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math/0103060 | We prove REF simultaneously by induction on MATH. In case MATH, they follow from a calculation involving REF or REF respectively, together with REF to commute MATH and MATH past MATH. The induction step is similar, noting that both MATH and MATH centralize MATH. |
math/0103060 | Recall the polynomial MATH from REF . Its image MATH is zero. Hence, MATH . But a calculation using REF shows that the left hand side equals MATH modulo terms of the given form. The lemma follows easily on multiplying on the right by MATH. |
math/0103060 | By REF , we know that MATH as a MATH-bimodule. Let MATH be the projection onto the first summand of this bimodule decomposition. We just need to show that if MATH has the property that MATH for all MATH, then MATH. Using REF , we may write MATH for MATH. Consider MATH. An application of REF reveals that this equals MATH, hence MATH. Next consider MATH. Using REF as well as REF this time, we get that MATH. Now consider similarly MATH, MATH, MATH in turn to deduce MATH. We have now reduced to the case that MATH . Now consider MATH. Note MATH so the terms of MATH with MATH yield terms of MATH which lie in MATH. Hence, by REF too, MATH where MATH is a term lying in MATH. Now multiplying MATH by MATH and applying MATH, as in the previous paragraph, gives that MATH. Hence in fact, by REF once more, we have that MATH and now one gets MATH on multiplying by MATH and applying MATH, again as in the previous paragraph. Continuing in this way gives that all MATH. Now repeat the argument in the previous paragraph again, this time considering MATH, to get that all MATH. Continuing in this way eventually gives the desired conclusion: MATH. |
math/0103060 | We show that there is an even isomorphism MATH of MATH-bimodules. The lemma then follows on applying the functor MATH: one obtains natural isomorphisms MATH . Note the existence of the second isomorphism here uses the fact that MATH is a projective left MATH-module, see CITE. To construct MATH, let MATH be as in REF , and define MATH to be the map MATH, for each MATH. One easily checks that MATH is then a well-defined homomorphism of MATH-bimodules. To see that it is an isomorphism, it suffices by dimensions to check it is injective. Suppose MATH for some MATH. Then for every MATH, MATH, that is, the left ideal MATH is contained in MATH. So REF implies MATH. |
math/0103060 | Proceed by induction on MATH. For the induction step, MATH applying REF and adjointness of tensor and Hom. |
math/0103060 | The fact that MATH is right adjoint to MATH is immediate from REF , since MATH is right adjoint to restriction by adjointness of tensor and Hom. But on finite dimensional modules, a standard check using REF shows that the functor MATH is also right adjoint to restriction. Now the remaining part of the corollary follows by uniqueness of adjoint functors. |
math/0103060 | Follow CITE. |
math/0103060 | It suffices to show that the eigenvalues of MATH are of the form MATH if and only if the eigenvalues of MATH are of the same form, for MATH. Actually, by an argument involving conjugation with the automorphism MATH, it suffices just to prove the `if' part. So assume that all eigenvalues of MATH on MATH are of the form MATH for various MATH. Let MATH be an eigenvalue for the action of MATH on MATH. We have to prove that MATH is also of the form MATH. Since MATH and MATH commute, we can pick MATH lying in the MATH-eigenspace of MATH so that MATH is also an eigenvector for MATH, of eigenvalue MATH say. By assumption, MATH for some MATH. Now let MATH be the element REF . By REF , MATH . So if MATH, we get that MATH so that MATH for some MATH by assumption. Else, MATH so MATH. So applying REF , we again get that MATH for some MATH. |
math/0103060 | If MATH, then the eigenvalues of MATH on MATH are of the form MATH for MATH, by definition of MATH. Hence MATH is integral in view of REF . Conversely, suppose that MATH is integral. Then the minimal polynomial of MATH on MATH is of the form MATH for some MATH. So if we set MATH, we certainly have that the element REF acts as zero on MATH. |
math/0103060 | By REF , MATH is spanned by elements MATH for MATH, in particular it is finite dimensional. Let MATH . By REF , it suffices to show that MATH annihilates MATH for sufficiently large MATH. Consider MATH for MATH. We may write MATH for MATH and MATH. Then, MATH commutes with MATH, so we just need to consider MATH. Now using the commutation relations, one checks that MATH can be rewritten as a MATH-linear combination of elements of the form MATH for MATH and MATH. Since MATH is integral by assumption, we can choose MATH sufficiently large so that each such term is zero. |
math/0103060 | This follows from REF with MATH. |
math/0103060 | This follows from REF with MATH and MATH. |
math/0103060 | The elements of MATH which are involved in the inequality act only on the positions MATH and MATH in the tensor product. So we may assume that MATH and MATH. Let MATH for MATH. Then MATH . Now the lemma follows from the inequality MATH for MATH. |
math/0103060 | We prove REF being similar. Note MATH, since by REF we know that MATH is a free right MATH-module on basis MATH. We first show that the eigenspace of MATH is a sum of the subpaces of the form MATH, where MATH is the subgroup of MATH generated by MATH. Well, any MATH can be written as MATH for some MATH and MATH. Note MATH for any MATH, by definition of MATH. Now the defining relations of MATH especially REF imply MATH where MATH stands for a sum of terms which belong to subspaces of the form MATH for MATH and MATH. Now assume that a linear combination MATH is an eigenvector for MATH. Then it must be annihilated by MATH. Choose the maximal MATH for which the coefficient MATH is non-zero, and for this MATH choose the maximal (with respect to the NAME order) MATH such that MATH is non-zero. Then the calculation above and REF show that MATH unless MATH. This proves our claim on the eigenspace of MATH. Now apply the same argument to see that the common eigenspace of MATH and MATH is spanned by MATH for MATH, and so on, yielding the first claim in REF . Finally, define MATH . It follows by induction from the calculation above and REF that MATH giving the second claim. |
