paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/0006021 | CASE: If MATH there is nothing to prove. So, let MATH. We show that we can assume MATH for the proof of MATH- compactness: Let MATH . Then MATH. We set MATH so that MATH . Since MATH is continuous and bounded MATH. If MATH is MATH-elliptic, that is, MATH, then we get MATH, that is . MATH is MATH-elliptic. Denote the sp... |
math-ph/0006021 | MATH is the spectral projection of a self-adjoint operator and therefore self-adjoint, and MATH-compact by REF . Thus the eigenspace MATH is the image of a closed adjointable projection MATH and therefore a closed complementable MATH- module. Since the projection MATH is MATH-compact MATH is algebraically finitely gene... |
math-ph/0006021 | Let MATH, so that MATH for MATH, that is . MATH has at least MATH components in MATH. Then MATH since all projections occurring in this sum are MATH-trace class by REF . |
math-ph/0006021 | CASE: Let MATH and MATH. By assumption there is MATH such that MATH . For all MATH we have MATH, and MATH. We set MATH. Then MATH CASE: by definition. CASE: because MATH for all MATH. CASE: Let MATH, MATH. Then there is MATH with finite spectrum so that MATH. If MATH is invertible then there is nothing to prove, otherw... |
math-ph/0006021 | CASE: As is well known, the spectrum of MATH is MATH, all irreducible representations MATH have dimension MATH. The canonical trace of MATH is MATH with the canonical trace MATH on MATH. Minimal projections have rank MATH in the fiber, and so the NAME constant is MATH. CASE: MATH has real rank zero by CITE. Since MATH ... |
math-ph/0006021 | REF . |
math-ph/0006021 | Let MATH. Since MATH is closed there is a neigborhood MATH of MATH and MATH with MATH, MATH. Then MATH and therefore MATH . Since MATH is smooth also, it leaves MATH invariant. |
math-ph/0006021 | This follows from associativity MATH and the projectivity REF : MATH . |
math-ph/0006021 | Let MATH. We choose MATH with MATH and set - a priori depending on MATH - MATH. If MATH, MATH, we get MATH that is, MATH depends on the value of MATH at the point MATH only; hence we omit MATH in the notation. The morphism property follows from the corresponding property of MATH, and from MATH. MATH induces MATH by con... |
math-ph/0006021 | Easy consequences of the cocycle property. |
math-ph/0006021 | For MATH we have by REF since a left Hilbert-MATH-module is a NAME space with conjugated scalar product (complex linear in the first argument, anti-linear in the second). For MATH we get after identifying MATH with MATH . This shows that the structures coincide. |
math-ph/0006021 | By REF we have for MATH . Therefore, the completion of MATH with respect to MATH is contained in the one with respect to MATH, that is, in MATH. But MATH is dense in MATH. We get the scalar product of the GNS representation with respect to MATH for MATH from MATH . Since MATH is dense the GNS representation space is ex... |
math-ph/0006021 | For tensor products of left NAME modules we have in general MATH . The statement about MATH is well known in the untwisted case since the opposite of left multiplication is right multiplication. It is easy to check that this holds in the twisted case also. |
math-ph/0006021 | Set MATH. Then MATH dense, we set MATH as operators on vector spaces. MATH is adjointable since MATH is symmetric and gauge-periodic: For MATH we have MATH . Finally, MATH is dense in MATH because MATH is essentially self-adjoint; therefore, MATH is regular. |
math-ph/0006021 | As a set MATH. We choose the topology so that the natural projection MATH is continuous and open: The topology is generated by the tubular neighborhoods MATH for open sets MATH, continuous fields MATH and MATH . It is easy to check that the tubular neighborhoods generate a topology on MATH with the desired properties. ... |
math-ph/0006021 | It is well known that right MATH-modules MATH and left MATH-modules MATH are in one-to-one correspondence. So we just have to verify the corresponding NAME module structures: Let MATH. We denote by MATH and MATH corresponding elements in MATH respectively, MATH. then MATH . Since MATH as NAME space we have MATH. Furthe... |
math-ph/0006032 | Diagonality of MATH was stated in REF . For MATH we first calculate MATH . Here we use the notation of REF. In the third line, the summation index was shifted and REF was used. It remains to show that writing coefficients MATH the coproduct MATH has got the following form: MATH . In the last line, the summation index w... |
math-ph/0006032 | The proof is completely analogous to that of REF . Here REF is used to deal with the coefficients MATH and with the shifts of the summation index which are necessary here. |
