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math/0103064 | CASE: This is clear from the definition of the action of MATH on MATH. CASE: That MATH follows from the fact that MATH. That MATH follows from REF . CASE: Let MATH and MATH. If MATH, then MATH. Thus, MATH. If MATH, then MATH. CASE: Follows from REF . CASE: We have MATH by the definition of MATH from REF. CASE: Using REF , we have MATH . CASE: From REF , MATH and MATH are linear combinations respectively of elements of form MATH and MATH where MATH. By REF , the statement to be proved then reduces to the same statement for MATH and MATH. But this follows from REF , and the associativity of action for unary polynomials REF . CASE: Follows from REF . CASE: Follows from REF , and REF , and the fact that by definition, MATH is the zero element of MATH, because it is the image under the pointed MATH-overalgebra homomorphism MATH of MATH. CASE: Follows from REF , and REF . CASE: Follows from the fact that MATH. |
math/0103064 | We have MATH . |
math/0103064 | Uses the fact that MATH is generated by the MATH. For MATH, we have MATH and MATH . Thus, the statement is true in that case. Now let MATH be a MATH-ary operation symbol, and let the statement be true for a MATH-tuple MATH of MATH-ary terms. That is, if for each MATH we write MATH for MATH, we will be assuming MATH for all MATH. We have MATH . |
math/0103064 | Let MATH and MATH be left MATH-modules and MATH a homomorphism. To prove that MATH is functorial, we must show that if MATH is a MATH-ary operation symbol, MATH, MATH, and MATH, then MATH . Let us write MATH for MATH. We have MATH . Now let MATH be a left MATH-module. If MATH is an identity of MATH, then by REF , we have MATH for each MATH. Thus, MATH is totally in MATH. |
math/0103064 | Let MATH be a MATH-module totally in MATH. We will show that for each MATH-ary operation symbol MATH, MATH, for all MATH and MATH. Let MATH. We have MATH . Thus, MATH. Now, let MATH be a left MATH-module. We must show that for all MATH, MATH, for all MATH, and for all MATH, we have the action of MATH on MATH the same in MATH as in MATH. It suffices to show this for MATH of the form MATH, since such elements generate MATH as a group, and the two actions are bilinear. Let MATH, and denote MATH by MATH. In MATH, we have MATH . However, this is MATH in MATH, by REF . Thus, MATH. |
math/0103064 | CASE: It suffices to show that given MATH, MATH, we have MATH. But this follows from the fact that MATH is a clone homomorphism. CASE: MATH is generated by the constant polynomials MATH, and MATH. Thus we can prove that MATH for all MATH, by showing that the generators satisfy this condition and that the subset of MATH satisfying the condition is closed under the basic operations. We have MATH for all MATH, and MATH. Finally, given MATH, and MATH, we have MATH. For, MATH by REF , so that we have MATH . CASE: Obviously, MATH is one-one and onto on objects. Let MATH, and MATH, and consider MATH. Let MATH. We have MATH . The range of MATH therefore contains all the generators of MATH, and thus is onto on all hom-sets. |
math/0103064 | Let MATH be a MATH-ary polynomial such that MATH. We will show that MATH. First, let MATH be a MATH-tuple of elements of MATH, and MATH a MATH-ary term, such that MATH. We have MATH . Also, by REF, we have for each MATH, MATH where in the second term, MATH appears as the MATH argument. Finally, we have MATH because the MATH-component of the natural map from MATH to MATH sends this element to MATH where MATH stands for MATH, and this is MATH. Thus, we have MATH where MATH for each MATH. Since the abelian group generators of MATH belong to the abelian group MATH, we have MATH. To show MATH, we will show that for each MATH, the elements of MATH (where MATH is the MATH-set defined in REF that generates MATH) are contained in MATH, and also that for each MATH, the MATH-tuple of abelian groups MATH is closed under action by elements of MATH, that is, that the MATH form a MATH-submodule of MATH. We start with the fact that MATH for each MATH. For, MATH. Next, let MATH be a MATH-ary operation symbol, MATH a MATH-tuple of elements of MATH, and MATH. We need to show that MATH belongs to MATH. Now, each MATH is an element of MATH, which is a subset of MATH, an algebra generated by MATH and elements of MATH. Thus, there are a MATH-tuple MATH of elements of MATH, and MATH-ary terms MATH, so that for each MATH we have MATH . For each MATH, define the MATH-ary polynomial MATH and let MATH. Then we have MATH proving that MATH. Now we must show that homomorphisms of the form MATH, where MATH is MATH-ary and MATH, send generators of MATH to elements of MATH. Consider the generator MATH, where MATH is MATH-ary and MATH. We have MATH where in the summation, MATH appears in the MATH position, and MATH is the MATH-ary polynomial MATH . Thus, MATH. |
math/0103064 | We have MATH . |
math/0103075 | Given MATH take any injective resolution MATH in MATH, and let MATH (compare CITE). |
math/0103075 | Let MATH be a cofinal inverse system in the filter MATH with all the left ideals MATH finitely resolved. Say MATH for some direct system MATH of MATH-modules. Since the left module MATH is finitely resolved we get MATH . Hence for any MATH we have MATH . |
math/0103075 | If MATH is an acyclic complex of MATH-flasque modules then MATH is also acyclic. Now use CITE. |
math/0103075 | Since MATH is flat over MATH, any injective MATH-module is also an injective MATH-module. |
math/0103075 | Let MATH be the filter of left ideals associated with MATH. Then MATH . Because MATH is flat over MATH each MATH is an injective MATH-module, and hence it is MATH-flasque. By REF the direct limit of MATH-flasque modules is MATH-flasque. |
math/0103075 | Take an injective resolution MATH in MATH. Then MATH, and MATH. According to REF , MATH is a complex of MATH-flasque MATH modules, so by REF , MATH. |
math/0103075 | It suffices (by ``way-out" reasons, see CITE) to consider a single module MATH. But by the stability assumption, the minimal injective resolution MATH is in MATH, so MATH. |
math/0103075 | For any bimodule MATH write MATH . It suffices by symmetry to prove that MATH. Since MATH, REF says that there is a functorial isomorphism MATH. Finally by REF there is a functorial isomorphism MATH. |
math/0103075 | If MATH is an acyclic complex of MATH-flasque modules then from the exact sequence of complexes MATH we see that MATH is also acyclic. Thus we can define MATH when MATH and MATH is a flasque resolution (compare proof of REF ). |
math/0103075 | Pick a MATH-flasque resolution MATH in MATH. The decreasing filtration MATH is bounded in the sense of CITE, that is . MATH and MATH for some MATH. Hence by CITE we get a convergent spectral sequence MATH . Now MATH and MATH. If MATH is a homomorphism between MATH-flasque complexes then there is a map between the two spectral sequences; and if MATH is a quasi-isomorphism then the two spectral sequences are isomorphic. |
math/0103075 | Choose an injective resolution MATH in MATH. Then MATH is a MATH-flasque resolution in MATH, the filtered complexes MATH and MATH coincide, and the spectral sequence defines both MATH and MATH. |
math/0103075 | Write MATH and MATH for the filtrations of MATH and MATH respectively, and apply REF . |
math/0103075 | This is really the implication MATH. We shall explain the minor modification needed in the proof to make it apply to our situation. Also we shall sketch the main ideas of the proof using our notation, so the interested reader can find it easier to consult the rather lengthy proof in CITE. The result in CITE refers to the abelian category MATH of sheaves of abelian groups on a topological space MATH. The space MATH has a filtration MATH by closed subsets, inducing a filtration MATH of MATH, with MATH. With this notation the proof involves homological algebra only, hence it applies to MATH as well. Here is the sketch. Let us abbreviate MATH. Define MATH to be the truncation in MATH. One shows by descending induction on MATH that there are (noncanonical) isomorphisms MATH such that the diagrams MATH commute. The horizontal arrows are the canonical ones. The starting point is that for large enough MATH, MATH. For such MATH one shows that MATH if MATH. Hence there is an isomorphism MATH. Also one shows that MATH if MATH, so that MATH. Since the modules MATH and MATH are canonically isomorphic in this case, we get an isomorphism REF for MATH. In the inductive step, depicted in REF , we have two canonical triangles (in which the morphisms MATH and MATH have degree MATH), canonical isomorphisms MATH and MATH (arising from the assumption (MATH)) and an isomorphism MATH that's already been constructed. The square on the left commutes because REF commutes. By definition of the Cousin complex it follows that MATH. Since MATH, REF implies that MATH . Therefore MATH, and hence, by the axioms of triangulated categories, there is an isomorphism MATH making diagram in REF commutative. Note that REF for MATH commutes too so the induction continues. |
math/0103075 | Invoke the theorem with MATH instead of MATH, and use REF . |
