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math/0011082 | There exists a primitive MATH which is MATH-invariant and whose image under MATH is MATH. Define the surjection MATH by MATH. Clearly, the kernel is exactly MATH, included via the standard inclusion on MATH and where MATH. NAME with MATH gives a MATH-equivariant isomorphism MATH, where in the second factor the map MATH is multiplication by MATH. Taking the quotient by MATH gives an isomorphism MATH as claimed. |
math/0011082 | Let MATH be the compact form of MATH. Viewing a point MATH as corresponding to a pair of elements MATH, where MATH is the maximal torus of MATH, let MATH be the centralizer of MATH in MATH. By CITE, the image of MATH is a singular point of MATH if and only if MATH. Let MATH be the set of roots MATH such that MATH is in the kernel of the homomorphism MATH defined by MATH, and let MATH be the subgroup of MATH generated by reflections in the elements of MATH. A formal argument (compare CITE) shows that there is an exact sequence MATH . Then the image of MATH in MATH is a singular point of MATH if and only if MATH. Suppose that MATH lies over a smooth point of MATH. Then MATH. Thus MATH is the stabilizer in MATH of MATH, and hence by REF is generated by the reflections in MATH which fix MATH. These generators fix MATH and act as reflections on the tangent space to MATH at MATH. Hence by NAME 's theorem MATH is smooth in a neighborhood of the image of MATH. |
math/0011082 | Begin with the central extension MATH . By construction, the bundle MATH lifts to a MATH-bundle. Let MATH be the MATH-bundle induced from the quotient map MATH. The inclusion MATH defines a surjective homomorphism MATH, and hence the lift of MATH is unique. |
math/0011082 | The reduction of the spectral cover MATH is the set of all points MATH such that there exists a lift of MATH to MATH and a weight MATH such that MATH. Define the map MATH by MATH, where MATH is the quotient morphism. Clearly, MATH descends to a morphism MATH, covering the projection to MATH, whose image is set-theoretically a component MATH of MATH. Since all weights in MATH have multiplicity one, MATH is generically reduced and hence is reduced. It will suffice to show that MATH has degree one. Since both MATH and MATH map to MATH with degree MATH, this is clear. |
math/0011082 | First suppose that MATH for some root MATH. Since the difference of two weights of an irreducible representation is a sum of roots, and the roots are primitive elements in the root lattice, MATH. The affine line MATH meets the orbit MATH at least at the two points MATH. Since every element of MATH has the same length (with respect to any NAME invariant inner product) and an affine line can meet a sphere in at most two points, we have proved REF . Conversely, suppose that MATH is not a multiple of any root. Then, for every root MATH, MATH and MATH are distinct hyperplanes in MATH. It follows that, for every root MATH, MATH meets MATH in a proper subset. Since there are only finitely many roots, REF follows. |
math/0011082 | An element MATH fixes a codimension one subset of MATH if and only if MATH fixes a codimension one subset of MATH if and only if MATH is the reflection in a root MATH. In this case, since MATH and MATH is primitive, the fixed set of MATH is MATH. Since by REF follows for an open dense subset of MATH. To prove REF , first suppose that MATH and that MATH vanishes along a component of MATH. Then, in MATH, MATH. Hence MATH is a multiple of MATH. Thus, if MATH are three distinct elements of MATH and MATH all agree along a component of MATH, then MATH and MATH are both multiples of MATH, and so MATH are all contained in an affine line in MATH. But this is impossible since, as MATH are all conjugate via MATH, they all have the same length, and a sphere meets an affine line in at most two points. To see REF , it is clear that MATH is regular since the regular bundle corresponding to MATH fails to be split if and only if a root vanishes on MATH (by CITE, in case MATH is smooth, and by the direct descriptions of REF in case MATH is singular). The vector bundle MATH is thus a direct sum of line bundles. |
math/0011082 | If MATH and MATH vanishes on a nonempty open subset of MATH, then MATH for some real number MATH. By the proof of REF , MATH, proving REF . REF and the first part of REF follow as in the proof of the previous lemma. The second part follows from the fact CITE that MATH is a weight whose length is the same as that of MATH if and only if MATH, and in this case by REF follow from the explicit description of the regular representative MATH given in CITE in the smooth case, and by the discussion of REF in the singular case. |
math/0011082 | Suppose that MATH for every MATH such that MATH. Given MATH which is the image of MATH, suppose that MATH. Then MATH for some MATH and MATH. It follows that MATH, so that MATH, that is, MATH and MATH have the same image MATH. Thus MATH. If MATH, then MATH, and so MATH, that is, MATH. Thus MATH, so that the morphism MATH is a local diffeomorphism near MATH. Since MATH, the morphism MATH must also be a local diffeomorphism onto its image near MATH. This proves REF . Now suppose that the hypotheses of REF hold and let MATH and MATH be as above. By REF or REF , if MATH is generic, then MATH is the unique element of MATH such that MATH. Suppose that MATH. If MATH, then necessarily MATH, and so MATH. Moreover, since MATH, MATH and the map MATH is a local diffeomorphism onto its image at both MATH and MATH. Hence the same is true for MATH. Since MATH, it is easy to see that the two divisors meet transversally at MATH. This proves REF . Finally suppose that we are in REF . In this case MATH and MATH. Thus MATH. It follows as in the previous case that, if MATH, then MATH. Thus MATH. For MATH generic, we can identify the tangent space to MATH at MATH with MATH and the kernel of the differential of the map from MATH to MATH can be identified with MATH. Since MATH, it follows that the differential of MATH at MATH is injective. This proves REF . |
math/0011082 | In the minuscule case, MATH. Note that the pullback of the NAME line bundle on MATH to MATH is simply MATH. Thus, it suffices to show that the normalization morphism MATH and the bundle MATH satisfy the hypotheses of REF . Let MATH be the union in MATH of the images of the divisors MATH, and let MATH be the dense open subset of MATH implicitly defined by REF . Note that, since MATH is minuscule, if the difference MATH of two weights is a multiple of a root MATH, then in fact this multiple is MATH, so that REF does not arise. If MATH does not lie in MATH, then MATH is smooth over MATH and MATH is regular. If MATH lies under a point of MATH satisfying the hypotheses of REF , then MATH satisfies REF (with MATH of rank one) or REF . Otherwise, we are in REF , and MATH lies under a generic point MATH for some root MATH. In this case, there is a connected subgroup MATH of MATH isomorphic to MATH or to MATH, the structure group of MATH reduces to MATH and lifts to MATH, and on the MATH factor it is the principal bundle corresponding to MATH. Under the induced action of MATH on MATH, MATH decomposes into MATH, where the MATH are irreducible, nontrivial representations of MATH, and the MATH are the trivial representation. Since MATH is minuscule, MATH by for example, CITE. Thus, for the action of MATH on MATH, MATH decomposes into MATH, where the MATH and MATH are characters of MATH. It then follows that the vector bundle MATH associated to MATH is of the form MATH. Here the MATH in the first summand are the line bundles of degree zero corresponding to MATH for MATH such that MATH for some MATH, and the MATH in the second summand are the line bundles of degree zero corresponding to MATH for MATH such that MATH if MATH, MATH. In particular, the line bundles corresponding to the various summands are all distinct. Thus, the vector bundle associated to MATH is regular. In this case, MATH fulfills the hypotheses of REF , completing the proof of the theorem. |
math/0011082 | Suppose that the multiplicity of the trivial weight in MATH is MATH. Let MATH and let MATH be the regular vector bundle whose NAME class corresponds to the image of MATH in MATH. We first claim that MATH for every MATH. Of course, if MATH is simply laced, then MATH is the adjoint representation, MATH, and the claim is essentially the definition of a regular bundle. In case MATH is not simply laced, the nonzero weights are the short roots. Let MATH and let MATH be the set of roots which are trivial on MATH. Let MATH be a principal nilpotent element for the reductive NAME algebra MATH which is generated by MATH and the root spaces MATH, MATH. By the recipe for constructing MATH given in CITE in case MATH is smooth, and by the discussion in REF in case MATH is singular, MATH is given by the kernel of MATH acting on MATH, where MATH is the sum of the weight spaces (including those for the trivial weight) which annihilate MATH. The trivial weight contributes a subspace of MATH of dimension MATH and the remaining weight spaces annihilating MATH are one-dimensional subspaces of MATH corresponding to the short roots in MATH. We may assume that we have chosen MATH so that there is a MATH such that MATH completes to a MATH-triple in MATH. Thus MATH completes to some MATH-triple in MATH. Since MATH is principal in MATH, for all MATH, MATH is an even nonzero integer. Thus the kernel of MATH is the weight zero subspace of MATH and so MATH. Since all of the eigenvalues of MATH are even, it follows from the classification of representations of MATH that MATH as well. Since MATH is constant, the direct image MATH is locally free of rank MATH and the induced map MATH is an isomorphism for every MATH. By semistability, the map MATH is injective for every MATH. It follows that the induced homomorphism MATH is injective on every fiber, and so the cokernel MATH is a subbundle as claimed. Clearly the spectral cover corresponding to MATH is just the trivial section of MATH with multiplicity MATH, and hence MATH. As in the proof of REF , we take MATH to be the union in MATH of the images of the divisors MATH, and MATH to be the dense open subset of MATH implicitly defined by REF . To complete the proof, we must analyze the corresponding bundles MATH, where either MATH or MATH. The arguments are very similar to those given in the proof of REF , and we shall be brief. The only essentially new case is when a weight, that is, a short root MATH, vanishes. In this case, for a generic MATH, there is a unique factor of MATH isomorphic to MATH, and the remaining summands of MATH are of the form MATH or MATH, where the MATH are pairwise distinct nontrivial line bundles, together with MATH copies of MATH. In this case, MATH has exactly the same summands, except that the MATH factor is replaced by a MATH and the remaining MATH copies of MATH are no longer present. Hence MATH is regular. |
