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math/9912059 | Denote by MATH the MATH-functor which corresponds to MATH by NAME. Then MATH for some MATH and some MATH. And necessarily, MATH (where the notation MATH means the smallest element). |
math/9912059 | We have MATH . But MATH is a natural transformation from MATH to MATH. By REF , we get MATH . |
math/9912059 | One has MATH . Therefore if MATH for some MATH-category MATH, then MATH for MATH and we obtain a fillable MATH-shell in the sense of REF . |
math/9912059 | We are going to show the formula by induction on MATH. The case MATH is trivial. If MATH, then MATH. And if MATH, then MATH and MATH . So the labelings MATH and MATH of MATH are the same ones. |
math/9912059 | It is an immediate consequence of REF and of the uniqueness of REF . |
math/9912059 | Set MATH where MATH and set MATH. By hypothesis, the equality MATH holds and let MATH . Take any thick MATH-functor MATH from MATH to MATH. Then MATH . By REF , it is clear that MATH induces a MATH-functor. |
math/9912059 | If MATH, then MATH and by construction of MATH, MATH for any MATH and MATH is MATH-dimensional for any MATH. For MATH equal to MATH, MATH or MATH, the converse is obvious. Suppose the converse proved for MATH and let us prove it by induction for MATH. By hypothesis, as soon as MATH, then MATH is MATH-dimensional. For M... |
math/9912059 | One has MATH. Therefore MATH . So one has MATH . And MATH . Consequently one has MATH . One has MATH. Therefore MATH . So MATH . And MATH therefore MATH . One has MATH. Then MATH . So MATH . One has MATH. Therefore MATH so MATH . Now let us calculate MATH. One has MATH . |
math/9912059 | Equalities REF are obvious. Equalities from REF to REF are immediate consequences of the definitions. With REF , one sees that MATH . For a given MATH, the above equalities are equalities in the free cubical MATH-category generated by MATH. Therefore, they depend only on the relative position of the indices MATH, MATH ... |
math/9912059 | It is obvious for MATH. We are going to make an induction on MATH. Let MATH and MATH. Then MATH . The equality MATH makes sense if MATH is a MATH-cube. And in this case, MATH is MATH-dimensional and MATH. This equality holds in the free cubical MATH-category generated by MATH, and therefore MATH . Now suppose that MATH... |
math/9912059 | Equalities REF are obvious. For the sequel, one can suppose MATH. In the cubical singular nerve of a MATH-category, two elements MATH and MATH of the same dimension MATH are equal if and only if MATH and for MATH and MATH, one has MATH. Now we want to prove Equality REF. Since MATH, then MATH for any MATH of dimension ... |
math/9912059 | It is easy to see that MATH. Now we suppose that MATH. Set MATH. We are going to prove that MATH by verifying that the second member satisfies the characterization of REF . Let MATH. REF implies that for MATH, the dimension of MATH is zero (or equivalently that it belongs to the image of MATH). With REF , one gets MATH... |
math/9912059 | First let us make the proof for MATH and MATH. Let us consider the following MATH-functor from MATH to MATH : MATH . Then MATH where MATH is a thin element. Therefore MATH and MATH are NAME. We claim that the above construction is sufficient to prove that MATH and MATH are NAME for any MATH and for any MATH. The labele... |
math/9912059 | It suffices to make the proof for MATH and MATH. And to consider the following thin MATH-cube MATH . Notice that the above MATH-cube is exactly MATH by REF . |
math/9912059 | It suffices to make the proof for MATH and MATH. Set MATH . One has already seen that MATH . It suffices to construct a thin MATH-cube MATH such that MATH and MATH. If the MATH-cube is conventionally represented by REF , the thin <<<<<<< coin.tex labeled MATH-cube of REF with MATH meets the requirement. ======= labeled... |
math/9912059 | Let us denote by MATH the following property : ``for any MATH and with MATH, for any MATH-morphisms MATH and MATH of any MATH-category MATH such that MATH exists, there exists a thin MATH-cube MATH and a thin MATH-cube MATH which lie in the cubical singular nerve of the free globular MATH-category generated by MATH and... |
math/9912059 | By definition, one has MATH. If MATH was equal to the globular sub-MATH-category generated by MATH then MATH still would exist in the cubical singular nerve. Therefore, MATH can be written as an expression using only the composition laws MATH of MATH and the variables of MATH and moreover, the variables MATH and MATH c... |