math/0103075 | Choose an injective resolution MATH in MATH. Denote by MATH the filtered complex with filtration MATH, and by MATH the filtered complex with filtration MATH. By the proof of REF , the homomorphism of filtered complexes MATH induces an isomorphism on the MATH pages of the spectral of the sequences from REF . Similarly for MATH. |
math/0103075 | CASE: Clear, since MATH and the same for MATH. CASE: Since MATH is a bounded below complex of injectives we have an isomorphism MATH . The purity implies that MATH and hence we get an isomorphism MATH induced by MATH. Finally given a morphism MATH in MATH we have MATH. |
math/0103075 | The first item is a slight variation of CITE and CITE. In REF the two actions of MATH on MATH correspond to the left action of MATH on MATH (and on MATH), and the right action of MATH on MATH. Since MATH these actions commute. Likewise in REF is CITE. |
math/0103075 | CASE: Choose a quasi-isomorphism MATH where MATH is a bounded above complex of finite free MATH-modules. Then the homomorphism of complexes MATH is bijective. CASE: With MATH as above we get a free resolution MATH as MATH-modules, and MATH is bijective. |
math/0103075 | CASE: This is proved in a special case (when MATH is commutative and MATH for a prime ideal MATH) in the course of the proof of CITE; but the same proof works here too. Among other things one gets that MATH in MATH and MATH in REF We consider MATH as a left (respectively, right) MATH-module via the outside (respectively, inside) action. Since MATH is noetherian and MATH is flat, by REF we obtain an isomorphism MATH in MATH. Now MATH in MATH. Finally using REF we get MATH . |
math/0103075 | First assume MATH, a polynomial algebra. Then the bimodule MATH is a dualizing complex. Because MATH and MATH for MATH it follows that the dualizing complex MATH is rigid. Next take any affine algebra MATH. Choose a finite homomorphism MATH. Let MATH be an injective resolution of the module MATH in MATH and define MATH. So MATH consists of central MATH-bimodules, and MATH. According to the calculations in the proof of CITE, MATH is a rigid dualizing complex. Finally suppose MATH is NAME and equidimensional of dimension MATH. Choose a noether normalization, that is a finite (and necessarily injective) homomorphism MATH, compare CITE. According to CITE, MATH is a projective MATH-module. Hence MATH is a rigid dualizing complex. If moreover MATH is NAME then the bimodule MATH is also a dualizing complex, and by the uniqueness of dualizing complexes over commutative algebras (compare CITE) we find that MATH must be an invertible bimodule. |
math/0103075 | Because of REF and the uniqueness of rigid dualizing complexes we may assume MATH is a complex of central MATH-bimodules. Now proceed as in the proof of CITE. |
math/0103075 | See CITE, noting that the morphism MATH constructed there satisfies axioms of rigid trace, as can be seen using the calculations done in the proof of CITE. |
math/0103075 | We may use the proof of CITE. |
math/0103075 | Since MATH consists of injective MATH-modules, MATH. So MATH. By the purity assumption MATH, so MATH . We see that MATH . The equality with MATH is proved symmetrically. |
math/0103075 | If we decide to forget the MATH-module structure of MATH, we may use the minimal injective resolution MATH of MATH as MATH-module to compute MATH. By the purity assumption MATH . We conclude that MATH as MATH-modules, and so MATH is an injective MATH-module, MATH-pure of dimension MATH. According to REF we see that the hypotheses of REF hold with MATH, MATH and MATH. This tells us that MATH as complexes of bimodules. By the previous paragraph applied to MATH instead of MATH, MATH is an injective MATH-module, MATH-pure of dimension MATH. REF says that REF holds here. We deduce the existence of an isomorphism MATH in MATH. According to REF there is some invertible element MATH such that MATH is multiplication by MATH. The isomorphism MATH has the desired property. |
math/0103075 | CASE: This is true because the kernel of this map is an essential submodule. CASE: The first equality is clear. The second follows from REF because factoring out MATH-torsion the complex MATH has zero coboundary maps. CASE: By assumption there is a nonzero left ideal MATH of MATH contained in MATH. Let MATH. Replacing MATH by MATH, we may assume MATH. Since MATH and MATH are non-torsion MATH-modules and MATH is finite, REF says there is a nonzero map MATH. We claim that MATH is nonzero in MATH. Otherwise there is a map MATH such that MATH. Let MATH. Then MATH is essential in MATH and hence MATH is a torsion MATH-module. Now MATH is a quotient of MATH, so it is a nonzero torsion MATH-module. This contradicts the fact that any nonzero submodule of MATH must be a torsion-free MATH-module. Therefore we proved that MATH is nonzero in MATH. Now let MATH be a regular element of MATH. Since MATH, MATH is a torsion MATH-module. Hence MATH is not contained in the kernel of MATH. This implies that MATH is nonzero. By the claim we proved in the last paragraph, we see that MATH is nonzero in MATH. So MATH is non-torsion in MATH. |
math/0103075 | Let MATH be the minimal injective resolution of MATH as complex of MATH-modules. First we will show that each MATH is essentially pure of MATH, meaning that MATH contains an essential submodule that's pure of MATH. It suffices to show that if MATH is a MATH-critical submodule of MATH, then MATH. The critical module MATH is uniform. Since MATH is FBN the injective hull of MATH is MATH for some prime ideal MATH. Replacing MATH by a nonzero submodule we can assume MATH is a left ideal of MATH, so it is a torsion-free MATH-module. By REF , MATH is a non-torsion MATH-module. In particular MATH and hence MATH. By REF we get MATH. From the weakly bifinite hypothesis, MATH is noetherian on both sides. Hence, by the symmetry of MATH, we have MATH. According to CITE we have MATH. We conclude that MATH. Again by REF we get MATH. Next we show that the MATH is a constant on the cliques of MATH. If there is a link MATH, then there is a nonzero MATH-bimodule MATH that is a subquotient of MATH and is torsion free on both sides. By REF and the symmetry of MATH we have MATH . The FBN ring MATH satisfies the second layer condition CITE. We know that MATH is constant on cliques. It follows from CITE that an indecomposable injective MATH-module has pure MATH (compare CITE). So the minimal injective resolution MATH is pure. All the above works also for the minimal injective resolution of MATH as a complex of MATH-modules. |
math/0103075 | Take a quasi-isomorphism MATH in MATH with each MATH injective over MATH, and a quasi-isomorphism MATH in MATH, where each MATH is injective over MATH and over MATH. Write MATH and MATH. Using REF we get a commutative diagram in MATH with the bottom row consisting of morphisms in MATH. The homomorphism MATH is actually bijective. And since MATH, MATH is an isomorphism, and hence so is MATH. The isomorphism we want is MATH. |
math/0103075 | CASE: According to CITE MATH in MATH. Therefore the complex MATH is an injective resolution of MATH in MATH. Choose elements MATH which generate MATH as a MATH-module. This gives rise to a surjection MATH of MATH-bimodules, and hence to an inclusion MATH. So MATH is MATH-pure of dimension MATH as a MATH-module. By CITE, the dualizing complex MATH is NAME, and MATH for any MATH-module MATH. Thus MATH is MATH-pure as MATH-module. We conclude that the injective resolution MATH is MATH-pure. But one easily sees that a pure injective resolution must be minimal. Symmetrically all the above applies to the right resolution MATH of MATH. CASE: Applying the functor MATH to the isomorphism MATH in MATH, and using the fact that MATH is a pure injective complex of MATH-modules, we obtain MATH in MATH. Thus in particular MATH is a complex of MATH-bimodules. By symmetry also MATH as complexes of bimodules. Denote by MATH and likewise MATH. We get filtrations MATH and MATH. Since MATH, REF tells us that MATH functorially for MATH. In particular MATH functorially for finite MATH-modules MATH. Passing to direct limits (using REF ) this becomes true for all MATH-modules. Hence if MATH is a MATH-flasque MATH-module, it is also MATH-flasque. By REF it follows that MATH functorially for MATH. CASE: Next we analyze the morphism MATH. By CITE, MATH so MATH for some (unique) MATH. We shall prove that MATH. If we forget the MATH-module structure, then MATH . From REF we get that there is a bijection MATH induced by MATH, and the inverse is induced by MATH. Hence we obtain MATH . This implies that MATH. CASE: By REF , if we apply the functor MATH to the homomorphism of complexes MATH we get a quasi-isomorphism MATH. But these are minimal injective complexes of MATH-modules, so it must actually be an isomorphism of complexes. By symmetry also MATH is an isomorphism of complexes. |
math/0103075 | Write MATH and MATH. By CITE we get a unique isomorphism MATH in MATH that's a rigid trace. According to REF above, MATH satisfies MATH, so it too is a rigid trace. |
math/0103075 | By REF we get an isomorphism of MATH-modules MATH for every MATH. This implies that MATH is annihilated by MATH on the right too, and hence MATH. By symmetry there is equality. |
math/0103075 | By CITE the morphism MATH is a rigid trace. According to REF the residue complex MATH exists, and MATH. |