math/0103060 | Denote MATH by MATH. CASE: Let MATH be a non-zero MATH-submodule of MATH. Then, MATH must contain a MATH-submodule MATH isomorphic to MATH. But the commuting operators MATH (or MATH if MATH) act on MATH as scalars, giving that MATH is contained in their common eigenspace on MATH. But by REF , this implies that MATH. This shows that MATH contains MATH, but this generates the whole of MATH over MATH. So MATH. To see that the type of MATH is the same as the type of MATH, the functor MATH determines a map MATH . We just need to see that this is an isomorphism, which we do by constructing the inverse map. Let MATH. Then MATH leaves MATH invariant by REF , so MATH restricts to a MATH-endomorphism MATH of MATH. Finally, to see that MATH is the only irreducible in its block, we have already observed using REF that MATH. Hence all composition factors of MATH are isomorphic to MATH. Now apply NAME reciprocity and the fact just proved that MATH is irreducible. CASE: That all composition factors of MATH are isomorphic to MATH follows from the parabolic analogue of REF . To see that MATH is simple, note that the submodule MATH of MATH is isomorphic to MATH. This module is irreducible, and so it is contained in the socle. Conversely, let MATH be an irreducible MATH-submodule of MATH. Then using REF as in the proof of REF , we see that MATH must contain MATH, hence MATH. CASE: By REF , MATH has a unique MATH-submodule isomorphic to MATH, namely, MATH. Since this is completely reducible on restriction to MATH, it follows that MATH. Conversely, take any irreducible MATH-submodule MATH of MATH. The common eigenspace of MATH (respectively, MATH if MATH) on MATH must lie in MATH by REF . Hence, MATH which completes the proof. |
math/0103060 | The assumption implies that MATH is the inflation of a MATH-module. So MATH all isomorphisms being the natural ones. |
math/0103060 | The definitions imply immediately that MATH. For the reverse inequality, REF and the irreducibility of MATH gives that MATH is a quotient of MATH. So applying the exact functor MATH, MATH is a quotient of MATH. In particular, MATH. Now one gets that MATH applying the Shuffle Lemma REF . |
math/0103060 | CASE: Clearly a copy of MATH appears in MATH. But by the Shuffle Lemma and REF , MATH, hence MATH . CASE: By REF , a copy of MATH appears in MATH for any non-zero quotient MATH of MATH, in particular for any constituent of MATH. But by REF , MATH only appears once in MATH, hence MATH must be irreducible. CASE: We have shown that MATH. Hence, MATH for any other composition factor of MATH by exactness of MATH. |
math/0103060 | Pick an irreducible submodule of MATH. In view of REF , it must be of the form MATH for some irreducible MATH-module MATH. Moreover, MATH by REF . By REF and the irreducibility of MATH, MATH is a quotient of MATH. Hence, MATH is a quotient of MATH. But this is isomorphic to MATH by REF . This shows that MATH . |
math/0103060 | Let MATH. By REF , we have that MATH for an irreducible MATH with MATH. By REF and the irreducibility of MATH, MATH is a quotient of MATH. So the transitivity of induction implies that MATH is a quotient of MATH. Now everything follows from REF . |
math/0103060 | Let MATH. Suppose MATH is a submodule of MATH, for irreducibles MATH. By REF , we have MATH, hence MATH for some irreducible MATH, for each MATH. This shows that MATH contains MATH as a submodule. But according to REF , MATH has irreducible socle, so this is a contradiction. Hence, MATH is irreducible. Now certainly MATH for some irreducible MATH-module MATH, by REF . It remains to show that MATH has the same type as MATH. Note by REF , MATH has irreducible cosocle, necessarily isomorphic to MATH by NAME reciprocity. So applying REF we have that MATH which implies the statement concerning types. |
math/0103060 | Let MATH if MATH is of type MATH and MATH, MATH otherwise. By REF , the socle of MATH is isomorphic to MATH for some irreducible MATH-module MATH, and MATH indeed it is exactly as in the statement of the corollary. Now take any irreducible submodule MATH of MATH. Consider the MATH-submodule MATH, where MATH is the subalgebra generated by MATH. All composition factors of MATH are isomorphic to MATH. In particular, the socle of MATH is isomorphic to MATH which implies MATH. We have now shown that MATH for some MATH. By REF , MATH where MATH, and MATH is irreducible by REF . So at least MATH copies of MATH appear in MATH . But MATH for some irreducible MATH, so applying REF and the facts about type in REF , MATH . Hence MATH too. |
math/0103060 | CASE: If MATH, then both parts in the equality above are zero. Let MATH. Clearly, MATH is a submodule of MATH. Hence, MATH is a submodule of MATH. So NAME reciprocity gives that MATH is a submodule of MATH. Now we are done by REF . CASE: By exactness of induction, MATH is a quotient of MATH. Now the result follows from the simplicity of the cosocle, see REF . |
math/0103060 | Suppose MATH. Then by REF , MATH so MATH appears in the socle of MATH. This means that MATH, whence MATH. The converse is similar. |
math/0103060 | We need to show that MATH is a linearly independent set in MATH. Proceed by induction on MATH, the case MATH being trivial. Suppose MATH and MATH for some MATH. Choose any MATH. We will show by downward induction on MATH that MATH for all MATH with MATH. Since every irreducible MATH has MATH for at least one MATH, this is enough to complete the proof. Consider first the case that MATH. Then, MATH except if MATH, by REF . Since MATH can be worked out just from knowledge of MATH using REF , we deduce on applying MATH to the equation that the coefficient of MATH is zero. Thus the induction starts. Now suppose MATH and that we have shown MATH for all MATH with MATH. Apply MATH to the equation to deduce that MATH . Now each such MATH is irreducible, hence isomorphic to MATH, according to REF . Moreover, for MATH, MATH by REF . So now the induction hypothesis on MATH gives that all such coefficients MATH are zero, as required. |
math/0103060 | Since MATH, MATH leaves characters invariant. Hence it leaves irreducibles invariant since they are determined up to isomorphism by their character according to the theorem. |
math/0103060 | Proceed by induction on MATH, the case MATH being trivial. If MATH, let MATH be an irreducible MATH-module with central character MATH. Choose MATH so that MATH. By definition, MATH has central character MATH. So by the induction hypothesis, MATH is of type MATH if MATH is odd, type MATH otherwise. But by general theory REF , MATH is of the opposite type to MATH if MATH, of the same type if MATH. Hence, MATH is of type MATH if MATH is odd, type MATH otherwise. Finally, the proof is completed by REF , since this shows that MATH has the same type as MATH. |
