paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0007029 | We first identify MATH with MATH as follows: for MATH define MATH by MATH it is straightforward to check that this defines a degree - preserving functor and thus an element of MATH. Under this identification MATH for all MATH, MATH. As in the proof of CITE define a map MATH as follows: for MATH with MATH set MATH where... |
math/0007029 | Since MATH is defined to be the universal MATH - algebra generated by the MATH's subject to the relations REF and MATH preserves these relations it is clear that it defines an action of MATH on MATH. The rest of the proof follows in the same manner as that of CITE (see CITE). |
math/0007029 | For every MATH let MATH be the closed linear span of elements of the form MATH with MATH. Fix MATH, MATH with MATH we claim that MATH if MATH. If MATH then by REF there are MATH, MATH with MATH and MATH such that MATH; but then we have MATH . Thus MATH and hence by the factorisation property MATH. Consequently MATH by ... |
math/0007029 | Observe that the map MATH given by MATH satisfies MATH . The first part of the result then follows from REF. To show that MATH is amenable we first observe that MATH is amenable. Since MATH is amenable, we may apply CITE to deduce that MATH is amenable. Since MATH is strongly NAME equivalent to the crossed product of a... |
math/0007029 | The first statement follows from the second with MATH; indeed, by REF there is a functor MATH such that MATH in an equivariant way. The second statement follows from applying CITE to the natural MATH-action on MATH. The final statement follows from the identifications MATH and CITE. |
math/0007032 | Let MATH be a normal arc. Let MATH be the two regions of MATH that contain MATH. Let MATH be a REF - normal surface. Since MATH, the number of components of MATH containing MATH and the number of components of MATH containing MATH coincide. Thus MATH satisfies the matching equation associated to MATH. We refer to these... |
math/0007032 | In Construction REF, we have MATH by REF . The iteration stops after MATH steps, thus MATH using MATH. Since MATH equals twice the number of edges of MATH, we have MATH, and the lemma follows. |
math/0007032 | It is to show that any REF - normal sphere MATH is isotopic mod MATH to a component of MATH. Let MATH be the component of MATH that contains MATH. If MATH contains a REF - normal fundamental projective plane MATH, then MATH by Construction REF. Thus MATH, which is isotopic mod MATH to a component of MATH. Hence we can ... |
math/0007032 | Let MATH be a REF - normal sphere with exactly one octagon. If MATH contains a REF - normal fundamental projective plane MATH, then MATH by Construction REF, and any pre-normal surface in MATH is a multiple of MATH, that is, is REF - normal. Thus since MATH is not REF - normal, there is no REF - normal fundamental proj... |
math/0007032 | Set MATH. Let MATH be a lower compressing disc for MATH. One can assume that a collar of MATH in MATH is contained in MATH. Then, since by REF, any point in MATH is endpoint of an arc in MATH. Therefore there is a sub-disc MATH, bounded by parts of MATH and of arcs in MATH, that is a lower compressing disc for one of M... |
math/0007032 | If MATH comprises more than a single point then the string of MATH is contained in the string of MATH. Thus MATH contains an arc different from the base of MATH, bounding in MATH a lower compressing disc, that forms with MATH a pair of nested upper and lower compressing discs for MATH. Assume that a component MATH of M... |
math/0007032 | Let MATH be a compressing disc for MATH. Choose MATH and MATH up to isotopy of MATH mod MATH so that MATH is almost REF - normal and MATH is minimal. Choose an innermost component MATH, which is possible as MATH. There is a closed tetrahedron MATH of MATH and a component MATH of MATH that is a disc, such that MATH. Let... |
math/0007032 | We choose MATH up to isotopy of MATH in MATH so that MATH is minimal in lexicographic order. Assume that MATH. Then, there is a tetrahedron MATH, a REF - simplex MATH, a component MATH of MATH, and a component MATH of MATH with MATH. Since MATH is REF - normal, the closure MATH of one component of MATH is a disc with M... |
