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math/0007186 | The integrability of MATH follows from the NAME theorem. The exactness of the NAME complex for MATH is a theorem due to NAME REF . |
math/0007186 | For any MATH, consider the closed form MATH. By REF , there is a function MATH such that MATH. By REF , MATH satisfies the lemma. |
math/0007186 | Let us choose MATH such that the NAME complex and the usual MATH-NAME complex MATH are exact. On the NAME complex MATH consider the NAME filtration generated by MATH. Considering the ``stupid" filtration on MATH we have that the natural inclusion MATH is a filtered quasiisomorphism, that is, induces a quasiisomorphism on the associated graded complexes. This follows from the fact that, since MATH is strong, the NAME complex is exact. Now our lemma follows from the exactness of MATH. |
math/0007186 | By REF one can choose MATH such that MATH. For any functions MATH from MATH the functions MATH also satisfy REF . Let us find MATH such that MATH. We find MATH from the equation MATH . Note that MATH is a closed REF-form of the MATH-NAME complex and the equation can be rewritten as MATH where MATH. By REF , such MATH exists. |
math/0007186 | REF follows from the existence of functions MATH satisfying REF . CASE: One needs to construct, locally, a NAME algebra morphism MATH such that MATH. Let us choose functions MATH and MATH, MATH, as in REF and assign MATH. Since the elements MATH form a local basis in MATH, this defines a map of MATH modules, MATH. One has MATH and also MATH, which proves that MATH is a NAME algebra morphism. |
math/0007186 | For any point of MATH there is an open neighborhood, MATH, such that on it MATH, where MATH is an REF-form. Let MATH be the vector field on MATH corresponding to MATH by the isomorphism MATH determined by MATH. Then, MATH is a formal automorphism of MATH which transforms MATH to MATH. Hence, one may suppose in the proposition that MATH and MATH and MATH are two polarizations of the same form MATH. Let MATH be a neighborhood of MATH such that over MATH the NAME complexes corresponding to both MATH and MATH are exact. Let functions MATH, MATH on MATH be such that MATH and MATH form local basises in MATH and MATH, respectively. Suppose MATH mod MATH. Then, we may assume that MATH mod MATH, that is, MATH for some functions MATH on MATH. Let MATH denote the NAME bracket inverse to MATH. One has MATH, hence MATH. It follows from the exactness of NAME complex that there exists a function MATH on MATH such that MATH for all MATH. Let MATH be the Hamiltonian vector field on MATH corresponding to MATH. Then, the automorphism MATH of MATH leaves MATH on the place and transforms MATH to a polarization, MATH, such that MATH mod MATH. Proceeding by iteration proves the proposition. |
math/0007186 | The operators MATH are pairwise commuting derivations of MATH restricted to an open set MATH. Let MATH be the free MATH module over MATH spanned on MATH. Denote by MATH the basis dual to MATH. Let us define in the obvious way the complex MATH, where MATH . Since MATH and MATH, this complex is a deformation of the NAME complex MATH over MATH with MATH defined by REF . Since MATH is strong, the NAME complex is exact, therefore the complex MATH is exact, too. Now we proceed as in the proof of REF . Namely, for any MATH, consider the closed form MATH. Since the complex MATH is exact, there is a function MATH such that MATH. By REF , MATH satisfies the lemma. |
math/0007186 | The same as of REF . |
math/0007186 | The same as of REF . |
math/0007186 | By REF one can suppose that MATH and MATH. So, MATH, where MATH is the algebra of functions constant along MATH. Let MATH and MATH be a sufficiently small neighborhood of MATH. Let MATH be functions on MATH such that MATH form a local basis in MATH and MATH be functions such that MATH, where MATH is the NAME bracket inverse to MATH. Let MATH and MATH be multiplications in MATH over MATH corresponding to MATH and MATH and consisting of bidifferential operators. So, MATH and one may suppose that MATH. Suppose that there exists a differential operator MATH which transforms MATH to MATH modulo MATH and such that MATH, MATH, take MATH to zero. This means, in particular, that we may suppose that MATH modulo MATH, so that MATH . We are going to prove that there exists a differential operator of the form MATH, where MATH is a vector field on MATH, which transforms MATH to MATH modulo MATH and such that MATH. It is easy to check that MATH is a NAME cocycle in the algebra MATH. Hence, MATH is a bivector field and there is the representation: MATH, where MATH is a differential operator and MATH the NAME differential. It is also easy to check that MATH, where MATH denotes the NAME bracket of polyvector fields. So, there exists a vector field MATH such that MATH. This means that for any MATH . Recall now that MATH for MATH. It follows that for any MATH . In particular, we have MATH for any MATH of our basis. It follows from the exactness of NAME complex that there exists a function MATH such that MATH for all MATH. Denote MATH, the Hamiltonian vector field corresponding to MATH. Put MATH. Then, since MATH, we get MATH. Since MATH for all MATH, we get MATH. If we transform the multiplication MATH by the operator MATH, we obtain the multiplication MATH of the form MATH, where MATH. Since MATH for all MATH, we have that MATH is a derivation from MATH to MATH. This derivation can be extended to a derivation of MATH. Indeed, put MATH. Then the operator MATH is such an extension. Now put MATH. One has MATH. It is easy to see that the operator MATH transforms MATH to MATH modulo MATH. It follows that the operator MATH transforms MATH to MATH modulo MATH. |
math/0007186 | Follows from the previous proposition. |
math/0007186 | Choose an open covering MATH of MATH such that, for each open set MATH there exist flat connections MATH on MATH and MATH on MATH. Let MATH denote the splitting of the extension MATH given by the negative of the NAME derivative. This splitting is a NAME algebra homomorphism, but is not MATH-linear. Instead, for MATH and MATH, MATH. We denote the restriction of this splitting to MATH by the same letter. Since both MATH and MATH are splittings of the same sequence, their difference MATH is a map MATH which is not MATH-linear, but satisfies MATH. Note that MATH. Let MATH. We claim that MATH is a (locally defined) flat connection on MATH. To this end we check various properties. The calculation MATH shows that MATH is a (local) splitting. The calculation MATH shows that this splitting is MATH-linear. One verifies other properties similarly. Hence one can use the locally defined flat connections MATH to calculate the characteristic class of MATH. The desired formula follows from MATH . |
math/0007186 | By REF , there is an open covering MATH of MATH such that there exist isomorphisms of polarized quantizations MATH . These isomorphisms induce isomorphisms MATH over each MATH. We may suppose that both MATH and MATH admit on each MATH flat connections MATH, MATH, such that MATH. Since MATH mod MATH, we may choose the connections in such a way that MATH mod MATH. Let MATH be functions on MATH such that MATH. Since MATH mod MATH, we may choose MATH such that MATH mod MATH. On MATH one has MATH. Since MATH, MATH is an automorphism of MATH on MATH. According to NAME 's definition, MATH is equal to MATH. |
math/0007186 | First, we assume that MATH corresponds to the trivial deformation MATH. By REF , there is an open covering MATH of MATH such that MATH are isomorphic to NAME polarized quantizations. Let MATH be MATH linear derivations of MATH such that for MATH one has MATH. Such MATH exist because they obviously exist for the NAME star-products. Then, MATH. Let the covering MATH be fine enough and on each MATH there exists a flat connection MATH. Then, on MATH one has MATH for MATH. Let MATH be functions not depending on MATH such that MATH form a local basis in MATH, and MATH be the dual basis in MATH. Put MATH. One has MATH. Hence, since MATH does not depend of MATH, MATH. This means that MATH for some MATH. It follows from MATH that MATH. It follows from the exactness of MATH-NAME complex that there exist MATH such that MATH. So, MATH. Taking MATH instead MATH we obtain that the new MATH satisfy MATH. This equations may be rewritten in the form MATH . Further, one has MATH . Using REF , we have MATH . Since derivations of MATH are completely defined by their values on MATH, it follows that MATH . According to NAME 's definition, this means that MATH represents MATH. Now, let MATH be arbitrary. Denote by MATH the class of a quantization corresponding to MATH and by MATH the class of a quantization corresponding to the trivial deformation MATH. The same meaning has the notation MATH, MATH and MATH, MATH. By REF we have MATH, hence MATH . Applying MATH to the left hand side of this equation, we obtain zero (by REF and what we have just proved). Hence, applying MATH to the right hand side and using REF , we have MATH which proves the proposition. |