math/0103060 | CASE: Proceed by induction on MATH. For the induction step, let MATH. By NAME reciprocity, there is a non-zero (hence necessarily injective) MATH-module homomorphism from MATH to MATH. Hence by induction we get a copy of MATH in MATH. CASE: Use REF and NAME reciprocity. CASE: We have just shown that MATH appears in MATH. But for any other MATH with the same weight as MATH, MATH has the same character as MATH, hence they have the same set of composition factors thanks to REF . Hence, MATH appears in MATH. |
math/0103060 | Suppose for a contradiction that MATH is reducible. Then we can find a proper irreducible submodule MATH, and set MATH. By NAME reciprocity, MATH appears in MATH with non-zero multiplicity. Hence, it cannot appear in MATH by REF . But REF show that MATH. Hence, MATH also has a quotient isomorphic to MATH, and the NAME reciprocity argument implies that MATH appears in MATH. |
math/0103060 | By REF and the Shuffle Lemma, it suffices to show that MATH . By the NAME Theorem, MATH contains MATH as a summand with multiplicity one, all other constituents lying in different blocks. Hence by NAME reciprocity, there exists a non-zero homomorphism MATH . Every homomorphic image of MATH contains a MATH-submodule isomorphic to MATH. So, by REF , we see that the image of MATH contains a MATH-submodule MATH isomorphic to MATH. Now we claim that the image of MATH also contains a MATH-submodule isomorphic to MATH. To see this, consider MATH. Pick a common eigenvector MATH for the operators MATH and MATH; then MATH and MATH. So according to REF , MATH acts on MATH by a scalar, and the assumption that MATH combined with REF shows that this scalar is necessarily non-zero. Thus, MATH, so by REF it is an irreducible MATH-module, namely, MATH . Next apply MATH to move MATH to the first position, and continue in this way to complete the proof of the claim. We have now shown that the image of MATH contains MATH . But by the Shuffle Lemma, all such composition factors of MATH necessarily lie in the irreducible MATH-submodule MATH of the induced module. Since this generates all of MATH as a MATH-module, this shows that MATH is surjective. Hence MATH is an isomorphism by dimension, which completes the proof. |
math/0103060 | The second statement of the theorem is an easy consequence of the first and REF . For the first statement, proceed by induction on MATH, the case MATH being trivial. For MATH, we may assume by REF that there exists MATH that appears in both the tuples MATH and MATH. Note then that for every MATH, either MATH or MATH, and similarly for every MATH, either MATH or MATH. So by the induction hypothesis, we have that MATH for some MATH, where MATH are tuples with no entries equal to MATH. By REF , MATH . So using REF and transitivity of induction, MATH . Finally this is irreducible by the induction hypothesis and REF . |
math/0103060 | We proceed by induction on MATH to show that MATH this being immediate in case MATH. For MATH, let MATH. We know by the inductive hypothesis and the Shuffle Lemma that MATH . Now consider the MATH-submodule MATH of MATH. The key point is that MATH is stable under the action of MATH, hence all of MATH. Although this is in principle an elementary calculation, it turns out to be extremely lengthy. It was carried out by hand for MATH but for the cases MATH (when MATH necessarily equals MATH), we had to resort to a computer calculation using the Gap computer algebra package. This proves the existence of an irreducible MATH-module MATH with character MATH. This must be MATH, by REF , completing the proof of the induction step. Now we explain how to deduce the characters of the remaining irreducibles in the block. In the argument just given, the quotient module MATH has character MATH, so must be MATH. Twisting with the automorphism MATH proves that there exist irreducibles with characters MATH and MATH, which must be MATH and MATH respectively by REF once more. This covers everything unless MATH, when we necessarily have that MATH and MATH. In this case, we have shown already that there exist four irreducibles with characters MATH . So by REF , there must be exactly one more irreducible module in the block, namely MATH, since none of the above involve the character MATH. Considering the character of MATH shows that MATH is either MATH or MATH. But the latter is not MATH-invariant so cannot occur as there would then be too many irreducibles. Now that the characters are known, it is finally a routine matter using the Shuffle Lemma and REF to prove the existence of the required non-split sequence. |
math/0103060 | Let MATH. We first claim that MATH is irreducible. To prove this, arguing in the same way as the proof of REF , it suffices to show that the MATH-submodule MATH of MATH is not invariant under MATH. Again this was checked by an explicit computer calculation. Hence, there is an irreducible MATH-module MATH with character MATH . Hence MATH and MATH by REF . So we deduce that MATH thanks to REF . Now consider the remaining irreducibles in the block. There are at most MATH remaining, namely MATH for MATH with MATH. Considering the known characters of MATH and arguing in a similar way to the second paragraph of the proof of the preceeding lemma, the remainder of the lemma follows without further calculation. |
math/0103060 | We first claim that MATH . This is immediate from REF in case MATH. If MATH, then transitivity of induction and REF give that MATH and now repeating this argument MATH more times gives the claim. Hence, by REF , MATH is self-dual. Now suppose for a contradiction that MATH is reducible. Then we can pick a proper irreducible submodule MATH of MATH, and set MATH. Applying REF , MATH . By NAME reciprocity, MATH contains a MATH-submodule isomorphic to MATH. So by REF , the irreducible MATH-module MATH appears in MATH with non-zero multiplicity, hence in fact by REF it must appear with multiplicity MATH (viewing MATH as a module over MATH). It follows that MATH is not a composition factor of MATH. But this is a contradiction, since as MATH is self-dual, MATH is a quotient module of MATH hence must contain MATH by the NAME reciprocity argument again. |
math/0103060 | By the argument in the proof of REF , it suffices to prove this in the special case that MATH. Let MATH. Recall from the previous lemma that MATH is an irreducible MATH-module. Moreover by REF , MATH. So since MATH and MATH, the NAME Theorem and a block argument shows that MATH for some MATH-module MATH all of whose composition factors lie in different blocks to those of MATH. Now let MATH. It follows from above that MATH where MATH is some quotient module of MATH. Then: MATH . Since MATH is completely reducible and MATH is irreducible, this implies that MATH is irreducible too, as required. |
