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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 and MATH. Note that MATH and hence MATH. The rest of the proof proceeds as in CITE mutatis mutandis. |
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 cancellation and the claim is established. It follows that for each MATH with MATH the elements MATH with MATH form a system of matrix units and two systems associated to distinct MATH's are orthogonal (see REF). Hence we have MATH . By an argument similar to that in the proof of REF , if MATH then MATH (see REF ); our conclusion now follows. |
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 an AF algebra by a MATH - action, it falls in the bootstrap class MATH of CITE. The final assertion follows from the NAME classification theorem (see CITE). |
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 equations as MATH - matching equations. We will transform the system of MATH - matching equations by eliminating the components of MATH that do not belong to MATH. Let MATH be normal arcs, and let MATH be two different regions of MATH with MATH and MATH. Assume that MATH is a parallelity region of MATH. Then MATH, thus the matching equation associated to MATH is of the form MATH. Hence we can eliminate MATH in the MATH - matching equations. For any region MATH of MATH and any normal arc MATH, the elimination transforms the matching equation associated to MATH into the matching equation associated to MATH. We iterate the elimination process. Since any component of MATH contains a region that is not a parallelity region, we eventually transform the system of MATH - matching equations to a system MATH of matching equations that concern only regions of MATH of types I and II. Let MATH be a solution of MATH. By the elimination process, there is a unique extension of MATH to a solution MATH of the MATH - matching equations. If MATH satisfies the compatibility condition then so does MATH, since a parallelity region contains at most one class of normal squares. Now the lemma follows by REF , that is proven in CITE. |
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 assume that MATH does not contain a REF - normal fundamental projective plane. We express MATH as a sum MATH of fundamental surfaces in MATH. Since MATH and the NAME characteristic is additive, one of the fundamental surfaces in the sum, say, MATH with MATH, has positive NAME characteristic. It is not a projective plane by the preceding paragraph, thus it is a sphere. By construction of MATH, the sphere MATH is isotopic mod MATH to a component of MATH, thus it is parallel to a component of MATH. Hence MATH is disjoint to any REF - normal surface in MATH, up to isotopy mod MATH. Thus MATH is the disjoint union of MATH and MATH. Since MATH is connected, it follows MATH. Thus MATH is isotopic mod MATH to a component of MATH. |
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 projective plane in MATH. We write MATH as a sum of REF - normal fundamental surfaces in MATH. Since MATH has exactly one octagon, exactly one summand is not REF - normal. Since any projective plane in the sum is not REF - normal by the preceding paragraph, at most one summand is a projective plane. Since MATH and the NAME characteristic is additive, it follows that one of the fundamental surfaces in the sum is a sphere MATH. Assume that MATH is REF - normal, that is, MATH. The construction of MATH implies that MATH is isotopic mod MATH to a component of MATH. Thus it is disjoint to any REF - normal surface in MATH. Therefore MATH is a disjoint union of a multiple of MATH and of a REF - normal surface with exactly one octagon, which is a contradiction since MATH is connected. Hence MATH contains the octagon of MATH. We have MATH by REF . The claim follows with REF and MATH. |
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 MATH. |
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 MATH is a circle. Then there are discs MATH and MATH with MATH. Since MATH , MATH does not contain arcs of MATH. Thus if we choose MATH innermost in MATH, then MATH. By cut-and-paste of MATH along MATH, one reduces the number of circle components in MATH. Therefore we assume by now that MATH consists of isolated points in MATH and of arcs that do not meet MATH. Assume that there is a point MATH. Then there is an arc MATH with MATH. Without assumption, let MATH. Let MATH be the closure of the component of MATH whose boundary contains arcs in both MATH and MATH. Define MATH, that is to say, MATH is the connected sum of MATH and MATH along MATH. By construction, MATH, and MATH. In that way, we remove all points of intersection of MATH. Thus by now we can assume that MATH consists of arcs in MATH that do not meet MATH, and possibly of a single point in MATH. Let MATH be an outermost arc in MATH, that is to say, MATH bounds a disc MATH with MATH. We move MATH away from MATH by an isotopy mod MATH and obtain a compressing disc MATH for MATH with MATH. In that way, we remove all arcs of MATH and finally get a pair of independent upper and lower compressing discs for MATH. |
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 be the closed REF - simplex of MATH that contains MATH. We obtain three cases. CASE: Let MATH be a circle, thus MATH. Then there is a disc MATH with MATH and a ball MATH with MATH. By an isotopy mod MATH with support in MATH, we move MATH away from MATH, obtaining a surface MATH with a compressing disc MATH. If MATH is almost REF - normal, then we obtain a contradiction to our choice as MATH. CASE: Let MATH be an arc with endpoints in a single component MATH of MATH. Since MATH has no returns, MATH is not the string of MATH. We apply to MATH an isotopy mod MATH with support in MATH that moves MATH into MATH, and obtain a surface MATH with a compressing disc MATH. If MATH is almost REF - normal, then we obtain a contradiction to our choice as MATH. CASE: Let MATH be an arc with endpoints in two different components MATH of MATH. If both MATH and MATH are normal arcs, then set MATH, MATH and MATH. If, say, MATH is a circle, then we move MATH away from MATH by an isotopy mod MATH with support in MATH. If the resulting surface MATH is still almost REF - normal, then we obtain a contradiction to the choice of MATH. In either case, MATH is not almost REF-normal, that is, the isotopy introduces a return. Therefore there is a component of MATH with closure MATH such that MATH connects two normal arcs MATH of MATH. Let MATH. Up to isotopy of MATH mod MATH that is fixed on MATH, we assume that MATH. There is an arc MATH contained in an edge of MATH with MATH. For MATH, there is an arc MATH that connects MATH with MATH. The circle MATH bounds a closed disc MATH. Eventually MATH is a compressing disc for MATH contained in a single tetrahedron. |
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 MATH. By choosing MATH innermost in MATH, we can assume that MATH. An isotopy of MATH in MATH with support in MATH, moving MATH away from MATH, reduces MATH, in contradiction to our choice. Thus MATH. Now, assume that MATH. Then, there is a tetrahedron MATH, a REF - simplex MATH, and a disc component MATH of MATH, such that MATH is a single circle. There is a ball MATH bounded by MATH and a disc in MATH. An isotopy of MATH with support in MATH, moving MATH away from MATH, reduces MATH, in contradiction to our choice. Thus MATH is contained in a single tetrahedron MATH. Since MATH is REF - normal, MATH bounds a disc MATH in MATH. An isotopy with support in MATH that is constant on MATH moves MATH to MATH, which yields the lemma. |