math/0007032 | Let MATH be a graph isomorphic to MATH. Since MATH is a copy of MATH, there is an embedding MATH with MATH, MATH, and MATH is the union of the normal arcs in MATH. Let MATH be a circle that does not meet MATH. Then, MATH bounds a disc MATH. The two components of MATH are discs. One of them is disjoint to MATH, since ot... |
math/0007032 | By hypothesis, MATH, and MATH has no lower compressing discs. Thus by REF , MATH has no lower compressing discs. Therefore its upper reduction MATH has no lower compressing discs. Assume that MATH is not almost REF - normal. Then MATH has a compressing disc MATH that is contained in a single tetrahedron MATH (see CITE)... |
math/0007032 | By the previous lemma, MATH is almost REF - normal. Recall that one obtains MATH up to isotopy mod MATH by splitting MATH along splitting discs that do not meet MATH. Assume that there is an arc MATH joining MATH with MATH. Let MATH be the component of MATH that contains MATH. By hypothesis, MATH is connected. Thus MAT... |
math/0007032 | If MATH, then the sphere MATH is REF - normal. It is not isotopic mod MATH to a component of MATH, since MATH. This contradicts the maximality of MATH. |
math/0007032 | Let MATH be REF - normal. We first consider the case where there is an almost REF - normal sphere MATH in MATH that has a compressing disc MATH, with string MATH and base MATH. We choose MATH innermost, so that MATH. In particular, MATH. Assume that MATH. Since MATH, there is an arc MATH that connects the endpoints of ... |
math/0007032 | By REF , there is an arc MATH that connects two components of MATH, say, MATH with MATH. By REF , MATH is contained in an edge of MATH. By REF - normal surfaces MATH have no lower compressing discs. Let MATH be a system of MATH arcs, such that the tube sum MATH of MATH along MATH is a sphere and an upper reduction MATH... |
math/0007032 | We apply REF to MATH, and together with the elementary reduction along MATH we obtain a sphere MATH with an almost REF - normal upper reduction MATH. One concludes MATH and the existence of MATH as in the proof of the previous lemma. |
math/0007032 | We apply REF with MATH to lower reductions of MATH, which is possible by symmetry. Thus, together with the elementary reduction along MATH, there is a lower reduction MATH of MATH, and MATH by REF . Since MATH and MATH, it follows MATH. Since MATH separates MATH from MATH, it follows MATH odd, thus MATH. |
math/0007032 | We give here just an outline. A complete proof can be found in CITE. Let MATH be an impermeable surface. By definition, it has upper and lower bonds with strings MATH. By isotopies mod MATH, one obtains from MATH two surfaces MATH, such that MATH has a return MATH with MATH, for MATH. A surface that has both upper and ... |
math/0007032 | Two octagons in different tetrahedra of MATH give rise to a pair of independent upper and lower compressing discs for MATH. Two octagons in one tetrahedron of MATH give rise to a pair of nested upper and lower compressing discs for MATH. Both is a contradiction to the impermeability of MATH. |
math/0007032 | By hypothesis, MATH has both upper and lower bonds. Assume that MATH does not contain octagons, that is, it is almost REF - normal. We will obtain a contradiction to the impermeability of MATH by showing that MATH has a pair of independent or nested compressing discs. According to REF , we can assume that MATH has a co... |
math/0007035 | REF follow from the definitions while REF follows from CITE or CITE. |
math/0007035 | The beginning of the proof is the same in all three cases. Note that MATH has a good filtration, simply because MATH for all MATH. Thus MATH also has a good filtration and so this induces a good filtration MATH on MATH such that MATH for all MATH. By REF , there exists MATH such that MATH for all MATH. Similarly, the r... |
math/0007035 | Let MATH be a left and right NAME filtration of MATH. Then REF implies that the induced filtration MATH on MATH is left and right NAME and so MATH is left (and right) noetherian. By CITE the growth of MATH is subexponential. By REF , MATH induces a good filtration on MATH as a left and a right MATH-module, which contra... |