math/9912059 | We need, only for this proof, the operator MATH introduced in CITE. One has MATH and MATH . Therefore MATH exists and is NAME to MATH by REF . If we work in the MATH-category generated by MATH and MATH, then MATH is a well-defined element of the branching simplicial nerve of MATH. And MATH is the free MATH-category gen... |
math/9912059 | Obvious. |
math/9912059 | One has in the reduced branching complex MATH and MATH therefore MATH induces a linear map from MATH to MATH. And MATH. Since MATH is the identity map on MATH, then MATH is generated by the MATH where MATH runs over MATH. Therefore the induced morphism of chain complexes is surjective. |
math/9912059 | Since MATH induces the identity map on MATH, then MATH. Therefore MATH . |
math/9912059 | One has MATH and MATH . Now suppose that MATH. Then MATH . So MATH. In the same way, one has MATH . |
math/9912059 | For all MATH, we have seen that MATH induces the identity map on the reduced branching complex. Therefore for all MATH, MATH. The latter equality can be translated into MATH . If the above equality was in MATH, the proof would be complete. Unfortunately, we are working in the reduced branching chain complex, and so the... |
math/9912059 | Consider the case of the reduced branching homology. Let MATH. If MATH then MATH and MATH are two thin elements of MATH. Therefore MATH in the reduced branching complex of MATH. Now suppose that MATH . By hypothesis, there exists MATH such that MATH. Therefore in the reduced branching complex, one has MATH. So MATH is ... |
math/9912059 | Set MATH for any MATH. It is clear that MATH in MATH. Now suppose that MATH for some MATH. We already mentioned that MATH is the free MATH-category generated by a composable pasting scheme in the proof of REF . It turns out that MATH and MATH belong to MATH and it is possible thereby to use the explicit combinatorial d... |
math/9912059 | The assertions concerning the formal branching homology are obvious. Since the negative folding operator induces the identity on the reduced branching complex, then MATH is equal to MATH for MATH and is generated by MATH and MATH for MATH. The point is to prove that there is no relations between MATH and MATH for MATH,... |
math/9912059 | Analogous to the previous proof. |
math/9912059 | We know that MATH . And the differential maps is also completely known. In the formal branching complex, one has MATH where MATH are the MATH's appearing in the word MATH with MATH. It follows that this chain complex can be splitted depending on the position and the number of the MATH's, and that these positions and nu... |
math/9912065 | Of course, an isomorphism between two MATH-framed surfaces must be a MATH-framing of an isomorphism between two ordinary surfaces in the category of ordinary three-dimensional cobordisms, and thus the surfaces must be diffeomorphic. Suppose MATH and MATH are MATH-framed surfaces which are MATH-framings of the same surf... |
math/9912065 | The functor from MATH to MATH is just the inclusion functor. For the other direction choose for each object of MATH which is connected an isomorphism to the chosen object of the same genus (using REF ). An arbitrary object MATH of MATH is a union of connected pieces, so it is the domain of an isomorphism MATH which is ... |
math/9912065 | Let MATH be the equivalence functor MATH and MATH be the functor MATH . Then of course MATH is a symmetric monoidal functor MATH and hence a TQFT. If MATH is a TQFT whose restriction to MATH is MATH and MATH is the natural isomorphism from MATH to the identity functor on MATH then MATH is an isomorphism from MATH to MA... |
math/9912065 | If MATH define MATH . For each permutation morphism MATH define MATH to be the corresponding permutation map. Because the permutation groupoid embeds into MATH we have MATH and because of the symmetry axiom we have MATH when MATH is basic. If MATH is not basic because it has an underlying manifold which is not connecte... |
math/9912065 | Of course two basic MATH-framed cobordisms are the same if and only if there is they have the same signature and there is a diffeomorphism of the boundaries of the underlying four-manifold intertwining the embeddings of each MATH in CITE NAME proves that given two compact connected three-manifolds with boundary and a d... |
math/9912065 | Of course invariance under the biframed NAME moves guarantees MATH is an invariant of the cobordism. We are given that it satisfies the Symmetry and Nondegeneracy Axioms of REF , and to see it satisfies Sewing and Mending it suffices to check that the embeddings and link shown in REF (shown here only for a genus two MA... |