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 otherwise the disc MATH would give rise to a splitting disc for MATH that is not isotopic mod MATH to a sub-disc of MATH, in contradiction to the preceding lemma. Thus by cut-and-paste along sub-discs of MATH, we can assume that additionally MATH. Let MATH be a circle so that MATH is contained in the boundary of a tetrahedron of MATH. Since MATH is REF - normal, MATH bounds an open disc in MATH. By the same argument as in the preceding paragraph, MATH bounds an open disc in MATH. One easily verifies that these two discs together with MATH bound a ball in MATH disjoint to MATH. Hence MATH is a disjoint union of balls in MATH, and this implies the existence of MATH. |
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), with string MATH and base MATH. Since MATH has no lower compressing discs, MATH is upper and does not contain proper compressing sub-discs. Thus MATH, that is, all components of MATH are circles. Since MATH is pre-normal, MATH is a disjoint union of discs. Therefore, since MATH is contained in a single tetrahedron, we can assume by isotopy of MATH mod MATH that MATH consists of arcs. We have MATH. It follows MATH, since otherwise a sub-disc of MATH is a compressing disc for MATH, which is impossible as MATH is REF - normal by hypothesis. Thus MATH and MATH. By an isotopy with support in MATH that is constant on MATH, we move MATH to MATH, and obtain from MATH a surface MATH that has MATH as upper bond. This is shown in REF , where MATH is indicated by plus signs and MATH is bold. The isotopy moves MATH to a system of arcs MATH and moves MATH to MATH with MATH. Since MATH, there is a homeomorphism MATH that is constant on MATH with MATH. One obtains MATH by a sequence of elementary reductions along bonds of MATH that are contained in MATH. These bonds are carried by MATH to bonds of MATH. Thus MATH is an upper reduction of MATH. Since MATH admits an elementary reduction along its upper bond MATH, we obtain a contradiction to the minimality of MATH. Thus MATH is almost REF - normal. |
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 MATH is connected, and there is an arc MATH with MATH. Since MATH split MATH, the set MATH is the only component of MATH with boundary MATH. Thus there is a component MATH of MATH connecting MATH with MATH. There is a splitting disc MATH of MATH contained in a single tetrahedron with MATH. By choosing MATH innermost, we assume that MATH is a single point in MATH. Since MATH is pre-normal and MATH is contained in a single tetrahedron, we can assume by isotopy of MATH mod MATH that MATH, thus MATH. Choose a disc MATH so that MATH and MATH. Then MATH, since MATH. We split MATH along MATH, pull the two components of MATH along MATH, and reglue. We obtain a surface MATH with MATH as an upper bond. Since a small collar of MATH in MATH is in MATH, there is a homeomorphism MATH that is constant on MATH. Set MATH. Then MATH with MATH. As in the proof of the previous lemma, MATH is an upper reduction of MATH, and MATH admits an elementary reduction along MATH. This contradiction to the minimality of MATH yields the lemma. |
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. The sub-disc MATH bounded by MATH is a compressing disc for REF - normal surface MATH, in contradiction to REF . By consequence, MATH. Assume that MATH is contained in a single component of MATH, say, in MATH. By REF , MATH is not a compressing disc for MATH, hence MATH. Thus there is a closed line in MATH that separates MATH on MATH, but not on MATH. This is impossible as MATH is a sphere. We conclude that if MATH has a compressing disc, then there is an arc MATH that connects different components of MATH. It remains to consider the case where no sphere in MATH contained in MATH has a compressing disc. We will show the existence of an almost REF - normal sphere in MATH with exactly one octagon, using the technique of thin position. This contradicts REF of MATH (see the begin of this section), and therefore finishes the proof of the lemma. Let MATH be a MATH - NAME embedding, such that CASE: MATH, CASE: MATH (or MATH, in the case MATH), CASE: MATH, and CASE: MATH is minimal. Define MATH. Assume that for some MATH there is a pair MATH of nested or independent upper and lower compressing discs for MATH. We show that we can assume MATH. Since MATH is REF - normal, it has no compressing discs by REF . Thus MATH consists of circles. Any such circle bounds a disc in MATH by REF . By cut-and-paste of MATH, we obtain MATH, as claimed. Now, one obtains from MATH an embedding MATH with MATH by isotopy along MATH, see CITE, CITE. The embedding MATH meets REF in the definition of MATH. Since MATH has no compressing discs by assumption, MATH consists of circles. Thus MATH is split equivalent to MATH. So MATH meets also REF , MATH, in contradiction to the choice of MATH. This disproves the existence of MATH. In conclusion, if MATH has upper and lower bonds, then it is impermeable. Let MATH be the greatest critical parameter of MATH with respect to MATH in the interval MATH. We have MATH by REF . Hence the critical point corresponding to MATH is a point of tangency of MATH to some edge of MATH. By assumption, MATH has no upper bonds, thus MATH for sufficiently small MATH. Let MATH be the smallest critical parameter of MATH with respect to MATH. By REF , MATH has no bonds, thus MATH. Therefore there are consecutive critical parameters MATH such that MATH . Thus MATH has both upper and lower bonds, and is therefore impermeable by the preceding paragraph. One component of MATH is a REF - normal sphere in MATH with exactly one octagon, by REF . The existence of that REF - normal sphere is a contradiction to the properties of MATH, which proves the lemma. |
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 of MATH minimizes MATH. We have MATH, since it is possible to choose MATH so that MATH has an upper bond with string MATH. Since MATH and by REF , MATH is almost REF - normal. By the maximality of MATH, it follows MATH with non-negative integers MATH. Moreover, MATH for MATH by REF . Since MATH separates MATH from MATH, so does MATH. Thus any path connecting MATH with MATH for some MATH intersects MATH in an odd number of points. So alternatively MATH and MATH for all MATH, or MATH and MATH for all MATH. Since MATH, it follows MATH and MATH for MATH, thus MATH. The existence of a MATH - NAME embedding MATH with the claimed properties follows then by REF . |
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 lower returns admits an independent pair of upper and lower compressing discs, thus is not impermeable. By consequence, under the isotopy mod MATH that relates MATH and MATH occurs a surface MATH that has no returns at all, thus is almost MATH - normal for some natural number MATH. If there is a boundary component MATH of a component of MATH and an edge MATH of MATH with MATH, then there is an independent pair of upper and lower compressing discs. Thus MATH. |
