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math/0104253 | By REF we have natural isomorphisms MATH for all MATH. With the notation of the proposition, MATH is the zero-differential complex MATH where MATH is the direct sum of MATH copies of MATH for MATH, zero otherwise. Consider the Cartesian square MATH . By flat base change around the square we get MATH . In order to compute MATH on MATH, we consider the Cartesian square MATH . Now by the general base-change REF we have MATH . But MATH, the skyscraper sheaf at MATH shifted MATH places to the right (see the proof of the theorem of REF). Notice that MATH is a regular embedding; using NAME resolutions it follows that MATH, the zero-differential exterior-algebra complex of the conormal sheaf of MATH in MATH, concentrated in degrees MATH to MATH. Similarly MATH, the exterior algebra of the fibre at MATH, whence the lemma follows immediately taking into account the shift by MATH. Using the projection formula we have MATH . The proposition now follows from the lemma because hypercohomology commutes with direct sums. |
math/0104253 | CITE Corollary (REFEF.). Notice that MATH automatically for MATH because MATH and the length of the complex MATH is MATH. |
math/0104253 | Let MATH be a subbundle stable under MATH. Then MATH is a subbundle of MATH stable under MATH. But tensoring with MATH affects neither the ranks nor the degrees of MATH and MATH, and hence the lemma follows from the stability of the NAME bundle MATH. |
math/0104253 | Let MATH be a subbundle stable under MATH. Let MATH. Then MATH. But MATH too, and hence MATH. Thus MATH is stable under MATH, and the lemma follows from the stability of MATH. |
math/0104253 | By REF we are reduced to showing that MATH is MATH with respect to MATH. We consider two cases. Let MATH denote the open subset MATH in MATH. CASE: Let MATH. Then (using the notation of REF) MATH and we need to show that MATH for MATH. But this follows from REF. Notice that for a rank-MATH . NAME bundle MATH one of the bundles MATH would be trivial, and the vanishing REF would fail. CASE: Let MATH. We consider the second hypercohomology spectral sequence: MATH . But MATH for a MATH-form MATH, determined up to multiplication by a non-zero scalar . Now MATH is clearly an injective map of sheaves; let MATH be its cokernel. Thus the MATH-terms of the spectral sequence are MATH . But MATH is a direct sum of skyscraper sheaves supported on the divisor of zeroes of the one-form, and since skyscraper sheaves are flasque, we have MATH. Hence MATH. |
math/0104253 | By the proof of REF MATH is MATH. Now the proposition follows from REF applied to the immersion MATH. |
math/0104253 | It follows from REF that MATH. Consider the first hypercohomology spectral sequence MATH . The MATH-terms of the sequence are: MATH . The sequence clearly degenerates at MATH, that is, MATH, and hence MATH . But these hypercohomologies vanish by REF, and thus MATH is injective and MATH is surjective. On the other hand, MATH and hence MATH . But as MATH, the NAME theorem gives MATH whence the result follows immediately. |
math/0104253 | Let MATH be a base point giving an embedding MATH, and denote by MATH the embedding MATH of the fibre MATH. Then by REF MATH . We apply the first hypercohomology spectral sequence MATH . The MATH-terms are given by MATH . The standard results on the cohomology of a projective space (CITE III. REF) show that the MATH. Furthermore, it is clear from REF of MATH that MATH is an injection. Thus we see that MATH . Thus in the direct sum of REF we have non-zero cohomology only when MATH, and the result follows immediately. |
math/0104253 | This is an easy application of the NAME formula. |
math/0104253 | We show this by actually exhibiting a process of recovering a NAME bundle MATH from its total NAME transform MATH. CASE: Choose a base point MATH as in REF, and let MATH be the corresponding embedding. Denote by MATH the immersion MATH. Then by REF MATH. By REF MATH is a category equivalence; let MATH be its inverse. Now by REF, and hence MATH . The differential MATH of the complex MATH is injective. Let MATH be an open subset and MATH a non-zero section. There is a point MATH for which MATH. Because MATH is locally free, it follows (using NAME 's lemma) that there is an open neighbourhood MATH of MATH such that MATH for MATH. If MATH, it follows from the definition of MATH that there is a point MATH with MATH, and in particular MATH. But this shows that MATH is injective as a morphism of presheaves and hence as a sheaf morphism too. Thus the lemma is proved. By the lemma there is an exact sequence MATH and consequently MATH is quasi-isomorphic to MATH. It follows from this that MATH in MATH. Since MATH is an honest sheaf, MATH also in MATH. This means that we can recover the cokernel MATH of MATH on MATH as MATH. CASE: Tensor REF with MATH and obtain the exact sequence MATH . We shall use the long exact MATH-sequence associated to REF. By the projection formula MATH . Now it follows from base change and the standard formulas for the cohomology of projective spaces that MATH . It follows then from the long exact sequence that MATH, and that we may consequently recover the underlying sheaf MATH of MATH from MATH by twisting by MATH, projecting down to MATH, and twisting by MATH. CASE: It remains to recover the NAME field MATH. This will be done after discarding much of the information contained in MATH. We choose a non-zero MATH, and we let MATH be an open affine subscheme of MATH over which MATH does not vanish; then MATH gives a trivialisation of MATH on MATH. Clearly it is enough to recover MATH over MATH. Let MATH be the subvectorspace of MATH generated by MATH. We can consider MATH as a closed subscheme of the open subscheme MATH of MATH. Furthermore, we consider MATH as a subscheme of MATH, and let MATH be the restriction of MATH to MATH; it is just the cokernel of MATH restricted to MATH. Notice that MATH. On MATH the underlying sheaf MATH of MATH corresponds to a MATH-module MATH and MATH corresponds to an endomorphism MATH of MATH. Furthermore, the pull-back of MATH to MATH corresponds to MATH. By the definition of MATH REF, MATH corresponds to the MATH-linear map MATH . But MATH fits into the exact sequence MATH where MATH is the MATH-module with MATH acting on MATH as MATH (compare REF no. REF). Hence MATH. But the structure of MATH-module of MATH determines MATH and hence MATH. |
math/0104253 | Let MATH and MATH be NAME bundles on MATH and let MATH and MATH be the cokernels of MATH and MATH respectively. Because the relative NAME transform is an equivalence of categories, we have MATH . Thus faithfulness is clear. On the other hand, let MATH; using the notation of the proof of the theorem, the previous remark shows that MATH gives a morphism of NAME bundles MATH. But as the genus of MATH is at least MATH, the canonical linear system MATH has no base points. Hence we can cover MATH by open sets like MATH; it is clear that the morphisms thus obtained glue to give a morphism MATH. |
math/0104260 | For MATH, it is quite easy to see MATH . By the NAME inequality, we get the inequality MATH . This shows that the MATH converges absolutely for all MATH. |
math/0104260 | We adopt the same idea as in the proof CITE. MATH . Note that we have used the MATH-Gamma function CITE, MATH . |
math/0104260 | As we have seen in REF , MATH . In addition, we derive by REF , MATH . Therefore, we finish the proof. |
math/0104260 | By the recurrence REF , we derive MATH . |
math/0104260 | The proof is done by REF , and REF . |
math/0104261 | If MATH, then by REF we see that MATH is also nonexpansive. Conversely, let MATH and MATH, MATH be the two half-spaces with boundary MATH. Suppose that both MATH and MATH are expansive for MATH. We prove that MATH is also expansive for MATH, which will complete the proof. Since MATH has an expansive half-space, it is an expansive action. Let MATH be an expansive constant for MATH. Let MATH denote the ball of radius MATH in MATH, and MATH be the line segment joining MATH to MATH. A ``finite" version of the expansiveness of MATH, entirely analogous to CITE, is that there is a MATH such that MATH codes MATH. Similarly, there is a MATH such that MATH codes MATH. Hence if MATH, then MATH codes MATH, which by the same argument codes MATH, and so on. Thus MATH codes MATH, which means that MATH is expansive. |
math/0104261 | Suppose that MATH is MATH-Noetherian. If MATH, then there is a MATH such that MATH. By definition, every MATH-submodule of MATH is MATH-Noetherian, and in particular so is MATH. Hence MATH is MATH-Noetherian for every MATH. Conversely, suppose that MATH is MATH-Noetherian for every MATH. Since MATH is MATH-Noetherian, there is a chain of MATH-submodules MATH with MATH for MATH, where each MATH is a prime ideal in MATH that contains some MATH CITE. The surjection MATH shows that MATH is again MATH-Noetherian. Recall the fact from module theory that over an arbitrary ring if MATH are modules, then MATH is NAME if and only if both MATH and MATH are NAME. Repeated application of this fact shows successively that MATH, MATH, MATH, and finally MATH are MATH-Noetherian. |
math/0104261 | Consider the matrix MATH acting on column vectors in MATH. If MATH has MATH-th entry MATH, then MATH. Multiplying this equation by the adjugate matrix of MATH gives MATH. Hence MATH for MATH, so that MATH annihilates MATH. |
math/0104261 | NAME modules over arbitrary rings are always finitely generated. Conversely, suppose that MATH is finitely generated over MATH, say by MATH, MATH, MATH. Fix MATH. Since MATH is a MATH-module, we can find MATH such that MATH . If MATH denotes the MATH identity matrix and MATH, then REF shows that MATH annihilates MATH. Multiplying this determinant by MATH shows that there is an element in MATH of the form MATH, with MATH, that annihilates MATH. Suppose that MATH is a MATH-submodule of MATH. For every MATH we have that MATH, so that MATH. Hence MATH is closed under the subring of MATH generated by MATH and MATH, which is all of MATH. Thus every MATH-submodule of MATH is also a MATH-submodule. Since MATH is finitely generated over MATH, it is finitely generated over MATH, hence MATH is MATH-Noetherian. It then follows that MATH is also MATH-Noetherian (since every MATH-submodule is also a MATH-submodule). |