math/0007035 | For REF , repeat the proof of REF , but with REF replaced by REF . For REF , use the proof of REF , applied to the opposite ring MATH, with REF replaced by REF . |
math/0007035 | Suppose that such a filtration MATH exists. By CITE the filtration MATH is also complete (in the natural sense that NAME sequences modulo the MATH should converge - see CITE). Thus CITE implies that the filtration is right NAME and the result follows REF . |
math/0007035 | Assume that MATH is not torsion. Replacing MATH by MATH, for some MATH, we may assume that MATH. Thus, REF implies that there exists MATH such that MATH for all MATH. Since MATH for MATH, this implies that MATH. Now consider MATH. Since MATH is a good filtration, MATH is finitely generated. We claim, for all MATH, that... |
math/0007035 | Suppose that such a filtration MATH exists. Let MATH be the induced filtration on MATH and MATH the induced filtration on MATH. Then MATH is a good left filtration and so MATH is finitely generated. The torsion submodule MATH of MATH is a MATH-bimodule and so we may pass to MATH without affecting the hypotheses (althou... |
math/0007035 | The diagonal matrices of the form MATH clearly form a subring MATH of MATH isomorphic to MATH. Since MATH is finitely generated as a left or right MATH-module, it follows that MATH is a finitely generated noetherian PI algebra. It is prime since it contains a nonzero ideal of the prime ring MATH. Suppose that MATH is a... |
math/0007035 | Notice that MATH, the ring from REF, and so MATH has no left noetherian finite filtration. Since MATH, the same is true on the right. |
math/0007035 | The proof that MATH is prime noetherian PI algebra over MATH is analogous to that for REF and is left to the reader. Clearly MATH is a complete local ring. In order to complete the proof it suffices, by REF , to work in a factor ring and we chose: MATH . The nilradical MATH of MATH is the set of strictly upper triangul... |
math/0007035 | The ring MATH is just the ring MATH from REF, and so does satisfy the hypotheses of the first paragraph. Set MATH . It is easy to see that MATH is generated by MATH and MATH and hence one has the standard MATH-filtration MATH, MATH and MATH for MATH. Let MATH and MATH be the images of MATH and MATH in MATH. Then, MATH ... |
math/0007035 | It is clear that MATH is a dualizing module for itself. By CITE one has natural isomorphisms: MATH for any finitely generated left MATH-module MATH. This implies that the injective dimension of MATH is bounded by the injective dimension of MATH. A similar assertion holds for MATH. It follows from CITE that MATH and MAT... |
math/0007035 | We check that the hypotheses of the lemma are satisfied for MATH. Since MATH is a free left MATH-module of rank two and a free right MATH-module of rank three, the proof is a routine computation which we will only outline. One first checks that, under the natural module structures, MATH where MATH is the matrix unit. I... |
math/0007036 | It holds that MATH by REF (compare CITE). This implies that MATH with MATH . It is easy to see that if MATH then MATH . |
math/0007036 | The assignment which sends a monomial MATH in MATH to MATH injects the union of the monomial bases in each MATH onto the monomials of degree MATH which are divisible by some MATH . It is easy to see that the cardinality of the set of complementary monomials of degree MATH is precisely MATH, where MATH denotes the dimen... |
math/0007036 | It is enough to mimic for the matrix MATH the proof performed by NAME in CITE to show that the determinant of the matrix MATH is an inertia form of the ideal MATH (that is, a multiple of the resultant). We include this proof for the convenience of the reader. Let MATH . Given an algebraically closed field MATH and MATH... |
math/0007036 | MATH is square if and only if MATH . (compare CITE). In this case, MATH because of NAME 's formula (compare CITE), we have that the right hand side equals MATH and the conclusion follows easily. Suppose now MATH . As in the introduction, let MATH denote the dimension of the MATH-vector space of elements of degree MATH ... |