math/0011082 | Write MATH, where MATH and we may assume that MATH, in fact that MATH. The proof of existence of MATH is by induction on MATH. If MATH is sufficiently negative, then MATH and we can take MATH. Suppose that we know the result for all MATH such that MATH, and let MATH be an element such that MATH. Let MATH. Then MATH . We can write this as MATH, where MATH. Let MATH, where MATH and MATH. Since MATH is MATH-invariant, MATH and thus MATH. We can thus assume that MATH. Thus by induction there exists a MATH such that MATH. Since MATH and MATH lie in the unipotent group MATH, MATH is defined and clearly MATH. This completes the inductive step and proves the existence of MATH. To see the uniqueness, suppose that MATH for MATH. Writing MATH, we see that it suffices to prove: if there exists MATH and MATH such that MATH, then MATH. If MATH, write MATH, where MATH is homogeneous of degree MATH and MATH. As before, we write MATH where MATH. Since MATH and MATH is a vector subspace of MATH, it follows that MATH. Since MATH is invariant under the action of MATH, each homogeneous component of MATH lies in MATH. But as MATH, it follows that MATH. Since MATH is a complement to the image of MATH, MATH, and hence MATH since MATH and MATH. This is a contradiction. Hence MATH. |
math/0011082 | Let MATH be the coordinate ring of MATH. Applying REF to the inclusion of MATH in MATH, which corresponds to an element of MATH, we see that there is a unique morphism MATH with MATH for all MATH. Clearly MATH by uniqueness. Define the morphism MATH by MATH. Symmetrically, there is a unique morphism MATH such that, for all MATH, MATH. If we set MATH, then MATH is a morphism from MATH to MATH. By the uniqueness statements, it follows that MATH for all MATH, and hence that MATH. Symmetrically, MATH as well, so that MATH and MATH are inverse to one another. |
math/0011082 | Clearly MATH is conjugate to MATH, and hence by REF maps to the same point of MATH as MATH. Since there is a unique such point in MATH with this property, namely MATH, it follows that MATH. |
math/0011082 | This is an immediate consequence of the fact that MATH, MATH, and MATH are regular and that the morphisms in question are finite and surjective. |
math/0011082 | This is immediate from REF . |
math/0011082 | Since MATH and MATH are conjugate in MATH (via MATH), it will suffice to prove the same statement with MATH replaced by MATH and MATH replaced by MATH . In this case, MATH if and only if MATH. Fix a basis MATH of MATH such that, for every MATH, MATH. We may assume that, for MATH, MATH. In particular, MATH. In this basis, the element MATH is a lower triangular matrix whose diagonal entries are the weights of MATH, viewed as elements of MATH. Clearly, then, MATH is identified with the set of all MATH which vanish on all weight spaces except for MATH. This space is identified with MATH, where MATH is the inclusion dual to the surjection MATH. Thus, MATH is a free rank one MATH-module with basis MATH. It follows that MATH is a basis for MATH. |
math/0011082 | By the previous proposition, MATH is an eigenvector for the action of MATH on MATH, with eigenvalue MATH, and the corresponding eigenspace is free of rank one as a MATH-module. Since MATH is MATH-invariant, MATH is an eigenvector for MATH with eigenvalue MATH. In particular, if MATH, then MATH for some MATH. Since MATH, MATH. Thus, MATH is the identity. On the other hand, given MATH and MATH, clearly MATH as well, so that MATH. It follows that MATH, so that MATH. Hence MATH is MATH-invariant. |
math/0011082 | We have a natural inclusion MATH. The element MATH lies in MATH, and so this inclusion is an isomorphism. Thus, MATH is free of rank one. Since MATH is faithfully flat over MATH, it follows that MATH is a free rank one MATH-module and that MATH is a generator. |
math/0011082 | If MATH is another slice for MATH, then by REF there is a map MATH with MATH such that MATH for all MATH. Let MATH and MATH, where MATH are the maps given by REF for the slices MATH and MATH, respectively. Then MATH . By uniqueness, it follows that MATH . Now MATH is the exponential of a MATH-invariant map from MATH to MATH, and thus MATH. Hence, MATH. |
math/0011082 | By REF the MATH-module map MATH makes the analogous diagram commute, where MATH is replaced by MATH and MATH by MATH, and the set of all such maps MATH with this property is a free MATH-module with MATH as a generator. Let MATH. Since MATH is MATH-invariant we see that the diagram as defined in the theorem is commutative. It remains to show that MATH is surjective in the quasiminuscule case and an isomorphism in the minuscule case. Let MATH be a cuspidal curve of arithmetic genus one. We consider the MATH-bundle over MATH which is trivial on MATH and which is given by the NAME section MATH under the correspondence of REF . Let MATH be the vector bundle induced from this MATH-bundle by MATH. This vector bundle is also trivialized on MATH and is given by the map MATH. Applying REF and using the fact that every line bundle over MATH is trivial since MATH is isomorphic to an affine space, we see that there is a surjection from MATH to the vector bundle MATH where MATH is the covering projection and MATH is the map induced by the MATH-invariant weight MATH. By REF , MATH pulls back to the trivial line bundle on MATH and hence is given by a map MATH. This map is the fixed identification MATH. Thus, when we use this identification to produce an isomorphism of the coordinate ring of MATH with MATH, MATH becomes the element MATH. Hence, MATH is identified with the element MATH which is the element MATH. In this way we identify MATH with the element MATH viewed as a MATH-valued endomorphism of MATH. Applying MATH produces the MATH-module MATH with the MATH-linear endomorphism given by multiplication by MATH. REF now implies that there is a surjective MATH-linear homomorphism MATH making the following diagram commute: MATH . Its extension to a map MATH is an element of the free module MATH and hence MATH for some MATH. Since MATH is surjective, MATH is invertible. Restricting to MATH, we see that MATH, and, since MATH is surjective, so is MATH. Finally, if MATH is minuscule, then MATH is an isomorphism and hence MATH is an isomorphism as well. |
math/0011082 | Suppose that we can show that there exists some NAME subgroup MATH and a morphism MATH such that, for all MATH, MATH. Then after a further conjugation we may assume that MATH contains MATH. In this case, we can write MATH for some morphism MATH, where MATH and MATH lie in MATH and have the same image in MATH. It follows that there is a MATH such that, for an open dense set of MATH, MATH. Hence we can assume that MATH for all MATH. After further conjugating MATH by a representative for MATH in the normalizer of MATH in MATH, we can then assume that MATH for some morphism MATH as desired. To find the morphism MATH, we first claim that there exists a morphism MATH from MATH to the space MATH of NAME subgroups of MATH, such that, for all MATH, MATH. To see this, fix a NAME subgroup MATH and let MATH be the incidence variety: MATH . Identifying the set of all NAME subgroups of MATH with MATH, MATH is the set of pairs consisting of an element MATH of MATH and a NAME subgroup containing MATH. Clearly MATH is a closed subvariety of MATH. The morphism MATH defines an isomorphism from MATH to MATH, where MATH operates on MATH by right multiplication and on itself by conjugation, with inverse MATH (compare CITE). Let MATH be the inverse image of MATH under the projection MATH. There is a morphism MATH defined by the composition MATH where the morphism MATH is induced by the homomorphism MATH, and the morphism MATH is projection onto the second factor. It is straightforward to verify that the following diagram is commutative: MATH . By REF, this diagram identifies MATH with a simultaneous resolution of the the morphism MATH, in the terminology of CITE. Since the adjoint quotient morphism MATH is smooth along MATH, the above diagram identifies MATH with the fiber product MATH. Using the morphisms MATH and MATH, we can take the fiber product MATH. By the above remarks (all products below are fiber products), MATH . The isomorphism MATH together with the projection MATH identify an element MATH of MATH with a pair MATH, where MATH is the image of MATH and MATH is a NAME subgroup containing MATH. Here MATH is the image of MATH under the morphism MATH, which is the composition MATH and we identify MATH with the variety of all NAME subgroups of MATH. Now suppose that we can lift the morphism MATH to a morphism MATH. It then follows that, for all MATH, MATH as claimed. To see that such a lift is possible, let MATH be the sheaf of morphisms from MATH to MATH, and similarly for MATH and MATH. Then the following is an exact sequence of sheaves of sets in the étale topology: MATH . In particular, we have the following long exact sequence (of pointed sets): MATH . On the other hand, the solvable group MATH has a filtration by normal subgroups, such that the successive quotients are either MATH or MATH. Since MATH is a torus, all line bundles on it are trivial, and since it is affine, all higher coherent sheaf cohomology vanishes. Thus MATH is trivial. It follows that every morphism from MATH to MATH lifts to a morphism from MATH to MATH, as claimed. This concludes the proof. |