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 compressing disc MATH with string MATH that is contained in a single closed tetrahedron MATH. Choose MATH innermost, that is, MATH. Without assumption, let MATH be upper. Since MATH has no octagon by assumption, MATH connects two different components MATH of MATH. Let MATH be a lower bond of MATH. Choose MATH, MATH and MATH so that, in addition, MATH is minimal. Let MATH be the closure of an innermost component of MATH, which is a disc. There is a closed tetrahedron MATH of MATH and a closed REF - simplex MATH of MATH such that MATH is a single component MATH. We have to consider three cases. CASE: Let MATH be a circle, thus MATH. There is a disc MATH with MATH and a ball MATH with MATH. We move MATH away from MATH by an isotopy mod MATH with support in MATH, and obtain a surface MATH with a lower bond MATH. As MATH is a bond, MATH consists of circles. Therefore the normal arcs of MATH are not changed under the isotopy, and the isotopy does not introduce returns, thus MATH is almost REF - normal. Since MATH and MATH, it follows MATH. Thus MATH is an upper compressing disc for MATH, and MATH in contradiction to our choice. CASE: Let MATH be an arc with endpoints in a single component MATH of MATH. By an isotopy mod MATH with support in MATH that moves MATH into MATH, we obtain from MATH and MATH a surface MATH with a lower bond MATH. Since MATH is a bond, the isotopy does not introduce returns, thus MATH is almost REF - normal. One component of MATH is isotopic mod MATH to the component of MATH that contains MATH. Thus up to isotopy mod MATH, MATH is an upper compressing disc for MATH, and MATH in contradiction to our choice. CASE: Let MATH be an arc with endpoints in two different components MATH of MATH. Assume that, say, MATH is a circle. By an isotopy mod MATH with support in MATH that moves MATH into MATH, we obtain from MATH and MATH a surface MATH with a lower bond MATH. Since MATH is a bond, the isotopy does not introduce returns, thus MATH is almost REF - normal. There is a disc MATH with MATH. Let MATH be the component of MATH that contains MATH. One component of MATH is isotopic mod MATH either to MATH or, if MATH, to MATH. In either case, MATH is an upper compressing disc for MATH, up to isotopy mod MATH. But MATH in contradiction to our choice. Thus, MATH and MATH are normal arcs. Since MATH is almost REF - normal, MATH, MATH are contained in different components MATH of MATH. Since MATH is a lower bond, MATH. There is a sub-arc MATH of an edge of MATH and a disc MATH with MATH and MATH. The disc MATH is a lower compressing disc for MATH with string MATH, and MATH. At least one component of MATH is a disc that is disjoint to MATH. Let MATH be the closure of a copy of such a disc in the interior of MATH, with MATH. By construction, MATH is a single point and MATH. Thus by REF , MATH has a pair of independent or nested upper and lower compressing discs and is therefore not impermeable. |
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 right MATH-module structure provides an induced good filtration MATH of MATH such that MATH for all MATH. By REF , there exists MATH such that MATH for all MATH. Hence, for MATH, MATH . We now consider the three cases separately. Under REF , we see that MATH for MATH. Thus, MATH. But, by construction, MATH and MATH and so MATH . Since MATH grows subexponentially this forces MATH. Under REF we have MATH for MATH. Thus, MATH and repeating the analysis of the last paragraph shows that MATH. Finally, assume that REF holds. In this case, MATH and so MATH. Thus there exists MATH such that MATH. Now a minor variant of the penultimate paragraph shows that MATH and hence, by symmetry, that MATH. |
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 contradicts to REF . |
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. Once this has been proved, then the last paragraph implies, for some MATH, that MATH. This contradicts the fact that MATH grows subexponentially CITE and proves the lemma. In order to prove the claim we ignore the right-hand structure of MATH and so, by induction, it suffices to prove it for a cyclic module MATH. Since MATH is torsion and MATH is NAME, MATH contains a regular element, MATH say, of MATH. We will still write MATH for the good filtration on any subfactor of MATH induced from MATH. Now, for any MATH, MATH is a homomorphic image of MATH and so MATH. Since NAME series are additive on short exact sequences, this implies that MATH, for any MATH. |
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 (although MATH may decrease). By REF MATH is also torsion-free. If MATH, replace MATH by MATH; thus MATH and MATH. Let MATH, respectively MATH, be the filtrations of MATH and MATH induced from MATH. Since MATH equals the direct sum MATH of MATH copies of MATH, certainly MATH. Similarly, MATH. Since MATH is a good left filtration and MATH is a good right filtration, by REF there exists MATH such that MATH and MATH for all MATH. Hence, MATH for all MATH. Since MATH grows subexponentially, this forces MATH. |
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 surjective homomorphism of algebras and that MATH does not admit a left noetherian MATH-filtration. Then an immediate consequence of REF is that MATH also has no such filtration. Thus, it suffices to prove the final assertion for a factor ring, and we use MATH . Notice that this is the ring from REF. The nilradical MATH is just the set of strictly upper triangular matrices. It is routine to check that MATH is a free left MATH-module of rank one but a free right MATH-module of rank two. Hence, by REF , neither MATH nor MATH can have a left noetherian finite MATH-filtrations. |
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 triangular matrices and is free of rank MATH as a left MATH-module and free of rank MATH as a right MATH-module. Now apply REF . |
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 is generated by MATH and MATH and satisfies the relations MATH. Thus MATH is spanned by the elements MATH. As such, MATH is a finitely generated right MATH-module, and so is right noetherian. As in the proof of REF , this suffices to prove that MATH is right NAME. For MATH we take the ring MATH from REF. We define MATH and MATH by REF and observe that MATH. Now use the MATH-adic filtration MATH defined by MATH if MATH but MATH if MATH. Then MATH. Hence in the associated graded ring one finds that the images of these elements satisfy MATH. The argument of the last paragraph shows that MATH is left noetherian and hence, by CITE, that MATH is left NAME. |
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 MATH are finitely generated. Therefore MATH is a dualizing module. |
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. It follows that MATH . Computing this out one finds that MATH . Finally, under this isomorphism MATH one finds that the modules MATH and MATH become isomorphic, as is required to prove the proposition. |
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 dimension of the MATH-graded piece of the quotient of the polynomial ring over MATH by the ideal generated by a regular sequence of homogeneous polynomials with degrees MATH (compare CITE). Moreover, using the assignment MATH it follows that MATH . We can compute explicitly this NAME function by the following formula (compare CITE): MATH . Then, MATH . Similarly, MATH . Therefore, MATH is square of size MATH . Since MATH the size of MATH equals MATH . |
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 a point in MATH we denote by MATH the polynomials obtained from MATH when the coefficients are specialized to MATH, and similarly for the coefficients of the NAME. Because of the irreducibility of MATH it is enough to show that for all MATH such that MATH have a non trivial solution in MATH the determinant of the specialized matrix MATH is equal to MATH . Suppose that this is case, and let MATH be a non trivial solution. Without loss of generality, we can suppose MATH . One of the rows of MATH is indexed by MATH . Replace all the elements in that row as follows: CASE: if the element belongs to a column indexed by a monomial MATH then replace it with MATH CASE: if it belongs to a column indexed by a monomial MATH replace it with MATH . It is easy to check that, the determinant of the modified matrix is equal to MATH . Now, we claim that under the specialization MATH the determinant of the modified matrix will be equal to zero if and only if MATH . In order to prove this, we will show that the following submatrix of size MATH has rank less or equal than MATH . This, combined with a NAME expansion of the determinant of the modified matrix, gives the desired result. If the rank of the block MATH is less than MATH then the claim follows straightforwardly. Suppose this is not the case. Then