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math/9912180 | It is immediate that MATH is a homomorphism. Since MATH is compact we may assume that MATH is a unitary representation of MATH, that is, the image of MATH is contained in the unitary group MATH. Then MATH is a constant multiple of a matrix in MATH so that MATH is contained in MATH for all MATH. For the continuity of MATH it suffices to show that the graph of MATH in MATH is closed, since MATH is a compact NAME space. Consider the family of continuous maps MATH for each MATH given by MATH. Then the set MATH is the inverse image of the graph of MATH in MATH by the canonical projection MATH, which is obviously closed in MATH. Therefore the graph of MATH is also closed in MATH. |
math/9912180 | The necessity is obvious so we prove the sufficiency. Define a function MATH by MATH for MATH and MATH. It is immediate that MATH on MATH. In this proof we shall use the symbols MATH and MATH for elements in MATH and MATH, respectively. Claim: MATH is well-defined. If MATH, then MATH. Then REF implies that MATH and thus MATH. Claim: MATH is a homomorphism. For MATH, REF implies that MATH since MATH and MATH. Claim: MATH is continuous. The map MATH sending MATH is a continuous surjection. Since both MATH and MATH are compact, MATH is a closed map so that MATH has the quotient topology induced by MATH. MATH . Then the continuity of MATH follows from the universal property of the identification map MATH since the composition MATH sending MATH is continuous. |
math/9912180 | Let MATH denote the identity component of MATH. Since the canonical projection MATH is open and closed, MATH is a connected component of MATH so that MATH. It is well known in NAME group theory CITE that MATH, where MATH is the identity component of the center of MATH, which is a torus and MATH is the commutator subgroup of MATH. Then MATH since MATH is abelian, and thus MATH. Using the isomorphism MATH we may view MATH as a one-dimensional unitary representation of the torus MATH. It is elementary in representation theory that there exists a circle subgroup MATH such that MATH. Therefore MATH and, furthermore, the proper subgroup MATH of the circle group MATH is finite cyclic. |
math/9912180 | Since any maximal torus MATH in MATH is conjugate to the subgroup MATH of diagonal matrices MATH it suffices to show that the exact sequence MATH splits. But the splitting is immediate because of the homomorphism MATH mapping a diagonal matrix MATH to the constant multiple MATH. |
math/9912180 | Let MATH be a given representation. Since MATH is compact, we may assume that all the images of MATH are contained in MATH. Moreover, it is enough to prove the case that MATH is irreducible. Since MATH is connected, MATH is MATH-invariant so that the associated projective representation MATH exists by REF . From REF we can choose a circle subgroup MATH of MATH such that MATH and MATH is finite cyclic. We shall find a lifting homomorphism MATH of MATH over MATH. Since MATH is compact, connected, and abelian, it is a torus in MATH. Note that every maximal torus in MATH has the form MATH for some maximal torus MATH of MATH CITE. Choose a maximal torus MATH of MATH such that MATH. By REF the exact sequence MATH splits, that is, the canonical projection MATH has a continuous section (homomorphism) MATH such that the composition MATH is the identity map of MATH. Then MATH is a desired lifting homomorphism of MATH over MATH. MATH . Let MATH denote a generator of the finite cyclic group MATH. Since MATH on MATH, MATH for some constant MATH. Note that MATH is a MATH-th root of unity, where MATH is the order of MATH. So it is possible to choose a one-dimensional unitary representation MATH of the circle group MATH such that MATH. Then the unitary representation MATH satisfies REF . |
math/9912180 | Since MATH is compact, connected, and abelian, it is isomorphic to a torus. So we have a finite chain of subgroups MATH such that MATH is normal in MATH and MATH. Applying REF inductively, any representation of MATH is extendible to MATH. |
math/9912180 | The necessity is obvious, and the sufficiency follows from REF since the factor group MATH is compact, connected, and abelian, that is a torus. |