math/0007036 | In CITE it is shown that MATH . This implies that MATH . In order to prove that MATH divides MATH we use the following trick: consider the ring MATH where MATH are new variables, and the polynomials MATH . Let MATH the matrix of the linear transformation MATH determined by REF but associated with the sequence MATH inst... |
math/0007036 | Set MATH . From the definitions of MATH and MATH it is easy to check that, if MATH is a maximal minor of MATH . Using NAME expansion, it is easy to see that MATH may be expanded as follows MATH where MATH, MATH is a maximal minor of MATH and MATH is a minor of size MATH in MATH . As each entry of MATH has degree MATH i... |
math/0007036 | We will prove this result by induction on MATH. The case MATH is obvious since MATH and MATH for any MATH . Suppose then that the statement holds for MATH variables and set MATH . Let MATH . We want to see that MATH is non negative. Recall from REF that, for every MATH, MATH equals the cardinality of the set MATH . The... |
math/0007036 | By REF , MATH has a maximum at MATH over MATH because MATH has a maximum and MATH has a minimum. If MATH we have that MATH . For MATH in this range, it is easy to check that MATH . Then, MATH . |
math/0007036 | When MATH is even, MATH and when MATH is odd, MATH . In both cases, MATH . Since MATH we deduce that MATH as wanted. |
math/0007036 | For MATH we get the NAME complex in degree MATH and so the specialized complex at a point MATH is exact if and only if MATH is a regular sequence, that is, if and only if the resultant does not vanish. The fact that the determinant of this complex equals the resultant goes back to ideas of NAME; for a proof see CITE, C... |
math/0007036 | Since MATH is a regular sequence in MATH the dimensions of the graded pieces of the quotient MATH in degrees MATH and MATH are MATH and MATH respectively. We can then choose blocks MATH and MATH as in REF such that MATH and MATH have maximal rank. Suppose without loss of generality that the blocks MATH and MATH have re... |
math/0007036 | The determinant of the resultant complex provides a determinantal formula for MATH precisely when MATH . This is respectively equivalent to the inequalities MATH and MATH from which the lemma follows easily. We have decreased the right hand side of REF by a unit in order to allow for a natural number MATH satisfying RE... |
math/0007036 | In order to prove that MATH is square, we need to check by REF that MATH . If there exists a determinantal formula, then the inequalily REF holds, from which it is straightforward to verify that MATH . To see that in fact REF holds, it is enough to check that MATH . But if the equality holds, we would have that MATH wh... |
math/0007036 | Denote MATH the homogeneous polynomial of degree MATH in MATH variables obtained by dividing the following determinant by MATH where MATH and MATH . It is easy to check that MATH and that MATH . We are looking for the elements in MATH of degree less or equal than MATH in the variables MATH . But it is easy to check tha... |
math/0007036 | Assume MATH . If REF is verified for MATH we deduce that MATH and so MATH . This equality cannot hold for any natural number MATH unless MATH . It is easy to check that for MATH there exist a determinantal NAME formula for any value of MATH . In case MATH, REF implies that MATH . Then, MATH as claimed. |
math/0007038 | By the uniformization theorem for super-Riemann surfaces, REF , any supersphere is superconformally isomorphic to the supersphere MATH. Let MATH be a supersphere with MATH tubes and MATH a superconformal isomorphism. We have a supersphere with MATH tubes MATH which is superconformally equivalent to REF . Let MATH . We ... |
math/0007038 | Let MATH be a superconformal equivalence from REF to REF . The conclusion of the proposition is equivalent to the assertion that MATH must be the identity map on MATH. By REF is a superconformal automorphism of MATH, that is, a superprojective transformation. Also by definition we have MATH . From REF and the fact that... |
