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math/0107132 | We first remark that the positive NAME property may be characterised as in REF ; the proof is verbatim the same as in CITE. NAME lattice has the positive NAME property if and only if for every positive MATH, every positive MATH and every MATH there is some positive MATH such that MATH and MATH. Now suppose that MATH has the NAME property; we shall verify the condition of REF . Note that MATH can be represented by all sequences MATH such that MATH and MATH can be represented by all sequences MATH, MATH, such that MATH . Let MATH and MATH be positive elements and let MATH. Pick MATH and MATH such that MATH, MATH and put MATH, MATH; then MATH. Since MATH has the NAME property, we can find MATH such that MATH and MATH; compare REF . Write MATH and MATH; then MATH and MATH where we have used the fact that the norm of MATH is monotonic in each variable. Hence MATH has the positive NAME property. (Incidentally, the assumption that the MATH have the NAME property did not enter this part of the proof.) Conversely, suppose that MATH has the positive NAME property. Let MATH and MATH, define MATH and MATH. Given MATH, find using REF some MATH such that MATH and MATH. Since MATH has the NAME property, one can find MATH such that MATH, MATH and MATH; just note that MATH. Therefore MATH satisfies MATH and MATH . Hence MATH has the NAME property. |
math/0107132 | To deduce REF from REF one just has to notice that if the set of all MATH for which MATH belongs to the ultrafilter MATH, then MATH. So for every MATH, MATH tends to REF when MATH tends to infinity, which proves the NAME property for MATH. To deduce REF from REF let us argue ad absurdum. Suppose there are MATH and MATH such that for every MATH the set MATH belongs to the ultrafilter MATH. Denote MATH. Let us construct two elements MATH and MATH of MATH in such a way that MATH and for every MATH the distance from MATH to MATH is bigger than MATH. The MATH-hull of a set is increasing when MATH is increasing, so for every MATH and every MATH the distance from MATH to MATH is bigger than MATH. This means in turn that for every MATH, MATH, so MATH, which contradicts the NAME property of MATH. |
math/0107132 | Suppose MATH. Fix MATH. There exist MATH, MATH, and MATH, for which MATH. Define elements MATH. Then MATH, MATH, so MATH. |
math/0107132 | Suppose MATH and let us take arbitrary points MATH and MATH from MATH. There is a partitioning of MATH into sets MATH such that MATH. Define functions MATH by MATH if MATH, and MATH if MATH. Then MATH, where MATH. On the other hand, if MATH, then MATH . So, MATH. If MATH, then proceeding as above, with MATH we get a decomposition MATH. Let us arrange the MATH's in decreasing order and take the first MATH of them. Then MATH . We need to prove that MATH. Assume the opposite. Then MATH hence MATH and MATH. Thus, MATH which is a contradiction. So, MATH and the proof of the lemma is finished. |
math/0107132 | Let MATH and MATH be arbitrary. Without loss of generality, assume that MATH attains the value MATH. Take an open neighbourhood MATH such that MATH for all MATH. Now pick MATH disjoint subneighbourhoods MATH inside MATH. For each of them choose a positive function MATH supported on MATH such that MATH, MATH and MATH attains the value MATH in MATH. Obviously, MATH, hence, MATH. On the other hand, MATH which proves the lemma. |
math/0107132 | Suppose MATH for some MATH. This means in particular that for a fixed element MATH (taking MATH) there are elements MATH, MATH and MATH, MATH, for which MATH. Without loss of generality we may assume that MATH (otherwise just change the enumeration). Plugging in MATH and MATH we obtain MATH which is a contradiction. |
math/0107134 | REF for the functor MATH is proved in CITE and CITE, and follows for MATH by taking projective limits. REF follows from REF and the universal property defining MATH. |
math/0107134 | We choose a formal model MATH of MATH such that we are in the situation of REF . The smooth locus MATH of MATH is quasi-compact and is a weak NAME model of MATH, since, by CITE , every MATH-valued point of MATH extends uniquely to a MATH-valued point of MATH. Also, it follows from REF that, if MATH is a weak NAME model of the rigid MATH-space MATH, then for every field MATH containing MATH, the formal MATH-scheme MATH is a weak NAME model of the rigid MATH-space MATH. |
math/0107134 | Indeed, it follows from REF that MATH is equal to the image of MATH in MATH. The morphism MATH being of finite type, first the statement follows from NAME 's Theorem. For the second statement, one may assume MATH, and the proof proceeds as before. |
math/0107134 | Since MATH is quasi-compact, this follows from NAME REFEF. |
math/0107134 | Follows directly from the definitions. |
math/0107134 | Let us prove REF . Write MATH, with MATH in MATH and MATH. Since MATH is smooth, MATH is locally free of rank REF and MATH is isomorphic to a principal ideal sheaf MATH, with MATH in MATH. Furthermore, the function MATH coincides with the function MATH which to a point MATH of MATH associates MATH. The fibres of MATH are stable cylinders. Since MATH is a gauge form, MATH induces an invertible function on MATH, hence, by the maximum principle (compare CITE), the function MATH takes only a finite number of values. To prove that MATH in MATH does not depend on the model MATH, it is enough to consider the case of another model MATH obtained from MATH by an admissible formal blow-up MATH. We may also assume MATH contains as an open dense formal subscheme a weak NAME model MATH of MATH. The equality MATH then follows from REF follows similarly from REF . |
math/0107134 | Let us prove the first statement, the proof of the second one being similar. It is enough to consider the case MATH. Choose a MATH-model MATH containing a weak NAME model MATH of MATH as an open dense formal subscheme and such that the covering MATH is induced from a covering MATH by open formal subschemes. It is sufficient to prove that MATH which follows from the fact that for every open formal subscheme MATH of MATH the function MATH restricts to MATH on MATH and the equalities MATH and MATH which follow from REF . |
math/0107134 | Let us prove the first assertion, the proof of the second one being similar. Choose MATH-models MATH and MATH of MATH and MATH respectively containing a weak NAME model MATH of MATH and MATH of MATH as an open dense formal subscheme. Also write MATH and MATH, with MATH and MATH in MATH and MATH, respectively. It enough to check that MATH is equal to MATH, which follows the fact that on MATH, the functions MATH and MATH are equal. |
math/0107134 | This follows from REF . |
math/0107134 | Denote by MATH the irreducible component of MATH with special fibre MATH. Since MATH is the disjoint union of the sets MATH, we may assume MATH is a smooth irreducible formal MATH-scheme of dimension MATH. Let MATH be a section of MATH which generates MATH at the generic point of MATH and induces a gauge form on the generic fibre. Let us remark that the function MATH is identically equal to MATH on MATH. Indeed, after shrinking MATH, we may write MATH with MATH a generator MATH a every point and MATH in MATH. By REF is a unit at the generic point of MATH. Assume at some point MATH of MATH, MATH ; it would follow that the locus of MATH is non empty in MATH, which contradicts the assumption that MATH induces a gauge form on the generic fibre. Hence, we get MATH, and the result follows. |
math/0107134 | Let MATH be a gauge form on MATH. By REF , the right hand side of REF is equal to MATH which does not depend on MATH. |
math/0107134 | Since any smooth rigid MATH-variety of dimension MATH admits a finite admissible covering by affinoids MATH, with MATH trivial, the morphism MATH is surjective. Hence it is enough to show the following statement: let MATH be a smooth formal MATH-scheme of relative dimension MATH with MATH trivial, and let MATH and MATH be two global sections of MATH inducing gauge forms on the generic fibre MATH, then MATH belongs to MATH. To prove this, we take MATH a global section of MATH such that MATH. If MATH is any global section of MATH, write MATH with MATH in MATH. By the maximum principle, the function MATH takes only a finite number of values on MATH. It follows we may write MATH as a disjoint union of the subsets MATH where MATH takes the value MATH. These subsets are stable cylinders and only a finite number of them are non empty. Hence the equality MATH holds in MATH and the statement follows. |
math/0107134 | By taking an appropriate admissible cover, we may assume there exists a gauge form on MATH, in which case the result follows from REF , since MATH. (In fact, one can also prove REF that way, but we prefered to give a proof which is quite parallel to that of NAME in CITE.) |
math/0107134 | One reduces to showing the following: let MATH be a smooth formal MATH-scheme of dimension MATH and let MATH be a function in MATH which induces a non vanishing function on MATH, then MATH with MATH the MATH-adic measure on MATH. It is enough to check that MATH is equal to the MATH-adic measure of the set of points MATH of MATH with MATH, which follows from REF . |