math/0103075 | Recall that the first NAME algebra is MATH. According to CITE, MATH is a rigid NAME dualizing complex over MATH, and MATH. The ring of fractions MATH is a division ring, and the global dimension of MATH is MATH. Therefore the minimal injective resolution of MATH in MATH is MATH and MATH. We see that MATH is pure of MATH. Since MATH is a torsion MATH-module we get MATH; but since there are no MATH-modules of MATH it follows that MATH is pure of MATH. So MATH has a pure injective resolution on the left. The same is true on the right too. We see that MATH has pure minimal injective resolutions on both sides, so MATH is a residue complex. Moreover MATH and MATH. |
math/0103075 | We may assume MATH is not commutative, so MATH is generated by MATH with MATH central and MATH. By CITE and CITE the complex MATH is a rigid NAME dualizing complex, and MATH. Consider a minimal injective resolution MATH of MATH in MATH. For any MATH consider the ideal MATH. The localizing subcategory MATH is stable (compare REF ). We get a direct sum decomposition of MATH-modules indexed by MATH: MATH . For any MATH, MATH is the rigid dualizing complex of MATH, and MATH is its minimal injective resolution as complex of left modules. Since the algebra MATH is isomorphic to either the commutative polynomial algebra REF or to the first NAME algebra, we see that MATH is pure of MATH, for MATH. Fix MATH and MATH. Introduce a filtration MATH on MATH by MATH. Then for MATH multiplication by MATH is a bijection MATH. It follows that MATH is pure of MATH. Therefore also MATH is pure of MATH. The direct sum complement MATH is a MATH-module, where MATH. In fact MATH is a minimal injective resolution of MATH in MATH. But MATH is isomorphic the first NAME algebra over the field MATH. Therefore MATH is pure of MATH, and MATH is pure of MATH. We conclude that each MATH is pure of MATH. By symmetry the same is true on the right too. So MATH is a residue complex over MATH. |
math/0103075 | By the NAME Theorem CITE and CITE we may find MATH, MATH such that MATH is an NAME algebra over its center MATH. The commutative prime MATH-algebra MATH is affine, and hence by CITE there is a nonzero element MATH such that the localization MATH is regular. By NAME 's Theorem CITE the fraction fields coincide: MATH. Because MATH is affine we may find MATH (the product of the denominators of a finite set of MATH-algebra generators of MATH) such that MATH. Hence MATH is also regular, affine over MATH, and MATH is NAME with center MATH. |
math/0103075 | CASE: By CITE the algebra MATH is noetherian. Therefore according to REF the complex MATH is a rigid dualizing complex over MATH. Moreover, MATH in MATH. By CITE we have MATH, and hence MATH (NAME dimension). According to REF the rigid dualizing complex of MATH is MATH with MATH a projective MATH-module of rank MATH. From REF we see that MATH is a rigid dualizing complex over MATH. Finally the reduced trace MATH induces a bimodule isomorphism MATH. Therefore MATH in MATH. CASE: Follows from REF . |
math/0103075 | Let MATH be a noetherian connected filtration of MATH. Then MATH is a noetherian connected graded PI MATH-algebra. By CITE MATH has an NAME dualizing complex MATH, and MATH (NAME dimension) on the categories MATH and MATH. Since MATH is symmetric (see CITE), so is MATH. Now take a bimodule MATH that's a subquotient of MATH. Then MATH admits a two-sided good filtration MATH (that is, MATH is a finite module over MATH on both sides), and by CITE we get MATH as bimodules. Hence this bimodule is finite on both sides. We conclude that MATH is weakly bifinite. |
math/0103075 | CASE: Is immediate from REF . CASE: Follows from REF . |
math/0103075 | Apply REF to the ideals MATH where MATH, using REF . |
math/0103075 | CASE: Let MATH, and define NAME filters MATH and MATH like in REF . According to CITE these two filters are equal. Hence (compare REF ) for each MATH there is an exact sequence MATH . By symmetry there is an exact sequence MATH . But by REF , MATH, which implies MATH. Finally use the fact that MATH. CASE: This is proved just like CITE (compare proof of REF ). |
math/0103075 | This proof was communicated to us by NAME. Choose orthogonal idempotents MATH lifting the central idempotents in MATH, so that MATH . We have MATH and likewise on the right, so each element of MATH comes from some element in MATH. But MATH. We see that the canonical homomorphism MATH is surjective. Obviously MATH. On the other hand MATH so MATH. Thus MATH. Finally the isomorphism is obtained by the decomposition MATH . |
math/0103075 | Within the proof of REF it is shown that MATH is constant on cliques in MATH. |
math/0103075 | Since MATH is pure of MATH it is a torsion-free MATH-module, and it follows that MATH. On the other hand for MATH, MATH is pure of MATH, so it is a torsion MATH-module and MATH. We see that MATH. But by REF we get MATH. |
math/0103075 | CASE: Let MATH be an indecomposable injective module with associated prime MATH, where MATH is a clique in the MATH-skeleton of MATH. Since MATH satisfies the second layer condition, we get MATH and also MATH. It follows that MATH is bijective. Therefore for any other clique MATH in the MATH-skeleton of MATH we must have MATH. Because MATH is an injective module, pure of dimension MATH, we get the left module decomposition MATH. By REF and symmetry this is a bimodule decomposition. CASE: Clearly MATH is an injective MATH-bimodule, and MATH is essential. Write MATH; then MATH is a semi-simple (left and right) MATH-module. CASE: This is because the injective hull of MATH as MATH-module is MATH. CASE: Assume by contraposition that MATH as bimodules, with MATH. Then the socle MATH of MATH (as left or right MATH-module) also decomposes into MATH with MATH and MATH, MATH. Take MATH such that there is a second layer link MATH. Recall that this means there is a bimodule surjection MATH with MATH a nonzero torsion-free module over MATH and MATH. Then replacing MATH with MATH we retain the link, only now MATH and MATH as bimodules. Let MATH. According to REF , MATH is a dualizing complex over MATH. As in the proof of REF we get MATH for all MATH, and MATH. Hence MATH. By REF MATH as rings. Take MATH to be the projection MATH. So MATH is left multiplication by a central idempotent MATH. Since MATH we see that MATH and MATH. Now the bimodule MATH . Being a subquotient of MATH, MATH centralizes MATH. We get a contradiction MATH, MATH. CASE: First assume there is specialization, and choose prime ideals MATH as evidence. Then the algebra MATH is nonzero, having MATH as a quotient. Thus MATH is prime. The complex MATH, with MATH, MATH and MATH otherwise, is dualizing by REF . If MATH were zero this would imply that MATH is decomposable as bimodule, contradicting it being a prime ring. Conversely assume MATH and pick some MATH s.t. MATH is nonzero. By REF we can find MATH such that MATH for some MATH. On the other hand there are primes MATH s.t. MATH, which implies that MATH. We conclude that MATH, and therefore MATH for some MATH. |
math/0103077 | Firstly, by assuming the existence of the finite-dimensional invariant space MATH, we derive the properties of the space, and later we show its existence. Let MATH be a finite-dimensional invariant space in MATH and MATH be operators on MATH defined by MATH. From the definitions of the operator MATH and the periodicity of the function which belongs to MATH, it follows that MATH, MATH, MATH and MATH on MATH. Set MATH then the space MATH is finite-dimensional and invariant under the actions of MATH and MATH. From the commutativity of the operators MATH, we have the decomposition MATH where MATH. Let MATH be an element of MATH. From the definition of MATH and MATH, the function MATH is odd or even. Hence MATH for some MATH. The exponents of the operator MATH at MATH are MATH and MATH. Let MATH be one of the exponents at MATH such that MATH is divisible by MATH. If MATH for some MATH such that MATH then the functions MATH are linearly independent and there is a contradiction with finite-dimensionality. Therefore if MATH then MATH. Conversely assume MATH for all MATH and MATH such that MATH. Then the function MATH is expressed as MATH and MATH for MATH such that MATH, because MATH is an exponent of the operator MATH at MATH. Therefore the space MATH consists of functions whose coefficients of the expansion (see REF ) at MATH satisfy MATH for all MATH and MATH such that MATH. From decomposition REF and the condition MATH, the number MATH must be even. Now we show that the non-zero finite dimensional vector space MATH exists iff MATH. Let MATH and MATH where MATH are the co-sigma functions which are defined in the appendix. Then the space MATH is maximum, that is, MATH and it is seen that MATH. Note that the dimension of the space MATH is MATH. Set MATH then the space MATH is the maximum finite-dimensional space spanned by functions in MATH and is invariant under the action of the Hamiltonian MATH. In particular, we have MATH. The dimension of the space MATH is the summation of the dimension of the spaces MATH for MATH. If MATH is even, then the dimension of the space spanned by all finite dimensional invariant spaces in MATH is MATH . If MATH is odd, then the dimension of the space spanned by all finite dimensional invariant spaces in MATH is MATH . Hence the theorem is obtained. |