math/0011082 | By REF , the module of such extensions is isomorphic to MATH, where MATH is the NAME transform of MATH. Thus it suffices to show that MATH. By REF , MATH is the direct image under MATH of the pullback of a line bundle on MATH, necessarily trivial, tensored with MATH. Applying duality for the finite flat morphism MATH, and using the fact that the relative dualizing sheaf MATH is a line bundle on MATH, and hence is trivial as well, it follows that MATH is the direct image under MATH of MATH. Note that reflection in the root MATH defines an element of MATH of order two normalizing MATH, and hence an involution MATH of MATH such that MATH and MATH. Thus MATH . On the other hand, from the symmetric definition of MATH, it follows that MATH . Thus MATH, and in particular MATH. |
math/0011082 | The bundle MATH is a universal MATH-bundle over MATH. It follows that the NAME homomorphism MATH is an isomorphism. (Of course, this is just a restatement of the result of CITE that the adjoint quotient morphism is smooth at a regular element.) Using relative duality, the isomorphism MATH given by the Killing form, and the fact that the relative dualizing sheaf MATH is trivial, it follows that MATH. Thus MATH, as claimed. |
math/0011082 | The sheaf MATH is the coherent sheaf associated to the MATH-module MATH. By REF (see for example CITE), MATH. But MATH, and the invariant bilinear form defines a MATH-equivariant isomorphism from MATH to MATH. Thus MATH. |
math/0011082 | We begin by verifying that, at a generic point MATH of MATH, the corresponding extension class lies in the image of the conormal line MATH in MATH. For such a MATH, we may assume that MATH and MATH are smooth at MATH. Let MATH be the MATH-bundle over MATH corresponding to MATH. By the proof of REF , MATH is described as follows: let MATH be the subtorus of MATH given by MATH and let MATH be the subgroup of MATH corresponding to the NAME algebra spanned by MATH, MATH, and MATH. Then there is the natural embedding MATH. Let MATH be the MATH-bundle corresponding to MATH. Then there exists a generic MATH-bundle MATH such that, if MATH is the bundle on MATH induced by MATH on MATH, then MATH is the induced MATH-bundle. Note that MATH and that MATH. There is the induced injection MATH. The arguments of REF imply that the corresponding homomorphism on MATH is an isomorphism. In particular, MATH . Clearly, MATH is the image of the tangent space MATH to MATH at MATH, and hence the normal bundle sequence is split at MATH. Another way to give the same splitting on the normal bundle sequence is as follows. The Killing form induces a quadratic form on MATH, which is easily seen degenerate with radical equal to MATH, and this splits the sequence. Dualizing, there is a direct sum decomposition MATH . Under the identification of MATH with MATH, the above is a splitting of the conormal sequence such that the natural morphism MATH is the projection to MATH. The extension MATH restricted to the slice MATH gives an extension MATH and hence defines an extension class MATH. By construction, the part of MATH supported at MATH is isomorphic to MATH and so MATH. Thus up to a scalar we can view MATH as an element of MATH. We claim that, if MATH is defined via the commutative diagram MATH then the corresponding extension MATH is split. There are many ways to see this. For example, it follows from the explicit knowledge of MATH given in the proof of REF . Another way is to use that fact that orthogonal projection from MATH to MATH is equivariant with respect to the action of the subgroup MATH. Thus there is an induced homomorphism of vector bundles from MATH to MATH. In turn, this homomorphism induces a surjection from MATH to MATH which splits the extension. In any case, we see that MATH lies in the kernel of the projection MATH, and hence MATH lies in the image of the conormal line MATH. Thus, at a generic point of MATH, the extension class MATH is equal to a nonzero multiple of MATH at MATH and hence is a nonzero multiple of MATH at MATH. Suppose that MATH is the section of MATH corresponding to the extension MATH. The above shows that there exists a regular function MATH on MATH such that MATH. Fixing a basis for MATH, we can write MATH. To complete the proof of REF , it suffices to show the following: for all MATH, there exists a MATH such that MATH. This will show that MATH is a nowhere vanishing function on MATH and hence a constant, and that MATH is a nowhere vanishing section of MATH. To see this, let MATH lie under MATH, let MATH be the corresponding MATH-bundle, and let MATH be the corresponding extension, where we use the given basis of MATH to trivialize the subbundle MATH of MATH. The fact about MATH that we shall need is that MATH and hence that the coboundary map MATH is injective. Write MATH, where MATH. By REF , we have identified MATH with MATH and MATH with MATH. Under this identification, the MATH component MATH of MATH corresponds to the MATH component of MATH, namely MATH. Thus, if MATH for every MATH, then MATH lies in the subspace MATH of MATH, which is a proper subspace since MATH, that is, MATH is not the unit ideal. This would imply that all of the MATH lie in a proper subspace of MATH, and hence that they have a common kernel. This contradicts the injectivity of MATH. Thus MATH for some MATH, proving that both MATH and MATH are units at MATH. |
math/0011082 | If MATH is not a weight of MATH, then MATH for each MATH such that MATH is semisimple. Thus MATH, and hence it is free of rank MATH. Thus we may assume that MATH is a weight of MATH. In this case, the proof uses the following result of CITE: Suppose that MATH, the multiplicity of the weight MATH in MATH, is positive. Let MATH be a regular element of MATH, let MATH denote the centralizer of MATH, and define MATH . Then MATH, and in particular it is independent of the choice of MATH. For each MATH, we just write MATH for MATH. Then MATH is a MATH-dimensional subbundle MATH of the trivial bundle MATH. Since MATH is a polynomial algebra, the vector bundle MATH corresponds to a free MATH-submodule of MATH, with basis MATH (although the main point is rather that the submodule is projective). Let MATH. Since MATH is independent of MATH, it follows that MATH is a projective, and hence free, submodule of MATH, clearly saturated. As MATH for every MATH, MATH. Finally, if MATH is the image of a regular semisimple element, then it is easy to see that MATH, and since MATH is saturated, we must have equality everywhere. Thus MATH, and in particular MATH is free. |
math/0011083 | For every element MATH of MATH, denote by MATH the image of MATH in MATH. Let MATH, MATH and MATH be elements of MATH such that MATH. By definition, there exists MATH such that MATH. Since MATH satisfies the NAME decomposition property, there are MATH and MATH such that MATH. Therefore, MATH, and MATH for MATH. |
math/0011083 | CASE: Inverse isomorphisms are given as follows: map each ideal MATH of MATH to MATH, and map each ideal MATH of MATH to MATH. CASE: We already know that MATH and MATH are complete lattices. Any ideal MATH of MATH is the supremum of the principal ideals MATH for MATH, and each of these principal ideals is in MATH. This shows that MATH is algebraic. One can argue similarly that MATH is algebraic, or just apply REF . CASE: It follows directly from NAME decomposition that for any two ideals MATH and MATH of MATH, the sum MATH is again an ideal. Hence, finite suprema in MATH are given by sums. It is clear that MATH for all MATH, MATH, MATH, and therefore MATH is distributive. Observe that elements MATH, MATH satisfying MATH generate the same principal ideal of MATH. Hence, there is a map MATH such that MATH for all MATH. Observe that MATH if and only if MATH, if and only if MATH. Hence, MATH is an order embedding. Any ideal MATH has an order-unit, say MATH, and so MATH. Therefore MATH is an order-isomorphism of MATH onto MATH, hence also a semilattice isomorphism. That MATH now follows from REF . CASE: By REF above, we have that MATH is distributive (compare CITE) and MATH. Now MATH by REF . |
math/0011083 | We mimic the proof presented in CITE. Put MATH. (This is just to ensure that we base our indexing on an infinite set, to cover the possibility that MATH might be finite.) Put MATH, the set of all nonempty finite subsets of MATH, ordered under inclusion. We construct inductively objects MATH and homomorphisms MATH for MATH, and transition homomorphisms MATH for MATH in MATH (where MATH denotes strict inclusion). If MATH, where MATH, choose, by REF , a MATH and a homomorphism MATH such that MATH. Now the induction step. Suppose that MATH has at least two elements, and suppose that we have constructed objects MATH for MATH in MATH, homomorphisms MATH for MATH in MATH, and MATH for MATH in MATH, satisfying the following conditions: CASE: If MATH in MATH, then MATH. CASE: If MATH in MATH, then MATH. CASE: If MATH in MATH, then MATH. Put MATH, where MATH denotes the coproduct in MATH. For all MATH in MATH, denote by MATH the canonical homomorphism from MATH to MATH. By the universal property of the coproduct, there exists a unique homomorphism MATH such that MATH for all MATH in MATH. By assumption, there exist an object MATH and homomorphisms MATH and MATH such that MATH and MATH. For all MATH in MATH, define MATH. The construction may be described by the commutative diagram below: We verify points REF to REF listed above for the larger set of all MATH such that MATH. CASE: It suffices to verify that, for MATH, we have MATH, that is, MATH. Since MATH, it suffices to prove that MATH, that is, MATH, which is indeed the case by the induction REF . CASE: It suffices to verify that, for MATH in MATH, we have MATH. This is a direct calculation: MATH . CASE: It suffices to verify that, for MATH in MATH, we have MATH. Let MATH, MATH. Then MATH if and only if MATH. Since MATH, this is equivalent to MATH, that is, MATH. Therefore, we have constructed a direct system MATH and homomorphisms MATH such that MATH and MATH for all MATH in MATH. Further, for each MATH we have, for MATH, that MATH and MATH. Now if MATH, together with limiting maps MATH, is the direct limit of the system MATH in MATH, there exists a unique homomorphism MATH such that MATH for all MATH, and MATH is surjective. To see that MATH is injective, let MATH. Then there exist MATH and MATH, MATH such that MATH for MATH, and MATH. Since MATH is infinite, there exists MATH such that MATH, and MATH by construction, whence MATH. Therefore MATH is an isomorphism. |