math/0107134 | Write MATH, with MATH a constructible subset if MATH. By REF. On the other hand MATH being smooth, the morphism MATH is surjective and its fibres are balls of radius MATH. It follows that the MATH-adic volume of MATH is equal to MATH. |
math/0107134 | By REF, the image of MATH in MATH converges to MATH in MATH. Since MATH by REF, one deduces the fact that MATH has a limit. To conclude we first remark that MATH for any non empty open subset MATH in MATH. Hence we may assume that MATH is locally free on a neighborhood of MATH. It then follows from REF, or rather from its proof, that MATH in MATH. Hence MATH belongs to MATH and not to MATH, and the result follows. |
math/0107134 | Clearly the mapping MATH is nondecreasing. Let MATH be a weak NAME model of MATH, with irreducible components MATH, MATH. Since MATH is smooth and irreducible, MATH is also smooth and irreducible, hence the NAME closure of MATH in MATH is irreducible. Since MATH is the union of the subschemes MATH, it follows that MATH is bounded by MATH. Now if MATH is not an essential component, there exists some weak NAME model of MATH, MATH, such that, if we denote by MATH the image of MATH in MATH, MATH is contained in the NAME closure of MATH. It follows that MATH is contained in the closure of MATH. The bound MATH follows. |
math/0107134 | Fix an integer MATH. By REF , for MATH, MATH . On the other hand, it follows from REF that MATH when MATH. |
math/0107136 | Suppose that MATH is a reflection about a hyperplane MATH, MATH and MATH are not in the adjacent alcoves. Then MATH can be written as a composition of reflections MATH such that MATH and each two weights MATH lie in adjacent alcoves. The number of reflections MATH is odd, and therefore any MATH belongs to MATH. By the definition of MATH for MATH, we have MATH if MATH is a reflection about one of the hyperplanes through MATH, then the condition that MATH can be dropped, and we get the statement of the Corollary. |
math/0107136 | Define a partial ordering in MATH by MATH if there exists a weight MATH such that MATH in the usual ordering in MATH. Then the basis of tilting characters is connected to MATH by a triangular transformation with respect to the ordering MATH. |
math/0107136 | Consider the graded version MATH of MATH with MATH, as defined in for example, REF, in which the toric part coincides with that of MATH. Then two simple modules MATH and MATH both are composition factors in an indecomposable MATH-module if and only if MATH REF. In the non-graded setting this means that any two simple composition factors of an indecomposable module have their highest weights in a MATH-orbit restricted modulo MATH, which coincides with the definition of an orbit of MATH-action. In particular, this means that the set MATH enumerates the blocks of the category MATH. |
math/0107136 | Use the notations introduced after REF . First we will show that for any MATH and a reflection MATH, the element MATH is in MATH. Decomposing the NAME characters in terms of simple composition factors over MATH, we can write MATH for some coefficients MATH. For each MATH the preimage of MATH under the homomorphism MATH contains a scalar multiple of an element of the form MATH for some MATH. Then MATH can be written as MATH, or in the basis of tilting characters, MATH for some coefficients MATH. We have MATH where MATH denotes the NAME denominator. For any MATH and MATH, we have MATH, and therefore MATH . If MATH is singular with respect to the MATH action, then the identity always holds. If MATH is regular, then it requires MATH. In both cases MATH, and therefore MATH. Now we will show that MATH cannot be bigger than required in the Theorem. Choose representatives for regular MATH-orbits in MATH so that MATH, and consider the characters MATH for MATH. Suppose that for some coefficients MATH there exists a linear combination MATH of characters of restrictions of tilting modules, which belongs to MATH. Then its pre-image under the homomorphism MATH lies in MATH. Since MATH correspond to pairwise distinct MATH-orbits in MATH, each of their pre-images MATH lies in MATH. Then MATH for each MATH, and hence MATH, which by construction of MATH requires each MATH. Therefore, no linear combination of characters of the restrictions of MATH lies in MATH, and MATH. On the other hand, MATH and the theorem follows. |
math/0107136 | Clearly, MATH is the fundamental domain for the shifted action of MATH on MATH, and MATH is the set of the regular weights with respect to this action. The ideal MATH of MATH is the image of MATH under the homomorphism MATH. The ideal of MATH spanned by MATH contains MATH, and therefore its image in MATH contains MATH. Compare the dimensions: the elements MATH form a basis in MATH. Therefore, MATH, which means that the algebras are isomorphic. |
math/0107136 | Any MATH can be written as a linear combination of characters of simple modules in MATH. Since MATH is projective, any element of the form MATH is a linear combination of characters of projective modules. Then we have MATH where MATH is the MATH-symmetric function REF , and MATH denotes the image in MATH of the character of the projective module MATH with highest weight MATH. Note that in the decomposition of MATH into a linear combination of projective characters, only the highest weights MATH can occur. Therefore, the two sets are related by a lower triangular transformation of bases, which provides a bijection between the span of indecomposable projective characters and MATH. |
math/0107136 | We will use the isomorphism MATH and prove the corresponding statement for MATH. The ideal of projective characters coincides with the linear span of MATH, thus all projective characters are in MATH. On the other hand, for any MATH, the element MATH is either zero, if MATH for some MATH, or is equal to MATH . If MATH, MATH is a projective and simple module CITE, and therefore MATH lies in the linear span of projective characters in MATH for any MATH. For nondominant weights, MATH is either zero or equals MATH for some MATH, and the Proposition follows. |
math/0107136 | This follows from REF and the factorization of MATH to the restricted weights REF . |
math/0107136 | For any MATH, the characters of any two projective MATH-modules with no composition factors of highest weight outside the orbit MATH, are proportional in MATH. The radical of the algebra MATH is spanned by the elements which annihilate MATH for any MATH. It coincides with the linear span of the differences MATH where MATH, MATH and we used the non-shifted MATH-action of MATH on the restricted weights. By REF , the character of the NAME module MATH annihilates the radical of MATH, and we have MATH for any MATH, MATH. By REF any projective character with composition factors of highest weights in one orbit MATH can be written as a linear combination MATH for some coefficients MATH, which by the above equals to MATH . This shows that any two projective modules in the same block of the category MATH have equal characters up to a scalar multiple. By REF , there exists at least one projective module in MATH with composition factors of highest weights in each MATH-orbit in MATH. Characters of modules in different blocks of category MATH are linearly independent in MATH, and we get the statement of the Theorem. |
math/0107136 | REF was observed in CITE: for MATH we have MATH if and only if MATH, which coincides with the set of highest weights of tilting modules in MATH. For REF , we can use the description of MATH in CITE. For each pair of nearest parallel hyperplanes of affine reflections in P, define the strip MATH to be the set of weights between these hyperplanes, and let MATH denote the set of all strips in MATH. Then by CITE, MATH . Therefore, MATH, which coincides with the set MATH. This is exactly the set of highest weights of the projective modules in MATH. |
math/0107136 | The character of the NAME module annihilates the radical of MATH and is therefore an element of MATH. Since MATH generates the algebra MATH over MATH, any element of MATH belongs to MATH. Now compare the dimensions: MATH where the first equality holds by REF , and the last follows from the block structure of MATH. Therefore, MATH. |
math/0107136 | Follows from the definition of MATH action on MATH. Note that here in the definition of the ideal, a reflection MATH can be replaced by any element MATH. |
math/0107136 | By REF , the character of MATH in MATH can be written as MATH for some coefficients MATH. As a character of an indecomposable projective module over MATH, it contains only NAME characters with highest weights in MATH. This requires MATH, which cannot be satisfied for MATH. Therefore, we have that MATH is proportional to MATH, and the normalization can be checked by evaluating the corresponding characters at MATH. |