math/9912066 | Without loss of generality, we may replace MATH with MATH. Let MATH be elements of the MATH-vector space generated by the elements MATH. Now, if MATH is a permutation of MATH, we claim that MATH where MATH is the degree of the left hand side. Since the elements MATH commute and MATH and MATH are at most linear in the v... |
math/9912066 | Since MATH is positive, MATH is positively graded commutative MATH-algebra and REF implies there are positive integers MATH and MATH such that MATH and MATH. Hence, we have MATH which proves REF follows immediately from REF by applying REF. |
math/9912066 | CASE: Every element MATH can be written uniquely as a finite sum of homogeneous components; MATH where MATH. Let MATH be defined by MATH where MATH. The definition of the MATH-module structure on MATH insures that MATH is a MATH-module homomorphism and the image is a MATH-module. It is clearly surjective. Now, if MATH ... |
math/9912066 | CASE: Let MATH, MATH and MATH. By REF , MATH is a subvector space of MATH which implies MATH and MATH. On the other hand, REF also states that the quotient MATH is a MATH-torsion module. Since MATH is a finitely generated module, there exists an integer MATH such that MATH. Thus, we have MATH which implies MATH. Combin... |
math/9912066 | We begin by observing the following: MATH . Next, notice that MATH. Since MATH is a nonzero-divisor of degree MATH on MATH, we have MATH. Thus, applying the formula MATH for a monotonically increasing function MATH and the function MATH, completes the proof. |
math/9912066 | By definition, irreducible components of MATH correspond to minimal primes in the support of MATH. Hence, if there exists an irreducible component of MATH of dimension MATH, we have a minimal prime MATH in the support of MATH of dimension MATH. Each minimal prime MATH in the support of MATH corresponds to a graded subm... |
math/9912066 | By definition, MATH is closed under addition, so it is enough to show that it is closed under left multiplication by the elements MATH. Consider the element MATH in MATH. One clearly has MATH where MATH is the MATH-th standard basis vector. Similarly, because MATH belongs to MATH, we obtain MATH where MATH has an initi... |
math/9912066 | See REF is CITE. |
math/9912066 | REF are analogous to REF. REF follows from the fact that the exponent vectors appearing in MATH for MATH form a subset of the exponent vectors appearing the standard expression of MATH in MATH. |
math/9912066 | See REF. |
math/9912066 | Since MATH, the set MATH generates MATH if and only if MATH generates MATH. Thus, it suffices to study the initial ideals. If MATH then we have MATH. Since MATH is a NAME basis, it follows that MATH for some MATH and MATH. NAME, we obtain MATH which implies MATH as required. |
math/9912066 | See REF. |
math/9912066 | Consider homogenization MATH of MATH with respect to a positive vector MATH. By REF there exists only finitely many distinct reduced NAME basis for MATH with respect to term orders. Let MATH be their union. Fix a term order MATH on MATH and let MATH denote the multiplicative order on MATH obtained from the multiplicati... |
math/9912066 | See REF. |
math/9912066 | Let MATH be any term order and let MATH be the multiplicative monomial order on MATH defined as follows: MATH . Next, fix a universal NAME basis MATH for MATH and choose MATH small enough so that the following assertions hold: MATH and for all elements MATH in MATH, the standard form of MATH breaks into four pieces MAT... |
math/9912066 | See REF. |
math/9912066 | Given the above lemmas and propositions, the proof is now identical to REF. |
math/9912066 | By subsection REFEF, we know that if MATH has a good finite dimensional filtration, then MATH. Moreover, the NAME dimension and NAME dimension of MATH are equal by REF . Since the NAME dimension of a finitely generated module is the NAME dimension of its support, it suffices to consider ideals. In particular, one reduc... |
math/9912066 | Fix a positive vector in MATH and let MATH and MATH be associated NAME ring and module. Now, suppose MATH has an irreducible component of dimension MATH. Since MATH this means that the MATH-ideal MATH has a minimal prime MATH of dimension MATH. By REF , we have MATH . Because MATH is a nonzero-divisor on MATH, MATH pul... |
math/9912066 | If the characteristic variety MATH has an irreducible component of dimension strictly less than MATH, then REF implies that there exists a submodule MATH of MATH with NAME dimension strictly less than MATH. However, this contradicts our hypothesis. |
math/9912066 | This is immediate from REF . |
math/9912069 | We use standard facts about composita of NAME extensions, compare CITE or CITE. Since the right-hand sides of the defining equations are linearly independent over MATH, the field MATH is a NAME extension of MATH with NAME group MATH, and the genus of MATH is the sum of the genera of all the degree-MATH subextensions of... |