math/0011083 | It is obvious that MATH is closed under multiplication by rational scalars. For all MATH, MATH, both MATH and MATH have finite range, thus so does MATH. Furthermore, for all MATH, we have MATH where MATH denotes the range of MATH, and thus MATH. Hence, MATH. Similarly, the product MATH and the element MATH belong to MATH. |
math/0011083 | It is easy to verify that MATH is the positive cone of a structure of partially ordered vector space on MATH: one has to verify that MATH is an additive submonoid of MATH, closed under multiplication by positive rational numbers, and that MATH; this is straightforward. Denote by MATH the pointwise ordering of MATH, and by MATH the ordering of MATH with positive cone MATH. Every element MATH is majorized (for MATH) by some MATH, where MATH and MATH (here MATH denotes the characteristic function of a subset MATH of MATH). Therefore, the support of MATH is equal to MATH, so that MATH. Hence, the partial ordering MATH is directed. It remains to verify interpolation. It is convenient to use REF , that is, to verify that MATH satisfies the refinement property. Thus let MATH, MATH, MATH and MATH be elements of MATH such that MATH. Put MATH. For all MATH, MATH, put (with the notation of REF ) MATH . By REF , MATH belongs to MATH. It is obvious that MATH. To prove that MATH, note that MATH. But MATH is closed under finite intersection, whence MATH. Thus MATH. Finally, it is obvious that MATH and MATH for MATH, MATH. |
math/0011083 | REF are trivial. Assume REF . Since MATH has finite range, it is majorized by some positive integer MATH. Let MATH; we prove that MATH. This is trivial when MATH. If MATH, that is, MATH, then, by assumption, MATH, thus MATH; but MATH, so that MATH. It follows easily that the support of MATH is equal to the support of MATH; whence MATH. |
math/0011083 | By NAME 's Theorem (see CITE), there exists a set MATH such that MATH embeds into MATH. Since MATH has a zero, the embedding can be arranged in such a way that its range includes MATH. Thus, without loss of generality, we may assume that MATH is a sublattice of MATH containing MATH. Put MATH. By REF , MATH is a dimension vector space. By REF , the maximal semilattice quotient of MATH is isomorphic to MATH (via the support map). |
math/0011083 | Denote by MATH the canonical basis of MATH, where MATH, and by MATH the canonical basis of MATH. Let MATH be the unique linear map from MATH to MATH defined by the formula MATH (recall the notation MATH). Let MATH, written as MATH, where all MATH are elements of MATH. Then we have MATH where we put MATH and MATH for all MATH. It is obvious that all MATH belong to MATH. Put MATH; by assumption, MATH belongs to MATH. If MATH, then, since MATH and since MATH is a MATH-semilattice homomorphism, there exists MATH such that MATH, that is, MATH. Since MATH, we have MATH, whence MATH. Conversely, if MATH, then, for all MATH, we have MATH and thus MATH; hence, MATH. This shows that MATH. It follows that MATH is a positive homomorphism, and that, for all MATH, we have MATH . Hence, MATH satisfies the required condition. |
math/0011083 | Let MATH be a countable distributive MATH-semilattice. By NAME 's result, one can write MATH as a countable, increasing union MATH, where all the MATH are finite distributive subsemilattices of MATH, containing MATH. Then each MATH is a distributive lattice. Denote by MATH the set of all (nonzero) join-irreducible elements of MATH, ordered by the restriction of the ordering of MATH, and by MATH the natural isomorphism from MATH onto MATH. Put MATH. By REF , there exists a positive homomorphism MATH such that MATH. The information can be partly visualized in the following commutative diagram: MATH . Consider the direct system MATH of partially ordered MATH-vector spaces whose objects are the MATH, for MATH, and whose morphisms are the maps MATH, for MATH. By REF , if MATH denotes the direct limit of MATH, then MATH is isomorphic to the direct limit of the MATH with the inclusion maps, that is, to MATH. Since all the MATH are dimension vector spaces (by REF ), so is MATH. It is clear that MATH is countable. Finally, suppose that MATH is bounded. Then MATH has a largest element, call it MATH. Let MATH be an element whose MATH-class is MATH. Hence, all elements MATH satisfy MATH, and so MATH is an order-unit of the monoid MATH. Since MATH is directed, MATH must also be an order-unit for MATH. |
math/0011083 | It suffices to prove the lemma for MATH. Since MATH is distributive, there are MATH and MATH such that MATH (for all MATH). Since MATH, there are, further, MATH and MATH such that MATH. Therefore MATH for all MATH, and MATH . |
math/0011083 | This is an immediate consequence of CITE. However, we present here a self-contained proof. By REF , for all MATH, there exists MATH such that MATH and MATH for all MATH. Then MATH satisfies the required conditions. |
math/0011083 | Let MATH be a homomorphism from MATH to MATH. We consider first the case where there exists MATH such that MATH is generated by MATH. Therefore, MATH . Let MATH list all the pairs MATH of elements of MATH such that MATH, and let MATH list all elements MATH of MATH such that MATH. For all MATH and all MATH, we have MATH, and thus MATH. Therefore, by REF , there exists MATH such that MATH . Let MATH list all elements MATH of MATH such that MATH. Then MATH for all MATH and MATH. Set MATH (MATH is defined as being equal to MATH if MATH), and observe that MATH . It follows from REF that MATH . By REF , MATH extends to a well-defined map MATH such that MATH and MATH for all MATH. Since MATH is a homomorphism, it follows easily that MATH is a homomorphism. In the general case, there exists a finite chain of subsemilattices MATH such that each MATH is generated by MATH for some MATH. Thus, we conclude by an easy induction argument. |
math/0011083 | If MATH is a finite semilattice, let MATH be the powerset semilattice of MATH, and embed MATH into MATH via the map MATH defined by the rule MATH . |
math/0011083 | Put MATH, and denote by MATH the quotient map. There exists a unique homomorphism MATH such that MATH. By REF , there exists an embedding MATH from MATH into some finite Boolean semilattice MATH. By REF , there exists a homomorphism MATH such that MATH. Put MATH. The situation can be described by the following commutative diagram: We obtain the following: MATH . Furthermore, MATH is one-to-one, and thus MATH. |
math/0011083 | We consider the first-order language consisting of one binary operation symbol MATH, the variety MATH of semilattices, and the subclass MATH of finite Boolean semilattices. Since the class of finite (not necessarily Boolean) semilattices is closed under finite coproducts (because every finitely generated semilattice is finite), REF is, by REF , satisfied. Since REF is trivially satisfied, the theorem follows. |
math/0011083 | We prove, for example, REF . The proof for REF is similar. Let MATH be a distributive MATH-semilattice. By REF , MATH is a direct limit of a direct system MATH where MATH is a directed set, the MATH are finite Boolean semilattices and the MATH are semilattice homomorphisms, with respect to limiting homomorphisms MATH. Without loss of generality, MATH has a least element, denoted by MATH, and MATH. For all MATH, put MATH and MATH. Then MATH is a finite Boolean semilattice, and MATH maps MATH to MATH for MATH. Furthermore, the least element of MATH is MATH, and MATH for MATH. Thus each MATH restricts to a MATH-preserving semilattice homomorphism MATH. Finally, MATH is the direct limit of the system MATH . |
math/0011083 | This was first proved by NAME in the unital case CITE; his argument easily extends to the non-unital situation, as noted in CITE. |
math/0011083 | That MATH is a conical refinement monoid follows from REF (the fact that MATH does not necessarily have a unit does not affect the proof). By REF , MATH is a distributive semilattice. |
math/0011083 | The semilattice isomorphisms follow from CITE. Again, the fact that MATH does not necessarily have a unit does not affect the proofs. Then by REF , these semilattices are distributive. (Recall that more generally, the semilattice of all compact congruences of any lattice is distributive, see REF). Now MATH. By REF , MATH is distributive, and the proposition is proved. (One can also prove directly that MATH is distributive; this is well known and easy.) |
math/0011083 | REF , and REF. |
math/0011083 | By REF , it suffices to solve the following problem. We fix a countable distributive MATH-semilattice MATH and a field MATH; we must find a locally matricial MATH-algebra MATH of countable dimension such that MATH. In view of REF , this is the same as to arrange for MATH. Further, if MATH is bounded, we must find a unital such MATH. By REF , there exists a countable dimension vector space MATH such that MATH. By REF , there exists a locally matricial MATH-algebra MATH of countable dimension such that MATH; therefore MATH. In addition, if MATH is bounded, then MATH has an order-unit, and thus, by REF , one can choose MATH unital. |