math/0007186 | Follows from the NAME construction, see also CITE. |
math/0007186 | By REF we have the equalities MATH . By REF we have MATH and MATH. Combining these with the preceeding identity we obtain MATH . By REF both differences in the left hand side of the last identity are equal to the differences of the respective constant terms, hence coincide (since the constant terms are not affected by MATH). Thus the above expression says MATH . The theorem follows from the fact that MATH. |
math/0007190 | Let MATH be an irreducible MATH module, recalling that any such module is unique up to equivalence CITE, with MATH as a complex algebra acting on MATH by the standard representation. (We use MATH to denote the algebra of complex MATH matrices.) Hence, the only MATH-module endomorphisms of MATH are given by complex scalar multiplication and we have an isomorphism of complex vector spaces, MATH, given by MATH. Therefore, we obtain an isomorphism of complex vector spaces MATH where MATH is identified with the MATH-module homomorphism MATH given by MATH. Indeed, the map REF is surjective because we can compose any homomorphism MATH with projection onto each factor and then use the fact that MATH. Moreover, the map REF is injective by construction. Given a complex NAME module MATH of rank four, define MATH . For every MATH and isomorphism MATH of complex NAME algebras, there are isomorphisms of complex NAME modules, MATH and MATH. The isomorphism REF then implies that MATH is a rank-two complex vector bundle over MATH. The map MATH given by MATH, where MATH and MATH for some MATH, is a MATH-module isomorphism since it is fiberwise injective, the ranks of the two bundles agree, and it is a complex NAME module homomorphism by construction. To see that the map MATH is injective, observe that if MATH, then either MATH or MATH and MATH has a non-trivial kernel, in which case one can see from the explicit form of the map REF that MATH must then be zero. Next we show that MATH is unique up to isomorphism. From REF below, we have MATH . From REF below we see that the bundle MATH determines the subbundle MATH up to isomorphism, independently of MATH, and hence MATH uniquely determines MATH. Then MATH and MATH is determined up to isomorphism by MATH and MATH. Finally, if MATH and MATH are Hermitian then MATH is Hermitian and the isomorphism MATH can be taken to be Hermitian. |
math/0007190 | The isomorphism MATH is given explicitly by CITE MATH if MATH is a local oriented, orthonormal frame for MATH. Using this isomorphism and the fact that MATH when MATH, it is easy to see that the left-hand side of REF is mapped into MATH, the skew-Hermitian endomorphisms of MATH. The elements of MATH and MATH give isomorphisms in MATH and their block-matrix representations in MATH have zero components in MATH, so they have zero trace. As remarked earlier, the elements of MATH are known to have zero trace. Next, MATH by definition of MATH and thus MATH has zero trace. Hence the left-hand side of REF is also mapped into MATH, the traceless endomorphisms of MATH. The map MATH is an isomorphism CITE of NAME modules and in particular is injective. Since the sum of the ranks of the bundles on the left-hand side of REF is MATH, which is equal to the rank of MATH, we see that the map REF must be an isomorphism, as claimed. |
math/0007190 | The connection MATH is spin by the remarks preceding the statement of the lemma. From REF one can see that MATH for all MATH and MATH, so MATH preserves the subspace MATH, inducing the NAME connection MATH. Recall that sections MATH of MATH can be characterized as sections of MATH having zero commutator with all MATH, for MATH. Thus, for any such MATH, MATH, and MATH, we have MATH and MATH . Hence, MATH for all MATH and therefore MATH preserves the subspace MATH, on which it induces an orthogonal connection MATH. Sections of the subbundle MATH are linear combinations of sections of the form MATH, where MATH and MATH. Since MATH is a NAME module derivation and induces the connections MATH and MATH on the subbundles MATH and MATH, respectively, we have MATH so the orthogonal connection induced by MATH on MATH is given by MATH. |
math/0007190 | Plainly, the map is an injective homomorphism, so it remains to show that it is surjective. Suppose MATH. For any MATH, the remarks preceding REF imply that we may write MATH, where MATH and so we have MATH for some MATH. But MATH and as MATH, we must have MATH and thus MATH, as desired. |
math/0007190 | Suppose that MATH is a reducible spin connection on MATH with respect to the splitting MATH. There is a unique complex line bundle MATH over MATH such that MATH. If MATH is a unitary connection on MATH, then both MATH and MATH are spin connections on MATH. But any two spin connections on MATH differ by an element of MATH, via the action REF, and thus we can write MATH for some unitary connection MATH on MATH. The unitary connection MATH on MATH induces the MATH connection MATH on MATH. To see this, we pass to a local trivialization of MATH and view MATH as a connection matrix one-form, MATH . The matrix MATH acts on a section MATH of MATH via the adjoint representation, MATH . Thus, MATH, with respect to the isomorphism MATH of REF and so, as a connection, we see that MATH. Therefore, the spin connection MATH on MATH induces the MATH connection MATH on MATH via the decomposition REF, MATH . The connections MATH on MATH and MATH on MATH induce the connection MATH on MATH. By convention, our spin connections MATH on MATH induce the connection MATH on MATH, and thus MATH. Conversely, if MATH is a MATH connection which is reducible with respect to the splitting MATH, then MATH lifts to a MATH connection MATH on MATH which is reducible with respect to the splitting MATH, and lifts to a spin connection MATH on MATH which is reducible with respect to the splitting MATH. |
math/0007190 | We closely follow the method described in CITE. Our NAME space of parameters is given by MATH . We define an extended MATH-equivariant map, MATH by setting MATH . The parametrized moduli space MATH is then MATH and MATH. The map MATH has differential at the point MATH given by MATH where MATH. Suppose MATH is MATH orthogonal to the image of MATH. We may assume without loss of generality that MATH is a MATH representative of the point MATH (see, for example, CITE) and so, as MATH, elliptic regularity implies that MATH is MATH. Then MATH for all MATH, which yields the pointwise identity MATH . If MATH for some MATH (and thus MATH on a non-empty open neighborhood in MATH), then REF implies that MATH on the non-empty open subset MATH. NAME 's theorem then implies that MATH on MATH, contradicting our assumption that MATH is a point in MATH. Hence, we must have MATH on MATH, so REF now yields MATH for all MATH. Since MATH, this implies that MATH, just as in the proof of CITE. Hence, MATH at each point MATH in MATH and so the parametrized moduli space MATH is a smooth NAME submanifold of MATH. Therefore, the NAME theorem, in the form of CITE (see the proof of REF), implies that there is a first-category subset of MATH such that for all MATH parameters MATH in the complement of this subset, the zero locus MATH is a regular submanifold of MATH. Lastly, we can constrain the parameter MATH to be MATH (and not just MATH) by the argument used in CITE - see CITE. |
math/0007190 | If MATH denotes the NAME space of MATH perturbation parameters MATH, then a generic, smooth path in MATH joining MATH to MATH induces a smooth, oriented cobordism in MATH joining MATH to MATH. This proves REF . If the compact space MATH is contained in the open subset MATH, then the same holds for MATH, that is, the latter subspace also contains no zero-section pairs, for any triple MATH which is MATH-close enough to MATH. This proves REF . |
math/0007190 | Any MATH spin connection on MATH can be written as MATH, where MATH. An element MATH of the gauge group MATH acts on this representation of MATH by sending MATH to MATH. The argument in CITE (also see CITE) implies that there is a solution of the equation MATH, which is unique up to a harmonic gauge transformation, so the map on quotients is surjective. If MATH, MATH, are pairs in MATH such that MATH and MATH satisfies MATH, that is MATH then MATH implies MATH, so MATH is harmonic and the map on quotients is injective. |
math/0007190 | The class MATH is the first NAME class of MATH restricted to MATH. REF then follows from the splitting MATH. |
math/0007190 | From the NAME decomposition, MATH where MATH denotes the harmonic, imaginary-valued one-forms on MATH, we see that MATH . Thus, we can write the pre-configuration space MATH as MATH . As usual, the factor MATH of the harmonic gauge group MATH acts trivially on MATH and by complex multiplication on MATH. An element MATH acts on an element MATH of MATH (where MATH and MATH) by MATH . Thus, replacing MATH with MATH, defines a MATH-equivariant retraction of MATH onto MATH . This yields a deformation retraction of MATH onto the projectivization of the complex vector bundle MATH . By NAME 's theorem (see REF ) this vector bundle is trivial, which completes the proof. |