math/9912180 | It is immediate that MATH commutes with MATH, since the map MATH sending MATH is a homomorphism. To prove the sufficiency, it is enough to choose the trivial representation MATH of MATH, that is, MATH for all MATH. Since MATH commutes with MATH, the two REF are satisfied immediately. On the other hand, suppose MATH is a MATH-extension of MATH. Since MATH commutes with MATH, we have MATH for all MATH and MATH. Then the NAME 's lemma implies that MATH is constant for all MATH, so we may view the restriction MATH as a one-dimensional complex representation of MATH. Since semisimple NAME groups have no nontrivial abelian factor group, the trivial representation is the unique one-dimensional complex representation of MATH. Therefore, MATH is trivial on MATH, in particular, on MATH. |
math/9912180 | The sufficiency is obvious so we prove the necessity. If MATH is not a direct summand of MATH, then MATH in REF contains a nontrivial element, say MATH. Since a faithful representation of MATH always exists CITE, we can choose an irreducible sub-representation MATH of MATH such that MATH is not trivial. Then MATH does not extend to a representation of MATH by REF . |
math/9912180 | Since MATH, the restriction of the canonical projection MATH on MATH is surjective and its kernel MATH is discrete. It follows that MATH is a covering homomorphism of MATH. From the uniqueness of the universal covering homomorphism MATH, there exists a covering homomorphism MATH such that the diagram MATH commutes (compare with CITE). Since MATH and MATH is isomorphic to MATH, we have a surjective homomorphism of MATH onto MATH. |
math/9912180 | Since the factor group MATH is semisimple and connected, the theorem follows immediately from REF . |
math/9912180 | We claim that MATH, the torsion subgroup of MATH, is isomorphic to MATH. Denote by MATH the torus MATH. Then the homotopy exact sequence of the fibration MATH implies that MATH, since the second homotopy group of a compact NAME group vanishes, see CITE. Since MATH is semisimple, MATH is finite CITE so that it is isomorphic to MATH as we claimed. Therefore, the condition of MATH being torsion free is equivalent to MATH being simply connected. By REF , MATH for some semisimple connected closed normal subgroup MATH in MATH and there is a surjective homomorphism MATH. Therefore, if MATH is simply connected, then MATH so that MATH is a direct summand of MATH. |
math/9912181 | The first part follows from the considerations above and the fact that the map MATH is a bijection on the space of antisymplectic endomorphisms of MATH. MATH commutes with MATH since MATH. Also MATH. Conversely, if MATH, given MATH commuting with MATH and such that MATH, let MATH; then MATH . |
math/9912181 | MATH and MATH both have MATH as complexification. |
math/9912181 | This follows immediately from the list in REF . The only case where MATH is compact is when MATH and MATH. |
math/9912182 | Assume REF and let MATH be Hermitian and let MATH be an index of the approximate identity such that MATH and choose a positive linear functional MATH with MATH. Then MATH shows that MATH in the GNS representation corresponding to MATH proving REF. Assume REF. Then the orthogonal sum over all GNS representations MATH is faithful: it is clear that MATH for MATH if MATH is Hermitian or anti-Hermitian. Let MATH be not anti-Hermitian. Then MATH and thus MATH since MATH is Hermitian. Thus MATH is also non-zero proving REF. Finally, let MATH be a faithful MATH-representation. Thus it is sufficient to prove REF for (a MATH-subalgebra of) MATH for an arbitrary pre-Hilbert space MATH. Let MATH be such that for all MATH we have MATH. Then by the usual polarization argument and the fact that MATH in MATH we conclude MATH for all MATH. Hence, by the non-degeneracy of the Hermitian product, MATH follows. Thus for a non-zero Hermitian MATH there exists a vector MATH with MATH. Then MATH is the desired positive linear functional proving the equivalence of the first three properties. Now assume that they are fulfilled. Then REF follows immediately from the fact that one has a faithful MATH-representation. Now let MATH be Hermitian and MATH as above and MATH a positive linear functional with MATH. By the NAME inequality we have MATH whence MATH. By induction we conclude that MATH and thus MATH cannot be nilpotent. This proves REF for Hermitian elements. Together with REF, it also follows for normal elements. Finally, for REF pass to a faithful MATH-representation and take expectation values. |