math/9912069 | We prove the lemma by induction on MATH. Throughout the proof, if MATH is an integer, then MATH denotes the unique integer such that MATH and MATH (mod MATH). For MATH we must find positive integers MATH and MATH which are coprime to MATH such that MATH (mod MATH) and MATH. This is easily done: if MATH, then set MATH a... |
math/9912069 | Fix a finite field MATH of characteristic MATH. Let MATH be a monic irreducible polynomial of even degree. For any MATH, consider the system of REF where both MATH and MATH are increasing sequences of positive integers coprime to MATH. This is the fiber product of a degree-MATH cover MATH (as discussed in REF ) with a ... |
math/9912069 | Let MATH be an increasing sequence of odd positive integers, and consider the system of REF . Just as in the proof of REF , we see that these equations define a curve MATH over MATH such that MATH and the genus of MATH is MATH . For a fixed positive integer MATH, we can find sequences MATH as above yielding curves of a... |
math/9912069 | For any MATH, there is a genus-MATH curve which has at least one MATH-rational point and which satisfies REF - REF : for instance, let MATH be squarefree of degree MATH, and take MATH for even MATH and MATH for odd MATH. Thus, it suffices to show that for sufficiently large MATH there exist curves in toric surfaces who... |
math/9912069 | A toric surface is a two-dimensional normal variety MATH, equipped with the action of the two-dimensional algebraic torus MATH and a dense equivariant embedding MATH. The combinatorial object attached to MATH is a fan, which we denote MATH, consisting of cones in a two-dimensional lattice MATH; the dual lattice MATH (u... |
math/9912069 | We use the notation of the proof of REF . Fix MATH, and let MATH be a curve of genus MATH which embeds in a toric surface MATH over MATH. Without loss of generality, we may assume MATH is nonsingular projective, and has fixed point set (for the torus action) disjoint from MATH. Since every rational point of MATH is eit... |
math/9912069 | Fix MATH, and let MATH be the least prime not dividing MATH. For any MATH, define MATH to be the least integer such that MATH . Note that the right side might be only slightly larger than the left, or it might be larger by a factor of as much as (slightly more than) MATH. We will modify the set of primes under consider... |
math/9912069 | Let MATH be the maximal tamely ramified abelian cover of MATH which is unramified outside MATH and in which MATH splits completely. By Class NAME Theory for MATH, the NAME group MATH of MATH fits in a short exact sequence MATH where MATH is the diagonal embedding into the product. Moreover, the inertia group over the p... |
math/9912069 | Assume MATH is odd, so MATH. Let MATH be positive integers which will be specified later. Put MATH and MATH for MATH; set MATH for all MATH. Note that MATH (mod MATH) for all MATH. An easy count shows that, for any MATH, there are at least MATH finite places on MATH of degree MATH; since MATH, it follows that we can ch... |
math/9912069 | Fix MATH. For any MATH, put MATH and let MATH be places of MATH of degrees MATH and MATH. Let MATH be the maximal tame abelian cover in which all places besides MATH are unramified and in which MATH splits completely. Then MATH is a cyclic cover of degree MATH in which MATH is totally ramified. Moreover, since MATH spl... |
math/9912070 | See for instance CITE REF |
math/9912070 | Let MATH be an indecomposable vector bundle. Then for any base of MATH and MATH, MATH, otherwise MATH for some vector bundle MATH. Let MATH of dimension MATH: if MATH and MATH is such that MATH, then the restriction of MATH in MATH defines a morphism of vector bundles MATH. Let us suppose MATH, then MATH: REF implies t... |
math/9912070 | Let us suppose that REF holds, then MATH and MATH have the same degeneracy locus and this implies: MATH . Since MATH, MATH, MATH and MATH are irreducible, it results MATH and MATH or MATH and MATH, thus REF holds. NAME let us suppose MATH and MATH and let MATH be the respective base change matrices. Then MATH . Otherwi... |
math/9912070 | Let MATH. Then MATH and thus if we consider REF-dimensional parameter subgroup defined by the weights MATH and MATH, it results: MATH . Thus the points of MATH are in one-one correspondence with the orbits of the matrices MATH by the action of MATH. The previous lemma implies the isomorphism in REF . |