math/0107136 | REF implies that we have MATH in MATH for all MATH, and we know that MATH is given by the NAME character formula. Multiplication gives MATH . All characters of projective MATH-modules in one MATH-orbit are proportional, and MATH is the fundamental domain for MATH-action which corresponds to the MATH-action after multiplication by MATH. Therefore, for each MATH there exists a unique projective character MATH such that MATH, and using the algebra isomorphism MATH, we obtain the Corollary. |
math/0107141 | According to NAME and NAME, CITE, every NAME polynomial occurs as the NAME polynomial of an unknotting number one knot. Hence, the proof is completed by finding irreducible NAME polynomials of arbitrarily high degree. Such examples include the cyclotomic polynomials MATH with MATH an odd prime. It is well known that cyclotomic polynomials are irreducible. We have that MATH . This is an NAME polynomial since MATH is symmetric and MATH. Hence, the unknotting number one knot with this polynomial has MATH but MATH. |
math/0107141 | Because MATH is algebraically slice, with respect to some generating set its NAME matrix is of the form MATH for some MATH matrices MATH, MATH, and MATH. Hence, MATH has homology presented by MATH, which is of the form MATH for other matrices, MATH and MATH, where MATH has nonzero determinant. The order of MATH is det-MATH. This presentation matrix corresponds to a generating set MATH. We claim that the set MATH generates a metabolizer. First, to see that it is self - annihilating with respect to the linking form, we recall that with respect to the same generating set the linking form is given by the matrix MATH . That this is the correct inverse can be checked by direct multiplication. The lower right hand block of zeroes implies the vanishing of the linking form on MATH. We next want to see that MATH generate a subgroup of order det-MATH. Clearly the MATH satisfy the relations given by the matrix MATH. What is not immediately clear is that the relations given by MATH generate all the relations that the MATH satisfy. To see this, note that any relations satisfied by the MATH are given as a linear combination of the rows of MATH. But since the block MATH has nonzero determinant, any such combination will involve the MATH unless all the coefficients corresponding to the last MATH rows of MATH vanish. This implies that the relation comes entirely from the matrix MATH. |
math/0107141 | A proof of the corresponding theorem for bilinear forms on vector spaces appears in CITE. A parallel proof for finite groups and linking forms can be constructed in a relatively straightforward manner. One such proof appears in CITE. Since all metabolizers split over the MATH - primary summands, the results follow for these summands. |
math/0107141 | The proof is straightforward. If MATH then MATH and MATH. Hence, MATH, so MATH. Similarly, if MATH and MATH, then MATH and MATH, so MATH and MATH. |
math/0107141 | It must be checked that this map is well-defined. Suppose first that MATH. Then for any MATH, MATH and MATH. Taking differences, we have that MATH, implying that MATH as desired. That this map is a homomorphism is trivially checked. To check injectivity, we need to show that for all MATH, MATH. But MATH since MATH by the definition of MATH. Since MATH, MATH is the identity coset, as needed. |
math/0107141 | If MATH is trivial we would have, by REF , an injection of MATH into MATH. But by REF the metabolizer MATH can be chosen so that it has rank less than MATH. It follows that a quotient will also have rank less than MATH. Hence, it cannot contain a subgroup of rank MATH. |
math/0107141 | Since MATH is slice, we let MATH be the metabolizer given by REF . We also have that MATH is algebraically slice, so we let MATH be an arbitrary metabolizer for MATH with rank-MATH and we let MATH be the metabolizer constructed above. We also let MATH be the nontrivial subgroup of MATH described above. Let MATH and MATH be characters on MATH vanishing on MATH. We are assuming further that MATH and MATH are in the same coset of MATH: MATH and MATH are both in MATH for some MATH. We want to show that MATH. Since MATH, we have that MATH and MATH. Hence, by REF , MATH . The result now follows immediately from the additivity of NAME - NAME invariants. |
math/0107141 | It is clear that MATH. However, since the rank of MATH is MATH, MATH. We must now show that MATH. This is based on the observation that the curves MATH form a strongly slice link: That is, they bound disjoint disks in MATH. To see this, note that by replacing the components of the MATH with copies of the complements of MATH and MATH, we have arranged that the MATH have become the connected sums of pairs of the form MATH, and such a connected sum is a slice knot. To build a genus REF surface in REF - ball bounded by MATH, simply surger the NAME surface using these slicing disks. |
math/0107141 | The difference MATH is given by: MATH . The set of values of MATH is a finite set, so is bounded above by a constant MATH. Pick MATH so that MATH. Pick MATH so that MATH. Finally pick each following MATH and MATH so that at each step the MATH signature has at least doubled over the previous choice. With this choice of knots the claim follows quickly from an elementary arithmetic argument. |
math/0107141 | Consider a MATH with MATH. Write MATH as a linear combination of the MATH and MATH. If MATH or some MATH has a nonzero coefficient, then MATH will link nontrivially with MATH or some MATH. In this case, either MATH or some MATH will be nontrivial. (Recall that the MATH and MATH are duals with respect to the linking form.) |
math/0107141 | To prove this, we have seen that we just need to show that one of the coefficients, either a MATH, MATH or MATH, is nonconstant on some coset. In the previous proof we used the MATH and MATH. We now focus on the MATH. Using the previous lemma, without loss of generality we can assume that MATH contains an element MATH for some set of coefficients MATH. The metabolizer MATH is of order MATH, so it must contain an element not in the span of MATH. Adding a multiple of MATH if need be, we can hence assume that MATH contains an element MATH, with some MATH or MATH nonzero. In fact, by changing sign, and adding a multiple of MATH, we can assume that one of the nonzero coefficients is REF. We can now select an element from the set MATH on which MATH evaluates to be REF. Denote that element MATH. We consider the MATH representing the pair MATH and MATH. In this case we have the following: CASE: MATH CASE: MATH CASE: MATH CASE: MATH . Using REF we have, for the corresponding MATH and MATH that: CASE: MATH CASE: MATH CASE: MATH CASE: MATH . Finally, from the definition of MATH we have that MATH and MATH. Hence, the NAME - NAME invariants cannot be constant on the coset and the proof is complete. |
math/0107145 | For a complex number MATH different from MATH, one checks immediately by induction on MATH, and determinant expansion of the first row, that MATH . Now, for MATH, taking MATH shows that MATH is an eigenvalue of MATH. Since we have MATH distinct eigenvalues for MATH, they all have multiplicity one. |
math/0107145 | Note that MATH, and this is a projection. Thus MATH . This is a family of pairwise orthogonal projections, since, if MATH, then either MATH, or the product of MATH and MATH is zero since it contains a factor MATH for at least one MATH. Since MATH is invertible, the result follows. |
math/0107145 | Let MATH and MATH. Here MATH . Thus MATH . It follows that the MATH are pairwise orthogonal. Moreover, MATH . Now, MATH since, for MATH, MATH. Also, MATH . Thus MATH, and, on right multiplying by MATH, we see MATH. |
math/0107145 | By REF , there is a complete family MATH of pairwise orthogonal projections in MATH, such that, if MATH, then MATH, and, by REF , MATH. Similarly, there is a complete family MATH of pairwise orthogonal projections in MATH such that, if MATH, then MATH, and MATH. By REF , there is a complete family MATH of pairwise orthogonal projections in MATH, such that, if MATH and MATH then MATH and MATH . Thus MATH is the NAME sum of the subspaces of the form MATH where MATH acts as multiplication by the scalar MATH. Hence MATH is the NAME sum of the MATH such that MATH. Therefore, MATH where MATH is the number of pairs MATH such that MATH, MATH, and MATH. But such pairs correspond bijectively to the fractions of the form MATH, MATH. Thus MATH. Hence MATH . Since MATH, the result follows. |