math-ph/0006032 | First we show that all factors in the denominator of the triangular part of MATH cancel with the numerator of MATH. Poles can be present only if MATH, but this implies MATH. The denominator of the triangular part, see REF, applied to MATH is MATH . Since MATH, it cancels with the numerator of MATH as given in REF. The ... |
math-ph/0006032 | The denominator of the triangular part of MATH acting on MATH is given by MATH where MATH. Some factors cancel with the numerator of MATH (see the expression for MATH in REF). The remaining denominator of the triangular part is MATH since MATH and MATH. The denominator of the diagonal part MATH is MATH . Poles in MATH ... |
math/0006001 | Consider the product MATH and observe the condition of vanishing even-even block, which gives MATH, and other conditions follows obviously. |
math/0006001 | It simply follows from supermatrix multiplication law and general previous considerations. |
math/0006001 | In our case MATH . |
math/0006001 | Follows from REF . |
math/0006001 | For the difference using the band REF we have MATH. Then we expand in NAME series around MATH and obtain MATH, where MATH denotes MATH-th derivative which is a finite series for the power-type REF . |
math/0006001 | To find the difference MATH we use the expansion REF and the band conditions for components REF - REF . |
math/0006003 | The proof is standard. Consider a REF-complex MATH associated to a presentation of MATH with the minimal number MATH of generators. By general position there exists an embedding MATH inducing an isomorphism of fundamental groups. Then a regular neighborhood of MATH in MATH has a handlebody decomposition with MATH REF-h... |
math/0006003 | Reversing the handlebody decomposition of MATH one finds a decomposition from MATH without REF-handles. One slides the handles to be attached in increasing order of their indices. Using NAME Theorem it follows that MATH, where MATH is the number of REF, and thus MATH. |
math/0006003 | For MATH this is clear. Thus we suppose MATH. For any compact MATH choose some MATH such that MATH. Consider a loop MATH. Then MATH bounds an immersed (for MATH embedded) REF-disk MATH in MATH. We can assume that MATH is transversal to MATH. Thus it intersects MATH along a collection of circles MATH. Since MATH one is ... |
math/0006003 | Actually the following more general engulfing result holds: If MATH is a REF-dimensional polyhedron whose boundary MATH is contained in MATH then there exists an isotopy of MATH (with compact support), fixing MATH and moving MATH into MATH. Suppose that MATH. One reverses the handlebody decomposition of MATH and obtain... |
math/0006003 | For MATH it is well-known that g.s.c. is equivalent to w.g.s.c. which is also equivalent to s.c.i. if the manifold is irreducible. For MATH this is a consequence of NAME 's result stating the equivalence of g.s.c. and simple connectivity in the compact case (see CITE). If MATH is w.g.s.c. then it has an exhaustion by c... |
math/0006003 | There exists at least one group MATH, for instance MATH. Further if MATH and MATH verify the condition MATH, then their product MATH does. Thus, NAME 's Lemma says that a maximal element for the lattice of subgroups verifying this property (the order relation is the inclusion) exists. |
math/0006003 | First, MATH satisfies the condition MATH otherwise the minimality will be contradicted. Pick an arbitrary MATH satisfying this condition. If MATH it follows that MATH, hence by a transfinite induction we derive our claim. |
math/0006003 | We set MATH in the sequel. We establish first: If the pair MATH is stably compressible then it is MATH-compressible. We will use a transfinite recurrence with the inductive steps provided by the next two lemmas. Set MATH for the morphism making the pair MATH strongly compressible. If MATH and MATH then MATH. By REF. Al... |
math/0006003 | Since MATH is w.g.s.c. there exists a compact REF-connected submanifold MATH of MATH such that MATH. We can suppose MATH, without loss of generality. From now on we will focus on the pair MATH and suppress the index MATH, and denote it MATH, for the sake of notational simplicity. The pair MATH is stably compressible. L... |
math/0006003 | One can realize the homomorphism MATH by a disjoint union of bouquets of circles MATH. There is one bouquet in each connected component of MATH. One joins each wedge point to the unique connected component of MATH for which that is possible by an arc, and set MATH for the manifold obtained from MATH by adding a regular... |
math/0006003 | We first consider a simpler case: The result holds in the case of a manifold MATH of dimension at least MATH with one end. In this case MATH has an exhaustion MATH with MATH connected for all MATH. MATH has an exhaustion such that the map MATH induced by inclusion is a surjection for each MATH. As MATH is simply connec... |