the family MATH is a basis of the piece of degree MATH of the generated ideal MATH . We will show that in this case the polynomial MATH belongs to MATH, which proves the claim. Because of REF , the polynomial MATH lies in the ideal MATH. Specializing MATH we deduce that MATH is in the graded ideal MATH . This, combined with the fact that MATH proves that MATH for all 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 in the ideal generated by a regular sequence of MATH polynomials with degrees MATH . Then MATH has MATH rows and MATH columns, and there is a bijection between the family MATH of MATH monomials of degree MATH and the maximal minors MATH of MATH . Namely, MATH is the determinant of the square submatrix made by avoiding all rows indexed by monomials in MATH . It is not hard to check that MATH is the determinant MATH which is used in CITE, for computing the subresultant associated with the family MATH . Now, using the generalized NAME 's formula for the subresultant (compare CITE), we have that MATH where MATH is the subresultant associated with the family MATH . It is a polynomial in MATH which vanishes under a specialization of the coefficients MATH if and only if the family MATH fails to be a basis of the MATH graded piece of the quotient MATH (compare CITE). Let MATH be the complementary minor of MATH in MATH (that is, the determinant of the square submatrix of MATH which is made by deleting all rows and columns that appear in MATH). By the NAME expansion of the determinant, we have that MATH with MATH . Setting MATH we have the desired result. |
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 instead of MATH . Set MATH, and consider the matrix MATH with coefficients in MATH given by MATH . It is easy to see that MATH, and because of REF , we have that MATH divides MATH in MATH . Transposing MATH and using a symmetry argument, again by the same lemma, we can conclude that MATH divides MATH in MATH where MATH has the obvious meaning. The ring MATH is a factorial domain and MATH and MATH have no common factors in MATH because they depend on different sets of variables. So, we have MATH for some MATH . Now, specialize MATH . The fact that MATH is a multiple of the resultant has been proved in REF (see also CITE) for MATH and in CITE for MATH . On the other side, since MATH is irreducible and depends on all the coefficients of MATH while MATH and MATH do not depend on the coefficients of MATH we conclude that MATH divides MATH . Moreover, the following lemma shows that they have the same degree. Then, their ratio is a rational number MATH . We can see that MATH considering the specialized family MATH . |
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 in the coefficients of MATH the lemma will be proved if we show that MATH . Now, as already observed in the proof of REF , MATH can be computed as the cardinality of the following set: MATH and MATH is the cardinality of MATH . In order to prove REF it is enough to exhibit a bijection between MATH and the disjoint union MATH . This is actually a disjoint union for all MATH unless MATH . But what follows shows that the bijection is well defined even in this case. Let MATH with MATH . If MATH then there exists a unique MATH such that MATH verifies MATH . If MATH then we send MATH to MATH . Otherwise, we send it to MATH . If MATH let MATH denote the multiindex MATH . Then, MATH and there exists a unique MATH such that the multiindex MATH defined by MATH has degree MATH . We can send MATH to MATH provided that MATH . Suppose this last statement does not happen, this implies that the monomial with exponent MATH has degree MATH contradicting the fact that MATH . With these rules, it is straightforward to check that we obtain the desired bijection. |
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 . Then, it can also be computed as MATH which gives the equality MATH . It follows that MATH . If MATH we deduce that MATH by inductive hypothesis. Suppose then that MATH is in the range MATH . As MATH it is enough to show that MATH and MATH which can be easily checked, and the result follows again by inductive hypothesis. |
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, CITE or CITE. Suppose MATH . Since MATH it is easy to see that REF is a complex. Set MATH . Note that the open set MATH is non void because the vector of coefficients of MATH lies in MATH since in this case MATH . For any choice of homogeneous polynomials MATH with respective degrees MATH and coefficients MATH in MATH, the resultant does not vanish by REF and then the specialized NAME complexes in REF are exact. Then, the dimension MATH of the image of MATH equals MATH . Similarly, MATH . Therefore, MATH . On the other side, the fact that MATH is non singular of size MATH implies that MATH . Therefore, MATH and the complex is exact at level MATH . In a similar way, we can check that the complex is exact at level MATH, and so the full specialized complex REF is exact when the coefficients MATH lie in MATH . In order to compute the determinant of the complex in this case, we can make suitable choices of monomial subsets in each term of the complex starting from the index sets that define MATH to the left and to the right. Then, MATH where MATH (respectively, MATH) is a quotient of product of minors of the morphisms on the left (respectively, on the right). Taking into account REF , it follows from CITE that MATH and so by REF we have MATH for all families of homogeneous polynomials with coefficients MATH in the dense open set MATH, and since MATH and the resultant are rational functions, this implies REF , as wanted. Moreover, it follows that the complex is exact if and only if the resultant does not vanish. The fact that MATH is the greatest common divisor of all maximal minors of the matrix representing MATH has been proved in REF . |
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 respectively the form MATH and MATH, where MATH and MATH are square invertible matrices of maximal size. We are going to prove that, with this choice, the matrix MATH is invertible. Our specialized matrix will look as follows: MATH . Applying linear operations in the rows and columns of MATH it can be transformed into: MATH where the block MATH is square and of size MATH . But it is easy to check that this MATH corresponds to the components in degree MATH of another NAME MATH (in the sense of REF ). This is due to the fact that each of the linear operations performed on MATH when applied to the block MATH can be read as a polynomial combination of MATH and MATH applied to the bezoutian MATH . Using the fact that the polynomials MATH read in the columns of MATH generate the quotient in degree MATH and they are as many as its dimension, we deduce that they are a basis and so MATH which completes the proof of the claim. |
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 REF . |
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 which is a contradiction. According to REF , we also know that MATH has the smallest possible size. |
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 that MATH . This, combined with the equality given in REF , implies that, for each MATH is equal to the piece of degree MATH in the variables MATH of the polynomial MATH . Besides, this polynomial does not depend on MATH so the following formula holds for every pair MATH such that MATH . This allows us to compute MATH for every MATH in terms of the homogeneization of MATH . From REF , the claim follows straightforwardly. |