math/0103060 | Let MATH and write MATH for irreducible MATH with MATH. It suffices to prove REF for any fixed choice of MATH, the conclusion for all other MATH then following immediately by REF . So take MATH. Note that MATH is a quotient of MATH which by REF has a filtration with factors isomorphic to MATH . So MATH is a quotient of some such factor, and to prove REF it remains to show that MATH for any irreducible quotient MATH of MATH. The inequality MATH is clear from the Shuffle Lemma. On the other hand, by transitivity of induction and REF , MATH. So by NAME Reciprocity, the irreducible module MATH is contained in MATH. Hence MATH. For REF , by REF , we also have MATH, and by the Shuffle Lemma, the only irreducible factors MATH of MATH with MATH come from its quotient MATH . Finally, in case MATH, the cosocle of the last module is precisely MATH. |
math/0103060 | We know that MATH and MATH are summands of MATH and MATH respectively. Moreover, MATH-duality leaves central characters invariant because MATH for each MATH. Now everything follows easily applying REF . |
math/0103060 | For MATH, it suffices to consider the effect on MATH for MATH with MATH, both sides of what we are trying to prove clearly being zero if MATH. Then, for all sufficiently large MATH, REF (in the special case MATH) implies that MATH . Hence, MATH . This proves the lemma for MATH. To deduce the statement about induction, it now suffices by uniqueness of adjoint functors to show that MATH is left adjoint to MATH . Let MATH and MATH. First observe as explained in the previous paragraph that the direct system MATH stabilizes after finitely many terms. We claim that the inverse system MATH also stabilizes after finitely many terms. To see this, it suffices to show that the dimension of MATH is bounded above independently of MATH. Well, each MATH is generated as a MATH-module by a subspace MATH isomorphic (as a vector space) to the cosocle MATH of MATH. Then MATH is generated as a MATH-module by the subspace MATH, also of dimension independent of MATH. Finally, MATH is a quotient of the vector space MATH, whose dimension is independent of MATH. Now we can complete the proof of adjointness. Using the fact from the previous paragraph that the direct and inverse systems stabilize after finitely many terms, we have natural isomorphisms MATH . This completes the argument. |
math/0103060 | To see that MATH has irreducible socle MATH whenever it is non-zero, combine REF with REF . The remaining facts follow since MATH is self-dual by REF , and MATH commutes with duality by REF . |
math/0103060 | CASE: This is immediate from REF . CASE: We deduce this from REF by an adjointness argument. Let MATH be an irreducible MATH-module, and MATH be an irreducible MATH-module. Let MATH equal MATH if MATH and MATH is of type MATH, MATH otherwise, and define MATH similarly. Then, by REF , MATH . By REF , the latter is zero unless MATH, or equivalently MATH by REF . Taking into account the superalgebra analogue of NAME 's lemma using REF , one deduces that MATH. Finally, note MATH is self-dual by REF so everything else follows. |
math/0103060 | Using REF and the definitions, it suffices to show that MATH . For MATH, this follows from REF in exactly the same way as in the proof of REF . Now MATH is left adjoint to MATH, so the statement for induction follows from uniqueness of adjoint functors on checking that the functor MATH is left adjoint to MATH . The latter follows as in the proof of REF . |
math/0103060 | Let us prove in the case MATH. The proof for MATH is the same idea, though the details are more delicate (one needs to use REF too). By REF , we have that MATH as operators on the NAME group. By REF , we have that MATH . This is enough. |
math/0103060 | Let MATH and MATH. CASE: By REF and NAME reciprocity, there is a short exact sequence MATH . Moreover, all composition factors MATH of MATH have MATH by REF . Applying the exact functor MATH, we obtain an exact sequence MATH . By the NAME Theorem, MATH . By considering characters MATH. Hence, MATH . Using REF again, the cosocle of MATH is MATH, and all other composition factors of this module are of the form MATH with MATH. Moreover, all composition factors of MATH are of the form MATH with MATH. So we have now shown that MATH for irreducibles MATH with MATH. The conclusion follows on applying REF . CASE: We give the argument for the case MATH, the case MATH being similar but using REF instead of REF . We know that MATH. So, applying the automorphism MATH to REF , we deduce that the maximal size of a NAME block of MATH on MATH is MATH. Hence the maximal size of a NAME block of MATH on MATH is at least MATH. On the other hand, the argument given above in deriving REF shows that the module MATH has a filtration with MATH factors, each of which is isomorphic to MATH. Since MATH annihilates MATH, it follows that MATH annihilates MATH. So certainly MATH annihilates its quotient MATH. So the maximal size of a NAME block of MATH on MATH is at most MATH. CASE: Let MATH if MATH and MATH if MATH. Consider the effect of left multiplication by MATH on the MATH-module MATH . Note MATH is equal to either MATH or MATH, by REF . In the latter case, MATH (respectively, MATH) acts as a scalar on MATH, hence it leaves the two indecomposable summands invariant. This shows that in any case left multiplication by MATH (which centralizes the subalgebra MATH of MATH) induces a MATH-endomorphism MATH. But by REF , MATH and MATH. Hence, MATH give MATH linearly independent even MATH-endomorphisms of MATH. In view of REF , we automatically get from these MATH linearly independent odd endomorphisms in case MATH is of type MATH, so we have now shown that MATH . On the other hand, MATH has irreducible cosocle MATH, and this appears in MATH with multiplicity MATH by REF , so the reverse inequality also holds. |
math/0103060 | Suppose there is a non-zero homomorphism MATH. Then, since MATH has simple head MATH, we see that MATH has MATH as a composition factor. Hence, by REF , MATH. Dualizing and applying the same argument gives the inequality the other way round, hence MATH. But then, MATH is a composition factor of MATH with MATH, hence by REF again, MATH. But this contradicts REF . |
math/0103060 | The statement involving MATH is obvious. For MATH, note that MATH and MATH for every MATH. Hence by REF , MATH if and only if MATH. This implies that MATH. |
math/0103060 | Follows from the Shuffle Lemma. |