math/0103077 | By shifting the variable MATH and rearranging the periods, it is sufficient to show the case MATH and MATH. Set MATH, where MATH or MATH for each MATH, and MATH. If the function MATH satisfies REF , then MATH . Note that the exponents of REF at MATH are MATH. Write MATH, then the recursive relations for coefficients MATH are given as follows: MATH where MATH, MATH, MATH and MATH. Either MATH or MATH is an integer. Let this integer be denoted by MATH. The coefficients MATH are polynomial in MATH of degree MATH. We denote MATH by MATH. If MATH is a solution to the equation MATH for the positive integer MATH, then it is seen that MATH from recursive relation REF , and also MATH for MATH. For the value MATH, the power series MATH is a finite summation and the function MATH is meromorphic and doubly periodic. If MATH is even, it occurs that MATH, MATH, MATH, MATH and the degrees of the polynomials MATH are MATH, MATH, MATH, MATH respectively. We denote these polynomials by MATH and MATH. Since MATH belongs to MATH, the function MATH is doubly periodic up to signs. Let MATH (respectively, MATH) be a solution to MATH (respectively, MATH) for some MATH (respectively, MATH) and let MATH (respectively, MATH) be the corresponding solution to REF which is determined by recursive relations REF . The periodicity of the functions MATH and MATH , (that is, the signs of MATH and MATH) are different if MATH. Assume that MATH is a common solution to MATH and MATH for MATH. Let MATH (respectively, MATH) be the solution to differential REF related to the value MATH. Then the functions MATH and MATH are the basis of solutions to REF . However, this contradicts the periodicity of the solutions when considering exponents. Hence, we proved that the equations MATH and MATH do not have common solutions. The degree of the polynomial MATH is equal to the dimension of the space MATH (see REF ). If we show that the roots of the equation MATH are distinct for generic MATH and MATH, the theorem is proved. It is enough to show that if MATH are real and MATH then the roots of the equation MATH are real and distinct for the case MATH, MATH, MATH, MATH. We will show that the polynomial MATH has MATH real distinct roots MATH such that MATH by induction on MATH. The case MATH is trivial. Let MATH and assume that the statement is true for MATH. From the assumption of the induction, MATH. It is immediate from REF that the leading term of the polynomial MATH in MATH is positive. Since MATH and MATH satisfy the equation MATH and MATH, the sign of the value MATH is MATH. By recursion relation REF and the equation MATH, the sign of the value MATH is opposite to that of MATH. Combined with the asymptotics of MATH for MATH and MATH and MATH, it follows that the polynomial MATH has MATH real distinct roots and the inequality MATH is satisfied. For the case that MATH is odd, the proof is similar. |
math/0103077 | It is sufficient to prove for the case MATH. If MATH are real then the condition MATH is equivalent to (MATH or MATH). For the case MATH, the proposition is proved in the proof of the previous theorem. For the case MATH, the proof is similar. |
math/0103077 | Each solution to REF has singular points only at MATH. Due to periodicity, it is sufficient to consider the case MATH. The exponents at the singular point MATH are MATH and MATH, hence there exist solutions of the form MATH and MATH, because REF lacks a first-order differential term and because the functions MATH are even and because the numbers MATH are odd. Then it is obvious that the monodromy matrix for the functions MATH and MATH around the point MATH is a unit matrix, hence the local monodromy matrix of REF around each singular point is also a unit matrix. |
math/0103077 | Recall that the function MATH is even and doubly periodic, and it satisfies REF . Here we show the following lemma. Suppose MATH and fix MATH. If REF does not have solutions in MATH, then the dimension of the space of even doubly periodic functions which satisfy REF is one. Since the dimension of even (or odd) functions (not necessarily doubly periodic) which satisfy REF is one and solutions to REF are spanned by products of two solutions to REF , the dimension of even functions which satisfy REF is two. Suppose that the function MATH is proportional to MATH, then by definition one has MATH and MATH, that is MATH, and it contradicts to the assumption of the lemma. Therefore the functions MATH and MATH are linearly independent. If the dimension of even doubly periodic functions which satisfy REF is no less than two, all even solution to REF must be doubly periodic. Hence the function MATH is doubly periodic. Since MATH, it follows that MATH or MATH. For example, suppose MATH. From the periodicity of MATH, MATH which contradicts the linear independence of MATH and MATH. Therefore the dimension of even doubly periodic functions which satisfy REF is less than two. Since the function MATH is an even doubly periodic function that satisfies REF , we obtain the lemma. We continue the proof of REF . Since the function MATH is an even doubly periodic function that satisfies the differential REF and the exponents of REF at MATH are MATH , it admits the following expansion: MATH . Substituting REF for REF , we obtain relations for the coefficients MATH and MATH. On the relations, the coefficient MATH is expressed as a polynomial in MATH and MATH, and the coefficient MATH is expressed as a polynomial in MATH and MATH. Through a multiplicative change of scale given by MATH we can choose the coefficients in such a way that MATH and MATH are polynomials in MATH, they do not have common divisors, and the polynomial MATH is monic. If there is a function in MATH that satisfies MATH, we have MATH (see REF ) for some MATH, and the value MATH is an eigenvalue of MATH on the space MATH. From REF , any even doubly periodic function that satisfies REF is determined uniquely up to constant multiplication except for finitely many MATH. Hence the polynomials MATH and MATH are determined uniquely. Let MATH, where MATH is the maximum of the degrees of MATH and MATH in MATH. Substituting this into REF and computing the coefficient of MATH yields MATH, and therefore we have MATH, MATH, and MATH. |
math/0103077 | It is sufficient to show the case MATH and MATH. The function MATH admits REF , but we now use the other expression, MATH where MATH. We are going to obtain a recursive relation for the coefficients MATH. Since MATH satisfies REF , we find that MATH where MATH and MATH. By recursive relation REF , we have REF All coefficients MATH are divisible by MATH as a polynomial in MATH. Hence we may set MATH. CASE: The coefficient MATH is a polynomial in variables MATH and the homogeneous degree of MATH with respect to MATH is MATH. CASE: MATH. CASE: If MATH then MATH. Put MATH in REF , then MATH. By putting MATH in REF , we also obtain MATH. It follows that MATH for all MATH, MATH, and the coefficient of MATH in MATH does not depend on MATH, MATH, MATH. From relation REF , it is seen that MATH and the coefficient of MATH in MATH does not depend on MATH, MATH, MATH. From REF has a solution in MATH iff the value MATH satisfies the equation MATH, and from REF , MATH (see REF ). From REF , the roots of the characteristic polynomial of the operator MATH (see REF ) on the space MATH are distinct for generic MATH and MATH. Hence the characteristic polynomial of the operator MATH on the space MATH coincides with the polynomial MATH up to constant multiplication. Therefore we obtain the proposition. |
math/0103077 | CASE: We show MATH up to periods of elliptic functions for each MATH and MATH. Since the function MATH is an even doubly periodic function and satisfies REF , it has a pole of degree MATH or a zero of degree MATH at MATH. Suppose that the function MATH has a zero of degree MATH at MATH for some MATH. Then the functions MATH and MATH have zeros of degree MATH at MATH, because MATH is proportional to the product MATH, MATH and MATH satisfy differential REF and their exponents at MATH are MATH and MATH. From the condition MATH, the functions MATH and MATH are linearly independent and form a basis of solutions to REF . It follows that every solution to REF has a zero of degree MATH at MATH and this contradicts the fact that one of the exponents at MATH is MATH. Therefore the function MATH has a pole of degree MATH at MATH. If MATH or MATH for some MATH and MATH, then from REF it contradicts the degree of the pole at MATH. CASE: Now we show MATH and MATH up to the periods of elliptic functions. If MATH, then we have MATH for some MATH from the condition MATH mod MATH and it is reduced to REF . Assume (MATH or MATH) and MATH, then the function MATH has zeros of degree no less than REF at two different points MATH and MATH. The degree of zeros of the functions MATH and MATH at points MATH and MATH are MATH or MATH, because MATH and MATH satisfy differential REF which is holomorphic at the points MATH. Since MATH is proportional to MATH, there is a contradiction with the degree of zeros of the function MATH. Therefore the proposition is proved. |