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 have three cases: CASE: If MATH, let MATH . Then MATH uniquely determines MATH. In this case, let MATH such that MATH . CASE: If MATH, that is, MATH with MATH, let MATH. Then MATH uniquely determines MATH. In this case, let MATH . CASE: If MATH, that is, MATH with MATH, let MATH. Then MATH uniquely determines MATH. In this case, let MATH . Then in each REF - REF , MATH uniquely defines MATH by REF - REF , and the supersphere with tubes REF is superconformally equivalent to MATH where MATH for MATH. Choose MATH such that MATH . Then the supersphere with tubes REF is superconformally equivalent to MATH where MATH satisfy REF , and REF , respectively. |
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 is a superprojective transformation, we obtain MATH for some MATH. Let MATH and MATH. From REF , we know that MATH . Thus by REF MATH that is, MATH. Thus MATH must be the identity map of MATH. |
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 MATH such that MATH, and there exists MATH satisfying MATH, and MATH where MATH satisfies REF , that is, has a power series expansion of the form REF . Let MATH where MATH is the even coefficient of MATH and MATH is the odd coefficient of MATH in the power series expansion of MATH about infinity. Then MATH has a power series expansion of the form REF with the even coefficient of MATH and the odd coefficient of MATH equal to zero, and there exists some MATH such that MATH is convergent in MATH. |
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 where MATH for some MATH with MATH. Using the automorphism REF , we have MATH as desired. |
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 proposition follows immediately. |
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 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. Therefore MATH . Thus writing MATH, we have MATH which proves that MATH satisfies the superconformal REF . The uniqueness follows from REF , the uniqueness of MATH on MATH, and the fact that if we write MATH then from REF , we can directly observe that MATH is of the form MATH (terms in powers of MATH strictly greater than REF), that is, MATH. |
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 power series in MATH. |
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 . Thus the result is true for MATH, MATH. CASE: For MATH, MATH, we note that by REF , and the above cases for MATH, MATH, and MATH, MATH . Since MATH, for MATH and MATH, the result follows by linearity. |
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 . Letting MATH be the sequence consisting of all zeros in MATH, it is clear that MATH and MATH are subgroups of MATH. |
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 has a unique inverse MATH with MATH . Setting MATH, and since MATH, we have MATH . Moreover, by REF , with MATH and MATH in the Proposition replaced by MATH, we have MATH which gives REF . |
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 definition, for MATH and thus the right-hand sides of REF must exist in MATH for MATH and MATH, respectively. But then MATH and MATH must also exist in MATH for MATH and MATH, respectively. Now note that REF hold if we replace MATH by MATH, and that these equations exist in MATH for MATH and MATH, respectively. In this REF also exist in MATH if we replace MATH by MATH, that is, if MATH then MATH exists in MATH, and if MATH then MATH exists in MATH. But since MATH and MATH, by the automorphism property REF , we have that MATH which exists in MATH, and thus MATH and MATH exist in MATH for MATH and MATH, respectively. Therefore MATH exists in MATH. In the case that MATH, MATH and MATH exist in MATH, by reversing the steps of the argument above, we conclude that MATH exists in MATH. |
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 MATH in the MATH's, degree MATH in the MATH's, and degree MATH in the MATH's, for MATH, and where MATH satisfies MATH for MATH. (Of course MATH for MATH even, MATH for MATH odd, and similarly for MATH and MATH.) Thus letting MATH, REF gives MATH from the boundary REF, we have MATH . These equations give MATH, and MATH uniquely for all MATH. By the superconformal condition MATH for MATH and the boundary conditions, we see that MATH where the square root is defined to be the NAME series expansion about MATH and MATH with MATH. Note that each MATH is determined by MATH for MATH, MATH, MATH, and MATH. Thus the MATH, and MATH that we have determined from the boundary conditions for all MATH, uniquely determine MATH and MATH for all MATH. from REF , it is clear that the MATH term on the right-hand side only depends on MATH, MATH, and MATH for MATH, MATH, MATH, and MATH. Note that for all MATH, MATH, MATH, and MATH the coefficient of a given term MATH or MATH is in MATH, that is, MATH . Thus by REF , we have MATH . And similarly for MATH, MATH, MATH, and in fact for any MATH . Now we can solve REF for MATH and MATH by induction on MATH, MATH, MATH, and MATH. In fact, if we compare terms which are 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, on both sides of REF , we have MATH where MATH and MATH are 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, and depend only on MATH and MATH for MATH, MATH, MATH, and MATH, where MATH. Assume MATH and MATH (and thus MATH) have already been obtained for MATH, MATH, MATH, and MATH, where MATH, and assume that we have shown that each MATH and MATH is in MATH. Then by using REF on each polynomial coefficient of each MATH (as we did for MATH), we can determine MATH and MATH. Then from REF , we have MATH where MATH and MATH such that MATH and MATH . By the principle of induction, we obtain MATH and MATH for all MATH, and thus we obtain MATH and MATH. It is clear from the procedure to solve REF that the solutions MATH and MATH are unique. Furthermore note that the even coefficient of MATH in MATH is one and the even coefficient of MATH in MATH is MATH. Let MATH and MATH be the unique formal superconformal series satisfying MATH and MATH such that the even coefficient of MATH in MATH is one and the even coefficient of MATH in MATH is MATH. To complete the proof, we note that MATH . Thus MATH and MATH . We note that MATH . Thus MATH and MATH . We note that MATH . Thus MATH and MATH . And finally, we note that MATH . Thus MATH and MATH . Substituting these into REF , we obtain REF . |
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 also in MATH respectively. Since MATH was arbitrary, REF are in MATH . Furthermore, they are obviously even. To prove REF , we note that using REF repeatedly and by REF , we have MATH in MATH. Using REF again, we see that for MATH, the expressions MATH and MATH exist in MATH and are equal. Thus by REF , we have MATH . Taking MATH to be MATH where MATH is any element of MATH, REF becomes MATH . Since MATH is an arbitrary element of MATH, we obtain REF . |
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 formal variables and MATH another even formal variable. Let MATH be the MATH-graded vector space MATH. Consider the subspace MATH spanned by the elements MATH and MATH . Then MATH, that is, MATH contains only even elements. In other words, letting MATH, then MATH is in the MATH-envelope of MATH which is a NAME algebra; see REF . Thus the same procedure used to prove REF is valid for MATH where we simply use REF instead of NAME 's REF, and the NAME algebra relations and corresponding universal enveloping algebra instead of just the NAME algebra relations and corresponding universal enveloping algebra. |
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 be obtained from type REF , and REF superspheres for MATH. Assume this is true for MATH. Let MATH be a closed NAME curve on the body of MATH, denoted MATH, such that REF and MATH are in one connected component of MATH (which we will call the interior of MATH) and MATH are in the other connected component of MATH (which we will call the exterior of MATH). By the NAME mapping theorem there exists a conformal map MATH from the interior of MATH to the open unit disc in MATH, and we can require that MATH. Then by expanding MATH about REF and choosing MATH such that MATH, we can let MATH be the unique formally superconformal power series such that MATH is of the form REF with even coefficient of MATH equal to MATH. Then MATH and MATH also satisfy REF for MATH. Let MATH . Then from the definition of the sewing operation, we have MATH which is obtained from sewing MATH, a supersphere of type REF , and a supersphere in MATH. By our inductive assumption, the result follows. |