math/0103060 | Let MATH and MATH. By REF , there exist unique MATH with MATH such that MATH . We need to show that MATH, which follows if we can show that MATH for all MATH. We claim that MATH for all MATH. Given the claim, we know by the definition of MATH, REF that MATH for all MATH. So the claim implies that MATH for all MATH, hence by REF once more, MATH as required. To prove the claim, note that MATH, so MATH. Hence, REF shows that there is a surjection MATH . By REF , MATH . Hence by NAME reciprocity, there is a surjection MATH . Combining, we have proved existence of a surjection MATH . Hence by NAME reciprocity there is a non-zero map MATH . Since the left hand module has irreducible cosocle MATH, we deduce that MATH has a constituent isomorphic to MATH on restriction to the subalgebra MATH. This implies the claim. |
math/0103060 | Proceed by induction on MATH, the case MATH being immediate by REF . For MATH, we may write MATH for some irreducible MATH-module MATH with MATH. By induction, MATH . The conclusion follows from REF . |
math/0103060 | CASE: Take any MATH. Since MATH is right exact, the natural surjection MATH and the obvious embedding MATH (see REF) induce a commutative diagram MATH where the rows are exact. Note if MATH then MATH. It follows that if MATH is an isomorphism so is MATH for every MATH. So by definition of MATH, the maps MATH are not isomorphisms but all other MATH are isomorphisms. Now to prove REF , we show by induction on MATH that MATH where the lower terms are irreducible MATH-modules MATH with MATH. This is vacuous if MATH. For MATH, MATH is not an isomorphism, so MATH. Hence, by REF , the image of MATH contains a copy of MATH plus lower terms. Now the induction step is immediate. CASE: Take MATH. One easily shows using the explicit construction of MATH in REF that there is an even endomorphism MATH of MATH-modules, such that the image of MATH is MATH for each MATH. NAME reciprocity induces superalgebra homomorphisms MATH . So MATH induces an even MATH-endomorphism MATH of MATH, such that the image of MATH is MATH for MATH. Now apply the right exact functor MATH to get an even MATH-endomorphism MATH induced by MATH. Note MATH and MATH because its image coincides with the image of the non-zero map MATH in the proof of REF . Hence, MATH are linearly independent, even endomorphisms of MATH. Now the proof of REF is completed in the same way as in the proof of REF . |
math/0103060 | Repeat the argument in the proof of REF , but using MATH and REF in place of MATH and REF . |
math/0103060 | By NAME reciprocity and REF , we have that MATH . Hence, by NAME 's lemma, MATH . Now if MATH is of type MATH, MATH by REF . Hence by REF , MATH and the conclusion follows in this case. The argument for MATH of type MATH is similar. |
math/0103060 | The statement about composition multiplicities follows from REF , taking into account how MATH and MATH are related to MATH and MATH as explained in REF . Now consider the statement about socles. We consider only MATH, the other case being entirely similar but using results from REF instead. By adjointness, it suffices to be able to compute MATH for any irreducible MATH-module MATH. But in view of REF , this can be computed from knowledge of MATH which is known by REF (if MATH) and REF (if MATH). The details are similar to those in the proof of REF , so we omit them. |
math/0103060 | Let MATH and MATH be the quotient map. Let MATH denote the subalgebra of MATH generated by MATH, and set MATH (respectively, MATH if MATH). Recall from REF that viewed as a MATH-module, we have that MATH. In particular, MATH is a cyclic module generated by the image MATH of MATH. We first observe that for any MATH, MATH annihilates the vector MATH for any MATH. This follows from the relations in MATH, for example, in case MATH one ultimately appeals to the facts that MATH annihilates MATH (see REF ) and MATH annihilates MATH. It follows that the unique MATH-homomorphism MATH such that MATH for each MATH factors to induce a well-defined MATH-module homorphism MATH . We then get from NAME reciprocity an induced map MATH for each MATH. Each MATH factors through the quotient MATH, so we get an induced map MATH . It remains to show that the composite of MATH with the canonical epimorphism from MATH to MATH is surjective. Let MATH. By NAME Theorem there exists an exact sequence MATH . In other words, there is a MATH-isomorphism MATH for MATH, where MATH is embedded into MATH as MATH. Recall from REF that MATH where MATH is as in REF . Hence, applying REF , MATH . So applying the exact functor MATH to the isomorphism above we get an isomorphism MATH . It follows that there is a surjection MATH such that the diagram MATH commutes for all MATH, where MATH is the map from REF and the left hand arrow is the natural surjection. Now surjectivity of MATH immediately implies the desired surjectivity of the composite. |
math/0103060 | Let us first show that MATH . Recall MATH if and only if MATH. Suppose first that MATH, when MATH. Then, MATH and MATH, so the conclusion certainly holds in this case. So we may assume that MATH, hence MATH. Note by REF , MATH . In particular, the map MATH in REF is non-zero. Now REF implies that the multiplicity of MATH in MATH is strictly greater than MATH, since at least one composition factor of MATH must be sent to zero on composing with the second map MATH. Using another part of REF , this shows that MATH and REF follows. Now using REF , we see that in the NAME group, MATH with equality if and only if equality holds in REF . By central character considerations, for MATH, MATH. So using REF we deduce that MATH with equality if and only if equality holds in REF for all MATH. Now REF shows that the right hand side equals MATH, which does indeed equal the left hand side thanks to REF . |
math/0103060 | We proceed by induction on MATH, the conclusion being known already in case MATH. For the induction step, take an irreducible MATH-module MATH with MATH, so MATH where MATH is an irreducible MATH-module with MATH, MATH. Then by REF and the induction hypothesis, MATH . This completes the induction step. |
math/0103060 | In view of REF , it suffices to show that MATH . But this is immediate from REF . |
math/0103060 | CASE: Since MATH by REF , we know by REF that MATH where the MATH are irreducibles with MATH. Suppose that MATH, for MATH. Then MATH and each MATH have central character MATH, since they are all composition factors of MATH. So by REF , MATH . It follows that MATH too. CASE: This is just the definition of MATH combined with REF . CASE: This follows from REF . |
math/0103060 | It just remains to check that MATH is an algebra homomorphism, which follows using the NAME Theorem. Note in checking the details, one needs to use REF to take the definition of MATH into account correctly. |