math/9912182 | This follows since MATH can be written as complex linear combination of the Hermitian elements MATH and MATH. |
math/9912182 | The first implication is obvious so let us prove the second. Let MATH be a MATH-representation of MATH and consider MATH. Define MATH which is a MATH-submodule of MATH for all MATH and clearly MATH though the sum may not be direct. NAME if the sum decomposition of MATH were direct the identifications in the MATH-balanced tensor product could make the sum decomposition of MATH non-direct. Nevertheless they are orthogonal as one immediately can verify using PC. To show that MATH is positive semi-definite we may restrict to MATH for fixed MATH due to their orthogonality. Let MATH with MATH and MATH for MATH, then there is a MATH such that MATH and hence we find MATH such that MATH for MATH. Thus MATH with MATH. Hence by REF the positivity MATH easily follows proving P. |
math/9912182 | This is a simple consequence of P obtained by passing to the pre-Hilbert space MATH. |
math/9912182 | The first part is trivial. For the second part consider MATH, MATH with MATH. Then MATH where the matrices MATH are defined by their matrix elements MATH and MATH. Then MATH and MATH are Hermitian and positive since for MATH one clearly has MATH where MATH and analogously for MATH. Then MATH by REF . |
math/9912182 | It remains to check the last part. It is straightforward to verify that MATH satisfies PREF - PREF with the canonically induced direct sum and the corresponding tensor products of the pseudo-cyclic vectors as pseudo-cyclic vectors for the Cartesian product of the corresponding index sets. Using the pseudo-cyclicity as well as XREFa for each of the given bimodules one finally verifies XREFa for the new bimodule MATH by a lengthy but easy computation. |
math/9912182 | The verification of REF - XREF is trivial. Thus it remains to show P. First notice that any MATH-representation MATH of MATH on a pre-Hilbert space MATH is of the form MATH where MATH is a Hermitian projection, and also that any such projection yields a MATH-representation of MATH. Consider MATH where the tensor product is now constructed using MATH. Moreover, let MATH and MATH. Using that MATH, we then have MATH where MATH are defined by MATH and MATH. As in the proof of REF we notice that MATH as well as MATH are Hermitian and positive, whence MATH by REF . Thus P is shown. |
math/9912182 | Consider the MATH-bimodule MATH as defined in REF and define on this bimodule a MATH-valued inner product given by MATH. Then, just as in REF , one can show that REF - REF - REF hold, as well as YREF - YREF, YREFa, YREF, YREF and QREF - QREF. Finally, a simple computation shows that EREF also holds. |
math/9912182 | Let MATH be a positive functional. Fix MATH and consider the linear functional MATH on MATH, defined by MATH. It is clear that MATH (by REF) for all MATH and hence MATH is positive. So if MATH is positive, MATH for all MATH positive and the proof is complete. |
math/9912182 | Let MATH be the (MATH balanced) tensor product of MATH and MATH. It has a natural (MATH)-bimodule structure, and we denote it by MATH. Note that the fomula MATH uniquely defines a map MATH satisfying XREF,XREF, XREF and XREF. Similarly, MATH uniquely defines a map MATH satisfying YREF, YREF,YREF and YREF. Let's show that MATH satisfies XREF. Recall that since MATH satisfies PREF - PREF, any MATH can be written as MATH . But following REF - REF , we know that for each MATH, there exists a MATH such that MATH. So, we can find MATH such that MATH and hence MATH where MATH. Therefore, we have MATH . But since MATH for all MATH with respect to MATH, it follows that MATH and hence MATH by REF . Similarly, we can use that MATH satisfies QREF, QREF and QREF to show that MATH is positive semi-definite. So we conclude that MATH satisfies XREF - XREF and MATH satisfies YREF - YREF. We also observe that it follows from REF that the actions of MATH on MATH and MATH are strongly non-degenerate and an easy computation, like in the case of MATH-algebras, shows that this implies the fullness REF . A straightforward computation, also similar to the MATH-algebra setting, shows that the compatibility REF is also satisfied. So it only remains to check EREF to conclude the proof. Let MATH be a MATH-representation of MATH. Then, since MATH satisfies P, we can define a positive semi-definite Hermitian product on MATH by MATH . But now MATH is a MATH-module with a positive semi-definite Hermitian product and MATH acts on it by adjointable operators. So due to REF , we can define a positive semi-definite Hermitian product on MATH by setting MATH . Finally note that the last expression is just the definiton of the Hermitian product induced on MATH by the bimodule MATH. So MATH satisfies property P. Analogously, MATH also satisfies P, for we can identify MATH. |