math/0104261 | If MATH is not MATH-Noetherian, then by REF it is not finitely generated over MATH. Hence there is a strictly increasing sequence of proper MATH-submodules MATH with MATH. Let MATH. Then each MATH is compact, MATH, and MATH. Hence MATH. Furthermore, since MATH is a MATH-module, MATH for every MATH. Let MATH. Choose MATH such that MATH, and pick MATH. Then MATH for all MATH, so that MATH. Since MATH was arbitrary, we see that MATH is not expansive along MATH. This contradiction proves that MATH is MATH-Noetherian. |
math/0104261 | Let MATH denote the MATH-shift action on MATH, where MATH. For MATH put MATH. Then MATH is the closed shift-invariant subgroup of MATH given by MATH . If MATH and MATH, the duality pairing is given by the formula MATH . First suppose that MATH is expansive along MATH. REF shows that MATH is MATH-Noetherian. If MATH, choose MATH such that MATH for all MATH. Consider MATH together with the MATH-shift action MATH, and the point MATH defined by MATH. For every MATH we have MATH, and so MATH for all MATH. Fix MATH. Then MATH for every MATH. Define MATH by MATH. Then clearly MATH, and MATH for all MATH. Since MATH was arbitrary, this contradicts expansiveness of MATH along MATH. Hence MATH. Conversely, suppose that MATH is MATH-Noetherian. REF shows that there is a polynomial MATH in MATH of the form MATH, where MATH for all MATH. Now MATH is finitely generated over MATH, say by MATH, MATH, MATH. For MATH put MATH. Define MATH. For MATH let MATH. We will prove that if MATH and MATH for all MATH, then MATH, which will show that MATH is expansive. There are two cases to consider: CASE: MATH for all MATH, and REF MATH for some MATH. In REF , choose MATH so that MATH for all MATH. If MATH with MATH then MATH for MATH. Now MATH, so that MATH since MATH for all MATH by assumption. Hence MATH for all MATH with MATH. Repeating this argument shows that MATH if MATH, then if MATH, and so on, which proves that MATH. We now turn to REF . We will show that there is a point in MATH. Consider the NAME space MATH of all bounded complex-valued functions on MATH. For every MATH the operator MATH defined by MATH maps MATH to itself, and clearly MATH. For MATH define the operator MATH. Put MATH . Clearly MATH is closed and mapped to itself by MATH for every MATH. We claim that MATH is nontrivial. For define MATH by taking MATH to be the unique number in MATH for which MATH. Since MATH if follows that MATH for all MATH. Our size condition on MATH coupled with MATH shows that MATH for MATH and MATH. The same argument works for MATH, proving that MATH is a nonzero element of MATH. For each MATH let MATH be the restriction of MATH to MATH. Let MATH be the commutative NAME algebra of bounded operators on MATH generated by MATH. The theory of commutative NAME algebras shows that there is a (nonzero) complex homomorphism MATH, and that MATH for all MATH. Let MATH for all MATH. Then MATH is a homomorphism from the monoid MATH to the multiplicative monoid MATH. It follows that there is MATH such that MATH for all MATH. Also, MATH on MATH, so applying MATH shows that MATH. We claim that MATH for MATH. Since MATH, there is a MATH such that MATH. Suppose that MATH for all MATH with MATH. If MATH, then MATH, so that MATH is also zero on MATH, a contradiction. Hence there is a MATH with MATH and MATH. It then follows from multiplicativity of the MATH that MATH for all MATH. Hence MATH for all MATH, and so MATH. Finally, for all MATH we have that MATH so that MATH, proving that MATH. |
math/0104261 | The hypothesis shows that there is a neighborhood MATH of MATH in MATH such that MATH and MATH. These imply that MATH, so that MATH is expansive. |
math/0104261 | REF shows that MATH. MATH: There is a chain of MATH-submodules MATH such that MATH, where MATH is a prime ideal in MATH containing some MATH. Since MATH, we see that MATH is expansive along MATH. Put MATH. Then MATH, and MATH. Repeated application of REF shows successively that MATH, MATH, MATH, MATH are all expansive along MATH. MATH: Suppose that MATH is expansive along MATH. By REF , MATH is MATH-Noetherian, so that by REF we have that MATH is MATH-Noetherian for every MATH. Suppose that there is a MATH for which MATH. Fix MATH with MATH for some MATH. Consider MATH as a MATH-module using the action MATH defined by MATH for all MATH. We will construct a MATH-homomorphism MATH. To construct MATH, first note that MATH is MATH-Noetherian by assumption, hence finitely generated over MATH. Choose generators MATH, MATH, MATH, MATH, and define the surjective map MATH by MATH. Let MATH. Define MATH by MATH. We claim that the dimension of the complex vector space generated by MATH is strictly less than MATH. For if not, there are elements MATH, MATH, MATH in MATH such that MATH, MATH, MATH are linearly independent in MATH. Denote the MATH matrix MATH by MATH. Since each MATH, REF shows that MATH annihilates MATH, and in particular MATH. Thus MATH contradicting linear independence of the MATH. This proves that MATH generates a proper complex vector subspace MATH of MATH. Since MATH generates all of MATH, it follows that the corresponding quotient map MATH is nonzero. We can therefore compose MATH with a projection of MATH to a REF-dimensional subspace so that the composition is still a nonzero MATH-homomorphism. The result is the desired MATH. By construction, MATH. After multiplying (if necessary) by a constant we can assume that MATH and that the point MATH defined by MATH is not trivial. Since MATH is close to MATH for all MATH, the half-space MATH is not expansive. |
math/0104261 | REF proves one direction. For the other, suppose MATH is expansive along MATH. Then trivially MATH is expansive along MATH. Let MATH, so that MATH. Now MATH, so REF shows that MATH is expansive along MATH. |
math/0104261 | We could proceed along the lines of the previous examples, which is direct but technically complicated. However, there is a short proof using the notion of homoclinic point (which is treated in detail in REF). If MATH, then MATH is not expansive by REF , so that MATH and we are done. So we may suppose that MATH. We represent MATH as MATH, so that MATH is the shift-action on MATH. For MATH let MATH. A point MATH is homoclinic if MATH as MATH. Consider the function MATH on MATH defined by MATH. Then MATH does not vanish on MATH by our assumption on MATH. Let MATH be the NAME transform of MATH at MATH. Define MATH by setting MATH to be the reduction mod REF of MATH. By CITE, MATH, and MATH as MATH by the NAME Lemma. Hence MATH is a homoclinic point. Let MATH be arbitrary. Let MATH and let MATH. Choose MATH so that MATH for MATH. Pick MATH such that MATH, and put MATH. Then MATH for all MATH. Since MATH was arbitrary, MATH for every MATH. |
math/0104261 | Let MATH and choose MATH with MATH. The natural surjection MATH shows that, for every half-space MATH, if MATH is not MATH-Noetherian, then MATH is not MATH-Noetherian. Hence MATH, establishing the first equality in REF. For the second equality, observe that if MATH, then MATH. |
math/0104261 | The theory of NAME bases applies to polynomials rather than NAME polynomials. In order to apply it here, we will subdivide MATH into the MATH orthants corresponding to the signs of the entries, and use a universal NAME basis in each of these orthants. Let MATH. For each subset MATH define MATH if MATH and MATH if MATH. Let MATH, and put MATH, an ideal in MATH. For each MATH we will construct a finite set MATH, and show that MATH satisfies REF. For notational simplicity we give the construction for MATH, and indicate the modifications needed for general MATH. Here MATH. Choose a finite set MATH of generators for MATH over MATH. Let MATH and define MATH (for general MATH the last generator is MATH). Construct a universal NAME basis MATH for MATH. Define MATH by MATH and MATH (for general MATH let MATH). Clearly MATH. Define MATH. We show that if MATH and MATH, then there is a MATH-monic polynomial MATH in MATH if and only if some element in MATH is also MATH-monic. (For general vectors MATH, use the above construction for MATH.) First, multiply MATH by a monomial so that MATH, and also that the MATH-exposed vertex of MATH has the form MATH, where MATH. Hence MATH, where MATH . Therefore MATH. Define a term order on monomials in MATH by declaring that MATH if and only if MATH, and in case of equality that MATH, and in case of equality there, that MATH is lexicographically less than MATH. Then MATH by REF. Since MATH is a NAME basis for MATH, there is a MATH such that MATH divides MATH. Hence MATH for some MATH. Let MATH where MATH. We claim that MATH is a polynomial in MATH that is MATH-monic. This follows since every monomial in MATH that could give rise to a term in MATH strictly larger than MATH would have to already be greater than MATH with respect to MATH. |
math/0104261 | Suppose that MATH is generated over MATH by MATH elements. Let MATH, and suppose that MATH is MATH-Noetherian. By REF , there is a MATH-monic polynomial MATH. Then REF shows that MATH, and MATH is clearly MATH-monic. Hence another application of REF implies that MATH is MATH-Noetherian. Thus MATH. Conversely, suppose that MATH is MATH-Noetherian. Since MATH for every MATH, we see that MATH is MATH-Noetherian for every MATH. Hence MATH proving that MATH. For the second equality, observe that since MATH for every MATH, MATH . To prove the reverse inclusion, take MATH. Then there is a MATH such that MATH. Suppose that MATH for every MATH. Then there are polynomials MATH with MATH. Put MATH. Then MATH, and so some power MATH (to see this, either use a prime filtration of MATH, or observe that the radical of MATH is MATH). By REF we see that MATH. But this contradicts MATH. Hence there is a MATH such that MATH. |