math/9912070 | By the proof of REF we have that for any MATH, MATH, where MATH is the image of MATH as a subspace of MATH. Thus we have, up to a base change, MATH where MATH and MATH are linearly independent. It is easily checked that MATH and since MATH, it must be MATH: this implies that MATH. Moreover MATH for any MATH otherwise M... |
math/9912070 | See CITE and CITE. |
math/9912070 | It is just a direct computation. |
math/9912070 | By REF , it follows that MATH for any MATH defined as in REF . Thus, by REF , the points represented by the matrices MATH, do not give any contribute to MATH. Moreover it is not difficult to see that, if MATH, the only matrix MATH, as in REF , such that MATH is given by MATH and MATH, where MATH. |
math/9912074 | To prove the proposition, it is enough to show that over an algebraic closure MATH of MATH, the category MATH is semi - simple and every irreducible object in it is MATH - equivariant. Indeed, this would imply that every object of MATH has a form MATH, where MATH is absolutely irreducible (and hence MATH - equivariant)... |
math/9912074 | By taking NAME expansions of the map MATH, we obtain a map MATH. Consider the map MATH obtained from the first map MATH and MATH . Our two MATH - points of MATH are the compositions of the above map followed by the projection MATH in the first case, and by the action map MATH in the second one. Hence, the proposition f... |
math/9912074 | Let MATH be a MATH - point of MATH with MATH. Then the data of MATH trivializes MATH. Therefore, MATH is a reduction to MATH of the trivial MATH - bundle on MATH, that is, a map MATH. This defines a map MATH and it is straightforward to see that it is an isomorphism. The fact that MATH follows immediately from REF. |
math/9912074 | Let MATH be the support of MATH in MATH. Choose an integer MATH so that MATH acts on MATH through the quotient MATH. Thus, we have to show that MATH is MATH - equivariant. Let MATH be the sheaf of groups (on the category of all schemes with NAME topology), which attaches to a scheme MATH the group of maps from the loca... |
math/9912074 | We will prove that if MATH is affine, the map MATH is a surjection. The proof goes by induction. The assertion is obvious for MATH, since MATH. Therefore, it suffices to show that any map MATH whose restriction to MATH is trivial, can be lifted to a map from the localization MATH to MATH. However, the group of maps MAT... |
math/9912074 | For two integers MATH we introduce a (relative) NAME scheme MATH, where for a scheme MATH, MATH consists of a map MATH and a coherent subsheaf MATH of MATH over MATH, such that the quotient MATH is MATH - flat. For MATH, there is a natural closed embedding MATH. It is easy to see that for MATH, MATH identifies naturall... |
math/9912074 | The fact that MATH and MATH follows immediately from REF. Hence, we must analyze the situation over MATH. We will prove the assertion for MATH, since the proof for MATH is the same. We will construct canonical morphisms in both directions between the corresponding functors. CASE: Let MATH be as above with MATH. First, ... |
math/9912074 | We will apply REF to the map MATH. We have: MATH . However, MATH. Hence, it remains to show that MATH, that is, that the vanishing cycles functor MATH applied to MATH yields zero. This follows almost immediately from the fact that MATH was obtained by the twisted external product construction: Let MATH be a closed MATH... |
math/9912074 | First, by the construction of the functor MATH, we have a MATH - action on MATH for every MATH. Moreover, REF implies that the isomorphisms MATH and MATH (for MATH) are compatible with this MATH - action. Therefore, for MATH we have the isomorphisms MATH . In addition, REF applied to the map MATH implies that MATH. Let... |
math/9912074 | MATH REF . Let us first prove the assertion when MATH. In this case for an affine MATH, MATH consists of pairs MATH, where MATH is a projective rank MATH module over MATH and MATH is an isomorphism MATH, where MATH is the standard MATH-dimensional vector space. For an integer MATH consider the MATH - dimensional vector... |
math/9912083 | Let MATH be a tubular neighborhood of MATH in MATH, and MATH its preimage in MATH. Then MATH has the structure of MATH. Let us denote by MATH and by MATH the two boundary components of MATH. Let MATH be a function on MATH which is constant and equal to MATH in a neighborhood of MATH and is constant and equal to MATH in... |
math/9912083 | We can pull back the integrand form in REF to MATH and integrate it over there. Using the fact that the pullback of MATH is exact, by NAME 's theorem we can rewrite REF as MATH . The codimension-one faces in MATH are labeled by subsets of MATH. Denote by MATH any of these subsets. Assume now that the cardinality of MAT... |