math/0107145 | By multiplying the numerator and denominator of MATH by a sufficiently high power of MATH, we see that MATH has an expression of the desired type. Now consider any expression MATH where MATH with MATH odd and MATH. We first show, by induction on MATH, that, if MATH, then MATH has an ideal whose dimension over MATH is MATH. Since the orthogonal complement is then an ideal of dimension MATH over the rationals, it amounts to the same if we consider only MATH. For MATH, we can take the zero ideal; thus, we may assume that MATH and the result holds for smaller MATH. Now MATH has a projection MATH; this is MATH for the subgroup MATH of order MATH in MATH. As rings MATH . By the induction hypothesis, the latter has an ideal of dimension MATH over MATH, and viewed in MATH this is an ideal of MATH. This completes the proof by induction. Hence, if MATH, then MATH has a projection MATH with MATH. Let MATH, so MATH, and MATH. By identifying MATH we see that, for MATH, we have projections MATH and MATH in MATH, with traces MATH and MATH, respectively, by REF . We claim there exist integers MATH, MATH with MATH such that MATH. We know that MATH. If MATH, then MATH, and we can take MATH and MATH. If MATH, then, by the division algorithm, MATH with MATH, and MATH. This proves the claim. Now let MATH, a sum of orthogonal projections. Thus, MATH is a projection and MATH as desired. It remains to show that MATH lies in MATH, but it is well known that this holds for all the idempotents of MATH. Alternatively, it is straightforward to check that all the projections involved in the foregoing proof have the right denominators. |
math/0107145 | Let MATH, and suppose that MATH satisfies a non-trivial differential equation over MATH, MATH where MATH, not all zero. By multiplying through by a common denominator, we may assume that all the MATH lie in MATH. (Notice it is natural not to have a ``constant term" on the right-hand side of REF since it could be eliminated by iterated derivation of the equation.) Viewing REF as a collection of equations, one for each power MATH, we see that there exists some MATH, and polynomials MATH such that MATH . Choose MATH, with MATH, and MATH such that MATH for all MATH, and all MATH with MATH. In other words, MATH eventually dominates all the MATH, MATH. It follows from the hypothesis on the MATH that there exists MATH such that MATH, and MATH for all MATH with MATH. Now take MATH. Then MATH, and MATH . Thus MATH . This contradiction shows that MATH does not satisfy any non-trivial differential equation over MATH, so, by REF , MATH is not algebraic over MATH. |
math/0107145 | In the following, MATH means MATH, and MATH means MATH. CASE: By NAME 's formula, MATH, and the latter is MATH, since MATH. One can argue directly that MATH, so MATH so MATH since MATH. CASE: By the Prime Number Theorem, MATH; see CITE. CASE: By NAME 's Theorem, MATH, where MATH is NAME 's constant; see CITE. By the Prime Number Theorem, MATH, so MATH. Since MATH, we see that MATH. The result now follows. |
math/0107145 | For each positive integer MATH, let MATH. Thus MATH . Now MATH so MATH . Thus MATH and MATH differ by an element of MATH, so it suffices to show that MATH is transcendental over MATH. By REF , it suffices to show that, for each MATH, there exist infinitely many MATH such that, whenever MATH satisfies MATH, MATH . We may suppose that MATH is fixed. Remember the MATH is the MATH-th prime number. For each MATH, let MATH . We may now suppose that MATH is fixed with MATH, and it suffices to show that MATH . We use the notation of REF , concerning MATH. Let MATH . Now suppose that MATH is an integer with MATH, let MATH and let MATH . We wish to bound MATH from below. Recall that, for any positive integer MATH, MATH, where the product is over the distinct prime divisors MATH of MATH. Thus MATH, where MATH denotes the number of divisors MATH of MATH. Also, MATH, which, by REF, is at least MATH. Thus MATH. Notice that MATH, since MATH divides MATH. From the definition of MATH, we see that MATH. Thus MATH . We next wish to bound MATH from above. Let MATH be the number, counting multiplicity, of prime factors of MATH, and let MATH be the factorization of MATH into prime factors. Then MATH, and MATH . Consider MATH. If MATH, then MATH divides MATH so MATH divides MATH. But MATH, so MATH, so MATH. Hence MATH divides MATH, but MATH, so MATH cannot divide MATH, so cannot divide MATH. Thus, the number of MATH which are less than MATH is at most MATH. Let MATH denote the number of MATH such that MATH, so MATH and MATH . Thus MATH . Hence it remains to show that MATH, or equivalently, MATH . Since MATH is the number, counting multiplicity, of prime factors MATH of MATH with MATH, MATH . We can write MATH by REF. Thus MATH . Hence MATH so MATH, and MATH . It follows that MATH as desired. |
math/0107147 | We think of MATH as a two-dimensional MATH-module, where MATH is the ring of real multiplications. Let MATH be the ramification index of MATH over MATH. The decomposability of MATH into an extension of MATH by MATH is a general fact about ordinary abelian varieties. Now the action of the inertia group MATH on MATH is an extension of the trivial character by the cyclotomic character. Since the ramification index of MATH is odd, these two characters are distinct. Therefore, MATH and MATH are also distinct. |
math/0107147 | CITE . |
math/0107147 | One checks that MATH is isomorphic over MATH to the NAME quadric threefold MATH. We also know (see CITE) an explicit MATH-parameter family of lines on MATH, which is to say a map MATH moreover, MATH is an isomorphism over any algebraically closed field. Composing MATH with an isomorphism between MATH and MATH yields a map MATH which is an isomorphism over any algebraically closed field. The set of MATH such that MATH consists of MATH distinct MATH-points is NAME. To check that it is not empty, we need only exhibit a single such line MATH in MATH . One such line is MATH . One checks that the restriction of MATH to MATH is MATH, which indeed has MATH distinct roots over MATH. Let MATH be the closed subscheme of MATH where the form MATH vanishes. Then MATH is a curve. Moreover, if MATH is a point in MATH, the subscheme of MATH parametrizing lines passing through MATH is one-dimensional. So the subscheme of MATH parametrizing lines intersecting MATH is at most two-dimensional. We may thus choose a point MATH such that MATH consists of four distinct MATH-points, none contained in MATH. Now let MATH be a lift of MATH to MATH. Then MATH is a line contained in MATH whose intersection with MATH consists of four distinct points defined over some unramified extension of MATH. Let MATH be the compositum of this extension with MATH. Since MATH is one-dimensional, we may choose MATH such that MATH is disjoint from MATH, by the same argument as above. Let MATH be a point in MATH, and choose integral coordinates for MATH so that at least one coordinate has non-positive valuation. Then MATH has non-positive valuation, so the third desired condition on MATH is satisfied. This completes the proof. |
math/0107147 | Let MATH be the determinant of the pushforward of the relative cotangent sheaf of MATH. Then MATH is a free rank MATH-module. Let MATH be a section generating MATH. Then every modular form MATH with coefficients MATH has a well-defined value MATH. Suppose MATH. Then by the hypothesis of the theorem, we have also that MATH and MATH. But this is impossible, as we show in the following paragraph. Let MATH be a prime, let MATH be an arbitrary full level MATH structure on MATH, and let MATH be a modular form of weight MATH and full level MATH. Since every even-weight modular form of full level MATH is integral over MATH we have MATH. But this is impossible, because for MATH sufficiently large, the sheaf MATH is very ample on the level MATH moduli scheme MATH. We conclude that MATH. So the mod MATH reduction MATH is not equal to MATH. Since the reduction mod MATH of MATH is the NAME invariant, MATH has good ordinary or multiplicative reduction by CITE. |
math/0107147 | We define new coordinates MATH on MATH by the rule MATH where MATH is a cube root of unity. With these coordinates, one checks that MATH is defined by MATH and MATH by MATH . So a family of lines in MATH is given by MATH . One checks that the equation for MATH is given by MATH where MATH are elements of MATH whose class in MATH is determined by MATH. The equation for MATH restricted to MATH is of the form MATH . Suppose that MATH and MATH are approximately equal and that both are much greater than MATH, which is in turn much greater than MATH. Then one checks that MATH . It is an easy calculation that MATH then factors over MATH into a constant and two quadratics, one of which has discriminant with valuation equal to that of MATH, the other of which has discriminant with valuation equal to that of MATH. In particular, if MATH and MATH are chosen so that MATH has the same parity as MATH, and MATH the same parity as MATH, the points of MATH are defined over an unramified extension of MATH. Note that since MATH and MATH are much larger than MATH, we know that MATH is very MATH-adically close to the cusp line MATH. Moreover, the form for MATH given above shows that the four points in MATH are MATH-adically close to MATH and MATH. These points do not lie in the image of any non-cuspidal point of the NAME modular surface MATH CITE. So the abelian varieties MATH corresponding to these four points to the points of MATH have potentially multiplicative reduction at MATH, as desired. |
math/0107147 | See CITE. |