math/0104261 | MATH : Suppose that MATH is finitely generated as an abelian group. Then MATH is trivially MATH-Noetherian for every MATH, so that MATH by REF . MATH : Our proof uses an algebraic analogue of the proof of REF. Since MATH, by REF for every MATH there is a polynomial of the form MATH that annihilates MATH and such that MATH for every MATH. This shows that there are MATH, MATH, and a neighborhood MATH in MATH such that if MATH with MATH and MATH, then MATH. The collection MATH is an open cover of MATH, so by compactness there is a finite subcover MATH. Put MATH and MATH. Let MATH be generated over MATH by MATH, MATH, MATH. For MATH let MATH denote the abelian group generated by MATH. We claim that MATH, so that MATH is finitely generated as an abelian group. We prove this by showing successively that MATH, so that MATH. Let MATH. Then MATH for some MATH. By our construction, there is a polynomial MATH that annihilates MATH, and such that MATH. Hence for MATH we see that MATH . This shows that MATH, and the reverse inclusion is trivial. The same argument applied to MATH shows that MATH, and so on, completing the proof. MATH : This equivalence is standard from duality. MATH : By assumption, MATH for every MATH. Hence MATH for every MATH by NAME 's NAME. There is a prime filtration MATH with MATH, where MATH is a prime ideal containing some MATH. Then the integer MATH annihilates MATH, so that MATH is a torsion abelian group. MATH : Suppose that MATH is a torsion abelian group. Let MATH. Then there is a MATH with MATH. Since MATH must also be a torsion abelian group, it follows that MATH. In particular, MATH contains a nonzero constant, so that MATH. Hence MATH. MATH : This equivalence is standard from duality. |
math/0104261 | Let MATH. It follows from CITE that if MATH is a MATH-plane, then MATH. A standard argument now shows that if MATH is a rational subspace of dimension MATH then MATH. Hence MATH. |
math/0104261 | Let MATH. Since the expansive set is open, there is a rank MATH lattice in MATH for which the restriction of MATH is expansive. By a simple change of variables, we may assume that this lattice is generated by the first MATH unit vectors. That is, the subaction dual to multiplication by MATH, MATH, MATH is expansive. We thus consider MATH as a MATH-module. We claim that the images of the variables MATH, MATH, MATH in MATH must satisfy a polynomial relation, and also that each MATH for MATH must be algebraic over MATH, MATH, MATH. For the first claim, observe that if the images of MATH, MATH, MATH in MATH do not satisfy any polynomial relation, then the natural map MATH is injective, so that MATH is a prime ideal associated to the MATH-module MATH. But by REF this contradicts the assumption that the restriction of MATH to the first MATH variables is expansive. If some MATH were not algebraic over MATH, MATH, MATH, then MATH would not be a NAME MATH-module, again contradicting the expansiveness assumption. It follows that MATH. Now let MATH, and assume first that MATH. Then any MATH monomials must satisfy two coprime irreducible polynomial relations, so that the entropy of the corresponding MATH-action is zero. Hence MATH. Now assume without loss of generality that the images of the variables MATH, MATH, MATH in MATH are algebraic over the images of MATH, MATH, MATH, and that MATH, MATH, MATH satisfy only one polynomial relation. If this relation is not a generalized cyclotomic polynomial, then the corresponding MATH-subaction has positive entropy CITE, so that MATH. So we may assume that this single polynomial relation is an irreducible generalized cyclotomic polynomial. After a suitable change of variables, this implies that MATH in MATH for some MATH. Since the original system is expansive, we may find among the variables MATH, MATH, MATH a variable MATH which is not a root of unity. Now the same argument may be applied to the set of variables MATH, MATH, MATH, MATH. Continuing, we either arrive at a MATH-subaction with positive entropy or a contradiction. We deduce that MATH, and so MATH. Finally, suppose MATH, and let MATH. Then MATH is a ring extension of MATH. By NAME normalization CITE we may choose variables so that MATH, MATH, MATH do not satisfy a polynomial relation, and the variables MATH, MATH, MATH are integral over MATH, MATH, MATH. This shows that there is an expansive, positive-entropy MATH-subaction. Hence MATH. |
math/0104261 | REF shows that MATH is closed. First suppose that MATH. Let MATH, and MATH be the image of MATH under the map MATH that reduces coefficients mod MATH. Then MATH. Hence MATH, so that MATH. Choose MATH. Then MATH is contained in MATH, which by CITE equals the MATH-skeleton of the spherical dual of the mod MATH . NAME polyhedron of MATH. Hence MATH has empty interior. Finally, suppose that MATH. Recall from REF the meaning of support and of MATH-exposed vertex. Let MATH be an arbitrary nonempty open set in MATH. Choose MATH minimal so that there is a MATH and MATH having a MATH-exposed vertex MATH with MATH. Note that MATH since MATH, so that MATH exists. We claim that MATH. If so, then REF shows that MATH contains a MATH for which MATH is MATH-Noetherian, and so MATH is nowhere dense. Suppose that MATH. Choose MATH and MATH such that MATH has a MATH-exposed vertex MATH with MATH. We may clearly assume that MATH is irreducible since any factor of MATH would also have a MATH-exposed vertex. Since MATH-exposure is an open condition and rational directions are dense, we may also assume that MATH is rational, that is, that there is a MATH with MATH. Adjusting MATH by a monomial if necessary, we may assume that MATH and that MATH, so that MATH. Next, we claim that there is a MATH with MATH. Equivalently, if MATH denotes MATH, then MATH. For suppose that MATH. Since MATH is irreducible and MATH has NAME dimension MATH, it follows that MATH is a chain of prime ideals in MATH. Let MATH denote the multiplicatively closed subset MATH of MATH. Since MATH, it follows that MATH is a chain of prime ideals in MATH. But MATH for a suitable monomial MATH, and MATH is a field, so that MATH, contradicting the existence of a chain of primes of length two. Hence MATH. Let MATH. Since MATH, then MATH. If all the coefficients of MATH are divisible by MATH, then primality of MATH shows that MATH is also in MATH. Hence we can find MATH not all of whose coefficients are divisible by MATH. Let MATH . Since MATH, MATH, and MATH, it follows that MATH, MATH, MATH, and MATH for every MATH. Hence there is a small perturbation MATH of MATH for which MATH has a MATH-exposed vertex MATH with MATH. This contradicts minimality of MATH, proving that MATH, and completing the proof. |
math/0104261 | The case MATH is the implication MATH in REF . The case of general MATH uses an entirely analogous adaptation of the proof of CITE. If MATH is rational, let MATH. Then MATH is a NAME ring. If MATH is generated as a group by MATH, then it is a finitely-generated MATH-module, hence NAME over MATH. Conversely, if MATH is NAME over MATH and MATH contains MATH, then trivially MATH is MATH-Noetherian as well, so that MATH is NAME along MATH. |
math/0104261 | MATH : The case MATH is exactly REF . We may therefore assume that MATH. By CITE, MATH is not expansive along MATH if and only if there is a MATH containing MATH such that MATH is not expansive along MATH. By REF , this occurs if and only if there is a MATH containing MATH. REF shows that this happens if and only if MATH for some MATH. Reversing the chain of equivalences, this time for MATH, shows that this occurs if and only if MATH is not expansive along MATH for some MATH. MATH : As in the proof of MATH , MATH is not expansive along MATH if and only if there is a MATH containing MATH. REF shows that this occurs if and only if either MATH is not MATH-Noetherian or MATH, where MATH is the outward unit normal for MATH. By definition, MATH is not NAME along MATH if and only if there is a half-space MATH containing MATH for which MATH is not MATH-Noetherian. If MATH, then MATH. Hence there is a MATH containing MATH with MATH if and only if MATH. |
math/0104261 | MATH is closed by REF . The case MATH is trivial. Suppose that MATH. Let MATH be MATH. By definition, MATH. Since MATH and each MATH is nowhere dense in MATH by REF , it follows that MATH is nowhere dense. We prove the general case by downward induction on MATH. Assume the result is established for MATH. Let MATH be an open set in MATH. Then there is an open set MATH in MATH such that every MATH contains a MATH. By the induction hypothesis, MATH contains a rational MATH-plane MATH along which MATH is NAME. By REF , MATH is NAME over MATH. We are now in the codimension one situation of the first part of the proof. The associated primes of MATH as a MATH-module are MATH for MATH. Then MATH is a subring of MATH, hence the hypothesis on NAME dimension is satisfied for this situation. Thus there is a MATH along which MATH is NAME. This proves that MATH is nowhere dense in MATH. |
math/0104261 | This can be proven exactly as in CITE. |
math/0104261 | We show that MATH is locally constant on MATH, hence constant on MATH by connectedness. Let MATH and MATH be an expansive constant for MATH. Recall REF of coding for subsets of MATH. Let MATH be the open ball of radius MATH in MATH. For MATH and MATH put MATH. It follows from CITE that there is a compact neighborhood MATH of MATH in MATH and positive numbers MATH such that for every MATH and every MATH we have that MATH codes MATH. Suppose that MATH and MATH. Then there is a MATH such that MATH for every MATH. Hence by coding, this inequality also holds for every MATH. Let MATH. A simple compactness argument using expansiveness of MATH shows that there is a MATH such that MATH implies that MATH. Choose MATH large enough so that if MATH and MATH, then MATH. It follows that MATH for every MATH with MATH. This proves that MATH, and so MATH. Interchanging the roles of MATH and MATH gives the reverse inclusion. |
math/0104261 | By REF , MATH has positive entropy if and only if MATH, and by REF we have MATH for all MATH, establishing the positive entropy portion. The proof for completely positive entropy is similar, using density rather than nontriviality of the homoclinic group. Finally, completely positive entropy for an algebraic MATH-action is equivalent to NAME (see CITE or CITE). |