math/0007038 | Given a supersphere with one tube MATH by the uniformization theorem for super-Riemann surfaces, REF , there exists a superconformal isomorphism MATH such that MATH is superconformally equivalent to REF . By REF in the proof of REF , we know there exists a (non-unique) superprojective transformation MATH from REF to MA... |
math/0007038 | Let MATH. Set MATH and let MATH denote the iterated bracket starting with the sequence of MATH of the MATH's then MATH of the MATH's, etc. Then, if well defined, the NAME formula gives MATH where MATH (compare CITE). For any MATH, let MATH be left multiplication by MATH. Then for MATH . Thus for MATH, we have MATH and ... |
math/0007038 | Given MATH consider the infinite system of REF for the unknown sequence MATH. Define MATH, and MATH for MATH. Then the infinite system of REF becomes a system of equations for MATH, and using REF , it is easy to show by induction on MATH, that this system of equations has a unique solution, that is, MATH . The proposit... |
math/0007038 | Using REF and the fact that MATH is a bijection, we have MATH . |
math/0007038 | Let MATH . Then MATH, that is, MATH is even, and thus MATH . Let MATH . Then MATH, that is, MATH is even, and we have MATH . Thus MATH . By REF , MATH . But in this case, the coefficient of MATH for a fixed MATH has terms with powers of MATH greater than or equal to MATH. Thus we can set MATH, and each power series in ... |
math/0007038 | Writing MATH, we have MATH . |
math/0007038 | For MATH, the term MATH has powers in MATH greater than or equal to MATH for MATH, and MATH has powers in MATH greater than or equal to MATH for MATH. Thus setting MATH in REF applied to this case, each power series in REF has only a finite number of MATH terms for a given MATH, that is, each term is a well-defined pow... |
math/0007038 | Write MATH. REF is trivial for MATH, MATH, and MATH. CASE: We prove the result for MATH, with MATH and MATH, by induction on MATH. Assume MATH for MATH, MATH. Let MATH. Then by REF , MATH . CASE: We next prove the result for MATH, MATH. Again by REF , MATH . Thus MATH . Assume MATH for MATH, MATH. Let MATH. Then by REF... |
math/0007038 | from REF , we have MATH . Similarly, MATH . Since the formal composition of two formal superconformal power series is again superconformal, by REF , MATH is superconformal. |
math/0007038 | Since the set of all formal superconformal power series of the form REF is a group with composition as its group operation, it is obvious from the definition of MATH that MATH is a group with this operation. Let MATH. By REF MATH . Or equivalently, MATH . But then from the definition of MATH, MATH . Thus we obtain REF ... |
math/0007038 | Since MATH is superconformal satisfying REF with even coefficient of MATH equal to MATH, the power series MATH is superconformal with MATH and with the even coefficient of MATH equal to one. Thus by REF , we have MATH . Write MATH. By the chain rule MATH and MATH . Therefore MATH which gives REF . By REF , we know that... |
math/0007038 | For MATH, and MATH, each MATH has powers in MATH less than or equal to MATH for MATH. Thus setting MATH in REF applied to this case, each power series in REF has only a finite number of MATH terms for a given MATH, that is, each term is a well-defined power series in MATH. |
math/0007038 | The proof is identical to REF , and REF in the proof of REF . To finish the proof, we only need note that since MATH for MATH and MATH, the result follows by linearity. |
math/0007038 | We need only define the existence of the necessary expressions since the proof of equality is the same as that for REF . Suppose MATH, MATH and MATH exist in MATH. Writing MATH for MATH, then MATH . Thus since MATH exists in MATH, and MATH this implies that MATH, for MATH, and MATH for MATH, exist in MATH. By definitio... |
math/0007038 | Let MATH from REF , we see that MATH. Then by REF , we have MATH . Therefore MATH . Using REF , we obtain REF . |
math/0007038 | The result follows from REF . |