math/0107147 | First of all, MATH is a character of NAME which annihilates complex conjugation, since MATH is odd. We thus have a totally real abelian extension MATH defined by MATH. Since MATH is trivial, MATH lies in the kernel of MATH, and MATH is unramified at MATH. Likewise, MATH is trivial, so MATH is unramified at MATH. By applying a quadratic twist, we may assume that MATH . Now the conditions on MATH and MATH imply the corresponding local conditions in REF . We may now choose an extension MATH and an abelian variety MATH satisfying the four hypotheses given in that Proposition. From here, we proceed along the lines of CITE. First, we claim that the irreducible representation MATH induced by the torsion subscheme MATH is modular. Now it follows from the discussion in REF that either MATH or MATH has multiplicative reduction at all primes of MATH over MATH. In case it is MATH which is semistable, return to the beginning, replace MATH by MATH, and start over. We may now assume MATH has multiplicative reduction at all primes of MATH over MATH. Now the subgroup MATH is unipotent. Thus, we can find a totally real solvable extension MATH, unramified over MATH and with odd ramification degree at every prime over MATH, such that MATH acts trivially on MATH for every prime MATH of MATH dividing MATH. Then the twist of the modular curve MATH by MATH is isomorphic to MATH when base changed to any MATH-adic completion of MATH. In particular, there exists an elliptic curve MATH such that MATH and MATH has good ordinary reduction at each prime of MATH over MATH. By another use of CITE, we can assume that MATH is an absolutely irreducible MATH-module. Since MATH has odd absolute ramification degree at all primes over MATH, we have by REF that MATH is distinguished at all primes over MATH. The mod-MATH representation MATH is modular by the NAME theorem, and the MATH-adic NAME representation MATH is modular by REF . It follows that MATH is modular, and so MATH, whence also MATH, is modular. By hypothesis, MATH has good ordinary or multiplicative reduction at MATH, so MATH is an ordinary representation. Because MATH has odd ramification degree over MATH, MATH is MATH-distinguished for all primes MATH dividing MATH. Now MATH is modular by another application of REF . This implies that MATH is also modular, which in turn implies the modularity of MATH, which is the restriction to MATH of our original representation MATH. Recall that the restriction of MATH to the decomposition group MATH is of the form MATH . By results of NAME as refined by CITE, MATH can be lifted to a MATH-adic representation MATH such that MATH . Now MATH is an ordinary, MATH-distinguished MATH-adic representation of MATH whose reduction mod MATH is isomorphic to the modular representation MATH. Applying REF once more, using MATH as MATH, we have that MATH is modular. Now we argue by cyclic descent as in CITE. Let MATH be a subfield of MATH such that MATH is a cyclic NAME extension. Then the automorphic form MATH on MATH corresponding to MATH is preserved by MATH. Therefore, MATH descends to an automorphic form on MATH. Continuing inductively, one finds that MATH itself is associated to a modular form on MATH; therefore, its mod MATH reduction MATH is modular. |
math/0107147 | Let MATH be a prime of the field of real multiplication dividing MATH. If MATH is absolutely reducible, then the corollary follows from REF and CITE. So we may assume that MATH is absolutely irreducible. If MATH is split or ramified in the ring of real multiplication MATH, then the corollary follows from NAME applied to MATH followed by NAME 's refinement of REF and CITE. If, on the other hand, MATH is inert in MATH, then MATH yields a representation MATH, which is easily seen to satisfy the conditions of REF . So MATH is modular, and it follows, again by NAME 's theorem, that MATH is modular. |
math/0107148 | Let MATH be a basis of MATH consisting of homogeneous elements, say of degrees MATH. For any MATH define MATH by MATH. Then MATH is a homogeneous element of degree MATH of MATH, the set MATH is a basis of MATH, and MATH for any MATH. In particular MATH is a complete system of orthogonal idempotents of MATH. Let MATH be an isomorphism of MATH-graded algebras, and define MATH for any MATH. Denote MATH, which is a graded vector subspace of MATH. Since MATH is a complete system of orthogonal idempotents of MATH, we have that MATH and MATH acts as identity on MATH for any MATH. Combined with the relation MATH, this shows that MATH induces an isomorphism of degree MATH from MATH to MATH. In particular MATH in MATH for any MATH. We obtain that MATH, showing that MATH has dimension REF. Repeating this argument for the identity isomorphism from MATH to MATH, and denoting by MATH, we obtain that MATH. Since MATH and MATH are graded vector spaces of dimension REF, we have that MATH for some MATH. Hence MATH. For the other way around, it is easy to see that MATH as MATH-graded algebras. |
math/0107148 | The orbit of an element MATH has length MATH if and only if the stabilizer of MATH is a subgroup with MATH elements of MATH, thus equal to MATH. The result follows now from the definition of MATH. |
math/0107148 | We have that MATH. The result follows now from REF and by applying the principle of inclusion and exclusion. |
math/0107148 | The number of isomorphism types of the good gradings is the number of orbits of the action of MATH on MATH. We have seen that if the orbit of an element MATH has length MATH, then necessarily MATH. The result follows since the number of orbits of length MATH is MATH. |
math/0107148 | Let MATH be the matrix algebra endowed with a certain MATH-grading, and pick MATH a graded simple module. Then MATH is a MATH-graded algebra. Moreover, MATH is a crossed product when regarded as an algebra graded by the support of the MATH-grading of MATH. Thus MATH for a cyclic group MATH and a cocycle MATH. On the other hand MATH is a finite field extension of MATH, so MATH. Hence MATH is a crossed product of MATH, which is central in MATH, and the cyclic group MATH, so MATH is commutative. Since MATH is a semisimple algebra with precisely one isomorphism type of simple module, say MATH, we have that MATH as MATH-modules for some positive integer MATH. But then MATH, and the commutativity of MATH shows that MATH must be MATH. Then MATH, so there exists a graded MATH-module which is simple as a MATH-module. By CITE, the grading is isomorphic to a good one. |
math/0107148 | We choose coset representatives MATH of MATH in MATH, and let MATH . For all MATH, we can find an automorphism MATH of MATH of degree MATH, and this implies that MATH and MATH are isomorphic, and have the same dimension. This can also be stated in the following way: MATH and it follows easily that MATH. Now take MATH. There exists an isomorphism MATH of degree MATH, and MATH is also an isomorphism of degree MATH, and this implies that MATH. If MATH, then MATH sends MATH to some MATH, with MATH, implying that MATH, and contradicting MATH. As a consequence, MATH, and MATH. |
math/0107148 | A straightforward adaption of the proof exhibited in CITE gives an exact sequence MATH with MATH . Now we have seen that MATH, and it is clear that MATH if and only if MATH, and it follows that MATH. |
math/0107148 | We have already seen that MATH is injective, so it suffices to show that MATH is surjective. The second statement then follows from the fact that the isomorphism classes of gradings are in bijective correspondence with MATH, and isomorphism classes of MATH-Galois extensions are in bijective correspondence with MATH, compare REF. Consider a cocycle in MATH. This cocycle corresponds to a MATH-Galois extension MATH of MATH, and MATH splits this cocycle, that is, it is represented by some cocycle in MATH and this cocycle is exactly the identity map MATH. To prove that MATH is surjective, it suffices to find an inverse image in MATH. Equivalently, we may look for the corresponding descent datum (see REF ) MATH is a finite abelian group, so we can write MATH with MATH. We know MATH once we know all the MATH. Recall from REF that MATH so MATH . We define MATH as follows: MATH with MATH given by MATH that is, the matrix of MATH with respect to the basis MATH is MATH . Clearly MATH is MATH-semilinear, and MATH, which means that MATH is a descent datum. For all MATH, MATH is homogeneous of degree MATH, and the same thing holds for the corresponding cocycle MATH. This means exactly that MATH maps MATH to the identity map MATH, proving that MATH is surjective. |
math/0107150 | Injectivity on the left of REF is clear. To show exactness in the center, first any MATH maps to MATH in MATH. On the other hand, suppose MATH and MATH is inner, say MATH with MATH. Then MATH represents the same class as MATH in MATH. In the case that MATH is transcendental over MATH, we show surjectivity on the right. We suppose MATH and MATH, where MATH and MATH are nilpotent. Without loss of generality we can assume MATH and MATH are both upper triangular. The spaces of biderivations MATH and MATH are both naturally MATH-linear, and the map defining elements of MATH, MATH is MATH-linear and respects the grading by degrees in MATH. Now MATH . If MATH is the matrix in MATH with a MATH in the MATH-th entry and zeros elsewhere, then MATH is an ordered MATH-basis for MATH. Using this basis, it follows from REF and the fact that MATH and MATH are both upper triangular that the map in REF, restricted to MATH, is lower triangular with MATH along the diagonal. Since MATH is non-zero by our assumption on MATH, MATH is an isomorphism of MATH-vector spaces. Therefore, for every MATH, if MATH, then MATH is inner, and so the right-hand side of REF is surjective. |