math/0007190 | Since MATH and MATH are free, we can write MATH with respect to the decomposition given by the NAME formula: MATH . For MATH and MATH, the restrictions of MATH to MATH and to MATH are trivial so MATH. We now calculate the first NAME class of MATH restricted to the tori MATH. Let MATH be the path in MATH, starting at zero and covering MATH. Let MATH be an element of MATH satisfying MATH and MATH. Thus, the degree of the map MATH is MATH. The restriction of the line bundle MATH to MATH is given by taking the trivial bundle over MATH and making the identification MATH . Since the degree of the map MATH is MATH, we see MATH . This completes the proof. |
math/0007190 | Because MATH is a retraction, it suffices to compute the restriction of the universal line bundle MATH to the image of the retraction where MATH is given by MATH . The bundle MATH is the restriction of the framed configuration space, MATH to the image of MATH and thus has first NAME class MATH. Then the restriction of MATH in REF is isomorphic to the tensor product of the bundle REF with the pullback of the bundle MATH by the map MATH. Thus MATH. The computations of MATH then follow from the definition of the map MATH in REF and the computation of MATH in REF . |
math/0007190 | Suppose MATH is a fixed point of the MATH action on MATH. Consequently, there is an element MATH such that MATH and hence a gauge transformation MATH such that MATH . Thus, MATH is in the stabilizer of MATH, and hence that of MATH, and MATH. CASE: If MATH has the trivial stabilizer MATH in MATH, then MATH and so MATH because MATH. The connection MATH is irreducible CITE since it is a MATH connection with minimal stabilizer in MATH. The curvature equation in REF implies that MATH, so MATH is anti-self-dual. The pair MATH is fixed by the MATH action REF. CASE: If MATH has non-trivial stabilizer in MATH, then MATH is reducible with respect to a splitting MATH CITE, for some complex line bundle MATH, and takes the form MATH. The connection MATH has stabilizer MATH acting on MATH by complex multiplication and trivially on MATH. REF implies that MATH is reducible with respect to the splitting MATH, taking the form MATH. If MATH, then MATH just as in REF and so MATH would be a reducible, anti-self-dual MATH connection on MATH. But MATH and the Riemannian metric MATH on MATH is generic, so this possibility is excluded by REF (the four-manifold MATH does not need to be simply connected). Hence, we must have MATH and MATH. Because MATH stabilizes MATH, it induces an action on MATH via complex multiplication by some MATH and the trivial action on MATH, as in REF. Hence, with respect to the splitting MATH, the gauge transformation MATH takes the form MATH . Because MATH and writing MATH with respect to the splitting MATH, the section MATH is fixed by MATH . If MATH, then MATH and we must have MATH; conversely, if MATH, then MATH and we must have MATH. The two cases differ only by how the factors in the splitting of MATH are labeled, so we can assume that MATH and MATH. In particular, the pair MATH is fixed by the MATH action REF. |
math/0007190 | That MATH is closed under the MATH action follows directly from the definitions. It is clear that the function MATH is smooth on each stratum of MATH: we claim it is continuous on MATH. Let MATH be a sequence of points in MATH which converge to MATH. We may assume, by an appropriate choice of a sequence of MATH gauge transformations of MATH, that the sequence MATH converges to MATH in MATH on MATH, where we define MATH. Therefore, MATH where the second inequality follows from the universal a priori MATH bound for the sequence MATH and MATH given by CITE. Thus, MATH and so MATH as desired. A generic MATH is a regular value for the function MATH on each smooth stratum of MATH. For such a MATH, the preimage MATH is a smooth submanifold of each stratum and because the function MATH is continuous, these smooth submanifolds fit together to form a smoothly stratified subspace of MATH (see REF). |
math/0007190 | With respect to the decomposition MATH, any element MATH takes the form MATH where MATH, for MATH. Thus, MATH, after identifying MATH. If MATH, then MATH and MATH with MATH and MATH, so any element of MATH takes the shape MATH and the desired isomorphism MATH is given by MATH. We recall from CITE that the induced fiber inner product on MATH is defined by MATH. Thus, if MATH correspond to MATH, respectively, we see that MATH and so the isomorphism of MATH bundles is an isometry. |
math/0007190 | The map MATH is clearly a MATH embedding. Furthermore, MATH . Since MATH, we see that MATH acts on MATH as MATH (see also REF) and so MATH. Thus, MATH, as desired. Next, we characterize the image of the map MATH. Suppose MATH is fixed by the MATH action REF on MATH: this action descends to the action REF on MATH, which fixes the induced connection MATH. As in the proof of REF , the connection MATH must then be reducible with respect to this splitting, taking the form MATH on MATH and MATH on MATH. Thus, MATH is in the image of MATH. If MATH is an open subset, so it is easy to see that MATH is open in MATH with respect to the subspace topology induced by MATH. We now show MATH is injective and that MATH is injective when MATH. Suppose MATH and MATH are pairs in MATH and that MATH satisfies MATH, and thus MATH in MATH. With respect to the decomposition MATH and splitting MATH, we can consider MATH to be a gauge transformation of MATH via the isomorphism MATH in REF (and acting as the identity on MATH) and write MATH noting that MATH and MATH. As in the proof of REF , we may write MATH and MATH, where MATH and MATH are MATH connections on MATH which are reducible with respect to the splitting MATH. Let MATH denote the action of MATH on MATH by the trivial action on MATH and scalar multiplication on MATH. Then MATH fixes MATH and thus for all real MATH where MATH is the stabilizer of MATH in the space of unitary automorphisms of MATH. Consequently, for every real MATH there are real constants MATH and MATH such that MATH and we can assume MATH. Simplifying, this becomes MATH . If MATH, then we must have MATH and thus MATH because MATH. Since MATH, we see that MATH. Hence, MATH and MATH because MATH. It remains to consider the case MATH, for which we must then have MATH. First suppose MATH and observe that MATH so MATH on MATH and hence MATH, contradicting our hypothesis that MATH. Therefore the case MATH cannot occur in this situation. Thus, MATH is injective. Otherwise, if MATH, suppose MATH and observe that the automorphism MATH induces an isomorphism MATH. But MATH and MATH: by the hypothesis in the final statement of the lemma, MATH, so this contradicts MATH and therefore the case MATH cannot occur in this situation either. Thus, if MATH, the map MATH is injective. |
math/0007190 | Suppose that MATH solves the MATH monopole REF . With respect to the splitting MATH, REF implies that MATH, where MATH is the product connection on MATH and MATH is a unitary connection on the complex line bundle MATH. Let MATH and observe that MATH is a reducible unitary connection on MATH which is a lift of MATH on MATH. Then MATH and therefore, since MATH, we have MATH . If MATH, where MATH and writing MATH, we have MATH . Hence, projecting to MATH, we see that MATH . Furthermore, with respect to the splitting MATH, we clearly have MATH . Then, combining REF , and REF, and noting that MATH shows that the pair MATH solves MATH . Comparing REF with the NAME REF concludes the proof. |
math/0007190 | Given REF , we need only characterize the image of MATH. If MATH represents a point in MATH and is reducible with respect to the splitting MATH, with MATH, then REF implies that MATH, for some pair MATH. Then REF implies that MATH satisfies the NAME REF since MATH satisfies the MATH monopole REF . |
math/0007190 | The isomorphisms identifying the cohomology of the tangential deformation complex with that of the NAME complex follow immediately from a comparison of these two complexes. If MATH then MATH and REF implies MATH and so MATH. Moreover, if MATH, then MATH. |
math/0007190 | Let MATH be a point in MATH: by the slice theorem for MATH, the restriction of the projection map MATH to a small enough open neighborhood of MATH in the slice MATH gives a smooth parameterization of an open neighborhood of MATH in MATH. Similarly, by the slice theorem CITE for MATH, the restriction of the projection map MATH to a small enough open neighborhood of MATH in the slice MATH gives a smooth parameterization of an open neighborhood of MATH in MATH. Comparing the operators MATH in REF and MATH in REF , we see that the differential of the smooth embedding MATH of REF restricts to an isomorphism MATH . Hence, the differential MATH is injective and the induced maps MATH and MATH are smooth immersions. Since the map MATH is a topological embedding according to REF , and is a smooth immersion, the map MATH is a smooth embedding. |