math/0006003 | There is a natural projection map on the spine MATH, which is the fiber bundle projection of MATH (with fiber a REF-disk). When both MATH and MATH are fixed then such a projection map is also defined only up to isotopy. Set therefore MATH . Since MATH does not depend on the particular projection map (in the fixed isoto... |
math/0006003 | The proof here follows the same pattern as that given by NAME CITE for REF-dimensional case. Let us establish first the following useful property of the wrapping number: If MATH then MATH. This is a consequence of the two lemmas below: MATH. Consider MATH, where MATH is a very thin tube around MATH, and the two project... |
math/0006003 | It is sufficient to consider the case of the NAME manifolds since the pair of groups appearing in the product exhaustions are the same as this case. We start with REF-dimensional case, and take for MATH the classical NAME manifold. Recall that MATH is an increasing union of solid tori MATH, with MATH embedded in MATH a... |
math/0006003 | The proof given in CITE for REF-dimensional statement extends without any essential modification, and we skip the details. |
math/0006003 | Consider a connected compact submanifold MATH. Assume that there exists a compact polyhedron MATH with MATH and an immersion MATH such that MATH. One can suppose that MATH is a manifold. The polyhedron MATH is endowed with an immersion MATH into the manifold MATH. Among all abstract regular neighborhoods (that is, thic... |
math/0006003 | We have to reconsider the proof of REF . Everything works as above except that the disks MATH cannot be anymore embedded, but only (generically) immersed. They may have finitely many double points in their interior. Then the manifold MATH obtained by adding REF-handles along the MATH has a generic immersion MATH, whose... |
math/0006003 | We use the fact that degree-one maps are surjective on fundamental group. Given an exhaustion MATH of MATH, pull it back to MATH of MATH. Notice that MATH where MATH stands for the frontier. One needs then the following approximation by manifolds result. Given two MATH-complexes MATH there exist regular neighborhoods M... |
math/0006003 | Let MATH be the quotient map. Assume that MATH satisfies the NAME conjecture. Suppose that G violates the NAME conjecture. Then we have generators MATH and relations such that MATH is the trivial group. Let MATH be the map extending MATH by mapping MATH to MATH. This is clearly a surjection, and induces a surjective ma... |
math/0006003 | Suppose that MATH (with some REF-handle or REF-handle added if one boundary component is empty). It is well-known (see CITE) that the homology groups MATH are the same as those of a differential complex MATH, whose component MATH is the free module generated by the MATH-handles. Therefore this complex has the form: MAT... |
math/0006003 | Assume now that MATH admitted a proper handlebody decomposition without REF-handles. One identifies MATH with MATH. We can truncate the handle decomposition at a finite order to obtain a manifold MATH such that MATH, because the decomposition is proper. We can suppose that MATH is connected since MATH has one end. Then... |
math/0006003 | Since attaching MATH-handles and MATH-handles correspond to surgery and cutting along MATH-spheres respectively, we merely have to show that the surgery is REF-frame about a homologically trivial curve and the spheres along which one cuts are non-separating. First note that the absence of MATH-handles implies MATH, for... |
math/0006003 | If a surgery is performed along a curve MATH, this means that a REF-handle is attached along the curve in the MATH-manifold MATH. Hence MATH bounds a disk in MATH, which projects to a disk bounded by MATH in MATH. |
math/0006003 | We can find MATH so that MATH is entirely after MATH, and MATH so that MATH, because the handlebody decomposition is proper. We then define MATH by repeating this process once, that is, MATH, for some MATH for which MATH is entirely after MATH. Consider MATH which persists till MATH. This means that there is an annulus... |
math/0006003 | It follows from a transversality argument that the image of MATH intersects only REF-handles, along REF-disks. Further it is sufficient to see that the circles MATH are homotopic to REF-framing since in MATH homotopy implies isotopy for circles. If one circle is null-homotopic then it can be removed by means of an ambi... |
math/0006003 | Consider the diagram MATH where MATH is surjective, and the subscript MATH means abelianisation. Then it is automatically that MATH. Since MATH, and the strong compressibility is symmetric, the result follows. |