math/9912182 | The proof basically follows CITE. Let MATH and MATH. Also define MATH and MATH. Note that there is a linear map MATH uniquely defined by MATH . Since MATH is strongly non-degenerate and MATH is full, it immediately follows that MATH is onto. A simple computation using the definitions shows that MATH preserves the Hermitian products, and therefore it is unitary. It is also easy to check that MATH intertwines MATH and MATH. Thus the conclusion follows. The same argument holds for MATH. |
math/9912182 | This is also a simple computation using the definitions, that can be carried out just like in the MATH-algebra setting (see CITE). |
math/9912182 | Note that for a general MATH, we have MATH . But since MATH has an approximate identity, we can find MATH such that MATH and MATH. So, for MATH we get MATH and by non-degeneracy of MATH it follows that MATH. The same argument can be applied to right MATH-modules. |
math/9912182 | Suppose MATH. Since MATH has an approximate identity, there exists MATH such that MATH and using that MATH is full, we can write MATH. So MATH since we are assuming that MATH. The same argument applies to MATH and MATH. |
math/9912182 | Suppose MATH and let MATH. Then note that MATH. Since MATH is arbitrary, it follows that MATH and hence REF implies that MATH for all MATH. So MATH. We can then reverse the argument and conclude that MATH. It is not hard to check that MATH still carries a natural left MATH-action and a right MATH-action (since MATH is both MATH and MATH invariant). Moreover, we can also define MATH- and MATH-valued inner products on MATH in the natural way and a simple computation shows that all the properties of an equivalence bimodule still hold. |
math/9912182 | By REF , we can assume that MATH satisfies MATH for all MATH implies that MATH. Let MATH be a positive linear functional in MATH and MATH. Note that the map MATH defines a positive linear functional in MATH. Let MATH. To show that MATH has sufficiently many positive linear functionals, it suffices to show that if MATH, then there exists MATH and MATH such that MATH. To see that, suppose that for all MATH and MATH positive linear functional in MATH, we have MATH. Then since MATH has sufficiently many positive linear functionals, it follows that MATH for all MATH. But then, by polarization, it follows that MATH for all MATH, and hence MATH for all MATH since MATH is torsion-free (see REF ). But then we must have MATH for all MATH and hence by REF we conclude that MATH. This finishes the proof. |
math/9912182 | Note that given a positive linear functional MATH, we can define a positive semi-definite Hermitian product on MATH by MATH. It then follows from REF that MATH . So, if MATH=REF it follows that MATH for all positive linear functional MATH. Hence, by REF , we have that MATH for all MATH. The conclusion is now immediate and the same argument can be used for MATH. |
math/9912182 | We know that MATH and that MATH is a MATH-homomorphism such that MATH is an adjoint of MATH. Note that MATH and hence MATH. But since MATH is full, it then follows that MATH. It is also easy to check that MATH is a MATH-homomorphism. Finally, injectivity of MATH follows from REF . |
math/9912182 | It is clear that MATH as defined in REF satisfies YREF. Note that MATH implies YREF and since MATH for all MATH, YREF also holds. By our hypothesis MATH and fullness is immediate from the definition of MATH. So EREF, EREF hold. Finally, the compatibility REF is also easy to be checked. |
math/9912182 | Just note that given any MATH such that MATH (and one can always find such a MATH), then we can write MATH. But since MATH, it follows that MATH. If MATH is a field, then the last claim in the proposition follows from the invertibility of MATH. |