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 in a neighborhood of MATH. Thus by REF , we see that MATH and MATH are in MATH. By REF , MATH . Thus MATH. From the definition of the sewing operation, we see that MATH and MATH . This then gives REF . |
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 the superspheres. Then REF follow from the definition of the sewing operation on MATH. |
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 supersphere with tubes is superconformally equivalent to the super-Riemann sphere with the negatively oriented puncture still at MATH, the positively oriented puncture at MATH, and with standard local coordinates at these punctures, that is, the local coordinate at MATH is given by MATH and the local coordinate at MATH is given by the superconformal shift MATH. Call this non-canonical supersphere MATH. Then choosing MATH such that MATH, we can sew the MATH-th puncture of MATH to the MATH-st puncture of MATH via the sewing defined in REF. The resulting supersphere MATH is given by MATH where MATH if MATH that is, MATH and MATH. Then the uniformizing function which sends MATH to a canonical super-Riemann sphere must satisfy MATH where MATH if MATH, that is, MATH. In addition, we would like MATH to send MATH to MATH, send MATH to MATH, and keep the even coefficient of MATH of the local coordinate at MATH equal to zero in order for the resulting supersphere to be canonical. Thus MATH and MATH is a solution to the uniformizing function and is the unique solution sending MATH to a canonical supersphere. Therefore the canonical supersphere representative of the equivalence class of the sewing MATH is given by MATH but this is the canonical supersphere represented by MATH as desired. |
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 . Let MATH, and MATH, for MATH, be given by MATH . Then MATH . From REF , we have MATH for MATH. Using REF and the above formulas, we obtain REF . |
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 MATH and MATH are homomorphisms from MATH to MATH. In addition, MATH . Thus MATH is a homomorphism from MATH to MATH. |
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, respectively; and MATH and MATH. The positive numbers MATH and MATH satisfy all the properties needed. Conversely, if there exist MATH, with MATH such that MATH and MATH are convergent and single-valued in MATH and MATH, respectively, MATH for MATH, MATH, and MATH for MATH, then we can choose MATH to be any positive number between MATH and MATH, and we can let MATH and MATH . By assumption, MATH and MATH are the only punctures in MATH and MATH, respectively. In addition, MATH and MATH . Thus the MATH-th tube of MATH can be sewn with REF-th tube of MATH. This finishes the first statement of the proposition. Now assume that the MATH-th tube of MATH can be sewn with REF-th tube of MATH and MATH. By the definition of the sewing operation, the canonical supersphere MATH is in the same superconformal equivalence class as the supersphere with tubes MATH where the supersphere MATH is given by the local coordinate system MATH with the coordinate transition function given by MATH for MATH such that MATH, with coordinates MATH, respectively; MATH, with coordinates MATH, respectively; in terms of the coordinate MATH for MATH, MATH and in terms of the coordinate MATH for MATH, MATH . By REF , there exists a superconformal equivalence MATH from this supersphere with tubes to some canonical supersphere with tubes MATH . Let MATH for MATH, and let MATH for MATH. Then in MATH, we have MATH . It is clear that we can choose MATH and MATH to satisfy REF - REF . On the other hand, for MATH, MATH . By REF , MATH and MATH satisfy REF . The formula in REF follows immediately from the relation between canonical superspheres and elements of MATH. This finishes the proof for REF MATH and MATH up to uniqueness of MATH and MATH. REF - REF are proved similarly. To prove that MATH and MATH are unique, assume that there exists another pair of functions MATH and MATH which also satisfy REF - REF . If we define MATH then MATH is a superconformal equivalence from the supersphere with tubes REF to a canonical supersphere with tubes. Thus MATH is a superconformal equivalence from one canonical supersphere with tubes to another canonical supersphere with tubes. By REF , we see that MATH must be the identity map, or equivalently MATH. Therefore MATH and MATH. This proves the uniqueness of MATH and MATH. |
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, for MATH, we have the global sections of MATH given by MATH and MATH, for each MATH. Letting MATH vary, we obtain global sections we denote by MATH respectively. Then defining MATH we have that MATH and MATH are genus zero compact complex manifolds. Let MATH for MATH, and let MATH be the restriction of MATH to MATH, for MATH. Then MATH is a coordinate system for MATH, with coordinate transition function given by MATH for MATH. It is clear that MATH is complex analytically isomorphic to MATH. Let MATH for MATH and let MATH. Consider MATH and MATH. In CITE, NAME uses the NAME Theorem on the body component MATH to prove that it is locally trivial in MATH, for MATH. We will follow this argument in the more general case of MATH and MATH to prove that they are both locally trivial in MATH for all MATH. The reason we split the domain of MATH into MATH and MATH is that we will need certain bijective properties to form a manifold from the family MATH, and thus have to restrict the domain of MATH in order to insure that there is no double cover in the MATH coordinate. We will then use the local triviality properties of MATH and MATH to show that MATH and MATH restricted to the sections MATH are analytic in MATH for MATH. Using the family of sections MATH instead of MATH, the analogous argument for MATH and MATH shows that they are also locally trivial in MATH. Then the analogous argument can be used to prove that MATH and MATH restricted to the sections MATH are analytic in MATH for MATH. This will allow us to conclude that MATH and MATH themselves are analytic in MATH for MATH. Let MATH. Let MATH . Then MATH. Let MATH and MATH . Obviously MATH is an open set in MATH for MATH. We define MATH for MATH. It is clear that the MATH are bijections. Thus MATH is a local coordinate system for MATH with the coordinate transition function given by MATH for MATH. This gives a complex manifold structure to MATH. It is clear that the projection from MATH to MATH is complex analytic and the rank of the Jacobian is one. Thus MATH is a complex analytic family. Since all the MATH are complex analytically isomorphic to MATH, they are all complex analytically isomorphic to each other. By the NAME Theorem, MATH is locally trivial for each MATH. Thus given a section MATH, for every MATH, there exists a neighborhood of MATH, denoted MATH, such that complex analytically MATH . Thus there exist complex analytic isomorphisms MATH for any MATH such that if we use the local coordinates of MATH and MATH to express MATH, then MATH is also analytic in MATH. For every MATH, we already have a superconformal isomorphism MATH from MATH to MATH given by MATH . For MATH, define the superprojective transformation MATH by REF and MATH . Choose a point MATH. We can always find some function MATH from MATH to MATH such that in terms of the