math/0006003 | The pull-backs in MATH of spheres MATH are planar surfaces with boundary components being the loci of future surgeries. Further, after compressing the spheres MATH of MATH (hence arriving into MATH) we have a surjection MATH, thus the map MATH is also surjective. This means that there exist curves in the complement of ... |
math/0006003 | If not then there exists a curve MATH that represents a non-trivial element of MATH. Modifying by a homologically trivial element if necessary, we may assume that MATH. By the previous lemma MATH persists. The group MATH is the quotient of MATH by the relations generated by the surgery curves, which are homologically t... |
math/0006003 | Let MATH be the rank of MATH and MATH be the fundamental groups of the surfaces. Since the surfaces are disjoint, MATH is obtained by HNN extensions from the fundamental group MATH of the complement of the surfaces. Thus, if MATH are the gluing maps, we have MATH . Now, since MATH and MATH, MATH surjects onto MATH, the... |
math/0006003 | By construction the images of the seams (which are roughly speaking half the generators of the fundamental group) are null-homotopic. If the fundamental groups of the generalized NAME surfaces from the previous remark map to the trivial group, then after doing surgery on the seams we obtain surfaces MATH representing h... |
math/0006003 | We will express each surgery locus MATH as a product of commutators of the form MATH, with each MATH being conjugate to a surgery locus (possibly MATH itself). It then follows readily that MATH, as now if each MATH, then each MATH. Suppose now MATH is a surgery locus. Then the MATH-frame surgery along MATH creates homo... |
math/0006003 | For the first stage, take two surfaces of genus MATH, and let them intersect transversely along two curves (which we call seams) that are disjoint and homologically independent in each surface. Next, take as NAME surfaces for these curves once punctured surfaces of genus MATH intersecting in a similar manner, and glue ... |
math/0006003 | We will take a variant of the example in the last section. Namely, we construct an explicit handle-decomposition according to a canonical form. Start with a MATH-handle and attach to its boundary three MATH-handles along an unlink. the resulting manifold has boundary MATH obtained by MATH-frame-surgery about each compo... |
math/0006003 | If we did have such a sequence of surgeries, then MATH bounds a MATH-manifold MATH with MATH , with a half-basis formed by embedded spheres. Now glue this to a manifold with form MATH which is bounded by MATH to get MATH. We can surger out the disjoint family of MATH's from MATH to get a MATH-manifold with form MATH an... |
math/0006014 | By REF , we know that MATH is a free MATH-module for all MATH, hence, MATH. Recall that MATH . Now, since MATH is an isomorphism, we conclude that, for all MATH, MATH is also a free MATH-module. Therefore, one has: MATH and MATH . Every NAME invariant MATH can then be seen as a linear map from MATH to MATH which vanish... |
math/0006014 | Since MATH is a normal subgroup of MATH, it is straightforward to prove that MATH. So, it suffices to prove that MATH. The inclusion MATH is obvious, once we notice that MATH and that MATH. For the other inclusion, we must prove that for all MATH one has MATH. Suppose that MATH, with MATH; then MATH, so it suffices to ... |
math/0006014 | Consider the NAME graph of MATH, which is defined as follows. Its vertices are the elements of MATH, and its edges are labeled by MATH. For every vertex MATH and for every MATH, there is exactly one edge labeled by MATH, with source MATH and target MATH. In this graph, the normal form of an element MATH corresponds to ... |
math/0006014 | The case MATH is trivial, since MATH. Hence MATH is a free group of infinite rank. Suppose now that MATH. It is well known that the kernel of the homomorphism MATH is MATH (see CITE). Moreover, one can easily see that MATH lies in this kernel, namely MATH. Similarly, one has MATH. The homomorphism MATH which sends MATH... |
math/0006014 | We only need to verify that the action of the generators of MATH on the generators of MATH is trivial after abelianization. Moreover, let us see that it suffices to show the result for the action defined by any set-map section MATH of MATH. Indeed, if MATH is a section of MATH, then for every MATH, there exists an elem... |
math/0006014 | We can suppose, without loss of generality, that MATH. Suppose first that MATH is a single letter. If MATH and MATH is odd, then one can easily show by drawing the braids that the following equality holds in MATH: MATH . Hence MATH where MATH . If MATH and MATH is even, then one has MATH . Therefore, MATH where MATH . ... |