math/0107150 | An extension MATH of MATH by MATH splits if and only if MATH, that is, if there exists MATH such that MATH. If MATH, then we obtain an inner biderivation MATH such that MATH . If MATH is greater than MATH, we can repeatedly subtract REF from MATH, with MATH, MATH, to reduce the MATH-degree of MATH; eventually this degree will be MATH. Namely, any extension MATH is equivalent to an extension MATH with MATH, which we call the reduced representative of MATH. According to REF, the non-zero inner biderivations of least degree have degree MATH, so two extensions MATH and MATH satisfying MATH, MATH, and MATH, are inequivalent. Therefore the map MATH where MATH is the reduced representative of MATH, induces an isomorphism of MATH-modules. We now turn to the MATH-module structure on MATH. Recall from REF that multiplication-MATH on MATH is defined by MATH. Here we think of MATH as being an element of MATH. In order to see the action explicitly, it is enough to consider MATH: MATH . Using REF with MATH, MATH, we can rewrite the last identity of REF as MATH . Thus in terms of the elements MATH REF of MATH, the MATH-module structure on MATH can be expressed by the map MATH, defined by MATH . Comparing with NAME in CITE, it is clear that the MATH-module MATH is an extension of MATH by the MATH-module denoted there MATH. By REF and the characterization of MATH in REF, it is clear MATH is an extension of MATH by MATH. Thus MATH is the same as NAME 's MATH-module. Moreover, NAME shows CITE that MATH is a pure MATH-module isomorphic to MATH. |
math/0107150 | Consider the biderivation MATH of any extension MATH of MATH by MATH. The inner biderivations are of the form MATH for MATH. Explicitly, MATH defines an inner biderivation, MATH, if it has the form MATH . An inner biderivation MATH is said to be basic if MATH for all MATH and MATH with MATH; we write MATH. Explicitly written, these are MATH where for MATH the possible non-zero coordinates of MATH are MATH, MATH, and MATH. Every inner biderivation arises from an additive combination of basic MATH. Consider MATH consisting of elements MATH with MATH-degrees of MATH zero and the MATH-degree of MATH less than or equal one. In other words, MATH and MATH with MATH. Elements of MATH give rise via REF to biderivations which we will call reduced. The map MATH induces an isomorphism MATH of MATH-vector spaces. We need only to prove the first isomorphism by REF . Let MATH be an extension. We want to subtract appropriate MATH's in REF from MATH so that the resulting biderivation is reduced. In this process, we need to keep track of the MATH-degrees of the MATH's. We define the MATH-degree of MATH to be the vector MATH with MATH-degree of MATH. Given two vectors MATH and MATH with integer coefficients, we shall say MATH if MATH for all MATH. So our claim is that MATH can be reduced to a biderivation MATH such that MATH. Let MATH be the maximum of the integers MATH; one has MATH for all MATH. We can modify MATH by MATH for an appropriate MATH to obtain MATH such that MATH and MATH for MATH. Subtracting an appropriate MATH, we obtain MATH such that MATH and MATH for MATH. Repeating this for MATH using appropriate MATH, we obtain a MATH whose degree vector is less than or equal to MATH. If MATH, we are done. If not (MATH), we can subtract an appropriate MATH to obtain a vector MATH whose degree vector is less than or equal to MATH. We repeat the procedure in the two paragraphs above until we arrive at a vector MATH whose degree vector is less than or equal to MATH. We can now determine the MATH-module structure on MATH. By the previous lemma, it suffices to see this structure on MATH. Consider the elements MATH, MATH, defined as follows. For MATH, we take MATH to be the vector with MATH in the MATH-th coordinate and zeros elsewhere. We take MATH to be the vector with MATH in the last coordinate and zeros elsewhere. The structure of a MATH-module on MATH is completely described by the action of MATH on elements of the form MATH, MATH, since additive combinations of such elements give all of MATH. Consider MATH. The last is no longer an element of MATH, and we need its equivalent vector in MATH. Using the reduction procedure of REF , it is easily computed that MATH . Thus there is a MATH-module structure MATH on MATH, which is completely described by MATH . Here MATH, and the others are given by MATH, for MATH. Moreover, MATH is an extension of MATH by MATH. By REF , MATH is MATH-dimensional, which completes the proof. |
math/0107150 | If MATH is a NAME module, then a morphism MATH is represented by MATH such that MATH for all MATH. Then MATH induces a MATH-module homomorphism MATH, and on the level of biderivations, MATH . Also MATH takes MATH into MATH. Since NAME modules of different ranks have no non-zero morphisms between them, we can assume that the rank of MATH is the same as the rank of MATH. We continue with the considerations of the proof of REF . As in REF, we need to measure the effect of MATH on biderivations in MATH represented by MATH in REF. We see that for MATH, MATH . Using REF we can subtract suitable inner biderivations in MATH and find that in MATH, MATH where each MATH is a MATH-linear polynomial in MATH whose coefficients depend only on MATH. Thus MATH is represented by a matrix in MATH, and MATH restricts to a MATH-module morphism MATH. |
math/0107150 | Let MATH be the MATH matrix with a MATH in the MATH-th entry and zeros elsewhere. For MATH and MATH, by taking MATH for MATH in MATH, we define MATH . Since every MATH is a MATH-linear combination of matrices of the form MATH, biderivations arising from REF generate MATH as a MATH-vector space. Suppose that MATH is arbitrary and that MATH for some MATH. We will show that by subtracting matrices in REF from MATH in various ways we can replace MATH by a matrix MATH which has each entry of MATH-degree MATH and also MATH equivalent to MATH modulo MATH. We bootstrap our way through the entries of MATH in the following way. Let MATH . Define a function MATH from the set of subsets of MATH to itself. We set MATH to be those entries of MATH whose degrees in MATH can be decreased by subtracting an element of MATH, without increasing the degrees of the other entries. Our claim then is that MATH. The following containments can be easily checked: MATH where the last containment holds for all MATH. Therefore, we can assume that every entry of MATH is a constant from MATH. Now that the MATH-degree of each entry of MATH is MATH, one checks the containments in REF still hold with the exception that all sets on the right-hand side must have MATH removed. That is, we can adjust MATH so that it can be replaced by a matrix in MATH, but elements of MATH can be reduced no further. |
math/0107150 | Let MATH be given as in REF so that by REF MATH . Recalling the definition of MATH from the proof above, let MATH be basis vectors for MATH over MATH. Combining REF , MATH has a natural MATH-module structure which we now make explicit. For MATH, MATH and MATH . We note that as defined in REF, MATH and since this sum defines an inner MATH-biderivation, we subtract it from REF and find that MATH . Therefore, combining this with REF, we see that the MATH-module structure on MATH can be expressed as a MATH-module by the map MATH defined by MATH . Thus there is an exact sequence of MATH-modules, MATH where MATH is a MATH-dimensional iterated extension of MATH, and one checks that moreover MATH. |
math/0107150 | The proof here is essentially the same as the that of the existence of the exponential function, and in particular we can easily adapt the proof of REF to this situation. |
math/0107150 | This follows directly from REF , using the fact that MATH if and only if MATH for all MATH. |
math/0107150 | The map MATH on MATH is clearly well-defined and MATH-linear by REF . If MATH, then by REF, and so by REF , MATH. Therefore MATH is well-defined. Furthermore, from REF , it follows that MATH and so MATH, for all MATH. |
math/0107150 | Let MATH and MATH be MATH-bases for MATH and MATH respectively. Because MATH and MATH are both MATH-dimensional, MATH. As in REF, MATH . We claim that MATH. Suppose MATH. Then MATH . Because MATH, there is a non-trivial MATH-linear dependency MATH, and thus MATH. Since MATH are linearly independent over MATH, it follows that MATH. By the following argument, the image of MATH is discrete in MATH. We proceed by induction on MATH, for which the base REF is trivial. Furthermore, it suffices to continue with MATH, which we will now assume; the cases where MATH follow as straightforward consequences. The MATH-linear map MATH has kernel MATH. We need to show that MATH is a MATH-dimesional vector space over MATH. The typical element of MATH has the form MATH where MATH. Let MATH denote the MATH matrix in the above formula. By our assumption that MATH, the matrix MATH is square, and its determinant is MATH where MATH is the MATH minor of MATH with the MATH-th column removed. If MATH, then this implies that MATH, and since MATH are MATH-linearly independent, MATH for each MATH. Thus the rank of the matrix MATH is less than MATH, and we can rewrite the last row of MATH as a MATH-linear combination of the other rows, say MATH with MATH. The formulation in REF can be rewritten as MATH where MATH is the MATH matrix obtained by removing the last row of MATH. Again MATH, but then our induction hypothesis with MATH replaced by MATH allows us to conclude that MATH. |