math/0007190 | If MATH is a vector bundle over MATH with fiber a NAME space MATH, then REF yields a MATH trivialization MATH, that is, a MATH section MATH of the MATH . NAME bundle MATH over MATH which gives a linear isomorphism on each fiber. Now suppose that MATH is a MATH section of the bundle MATH. If MATH is chosen so that MATH is sufficiently small for each MATH, then MATH is an isomorphism on each fiber and gives the desired global, MATH trivialization. |
math/0007190 | By REF , there is a smooth trivialization of MATH given by a smooth isomorphism of complex NAME bundles, MATH as the space MATH (of MATH gauge transformations) is a subgroup of the unitary group of the NAME space MATH. We have an equality of MATH-orthogonal complements, MATH for every MATH, by REF and the fact that MATH by REF , for generic parameters MATH. Thus, the family of operators MATH defines quasi-subbundles MATH and MATH of MATH with fibers MATH . For each point MATH, we define MATH . Observe that there is an open neighborhood MATH with the property that for all MATH, the map MATH defined by MATH-orthogonal projection onto MATH is surjective. Since MATH is compact, it has a finite subcover MATH of such neighborhoods. If MATH, then MATH is a finite-dimensional, complex subspace of MATH. If we define MATH by setting MATH then MATH will be a complex, finite-dimensional, trivial subbundle of MATH such that fiberwise MATH-orthogonal projection onto the subbundle MATH is surjective when restricted to the quasi-subbundle MATH. We now extend the bundle MATH to a subbundle of MATH over an open neighborhood MATH of MATH in MATH, which does not contain any other reducible or zero-section pairs, and which is MATH equivariant with respect to the circle action REF. The space MATH is a smooth submanifold of the Riemannian manifold MATH by REF and so it has a MATH-equivariant normal bundle MATH and a MATH-invariant tubular neighborhood given by a MATH-equivariant diffeomorphism MATH from an open, MATH-invariant neighborhood MATH of the zero section of MATH onto an open, MATH-invariant neighborhood of MATH in MATH (see, for example, CITE). The bundle projection MATH and the diffeomorphism MATH define a MATH-equivariant, MATH retraction, MATH . The complex vector bundle MATH then extends to a vector bundle MATH, a subbundle of MATH, which are both MATH equivariant with respect to the circle action REF. Since the map MATH is a MATH-equivariant, MATH retraction, there is a MATH-equivariant, MATH isomorphism CITE, CITE, MATH . We obtain a MATH subbundle of the vector bundle MATH by setting MATH both MATH equivariant with respect to the action REF. Because MATH is isomorphic to MATH as a complex vector bundle, for some MATH, we obtain an isomorphism of MATH-equivariant vector bundles, MATH since the maps MATH and MATH are MATH equivariant; the circle acts non-trivially on MATH, except along the stratum MATH, and acts by complex multiplication on MATH. By construction, the fiberwise MATH-orthogonal projection MATH restricts to a surjective map MATH for any pair MATH representing a point in MATH, after shrinking MATH if necessary. Given a MATH pair MATH representing a point in MATH, our construction yields a subspace MATH . If MATH is an integer, it does not necessarily follow that MATH is contained in the subspace of MATH pairs when MATH is a MATH pair. However, for any MATH the heat operator MATH is a bounded, MATH-equivariant, MATH-equivariant, linear map and we can define MATH . For small enough MATH, the approximation properties of the heat kernel (see CITE for a similar application), ensure that CASE: MATH is a trivial, MATH-equivariant vector bundle, with the same rank as MATH, and CASE: MATH-orthogonal projection MATH restricts to a surjective map MATH for any pair MATH representing a point in MATH. We let MATH be the MATH bundle map defined by fiberwise MATH-orthogonal projection. This completes the proof. |
math/0007190 | The space MATH is the zero locus of the section MATH of a vector subbundle MATH of MATH constructed in REF , for some open neighborhood MATH of MATH in MATH. For any point MATH, we have MATH . (The differential of MATH does not appear here since MATH.) According to REF , the MATH-orthogonal projection MATH gives a surjective map MATH and thus MATH is surjective at all points MATH in the image of MATH. NAME is an open condition, so we may assume that MATH is surjective on the open neighborhood MATH of MATH, after shrinking MATH if necessary. The zero locus of MATH in this open set is regular and thus a smooth submanifold of MATH, which gives REF . By REF the space MATH is a smooth submanifold of MATH and as its image is contained in MATH, it is also a smooth submanifold of MATH and REF follows, as the zero locus of a MATH-equivariant section is MATH invariant. We observe that MATH is the normal bundle of MATH in MATH, since MATH . In the second equality above, we make use of the fact that the fibers of the vector bundle MATH are contained in MATH and so (by definition) MATH on MATH. Also, the splitting REF of tangent spaces corresponds to the splitting of NAME spaces MATH, defined by the subspaces REF. Hence, the isomorphism REF is equivariant with respect to the circle action REF on the subspace MATH of MATH, the trivial action on the subspace MATH of MATH, and complex multiplication on the subspace MATH of MATH. This proves REF . Because MATH is given by MATH, we see that MATH takes values in MATH on MATH and so MATH on MATH. According to REF , our transversality result for the section MATH of MATH defined by the MATH monopole REF , the section MATH vanishes transversely on MATH with zero locus MATH. This implies that for each MATH in MATH, the differential MATH is surjective. But we have the identifications MATH and so one has a surjective differential MATH for MATH. Thus, the sections MATH and MATH - which are equal on MATH - vanish transversely when restricted to MATH, proving REF . |
math/0007190 | From the final statement of REF , we see that MATH vanishes transversely on MATH and so MATH cuts out the zero locus, MATH, as a regular submanifold of MATH. Then for generic values of MATH, the zero locus will intersect MATH transversely. |
math/0007190 | As an element of MATH, the index bundle MATH depends only on the homotopy class of the leading symbol CITE. Thus, the index bundle of the family MATH is equivalent to that of the family MATH, where MATH. |
math/0007190 | The associated line bundles MATH are given by the quotients of MATH by the relation MATH, for MATH and MATH; under the same relation, MATH. A tensor product of a complex line bundle MATH with a complex vector bundle MATH is given by the quotient of the fiber product MATH by the relation MATH, for MATH, and MATH. Hence, MATH where in the second line above we can assume without loss that MATH. This proves REF follows trivially from this. |
math/0007190 | There are isomorphisms, MATH where, from REF , the action of MATH on MATH is given by MATH. Let MATH be the unit sphere bundle of MATH. If MATH acts on MATH by MATH - where MATH, MATH, MATH and MATH - then we obtain an isomorphism of complex line bundles MATH . REF then implies that MATH and REF gives MATH . The desired expression for MATH then follows from the isomorphism MATH given by REF . In the bundles MATH an element MATH acts on MATH by MATH, where MATH, as noted in the remark following REF and the argument yielding REF then imply that there are isomorphisms of complex vector bundles MATH . The isomorphisms in the conclusion of the lemma now follow from REF and the isomorphism MATH implied by REF . |
math/0007190 | For convenience in the proof, we write MATH and MATH, as in REF . The NAME identification REF of MATH, REF , and the homomorphism property of the NAME character (see, for example, CITE) imply that MATH . Recall that MATH denotes the index bundle of the family of operators obtained, as in REF , by twisting the NAME operator MATH by a family of connections on the bundle MATH. Note that MATH, where MATH. We now use REF to partly compute the NAME characters of the index bundle MATH. Bundles over MATH will be considered to be bundles over MATH; the pullback MATH will be omitted. MATH . Similarly, we use REF to partly compute the NAME character of the index bundle MATH: MATH . In the first lines of REF we simply rewrite the index bundles using the notation of REF . For the calculation of MATH in REF, we compute MATH using the expression for MATH in REF : MATH . Because MATH as noted before REF , we see MATH. We shall write MATH for MATH. Applying this to REF and noting that MATH, we obtain MATH and therefore, using MATH (see CITE) MATH completing the calculation of MATH. We now complete the calculation of MATH in REF. To compute MATH we observe that MATH while MATH and so, using this and the isomorphism MATH, we have MATH . Applying this to the terms involving MATH in REF and writing MATH yields MATH . Hence, using the preceding expression in REF and MATH, we see that MATH . Simplifying the preceding expression for MATH and recalling that MATH (from CITE), yields MATH and thus, MATH . The desired expression for MATH follows from REF, and REF. |