math/0103077 | For simplicity, we prove for the case MATH and MATH. The proofs for the other cases are similar. From the proof of REF , it follows that the polynomial MATH is obtained by a suitable limit of the characteristic polynomial of the space MATH (see REF ). For this case, we have MATH. Set MATH, then MATH as MATH. We will solve recursive relation REF for the case MATH. If MATH then the invariant subspace MATH is given as MATH. Then the characteristic polynomial of the space MATH for MATH is expressed as MATH where MATH is a constant. Similarly, the leading terms of the characteristic polynomials of the spaces MATH, MATH, MATH as MATH are given as MATH up to constant multiplications respectively. By multiplying the leading terms of these four cases, we obtain the proposition. |
math/0103077 | From REF , it follows that MATH for some MATH, where the polynomial MATH is common in REF . From the relation MATH, we have MATH. Hence MATH is written as MATH for some MATH. By a similar argument to that performed in REF, we obtain MATH where the value MATH satisfies MATH and the signs of MATH are suitably chosen. If MATH satisfies the differential equation MATH, MATH also satisfies it. From the trigonometric version of REF , it is seen that the functions MATH and MATH are linearly independent. By a similar argument to the proof of REF , the values MATH, MATH, MATH are mutually different. By setting MATH and MATH, we obtain the proposition. |
math/0103077 | By the assumption MATH and REF , there exists a solution to the equation MATH of the form MATH. Then the values MATH satisfy equations REF and we have MATH . By comparing the coefficients of MATH of the equation MATH (MATH), we obtain recursive relation REF for MATH. |
math/0103077 | From REF , for each MATH there exists MATH such that if MATH then there exists the eigenvalue MATH and the square-integrable eigenfunction MATH that converge respectively to the trigonometric eigenvalue and eigenfunction as MATH. Set MATH, then MATH and MATH are holomorphic in MATH near MATH and they satisfy MATH . These equations are the same as REF . From the uniqueness of the coefficients in MATH, we obtain MATH and MATH. From the holomorphy of MATH and MATH, the convergence of MATH and MATH in MATH is shown. |
math/0103079 | For all MATH we compute, using the zero weight properties of MATH and MATH and the dynamical twist equation for MATH : MATH . The verification of REF is straightforward. |
math/0103079 | The computation below is similar to that of REF and uses the zero weight properties of the functions involved : MATH . The counit identity is straightforward. |
math/0103079 | The statements are proved by direct verification. We will only prove the more difficult REF . One has MATH as desired. |
math/0103079 | Let MATH. It is easy to see that the condition for MATH to be a vertex-IRF transformation is MATH for all MATH. Setting MATH one gets MATH. Also, setting MATH, we find MATH where MATH and MATH. Putting in the last equation MATH, we find MATH, that is, MATH as desired. The converse statement is straightforward. |
math/0103079 | If MATH is realized by a vertex-IRF transformation MATH, then for all MATH the characters MATH and MATH are conjugate via MATH, and therefore induce equivalent representations of MATH. Conversely, if REF holds for a bijective function MATH then the collection MATH is a dynamical datum for MATH. For every element MATH of weight MATH the homomorphism MATH arising in the exchange construction CITE, is given by MATH, where the element MATH belongs to the normalizer of MATH in MATH. In every MATH-module MATH maps elements of weight MATH to elements of weight MATH, for all MATH. Clearly, MATH realizes MATH and the twist defined by the above action is equal to MATH. |
math/0103079 | The proof is contained in CITE. |
math/0103079 | The proof is by a direct calculation. |
math/0103079 | Let us first prove REF . Let MATH be given by REF . First of all, it is easy to check that MATH has zero weight. Namely, the zero weight condition coincides with the cross-derivative condition for MATH, which is equivalent to the condition that MATH for some MATH. It is also clear that MATH. So it remains to check the classical NAME equation for MATH. Let us write MATH with MATH. Then it is straightforward to check that the zero weight condition for MATH translates into MATH for all MATH. Using this identity we can write the classical dynamical NAME equation for MATH in terms of functions MATH as MATH . This equation is clearly satisfied, since MATH. So REF is proved. Let us now prove REF . Let MATH be any completely degenerate dynamical MATH-matrix. We have MATH. Clearly, MATH. Write MATH in the form REF . Since MATH has zero weight, the projection MATH of MATH on MATH also has zero weight. This implies that MATH in MATH for all MATH, therefore MATH, and hence MATH. Now, as we explained above, the classical dynamical NAME equation for MATH and the zero weight condition reduce to REF respectively. In particular, we have MATH . This implies that the differential equations MATH are compatible and have a unique solution MATH with the initial condition MATH. Let MATH be an arbitrary lift of MATH. Then MATH where the second sum is a MATH-form. It follows from the classical dynamical NAME equation that MATH is a closed form. Therefore, MATH is gauge equivalent to REF , as desired. This proves the first statement of REF . To prove the second statement of REF , it is sufficient to recall that gauge transformations are functions from MATH to the centralizer MATH of MATH in MATH. Now, the action of such a function MATH on MATH is given by MATH. This implies easily the second statement of REF , since the image of MATH in MATH is exactly MATH. Finally, let us prove REF . First of all, as we already mentioned in the proof of REF , the zero weight property of MATH implies that MATH for some function MATH. Now, for MATH define the element MATH of the group MATH for MATH. Then it is easy to check that MATH. Therefore, by REF, MATH is a vertex-IRF transformation of MATH. Hence MATH is a dynamical twist. It is easy to check directly that the quasi-classical limit of MATH is MATH. The theorem is proved. |
math/0103082 | This is REF. |
math/0103082 | Clearly, MATH . This implies easily that MATH where MATH. Since MATH is a proper subset of MATH, we have MATH . Assume that REF holds. Then there exists MATH such that MATH lies in MATH and the exact multiplicative order of MATH is MATH. This implies that MATH. Since MATH, we conclude that MATH. Therefore MATH . Now assume that REF holds. It follows from REF that there exists MATH such that MATH for some MATH with MATH. This implies that if we put MATH then MATH . |
math/0103082 | Recall ( CITE, p. REFEFc), that the adjoint representation of MATH in MATH splits into a direct sum of the trivial one-dimensional representation (scalars) and an absolutely simple MATH-module MATH of dimension REF (traceless operators). The kernel of the natural homomorphism MATH coincides with the center MATH which is either trivial or a cyclic group of order MATH depending on whether MATH or MATH. In both cases we get an embedding MATH . If MATH (that is, MATH) then MATH and one may use NAME character tables CITE in order to study absolutely irreducible representations of MATH in characteristic MATH. Notice (CITE, p. REF) that the reduction modulo MATH of the irrational constant bREF does not lie in MATH. Using the Table on p. REF, we conclude that there is only one (up to an isomorphism) absolutely irreducible representation of MATH defined over MATH and its dimension is MATH. This proves the assertion of the theorem in the case of MATH. So further we assume that MATH . Clearly, for each MATH with trace MATH the image MATH of MATH in MATH has trace MATH. In particular, if MATH with MATH then the trace of MATH is MATH . Now let us start to vary MATH in the MATH-element set MATH . One may easily check that the set of all MATH's consists of MATH elements of MATH. Since MATH, MATH . This implies that MATH satisfies the conditions of REF with MATH. In particular, none of representations MATH of MATH could be realized over MATH if MATH is a proper subset of MATH. On the other hand, it is known (CITE, p. REF ) that each absolutely irreducible representation of MATH over MATH either has dimension divisible by MATH or is isomorphic to the representation obtained from some MATH via MATH. The rest is clear. |