math/0103060 | This is immediate from REF . |
math/0103060 | In view of REF and coassociativity of MATH, this reduces to checking it on irreducible MATH-modules for MATH respectively. For this, the character information in REF is sufficient. |
math/0103060 | Let MATH be an irreducible module in MATH or MATH. Let MATH for irreducible MATH-modules MATH and irreducible MATH-modules MATH. By the definition of the action of MATH on MATH, it follows that MATH . Hence, since MATH, we get that MATH acts in the same way as MATH. So in view of REF , it just remains to check that MATH acts in the same way as MATH, which follows by REF . |
math/0103060 | Since MATH is a faithful MATH-module, REF imply that the operators MATH satisfy the same relations as the generators MATH of MATH. This implies existence of a unique such algebra homomorphism. The fact that MATH is a coalgebra map follows because MATH by the definition of the comultiplication on MATH. |
math/0103060 | Let MATH be an irreducible MATH-module and MATH be an irreducible MATH-module. We check that MATH in the special case that MATH, MATH is of type MATH and MATH is of type MATH. In this case, REF , MATH . All the other situations that need to be considered follow similarly. |
math/0103060 | Proceed by induction on MATH, the conclusion being trivial for MATH. So suppose MATH and that the result is true for all smaller MATH. Suppose for a contradiction that we can find an irreducible MATH-module MATH for which the result does not hold. Pick MATH with MATH. Since there are only finitely many irreducible MATH-modules, we may assume by the choice of MATH that the result holds for all irreducible MATH-modules MATH with MATH. Write MATH for coefficients MATH. By REF , MATH . In particular, MATH unless MATH. Moreover, if MATH for MATH with MATH, then by REF , we have that MATH, whence MATH and MATH. This shows: MATH for some MATH. But the inductive hypothesis and choice of MATH ensure that all terms on the right hand side are integral linear combinations of terms MATH, hence the same is true for MATH, a contradiction. |
math/0103060 | We show by induction on MATH that the map MATH is injective. This is clear if MATH, so assume MATH and the result has been proved for all smaller MATH. Suppose we have a relation MATH with not all coefficients MATH being zero. We may choose MATH and an irreducible module MATH such that MATH and MATH for all MATH with MATH. Apply MATH to the sum. By REF , we get MATH where MATH is a sum of terms of the form MATH with MATH. Now the inductive hypothesis shows that MATH and that MATH for each MATH with MATH. In particular, MATH, a contradiction. |
math/0103060 | It is non-degenerate by construction, being induced by the pairing MATH from REF . So we just need to check that it is symmetric. Proceed by induction on MATH to show that MATH is symmetric. In view of REF , any element of MATH can be written as MATH for MATH. Then for any other MATH, we have that MATH by the induction hypothesis. |
math/0103060 | We know the MATH satisfy the NAME relations on all of MATH by REF , so they certainly satisfy the NAME relations on restriction to MATH. Moreover, MATH and MATH are adjoint operators for the bilinear form MATH according to REF , and this form is non-degenerate by REF . The lemma follows. |
math/0103060 | Let MATH be an irreducible MATH-module. It follows immediately from REF (together with central character considerations in case MATH) that MATH appears in MATH with multiplicity MATH . Therefore, it suffices simply to show that MATH is a multiple of MATH. Let us show equivalently that MATH is a multiple of MATH. For MATH, we have a surjection MATH . Apply MATH to get a surjection MATH . By the NAME Theorem and REF , there is an exact sequence MATH where MATH is as in REF . For sufficiently large MATH, MATH. So on applying the right exact functor MATH and using the irreducibility of MATH, this implies that there is an exact sequence MATH for some MATH. Now let MATH be any irreducible MATH-module with MATH. Combining REF shows that MATH . Now summing over all MATH and using REF gives that MATH . But REF shows that equality holds here, hence it must hold in REF for all MATH. This completes the proof. |
math/0103060 | It makes sense to think of MATH as a MATH-module according to REF . The actions of MATH and MATH are locally nilpotent by REF . The action of MATH is diagonal by definition. Hence, MATH is an integrable module. Clearly MATH is a highest weight vector of highest weight MATH. Moreover, MATH by REF . This completes the proof of REF , and REF follows immediately from REF . For REF , we know already that MATH are dual lattices of MATH which are invariant under MATH. Moreover, REF again shows MATH. |
math/0103060 | Note by REF that the action of the MATH on MATH, hence on each MATH, factors through the map MATH. So if MATH, we have by the previous theorem that MATH acts as zero on all integrable highest weight modules MATH. Hence MATH and MATH is injective. To prove surjectivity, take MATH. It suffices to show that MATH for all MATH implies that MATH. Note MATH where the second equality follows because the right regular action of MATH on itself is precisely the dual action to the left action of MATH on MATH, and the third equality follows from REF . Hence, if MATH for all MATH, we have that MATH for all MATH. Now choose MATH sufficiently large so that in fact MATH. Then, it follows that MATH for all MATH, where MATH now is canonical pairing between MATH and MATH. Hence by REF , MATH for all MATH. But then REF implies that MATH. |
math/0103060 | We prove REF , the proof of REF being similar. By the Shuffle Lemma, we certainly have that MATH. Now let MATH. Then obviously, MATH. Hence, MATH. |