math/0007190 | For convenience, we write MATH. The NAME character determines the NAME polynomial as an element of rational cohomology (see the formula in CITE or CITE), so MATH determines MATH where the MATH are the NAME roots of MATH. REF implies that, with the constraint on MATH, MATH . Suppose MATH and MATH. If MATH, then MATH for MATH and MATH for MATH, while MATH for MATH. Then one can easily see that MATH is equal to the NAME character associated to the NAME polynomial MATH. The case where either MATH or MATH is negative follows from the observation that if MATH, then MATH while MATH. |
math/0007190 | We may assume without loss that MATH is primitive because a sublattice MATH is orthogonal to MATH if and only if it is orthogonal to MATH, where MATH and MATH is primitive. From the classification of indefinite, integral, unimodular forms (for example, see REF) we have MATH . By hypothesis, MATH, so if MATH then MATH is a sublattice of MATH. Let MATH denote the component of MATH in MATH. Because MATH is even, we have MATH for some MATH. Let MATH and MATH be bases for the two hyperbolic sublattices of MATH, so MATH and all other pairings of these four vectors vanish. Define MATH, so MATH and MATH is primitive. The hyperbolic sublattice MATH is orthogonal to MATH. According to REF, the orthogonal group of MATH acts transitively on primitive vectors of a given square. Hence, there is an automorphism MATH of MATH such that MATH for all MATH and MATH. Because MATH is contained in MATH, then MATH. Hence, the hyperbolic sublattice MATH is contained in MATH. |
math/0007190 | We consider the intersection form MATH to be a form on MATH. By NAME 's theorem (see REF), we know that MATH and so the classification REF of forms yields MATH with MATH. A result of CITE then implies that MATH, where MATH and MATH. If MATH and MATH, then MATH and so MATH; if MATH, then MATH and NAME 's theorem implies MATH and so again MATH. Therefore, MATH contains a hyperbolic sublattice by REF . |
math/0007190 | We may assume without loss that MATH is primitive for, if not, write MATH where MATH and MATH is primitive. Then MATH for all MATH since MATH is characteristic. If MATH were even we would have MATH for all MATH, contradicting our hypothesis that MATH is an odd form. Hence, MATH is odd and MATH for all MATH and thus MATH is also characteristic. Then a sublattice MATH is orthogonal to MATH if and only if it is orthogonal to MATH. The classification of indefinite, integral, unimodular forms shows that, because MATH is odd, we can find a basis MATH for MATH for which MATH, MATH, and MATH . For odd integers MATH yet to be determined, choose MATH . The element MATH is primitive since at least one basis coefficient is equal to one. By hypothesis, MATH and MATH, so we can define MATH and observe that MATH is hyperbolic and orthogonal to MATH. Then MATH and so MATH . Since the coefficients of MATH are odd, we have MATH for all MATH; thus MATH is characteristic and we have MATH. As MATH is odd, so we can write MATH for some MATH. The left-hand side of the preceding equation therefore yields MATH . Now any positive integer which is congruent to MATH can be written as the sum of four odd squares (see REF ). So we select any odd integer MATH for which MATH and then choose odd integers MATH so that MATH . Therefore REF give MATH . A result of NAME, (see CITE) implies that the orthogonal group of MATH acts transitively on the primitive, characteristic elements with a given square. Hence, we can find an orthogonal automorphism MATH of MATH with MATH. Then MATH is a hyperbolic sublattice of MATH which is orthogonal to MATH. |
math/0007190 | Let MATH be a positive integer. The set of squares modulo MATH is MATH. Hence, if MATH is the sum of four odd squares, then we must have MATH. Conversely, suppose that MATH. By NAME 's theorem we can write MATH . Since MATH is even, we must have one of the following possibilities (up to rearranging the terms on the right-hand side): CASE: MATH are even, and MATH are odd. CASE: MATH are all even. CASE: MATH are all odd (as desired). Now since MATH, it is not hard to check that REF cannot happen. In fact, it is easy to see that if REF is true, then MATH. To see that REF can occur, write MATH, where MATH is a non-negative integer, so MATH . NAME 's theorem tells us that a positive integer is the sum of three integral squares if and only if it cannot be written in the form MATH, for some MATH (see CITE for the statement and CITE for a proof). Now MATH is odd, and since MATH cannot be of the form MATH, we can express MATH as a sum of three squares. So we can write MATH . Since MATH is odd, and we have ruled out REF above, it follows that MATH are all odd. |
math/0007190 | Interchanging the role of MATH and MATH in the proof of REF takes care of REF . Thus we need only consider REF . Continuing the notation of the proof of REF , we choose MATH . Plainly, MATH is characteristic, primitive, and is orthogonal to the hyperbolic sublattice MATH of MATH, while (as MATH) MATH . Since MATH is odd, we cannot have MATH (which would give MATH) so MATH or MATH, which gives MATH or MATH, respectively, neither of which is divisible by MATH, MATH, unless MATH. Thus MATH is primitive. Just as in the proof of REF , NAME 's theorem implies that we can find an orthogonal automorphism MATH of MATH with MATH, since MATH are both primitive, characteristic, and have equal square. Then MATH is a hyperbolic sublattice of MATH. This takes care of REF and completes the proof of the lemma. |
math/0007190 | Let MATH be a compact, complex algebraic, simply connected surface with MATH. Suppose MATH is minimal. The NAME classification then implies that MATH is one of the following (see CITE): CASE: A MATH surface, CASE: An elliptic surface, CASE: A surface of general type. The cases where MATH is diffeomorphic to MATH, MATH, or MATH (when MATH has NAME dimension MATH, see CITE) are eliminated by our requirement that MATH. If MATH is elliptic, then it is diffeomorphic to MATH, for some MATH, MATH, MATH (see REF); if MATH is a MATH surface, then it is diffeomorphic to the surface MATH (see REF), which is included in this family as MATH (see CITE for the construction of this family). Let MATH be the homology class of a regular fiber of MATH and observe that MATH is a primitive, integral homology class. According to CITE (see CITE), the NAME classes of MATH are multiples of the NAME dual MATH. Thus, MATH is abundant by REF REF and the observation that MATH and MATH (see CITE), so MATH and MATH for MATH. If MATH is a minimal algebraic surface of general type, then its NAME classes are MATH by CITE, where MATH is the canonical class. If MATH is even, then MATH is abundant by REF . If MATH is odd, the NAME inequality, MATH (see CITE), implies that MATH since MATH and MATH. Thus MATH . Equality in REF, or MATH, holds only if the universal covering space of MATH is the closed unit ball in MATH by CITE (or see CITE and the discussion in CITE or CITE). Since MATH is simply-connected and closed by hypothesis, we must have MATH . If MATH, then the preceding inequality yields MATH while if MATH, it yields MATH. Hence, for MATH odd, MATH is again abundant by REF . This proves the theorem for minimal surfaces. If MATH is an abundant, smooth four-manifold, there is a hyperbolic sublattice MATH such that MATH is orthogonal to MATH if MATH is an NAME class. The blow-up MATH is also abundant, since we may view MATH as a hyperbolic sublattice of MATH, all NAME classes of MATH have the form MATH (see CITE), and such classes are again orthogonal to MATH. Hence, if MATH is abundant, all its blow-ups are abundant too. This completes the proof of the theorem. |
math/0007191 | This follows from CITE. |
math/0007191 | By CITE the ring of invariants MATH is generated by the MATH. Now MATH is a closed subvariety of MATH, so the restriction map on functions MATH is surjective. Since MATH is reductive and the base field MATH has characteristic zero, there is a NAME operator, and so it remains surjective on taking invariants. |
math/0007191 | Since MATH the vertex MATH must be loopfree. Now some composition factor must have dimension MATH with MATH. Then MATH by CITE. Since there is no loop at vertex MATH, the relevant composition factor is isomorphic to MATH. Now because MATH, the choice of a decomposition MATH induces an embedding MATH and hence a map MATH which by the observation above is a bijection. We want to prove that is is an isomorphism of varieties. For this it suffices to prove that it is a closed embedding. That is, that the map of commutative algebras MATH is surjective. Now it is easy to see that this map sends the trace function MATH for dimension MATH to the trace function MATH for dimension MATH. Thus the assertion follows from REF . |