math/0104264 | CASE: The volume part. Let MATH and choose MATH such that MATH . Let, for every MATH, MATH . By REF the functions MATH converge to MATH in MATH. Moreover, MATH . Indeed, by REF where MATH. Then from REF we get MATH, where we used also REF . This shows that MATH in MATH. On the other hand from the continuity of the trace with respect to the intermediate topology it follows that MATH . Then by REF we obtain MATH . Now REF with MATH implies that MATH . CASE: The surface part. As in CITE, it can be proved that MATH where MATH for all MATH, MATH, MATH, and where MATH . Using again REF we obtain MATH hence REF , concluding thus the proof. |
math/0104281 | If MATH is degenerate then we get MATH for some MATH, MATH for MATH. Then MATH. Conversely if MATH is nondegenerate we get a surjective natural map of vector bundles over MATH . Indeed, by our definition, MATH is surjective if and only if MATH is nondegenerate. We construct a vector bundle MATH over MATH whose dual MATH is the kernel of MATH so that we have the exact sequence MATH . After tensoring by MATH and taking cohomology we get MATH and we need to prove MATH . Let MATH. Since MATH and MATH from REF it follows that MATH . Hence, by taking the MATH-th wedge power of the dual of the sequence REF , and using NAME formula to calculate the cohomology as in CITE, the result follows. In order to prove the irreducibility of the subvariety MATH of degenerate matrices it is sufficient to construct the incidence variety MATH is a vector bundle over MATH, hence it is irreducible and its projection over MATH is MATH. |
math/0104281 | We first observe that the convolution of boundary format matrices MATH and MATH is also boundary format, then by REF its hyperdeterminant is the usual determinant of MATH . We put MATH . Since MATH and MATH are nondegenerate matrices, they define vector bundles MATH and MATH respectively over MATH and MATH which verify the following exact sequences MATH . MATH . Moreover the matrix MATH defines the sheaf morphism MATH . If the maps MATH are the natural projections and MATH, we can construct the following commutative diagram: MATH . The surjectivity of maps MATH and MATH induce the surjectivity of MATH and MATH, thus MATH is nondegenerate and MATH is a vector bundle Moreover, since MATH then MATH that is, by REF MATH as we wanted. |
math/0104283 | REF implies REF. Let MATH denote the multi-degree of MATH for MATH. Clearly, MATH is a basis of MATH as a left MATH - module. Thus, given MATH, the homogeneous element MATH has a unique representation as homogeneous standard left polynomial in MATH with coefficients in MATH. Thus, MATH where the MATH's are in MATH. Choose, for each MATH, an element MATH such that MATH. Let MATH denote the MATH matrix whose rows are MATH. Write the equality REF as MATH . Therefore, we can prove by induction on MATH that MATH where MATH. To deduce that MATH is a basis for MATH we only need to check the linear independence. Given a relation MATH we proceed by induction on MATH. The relation REF can be written as MATH which, in MATH, gives MATH . As the monomials MATH are MATH - linearly independent, we have that MATH for MATH. The remaining coefficients are zero by induction in view of REF. Let MATH and MATH. Since MATH and MATH we get MATH has degree MATH, that is, MATH. Write MATH. Then MATH . Since MATH is left noetherian and MATH is finitely generated as a left MATH - module, we have that MATH is a noetherian left MATH - module. Thus, we deduce from REF, in conjunction with REF, that MATH for some MATH, where MATH is a finite subset of MATH such that MATH for every MATH. On the other hand, for MATH, we have MATH which entails, by REF, MATH where MATH is a finite subset of MATH such that MATH for every MATH. Let MATH be the admissible order on MATH defined by MATH . Since MATH for every MATH, the relations REF can be written as MATH and MATH which gives REF implies REF. First, notice that, by hypothesis, the relations REF are satisfied. Let MATH where MATH and MATH. Clearly MATH is a finite subset of MATH whose maximum with respect to MATH is MATH. By CITE (see also CITE and CITE), there is MATH such that MATH for every MATH. This implies that the relations REF can be written as MATH and MATH . By CITE, the family MATH where MATH is the left MATH - module generated by the set MATH, is a MATH - filtration on MATH. Since MATH has no zero component, it follows that MATH is finitely generated as a left MATH - module for every MATH. For each MATH, define MATH, which is a finitely generated left MATH - module. A straightforward verification shows that MATH is a filtration on MATH. Clearly, MATH for every MATH. Finally, let MATH for MATH. By REF, MATH for every MATH and MATH. Moreover, since the monomials MATH are MATH - linearly independent, it follows that MATH is a left MATH - basis for MATH. It follows from CITE that MATH . Finally, REF implies REF obviously. |
math/0104283 | The first statement is a consequence of CITE. If MATH is NAME, then, by CITE, MATH is NAME. The result follows now from CITE. |
math/0104283 | Let MATH be the filtration on MATH given by REF with MATH. Since MATH for every MATH and the filtration MATH is standard, we get that MATH for every MATH and every MATH. Therefore, MATH is a MATH - bounded extension of MATH in the sense of CITE. Here, MATH with MATH, MATH. Let MATH be the admissible order defined by MATH . Write MATH. By CITE, these vector subspaces form a MATH - filtration for MATH. Let MATH. Since MATH it follows that MATH is a filtration on MATH. Moreover, the inclusion MATH is a strict filtered morphism, hence MATH can be viewed as a subalgebra of MATH. Therefore, MATH. Here, MATH denotes the graded automorphism induced by the homonymous filtered automorphism of MATH. Since MATH is a finitely presented and noetherian algebra, we see that MATH enjoys the same properties. Thus, the filtration MATH satisfies the hypotheses of CITE. Now every finitely generated left MATH - module is endowed with a filtration such that MATH is finitely generated. By CITE, MATH. In particular, MATH. On the other hand, from the proof of CITE we obtain that MATH. If we assume that MATH is NAME, then MATH whence MATH is NAME too. Lastly, if MATH is NAME, then MATH satisfies the hypotheses of CITE, which implies that it is NAME. Since the filtration MATH is finite-dimensional, we obtain that MATH is NAME. Thus, MATH is NAME by the foregoing argument. |
math/0104283 | Clearly, MATH is an iterated NAME extension of a NAME algebra, whence its global homological dimension is finite by CITE and CITE. On the other hand, MATH is NAME and NAME (see, for example, CITE). By CITE, MATH satisfies the NAME condition. Since the multiplicative set generated by MATH consists of monomials, which are local normal elements, we have, by CITE, that our algebra MATH is NAME. |
math/0104283 | Accordingly with CITE, MATH is endowed with a MATH - filtration MATH for some MATH and a lexicographical order MATH in such a way that the multi-graded associated algebra MATH for a certain multiplicatively anti-symmetric matrix MATH. By REF , MATH is NAME and NAME. Moreover, MATH, a commutative NAME polynomial ring. Filter MATH with the standard filtration obtained by giving degree MATH to MATH (MATH). Then MATH is a factor algebra of the commutative polynomial ring in MATH variables with coefficients in MATH. In particular, it is finitely presented and noetherian. Therefore, the hypotheses of REF are fulfilled and, hence, MATH is NAME and NAME. |
math/0104289 | Denote by MATH the value of MATH. Compare the transition functions MATH with the function MATH as MATH varies over the complex conjugate of MATH running over MATH. The effect of the former is this: MATH . Suppose MATH is a local expression of the transition function MATH as a power series (about the origin) in MATH. Then, the power series expressions for MATH comes by applying complex conjugation to the coefficients of MATH, and so the resulting function is analytic in MATH. |
math/0104289 | The map MATH induces MATH. Action of MATH preserves MATH. Thus, it induces a homomorphism of MATH into MATH where MATH goes to the automorphism MATH, MATH (as in REF). Conclude: Modulo inner automorphisms of MATH, MATH acts trivially, producing the desired homomorphism MATH . Now consider the explicit formulas. Most of the calculation is in REF: MATH cycles entries of MATH back REF, and conjugates all entries by the first entry's inverse. So MATH leaves entries of MATH untouched except for conjugating them by MATH . Also, MATH cycles the entries of MATH forward REF. The new first entry is the old REFth entry conjugated by the inverse of the product of the old first three entries. Thus, MATH and MATH act the same. Add that MATH maps to REF in MATH to see MATH . Let MATH. Extending the calculation above gives the next list: MATH . Add the relation MATH to deduce, in order, these relations in MATH: MATH . So, the image of MATH in MATH is REF. The kernel of MATH contains elements of MATH inducing inner automorphisms commuting with conjugation by MATH. Since MATH generates conjugations commuting with MATH, MATH generates the kernel of MATH. Generators of MATH act on MATH: Respectively, MATH induce conjugation by MATH. These conjugations on MATH form a free group. So MATH is a free group on these generators. |
math/0104289 | The image of MATH, MATH and MATH in MATH identify with classical generators of MATH REF . Each MATH has image an involution MATH in MATH. The product-one condition, MATH implies these elements generate a NAME REF-group MATH REF . The group MATH is MATH (in its regular representation). To check how MATH acts on the MATH, note the element of order MATH (image of MATH) acts nontrivially on MATH (by conjugation). Here is the effect (in MATH) of conjugating MATH by MATH: MATH . Since MATH, the action is nontrivial and MATH is MATH. To see these relations don't kill MATH, verify the normalizing action of MATH on MATH gives MATH acting trivially, MATH. Use conjugation by MATH as in REF. Finally, to see it is nonsplit, consider the element MATH whose image in MATH is MATH (image of MATH) of order REF. A splitting of MATH would give a lift of this to an element of order REF, contrary to REF giving no such element in MATH. |