math/9912083 | We work as in the proof of REF . The only difference is that we must distinguish between the cases when the collapse is at a point on MATH or otherwise. Notice that, since MATH does not belong to the image of MATH, there is no such term as MATH. By the same reason, when we consider a collapse at a point on MATH, we onl... |
math/9912086 | Note that MATH is a topologically trivial MATH-bundle over MATH. Hence by NAME formula we have MATH . Since the higher direct image sheaves MATH, vanish, it follows that MATH. We deduce that the NAME group MATH is generated by MATH and the subgroup of MATH generated by the irreducible components of MATH. Thus for MATH ... |
math/9912086 | CASE: Assume the contrary. Recall that MATH is a compact complex torus with universal covering MATH. We may regard the NAME class MATH as bilinear form on the vector space MATH. Suppose that MATH is not semi-positive definite. Let MATH with MATH and let MATH denote the orthogonal complement of MATH (that is, MATH). Let... |
math/9912086 | Assume the contrary. Then there exists a stratum MATH such that MATH. Clearly the closure MATH of MATH is likewise contained in MATH. But the closure of any stratum contains a minimal stratum, that is, contains a stratum of MATH. However, this is in contradiction to REF . |
math/9912086 | First, if MATH is invariant under one of the MATH direct factors of MATH in REF , we take the corresponding quotient. Thus we may assume that MATH does not contain anyone of the MATH coordinate factors of MATH. Assume that MATH. Let MATH be a subgroup of MATH isomorphic to MATH. Then there are integers MATH such that M... |
math/9912086 | Let MATH (respectively, MATH) be the universal covering of MATH (respectively, MATH), and MATH the preimage of MATH. We may assume that MATH . Let MATH be MATH-linear independent real numbers with MATH . Let MATH or MATH be an arbitrary integer as assumed in the proposition, and set MATH where MATH is a linear map such... |
math/9912088 | It is easy to see that MATH, the group algebra of MATH. We define a map MATH by noting that MATH extends to a non-zero MATH-algebra map MATH. We then take MATH, which is a maximal ideal of MATH. Since the domain and codomain of MATH both take products of groups to products of varieties, it suffices to check that MATH i... |
math/9912088 | Define MATH, and MATH the set of all connected components of the elements in MATH. Put a metric on MATH which yields its usual topology. Denote this metric by MATH. Let MATH. If MATH for all MATH (this is possible only when MATH is connected), then choose MATH such that MATH, and MATH is small. If, on the contrary, the... |
math/9912088 | MATH is the group of irreducible characters of MATH, so for MATH consider the complex vector bundle MATH over the classifying space MATH, where the map MATH is given by MATH. Then the first NAME class of MATH gives a natural isomorphism MATH. If MATH is a torus, we saw in REF that MATH can be identified with the dual o... |
math/9912088 | All the results we mention in this proof are from CITE. Identify the complex NAME algebra MATH with MATH. We denote by MATH the algebraic structure sheaf of MATH, and by MATH the analytic structure sheaf. Let MATH be the completion of the local ring MATH with respect to its maximal ideal. It is sufficient to show that ... |
math/9912088 | First we show that MATH is a sheaf of MATH-modules. If MATH is an open cover of a topological space MATH, denote by MATH, etc. Then a presheaf MATH is a sheaf if and only if for any MATH and any finite cover MATH the following sequence is exact MATH where MATH, MATH, etc. Denote this sequence by MATH. Since MATH is a s... |
math/9912088 | The case when MATH is trivial, so we assume the contrary. Then there exists a compact subgroup MATH of MATH such that MATH and MATH. Then REF from REF implies that MATH. But MATH, so we deduce that MATH. This means that for every MATH we have MATH. Now we will prove that whenever MATH, MATH becomes an isomorphism if te... |
math/9912088 | Since MATH, it follows that MATH. Let MATH be the connected component of MATH containing the identity. Because both MATH and MATH belong to MATH, REF from REF implies that MATH and MATH belong to the same connected component of MATH, or equivalently that MATH. Let MATH be the complexification of the NAME algebra of MAT... |
math/9912088 | MATH is fixed by MATH, so by REF we know that MATH, with MATH. Therefore it is enough to define a map MATH . But MATH was shown in REF to be a map of MATH-algebras, so we can just take MATH. More formally, we define MATH as the composition of the following isomorphisms: MATH. |
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