math/0007191 | Clearly any real root MATH in MATH must be indecomposable since MATH. Conversely, by CITE any indecomposable element is in MATH. If MATH is not MATH then MATH is a root with some positive component, hence a positive root. But MATH, contradicting indecomposability. |
math/0007191 | Say MATH are elements of MATH with MATH. By induction it suffices to find a subset MATH of MATH with MATH. We prove this by another induction: if MATH is a subset for which the sum is a root MATH, we show how to enlarge MATH so that the sum is a root MATH. Now MATH and MATH, so MATH. Thus MATH for some MATH. Clearly MATH, so MATH so MATH is a root. Moreover MATH, for otherwise MATH is a root (since MATH) with some positive component, hence a positive root. But then MATH, a sum of elements of MATH, which contradicts the fact that MATH. |
math/0007191 | Since MATH, the trace function MATH for a path which starts and ends at MATH involves the trace of a MATH matrix, which is just the unique entry of the matrix. The assertion thus follows from CITE. |
math/0007191 | By REF we know that MATH is surjective. Thus it suffices to prove that it is a closed embedding, that is, that the map on functions MATH is surjective. By REF the ring MATH is generated by the trace functions MATH for MATH a path in MATH starting and ending at REF. Since the ring is finitely generated, a finite number of paths MATH suffices. For MATH let MATH be the projection from the product of MATH copies of MATH onto the MATH-th factor. Thus the coordinate ring of this product is generated by elements MATH. There is a surjective map from the polynomial ring MATH to MATH sending MATH to MATH. This induces a surjective map MATH . Now by REF below, MATH is generated by the power sums MATH . Thus MATH is generated by the elements MATH . Since MATH we have MATH for any paths MATH which start and end at REF, so MATH where MATH is the path MATH. Thus MATH . This shows that MATH is the image under MATH of the trace function MATH for MATH. Thus the image of MATH contains a set of generators, so MATH is surjective, as required. |
math/0007191 | By CITE the ring of invariants is generated by polarizations of the elementary symmetric polynomials, so by elements of the form MATH where the sum is over all distinct MATH in the range MATH to MATH. Now the elementary symmetric polynomials can be expressed as polynomials in the power sums by NAME 's formulae, and on polarizing this expresses MATH as a polynomial in the MATH. For example polarizing the formula MATH with respect to the sets of variables MATH, MATH and MATH gives MATH and all sums on the right hand side are of the form MATH for suitable MATH. |
math/0007191 | We consider the pairs MATH which can be obtained from MATH by a sequence of such admissible reflections. Always MATH is positive, since it is in MATH by CITE. Thus we can choose a pair MATH with MATH minimal. Clearly we have MATH. For a contradiction, suppose that MATH. Since MATH is unchanged by these reflections, we have MATH. Also, for each vertex MATH we have MATH, for either there is a loop at MATH, in which case it is automatic, or MATH, in which case it follows from CITE, or there is an admissible reflection at MATH, and it follows from the minimality of MATH. We deduce that MATH for MATH, and MATH. Suppose first that MATH. Then MATH and all terms in the second sum are MATH. Thus exactly one of the terms is MATH, and all others are zero. That is, there is a vertex MATH in MATH with MATH and MATH, and MATH for all vertices MATH in MATH. This is impossible by CITE. Thus MATH. It follows that MATH, so since MATH is extended NAME we have MATH for some MATH. Now the decomposition MATH is easily seen to satisfy MATH . We have seen that MATH. Also MATH since MATH, so that MATH. This contradicts the fact that MATH. Thus MATH, as required. |
math/0007191 | Some sequence of admissible reflections at vertices MATH sends MATH to MATH. If MATH and MATH then by CITE the reflections send it to a positive root MATH, still with MATH. Thus MATH, and so MATH which is one of the inequalities. The other one is obtained by replacing MATH with MATH. |
math/0007191 | Apply CITE to MATH, and then consider the sequence of reflections as reflections for MATH. Of course non-admissible reflections can be omitted, for if MATH and MATH then MATH. |
math/0007191 | Any indecomposable element of MATH which vanishes at MATH is MATH, so it suffices to prove that if MATH is a vector with MATH and MATH, then either MATH for all MATH with MATH, or MATH for all MATH with MATH. Since MATH is an extending vertex for MATH we have MATH, and so by replacing MATH by MATH if necessary, we may assume that MATH. Now the equality MATH implies that MATH. By the dominance condition it follows that MATH for any vertex MATH with MATH. |
math/0007191 | First suppose that MATH. If MATH then the expression for MATH as a sum of coordinate vectors is a non-trivial decomposition into elements of MATH. Since MATH by REF , this contradicts the fact that MATH. Thus we may suppose that MATH. Replacing MATH by the pair MATH of REF , we may assume that the restriction of MATH to MATH is dominant. Observe that the reflections involved, at vertices in MATH, can change MATH, but they do not affect the dimension vectors MATH and MATH. The standing hypotheses on MATH still hold, as do the hypotheses of the proposition by CITE. Now the restriction of MATH to MATH is non-zero, for otherwise the condition that MATH implies that MATH, and then since MATH we have MATH. Thus MATH. By REF there is some integer MATH with MATH for all MATH with MATH. Let MATH. Of course MATH and for any vertex MATH with MATH we have MATH. Suppose that MATH is nonzero. Consider the restriction of MATH to a connected component of the quiver obtained from MATH by deleting all vertices MATH with MATH. It is actually a subquiver of MATH, so NAME. If MATH is a positive root for this connected component, then MATH, and MATH so REF implies that MATH. But this is impossible by REF below. Thus MATH, so MATH. Now since MATH we have MATH. |
math/0007191 | We cannot have MATH for all MATH, for otherwise MATH, so MATH since MATH is NAME. Embed MATH in an extended NAME quiver of the same type by adding an extending vertex MATH, and consider MATH as a dimension vector for this quiver. Let MATH be the minimal positive imaginary root. Since MATH is a root for MATH we have MATH. Now it is equal to MATH, and all terms in the sum are MATH, but not all are zero. Thus exactly one term is nonzero, say for MATH, and it is equal to MATH. This implies that MATH is an extending vertex, and MATH. Thus the vector MATH and the extending vertices MATH and MATH contradict CITE. |
math/0007191 | We can identify MATH with a MATH-stable closed subvariety of MATH (defined by the vanishing of all arrows with one end in MATH and the other end in MATH). The inclusion thus induces a closed embedding MATH and by the assumption on composition factors this is a bijection. |
math/0007191 | The statement does not depend on the orientation of the arrows in MATH, so we may suppose that the arrow connecting MATH and MATH is MATH. By CITE the condition that MATH is that there is a MATH-module of dimension MATH. Similarly for the other two conditions. Now if the module is given by an element MATH, then for any vertex MATH we have MATH . Taking the trace and summing over all MATH, all but one term cancels, leaving MATH. Since this is a MATH matrix we have MATH. It follows that the components of MATH corresponding to arrows with head and tail in MATH define a MATH-module of dimension MATH, and the remaining components of MATH define a MATH-module of dimension MATH. Clearly two such modules can also be used to construct a MATH-module of dimension MATH. |
math/0007191 | Because of the existence of a module of dimension MATH we have MATH, hence also MATH. For a contradiction, suppose there is a composition factor whose dimension MATH does not have support in MATH or MATH. Then MATH. Since the dimension vector MATH of any other composition factor must have support in MATH or MATH, and has MATH, we deduce that MATH. By REF we have MATH, and by symmetry also MATH. But clearly MATH, so that MATH, contradicting the fact that MATH. |
math/0007191 | Because of the existence of a module of dimension MATH, we have MATH. Since the field MATH has characteristic zero, we deduce that MATH. For a contradiction, suppose there is a composition factor whose dimension MATH does not have support in MATH or MATH. Then MATH. Since the dimension vector MATH of any other composition factor must have MATH, it has support in MATH or MATH, and since it has MATH, we deduce that MATH. Also MATH. Let MATH be the quiver obtained from MATH as in REF , and let MATH be the corresponding vector. Since MATH has support in MATH it can be considered as an element of MATH. By REF we have MATH and MATH. Now by REF is nonzero, so REF implies that MATH. By CITE this implies that there are nonzero MATH with MATH and MATH. Without loss of generality, MATH and MATH. Considered as a dimension vector for MATH we clearly have MATH. Also, REF applies to the dimension vector MATH, and shows that it belongs to MATH. Since also MATH we have MATH by CITE. A contradiction. |