math/0107150 | Given MATH REF as the quotient of MATH by lattices MATH, where each MATH may be taken to be have positive imaginary part, we see from CITE that MATH is naturally the quotient of MATH by the subgroup MATH generated by MATH and MATH. Let us suppose that an elliptic curve MATH has complex multiplication; let us think of MATH as a quotient of MATH by MATH. By the theory of complex multiplication CITE, we have the following: CASE: MATH lies in an imaginary quadratic field; CASE: MATH for some integers MATH and MATH; and REF if MATH (with MATH) is an elliptic curve isogenous to MATH, then MATH and MATH lie in the same imaginary quadratic field and MATH also has complex multiplication with the same NAME. If MATH, then we have MATH. By REF , we get that MATH is the lattice MATH. Thus we obtain that MATH is naturally identified with MATH. The last statement in REF follows from CITE. If MATH and MATH are isogenous curves with complex multiplication, then MATH and MATH lie in the same imaginary quadratic field MATH; in this case, MATH is isomorphic to a fractional ideal of an order of MATH. This proves REF . For REF , suppose at least one of MATH and MATH does not admit complex multiplication. By the fundamental theorem of complex multiplication, MATH and MATH are not both contained in one imaginary quadratic field. In other words, in this case MATH is a subgroup of MATH-rank greater than two, and so it is not a discrete subgroup of MATH. |
math/0107150 | The fact that MATH is fully faithful implies that MATH is injective in the following way. Suppose the images of MATH and MATH under MATH coincide. Pick representatives of MATH and MATH, that is, extensions MATH and MATH of MATH by MATH which satisfy MATH. We obtain a commutative diagram MATH where MATH is an isomorphism of MATH-modules. Since MATH is fully faithful, MATH is an isomorphism of MATH-motives. More precisely, MATH is induced by an isomorphism MATH of MATH-motives. It remains to show the surjectivity of MATH. For this, we have to show the following: given any extension MATH of MATH by MATH in MATH, the left MATH-module MATH is a MATH-motive, that is, REF it is free and finitely generated as a MATH-module and REF the associated primes of MATH, viewed as a module over the commutative ring MATH, consist only of the principal ideal MATH. Geometrically, we want the coherent sheaf associated to MATH on MATH to be supported only at the point MATH. REF is clearly satisfied by MATH by general properties of modules over the ring MATH CITE. For REF , consider an extension MATH of MATH by MATH where MATH and MATH are finitely generated MATH-modules. Every associated prime of MATH is an associated prime of either MATH or MATH. Also, if MATH is a quotient module of MATH, then every associated prime of MATH is an associated prime of MATH (this is easy to see via the geometric interpretation). Now, by assumption, the associated primes of the MATH-modules MATH and MATH consist of just the ideal MATH. The MATH-module MATH is an extension of MATH by MATH (MATH a quotient module of MATH). So we may apply the comments in the previous paragraph to the extension MATH to deduce that MATH satisfies REF . |
math/0107151 | Consider the product action of MATH on MATH in which each MATH factor acts by multiplication on the corresponding factor of MATH. This is a Hamiltonian action. If MATH is the moment map associated with MATH, then the moment map of this product action is MATH, with MATH . Let MATH. If MATH for every MATH, then MATH and MATH are equivariantly diffeomorphic, so the reduced space MATH is diffeomorphic to MATH. Moreover, there is another action of MATH on MATH, namely MATH coupled with the trivial action on MATH. Since this commutes with the product action, it induces a Hamiltonian action of MATH on MATH. In addition, one gets from MATH an involution MATH of MATH. This induces an anti-symplectic involution MATH on MATH. Thus, one can apply NAME 's theorem to MATH to get a formula for the cohomology of the space MATH in terms of the cohomology of the spaces MATH . Now MATH is obtained from MATH by attaching cells of dimension MATH and higher. So, for fixed MATH, the sequence MATH stabilizes as MATH grows large, and moreover is equal to the equivariant cohomology of MATH. Thus one obtains from REF the following real analogue: MATH where MATH. |
math/0107151 | REF tells us that the MATH-term of the NAME spectral sequence converging to MATH takes precisely the form that MATH itself is asserted to have. However, the results of the previous section tell us that the graded pieces of MATH have the same dimension as those of this MATH-term, and hence that the spectral sequence collapses. Therefore, MATH so MATH is a free module over MATH generated in dimension zero. Thus, its additive structure is as given by REF . |
math/0107151 | First of all, since the map MATH factors as MATH we know that for all MATH, the inclusion MATH holds. For the other direction, recall first that MATH is free; therefore, by REF the map MATH is injective and we may consequently view MATH as a submodule of MATH. Suppose MATH is a MATH-basis for MATH. By REF , there is a monomial MATH with MATH for every MATH. Thus, we may write MATH for unique MATH. Now, since MATH is a unique factorization domain, we may divide both sides of this identity by MATH and cancel common factors to obtain the formula MATH where the MATH are uniquely-determined elements of MATH and the MATH are uniquely-determined divisors of MATH such that MATH and MATH are relatively prime. Suppose now that MATH were actually in MATH. We may find a subset MATH such that no MATH kills MATH, and MATH annihilates the cokernel of the map MATH . Therefore, multiplying both sides of REF by MATH, we find that MATH with MATH. Thus, in REF , none of the weights MATH that divide the denominators MATH vanish on MATH. Hence, if MATH for all MATH, then MATH for all MATH, and so REF tells us that MATH and the proof is complete. |
math/0107151 | The result follows immediately from REF . |
math/0107151 | This follows immediately from REF . |
math/0107157 | CASE: Note that MATH acts on MATH . Set MATH, for MATH and MATH, otherwise. First we show REF. Since the fixed point algebra of MATH is the diagonal MATH clearly MATH . Now let MATH; then, if MATH where the last equality uses the translation invariance of NAME measure on MATH . Thus, the fixed point algebra of MATH which by assumption is MATH contains MATH . Thus MATH . Next, observe that MATH . Indeed, let MATH . Then, for MATH . Conversely, if MATH a similar calculation shows that MATH . Thus, REF is clear, as is REF. To show REF, let MATH be a continuous linear functional on MATH and suppose MATH . Fix MATH and consider the continuous function MATH on MATH . By the characterization of MATH above, we have MATH for all MATH . Thus, MATH and hence MATH . Since MATH was arbitrary, it follows that MATH . Thus, MATH is dense in MATH . Since by assumption the spaces MATH are MATH we have that MATH is dense in in MATH . REF implies that each MATH is a closed MATH-bimodule. Hence, by REF , MATH is spanned by the matrix units it contains. Write MATH . We claim: MATH . Let MATH and suppose MATH . Define a representation MATH of MATH as follows: let MATH be a NAME space with orthonormal basis MATH and define MATH on matrix units by MATH . By CITE MATH extends to a representation of MATH . By supposition, MATH for MATH hence for MATH . Since MATH is dense in MATH we have that MATH for MATH . That is impossible, since there is a matrix unit MATH with MATH hence MATH . This proves the claim. Next, MATH for MATH . Indeed, the intersection MATH is open, so if it is nonempty there is a matrix unit MATH such that MATH lies in the intersection. But then MATH contradicting REF. Define a cocycle MATH by MATH if MATH . Then MATH is well-defined, and continuous since MATH is open in MATH . Also, if MATH then by REF MATH so MATH that is, the cocycle condition is satisfied on MATH . MATH can be extended to a cocycle on MATH by setting MATH if MATH . One easily checks that the cocycle condition is satisfied on the groupoid MATH . CASE: Let MATH be the partial homeomorphism on MATH whose graph MATH is MATH . It is clear from REF that MATH form a nested sequence and that the graphs MATH have the required properties. CASE: Write MATH . We will define a MATH action on MATH first by defining it on matrix units. Here it will be convenient to regard MATH as MATH . Let MATH be a matrix unit in MATH then (by compactness) MATH for some MATH . Observe that MATH is both open and closed in MATH so that, setting MATH each MATH is a matrix unit or sum of matrix units in MATH . Now set MATH . MATH extends by linearity to (the dense subalgebra of) all finite linear combinations of matrix units. A short calculation shows that MATH is isometric, so it extends to an isometric map of MATH . By the nested property of the MATH the automorphism property MATH holds for MATH matrix units, hence for linear combinations of matrix units, and finally for arbitrary MATH . One can verify directly, or use CITE or CITE to get that MATH is the restriction of a star automorphism of MATH . Finally, one notes that the action MATH is continous in the pointwise-norm topology; that is, for each MATH the map MATH is norm continuous. |