math/0006014 | Recall the epimorphism MATH. One has MATH, thus MATH, where MATH is a word over MATH. By drawing the braids, one sees that MATH, for MATH. Moreover, since MATH, one has MATH, where MATH, for all MATH. Therefore, since MATH is a word over MATH, one has: MATH, where MATH. Hence, MATH where MATH, as we wanted to show. |
math/0006014 | Clearly, it suffices to show the lemma when MATH is a single letter. Besides, we can suppose that MATH. Then the result is a consequence of the following relations in MATH. MATH where MATH. These relations can be easily verified by drawing pictures. |
math/0006014 | This lemma follows from the well known congruences MATH (see, for example, CITE), together with the inclusion MATH. |
math/0006014 | We only need to prove that MATH is the inverse of MATH as a homomorphism of MATH-modules. For MATH, let MATH be the submodule of MATH consisting of the homogeneous polynomials of degree MATH. Consider as well MATH, where MATH, MATH and MATH. By Relations REF , seen as relations in the enveloping algebra MATH of MATH, o... |
math/0006014 | Recall that, by REF , the ideal MATH of MATH is isomorphic to MATH via MATH, for all MATH. Moreover, since MATH is a free MATH-module, one has: MATH . Hence, MATH via MATH. Now, MATH and both MATH and MATH are isomorphisms of MATH-modules, thus MATH is an isomorphism of MATH-modules. |
math/0006014 | Let MATH. Write MATH and MATH for MATH. Then MATH . So, in order to prove that MATH, it suffices to show that MATH . But MATH, thus there exists MATH such that MATH with MATH, hence, in MATH, MATH since MATH, as we wanted to show. |
math/0006014 | Write MATH and MATH, to simplify notation. Write as well MATH. We know that MATH is a MATH-algebra isomorphism, so MATH . On the other hand: MATH . Therefore, we need to show that, in MATH, MATH . Since the action by conjugation does not depend on MATH, we only need to verify the above formula when MATH is a generator ... |
math/0006014 | The first equation is a consequence of the following relations in MATH, which are easily verified. MATH . The second equation comes from REF , and from the following relations, where MATH and we have denoted MATH if MATH is odd, and MATH if MATH is even. MATH . Indeed, in this case, MATH and by REF , this is equivalent... |
math/0006022 | For MATH, we compute MATH where in the penultimate equality, we have used the fact that MATH is a homomorphism of NAME algebras and REF . This establishes REF , and REF follows from REF . |
math/0006022 | For MATH, we have MATH . Thus MATH is closed under the product MATH if and only if MATH is a homomorphism. The remaining assertions are clear. |
math/0006022 | It is straightforward to check that REF give the NAME identity for the bracket REF . That MATH is immediate from REF , and so the decomposition is reductive. From REF we have MATH and MATH for MATH, which proves the remaining assertion. |
math/0006022 | Indeed, for MATH, we have MATH using REF . This establishes the equivalence of REF . |
math/0006029 | See REF . |
math/0006029 | We can decompose MATH where the MATH's are irreducible representations. By what we have said before, there are integers MATH, MATH, with MATH, MATH, such that MATH is a direct summand of MATH. Our assumption on the action of MATH implies that MATH. Let MATH be a positive integer which is so large that MATH for MATH. Th... |
math/0006029 | We may assume that MATH is a very ample line bundle. Set MATH. The linearization MATH provides us with a representation MATH and a MATH-equivariant embedding MATH. Since obviously MATH for all points MATH and all one parameter subgroups MATH of MATH, we can assume MATH. Now, there are a basis MATH of MATH and integers ... |
math/0006029 | This theorem can be proved with the methods developed in CITE for MATH. A more elementary approach is contained in the note CITE. |
math/0006029 | Let MATH be any subbundle. By REF, MATH, so that MATH-semistability gives MATH that is, MATH so that the theorem holds for MATH. |
math/0006029 | By our assumptions, the sheaf MATH is locally free of rank MATH. Therefore, we can choose a covering MATH, MATH, of MATH, such that it is free over MATH for all MATH. For each MATH, we can choose a trivialization MATH so that we obtain a surjection MATH on MATH. Therefore, MATH is a quotient family of MATH-pairs of typ... |
math/0006029 | The two morphisms MATH and MATH provide us with quotient families MATH and MATH of MATH-pairs parameterized by MATH. By hypothesis, MATH, and we have isomorphisms MATH and MATH with MATH . In particular, there is an isomorphism MATH . This yields a morphism MATH and MATH. Let MATH be the fibre product taken with respec... |