math/0103082 | First, MATH is a simple non-abelian group, whose order is MATH (CITE, p. XVI, REF ; CITE, pp. REF). Second, notice that MATH acts transitively on MATH. Indeed, the classification of subgroups of MATH (REF and its proof, p. REF) implies that each subgroup of MATH has index MATH. This implies that MATH acts transitively on MATH. Third, we claim that this action is, in fact, doubly transitive. Indeed, the stabilizer MATH of a point MATH has index MATH in MATH. It follows easily from the same classification that MATH is (the image of) the stabilizer (in MATH) of a proper subspace MATH in MATH. If MATH is a plane then counting arguments imply that the restriction of the Hermitian form to MATH could not be non-degenerate and therefore MATH coincides with (the image of) the stabilizer of certain isotropic line MATH. (The line MATH is the orthogonal complement of MATH.) If MATH is a line then counting arguments imply that MATH is isotropic. Hence we may always assume that MATH is (the image of) the stabilizer of an isotropic line in MATH. Taking into account that the set of isotropic lines in MATH has cardinality MATH, we conclude that MATH is isomorphic (as MATH-set) to the set of isotropic lines on which MATH acts doubly transitively and we are done. By REF , the double transitivity implies that the MATH-module MATH is absolutely simple. Since MATH is surjective, the MATH-module MATH is also absolutely simple. Also, in order to prove that MATH-module MATH is very simple, it suffices to check that the MATH-module MATH is very simple. Recall that MATH. By REF , there no absolutely simple nontrivial MATH-modules, whose dimension strictly divides MATH. This implies that MATH is not isomorphic to a tensor product of absolutely simple MATH-modules of dimension MATH. Therefore MATH is not isomorphic to a tensor product of absolutely simple MATH-modules of dimension MATH. Recall that all subgroups in MATH different from MATH itself have index MATH. It follows from REF that the MATH-module MATH is very simple. |
math/0103082 | This is REF. |
math/0103083 | CASE: Core function MATH mod MATH REF, -REF mod MATH has MATH distinct residues for each MATH-REF, satisfying MATH mod MATH, with MATH mod MATH due to MATH. Apparently, including MATH we have: MATH mod MATH for each MATH-REF, with period MATH in MATH. And MATH of odd degree MATH has odd symmetry because: MATH mod MATH. CASE: Increment MATH mod MATH also has period MATH because MATH mod MATH. This yields residues REF mod MATH in extension group MATH. It is an even degree polynomial, with leading term MATH, and even symmetry: MATH, so MATH mod MATH for REF-complements: MATH mod MATH. CASE: Write MATH for MATH, the subgroup of MATH-th power residues mod MATH in units group MATH. Then subgroup closure MATH implies MATH, since MATH due to MATH in MATH for odd prime MATH-REF. So non-zero pairsum set MATH is the disjoint union of cosets of MATH in MATH, as generated by differences MATH. Due to REF : MATH, where MATH, it suffices to consider only differences REF mod MATH, hence in extension group MATH, that is: in MATH. This amounts to MATH distinct increments MATH, for MATH due to even symmetry REF , and excluding MATH=REF involving non-core MATH=REF. These MATH cosets of MATH in MATH yield: MATH, where MATH and MATH. |
math/0103083 | Core MATH mod MATH, here denoted by MATH as subgroup of units group MATH, satisfies MATH so the set of all core pairsums can be factored as MATH. Hence the nonzero pairsums are a (disjoint) union of the cosets of MATH generated by MATH. Since MATH with MATH=REF mod MATH, there are MATH cosets of MATH in MATH. Then intersection MATH of all residues REF mod MATH in MATH generates MATH distinct cosets of MATH in MATH. Due to MATH in core MATH we have MATH sothat MATH. View set MATH as function values MATH mod MATH, with MATH mod MATH. Then successive core increments MATH form precisely intersection MATH, yielding all residues REF mod MATH in MATH. Distinct residues MATH generate distinct cosets, so by definition of MATH there are for MATH cosets of core MATH generated by MATH mod MATH. |
math/0103083 | CASE: Write MATH for MATH then, as in REF : MATH. For residues mod MATH we seek intersection MATH of all distinct residues REF mod MATH in MATH that generate the cosets of MATH in MATH mod MATH. By MATH core extension MATH. Discard terms divisible by MATH (are not in MATH) then: MATH for each core extension. So MATH and MATH have the same coset generators in MATH, namely the core increment set MATH. CASE: Notice successive cores satisfy by REF mod MATH. In other words, each MATH-th power step MATH produces one more significant digit REF while fixing the MATH less significant digits (MATH). Now MATH mod MATH has MATH-th power residue MATH mod MATH, implying lemma REF by induction on MATH in MATH. This yields an efficient core generation method. Denote MATH, with MATH, then: CASE: MATH mod MATH, implying: CASE: MATH mod MATH, next core msd MATH mod MATH. Notice that by MATH mod MATH for all MATH, and MATH implies MATH mod MATH. CASE: In REF take MATH and notice that MATH mod MATH contains MATH distinct integer increments MATH . REF . which are REF mod MATH, hence in MATH. They generate MATH distinct cosets of core MATH in MATH mod MATH, although they are not core MATH increments. Repeated MATH-th powers MATH in constant MATH-digit precision yield increments MATH mod MATH, which for MATH produce the increments of core MATH mod MATH. Distinct increments MATH mod MATH remain distinct for MATH, shown as follows. For non-symmetric REF let increments MATH satisfy: CASE: MATH mod MATH for some MATH, and REF MATH mod MATH. Then for MATH the same holds, since MATH where MATH equals MATH and MATH respectively. Because in MATH each of the four MATH terms has form MATH mod MATH where the respectively, MATH yield REF , and the respectively, msd's MATH cause inequivalence REF . Then: CASE: MATH mod MATH mod MATH which depends only on MATH, and not on msd MATH of MATH. This preserves equivalence REF mod MATH for MATH, and similarly inequivalence REF mod MATH because, depending only on the respective MATH mod MATH, equivalence at MATH would contradict REF at MATH. Cases MATH and MATH behave as follows. For MATH the successive differences MATH mod MATH . REF vary with MATH from REF to MATH, and by REF the core residues MATH mod MATH settle for increasing precision MATH. So initial inequivalences mod REF , and more specifically mod REF , are preserved. And for all MATH the differences REF are some constant MATH mod MATH, again by REF . Hence by induction base REF and steps MATH : core MATH mod MATH has MATH distinct increments, so critical precision MATH. |
math/0103083 | MATH mod MATH implies inequality mod MATH. With MATH mod MATH, so each MATH-th power is in core MATH mod MATH, it suffices to show MATH mod MATH. Factorize MATH, with positive integer cofactors MATH and MATH. Then MATH mod MATH, so opposite signed cofactors MATH or MATH form an inverse pair mod MATH. Inverses in a finite group MATH have equal order REF in MATH, with order two automorphism MATH. So orders MATH and MATH are equal in MATH. Notice MATH is not in core MATH, where MATH mod MATH is the only core residue that is MATH mod MATH, since the MATH core residues MATH of MATH are distinct MATH mod MATH. In fact MATH mod MATH and no smaller exponent yields this. So MATH has order MATH in MATH, generating all MATH residues MATH mod MATH, with inverse pair MATH of equal order in MATH. Core MATH is closed under multiplication, so at most one cofactor of non-core product MATH can be in core. In fact neither is in core MATH, so both MATH and MATH are MATH mod MATH, seen as follows. By MATH: each MATH has product form MATH mod MATH of two components, with MATH in core MATH and MATH in extension group MATH. Then MATH mod MATH, where MATH and MATH as inverse pair in MATH have equal order, and each component forms an inverse pair of equal (and coprime) orders in MATH and MATH respectively. The latter must divide MATH, and discarding order REF (both MATH cannot be in core, as shown) their common order is MATH or REF . For any unit MATH the order of MATH divides that of MATH, so MATH dividing the common order of MATH and MATH implies MATH dividing also those of MATH and MATH, hence cofactors MATH and MATH of MATH are both outside core MATH. |
math/0103083 | Analysis mod MATH suffices, because each MATH mod MATH is reached upon multiplication by MATH, due to REF distributing over REF . Core MATH has order MATH for any MATH-REF, and MATH implies powersums MATH mod MATH to be sums of three core residues. Assume MATH mod MATH for some positive MATH with MATH. Such MATH generates all MATH residues in MATH mod MATH. And for each prime MATH there are many such coresums MATH with MATH mod MATH, seen as follows. Any positive triple MATH with MATH yields, by MATH, coresum MATH mod MATH, hence with a coresum MATH mod MATH. If MATH=REF then this solves MATH for residues mod MATH, for instance the cubic roots of REF mod MATH for each prime MATH mod REF, see CITE . NAME MATH is the dominant case for any prime MATH. In fact, normation upon division by one of the three core terms in units group MATH yields one unity core term, say MATH mod MATH hence MATH. Then MATH yields MATH mod MATH, where MATH. There are MATH distinct cosets of MATH in MATH (lemREF, lemREF), yielding as many distinct core pairsums MATH mod MATH in set MATH. |
math/0103083 | Analysis mod MATH suffices, by extension lemREF, and by REF all non-zero multiples of MATH are MATH, while MATH because MATH. Hence MATH covers MATH. Adding an extra term MATH yields MATH, which also covers MATH because MATH and MATH, so all of MATH is covered. |