math/0007038 | Write MATH where MATH is superconformal, and MATH are both homogeneous of degree MATH in the MATH's, degree MATH in the MATH's, degree MATH in the MATH's, and degree MATH in the MATH's, for MATH. The fact that MATH is superconformal implies that MATH where each MATH is homogeneous of degree MATH in the MATH's, degree M... |
math/0007038 | REF , and REF follow immediately from REF . By REF , we know that REF are in the algebra MATH. By definition, REF applied to MATH are in MATH respectively, and we have MATH . Thus for any MATH, we have MATH and MATH are in MATH . Then by REF is equal to MATH and by REF is equal to MATH . Thus the above expressions are ... |
math/0007038 | NAME 's theorem implies that for any MATH that is, MATH . Thus MATH . On the other hand, since MATH by REF , we have MATH . Thus MATH . By REF , this equality implies REF . |
math/0007038 | By NAME 's theorem MATH . On the other hand, since MATH by REF , we have MATH . Thus MATH . By REF , this equality implies REF . |
math/0007038 | We use NAME 's proof of REF generalized to the present situation. The idea is to use the NAME formula and compare the result with REF . However one must first establish the appropriate algebraic setting in which the NAME formula holds rigorously. Let MATH be a sequence of even formal variables, MATH a sequence of odd f... |
math/0007038 | The proof is the same as that for REF except that in this case we only consider the subalgebra with basis MATH, MATH, for MATH and use REF instead of REF . |
math/0007038 | The proof is the same as that for REF except that in this case we only consider the subalgebra with basis MATH, MATH, for MATH and use REF instead of REF . |
math/0007038 | Using the NAME modules MATH defined in REF , the proof is completely analogous to the proof of REF. |
math/0007038 | Since MATH is a positive energy module for MATH and by REF MATH for MATH, both the right-hand side and the left-hand side of REF are well-defined elements in MATH. Then by REF , they are equal. |
math/0007038 | The proof is analogous to the proof of REF except that we use REF instead of REF . |
math/0007038 | Since MATH is a positive energy module for MATH and by REF MATH for MATH, both the right-hand side and the left-hand side of REF are well-defined elements in MATH. Then by REF , they are equal. |
math/0007038 | The proof is analogous to the proof of REF except that we use REF instead of REF . |
math/0007038 | Since MATH is a positive energy module for MATH and by REF MATH for MATH, both the right-hand side and the left-hand side of REF are well-defined elements in MATH. Then by REF , they are equal. |
math/0007038 | The proof for MATH is by induction on MATH. If MATH, then MATH is a type REF supersphere sewn with a type REF supersphere. If MATH, then it is type REF . If MATH for MATH is given by MATH then MATH which consists of MATH type REF superspheres and the supersphere MATH sewn together. Thus we need only prove that MATH can... |
math/0007038 | Recall that in REF, we defined a group operation on infinite sequences in a superalgebra MATH; see REF . Letting MATH, and for MATH this operation is given by MATH . Let MATH . Since MATH, and MATH are in MATH, the superfunctions MATH and MATH are convergent in a neighborhood of MATH. Hence MATH and MATH are convergent... |
math/0007038 | Because each supersphere involved in the proposition has standard local coordinates at each puncture, the sewings depend only on the bodies of the superspheres. The sewing of the bodies is exactly the sewing given in REF. Thus as in REF there exist MATH which allow the above two sewings including the soul portions of t... |
math/0007038 | Note that MATH, and thus MATH . In fact, MATH represents the equivalence class of superspheres with MATH tubes with canonical supersphere having local coordinates (in terms of the coordinate chart about zero of the underlying super-Riemann sphere) at MATH given by MATH and at MATH given by MATH. This canonical supersph... |
math/0007038 | If MATH is given by REF , then MATH. Thus, the only thing we need to prove is REF . The canonical supersphere MATH is the super-Riemann sphere with one puncture at MATH and local coordinate at MATH given by MATH . Define MATH, and MATH, for MATH, and MATH by MATH and MATH . By definition of the sewing operation, MATH .... |