math/0007191 | We prove this for all MATH, MATH and MATH by induction on the maximum possible number of terms in an expression for MATH as a sum of elements of MATH. If MATH then the assertions are vacuous, so assume that MATH. By CITE and REF we can always apply a sequence of admissible reflections to the pair MATH. Let MATH be the set of CITE. If MATH then by applying a sequence of admissible reflections to MATH we may assume that there is a loopfree vertex MATH with MATH and MATH. Clearly in any decomposition of MATH as a sum of elements of MATH one of the terms, say MATH, has MATH. But by CITE this implies that MATH. Now MATH, and by the inductive hypothesis the assertions hold for MATH. If the decomposition is MATH then clearly MATH is a suitable decomposition of MATH. Moreover, if we have MATH then since MATH is just a point, any term MATH if it occurs, can be removed, and replaced by MATH without changing the product. Thus by REF we obtain the required expression for MATH. Thus we are reduced to the case when MATH. By applying a sequence of admissible reflections to the pair MATH, and then passing to the support quiver of MATH, we may assume that one of REF or REF holds. We deal with each of these in turn. CASE: Here MATH is extended NAME, MATH, and MATH for some MATH. By REF the decomposition MATH has the required properties. CASE: Here MATH decomposes as in REF . In the notation of REF we write MATH. Since MATH there is a MATH-module of dimension MATH. Since the dimension vector of any composition factor has support in MATH or MATH we deduce that MATH and MATH are in MATH. By the inductive hypothesis the conclusions of the theorem hold for MATH and MATH. Adding together the decompositions of MATH and MATH we obtain a decomposition of MATH. Obviously, since MATH and MATH have disjoint support, no summand occurs in both parts. The result thus follows from REF . CASE: Here MATH decomposes as in REF . We write MATH. Again MATH and MATH are in MATH and by the inductive hypothesis the conclusions of the theorem hold for them. This gives a decomposition of MATH which has the required properties by REF . |
math/0007191 | CASE: If MATH is a real root in MATH, then MATH is a point by CITE. CASE: If MATH is an isotropic imaginary root in MATH, then it is indivisible, for if MATH then MATH is a root, it has MATH since the base field MATH has characteristic zero, and the decomposition MATH has MATH, contrary to the definition of MATH. By CITE, some sequence of admissible reflections sends the pair MATH to a pair MATH with MATH in the fundamental region. Since it is isotropic imaginary we have MATH for any vertex MATH in the support of MATH. By CITE this implies that the support quiver MATH of MATH is extended NAME and NAME, its minimal positive imaginary root. Finally MATH by REF , and this is a deformation of the Kleinian singularity of type MATH by NAME 's work CITE. See for example CITE. CASE: Suppose that MATH is a non-isotropic imaginary root in MATH and MATH. If MATH is the set of CITE, then CITE implies that MATH, and hence also MATH. Now in CITE, REF cannot occur since MATH is non-isotropic, and REF cannot occur since all components of MATH are divisible by MATH. Thus MATH. |
math/0007192 | The first cohomology group of the product of two abelian varieties is the direct sum of the first cohomology groups of the individual abelian varieties. Moreover, the NAME cycles on the individual varieties pull-back to give NAME cycles on the product. Thus it follows that the NAME group of the product contains the product of the NAME groups. The result now follows from the above constructions. |
math/0007192 | Let MATH be the standard representation of MATH. Let MATH be its decomposition into isotypical components as a representation of MATH. Let MATH be the decomposition of MATH as a representation of MATH. Then each MATH is either MATH or MATH or MATH. The result follows by dimension counting. The lemma also follows from the fact that the quotient MATH of the NAME algebras is an irreducible module over MATH. |
math/0007192 | The first cohomology group of MATH decomposes as a direct sum of two (polarised) sub-Hodge structures. It follows that MATH is the product of two abelian subvarieties. Hence we have the result. |
math/0007192 | We begin with the case where the base curve has genus zero. In this case the NAME varieties are the NAME of the corresponding hyperelliptic double cover. The result is classical for elliptic curves which can be considered as the NAME varieties associated with double covers of smooth rational curves branched at REF points. By induction, let us assume that the result is known for hyperelliptic NAME of genus less than MATH. REF then shows that the NAME group of a general hyperelliptic curve of genus MATH contains MATH. By the above results we see that thus NAME group must be either MATH or MATH. In the latter case, the NAME group of the curve would have rank at least two but by NAME 's result we know that this is not true for the general hyperelliptic curve. Hence we see that the NAME group of a general hyperelliptic curve must be MATH where MATH is the genus of the curve. Now let us consider the case where the cover is unramified. Then we may assume that the base curve of the double cover has genus MATH at least REF (else the NAME variety is just a point). In this case the NAME variety has dimension MATH. By REF we know that the NAME group contains the NAME group of any curve of genus MATH. In particular, it contains the NAME group of any hyperelliptic curve of genus MATH and hence by the previous paragraph it contains (and is thus equal to) MATH. Now assume that the base curve of the double cover has genus MATH at least REF and the cover is ramified. We argue by induction on the genus of the base curve. We can begin the induction since we already know the result for the hyperelliptic curves. Let us assume that the result is known for base curves of genus less than MATH. By REF we know that the NAME group contains the product of the NAME group of an elliptic curve with the NAME group of the NAME variety of a the double of a curve of genus MATH; in other words it contains MATH by induction. Now as argued above, the three lemmas above imply that the NAME group must be MATH. |
math/0007193 | Write MATH . By induction on MATH we have that MATH for MATH, from which it is clear that MATH for MATH. We write MATH, or MATH . Then MATH so MATH which has positive entries for MATH. |
math/0007193 | We present an outline of the more detailed proof in CITE. Suppose that MATH is a hyperbolic fixed point of MATH. Let MATH be a generator of the cyclic group of matrices fixing MATH. MATH is determined up to inverses and MATH . We show in CITE that if MATH is a principal ideal domain, there is a uniquely determined positive number MATH with MATH. Then MATH is primitive and the unique primitive MATH-BQF for MATH is MATH . If MATH fails to be a principal ideal domain we put MATH. |
math/0007193 | Suppose, by way of contradiction, that MATH but MATH. By the first relation REF we have MATH, then by the second relation REF we have MATH for some MATH, MATH. Repeating this, we produce a sequence of poles MATH, none of which are real, with MATH and MATH for each MATH. REF and a geometric argument show that MATH for each MATH. But this is impossible, since MATH is a finite set. |
math/0007193 | Suppose that MATH. As in the proof of REF , MATH for some MATH, MATH. By REF we may take each entry of MATH to be non-negative, which implies that MATH. Repeating this process gives a sequence of poles MATH, with MATH and MATH for each MATH. Since MATH is finite we must have that MATH for some MATH. Thus we have a finite cycle of positive poles, MATH. Reversing REF we have that MATH, since MATH and MATH for each MATH. Reversing REF we have that MATH, since MATH and MATH. Hence MATH is a finite cyclic orbit of MATH, so MATH, where MATH is some MATH-equivalence class of MATH-BQFs. Hence MATH is hyperbolic and contains MATH, while MATH and MATH are both hyperbolic and simple. Since MATH is a positive pole, each element of MATH must also be positive pole, so MATH. This, with the first relation REF, implies that every element of MATH is a negative pole, so MATH. On the other hand, if MATH, then MATH, where MATH is a product of matrices, each one equal to MATH or to MATH, MATH. Thus MATH and MATH are MATH-equivalent numbers, hence MATH and MATH are MATH-equivalent BQFs, so MATH. If MATH, then MATH. If MATH, then MATH, and MATH. |
math/0007193 | A routine exercise showing containment in both directions establishes the Lemma. |