math/0006029 | First, suppose we are given a weighted filtration MATH, MATH, such that MATH is globally generated and MATH for MATH. Then, for MATH, MATH so that the claimed condition follows from MATH being MATH-(semi)stable. Next, recall that we have found a universal positive constant MATH depending only on MATH, MATH, and MATH, s... |
math/0006029 | To see the equivalence between REF. and REF., observe that MATH provides an equivariant embedding of MATH into MATH. Via the canonical surjection MATH, the latter space becomes embedded into MATH, so that we have an equivariant embedding MATH. Since for every point MATH and every one parameter subgroup MATH the claimed... |
math/0006029 | First, assume REF. Let MATH be a homomorphism. Call a sub vector space MATH-superinvariant, if MATH and MATH. Let MATH. Given a basis MATH of MATH and MATH, set MATH. Then REF MATH, if MATH is not MATH-invariant. CASE: MATH, if MATH is MATH-superinvariant and MATH. CASE: MATH in all the other cases. Now, let MATH be a ... |
math/0006033 | First note that MATH is a one-dimensional non-commutative MATH complex, as defined in CITE. Hence MATH is semiprojective by CITE and finitely generated by CITE. |
math/0006033 | Let MATH denote the MATH identity matrix for any non-negative integer MATH. For each MATH, define an integer matrix of size MATH by MATH . Let MATH denote the MATH matrix MATH and define for MATH, an integer matrix of size MATH by MATH . For MATH, define yet another MATH matrix by MATH . Finally, let MATH be the MATH m... |
math/0006033 | Define a *-homomorphism MATH by MATH . Via the identification MATH we define a *-homomorphism MATH by MATH . The short exact sequence MATH gives rise to a six-term exact sequence MATH where MATH denotes the exponential map. By NAME periodicity MATH is generated by MATH, where MATH is a unitary in MATH. Note that MATH a... |
math/0006033 | Let us start with a simple and well-known observation. Let MATH be a unitary in the MATH-algebra MATH such that the winding number of MATH is MATH. Then MATH can be connected to MATH via a continuous path MATH in MATH. If MATH we may assume that MATH for every MATH. Let MATH be the exceptional points of MATH, where MAT... |
math/0006033 | Let MATH denote the rank of MATH and let MATH be the exceptional points of MATH, where MATH. Since MATH divides MATH for MATH, it follows that MATH also divides MATH. Hence there is a projection MATH with the same trace as MATH. For each MATH there is a unitary MATH such that MATH . We may assume that MATH, MATH, and t... |
math/0006033 | As in the proof of REF we see that there exists a projection MATH of rank MATH if and only if MATH divides MATH. The conclusion follows. |
math/0006033 | Let MATH where MATH is a positive integer, MATH are non-negative integers, and MATH are prime numbers. Let MATH be prime numbers, mutually different as well as different from MATH. Define integers MATH and MATH by MATH . Set MATH. Then MATH by REF . MATH is unital projectionless by REF . |
math/0006033 | Choose MATH such that MATH. Set MATH . We have a pull-back diagram MATH where MATH are the restriction maps and MATH evaluation at MATH. Apply the NAME sequence CITE to get a six-term exact sequence MATH . Note that MATH and MATH. Thus the exact sequence becomes MATH . Since MATH is homotopic to evaluation at MATH in M... |
math/0006033 | Choose MATH such that MATH, MATH, are the exceptional points for MATH. Set MATH . Define a *-homomorphism MATH by MATH. Let MATH be evaluation at MATH. Let MATH denote the map MATH. Let MATH be the map MATH. We have a pull-back diagram MATH and hence by CITE a six-term exact sequence of the form MATH . MATH is generate... |
math/0006033 | By REF , or simply because homotopic *-homomorphisms MATH define the same elements in MATH, we have that MATH . From this the existence follows. To check uniqueness, assume MATH where MATH . Fix some MATH. By REF there exist integers MATH such that MATH, MATH. Therefore MATH for MATH. Finally, to prove the equations ab... |
math/0006033 | Let MATH. By REF there is an integer MATH, MATH, and integers MATH, MATH, such that MATH . Note that MATH for MATH, and MATH. By REF we see that for MATH, MATH . By REF there exists a unitary MATH such that the matrix MATH belongs to MATH for all MATH. Set MATH . Let MATH denote the exceptional points of MATH and let M... |
math/0006033 | By REF there exists an element MATH such that MATH. Let MATH have standard form MATH . By adding an integer-multiple of MATH we may assume that MATH for MATH. Define MATH and MATH, MATH, as in the proof of REF . Let MATH . Choose a positive integer MATH such that MATH. Choose for each MATH, a unitary MATH such that the... |
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