math/0103060 | Set MATH. Let MATH, so MATH is an irreducible MATH-module with MATH and MATH. For MATH, let MATH. REF imply that in the NAME group, MATH is some number of copies of MATH plus terms with strictly smaller MATH. In particular, MATH for all composition factors MATH of MATH, while MATH is consists only of copies of MATH. We will show by decreasing induction on MATH that there is a non-zero MATH-module homorphism MATH . In case MATH, MATH is a quotient of MATH and MATH is a quotient of MATH, so the induction starts. Now we suppose by induction that we have proved MATH exists for MATH and construct MATH. Consider MATH. By the NAME Theorem, this has a filtration MATH with successive quotients MATH where MATH is the obvious permutation. As MATH, NAME reciprocity implies that there is a copy of the MATH-module MATH in the image of MATH. Now MATH, so the MATH-eigenspace of MATH acting on MATH is non-trivial. We conclude that the map MATH is non-zero. If MATH, then it follows from the description of MATH and MATH above that MATH. So in this case, we necessarily have that MATH. Similarly if MATH, MATH, so if MATH we see that MATH factors to a non-zero homomorphism MATH . But this implies that MATH has a constituent MATH with MATH, which we know is not the case. Hence we have that MATH in the case MATH too. Hence, the restriction of MATH to MATH gives us a non-zero homomorphism MATH . Now finally as all composition factors of MATH are isomorphic to MATH, this implies the existence of a non-zero homomorphism MATH completing the induction. Now taking MATH we have a non-zero map MATH . But the left hand side has irreducible cosocle MATH while all composition factors of the right hand side are isomorphic to MATH. This completes the proof. |
math/0103060 | Since we can apply the automorphism MATH, it suffices to check just one of the implications. So suppose MATH. Clearly, MATH. For the reverse inequality, the preceeding two lemmas show that MATH . This shows that MATH. |
math/0103060 | Set MATH and MATH. By the Shuffle Lemma and REF , MATH . Hence by REF , MATH equals MATH plus terms MATH for irreducible MATH with MATH. On the other hand, MATH surjects onto MATH. So the assumption MATH implies MATH. |
math/0103060 | REF is REF or the definition in the affine case. REF is REF . The remaining properties are immediate. |
math/0103060 | Since MATH and MATH are restrictions of MATH from MATH to MATH, respectively, the first statement is immediate. The second is a restatement of REF . |
math/0103060 | Let MATH, MATH, and MATH. CASE: By twisting with MATH, it suffices to prove that (for arbitrary MATH) MATH . Define the weights MATH by taking the MATH-coefficient of MATH to be MATH and the MATH-coefficients of MATH for MATH to be MATH. Then MATH for any MATH, see NAME REF. Moreover, the same corollary implies that for MATH, MATH . Now, MATH and MATH. Moreover, by REF (twisted with MATH) we have MATH for any MATH. All of these applied consecutively to MATH imply: MATH for all MATH. For MATH, taking into account MATH, this gives REF . CASE: Observe by REF that MATH if and only if MATH, and that otherwise MATH. But MATH and MATH so REF follows if we show that MATH if and only if MATH. But by REF , MATH . Moreover, obviously MATH and MATH, and it remains to observe that MATH if and only if MATH, thanks to REF . CASE: This follows from REF (twisted with MATH). |
math/0103060 | REF - REF are immediate from our construction of MATH. The information required to verify REF is exactly contained in REF . Finally, REF is immediate from the definition of MATH, and REF holds because every such MATH has MATH for at least one MATH. |
math/0103060 | We note that for irreducible modules MATH in MATH with MATH, we have that MATH . This follows immediately from REF . Using this in place of CITE, the proof of the theorem is now completed by exactly the same argument as in CITE. |
math/0103060 | We have already discussed the first statement of the theorem, being a consequence of our main results combined with NAME 's REF . For the rest, REF follows from REF is a special case of REF combined with REF , and REF is a special case of REF . For REF , note that if MATH is projective then it is the only irreducible in its block, hence by REF , MATH. So either MATH, or MATH and MATH. Now if MATH then MATH is a MATH-bar core so the NAME form on the (MATH-dimensional) MATH-weight space of MATH is MATH (since MATH is conjugate to MATH under the action of the affine NAME group). Hence, MATH is projective by REF . To rule out the remaining possibilty MATH and MATH, one checks in that case that the NAME form on the MATH-weight space of MATH is MATH. |
math/0103060 | The existence of such modules MATH follows from REF , combined as usual with NAME 's REF . Uniqueness follows from NAME and the block classification from REF . For the remaining properties, REF follow from REF . Finally, REF follows from REF as MATH is a module with simple socle and cosocle both isomorphic to MATH. |
math/0103060 | The argument is the same as REF , but using REF . |
math/0103060 | This is obvious if MATH and easy to check directly if MATH. Now for MATH we proceed by induction using REF together with the observation that MATH. We consider the four cases MATH or MATH and MATH separately. Suppose first that MATH. Considering the crystal graph shows that MATH only for MATH and for one other MATH, for which MATH. By the induction hypothesis, MATH. Hence by REF , MATH can only contain MATH and MATH as composition factors. But the latter case cannot hold since by REF again, MATH is not isotypic. Hence all composition factors of MATH are MATH, and one easily gets that in fact MATH by a dimension argument. Next suppose that MATH. This time, MATH and all other MATH are zero. Hence, by the induction hypothesis and the branching rules, MATH only involves MATH as a constituent. But we have that MATH so in fact that MATH must have MATH as a constituent with multiplicty two. Further consideration of the endomorphism ring of MATH shows moreover that it is an indecomposable module. The argument in the remaining two cases MATH and MATH is entirely similar. |
math/0103060 | Most of these facts are proved in CITE, but we recall some of the details since we need to go slightly further. Let us consider the proof that MATH and MATH, assuming that MATH is odd. Let MATH be a basis for MATH with MATH being the identity and MATH being an odd involution. Then, there are natural isomorphisms MATH . The first is defined for each MATH-module MATH by MATH, MATH. The second is defined for each MATH-module MATH by MATH, MATH, where MATH for each MATH. The proof that MATH and MATH really are isomorphisms is similar to the argument in CITE. Now consider the composite natural transformations MATH where the unmarked arrows are the obvious even ones. Then, REF (respectively, REF) gives the unit and counit of the adjunction needed to prove that MATH is left (respectively, right) adoint to MATH. We leave the details to be checked to the reader, see for example, CITE. In case MATH is even, a similar but easier argument shows that MATH so that MATH and MATH are inverse equivalences, hence left and right adjoint to each other. Now the statements in REF about isomorphism classes of irreducible modules follow easily as in CITE. Let us next prove that MATH commutes with duality. The antiautomorphism MATH of MATH induces the antiautomorphism MATH of the subalgebra