math/0104289 | For MATH an integer, MATH . The minimal MATH with MATH is MATH. Further, the minimal MATH with MATH divides any other integer with this property. So MATH and if MATH is odd, MATH. From the above, if the orbit of MATH does not have length MATH, it has length MATH. Use the notation around REF. The expressions MATH and MATH are tautologies. If MATH is odd, then MATH. Assume this equals MATH, which is true if and only if MATH. The expression MATH and MATH are equivalent. Conclude MATH. So long as the order of MATH is not REF, this shows REF holds. If, however, MATH, then MATH, contrary to hypothesis. This reversible argument shows the converse: MATH follows from REF. This concludes the proof. |
math/0104289 | It only remains to show the last statement. REF gives this because REF shows MATH acts trivially on the NAME class. A classical argument, however, suffices in this case by contrast to REF. Given MATH, consider the MATH cover MATH. If MATH is isomorphic to MATH over MATH, this gives a regular MATH realization in the inner NAME class of MATH. Such an isomorphism occurs if and only if MATH has a MATH point. The image from origin in MATH is such a MATH point. |
math/0104289 | The opening statement follows because the quaternion group of order REF is only group of order REF with a NAME REF-group as quotient having all its elements lifting to have order REF. The last statement is a variant on that using REF . |
math/0104289 | With no loss assume one MATH orbit and two blocks, with orbit listings as MATH. As, however, there is one orbit, for some MATH, MATH for some MATH. This contradicts there being two blocks. |
math/0104289 | A version of this is in CITE. Consider any NAME cover MATH (over some algebraic closure MATH of MATH) in the equivalence class of MATH. Then, choose any MATH over MATH unramified in MATH. From the moduli property, for MATH, there is an isomorphism MATH commuting with the maps to MATH. Compose such a map with the unique automorphism of the NAME cover assuring MATH takes MATH to MATH. Apply NAME 's cocycle condition to MATH for an equivalent pair MATH over MATH CITE. We may, however, lose the automorphisms: MATH defines a cover in MATH where MATH is the regular representation of MATH. The cover, however, is special, for it has a rational point MATH over MATH. The proof works with a general curve replacing MATH. With MATH replacing MATH and MATH, this construction works uniformly to give a fine moduli space of geometrically NAME covers with a point over MATH. |
math/0104289 | Let MATH be the (normal) subgroup of MATH generated by its MATH elements. If MATH is a proper subgroup of MATH, then MATH is a nontrivial MATH-group, and any MATH-group has a MATH quotient. Conversely, given MATH, the kernel of MATH contains all the MATH elements of MATH. |
math/0104289 | The first part is a subset of REF . Assume MATH is centerless and MATH-perfect. We inductively show MATH also has these properties for all MATH. CITE shows MATH is centerless if the following hold. CASE: MATH has no center. CASE: MATH has no MATH subquotient of NAME type MATH. The module in (REF is distinct from MATH. It comes from a nontrivial representation of MATH of form: MATH. The map MATH is a homomorphism of MATH into the MATH. By hypothesis this doesn't exist. That leaves showing MATH has no quotient isomorphic to MATH, assuming MATH is centerless and has no such quotient. Suppose MATH is surjective with kernel MATH. Consider the map from MATH to MATH induced by the canonical map MATH. This is a NAME cover. So, MATH is not onto MATH. Then, MATH has image an index MATH normal subgroup of MATH. This is contrary to our assumptions. Finally, since MATH is centerless, MATH has a center if and only if MATH stabilizes some nontrivial element of MATH; if and only if MATH appears to the far left in the NAME display of MATH). |
math/0104289 | Suppose MATH factors through MATH. Then a point MATH has image MATH. The (geometric) decomposition group for MATH in the cover MATH is a subgroup of MATH. It must contain the image MATH since MATH is a subgroup of the fundamental group of MATH. The argument is reversible. |
math/0104289 | NAME class is given by MATH with MATH a faithful permutation representation. The equivalence depends on some subgroup of MATH, containing MATH and normalizing it. A cover in the NAME class is of degree MATH and has MATH the natural permutation representation associated to it. When, it is an inner class, we also attach an isomorphism between the group of the cover and the group MATH. The monodromy groups of the covers MATH and MATH are exactly the same. If MATH are the branch points for MATH, then MATH are the branch points of MATH, with MATH having the conjugacy class MATH attached to it. This shows MATH preserves NAME classes. The action of MATH mapping on s-equivalence classes of covers in a NAME class is continuous. Given MATH, map MATH to the s- equivalence class of MATH. Since MATH is connected, the MATH orbit of MATH lies in one connected component of the NAME space. The orbit contains MATH; the component is that of MATH. |
math/0104289 | The argument of REF shows how to identify the group of the cover MATH with the action of MATH on the points of MATH modulo a NAME REF-group. |
math/0104289 | The hypotheses are for the situation of a REF-branch point real cover of the sphere. The element MATH will be independent of the value of MATH. Since MATH is the image of MATH with MATH, REF says MATH. The formula for computing complex conjugation is the special case of REF where MATH and all branch points are real. The effect of complex conjugation on the MATH-s take them to their inverse. This induces the effect of taking MATH to its inverse MATH. The effect, however, of this image element on reduced NAME classes is an element of order REF. Since the permutation effect of complex conjugation is to conjugate by an involution MATH, this gives the formula MATH. Similarly, MATH. |
math/0104289 | Suppose MATH contains no involutions. Let MATH be two involutions with the same conjugation effect on MATH. Then MATH is an involution that centralizes MATH, and so it is trivial. Now assume nothing about MATH, that MATH and MATH have the same effect on MATH, both fix REF and G is transitive. Then, MATH fixes REF. From transitivity, for any MATH, there is MATH with MATH. Conclude: MATH, and MATH is the identify. The statement on MATH comes from identifying it with MATH. List the right cosets MATH of MATH in MATH. Elements of MATH that permute these by left multiplication on these cosets are in MATH. Those acting trivially are in MATH. Left action commutes with the right action of MATH, thus producing elements in MATH. CITE has complete details. |
math/0104289 | Once we know that MATH is NAME and NAME is a subgroup of MATH, the centralizer statement follows from NAME. CASE: That identification is in CITE. |
math/0104289 | Suppose MATH gets killed in the MATH quotient of all moduli spaces of MATH branch point covers. This happens if for every MATH, there exists MATH fixing MATH and inducing on MATH the same effect (modulo inner automorphisms) as does MATH. REF identifies the group MATH as the group of such MATH, and MATH as the quotient of MATH by a NAME REF-group, MATH. |
math/0104289 | The field of definition statement follows from MATH defined over MATH. Then, the divisor hypothesis produces an odd degree MATH divisor on MATH. Since MATH has genus REF and an odd degree MATH divisor, it is MATH isomorphic to MATH. If MATH is a finite field, a homogeneous space for any NAME variety always has a rational point (the NAME group of a finite field being trivial CITE). Let MATH. Let MATH and MATH be in the MATH orbit of MATH. Consider a given set of branch points MATH and classical generators MATH of MATH. Assume the image of MATH in MATH is different from REF or REF. Denote by MATH and MATH the covers from the homomorphisms sending MATH respectively to MATH and to MATH. Suppose for some MATH there exists an isomorphism MATH for which MATH. As in REF, MATH lies in a NAME REF-group identified with MATH. The b-fine hypotheses implies MATH is unique. So, the subgroup of MATH fixing MATH extends to a faithful action of MATH on MATH over MATH. Quotient action gives the total space representing w-equivalence classes of covers for points of MATH. The fine moduli hypothesis say MATH fixed on MATH over MATH or REF extends to MATH without fixed points over that value of MATH. For MATH, the b-fine property follows from REF. |
math/0104289 | The linear system MATH attached to MATH gives an isomorphism of the genus REF curve MATH with MATH. Take MATH to be the composition of MATH and this isomorphism. Apply each MATH to MATH to get MATH. The isomorphism here is that given by replacing MATH by MATH. Since MATH is a fine moduli space, there is a unique MATH and MATH satisfying MATH. The cocycle condition follows from the uniqueness conditions. Call this data a cocycle of NAME class covers attached to MATH. It is standard the cocycle is trivial if and only if MATH has a MATH point. Since MATH has genus REF, this is equivalent to MATH having a degree one MATH divisor. Since the canonical class on MATH is a degree -REF class over MATH, this is equivalent to MATH having an odd degree MATH divisor. Now suppose we have such a cocycle of NAME class covers. This produces MATH as MATH. We have only to check MATH is well-defined, independent of MATH. The cocycle condition guarantees this. Finally, if there is a MATH point on MATH, this gives MATH lying over MATH. Composing the cover MATH with elements of MATH gives the correspondence between the MATH points of MATH and the MATH points of MATH over MATH. |