math/0007193 | Fix MATH and put MATH and MATH. For REF we apply MATH to the second relation REF, use the first relation REF and rearrange to get MATH . We claim that the right hand side of this expression approaches MATH as MATH. To prove our claim it suffices to show that for MATH, MATH . Write MATH, so MATH and MATH . Now MATH as MATH, since MATH for MATH. We calculate that MATH for MATH, and MATH is parabolic, so MATH cannot be a pole of MATH, that is, MATH is bounded near MATH. Thus MATH as MATH, and the claim holds. If MATH had a pole at MATH of order MATH, then MATH where MATH and MATH. Then MATH where ``MATH" denotes a polynomial of degree less than MATH. Thus as MATH, MATH which contradicts the claim and establishes REF . REF follow immediately from REF and the first relation REF in the form MATH . |
math/0007193 | Fix MATH and put MATH. Note that if MATH has a pole at MATH and MATH is a linear fractional transformation, then MATH has a pole at MATH. In fact, MATH . By REF with MATH, we have MATH . In a similar way, we use REF with MATH, MATH an integer, to get MATH . Suppose that MATH. Then by the proof of REF , MATH is fixed by an element of MATH of the form MATH, where MATH for each MATH. Applying REF MATH times each, we get MATH and, since MATH, MATH . If MATH, then MATH is fixed by an element of MATH of the form MATH. Thus MATH is fixed by MATH. We apply REF MATH times each in this case as well, and MATH also satisfies REF when MATH. Let MATH be the order of the pole of MATH at MATH and suppose that MATH is normalized so that MATH, where MATH is a rational function with a pole at MATH of order less than MATH. Put MATH so MATH. We calculate MATH where MATH and MATH are rational functions, each with a pole at MATH of order less than MATH and a pole at MATH. We have used partial fractions, the fact that MATH fixes MATH, and the fact that MATH. Then by REF we have MATH where MATH is a rational function with a pole at MATH of order less than MATH. Thus MATH, so either MATH or MATH. But if MATH, then MATH is parabolic, a contradiction. We conclude that MATH. |
math/0007193 | Functions of the given form exist, since MATH has a pole of order MATH at MATH and can be normalized so that the coefficient of MATH is MATH. For uniqueness, suppose that MATH and MATH are functions satisfying the hypotheses of the Lemma, with MATH. Then there exist RPFs MATH and MATH and nonzero constants MATH and MATH such that MATH and MATH. But then MATH is an RPF with a pole at MATH of order less than MATH, a contradiction. |
math/0007193 | Write MATH. Let MATH be an element of MATH which fixes both MATH and MATH. We calculate MATH . We have used the fact that MATH. From this it follows that MATH satisfies REF, as does MATH. By an argument similar to the proof of REF , MATH is a nonzero multiple of MATH, which is in turn a multiple of MATH. Thus MATH is a multiple of MATH. Finally MATH so the proportionality constants are as given in the Lemma. |
math/0007193 | By REF we have that any RPF MATH of weight MATH on MATH has the form MATH where each MATH is a hyperbolic MATH-equivalence class of MATH-BQFs and MATH is given by REF. From REF we see that the coefficients MATH alternate in sign as MATH goes through any cycle MATH. Thus for each irreducible system of poles MATH there is a constant MATH such that MATH for each MATH and MATH for each MATH. As a result any RPF of weight MATH on MATH has the given form. |
math/0007193 | Suppose that MATH is an RPF of weight MATH on MATH with NAME MATH-symmetric irreducible systems of poles. Then MATH for each MATH-equivalence class of MATH-BQFs associated with poles of MATH. Thus the expression for MATH in REF simplifies to MATH where MATH is given by REF and MATH and MATH are all constants. A calculation shows that for any pole MATH, MATH where MATH is the discriminant of MATH. Since MATH is odd, MATH so MATH where MATH and MATH is the discriminant of MATH-BQFs in MATH. NAME proves in CITE that if MATH is odd, then MATH is an RPF of weight MATH on MATH. Thus the first sum in REF is an RPF of weight MATH on MATH. Since MATH and MATH are also RPFs, we must have that MATH is an RPF of weight MATH on MATH. But then we must have MATH for MATH or else we contradict REF . Thus MATH has the form REF. For the converse, assume that MATH has the form REF. Since MATH and MATH are both RPFs of weight MATH on MATH, MATH is also an RPF of weight MATH on MATH. The set of nonzero poles of MATH is MATH which is clearly NAME. |
math/0007194 | Immediately, by definitions and induction on MATH. |
math/0007194 | MATH . Let MATH; choose MATH such that MATH. Since MATH avoids MATH, we see that MATH for all MATH, and since MATH avoids MATH, we get that MATH, where MATH, and so on. MATH . Let MATH, and let MATH; choose MATH such that MATH. Similarly to MATH, MATH, therefore MATH which means MATH for all MATH, where MATH. Hence MATH for all MATH, or equivalently, MATH . Since MATH for all MATH, MATH, and MATH REF, we obtain that MATH for all MATH, and MATH . Hence, by REF , the theorem holds. |
math/0007194 | Let MATH; put MATH. Since MATH avoids MATH, we see that MATH contains MATH, since MATH avoids MATH, we get that MATH, and since MATH avoids MATH, we have two cases: either MATH for MATH, or MATH such that MATH. Now let us consider the two cases: CASE: Let MATH, where MATH, MATH. Similarly to the above, we have two cases for MATH: in the first case MATH for MATH, so there are MATH permutations like MATH. In the second case MATH, so there are MATH permutations, which means MATH. Besides, MATH (see REF), hence MATH. CASE: Let MATH, and let MATH such that MATH. Similarly to the above, we have two cases for MATH: in the first case MATH for MATH, so there are MATH permutations like MATH. In the second case MATH, so there are MATH permutations. Hence MATH. |
math/0007194 | MATH . Immediately, by REF , MATH where MATH, MATH. MATH . Again, by REF , for all MATH hence, this theorem holds. |
math/0007194 | MATH . Let MATH, and let MATH. Since MATH avoids MATH, we see that MATH contains MATH, and since MATH avoids MATH, we get that MATH, where MATH, and so on. MATH . Let MATH, and let MATH; similarly to MATH, MATH. Therefore, for MATH we have MATH if and only if MATH, and for MATH we have MATH if and only if MATH. Hence MATH which means that MATH . Let us define MATH for MATH, so MATH . If MATH, then immediately MATH, hence, by REF , this theorem holds. |
math/0007194 | By the proof of REF , MATH which means that MATH . Besides, MATH, and MATH for MATH (see REF). Hence, the corollary is true. |
math/0007194 | MATH . Let MATH; if MATH, then MATH contains either MATH or MATH, which means MATH contains either MATH or MATH, hence MATH or MATH, and so on. MATH . Let MATH; similarly to MATH, MATH or MATH. Let MATH, hence in the above two cases (MATH or MATH) we obtain MATH for all MATH. Besides, MATH, MATH, and MATH for all MATH (see REF). Hence, similarly to REF , the theorem holds. |
math/0007194 | Let MATH; since MATH avoids MATH and MATH we have either MATH or MATH. If MATH, then, since MATH avoids MATH, we see that MATH. Now we consider the two cases: CASE: Let MATH, MATH. Similarly to the above, either MATH, or MATH, so evidently MATH . CASE: Let MATH, MATH. Similarly to the above, either MATH, or MATH, so evidently MATH . |
math/0007194 | Let MATH; by REF , there are two cases: CASE: MATH. So MATH, MATH, and for all MATH . Besides, MATH for all MATH (see REF), where MATH is the MATH-th NAME number. Hence MATH . CASE: MATH. So MATH, MATH, and for all MATH, MATH . Besides, MATH for all MATH (see REF), where MATH is the MATH-th NAME number. Hence MATH . Hence, by the definitions and REF , the theorem holds. |
math/0007194 | MATH . Let MATH; put MATH. Since MATH avoids MATH, we see that MATH contains MATH, since MATH avoids MATH, we see that MATH, and since MATH avoids MATH, we get that MATH. MATH . Let MATH; similarly to MATH, MATH for MATH, hence MATH . |
math/0007194 | MATH . Let MATH and choose MATH such that MATH. Since MATH avoids MATH, we get that MATH for all MATH, since MATH avoids, MATH we see that MATH contains MATH, and since MATH avoids MATH, we get that MATH. MATH . Let MATH; similarly to MATH, MATH for MATH, hence MATH. |
math/0007194 | MATH . Let MATH; put MATH. Since MATH avoids MATH, we get that MATH contains MATH, and since MATH avoids MATH, we see that MATH, and since MATH avoids MATH, we get that MATH for MATH. MATH . Let MATH and MATH; similarly to MATH, MATH for MATH. Hence MATH. |
math/0007194 | MATH . Let MATH; put MATH. Since MATH avoids MATH, we get that MATH contains MATH, since MATH avoids MATH, we see that MATH, and since MATH avoids MATH, we get that MATH for MATH. MATH . Let MATH; similarly to REF , MATH for MATH. Hence MATH. |
math/0007195 | In any NAME loop, each MATH is a pseudo-automorphism with companion MATH, and each MATH is a pseudo-automorphism with companion the commutator MATH REF . In general, if MATH is a companion of the pseudo-automorphism MATH, then MATH is in the nucleus iff MATH is an automorphism. Thus all cubes and commutators are in the nucleus iff all inner mappings are automorphisms. |
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