math/0011083 | This proof does not use the results of NAME, or NAME, NAME, NAME and NAME. In fact, it uses nothing more than the countable case of REF . As in the first proof, we fix a countable distributive MATH-semilattice MATH and a field MATH, and we find a locally matricial MATH-algebra MATH of countable dimension such that MATH. According to REF , we may assume that MATH is the direct limit of a sequence MATH in the category of MATH-semilattices. Set MATH (the direct product of MATH copies of MATH), and observe that MATH. More precisely, if MATH are the primitive central idempotents in MATH (that is, the atoms of the finite Boolean algebra of central idempotents of MATH), then MATH are distinct atoms which generate MATH. Hence, if MATH are the distinct atoms in MATH, there exists an isomorphism MATH such that MATH for MATH. Let MATH be the distinct atoms in MATH. There are integers MATH such that MATH for all MATH. Choose a positive integer MATH for each MATH, and set MATH . Let MATH be the block diagonal MATH-algebra homomorphism with multiplicities MATH, that is, each component map MATH is given by MATH where the notation MATH means that MATH appears if MATH but not if MATH. Let MATH be the primitive central idempotents in MATH. Then there exists an isomorphism MATH such that MATH for all MATH, and we observe that MATH. Continuing in the same manner, we obtain a sequence MATH of matricial MATH-algebras and MATH-algebra homomorphisms together with MATH-semilattice isomorphisms MATH such that the following diagram commutes: MATH . Therefore, if MATH is the direct limit of the sequence REF , then we have MATH as desired. It remains to modify the proof for the case that MATH has a greatest element, say MATH. As before, we express MATH as the direct limit of the sequence REF ; in view of REF , we may now assume that the maps MATH preserve greatest elements. Thus MATH for all MATH, where MATH denotes the greatest element of MATH (the sum of all the atoms). Define MATH as before, and note that MATH maps MATH to MATH. Let MATH and the MATH be as before. Since MATH we must have MATH for all MATH, and so we can choose MATH. Now if MATH and MATH are defined as before, MATH is a unital homomorphism. Continuing as before, we can obtain a sequence REF in which all the homomorphisms MATH are unital, and therefore MATH is a unital algebra. |
math/0011083 | By REF , there is a countable distributive MATH-semilattice MATH such that MATH. If MATH is an AF C*-algebra, let MATH denote the semilattice of finitely generated closed ideals of MATH. This semilattice consists precisely of the compact elements of MATH, and hence MATH. Thus, it suffices to find an AF C*-algebra MATH such that MATH. For any AF C*-algebra MATH, the lattice MATH is isomorphic to MATH, see CITE, and, consequently, MATH is isomorphic to MATH. By REF , MATH, and hence MATH. Thus, to find an AF C*-algebra MATH with MATH is the same as to find a MATH with MATH. By REF , there exists a countable dimension vector space MATH such that MATH. By the NAME Theorem and the C*-algebra analogue of NAME 's Lemma, see CITE, there exists an AF C*-algebra MATH such that MATH. Therefore MATH, as desired. In addition, if MATH is bounded, then MATH has an order-unit, and then MATH can be chosen to be unital. |
math/0011083 | As above, we just need to find an AF C*-algebra MATH such that MATH is isomorphic to a given countable distributive MATH-semilattice MATH. The construction in our second proof of NAME 's Theorem yields a sequence REF of matricial MATH-algebras and MATH-algebra homomorphisms such that the direct limit of MATH of REF is isomorphic to MATH. Each MATH can be viewed as a finite-dimensional C*-algebra. Observe that the block diagonal maps MATH are *-homomorphisms. Hence, the C*-algebra direct limit of the sequence REF is an AF C*-algebra, say MATH. Since the functor MATH commutes with C*-algebra direct limits, we therefore have MATH, as desired. |
math/0011083 | The rule MATH defines an order-preserving map MATH from MATH to MATH. Since MATH is unital, there is a *-homomorphism MATH given by the rule MATH, and the set map MATH induces an order-preserving map MATH from MATH to MATH. Clearly MATH is the identity on MATH. Thus, to prove that MATH is a lattice isomorphism, it suffices to show that MATH is surjective. Let MATH, set MATH, and consider the algebraic (that is, uncompleted) tensor products MATH and MATH. Note that MATH is an ideal of MATH such that MATH. Since MATH is simple and unital, its center is a field as well as a C*-algebra, so the center of MATH is MATH. Consequently, MATH (see, for example, CITE). Thus the composition of the inclusion map MATH with the quotient map MATH induces a *-algebra embedding MATH. The composition of MATH with the quotient norm on MATH then defines a C*-norm, call it MATH, on MATH. (It is a norm, rather than just a seminorm, because MATH is injective.) By, for example, CITE, MATH is a cross norm on MATH. Because of our nuclearity assumption, MATH is the unique C*-cross norm on MATH, and so the completion of MATH with respect to MATH yields the C*-tensor product MATH. On the other hand, MATH is an isometry and the image of MATH is dense in MATH. Hence, MATH induces a *-isomorphism of MATH onto MATH. It follows that the kernel of the induced map MATH is precisely MATH, and therefore MATH, as desired. |
math/0011083 | REF . |
math/0011083 | Let MATH be unital; then MATH is a bounded, countable, distributive MATH-semilattice. The lattice MATH is a distributive algebraic lattice whose semilattice of compact elements is isomorphic to MATH and thus is countable. Further, the greatest element of MATH is compact. By REF , there exists a unital AF C*-algebra MATH such that MATH. Hence, MATH . Therefore we conclude from NAME 's classification theorem, see CITE that MATH. |
math/0011084 | This lemma will be proved in full generality in CITE. We give here in this section an ad-hoc proof for the case MATH, where MATH has a state MATH (as in the next lemma). Indeed the next step in the basic construction (as proved in the next lemma) is MATH. Here the algebra MATH is MATH and consequently MATH is MATH. By construction MATH is of the form MATH, for some fixed MATH in MATH. Let MATH be a matrix unit diagonalizing MATH. By cutting by a minimal projection, it follows that MATH takes values into MATH. Thus necessary, there exists real MATH numbers such that MATH is of the form MATH . But MATH has also the property that the conditional expectation from MATH onto MATH maps MATH into the scalars. This is only possible if MATH is an orthonormal basis for MATH, that is, if MATH are the inverses of the eigenvalues of MATH. |
math/0011084 | Indeed if MATH is a matrix unit, which also diagonalizes MATH, then the NAME 's projection is MATH . |
math/0011084 | This is basically proved in CITE,CITE (see also CITE). The only additional care comes from the fact that we are dealing with a state instead of a trace. But since MATH are chosen in the centralizer of the state it follows that MATH is a type MATH factor. The main part of the argument (the NAME of the trace) follows from the fact that the conditional expectation of the algebra MATH onto MATH is the algebra of scalars, which follows from the properties of the NAME basis. |
math/0011084 | We apply NAME 's construction for the inclusion MATH described in REF . Since this is the basic construction for MATH we can apply REF and one obtains an equivalent description of the subfactor MATH sitting in MATH. Since MATH is a NAME basis for the inclusion MATH, the result follows. That the subfactor inclusion is of type MATH is proved in a more general context in CITE. |
math/0011084 | Let MATH be a matrix unit for MATH that diagonalizes MATH. Assume that the unitary implementing the representation is represented as MATH . Note that we use a another indexing for the entries of the unitary than the one used in REF. Since as proved in CITE, MATH is the unitary corresponding to the conjugate representation it follows that the matrix MATH is a unitary too. Thus we have the following unitarity conditions : MATH . The corepresentation for the algebra MATH obtained by conjugation by the unitary MATH is given by MATH . The elements in the algebra MATH have an open and closing parenthesis structure (CITE,CITE). This proves that these elements are fixed points under the coaction, by recursively using (first equality in each of) the relations REF. For example the coaction on an element of the form MATH gives MATH . By applying first REF and then REF it follows that MATH . This proves that the coaction on MATH in REF gives MATH. The only other fixed point are obtained by recurrence. Note that the algebra MATH is in the fixed point algebra of the coaction. More precisely if MATH in the fixed point algebra and MATH is NAME measure, we need to show that MATH . This follows by using the modular properties of the NAME measure and using the relations REF. Once we have shown this and since MATH is obviously in the fixed point algebra the result follows from the definition of the algebra MATH as the minimal algebra containing MATH closed to the operation in the statement of REF . Let MATH be the subspace of MATH consisting of elements of zero trace. The fixed point algebra and NAME 's algebra admit a filtration given by the subspaces MATH corresponding to the number of occurrences of copies of MATH. Moreover the averaging argument (with respect to NAME measure) used in CITE or CITE to establish that the fixed point algebra in the NAME factor is the algebra generated by the NAME projections, allows to reduce the determination of the the fixed point algebra, to the determination of the intersection of the fixed point algebra with the subspaces in the filtration. Therefore, to show the equality, when MATH is MATH, of the algebra in NAME 's construction with the fixed point algebra, it is therefore sufficient to check that the intersections of these two algebras with the space MATH coincide. (This comes also to to the determine the fixed point algebra in spaces of the form MATH, by formally replacing elements in MATH with tensor product sign. Counting dimensions of fixed point algebra could probably give an alternative for the argument bellow). Since we allready have shown the reverse equality, it is sufficient to check that any element in the fixed point algebra intersected