math/0104289 | The action of MATH on MATH REF induces an action on any closed subgroup of MATH. For MATH denote its fixed field by MATH. Let MATH, so it defines a homomorphism MATH. Then, MATH is a projective system of points over MATH of MATH if and only if MATH normalizes the kernel of MATH, and the induced action of MATH on the quotient is trivial. Since levels of the Modular Tower are fine moduli spaces, this produces the desired sequence of NAME covers over MATH. The argument reverses. Further, MATH acts on MATH, a characteristic subgroup of MATH (on which MATH acts by hypothesis). The image of MATH in MATH is into MATH. This induces a MATH action on MATH. |
math/0104289 | A MATH-Sylow of MATH contains a MATH-Sylow of MATH. So, the latter is profree. CITE characterizes MATH as the minimal cover of MATH with pro-free MATH-Sylow. So, there is a natural map MATH commuting with the map to MATH. As MATH is a MATH-Frattini cover of MATH the map is surjective. Since the natural map MATH has a pro-MATH group as kernel, the natural map MATH produces MATH commuting with the projections to MATH. The composition MATH (commuting with the projections to MATH) is an endomorphism of MATH. The image of MATH is a closed subgroup of MATH mapping surjectively to MATH. So, from the NAME property, MATH is onto. An onto endomorphism of finitely generated profinite groups is an isomorphism CITE. In particular, MATH is an injection. The characteristic quotients have maps between them induced by MATH, and so MATH injects into MATH, inducing an injection of MATH. If (for MATH), MATH and MATH have the same dimension, they are isomorphic. As these groups characterize MATH and MATH, that implies they are equal. This gives an isomorphism of MATH and MATH in the special case. |
math/0104289 | By replacing MATH by MATH, with MATH its NAME subgroup, form MATH. The universal MATH-Frattini cover of MATH is the same as that of MATH. So, NAME. REF shows MATH occurring at some level for MATH implies it occurs at infinitely many levels for MATH. So, for our question, with no loss take MATH to assume MATH is an elementary MATH-group and a MATH module. Then, with MATH the projective indecomposable for MATH, consider it as a MATH (therefore a MATH) module. Since MATH, NAME 's criterion says it is the projective indecomposable for MATH CITE. Its most natural display might by MATH. In its NAME display MATH appears to the far right. Write the next NAME layer as MATH, with the MATH-s irreducible MATH REF modules. Forming MATH (* is the MATH dual) returns MATH. So the dual of the MATH-th socle layer of MATH is the MATH-th NAME layer of MATH. Conclude that to the far left there is MATH (REFst socle layer) with MATH immediately to the right of that (REFnd socle layer). Now we apply that MATH is an elementary MATH-group. The MATH module MATH is the group ring MATH. Let MATH be a basis for MATH and identify MATH in the vector space MATH as the space spanned by MATH. The action of MATH on MATH preserves this space modulo the second power of the augmentation ideal of MATH. Conclude: MATH as MATH (or MATH) modules. As in REF use the notation of CITE. So, the kernel MATH of MATH starts with MATH. Now consider a minimal projective MATH mapping surjectively to MATH. NAME 's criterion again implies the MATH module MATH, being projective for MATH, is also projective for MATH. The NAME layers of MATH are just the NAME layers of MATH tensored over MATH with MATH modules. NAME: a NAME layer of MATH tensored with MATH is semisimple. Therefore MATH. Then, MATH modulo projective summands (as in CITE, called MATH there). The REFnd socle layer of MATH is MATH with one copy of MATH removed from the end of MATH. The module MATH has exactly one appearance of MATH for each absolutely irreducible factor in MATH (NAME 's Theorem; since MATH acts trivially on MATH, the same is true of MATH). Thus, MATH appears in the NAME display of MATH as a MATH module unless MATH, and MATH is absolutely irreducible. Now suppose MATH. Instead of looking at MATH, look at MATH as a MATH module. By our hypotheses, MATH is not a cyclic module. Therefore, a computation of the rank of MATH comes directly from NAME 's formula for ranks of subgroups of pro-free groups see REF. The rank of MATH, as MATH increases, exceeds the degree of any irreducible MATH module. Replace MATH with a suitable MATH to revert to the case MATH is not absolutely irreducible. This completes showing the appearance of MATH. |
math/0104289 | The hypothesis says the morphism MATH, a priori a MATH module homomorphism, is actually a MATH module homomorphism. Here is why. Suppose MATH is a proper subgroup of MATH mapping surjectively to MATH. Then, the pullback MATH of MATH in MATH is a proper subgroup of the pullback of MATH in MATH. Further, MATH maps surjectively to MATH, contrary to MATH being a NAME cover. Conclude that MATH is also a NAME cover. So, there is a natural map from MATH inducing a surjective MATH module homomorphism MATH. Since MATH is the REFst characteristic quotient of MATH, universal for covers with elementary MATH-group kernel, there is a MATH module splitting of MATH. NAME 's Theorem CITE says, since MATH, a MATH splitting of this MATH map gives a MATH splitting. This is contrary to MATH being an indecomposable MATH module unless this is an isomorphism (CITE or CITE). Suppose a MATH-Sylow MATH of MATH has this property: Either MATH for each MATH. CITE shows MATH is an isomorphism (so both have dimension REF) guaranteeing REF. This holds for MATH and its REF-Sylows since they are distinguished by which integer from MATH each element in REF-Sylow fixes. |
math/0104289 | Suppose MATH is pro-free pro-MATH group on MATH generators. Let MATH be a surjective homomorphism, with MATH any (finite) MATH group. NAME 's construction gives explicit generators of the kernel of MATH CITE. Apply this with MATH and MATH, the NAME REF-group and MATH. Let MATH be a generator of MATH. For MATH and MATH generators of MATH let MATH act on MATH by mapping MATH to MATH. Use MATH as coset representatives for MATH in MATH. Form the set MATH of elements in MATH having the form MATH or MATH with MATH. Toss from MATH those that equal REF. Now consider the images of MATH and MATH on MATH. This produces MATH and MATH. Consider MATH on MATH. Recall: Modulo MATH any two elements in MATH commute. Apply this to get MATH . Apply MATH again to get MATH. The action of MATH on the MATH-s is the same as the action on the six cosets of an element of order REF. Denote a commutator of two elements MATH by MATH. Modulo MATH there are relations among the MATH-s: MATH; and MATH. So, the product of the MATH-s is REF. The proof follows from associating a MATH in MATH with a dihedral in MATH as in the statement of the proposition. |
math/0104289 | The action of MATH preserves cosets. So, orbits for conjugation by MATH are clear if MATH, MATH and MATH each consist of one MATH orbit. This follows from triple transitivity of MATH in the standard representation. The exact sequence REF comes from writing the right cosets of MATH in MATH. Use MATH, MATH, MATH, MATH, MATH and MATH. Then, MATH has generators MATH and MATH. Images of MATH, MATH and MATH generate MATH. Consider MATH of order REF. To be explicit assume MATH lifts MATH. Then, MATH fixes MATH, MATH and MATH, so it acts like MATH on the cosets. The centralizer of MATH in MATH is therefore MATH . An element in MATH is in MATH if and only if MATH. Note: MATH fixes MATH. Sum both with MATH to get two elements in MATH. The observation that MATH is a special case of REF . Now consider the sequence REF. Write MATH as MATH with MATH. First: MATH is a MATH module. As MATH sends MATH to MATH, it preserves the leftmost layer of the NAME display. In the quotient MATH, MATH and MATH generate a copy of MATH. Then, MATH fixes MATH and MATH generates a copy of MATH. The module MATH has NAME display MATH. If there is a subquotient of MATH having a NAME display MATH, it must be MATH. Clearly this is invariant under MATH. The action of MATH is to take MATH to MATH. Since MATH and MATH differ by an element of MATH, this shows MATH is invariant under MATH. |
math/0104289 | We review the NAME algebra setup used in CITE. Let MATH be the NAME algebra on MATH with generators MATH subject to relations MATH . In the NAME algebra, write MATH. Then, MATH and MATH. The collection of MATH under multiplication generate a subgroup MATH. Characterization: It is the central nonsplit extension of MATH whose restriction to transpositions splits, and whose restriction to products of two disjoint transpositions is nontrivial CITE. The map MATH appears from MATH. So, MATH if MATH. That MATH is nontrivial if MATH shows from lifts of certain elements of order REF. Example: MATH lifts to have order REF: MATH . Of course the order of a lift is conjugacy class invariant. Similarly, with MATH, MATH . By induction, the result is MATH: MATH has order MATH. |
math/0104289 | Everything has an explanation already except the automatic appearance of MATH in MATH for finitely many MATH given its appearance at level MATH. Apply REF with MATH replacing MATH. Then, use that the universal MATH-Frattini cover of both MATH and MATH is the same as that of MATH. So, the characteristic quotients for the universal MATH-Frattini of MATH (in place of MATH) are cofinal in the projective system MATH. Therefore the simple modules (including copies of the identity representation) appearing in the analog for MATH of the modules MATH also appear as simple modules for a cofinal collection from MATH. |
math/0104289 | With no loss assume MATH. The exists MATH braiding MATH to MATH in the same NAME class with MATH and MATH. From CITE, MATH. (More generally, MATH where MATH is the big braid invariant of a NAME class.) Apply the argument of REF to MATH to reduce computing MATH to computing MATH. |
math/0104289 | That there is only one MATH with MATH, MATH and MATH follows from the data in REF. The computation MATH comes from MATH in REF. |
math/0104289 | Write MATH with MATH and MATH a conjugate of MATH not in MATH. Choose MATH in one of the orbits MATH or MATH (for conjugation by MATH; REF) in MATH. If we choose MATH to be the orbit of MATH (in MATH), then conjugate by MATH to see MATH is also in MATH. The other MATH orbit of REF, represented by MATH, doesn't give an element in the NAME class: The product MATH is a REF-cycle. Calculate MATH by considering MATH. Then, MATH. Use NAME. REF, then NAME. REF to see MATH equals MATH . One last application of NAME. REF now shows MATH. |