math/0103084 | Let MATH be a general member of MATH and let MATH be the normalization of MATH. Notice that if we blow up MATH along MATH and let MATH be the resulting surface and MATH and MATH be the proper transforms of MATH and MATH, respectively, we have MATH, MATH and MATH. So we can simply start with MATH. Let MATH be a desingularization of MATH, MATH be the proper transform of MATH and MATH be the reduced total transform of MATH. Our assumptions on MATH guarantee that MATH. Also observe that MATH and MATH has normal crossing. So we may assume that MATH is smooth. Furthermore, if MATH meets MATH at a singular point MATH of MATH, there exists a series of blowups of MATH over MATH such that MATH does not meet MATH at any singular point over MATH on the resulting surface MATH. Therefore, we may assume that MATH meets MATH only at smooth points of MATH. It suffices to show that the dimension of the deformation space of the map MATH with MATH fixed is bounded from the above by MATH. Here by the deformation of the map MATH, we mean the deformation of the pair MATH. Let MATH be the normal bundle of the map MATH. Let MATH where MATH. Fixing MATH is equivalent to imposing certain tangency conditions at MATH. The dimension of the deformation space of such map MATH is at most MATH (see for example, CITE), where MATH is the torsion part of MATH. Since MATH and we are done. If MATH is nonempty, then MATH has at least the virtual dimension MATH . |
math/0103084 | There is nothing to do if MATH since we have REF. Suppose that MATH is rigid in MATH. Since MATH moves in the BPF linear system MATH for MATH, by NAME, we can choose MATH such that MATH meets MATH transversely and MATH. Therefore, MATH . Suppose that MATH is nonrigid. For each MATH, let us fix MATH. By REF, there are only countably many MATH such that the following fails MATH . Again, by NAME, we can choose MATH such that MATH meets MATH transversely, MATH for each MATH and MATH for all MATH that REF fails. In summary, we have either REF holds for some MATH, in which case we are done, or MATH meets MATH transversely and MATH, in which case we have REF. Either way we obtain REF. |
math/0103084 | Without the loss of generality, let us assume that MATH is smooth and irreducible. Suppose that it fails for MATH for some MATH. Since MATH is smooth, MATH is a NAME divisor of MATH and hence MATH is a divisor of MATH which is supported on a union of contractible components of MATH. Then MATH, which can happen only when MATH. But this contradicts with MATH. |
math/0103084 | We argue by induction on MATH. When MATH, we apply NAME directly to the projection MATH by observing that MATH . This yields MATH and hence REF follows. For MATH, we will use degeneration to bring down MATH. Since MATH is chosen to be generic in the linear series given in REF, it can be degenerated to a union MATH. Here MATH is the union of MATH fibers of MATH over the vanishing locus of MATH and MATH is a general member of the linear series MATH where MATH. Our induction hypothesis is MATH if MATH meets MATH properly. Let MATH and MATH be an effective divisor on MATH with components MATH REF satisfying CASE: MATH; CASE: MATH is a general member of the linear series REF for MATH; CASE: MATH and MATH for all MATH. So MATH. Let MATH be a flat family of curves over MATH with the commutative REF . We assume that MATH is smooth and irreducible, the map MATH is birational onto its image for MATH and MATH meets MATH properly in MATH. Our goal is to prove MATH for MATH, where MATH is the degree of the map MATH. As before, we may assume that MATH is a family of semistable maps with marked points MATH on MATH REF and let MATH be defined similarly for each component MATH. Since we still have REF, the proof of REF, like that of REF, again comes down to the estimation of MATH for each irreducible component MATH. Let MATH be the degree of the map MATH. If MATH, then MATH where MATH is the restriction of MATH to MATH. Our standard NAME argument shows MATH . If MATH, then MATH and by NAME, we have MATH . If MATH meets MATH properly and dominates MATH via MATH, then by the induction REF and NAME, MATH . In summary, the combination of REF produces MATH for each component MATH that dominates MATH via the map MATH of degree MATH. Therefore, MATH where we sum over all irreducible components MATH that dominate MATH via the map MATH. So it remains to show MATH where MATH runs over all irreducible components of MATH that are contractible under the map MATH, that is, MATH is supported on a fiber of MATH. The combination of REF will produce MATH and this implies REF thanks to REF. Let MATH be a maximal connected component of MATH . In order to prove REF, it suffices to prove MATH for all such MATH. We extend the notation MATH to MATH, for which MATH is the finite subset of MATH consisting of the marked points MATH and the intersections between MATH and the rest of the components of MATH. Then we have MATH . Obviously, MATH since MATH is connected and MATH consists of at least one point unless MATH, which is impossible since we assume that MATH dominates MATH under the map MATH. So MATH . Suppose that the equality in REF holds. Then we must have MATH and MATH. This means that every irreducible component of MATH is smooth and rational, the dual graph of MATH is a tree, MATH meets the rest of MATH at exactly one point and there are no marked points on MATH. In addition, MATH cannot be contracted to a point by MATH due to our minimality assumption on MATH and hence MATH is supported along a fiber of the projection MATH. For each component MATH (MATH), we write MATH where MATH consists of the sections of MATH, MATH is supported along MATH and MATH is supported along the rest of MATH. Since there are no marked points on MATH, MATH and MATH for MATH. Let MATH be the only irreducible component of MATH that intersects MATH. If MATH is supported on MATH, we have MATH and set MATH; otherwise, if MATH is not supported on MATH, we have MATH and set MATH. In any event, we have MATH, MATH and MATH. This is possible only if MATH. But then we have MATH, which contradicts the fact that MATH is supported along a fiber of MATH and hence MATH. Hence we must have MATH and REF follows. This finishes the proof of the proposition. |
math/0103084 | The first REF follows immediately from REF and the second REF follows from REF and NAME 's cone theorem, as mentioned before. It remains to verify the last statement, which requires us to show that MATH is NEF if MATH. By NAME 's cone theorem, it suffices to show that MATH for all rational curves MATH with MATH. Note that if such MATH exists, MATH is rational and covered by the curves in the numerical class of MATH. So it is enough to show that MATH is effective. To see this, let MATH be a minimal desingularization of MATH. Then MATH since MATH is canonical. So MATH is effective if and only if MATH is, where MATH. So it comes down to a NAME computation on MATH: MATH and we are done. |
math/0103084 | Every BPF divisor on MATH lies in the cone of numerically effective divisors (NEF cone) generated by MATH and MATH (BPF is equivalent to NEF on MATH). Hence each MATH is linearly equivalent to a sum of divisors MATH and MATH. Applying REF, we may take MATH, each MATH to be either MATH or MATH and MATH to be the smallest of the following numbers: MATH where MATH runs over all effective divisors of MATH; it is enough to check REF with MATH since the cone of effective divisors of MATH is generated by MATH and MATH. And REF follows immediately. |
math/0103084 | Again BPF is equivalent to NEF on MATH. The NEF cone of MATH is generated by divisors MATH with the property that MATH or MATH for each MATH. Applying REF, we may take MATH and MATH to be the smaller of the following two numbers: MATH where MATH runs over all divisors satisfying MATH or MATH for each MATH and MATH runs over all effective divisors on MATH. It is enough to check REF with MATH since the cone of effective divisors on MATH is generated by MATH and then it is obvious that MATH. |
math/0103084 | Let MATH be the blowup of MATH along MATH. The map MATH is a small morphism REF and it is an isomorphism away from MATH . The central fiber of MATH looks like MATH where MATH for MATH. Let MATH. It is not hard to see that MATH is the blowup of MATH along MATH. We let MATH . Obviously, MATH is the proper transform of MATH under the blowup MATH. The rational map MATH can resolved into a regular map MATH and we have the commutative REF . Let MATH be the proper transform of MATH under the blowup MATH. Obviously, MATH. Since MATH, MATH passes through the close point MATH of MATH. Apply REF to MATH locally at MATH and we are done. |
math/0103084 | The first two statements follow from a standard ``projection" argument. Let MATH be the incidence correspondence given by MATH, where we regard MATH as a homogeneous polynomial with generic coefficients. Obviously, MATH is a general fiber of MATH when it is projected onto the first factor MATH. To see that MATH has the stated properties, it suffices to justify that MATH has the same properties, that is, MATH is irreducible of codimension MATH in the total space and MATH consists of the points such that MATH. This can be shown by projecting MATH to the second factor MATH. The general fiber of MATH over a point in MATH is (almost) a generic determinantal variety, which is classically known to have the stated properties. We will let the readers to fill out the details. Choose linear forms MATH and specialize the MATH-th column of MATH to MATH. Since NAME polynomials are invariant under deformations, such degeneration does not change MATH. Let MATH be the hyperplane given by MATH. Then MATH, where MATH is the subvariety of MATH given by MATH where MATH's occupy the MATH-th column of the matrix and the rest of the matrix are the same as MATH. Observe that MATH and hence we may apply row operations to the matrix to eliminate its MATH-th row and MATH-th column. Therefore, it is easy to see that MATH and MATH are deformationally equivalent inside MATH. Especially they have the same numerical properties. For a general choices of MATH, MATH is of normal crossing with components MATH and MATH. Apply the inclusion-exclusion principle to MATH and the recursion REF follows. |