with one of the above finite dimensional spaces MATH is one of the elemnts that respects the open and closing paranthesis structure of the reccursive construction of the NAME 's algebra. So assume that for given scalars MATH we have an fixed point element of the form MATH . Then the fixed point condition comes to the following condition that has to hold for all MATH: MATH . By denoting MATH the matrix MATH, and by MATH the matrix MATH and by MATH the matrix MATH, the relation REF comes to the following equation: MATH . Let MATH and MATH be the matrices defined by MATH and MATH. Then REF correspond to MATH where MATH and MATH are considered as the matrices indexed as follows: MATH and MATH. As in CITE, CITE, the relations REF are the only relations defining MATH . Thus if a matrix MATH verifies the relation REF, then MATH is obtained by reccurrence (and linear span) from consecutive applications of the relations REF. To have a reduction we therefore have to use the relation MATH, or MATH . (the other two equtions in REF involve a summation index which is not appropriate for the corresponding sum). This means that the only possibility to have a simplification in REF, by using the defining relations of MATH, is if somewhere in the sum of REF, we have that MATH splits as a sum of elements the form MATH or in the form MATH . The symbol MATH corresponding to omission of the corresponding symbol. Moreover after applying such a simplification as in REF and respectively REF we obtain a sum in MATH involving a corresponding MATH or respectively MATH. Repeated application of the above two procedures outlined in REF to the sum in REF, will give, once we have finished the simplification procedure to the sum reducing REF to scalars, it follows that the fixed point element has exactly the open and closing paranthesis structure in the indices MATH that corresponds to an element in the algebra derived from NAME 's construction. |
math/0011084 | Indeed via the the identification of MATH this corresponds exactly to the subfactor constructed in REF, by reducing by the projection MATH the first component of MATH. |
math/0011084 | Let as in CITE MATH, MATH, MATH, . be the representations of MATH, with MATH the fundamental representation. Then the fundamental unitary for MATH is given by the unitary representing MATH. As MATH is MATH it follows that the coaction of MATH gives the coaction MATH of MATH on MATH, (for classical groups that means that the representation comes from a representation of the quotient). Since the unitary implementing both coactions is the same, it also follows that we have the same fixed point algebra, (in terms of groups representations if we have a representation that factors through the quotient, then we have the same fixed point algebra). |
math/0011084 | Let MATH. By the argument in the proof of REF (and using also REF, as pointed out in CITE) it is sufficient to prove that for any irreducible unitary coaction MATH of MATH, the corresponding spectral subspace (CITE,CITE, CITE) MATH is non-trivial. But, since MATH appears in a tensor product situation in MATH this follows from the fact that the tensor product of the representation MATH with itself contains any other representation (see CITE, CITE). |
math/0011084 | . Indeed the unitary implementing the adjoint corepresentation MATH is the same as the unitary implementing MATH. Since MATH has this property, the same will hold for MATH. |
math/0011084 | By REF is isomorphic to the NAME algebra generated in MATH by MATH and MATH. But MATH is contained in MATH, since MATH acting on MATH maps the matrix coefficients (viewed as elements of MATH) corresponding to a finite dimensional unitary representation MATH of MATH into a linear combination of the matrix coefficients (viewed as elements in MATH) of the representation MATH. We describe this map in classical setting and then explain the modifications needed for a quantum compact group. Indeed in a classical setting, if MATH is a compact group, let MATH be the fundamental unitary viewed as an element inin MATH defined by MATH, MATH, where MATH is the left regular representation of MATH into the unitary group on MATH. Assume MATH is a finite dimensional unitary representation of MATH on a NAME space MATH. Let MATH be the unitary in MATH given by MATH. Then in MATH the following holds (the absorption principle for the left regular representation): MATH . By using the fundamental unitary MATH defined above, in MATH, the last equality gives (by using NAME 's notations) MATH . Let MATH be a matrix unit for MATH and let MATH be the matrix coefficients for MATH in this matrix unit. For convolution operators in MATH the equality REF corresponds, for MATH in MATH to the following equality in MATH. MATH . This gives, using the decomposition of tensor products of irreducible unitary representations of MATH, a complete description of the inclusion MATH since we have described the inclusion MATH into MATH in terms of fusion rules of tensor by the representation MATH (on matrix coefficients). If we want to generalize the above statement for arbitrary quantum groups, we will use the corresponding absorption principle for arbitrary quantum groups described in REF which replaces REF. This formula now holds in MATH (where MATH is the representation in our statement), MATH is now one of the fundamental unitary in MATH replacing the MATH used for the classical case. In this context a convolution operator MATH is replaced in MATH by MATH . This is a generic element in a weakly dense subspace of MATH. (here we use MATH which is the flip of what is usually the fundamental unitary). REF now reads in MATH as follows MATH . By using REF in the paper CITE we get that this is further equal to (and using a matrix unit MATH in MATH with respect to which MATH has components MATH) MATH which is thus the following element in MATH, MATH . This shows that in MATH, the transformation MATH maps MATH onto MATH. |
math/0011084 | Denote MATH and MATH and let MATH be the dual algebras. The representation MATH is the restriction of the adjoint corepresentation induced by MATH. This holds because MATH belongs to MATH, since MATH is generated by the matrix coefficients of MATH. But then if MATH is the projection (in MATH) onto MATH, then MATH since MATH and since MATH acts on MATH as the NAME measure on MATH is the restriction of NAME measure on MATH CITE. |
math/0011084 | Indeed if MATH is not a trace then the modular group of MATH will be non-trivial on off diagonal elements. If the trace is finite this is exactly the content of REF (see also CITE for a different, more recent proof). If the trace is infinite, the arguments in the proof of the theorem mentioned above can obviously be modified to handle this case (by changing the principle of counting the blocks as in the construction of a one parameter group of automorphisms in CITE). |
math/0011084 | From REF it follows that MATH is isomorphic to MATH, where MATH is an infinite dimensional space. The result follows now from REF, since MATH is a faithful representation of MATH (no adjoint is required here). |
math/0011087 | For MATH and MATH, we define MATH by MATH. We fix a basis MATH of MATH and we write MATH for MATH, MATH. By REF, in order to determine MATH we have to assign the (reduced) building data, namely: REF for every nonzero MATH an effective divisor MATH; REF for every MATH a line bundle MATH in such a way that the following relations are satisfied: MATH . For MATH we denote by MATH the homomorphism defined by MATH. We define MATH to be the sum of the MATH such that MATH. Notice that the MATH are disjoint and that MATH is the union of the MATH that appear in MATH. If we write MATH, and we identify again MATH with the integers MATH, then it is not difficult to check that relations REF can be rewritten as: MATH . So REF can be solved uniquely by setting MATH, MATH. The corresponding cover MATH satisfies REF. In addition, MATH is smooth by REF, since MATH is smooth, and it is connected since the set of MATH such that MATH generates MATH. In order to complete the proof we have to check that for every MATH the eigensheaf MATH of MATH on which MATH acts via the character MATH is MATH. By REF, we have MATH. This equation can be rewritten as MATH, and thus MATH in MATH. The equality MATH follows since MATH. |
math/0011087 | If the base field is MATH, then the formula follows easily by topological considerations. We give an algebraic proof, valid for fields of characteristic MATH. Denote by MATH the branch divisor of MATH (which is the union of MATH disjoint nodal curves), and by MATH the ramification divisor. Consider the following exact sequence of sheaves on MATH: MATH where the cokernel MATH is a torsion sheaf supported on MATH. Consider a ramification point MATH and let MATH be the irreducible component of MATH containing MATH. The subgroup MATH consisting of the elements that induce the identity on MATH is isomorphic to MATH (compare REF). The surface MATH is smooth, since the fixed locus of MATH is purely MATH-dimensional, and MATH factorizes as MATH. Let MATH and MATH. The map MATH is étale in a neighbourhood of MATH, and thus MATH is an isomorphism locally near MATH. It follows that the inclusion MATH is an isomorphism locally around MATH. There exists an open neighbourhood MATH of MATH in MATH such that MATH is defined in MATH by the equation MATH, where MATH is a local equation for MATH and MATH is the affine coordinate in MATH. Notice that MATH is a local equation for MATH around MATH. Let MATH be a function on MATH such that MATH are local parameters on MATH around MATH. Then the map MATH of sequence REF can be written locally as MATH. It follows that the cokernel MATH is naturally isomorphic to the conormal sheaf of MATH, MATH. A standard computation with NAME classes gives: MATH. |