math/0104289 | Condition REF is the product one condition for MATH. Since MATH is an irreducible MATH module, given MATH, there is a unique solution for MATH. From the same argument, MATH if and only if the projection of MATH onto MATH has no kernel; otherwise the kernel is all of MATH since it is a nontrivial MATH invariant submodule of MATH. Conclude, MATH is a splitting of MATH into MATH. From NAME, it is conjugate to the canonical copy of MATH in MATH. Condition REF is exactly the computation for that. |
math/0104289 | The extending action of MATH to MATH is a special case of REF applied to MATH. The extending group MATH is the universal MATH-Frattini of MATH. If the action on MATH is trivial, then the trivial extension of MATH to MATH is an extending action. Consider the case MATH. We have the natural short exact sequence MATH. Form the (group) fiber product MATH . By the basic property of the MATH-Frattini cover, MATH is the minimal MATH-projective cover of MATH. Since MATH has MATH kernel, MATH is the minimal MATH-projective cover of MATH. So, by REF , it is MATH. |
math/0104289 | Once we know the rank of MATH exceeds REF, then so does the rank of MATH on MATH for all MATH. Since MATH is a MATH normal subgroup, there is a splitting of the pullback of MATH in MATH. That identifies MATH as the maximal normal MATH subgroup of MATH for each MATH. Given the NAME result for MATH therefore gives it for all MATH. |
math/0104289 | The operator MATH acts as an involution: MATH and MATH commutes with conjugation by MATH. So MATH acts trivially on MATH. The condition for MATH to have fine moduli is that the centralizer of MATH in MATH is trivial REF . |
math/0104289 | Recall the cross ratio of distinct points MATH: MATH. The basics are in CITE. Four points in complex conjugate pairs (or on the real line) lie on a circle and the cross ratio is real. The cross-ratio is invariant under a transform of the points MATH by MATH. Since there is a MATH that takes two complex conjugate pairs of points to four points in the reals, with no loss assume MATH has either two or four real points in its support. For these cases apply MATH to assume MATH and MATH. Then, MATH. In the former case MATH runs over the unit circle (excluding MATH) and in the latter case over all real numbers (excluding MATH, REF and MATH). The MATH value corresponding to MATH is MATH with CITE. (Classically this is without the MATH. We chose it so the ramified MATH-values are REF, MATH.) For MATH the connected range of MATH includes large positive values and is bounded away from REF. So the range of MATH for real MATH is MATH. For MATH in the unit circle (minus REF), the range of MATH includes both sides of REF. Also, for MATH close to REF, the numerator of MATH is positive and bounded, while the denominator is approximately MATH. Therefore the range is the interval MATH. |
math/0104289 | Suppose MATH is a local uniformizing parameter for MATH in a neighborhood of MATH. The argument for MATH is the same, so we do only the former case. In local analytic coordinates over MATH, choose MATH so MATH and there is a parametrization of the neighborhood of MATH using power series in MATH with real coefficients. If MATH is odd, then real points around MATH map one-one to real points around MATH. If MATH is even, then real points around MATH map two-one to the positive number side of MATH (no points falling on the negative side of MATH). That interprets the lemma's statement in local coordinates. |
math/0104289 | REF gives the component containing both H-M and near H-M reps. The statement on MATH is from REF . Nontriviality of MATH is from the discussion prior to REF , an inductive consequence of REF and that MATH. The complex conjugation operator for a near H-M rep. in this case gives conjugation by an element whose lift to MATH has order REF. From this, REF shows a near H-M MATH (regarding it in MATH) has no cover realizing it over MATH. If, however, MATH corresponds to an H-M rep., then regarding it as the inner class of a MATH cover, it has a trivial complex conjugation operator. So, we have MATH with group MATH over MATH and geometrically MATH (MATH unramified) with field of moduli MATH, and the group of MATH is MATH. Apply REF to MATH; the resulting cover MATH has degree REF. It is therefore NAME, and MATH produces the MATH realization over MATH. |
math/0104289 | From the above, MATH conjugates both MATH and MATH to their inverses. The unique element doing this is MATH. So there are three real points above MATH. Check: MATH and MATH. Now we show cover points account for all MATH points on MATH. The fixed points of MATH point to three covers in REF labeled MATH. Each attaches (by the MATH operator) to a distinct cusp over MATH, of respective widths REF. That, however, comes from choosing classical generators of MATH for some MATH and MATH. It doesn't show what the description of branch cycles will be for a set of paths on MATH supporting a complex conjugation operator of REF . Actual covers MATH having these real structures come from two types of MATH-s using the paths REF. For MATH with real entries: MATH has complex operator MATH; MATH has MATH and MATH has MATH. |
math/0104289 | Use MATH in REF to see that MATH contains a REF-cycle, MATH, so it must primitive. It also contains a REF- cycle MATH. A well-known argument says a primitive subgroup of MATH containing a REF-cycle is MATH: MATH. Apply CITE. The gist: As the branch points of the cover are rational, the arithmetic monodromy group contains the character field of the conjugacy class of any element of form REF . Elements that are products of distinct odd disjoint cycle lengths, form two conjugacy classes in MATH. The outer automorphism of MATH (conjugation by MATH) permutes these two conjugacy classes. For MATH, the degree REF cover breaks into a chain of degree REF and degree REF covers. The largest possible group for the geometric closure is MATH and for the arithmetic closure it is MATH. To complete the proof only requires showing the kernel of the geometric closure to MATH contains one of the factors of MATH. Since MATH generates such a factor, we are done. |
math/0104289 | Since MATH is conjugation of MATH by MATH, if MATH is centerless, and MATH fixes MATH, then MATH. So, MATH is an H-M rep. Squares of elements in MATH act trivially on an H-M rep. This shows MATH acts on the set MATH as a quotient of MATH. Assume MATH is an H-M rep. and MATH. Then MATH and MATH: Both are involutions. Similarly, if MATH, then MATH and MATH is cyclic (in particular abelian). Also, MATH implies MATH, contradicting MATH is centerless. Assume MATH. As MATH REF , for MATH, MATH . So, MATH depends only on the MATH orbit of MATH. If MATH, then MATH. This gives the statement on faithful action at level MATH. Now assume MATH and MATH is MATH-perfect and centerless, so these hypotheses apply at all levels REF . With no loss, on the general statement on a MATH orbit containing MATH, assume MATH. Suppose for MATH, MATH for some MATH. From the above, MATH is not the identity. As MATH acts trivially on MATH, MATH with MATH REF . This contradicts MATH being centerless. Now assume MATH. We show MATH is faithful on any MATH orbit MATH in MATH. Any orbit at level REF has elements lying over any element of the unique orbit at level REF. Anything in MATH above an H-M rep. MATH looks like MATH with MATH, and some MATH generating MATH. We handled when MATH is in the centralizer of MATH, so assume it is not. Apply the previous argument when MATH with MATH. Conclude: If MATH, then MATH lifts an element of MATH having order REF. From REF , MATH. Compute the first two entries of MATH to be MATH. The remainder of the argument uses REF . Let MATH be the generator of the centralizer of MATH. Multiply MATH if necessary by MATH to assume MATH. Then, MATH. Conclude MATH contrary to our assumption. For other elements in MATH, the argument is similar, though no easier. |
math/0104289 | We do all computations on inner NAME classes MATH. Suppose MATH and MATH or MATH (or with MATH replacing MATH) and MATH is even. As MATH commutes with even powers of MATH, this gives MATH: MATH. Suppose MATH with MATH odd. Apply MATH to both sides. From REF conclude MATH . So, MATH. Apply MATH to both sides of MATH. This gives MATH . Inductively, this shows MATH. So, MATH and MATH, or MATH. |
math/0104289 | That MATH has order REF is a special case of REF . An element of order REF in MATH acts on MATH by right multiplying cosets of a MATH in MATH REF . So MATH fixes one nontrivial element. It must be MATH. That MATH satisfies the product-one condition is exactly that MATH. Since MATH has no center, what a conjugation on MATH does to the REFst and REFth elements determines it. Therefore, the MATH orbit of MATH has length REF times the order of MATH: The orbit has length REF. |
math/0104289 | That REF holds is a simple check. The order of the product MATH is from REF . Apply REF to an H-M rep. cover MATH with respect to the MATH operator for two pairs of complex conjugate branch points to compute the effect of complex conjugation over MATH. It is given by MATH equal the identity. So all points on MATH over MATH are real. For a near H-M rep. the effect of complex conjugation is given by MATH as in the statement of the proposition. So, MATH moves all points over MATH. There are no real points on MATH. Now assume MATH and MATH satisfies MATH (automatically associated with MATH for MATH and MATH as complex conjugate pairs). Write MATH: MATH . From REF , assume MATH. Apply MATH to see MATH . The product-one condition for MATH is equivalent to MATH. Then, MATH gives four conditions according to the entries of MATH, with the last two automatic. The second gives MATH showing MATH REF : the first condition is automatic. The product-one condition says MATH: MATH generates the centralizer of MATH. |