math/0103084 | We do induction on MATH. If MATH or MATH, MATH is a complete intersection in MATH and REF follows easily from adjunction. Suppose that MATH. Let MATH in REF and we have MATH and MATH where MATH and MATH are defined as in REF and MATH is a hyperplane in MATH. By induction hypothesis, MATH and MATH. Therefore, REF follows from REF. |
math/0103085 | From the very definition of the determinant developed from the MATH-th row. |
math/0103085 | It is only necessary to consider that MATH . |
math/0103085 | Using the NAME logarithmic free resolution of the holonomic MATH-module MATH, we first compute a presentation of the right MATH-module MATH and then we prove that left MATH-module associated to MATH is MATH. The matrix of the MATH-th morphism in the resolution of MATH (see CITE) has components of the form MATH so it is enough to prove that MATH . In order to prove the last equality, we will show that MATH . Using REF, we obtain MATH . So we have collected in the first sum precisely (see REF) MATH. It remains to check that MATH . As MATH, we have MATH . |
math/0103085 | We show here how to read the original (topological) proof of CITE to give a differential proof of ``only if" part. Part ``if" is a consequence of CITE because any plane curve is a free divisor CITE. The problem is local. Suppose the local equation MATH of MATH is defined in a small open neighbourhood such that the only singular point of MATH is the origin. Denote MATH. Let us consider (see REF) the natural surjective morphism MATH where the last isomorphism follows by a result of NAME (that is, the local MATH-function MATH of MATH verifies MATH for any integer MATH, CITE). The kernel MATH of MATH is supported by the origin (because MATH is smooth outside MATH) and MATH. In particular MATH and MATH are regular holonomic (compare CITE) because as we said before MATH satisfies the hypothesis of REF . Let us denote MATH the solution complex of MATH. Assume LCT holds for MATH. Then we have MATH . Then both MATH-modules MATH and MATH have the same NAME complex and then the same characteristic cycle. In this case MATH and MATH. Finally, by CITE (or by CITE, see also CITE) MATH is weighted homogeneous in suitable coordinates. That proves the ``only if" part of the theorem. |
math/0103085 | By CITE MATH is NAME free and then MATH is holonomic and the dual of MATH is MATH (see REF). So, MATH is also holonomic. It is enough to prove that MATH is regular. To avoid confusion we will denote MATH to emphasize the divisor MATH. In fact we will prove, by induction on MATH, that the natural morphism MATH is an isomorphism. We follows here the argument of CITE. There is nothing to prove in the case MATH. We note that in dimension REF the result is proved in CITE (see REF). Suppose the result is true for any free, l.q-h. divisor in dimension MATH. Let MATH be a free, l.q-h. For any MATH there exists an open neighbourhood MATH of MATH such that for any MATH the germ MATH is isomorphic to MATH, where MATH is a free, l.q-h. divisor in MATH (see CITE). So, by induction hypothesis the morphism MATH is an isomorphism. Then, by applying the functor MATH (where MATH is the projection), we have that for any MATH, MATH, the morphism MATH is an isomorphism between MATH and MATH. We owe this argument to NAME. So, the kernel of MATH is concentrated on a discrete set and it is regular holonomic (here MATH is the MATH-module MATH, where MATH is a local equation of MATH). As MATH is regular holonomic we deduce the regularity of MATH. On the other hand, by CITE the logarithmic comparison theorem holds for MATH. So, by using duality REF and the natural quasi-isomorphism REF , we deduce (as in REF) that MATH and MATH are naturally quasi-isomorphic and therefore, by NAME correspondence, MATH and MATH are naturally isomorphic, that is, MATH is an isomorphism. Thus we have concluded the induction. |
math/0103087 | Let MATH be the algebraic set in MATH defined by all the bi-homogeneous equations in MATH and REF . We shall first prove that MATH as sets. Clearly, the coordinates of the points of MATH satisfy all the equations in MATH and REF , so as sets, MATH, hence MATH. To prove the reverse inclusion, let MATH (where MATH and MATH). The coordinates of MATH satisfy equations in MATH and the MATH minors of MATH, so by REF , and following the same argument as that of CITE, there exists a unique MATH such that the coordinates of MATH and MATH satisfy MATH, and MATH must have the form MATH for some MATH. This and the equations in MATH implies that MATH . Thus, if MATH then MATH for some MATH; and otherwise, if MATH, then MATH lies on the exceptional line corresponding the the blowup at MATH. Thus, MATH. We also note that the equations in MATH and the MATH minors of MATH are exactly the MATH minors of MATH. Thus, REF shows that for each MATH, there exists a unique MATH such that the coordinates of MATH and MATH satisfy the MATH minors of MATH. This implies MATH. Since the linear system MATH is very ample CITE, the projection map MATH defined by sending MATH to MATH is an isomorphism; and so, for each MATH, there exists a unique MATH, such that MATH. Moreover, the coordinates of every point on MATH satisfy MATH, so the coordinates of every point on MATH also satisfy MATH. Thus, the coordinates of MATH satisfy MATH, and so all the MATH minors of MATH. Therefore, MATH, that is, MATH. We have shown that MATH. Hence, MATH. In conclusion, the equations in MATH and the MATH minors of MATH describe MATH as a set. Furthermore, MATH is a matrix of indeterminates, so it is a well known fact that the MATH minors of MATH form a prime ideal. Similar to the last part of the proof of CITE, we consider the following sequence of surjective ring homomorphisms: MATH where MATH sends MATH to MATH, and sends MATH to MATH; and MATH sends MATH to MATH, and sends MATH to MATH. Then from the proofs of REF and MATH, we further deduce that the MATH minors of MATH form the kernel of MATH, and the images of equations in MATH through MATH form the kernel of MATH. Therefore, the MATH minors of MATH and the equations in MATH form the kernel of MATH, which is a prime ideal. Hence, the equations in MATH and the MATH minors of MATH form the defining ideal for MATH in MATH. The theorem is proved. |
math/0103087 | The first statement of the theorem follows from REF and the fact that the NAME algebra MATH is the bi-graded coordinate ring of MATH in MATH (compare CITE). For the second statement of the theorem, we observe that the defining ideal of MATH is the ideal of MATH minors of a matrix of linear forms of size MATH, so MATH. Furthermore, MATH is a surface in the product space MATH (after factoring out the linear forms in MATH), so its codimension is exactly MATH. This implies that the defining ideal of MATH is perfect, and has the same NAME numbers as that of the ideal of MATH minors of a generic MATH matrix. The result then follows. |
math/0103087 | The proof goes in the same manner of that of REF . First, since all the points on MATH satisfy the defining equations of MATH, we have MATH, whence MATH. To prove the reverse inclusion, suppose MATH, where MATH and MATH. By REF and follow the same argument as that of CITE, there exists a unique MATH such that the coordinates of MATH and MATH satisfy the equations in MATH, and MATH must have the form MATH for some MATH, if MATH, or MATH for some MATH, if MATH. Now, substituting the coordinates of MATH into the entries of the product matrix MATH and the equations in REF , we get MATH if MATH, or MATH if MATH. Thus, if MATH, then MATH if MATH, or MATH if MATH (for some MATH); and if MATH, then MATH belongs to the exceptional line corresponding the the blowup at MATH. Therefore, MATH. We note further that the equations in MATH are among the equations of MATH, so REF shows that for each MATH, there exists a unique MATH such that the coordinates of MATH and MATH satisfy the equations of MATH (when MATH), or of MATH and MATH (when MATH). Thus, MATH. Now, with our assumption that there are no MATH points of MATH lying on a line, the linear system MATH is very ample, so the projection that sends MATH to MATH is an isomorphism from MATH to MATH. Thus, for each MATH there exists a unique MATH such that MATH. Moreover, since MATH, the coordinates of MATH and MATH satisfy the equations of MATH (when MATH), or of MATH and MATH (when MATH). Hence, MATH. In other words, MATH. We just proved that MATH. Hence, MATH as sets. |
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