math/9912182 | Just note that MATH. |
math/9912182 | By REF and the remark above, everything is shown except for EREF. But this follows from an easy computation. |
math/9912182 | Note that REF in the definiton of a set of equivalence data hold by the compatibility REF . So it remains to show that MATH and MATH are bimodule isomorphisms (it is clear that they are homomorphisms). Observe that since MATH and MATH are full, it follows that MATH and MATH are surjective. The conclusion then follows from REF . |
math/9912182 | Suppose MATH. Then MATH and hence we can consider MATH such that MATH. So it follows that MATH . But we also have that MATH . Hence, since MATH, we conclude that MATH. But since MATH is unital and MATH is full, it follows that MATH, or MATH. In other words, MATH. |
math/9912182 | By the previous proposition, MATH and MATH are MATH-isomorphic. So the algebras MATH and MATH are also isomorphic and hence MATH and MATH are diffeomorphic (see CITE). |
math/9912182 | If MATH and MATH are NAME equivalent, then as we discussed before, there exists a full idempotent MATH (not necessarilly self-adjoint) so that MATH. Then it follows from CITE that there exists a projection MATH (that is, MATH) such that MATH and it is then easy to check that MATH is full, for so is MATH. Moreover, it follows from CITE that MATH and hence MATH is isomorphic to MATH as a MATH-algebra. But since MATH has a natural involution inherited from MATH (as a MATH-subalgebra), we can define an induced MATH-involution on MATH, so that MATH and MATH are MATH-isomorphic. But now it follows from REF that MATH and MATH are formally NAME equivalent. |
math/9912182 | It is clear that MATH defines a directed filtered system of submodules of MATH. Now let MATH be given then we find a sequence MATH converging to MATH in the MATH-adic topology. But MATH can only converge to MATH if there exists a MATH such that for all MATH we have MATH. Since on the other hand MATH for some MATH we conclude MATH for MATH whence MATH is shown. It remains to show the defining properties of the MATH which is straightforward. |
math/9912182 | Let MATH satisfy MATH and MATH. Then MATH shows that MATH since MATH has non-negative MATH-adic order which proves the first part since the other inclusion is trivial. The other statements are straightforward. |
math/9912182 | Let MATH then MATH according to REF . Thus MATH and MATH is a well-defined MATH-linear map. The second part is an easy computation. |
math/9912182 | The well-definedness is shown analogously to the last lemma and the MATH-representation properties are a straightforward computation. |
math/9912182 | REF - REF , and XREF are an easy check. Thus let us consider PREF where we assume MATH. Then clearly MATH coincides with the whole space MATH and the sum is also orthogonal. But from the above remark we conclude that the sum is also direct and hence PREF is valid for the classical limit. Finally PREF is obvious and PREF follows by taking MATH as pseudo-cyclic vectors for MATH. |
math/9912182 | Let MATH then we have to prove MATH for all positive linear functionals MATH. Choose a positive MATH-linear functional MATH with MATH which exists since MATH is a positive deformation. Then MATH by REF implies MATH. |
math/9912182 | Using the present results the well-definedness of MATH is easily established. The rest is a simple computation. |
math/9912182 | We can proceed almost analogously as in the proof of REF . First we need the following analogue of REF : let MATH and assume MATH for all MATH and let MATH satisfy MATH. Due to the topological fullness of MATH we find MATH such that MATH with some element MATH. Then MATH shows that MATH cannot have a zeroth order and hence MATH. Now let MATH and MATH then MATH implies that MATH satisfies MATH whence MATH. Thus MATH follows. Reverting the argument finishes the proof. |
math/9912182 | It remains to show EREF for the classical limit which is a simple computation. |
math/9912182 | One inclusion is trivial. For the other we consider MATH, then we have for all positive MATH-linear functionals MATH the inequality MATH. Hence we obtain in the classical limit MATH where MATH is the classical limit of MATH. If MATH then MATH follows. Since MATH is a positive deformation any positive linear functional of MATH occurs as classical limit of some MATH and since MATH has sufficiently many positive linear functionals and an approximate identity it follows from REF that MATH. |