math/0011087 | We have MATH, since MATH is unramified in codimension MATH and MATH is normal, and therefore MATH. The formula for MATH follows immediately. Since MATH is a birational invariant, it is enough to compute it for MATH. Then the formula for MATH follows from REF and NAME 's formula. If MATH, then we have MATH (MATH is separable since the characteristic of MATH is MATH). So assume that MATH and denote by MATH the minimal model of MATH and MATH. Then MATH acts biregularly on MATH. We denote by MATH the quotient map. The surface MATH has canonical singularities and it is birational to MATH and MATH. Denote by MATH the minimal resolution. We have MATH and thus MATH and MATH are nef. So MATH is minimal and, in addition, MATH iff MATH and MATH iff MATH. This remark shows that MATH. |
math/0011087 | Consider the cover MATH of REF associated to MATH and the corresponding cover of MATH, MATH. By REF , MATH is ruled and thus MATH. The result follows by using the formula for MATH of REF . |
math/0011087 | Let MATH be the cover of REF and let MATH be the corresponding cover of MATH. By REF , the surface MATH is ruled and has irregularity MATH. Denote by MATH the NAME pencil. By the canonicity of the NAME map, the group MATH preserves the divisor class of a fibre. Consider the canonical homomorphism MATH: if it is not injective, then there exists MATH that maps each fibre of MATH to itself. Hence a general fibre, being isomorphic to MATH, has MATH fixed points of MATH and the ramification locus for the action of MATH has components of dimension MATH, a contradiction since the MATH-cover is branched precisely over the singularities of MATH. Thus we have a commutative diagram: MATH where MATH is a MATH-cover. The general fibre of MATH is MATH, since it is isomorphic to the general fibre of MATH. Since the genus of MATH is equal to MATH, by the NAME formula the branch locus of MATH consists of MATH points (the inverse image of a branch point consists of MATH simple ramification points). The cover MATH is obtained from MATH by base change and normalization, thus the fibres of MATH over the branch points MATH of MATH are of the form MATH, MATH, and MATH contains all the nodes of MATH. We claim that each double fibre contains at least one node. Indeed, otherwise MATH would be contained in the smooth part of MATH and so it would be a NAME divisor with MATH, MATH, a contradiction to the adjunction formula. Set MATH. Then for every MATH, one can write MATH and it follows that for every choice of MATH the divisor MATH is divisible by MATH in MATH, namely it corresponds to a vector of MATH. Since the weights of MATH are all divisible by MATH, it follows easily that each MATH contains precisely MATH nodes of MATH. So it is possible to relabel the MATH in such a way that MATH for MATH, and that MATH is divisible by MATH in MATH for every choice of MATH. This shows that MATH is essentially isomorphic to the code MATH. |
math/0011087 | The group MATH is free abelian of rank MATH. The intersection form on MATH extends to a nondegenerate bilinear form of signature MATH on MATH. The subspace of MATH spanned by the classes of the MATH has dimension MATH and the intersection form is negative definite there, thus we get MATH. We start by proving REF . As before, we let MATH be the number of nodal curves that appear in the code MATH. Recall that the dimension MATH of MATH is MATH. So, for MATH, we have MATH and thus MATH by REF . Assuming then that MATH, we can apply REF . Thus MATH is even, say MATH, and there exists a morphism MATH such that the general fibre of MATH is MATH and MATH has MATH double fibres, occurring at points MATH of MATH. Each double fibre contains precisely MATH nodes of MATH, and the code MATH is MATH. So we have MATH, namely MATH. It follows that MATH is even and MATH. In particular, MATH, that is, all the MATH appear in MATH. Set MATH, denote by MATH the cohomology class on MATH of a fibre of MATH and let MATH . A basis of MATH is given by MATH and the classes of MATH, since these are independent classes and MATH. On the other hand, it is well known that, if one removes a component from each reducible fibre of MATH, then MATH and the classes of the remaining components of the reducible fibres are independent. It follows that the MATH, MATH are the only reducible fibres of MATH. As in the proof of REF , it is possible to relabel the MATH in such a way that for each MATH one has MATH , with MATH irreducible and such that MATH. From MATH, we get MATH, MATH, and thus MATH, namely MATH is a MATH-curve. The curve MATH has nonempty intersection with both MATH and MATH, since MATH is connected. So the equality MATH gives: MATH . Blowing down MATH one obtains a smooth surface ruled over MATH with precisely MATH reducible rulings, each consisting of two MATH-curves intersecting transversely. Blowing down a MATH-curve of each ruling, we obtain a ruled surface MATH. So MATH is the standard example. In order to complete the proof of REF , we have to describe the cases MATH. In addition we may assume MATH, since for MATH (and MATH) one can apply the argument above to show that MATH is the standard example. Since MATH all the elements of MATH have weight MATH and it is easy to check that the only (numerical) possibilities for the pair MATH are: MATH, MATH, MATH and MATH. One has MATH in all cases but the last one, where MATH. Consider the first three cases. Let MATH be the NAME cover considered in REF and MATH the corresponding cover of MATH. By REF , MATH is a surface satisfying MATH, MATH, MATH. So MATH is rational and MATH implies that MATH for some MATH. Denote by MATH the trace of MATH on the MATH-adic cohomology MATH. Since the class in MATH of the canonical bundle of MATH is MATH-invariant, MATH is either equal to MATH or MATH. Applying the (MATH-adic) NAME fixed point formula (see REF , compare the next section for the analogous statement for the complex cohomology) we see that MATH is impossible and hence MATH acts identically on MATH. In particular, given the ruling (or a ruling if MATH) MATH the action of the NAME group MATH of MATH descends to an action on MATH and there is an induced fibration MATH. The same argument as in the proof of REF shows that the action of MATH on MATH is faithful. Thus each element MATH of MATH fixes precisely two fibres of MATH, each containing two fixed points of MATH. Since MATH does not contain a subgroup isomorphic to MATH, we can rule out immediately the case MATH. In the remaining two cases, the cover MATH is branched over MATH points, and over each of these points MATH has an irreducible double fibre containing MATH nodes. It follows easily that MATH is the standard example. Finally consider the case MATH, MATH (the code is essentially isomorphic to the NAME code defined by the root lattice of type MATH). By REF the MATH-cover MATH is a smooth ruled surface with invariants MATH, MATH. Thus MATH is rational. The preimage of the nodal curve of MATH not appearing in MATH is a set MATH of disjoint nodal curves on which MATH acts transitively. The code MATH associated to MATH is acted on by MATH, and therefore all the nodes appear in MATH, namely MATH has MATH. Thus MATH is a standard example with MATH. If there is only one pencil with rational fibres MATH such that the MATH are contracted by MATH, then one argues as in case MATH and obtains a contradiction by showing the existence of a MATH-cover MATH. So assume that there are two pencils with rational fibres MATH, MATH such that the MATH are contracted both by MATH and MATH. Denote by MATH, MATH, the class in MATH of a smooth fibre of MATH. Considering the intersection form, one sees immediately that the classes of MATH, MATH, MATH are a basis of MATH. Consider a nonzero MATH. The surface MATH is a rational surface with MATH singular points of type MATH, that are the images of the fixed points of MATH on MATH. By the standard double cover formulas: MATH and so MATH. Denote by MATH the trace of MATH on the MATH-adic cohomology MATH. Applying again the NAME fixed point formula we get MATH. The action of MATH on MATH preserves the subspace MATH generated by the fundamental classes of the divisors MATH, and thus it preserves also its orthogonal subspace, which is spanned by the classes of MATH. The trace of MATH on MATH is zero. It follows that MATH is the identity on MATH, namely every MATH preserves both pencils. Thus we can apply again the argument above to one of the pencils and the proof of REF is complete. Finally we prove REF . Assume that MATH. The code MATH has length MATH, dimension MATH and all the weights divisible by MATH. Thus if MATH, then MATH by REF and one can argue as in REF and show that MATH is the surface constructed in the standard example and MATH is essentially isomorphic to MATH, with MATH. In particular, MATH, contradicting MATH. So assume MATH. If MATH, then MATH and so MATH is the minimal ruled surface MATH. If MATH, the only numerical possibility is MATH, MATH. Let MATH be a surface corresponding to this possibility. We have MATH. Up to a permutation, we may assume that MATH is an even set. The corresponding double cover MATH is a smooth rational surface (same proof as REF ), with MATH. The inverse images of MATH are MATH curves, while the inverse images of MATH are three pairs of disjoint nodal curves. Blowing down the MATH-curves, one obtains a rational surface MATH with MATH and containing MATH disjoint nodal curves. This is impossible, and the proof is complete. |
math/0011087 | Assume first that MATH is minimal. In this case we can apply NAME 's formula (CITE, section MATH): MATH, and REF follows immediately using MATH and NAME 's formula. Now assume that MATH is not minimal and let MATH be the minimal model of MATH. We use induction on MATH. Let MATH be an irreducible MATH-curve and let MATH be the surface obtained by blowing down MATH. If MATH does not intersect any of the MATH, then MATH contains MATH disjoint nodal curves and induction gives: MATH. So assume, say, MATH. Then the image MATH of MATH in MATH is an irreducible curve such that MATH, MATH. Now necessarily MATH. In fact suppose that MATH. Then MATH and therefore the image of MATH in the minimal model MATH of MATH is a curve MATH. Since MATH and MATH is nef because MATH, we have a contradiction. Therefore MATH is a MATH-curve. In addition, MATH for MATH, since otherwise MATH would contain a pair of irreducible MATH-curves with nonempty intersection, which is impossible again because MATH. Now blowing down MATH we obtain a surface MATH containing a set of MATH disjoint irreducible nodal curves. Using induction again, we have MATH and the proof is complete. |