math/0107157 | CASE: With the spectral subspaces MATH defined as in the proof of REF, the condition that MATH be semi-saturated implies that MATH . In particular, this implies REF. Otherwise the proof is the same as REF. Define an action of MATH on the algebraic direct sum of the MATH by MATH . MATH extends to an action of MATH on MATH which is dense MATH hence to an action of MATH . Define MATH on MATH by MATH . This gives rise to an isometric action on MATH and hence a star-action of MATH on the norm-closure of MATH which is MATH . REF implies that the action is semi-saturated. CASE: We use the same notation as in REF . Thus, MATH . By assumption, MATH so that MATH . So MATH is the graph MATH of a partial homeomorphism MATH of MATH and the graph MATH is disjoint from the diagonal set MATH . Also, repeated application of REF implies that MATH . Now as in the proof of REF , MATH so that MATH . Hence, MATH . CASE: Let MATH be the closed linear span of the set MATH . We claim that MATH . CASE: MATH . Of course if MATH it is true by assumption. If MATH for some MATH, let MATH . Then there is a MATH with MATH and MATH . But then MATH and MATH are both in MATH or MATH so that MATH and MATH contrary to hypothesis. CASE: MATH . Suppose MATH . Then MATH and MATH . But MATH so that MATH . Since by REF the graph of MATH is disjoint from the diagonal, this is impossible. Thus the claim is established. From the claim we have MATH for MATH . REF follows from the fact that MATH contains the algebra spanned by the matrix units of MATH . CASE: Follows from REF. CASE: Let MATH . This is open in MATH and disjoint from MATH . It follows that MATH and hence MATH so (with MATH), MATH . CASE: Observe that the condition that MATH is open in MATH implies that the sets MATH are disjoint and open. Furthermore, the orbit (or equivalence class) of a point MATH is given by MATH (where for MATH negative, MATH denotes the MATH-fold composite of MATH at MATH . ) In particular, each orbit has the order type of a subset of MATH so that MATH is finite on the groupoid MATH if and only if MATH . Thus the counting cocycle MATH is finite and continuous on MATH and MATH . |
math/0107157 | Each MATH is a discrete space, hence metrizable. Let MATH, with the product topology. Therefore, MATH is compact and metrizable and MATH. The topology of MATH is equal to that inherited from MATH. MATH is separable because it has a countable base MATH. |
math/0107157 | We prove only REF, as the proof of REF is similar. Let MATH. If MATH is finite, then MATH for some MATH. Therefore, MATH. So, we may assume that MATH is infinite. We are going to show that there exists a sequence MATH such that MATH converges to some MATH. Choose MATH such that MATH. Then we have MATH for sufficiently large MATH, and hence MATH. For each MATH, let MATH be the unique minimal path from MATH to MATH such that MATH. Let MATH be the point satisfying MATH for MATH and MATH for MATH. Then MATH for some MATH. Since MATH is infinite, MATH is also infinite. Let MATH be an accumulation point of MATH. Then MATH and there exists a subsequence MATH such that MATH converges to MATH. |
math/0107157 | This follows from the fact that MATH is a homeomorphism from MATH to MATH . |
math/0107157 | We show REF MATH REF. By the remark, we may assume that MATH for some clopen MATH. Suppose REF fails to hold. Then there exists MATH such that for all MATH . In particular, MATH so the forward orbit MATH is defined. Since MATH is open, the closed orbit MATH does not intersect MATH, and hence MATH . Let MATH . Thus there is a sequence MATH with MATH . By REF , MATH for some MATH . Thus, MATH . Since MATH is open, there is a MATH with MATH for which MATH . But then MATH . This is a contradiction, and the proof is complete. The implication REF MATH REF is analogous. |
math/0107157 | Assume to the contrary there is a periodic point MATH with period MATH . Then the orbit MATH is finite and evidently disjoint from MATH . Thus there is a clopen neighborhood MATH of MATH which is disjoint from MATH . But by REF of NAME systems there is a nonnegative integer MATH with MATH that is, MATH a contradiction. |
math/0107157 | The NAME diagram MATH to be constructed will have as its vertices the set MATH . For convenience of notation, define MATH. Let MATH. For each MATH and MATH, if MATH for some MATH, MATH, we put an edge MATH from MATH to MATH. We will indicate the correspondence between MATH and MATH by MATH. Define MATH and MATH. For two edges MATH, MATH with MATH, we put MATH if MATH. This defines an ordered NAME diagram MATH. Let MATH be the corresponding NAME partial dynamical system . We are going to show that MATH is conjugate to MATH . Since MATH, we have MATH for all minimal edges MATH. Suppose MATH is a maximal edge from MATH to MATH. Let MATH. Then we have MATH but MATH for all MATH. Since for each MATH, MATH or MATH, we have MATH for MATH. On the other hand, MATH. Hence, MATH, which implies that MATH. For MATH, let MATH be an edge from MATH to MATH, then MATH is a finite path in MATH from MATH to MATH. For each MATH, let MATH. Define MATH where MATH. Suppose MATH. Let MATH and MATH. We have MATH . Define a map MATH as follows. Let MATH be the infinite path MATH. Then let MATH. Clearly, MATH is a homeomorphism such that MATH and MATH. Let MATH such that MATH is an edge from MATH to MATH and MATH for all MATH. If MATH is not maximal, let MATH be the smallest integer such that MATH is not maximal. Let MATH be the smallest integer MATH such that MATH for some MATH, MATH. Since MATH for all MATH and MATH, we have MATH. For MATH, MATH because MATH is maximal. Therefore, MATH. Let MATH be the edge from MATH to MATH with MATH. Let MATH be the unique minimal path from MATH to MATH. Then MATH. For MATH, let MATH. We have MATH . It follows that MATH. |
math/0107157 | REF was proved in REF To begin, we need the fact that if a strongly maximal TAF algebra is the inductive limit of a system MATH then the spectrum MATH of MATH is the projective limit of the spectra of MATH . It follows in this case that the spectrum MATH of MATH is MATH . Suppose REF. is satisfied, and let MATH be a homeomorphism which induces a conjugacy of the two partial dynamical systems. Define MATH, MATH. Then MATH is a semigroupoid isomorphism, and so by CITE, MATH are isometrically isomorphic. CASE: Conversely, suppose there is a semigroupoid isomorphism MATH . Setting MATH note that MATH . Indeed, if MATH, then MATH forcing MATH . Thus there is a homeomorphism MATH so that MATH . Also, MATH from which MATH . Since MATH, MATH for some MATH . Now MATH is impossible, as MATH maps MATH into MATH. Now if MATH then MATH is the composition of two elements of MATH . Applying MATH we have that MATH is the composition of two elements of MATH . But this is impossible, as MATH is the immediate successor of MATH . Thus, MATH . In other words, MATH is a conjugacy of the two partial dynamical systems. |
math/0107157 | Define MATH by MATH . We are going to prove that CASE: MATH is bijective and MATH . CASE: MATH is a homeomorphism. CASE: MATH. Proof of REF. Suppose MATH and MATH. Let MATH . So MATH. If MATH, then we have MATH. Therefore, MATH and MATH, a contradiction. Therefore, MATH. Similarly, MATH and consequently, MATH. This proves that MATH is one to one. Let MATH and MATH. Then MATH for some MATH. Suppose MATH and MATH. Then MATH and MATH. Therefore, MATH. Hence, MATH and MATH. Proof of REF. Let MATH be a closed subset of MATH. We will show that MATH is closed. Suppose MATH and MATH. By choosing a subsequence if necessary, we may assume that for some fixed MATH, MATH for all MATH and MATH. Therefore, MATH. Suppose MATH and MATH. Then MATH for sufficiently large MATH. Hence, MATH and MATH. Therefore, MATH is closed. Since MATH is bijective, MATH is open for every open set MATH in MATH . The same proof shows that MATH is open for any open set MATH in MATH . Consequently, MATH is a homeomorphism. Proof of REF. Since MATH, we have MATH . Hence, MATH. Similarly, MATH. It follows that MATH is a NAME system. By construction, this system generates the same groupoid as that defined by the nest MATH. |
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