeta function. Using REF, p. REF we have the factorization MATH where MATH is the idele class character of MATH corresponding by class field theory to MATH. Multiplying both sides of this identity by MATH we obtain MATH . Since MATH, it follows that MATH . Recall that we have a functional equation MATH where MATH denotes the number of real places of MATH and MATH the number of complex places of MATH. It is easy to check that MATH and MATH for MATH. Comparing the factors in the functional equation on both sides of REF now shows that MATH and the identity follows. |
math/9906174 | We have MATH by REF with MATH and MATH the characteristic function of MATH. But MATH for all MATH and so MATH. |
math/9906174 | For the standard orbital representatives, MATH, at the infinite places we have required that MATH and so MATH. If MATH is a real place of MATH which splits in MATH then MATH is the union of two orbits with indices (sp) and (sp rm) respectively. From REF we see that MATH for both these orbits. In the product, the total contribution from these places is thus MATH. If MATH is a real place of MATH which ramifies in MATH then MATH is the union of two orbits with indices (rm) and (rm rm)* respectively. From REF we see that MATH for both these orbits. In the product, the total contribution from these places is thus MATH. Finally, if MATH is a complex place of MATH then MATH consists of a single orbit with index (sp) and, from REF , MATH for this orbit. The total contribution to the product from the complex places of MATH is thus MATH and the formula follows. |
math/9906174 | First suppose that MATH. Then every index corresponds to a single orbit, with the exception of (sp rm) and (in rm), which correspond to two orbits each. Using this and the values of MATH given in REF it is routine to check the given expressions. Now suppose that MATH. We have MATH where the sum now runs over a complete set of representatives for the MATH equivalence classes. The values of MATH are given in REF and using them one can easily establish the claim when MATH. We carry out the case MATH explicitly since it is rather more elaborate. First suppose that MATH with MATH. The indices which are possible with our assumptions are (rm), (rm ur), (rm rm)* and (rm rm ur), corresponding to one orbit each, and (rm rm rm), which corresponds to many orbits. Using REF , the contribution to MATH from the first four of these indices is MATH . Recall that the orbits with index (rm rm rm) have been grouped under MATH by MATH if MATH and by level if MATH. If MATH then either MATH with MATH or MATH. Using REF , we see that the contribution from these equivalence classes is MATH . If MATH then the level, MATH, runs from MATH up to MATH. By REF , the value MATH, although possible, corresponds to the orbit with index (rm rm ur) and so is excluded here. The contribution from the equivalence classes with MATH is thus MATH . Let us now collect all the terms from REF which have MATH as a visible factor. The result is MATH on expanding the factor in the square brackets. It remains to add REF , the first term of REF and the term MATH from REF to obtain MATH. This is easily done. The situation where MATH is similar, but simpler, and we leave it to the reader. |
math/9906174 | By REF we have MATH if MATH. Recall, from REF , that MATH and, from REF , that MATH where MATH is the number of real places of MATH and MATH the number of complex places of this field. Thus MATH . Let MATH and choose a MATH-tuple, MATH. According to REF , MATH exists and equals MATH . Making use of REF this quantity equals MATH . But MATH, MATH and MATH. Thus MATH and MATH and we have evaluated REF as MATH . Now MATH and so REF equals MATH . Now we sum REF over all MATH-tuples MATH which satisfy MATH for all MATH to obtain the statement of the theorem. |
math/9906174 | Let MATH be a quadratic extension and suppose that MATH contains a primitive MATH root of unity, MATH, for some MATH. Since MATH, it follows that MATH. But it is well-known that MATH as MATH and so there is some constant MATH, independent of MATH, such that MATH. We conclude that, for all but finitely-many quadratic extensions MATH of MATH, MATH. This finite list of exceptions may be ignored in the limit. Since MATH the corollary is now an immediate consequence of the theorem and the definition of MATH. |
math/9906174 | By REF , MATH. So MATH . As in the proof of REF , MATH and MATH except for a finite number of quadratic extensions MATH. Therefore REF follows from REF . |
math/9906174 | Examining the second equation in REF we see, in light of REF and the hypotheses, that every factor in the definition of the local orbital zeta function remains unchanged when we replace MATH by MATH. |
math/9906174 | If MATH then MATH, MATH and MATH for any vector MATH. This makes it clear that MATH in all cases. The rest of the argument will be case by case, but we make two observations which will be used repeatedly. First, it follows at once from the definition in every case that MATH and so to establish REF we need only prove the reverse inclusion. This will be done if we can show that given MATH with MATH we can find MATH such that MATH. Secondly, any MATH may be expressed as MATH, in the notation of REF - REF , and MATH if and only if all the components of MATH are units. Consider the cases (sp), (in) and (rm). We may assume, for simplicity, that MATH has been chosen to be the identity. Take MATH with MATH and let MATH be its NAME decomposition. Let MATH be as in REF or REF . By choosing MATH in the first case and MATH in the second, we may arrange that MATH. This proves REF . Moreover, if MATH and all the components of MATH are units then commuting MATH past the MATH and MATH factors in the NAME decomposition and absorbing it into the MATH factor shows that MATH also, which proves REF . For REF , observe that in the NAME decomposition, the MATH factor is unique up to multiplication of its diagonal elements by units. Thus if MATH and MATH with MATH then MATH. This proves REF . We next turn to case (sp ur). Let MATH be as in REF and MATH with MATH. Note that MATH and since MATH is unramified, this may be any even power of MATH. The same holds for MATH and the determinants of the components of MATH are the same as those of the corresponding components of MATH. It follows that we can arrange MATH for a suitable choice of MATH and this proves REF . If MATH then the determinants of each of its components are units and this makes REF obvious. Also, this argument shows that if MATH, MATH and MATH then MATH and MATH are units, which implies that MATH; hence REF . The case (sp rm) is very similar, with the one difference that since MATH is ramified, MATH can be any integer power of MATH. Next we treat (in ur). Let MATH with MATH and MATH be as in REF . Note that MATH is an eigenvector for the first component of MATH with eigenvalue MATH. So if MATH then MATH. Also, MATH. We are free to choose the pair MATH arbitrarily and so there exists MATH with MATH, proving REF . If MATH and MATH then MATH and MATH are units and so MATH and MATH, which proves REF . Also, if MATH, MATH and MATH then MATH, which implies that MATH and REF follows. Finally, cases (in rm) and (rm ur) are very similar to cases (sp rm) and (sp ur). Note that if MATH is as in REF then MATH. In case (in rm), MATH is ramified and so this takes every value in MATH. In case (rm ur), MATH is unramified and so MATH takes every value in MATH. The rest of the argument is identical to that in the cases already mentioned. |
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