math/0007038 | From the definition of the sewing operation defined above, MATH and MATH are closed under MATH. The identity for both MATH and MATH is MATH . In MATH, the inverse of MATH is MATH and in MATH, the inverse of MATH is MATH. From the definition of the sewing operation, REF , we have MATH . Therefore MATH and MATH . Thus MA... |
math/0007038 | If the MATH-th tube of MATH can be sewn with REF-th tube of MATH, then by definition there exist MATH satisfying MATH and NAME open neighborhoods MATH and MATH of MATH and MATH, respectively, such that: MATH and MATH are the only punctures in MATH and MATH, respectively; MATH and MATH are convergent in MATH and MATH, r... |
math/0007038 | MATH is represented by a supersphere with tubes MATH given by the local coordinate system MATH with the coordinate transition function given by MATH for MATH such that MATH and MATH . We know that for each MATH the coordinate transition function MATH is superconformal. From the uniformizing function at MATH, that is, f... |
math/0007038 | By REF , MATH and MATH are analytic in MATH for MATH, and therefore MATH is also analytic in MATH for MATH. Thus MATH and MATH can be expanded as power series in MATH. The functions MATH as polynomials in the MATH and MATH coefficients of MATH and MATH can also be expanded as power series in MATH. Since MATH and MATH s... |
math/0007038 | We prove the bracket formula for MATH for MATH, MATH, and MATH. The proofs for the other cases are similar. To simplify notation, we will write, for example, MATH instead of MATH . For MATH and MATH, MATH . Using REF, we have MATH . Also, we have MATH and similarly MATH . For simplicity we will write MATH, for MATH. Le... |
math/0007039 | We separately consider each of the eight cases in the statement of the proposition. CASE: Let MATH. Replacing MATH by a conjugate under MATH, we may assume that MATH is orthogonal to MATH; that is, MATH. Then it is clear that MATH. CASE: Let MATH. We have MATH and MATH . Therefore MATH. CASE: For any large MATH, let MA... |
math/0007039 | We separately consider each of the seven cases in the statement of the proposition. CASE: Because MATH for every MATH, it is clear that MATH for every MATH. CASE: We have MATH, so MATH and MATH whenever MATH. Thus, MATH. We have MATH. If MATH, then there is some nonzero MATH, such that MATH. Then, for MATH, we have MAT... |
math/0007039 | It suffices to show that MATH is described in either REF or REF . We may assume MATH (otherwise, REF holds). Because MATH is a quadratic form of signature MATH on MATH, then we must have MATH. Thus, we may assume MATH (otherwise REF holds). Assume MATH and MATH for every MATH (and MATH). We may assume MATH for every MA... |
math/0007039 | Let us begin by establishing that MATH for MATH. If not, then we may assume either that MATH or that MATH. Then, because MATH, we see from REF that MATH . This contradiction establishes the claim. Assume there is a nonzero MATH, such that MATH and MATH. Note, from the preceding paragraph, that MATH. Then, because MATH,... |
math/0007039 | We separately consider each of the five cases in the statement of the proposition. CASE: From REF , we have MATH for all MATH. CASE: Replacing MATH by a conjugate under MATH, we may assume that MATH (and MATH). Then, from the assumption of this case, we know that MATH is also MATH. Therefore, REF implies that MATH for ... |
math/0007039 | We separately consider each of the eight cases in the statement of the proposition. CASE: From REF, we know that the quadratic form MATH is anisotropic on MATH, so MATH . CASE: Because MATH, we have MATH so REF implies MATH. CASE: From either REF or REF (depending on whether MATH is MATH or not), we have MATH for every... |
math/0007039 | It suffices to show that MATH is described in either REF or REF . We may assume REF holds (otherwise, Conclusion REF holds). Assume MATH for every MATH. We may assume there exists MATH, such that MATH (otherwise REF holds). Furthermore, we may assume MATH for every such MATH (otherwise REF holds). Then we may assume MA... |
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