MATH with MATH for each MATH. Let MATH be a homogeneous isomorphism; note that it is not always possible to choose MATH to be even, since MATH in case MATH. We get a natural isomorphism MATH for each finite dimensional MATH-module MATH, where MATH for each MATH. This shows that MATH commutes with duality, that is, MATH. Hence, using REF and the fact that MATH is left adjoint to MATH, the composite functor MATH is right adjoint to MATH. But we already know that MATH is right adjoint to MATH, so uniqueness of adjoints gives that MATH. This shows that MATH commutes with duality too. It just remains to check the isomorphisms REF - REF . Well, REF follows from the definition on noting that MATH if MATH is even. Then REF follows from REF on composing on the left with MATH and on the right with MATH. Next, REF follows from the definition and an application of NAME reciprocity, using the observation that MATH if MATH is odd. As before REF then follows, composing with MATH and MATH. Finally, REF follow from REF respectively by uniqueness of adjoints. |
math/0103060 | It just remains to observe that REF - REF follow follow directly from REF - REF using REF . |
math/0103060 | The statement about dimension is immediate from REF . The final statement is easily proved by working in terms of MATH and using the explicit construction given in REF . |
math/0103060 | If MATH is odd, we simply define MATH for each MATH. If MATH is even, take MATH . We need to explain the notation MATH used in the last two cases: here, MATH admits an odd involution by REF , and also MATH has an odd involution since MATH is even. So in exactly the same way as in the definition of REF , we can introduce the space MATH . It is then the case that MATH. Equivalently, by REF MATH can be characterized by MATH if MATH is even and MATH, or MATH is odd and MATH. With these definitions, it is now a straightforward matter to prove REF - REF using REF . Finally, the uniqueness statement is immediate from NAME and the description of blocks from REF . |
math/0103060 | This is deduced from REF by similar argument to the proof of REF . |
math/0103061 | There exists a completely metrizable space MATH with MATH and a MATH-soft mapping MATH CITE. Consider a MATH-space MATH of dimension MATH containing MATH as a closed subspace CITE. Since the pair MATH is MATH-connected, we can extend the mapping MATH to a mapping MATH. |
math/0103062 | Using the fact that MATH we have MATH . |
math/0103062 | MATH . |
math/0103062 | In terms of the basis for the kernel of MATH we can rewrite REF as MATH . Observe that MATH . So the equation MATH has a solution only if MATH, in which case we can set MATH . To compute MATH we note MATH and (with the symmetry assumption), MATH . This means that MATH . |
math/0103062 | Let MATH be a neighborhood of MATH in which NAME coordinates MATH are valid, and MATH a cutoff function with supp-MATH and MATH in some neighborhood of MATH. Then we define the sequence MATH by MATH where MATH are the solutions obtained above, MATH, and MATH normalizes MATH. This could be written as MATH where MATH is a polynomial of degree MATH (with coefficients independent of MATH). Since MATH, we have that MATH . The concentration of MATH on MATH described in REF then follows immediately from REF . By virtue of the factor MATH, we can turn the formal considerations used to obtain the operators MATH into estimates. With cutoff, MATH could be considered an operator on MATH with support in MATH. By construction we have MATH where MATH is supported in MATH and vanishes to order MATH at MATH. We also have MATH . Combining these facts with the definition of MATH we deduce that MATH where MATH is supported in MATH and vanishes to order MATH at MATH. Using this order of vanishing we estimate MATH . Noting that MATH on MATH, we obtain the estimate REF . |
math/0103064 | The only nonobvious part is to show that if MATH, and MATH, MATH, MATH, then MATH, which, in case MATH is a MATH-tuple of abelian groups or pointed sets, requires showing that for each MATH, MATH is a homomorphism. However, MATH is the composition of two homomorphisms: MATH, and the homomorphism from MATH to the product MATH determined by the MATH. |
math/0103064 | For each MATH, and each MATH, we have MATH . |
math/0103064 | REF , and REF are straightforward. For REF , MATH is simply the insertion of the relatively free algebra into the coproduct. |
math/0103064 | All parts are straightforward, if not obvious, except perhaps REF , the key to proving which is to know that if MATH, then MATH sends MATH to MATH because it is a homomorphism, and then to use REF . |
math/0103064 | We use the fact that MATH is generated by the projections and the constants MATH. Certainly the conclusion is true for MATH one of these generators. Suppose it is true for the components of MATH, and let MATH. We have MATH, and MATH for all MATH. Thus, by REF , MATH is a MATH-operation on MATH over MATH. But, we have MATH for every MATH, by definition and by the fact that MATH is a clone homomorphism. Thus, the conclusion is true for MATH. |
math/0103064 | Let MATH, MATH, and MATH. For each MATH, MATH preserves the basepoint, because MATH, so that MATH . Also, because MATH, we have MATH . Thus, MATH is a homomorphism. Now let MATH, and MATH. We must show that there is a unique MATH-module homomorphism MATH such that MATH. By the construction of MATH, there is a unique MATH-module homomorphism MATH such that MATH; this shows that MATH is unique if it exists, because then MATH, and MATH is an epimorphism. To show MATH exists, we must show that for each MATH, MATH is zero on MATH. But, we have MATH, because MATH, and if MATH is a MATH-ary operation symbol, MATH, and MATH, we have MATH . |
math/0103064 | REF follows from REF , which follows from REF . To prove REF , first we prove that for each MATH-ary term MATH, MATH-tuple MATH of elements of MATH, MATH such that MATH, and MATH, we have MATH we will use the fact that MATH is generated by the MATH. Let MATH. We have MATH . Now, suppose MATH is a MATH-ary operation symbol, and REF is true for MATH-ary terms MATH, MATH, MATH. Then we will show it is true for the MATH-ary term MATH. We have MATH where each MATH, and MATH, MATH . Now, each MATH-operation MATH is determined by the actions of the MATH on elements of MATH. If MATH and MATH are MATH-ary terms such that MATH, then MATH because MATH is onto, and REF implies that values of MATH and MATH differ by elements of the submodule MATH. This implies that MATH. In other words, MATH satisfies any identity satisfied by MATH. |
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