coordinate atlas MATH of MATH restricted to the section MATH, the function MATH is analytic in MATH for MATH. Note that MATH is a family of analytic isomorphisms from MATH to itself. Hence it must be a family of linear fractional transformations depending on MATH. Any linear fractional transformation is determined by its values on three complex variables. Furthermore, it is clear that if the values at these three points depend analytically on MATH, then the value at any point depends analytically on MATH. Consider the three points MATH, MATH and MATH. Since MATH is analytic in MATH, and the value of any superfunction at MATH is the NAME expansion about the body MATH, we have that MATH is analytic in MATH. Furthermore, MATH is analytic in MATH, and by our choice of MATH, MATH is analytic in MATH. Thus MATH is analytic in MATH. This implies that MATH is analytic in MATH. By the definition of MATH and the normalization conditions that it satisfies, we have MATH . Since MATH is analytic in MATH, its coefficients in MATH and MATH are also analytic in MATH. Hence MATH is analytic in MATH. We conclude that MATH is analytic in MATH. Since MATH is an arbitrary complex number in MATH, we have that MATH is analytic in MATH for any MATH. Following the same argument above with MATH replaced by MATH, we conclude that MATH is analytic in MATH for MATH. Then following a similar argument using the sections MATH, we can prove that MATH restricted to MATH is analytic in MATH for MATH. Then since MATH we have that MATH is analytic in MATH for every point MATH. We conclude that MATH is analytic in MATH for MATH, and thus MATH and MATH are analytic in MATH for MATH. We now prove the second statement of the proposition. Let MATH and MATH be the even and odd superfunction components of MATH, and let MATH for MATH. For MATH, we define MATH . Then MATH is analytic in MATH for MATH. Let MATH . It is easy to see that the MATH-th tube of MATH can be sewn with REF-th tube of MATH when MATH. By REF , there exist MATH and MATH such that MATH and in some region MATH holds for MATH. Moreover MATH and MATH are unique. Using the same method as that used in the proof of the analyticity of MATH and MATH, we can show that MATH and MATH are analytic in MATH for MATH. But from MATH and MATH we see that MATH and MATH satisfy the above equation for MATH and MATH. By uniqueness, MATH that is, MATH. Since MATH and MATH are analytic in MATH not only when MATH, but also when MATH, it must be that MATH is a removable singularity of MATH and MATH. Finally, since MATH, and MATH, and for any MATH with MATH, the superfunction MATH is nonzero for MATH, we see that MATH is a second-order zero of MATH and a first-order zero of MATH. |
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 satisfy the sewing equation MATH or equivalently MATH and the obvious boundary conditions, the formal series MATH and MATH in MATH corresponding to the expansions of MATH and MATH, respectively, satisfy the equation MATH in MATH and the corresponding formal boundary conditions. Note that the coefficients of the right-hand side of REF are, in general, infinite sums. Thus MATH and MATH satisfying REF means that the coefficients of the right-hand side are absolutely convergent to the coefficients of the left-hand side. Note that REF and the corresponding formal boundary conditions can be obtained from the formal sewing equation and formal boundary conditions in REF by substituting MATH, MATH, MATH, MATH, and MATH for MATH, MATH, MATH, MATH, and MATH, respectively, for MATH. Since the solution of the formal sewing equation and the formal boundary conditions in REF is unique, the solution MATH and MATH to REF and the corresponding boundary conditions can be obtained by substituting MATH, MATH, MATH, MATH, and MATH for MATH, MATH, MATH, MATH, and MATH, respectively, for MATH, into the solution of the formal sewing equation and the formal boundary conditions given in REF . Thus we have MATH and MATH . By the definition of MATH, for MATH, the expansion of MATH is equal to MATH for MATH, that is, MATH is convergent to MATH for MATH. |
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. Let MATH . Then from the definition of MATH, we have MATH . Thus using REF , and REF , we have MATH . Substituting these calculations into REF , we have MATH . |
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 MATH. Clearly, we have MATH and MATH, so MATH. We have MATH . Therefore, MATH. CASE: For any large MATH, let MATH. Then MATH, so it is easy to see that MATH. We have MATH. CASE: Replacing MATH by a conjugate (under a diagonal matrix), we may assume that MATH. Then, by renormalizing, we may assume that MATH. Let MATH be the element of MATH with MATH. By subtracting a multiple of MATH from MATH, we may assume MATH. For any large MATH, let MATH, so MATH is real. We have MATH so MATH. CASE: For each large MATH, let MATH be an element of MATH, such that MATH is pure imaginary. (This exists because the sign of MATH is opposite that of MATH.) We note that MATH and MATH, but MATH and MATH. Thus MATH and MATH . CASE: Because MATH, we have MATH, so, for any large MATH, we may choose MATH, such that MATH. Thus MATH, but MATH and MATH. Then (because MATH) it is easy to verify that MATH. However MATH . So MATH. CASE: For any large MATH, choose MATH, such that MATH. (This is possible, because MATH.) Then we may choose MATH, such that MATH. Then MATH, MATH, and MATH, so we have MATH. |
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 MATH. CASE: Replacing MATH by a conjugate under MATH, we may assume that MATH, so MATH for every MATH, and MATH for every MATH (which means MATH). Therefore, the NAME reflection corresponding to the root MATH conjugates MATH to a subalgebra either of type REF or of type REF , depending on whether or not there is some MATH, such that MATH. CASE: By assumption, the quadratic form MATH is definite on MATH, so MATH. Therefore, MATH whenever MATH. Furthermore, MATH whenever MATH and MATH. Thus, MATH. CASE: For any sequence MATH in MATH, we write MATH for MATH, etc. We have MATH. If MATH, then MATH. (This completes the proof if MATH.) If MATH, then MATH, but MATH whenever MATH, and MATH whenever MATH and MATH. Therefore MATH . If MATH, then there is some (large) MATH with MATH (and hence MATH). Thus MATH. CASE: For any sequence MATH in MATH, we show that MATH. We write MATH for MATH, etc. If MATH, then MATH, but MATH whenever MATH, and MATH whenever MATH and MATH. Thus, MATH. We may now assume that MATH. Thus, there is some MATH, such that MATH and MATH. (Note that, because MATH, we have MATH.) Because MATH, we must have MATH, so MATH. In particular, MATH, so MATH. We have MATH, but MATH whenever MATH, and MATH whenever MATH and MATH. Thus, MATH and MATH. Furthermore, we have MATH because MATH, and the terms MATH and MATH cannot cancel (since they both have the same sign as MATH). We conclude that MATH. All that remains is to show MATH. If MATH, then MATH as desired. If MATH, then MATH so MATH, as desired. Thus, we may assume that MATH. Because MATH and MATH, we have MATH as desired. CASE: For MATH, we have MATH. For MATH with MATH, we have MATH. All that remains is to show MATH for every MATH. Note that MATH, and MATH. If MATH, then it is obvious that MATH. Thus, we may assume MATH. Then, because MATH, but MATH whenever MATH, and MATH whenever MATH and MATH, we have MATH . |