math/0011087 | The first fixed point formula gives MATH . Together with the second formula we obtain MATH . For the last part notice that we have MATH . Since by the topological fixed point formula MATH, one has MATH . Now MATH implies MATH and we obtain MATH . |
math/0011087 | Assume otherwise. Since MATH, we have MATH. REF gives MATH. Since the canonical class is invariant for MATH, we have MATH for some MATH. Then MATH yields MATH and MATH. Thus REF gives MATH and MATH, that is, MATH. So MATH contains MATH disjoint nodal curves and MATH. This is a contradiction in view of REF . |
math/0011087 | Since MATH, the possible values for the trace MATH are MATH and MATH. The case MATH does not occur. Indeed, assume otherwise. By REF , MATH so that MATH. Since MATH, the invariant part of MATH is one-dimensional and thus (because the canonical class is invariant for MATH), MATH for some MATH. Thus MATH and, hence MATH. REF gives MATH and MATH, and so by NAME 's formula MATH. Since MATH contains MATH disjoint nodal curves, we have a contradiction to REF . So MATH. Now we consider the case MATH, that is, the involution MATH acts identically on MATH. In this case MATH. If MATH, we get MATH and the surface MATH is a surface of general type with MATH and MATH. It contains an even set of four disjoint nodal curves MATH and thus it is minimal by REF . This is REF . The last case to consider is MATH and MATH. Since MATH, we have MATH, with MATH. Then REF gives MATH, so that in particular MATH is MATH and even, and MATH. Assume that MATH. Since MATH is a minimal surface by REF , MATH and so MATH. So either MATH or MATH. If MATH, MATH and so MATH is of general type and we have REF . If MATH, then MATH and thus MATH, being minimal, is not of general type. Since MATH, MATH is either an NAME surface or a surface of NAME dimension REF. The first case cannot occur. In fact since MATH and MATH we would have MATH, a contradiction. So MATH and MATH is a minimal properly elliptic surface. Denote by MATH the elliptic fibration and let MATH be a general fibre of MATH. Since MATH is numerically a rational multiple of MATH, we have MATH for every MATH, namely the MATH are mapped to points by MATH. Let MATH be a fibre containing, say, MATH and let MATH be the remaining irreducible components of MATH. It is well known that the classes of MATH in MATH are independent and span a subspace MATH on which the intersection form is seminegative. The classes of MATH are also independent and span a subspace MATH such that the intersection form is negative on MATH and MATH. Since MATH, we see that the only possibility is MATH. Looking at NAME 's list of singular elliptic fibres (see for example, CITE, pg. REF), one sees that the possible types of singular fibres containing some of the MATH are MATH, MATH and MATH. In addition, we have MATH, where MATH is the fibre of MATH over the point MATH and MATH denotes the topological NAME - NAME characteristic. It is easy to check that the only numerical possibility is that MATH has two MATH fibres, each containing MATH of the MATH, and that every other singular fibre is a multiple of a smooth elliptic curve. Up to a permutation we may assume that the MATH fibres of MATH are MATH and MATH. So MATH is divisible by MATH in MATH. Let MATH be the corresponding double cover. For a general fibre MATH of MATH, MATH is disconnected and the NAME factorization of MATH gives rise to an elliptic fibration MATH ``with the same fibres" as MATH. The inverse images of MATH, MATH are smooth elliptic curves. The inverse images of MATH are MATH curves contained in the fibres of MATH. Blowing these exceptional curves down, one obtains an elliptic fibration MATH whose only singular fibres are multiples of smooth elliptic fibres. Thus MATH has constant moduli, and therefore MATH and MATH have constant moduli too. This is REF . Finally, assume that MATH is a rational surface. Since MATH and MATH we can apply REF to obtain that MATH is as in the standard example. In particular there is a fibration MATH with general fibre MATH isomorphic to MATH. If we write MATH (hence MATH), then MATH has precisely MATH singular fibres of the form MATH, with MATH a MATH-curve and MATH. Denote by MATH the image of MATH on MATH and by MATH the line bundle of MATH such that MATH. The intersection number MATH is even. Thus we may write MATH, and the pre-image in MATH of the ruling on MATH is a pencil of hyperelliptic curves of genus MATH. Blowing down the curves MATH and then the images of the MATH, we obtain a birational morphism MATH onto a relatively minimal ruled surface. Let MATH be the image of MATH on MATH. Let MATH be the standard generators of MATH with MATH. We have MATH. The curve MATH has MATH singular points of type MATH, that are solved by the morphism MATH. Since MATH, we get MATH . This gives us a first equation: MATH . We also know that MATH. On the other hand, MATH and we get the second equation MATH . Comparing the two equations, we get MATH. This has the solutions MATH, which yield REF , respectively. |
math/0011090 | See for instance CITE. |
math/0011090 | If MATH is a NAME field along MATH and MATH is a smooth function vanishing at the endpoints, it is easily computed: MATH . Let MATH; if MATH, then we can find a NAME field MATH with MATH in a neighborhood MATH of MATH. If MATH or if MATH and MATH, then the field MATH can also be chosen orthogonal to MATH everywhere. From REF, it follows that MATH is negative definite in the space of fields MATH, where MATH is supported in MATH. |
math/0011090 | See CITE . |
math/0011090 | Let MATH be linear maps whose graphs in MATH are equal to MATH and MATH respectively; moreover let MATH be the linear map whose graph in MATH is MATH. Observe that MATH is invertible; it is easily computed: MATH . From REF, we have: MATH . From REF, we compute: MATH . Using REF, it follows MATH . The conclusion follows from REF. |
math/0011090 | It follows immediately from REF, considering that: MATH . |
math/0011090 | Let MATH be small enough so that MATH for all MATH; since MATH is the unique instant where MATH passes through MATH, then MATH. Using REF , we obtain MATH . The conclusion follows by applying twice REF to the above equation, once by taking MATH and again by taking MATH. |
math/0011090 | If MATH is the curve of NAME corresponding to a different choice of a parallel trivialization of MATH along MATH, then the relation between MATH and MATH is given by: MATH where MATH is a fixed symplectomorphism that preserves MATH. Namely, MATH is given by: MATH where MATH is the isomorphism that relates the two trivializations. The conclusion follows from the fact that composition with a fixed symplectomorphism that preserves MATH induces the identity in the relative homology group MATH CITE. |
math/0011090 | See CITE. |
math/0011090 | See CITE. |
math/0011090 | See CITE. |
math/0011090 | By possibly passing to a smaller MATH, we can assume the existence of a MATH-curve MATH of isomorphisms of MATH such that MATH carries MATH to a fixed subspace MATH of MATH. Define MATH as a bilinear form on the fixed space MATH. Then, MATH is a push-forward of MATH and the restriction of MATH to MATH is a push-forward of MATH. The conclusion follows by applying REF to the curve MATH in MATH. |
math/0011090 | Use REF twice, once to MATH and once to a backwards reparameterization of MATH. |
math/0011090 | Recalling REF, we define: MATH . Since MATH is not MATH-focal and MATH, it follows easily that MATH . Integration by parts in REF shows that the direct sum in REF is MATH-orthogonal, hence MATH . The restriction of MATH to MATH is obviously equal to the restriction of MATH to the same space, hence the first term on the right hand side of equality REF is computed in REF . The conclusion follows by observing that the isomorphism MATH carries the restriction of MATH to MATH. |
math/0011090 | See CITE. |
math/0011090 | See CITE. |
math/0011090 | The result is well known for NAME - NAME systems (see for instance CITE). The conclusion follows from REF . |
math/0011090 | It is a simple calculation based on the definition of the space MATH given in REF. |
math/0011090 | It is an easy calculation that uses REF. |
math/0011090 | Observe that the coefficient of MATH in the first equation of REF is positive definite. The conclusion follows from REF . |
math/0011090 | It follows directly from REF . |
math/0011090 | The index form of any symplectic differential system REF with the coefficient MATH positive definite is represented by a compact perturbation of a positive isomorphism of MATH (see CITE). The conclusion follows from REF . |
math/0011090 | We identify the space MATH with the space MATH by the map MATH; then using REF, the map MATH on MATH can be written as: MATH . Using the fact that the inclusion of MATH in MATH is compact, it is easy to see from REF that the restriction of MATH to MATH is a compact perturbation of the isomorphism MATH. Hence, the restriction of MATH to MATH is a NAME operator of index zero, and to prove the Lemma it suffices to show that MATH is injective on MATH. To this aim, observe that if MATH, then using REF we see that MATH is a solution of a second order homogeneous linear differential equation, and that MATH. This concludes the proof. |
math/0011090 | It follows easily from REF and the fact that MATH is the inverse image by MATH of the space MATH of constant functions, which is MATH-dimensional. |
math/0011090 | The proof is essentially the same as the proof of REF . Namely, using REF we show that the restriction of MATH to MATH is a NAME operator of index zero. The injectivity of this restriction is obviously equivalent to MATH. |
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