math/0104289 | Suppose MATH and MATH for some MATH and MATH. Reduce all expressions modulo MATH to conclude MATH fixes MATH. This is a contradiction. The same argument works for MATH. By inspection REF shows REF holds for MATH, and so at all levels in this MATH . Modular Tower. If MATH or REF follows from the opening statement of REF . The only other possibility is that MATH is an H-M rep., but MATH above it is not. REF says if MATH, then MATH for MATH over MATH. That completes the proof. |
math/0104289 | The references explain most of this lemma. Given MATH over MATH the ramification index of the respective cusps is exactly MATH which by previous comments is MATH as in REF. It is well-known that if a MATH is a covering of projective nonsingular curves with the lower curve of genus REF, then the genus of MATH is REF if and only if the cover is unramified. REF expresses the NAME formula applies to the relative curve covering MATH, when the latter has genus REF. |
math/0104289 | Apply REF to MATH. From REF , one MATH orbit on MATH contains all H-M and near H-M reps. Call this orbit MATH. REF says there are exactly eight MATH orbits with the following properties: CASE: REF divides MATH; CASE: they are in the MATH orbit of a near H-M rep. REF says the MATH cusp type of an H-M or near H-M rep. is MATH. So, a cusp of MATH lying over an H-M rep. of MATH has ramification index REF. Let MATH be the component of MATH corresponding to MATH. Conclude: Each cusp (eight total) of MATH with MATH divisible by REF has ramification index REF over the cusp below them at level REF. Together they contribute MATH to the right side of REF. Similarly, consider cusps of MATH over cusps of MATH with MATH equal REF. REF gives a similar conclusion about cusps at level REF with middle product REF. Cusps above them on MATH have ramification index REF (MATH does not shorten them). They also contribute MATH to the right side of REF. Now apply REF for the contribution of cusps on MATH over MATH applied to level REF H-M reps. This contributes MATH for the shift of H-M reps. and MATH for the others to the right side of REF. So, the right side of REF is REF. The expression MATH gives MATH. Let MATH be the collection of cusps at level REF not in MATH. First assume they all lie in one MATH component for the computation of the genus of this orbit. From REF there are eight cusps in MATH with width REF, and four with width REF. Similarly, there eight cusps in MATH with width REF, and four with width REF. Finally, REF gives REF type MATH orbits in MATH. To complete the calculation above for MATH, list respective contributions to the right of REF: Type MATH contribute REF; type MATH contribute REF; type MATH contribute MATH; type MATH contribute REF and type MATH contribute MATH. So, MATH gives MATH. Since it is true at level REF, it is also true at level REF that every MATH not in MATH is in the MATH orbit of an element MATH with MATH. From REF (and its notation) any cusp of MATH (respectively, MATH) connects to each cusp of MATH (respectively, MATH). So, to prove all elements of MATH lie in one MATH, it suffices to connect some cusp of MATH (respectively, MATH) to some cusp of MATH (respectively, MATH). This shows a component MATH containing a cusp of MATH has degree sixteen over a cusp at level REF with MATH. So MATH has degree sixteen everywhere. For the degree over every cusp to be REF forces including all cusps in MATH. To join something in MATH to something in MATH consider a MATH type MATH orbit in MATH. A representative for such an orbit has MATH. REF shows each type MATH orbit in MATH is such an element. This concludes the proof of the corollary. |
math/0104289 | There is an orbit shortening for MATH only if there is a MATH (lifting MATH) with MATH. This implies MATH is invariant under MATH. Given one such lift MATH, REF shows you get all others by multiplying this MATH by MATH as MATH runs over MATH. Given a lift MATH giving MATH, multiplying MATH by MATH produces MATH. Producing one such MATH is the final step. REF proof gives explicit action of MATH on the cosets of the MATH. Respectively: MATH . Consider a near H-M rep. as in REF: MATH with MATH centralizing MATH. For reasons coming up, we take MATH, a lift of MATH and MATH, a lift of MATH. Let MATH be the centralizer of MATH. Then, compute the shift of the complement of MATH to get MATH with MATH and MATH and MATH centralizing MATH. Since MATH, the desired shortening amounts to showing MATH and (using that by REF) MATH centralizes MATH. Explicit computations are reassuring: MATH and so MATH. Similarly, MATH, so MATH. Thus, MATH and MATH. The other check works as easily. |
math/0104289 | Each MATH with MATH is the shift of an H-M rep. MATH. Further, applying MATH to such a MATH gives another such element. By inspection MATH is stable on this set. So, according to REF , the shift applied to H-M reps. contributes a total of two length REF orbits for MATH on reduced classes. Now, consider the case MATH has MATH: MATH. With no loss, MATH and MATH. Further, assume MATH or MATH. Fix MATH. There are MATH choices of MATH, all lifts of MATH modulo conjugation by the centralizer of MATH. Then, there are MATH lifts of MATH, now determining MATH. So, there are MATH total MATH with MATH having MATH. Since MATH acts faithfully REF , there are MATH elements of MATH in MATH orbits of length REF or REF. REF completes showing all H-M and near H-M reps. fall in one MATH orbit. |
math/0104289 | We show the conclusion of REF first. Consider the projective system MATH in the Modular Tower for MATH. Let MATH be the second and REFrd entries of MATH. Suppose MATH exists so MATH. By assumption, MATH is a lift of MATH to MATH. Thus, REF says MATH. Inductively apply this for the conclusion of the lemma. For MATH the value of MATH is either REF or REF. If MATH, then MATH is an H-M rep. If MATH, then MATH, and REF-tuple MATH satisfies the product-one and genus REF (MATH implies MATH) conditions of REF . Conclude that MATH. Equivalently: If MATH are the lifts of MATH to MATH, then MATH. Similarly, if MATH, then MATH, the genus REF condition holds, and the conclusion from REF is that MATH has order REF. From REF , if MATH, then the length of the orbit of MATH on MATH is twice MATH with MATH written as a perturbation of an H-M rep. MATH, MATH. |
math/0104289 | The statement on MATH comes to noting the product-one condition is equivalent to MATH. Apply MATH to both sides to see MATH fixes MATH. REF shows MATH is the unique nontrivial element of MATH in MATH. REF shows MATH braids MATH to MATH. Apply MATH to MATH to get MATH . Use MATH to rewrite this as MATH with MATH. Apply the first part of this lemma with MATH: An element of MATH takes MATH to the H-M representative MATH. As in REF , MATH centralizes MATH. Use that MATH and its conjugates don't centralize MATH REF to see MATH and MATH give two H-M reps. at level REF in one MATH orbit lying over the same level REF H-M rep. We already know how to braid from MATH to MATH. Since, there are only two H-M reps. over MATH, it suffices that we have this braid from one to the other. So, REF gives the braiding between all the width two cusps. The last expression relating H-M reps. to complements of near H-M reps. comes from writing MATH as MATH to express this in standard generators of MATH. The relation between H-M reps. MATH and near H-M reps. given by MATH works at any level (with MATH). REF has already established that near H-M reps. have the complex conjugation properties stated here. The element MATH has the form MATH. Lift it to MATH by lifting MATH and MATH to (respectively) MATH and MATH of order REF in MATH. Then form MATH. Let MATH. The characterizing property of MATH implies MATH has order MATH or MATH has order REF. So, any lift of MATH to MATH has order REF. Since MATH is just a conjugate of MATH, it applies to MATH as well. |
math/0104289 | The first sentence follows if MATH is one-one. Consider MATH as an element in the group ring MATH. It suffices to show MATH is invertible on MATH. Use notation from the proof of REF : MATH and MATH on MATH cosets act by MATH, MATH and MATH. Since MATH, compute MATH to be MATH. Then MATH, in the group ring, acts on the six cosets generating MATH as follows: MATH or as MATH. Recall MATH in MATH. Check MATH on a basis for MATH: MATH . The range vectors form a basis. So, MATH is invertible on MATH. The remainder of the proof follows from the setup to this lemma. |
math/0104289 | Suppose MATH is an H-M or near H-M rep., or a complement of one of these. Then, MATH has MATH six if only if MATH. So, for values of MATH, there are eight distinct elements MATH with MATH six. We show no two are on the same MATH orbit. First consider level REF from REF and MATH and MATH. Conjugation by MATH takes one of the two MATH orbits to the other. So, it suffices to show MATH, MATH are in distinct MATH orbits. That requires showing MATH and MATH are in distinct MATH orbits. This is saying the complements shift to distinct MATH orbits. The computations are similar for H-M and near H-M reps., so we do the former only. For simplicity write MATH as MATH. With MATH and MATH, we show MATH and MATH are in distinct MATH orbits. By reducing modulo MATH, the only possibility is MATH is in the same reduced equivalence class as MATH. Check easily this isn't so. The level one lifting invariant of REF has value MATH on MATH applied to the MATH orbits of elements MATH with MATH six described above. From REF , such an element MATH is in the MATH orbit of an H-M or near H-M rep. REF also says that orbits MATH in the statement of the lemma have lifting invariant MATH. Since the lifting invariant is a MATH invariant, they cannot intersect the set from MATH applied to the MATH orbit of MATH. |
math/0104289 | Let MATH be an H-M rep. The relationship MATH already says MATH takes the complement of an H-M rep. to a near H-M rep. Now use MATH in MATH. Let MATH be MATH, and apply MATH to it, to get MATH is the shift of a near H-M rep. This shows why a type MATH orbit containing MATH applied to a complement of an H-M rep. also contains MATH applied to a near H-M rep. The remaining conclusions are similar. |
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