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 MATH, for, otherwise, REF holds. Then, from REF, we have MATH for every MATH. Thus, MATH. We may assume MATH for every MATH, for otherwise Conclusion REF holds. We conclude that REF holds. Assume MATH for every MATH, and there is some MATH with MATH. We may assume that MATH and MATH are linearly dependent over MATH for every MATH (otherwise REF holds). In particular, there exists MATH, such that MATH. Assume MATH. Assume there exists MATH, such that either MATH or MATH. We may assume there exists MATH, such that MATH (otherwise REF holds). Furthermore, by adding a small linear combination of MATH and MATH to MATH, we may assume that MATH and that either MATH or MATH. Because MATH and MATH are linearly dependent, there exists MATH REF such that MATH. (Then note that we must have MATH.) Then MATH (because MATH and MATH is linearly independent over MATH). This contradicts the fact that MATH and MATH are linearly dependent over MATH. Assume MATH, for every MATH. For each MATH, there exist MATH, such that MATH and MATH. Because MATH, we must have MATH, so MATH, which means that MATH is real. We must also have MATH, so MATH . Thus MATH, so MATH . Therefore REF holds. Assume MATH. We show that either REF or REF holds. Straightforward calculations show that REF are invariant under conjugation by MATH, so we may assume that MATH; that is, MATH. Thus, we may assume MATH for every MATH, for, otherwise, REF holds. Then we may assume MATH for every MATH, for, otherwise, REF holds; therefore MATH. We may now assume MATH for every MATH, for, otherwise, REF holds. Thus, REF holds (with MATH). Assume there exists MATH with MATH. We claim that MATH. If not, then there is some MATH, such that either MATH or MATH. If MATH, then MATH, so REF holds. On the other hand, if MATH, then, letting MATH, we have MATH and MATH, so REF holds once again. Assume MATH for every MATH. We may assume that there is some MATH, such that MATH (otherwise, either REF or REF holds). Then we may assume MATH (otherwise, REF holds). We claim that REF holds. If not, then there is some MATH, such that MATH and MATH. Then MATH, which contradicts the assumption that MATH. Assume there is some MATH, such that MATH. Assume MATH. Suppose, for the moment, that there exists MATH with MATH. We may assume that MATH (otherwise, REF holds). Therefore MATH, so REF holds. We may now assume that MATH for every MATH. This implies that MATH, MATH, MATH, and MATH are functions of MATH; in particular, MATH. Also, because MATH and MATH, we must have MATH. We claim MATH (so REF holds). If not, then MATH, so there exist MATH, such that MATH and MATH. Because MATH, we must have MATH. Therefore MATH, so MATH. Furthermore, MATH so MATH. Then MATH is also MATH. This implies MATH for every MATH. This contradicts the fact that MATH. Assume MATH. REF below implies that either REF or REF holds. |
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, we must have MATH. Therefore, REF hold, so MATH, so MATH is a NAME subgroup. Assume there does not exist such an element MATH. Then MATH is described in REF and in REF . |
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 all MATH. CASE: From REF , we have MATH for all MATH. CASE: For any large MATH, choose MATH, such that MATH. Note that MATH, so MATH, but MATH whenever MATH, and MATH whenever MATH or MATH. From the choice of MATH, we have MATH so it is not difficult to see that MATH. CASE: Replacing MATH by a conjugate, we may assume MATH. (First, conjugate by an element of MATH to make MATH. Then conjugate by an element of MATH to make MATH orthogonal to MATH. Then conjugate by an element of MATH that centralizes MATH, to make MATH. Then conjugate by an element of MATH to make MATH. Then conjugate by an element of MATH to make MATH.) Then, by assumption, we must have MATH, because MATH and MATH. Furthermore, replacing MATH by a conjugate under a diagonal matrix (that belongs to MATH), we may assume that MATH and MATH are real. Then MATH must also be real (because MATH is real). Thus, we see that MATH. So CITE implies that MATH is a NAME subgroup. |
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. CASE: From REF , we have MATH. From REF , we see that MATH for MATH. From REF , we see that MATH for MATH. CASE: See REF . CASE: Because MATH, it is easy to see that MATH. CASE: Replacing MATH by a conjugate, we may assume MATH and MATH. Therefore, MATH and MATH for every MATH. Thus MATH so MATH. From REF, we know MATH. We have MATH (see REF ). Because MATH is a real multiple of MATH, we may let MATH be a large element of MATH, such that MATH. (So MATH and MATH.) Then MATH . It is clear that all other matrix entries of MATH are MATH. Thus, we have MATH. Now suppose there is a sequence MATH in MATH with MATH. Assume MATH. We have MATH, so MATH . This is a contradiction. Assume MATH. We have MATH, so MATH . This is a contradiction. Assume MATH. We have MATH, so MATH . This is a contradiction. CASE: See REF . |
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 MATH for every nonzero MATH (otherwise REF holds). The argument in NAME REF of the proof of REF implies there exists MATH, such that, for every MATH, we have MATH (or vice-versa: for every MATH, we have MATH). Thus, replacing MATH by a conjugate under MATH, we may assume MATH for every MATH. If MATH, then there is some nonzero MATH, such that MATH. This contradicts the fact that MATH. Thus, we conclude that MATH, so REF holds. Assume the projection of MATH to MATH is one-dimensional. Replacing MATH by a conjugate under MATH, we may assume MATH is real for every MATH. Fix some MATH, such that MATH. We may assume that MATH (otherwise Conclusion REF holds). Therefore MATH must be zero, so MATH and MATH is a nonzero real, for every nonzero MATH. (This implies MATH.) Assume MATH for every MATH. We may assume Conclusion REF does not hold. We claim that MATH. Suppose not. Then there is some MATH, such that MATH and MATH. Because Conclusion REF does not hold, we must have MATH. Then MATH is a nonzero element of MATH. (This can be seen easily by replacing MATH with a conjugate, so that MATH.) This contradicts our assumption that MATH. If MATH, then either Conclusion REF or REF holds (depending on whether MATH is MATH or not). If MATH, then REF or REF holds. Assume the projection of MATH to MATH is nontrivial. Then we may assume MATH. Assume there are nonzero MATH and MATH, such that MATH, MATH, and MATH. We may assume that Conclusion REF does not hold. Therefore, for every real MATH, we must have MATH . Thus, the coefficient of MATH must vanish, which (using the fact that MATH is real and nonzero) means MATH . We have MATH, so MATH is real. Thus, MATH . Comparing this with REF, we conclude that MATH. Therefore MATH, so Conclusion REF holds (for the element MATH). Assume there do not exist nonzero MATH and MATH, such that MATH, MATH, and MATH. We must have MATH (Otherwise, we obtain a contradiction by setting MATH and MATH.) We may assume MATH (Otherwise, Conclusion REF holds.) We claim MATH. If not, then there exist linearly independent MATH, such that MATH. From REF , we know that MATH. By replacing with a linear combination, we may assume MATH. Then, from REF , we know that MATH, so MATH. Because MATH is (at most) one-dimensional, but MATH and MATH are linearly independent, we know that MATH, so MATH. This contradicts the assumption of this subsubcase. We may now assume MATH (otherwise Conclusion REF holds). Choose a nonzero MATH, such that MATH. If MATH, then Conclusion REF holds. If MATH, then MATH, so either Conclusion REF or REF holds. Assume the projection of MATH to MATH is two-dimensional. We may assume MATH (otherwise, MATH, so Conclusion REF holds). We may assume MATH for every MATH otherwise REF implies that either REF or REF applies. Therefore MATH, so MATH is abelian. Let MATH with MATH and MATH. Then MATH so MATH. Then, for every MATH, we have MATH, so MATH. We may assume MATH for every MATH (otherwise Conclusion REF holds). This implies MATH (otherwise, there is some MATH such that MATH and MATH, and then REF does not hold for MATH when MATH is sufficiently large). Thus, Conclusion REF holds. |
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