math/9912182 | If MATH for all MATH then consider MATH whence MATH for all MATH follows. If on the other hand MATH for all MATH then choose MATH with MATH for MATH. Then MATH. But since MATH has characteristic zero and MATH we conclude MATH. |
math/9912182 | The proof is obtained by observing that for finitely many elements MATH, written as fractions, we can find a common denominator which we can choose real and positive. |
math/9912182 | This is standard, see for example, CITE, where MATH follows from MATH. |
math/9912182 | If MATH is positive then clearly MATH for all MATH since the functional MATH is a positive linear functional. For the other direction we have to show MATH for all positive linear functionals MATH. Due to REF we have to show MATH for all density matrices MATH, and REF allows us to consider MATH instead of MATH. Then MATH proves the proposition. |
math/9912182 | Let MATH then MATH where MATH are given by their matrix elements MATH and MATH. Clearly MATH, MATH are positive matrices since MATH where MATH and MATH, and similar for MATH. Then MATH by REF . |
math/9912182 | Let MATH then MATH is non-negative whence for all MATH the function MATH is strictly positive. Thus the square root is still smooth and contained in MATH whence MATH. Thus MATH follows and with the NAME inequality MATH the proof is finished. |
math/9912182 | The case MATH was shown in CITE. Since any positive linear functional of MATH is given by a positive linear functional of MATH having compact support and since the construction in CITE does not increase the support, the corollary follows. |
math/9912183 | The free MATH-module functor MATH is a left adjoint, so preserves all colimits, and thus MATH (we need MATH to be cofibrant in order for MATH to be meaningful). If MATH is the NAME equivalence functor, then for any MATH, MATH is the homology of the chain complex MATH, so it suffices to show that for any cosimplicial chain complex MATH the map MATH is an isomorphism for all MATH. Now for a cosimplicial object MATH over any abelian category, we can define the NAME cochain complex by MATH (with differential MATH). We claim that the composite MATH is an isomorphism MATH. First note that MATH, since if MATH (which we may assume by induction on MATH) and MATH for MATH, then MATH, so MATH. Thus MATH is in fact of the form MATH - which implies that MATH (for MATH). This shows that MATH is one-to-one. Next, given MATH with MATH for MATH (which we again assume by induction on MATH), then there is a MATH with MATH such that MATH for MATH - namely, MATH. This shows that MATH is onto. Thus the Proposition will follow if we show MATH is an isomorphism: given MATH represented by MATH with MATH, (where MATH is the boundary map of the chain complex MATH), we assume that MATH for MATH, and that there are MATH such that MATH for MATH. Replacing MATH by MATH, we see by induction on MATH that we may choose a representative for MATH with MATH for all MATH. Thus MATH is surjective. Finally, if MATH for MATH, there is a MATH such that MATH, with MATH for MATH; setting MATH, we see that we can assume MATH, so MATH is one-to-one. |
math/9912183 | Start with MATH, and note that REF implies (by induction on MATH) that MATH for some MATH. Now because MATH has the ``underlying structure" of an abelian category, we may define a homomorphism of the underlying abelian groups MATH by MATH where MATH (and we let MATH). It then follows from the cosimplicial identities REF that MATH for MATH. Since the definition of MATH depends only on MATH, we see that MATH - and in fact even MATH - are epimorphisms. Moreover, given MATH such that MATH for MATH, the identities REF imply that MATH so by induction on MATH we see MATH - and thus MATH is one-to-one, so in fact it is an isomorphism of unstable coalgebras. We thus have MATH for some MATH, where MATH denotes the disjoint union, and we may assume MATH. Finally, for each MATH, set MATH, and define MATH inductively by MATH (MATH). We see that MATH for MATH, so MATH has MATH for all MATH, that is, MATH. Moreover, MATH, so if we set MATH, we get the required CW basis. |
math/9912183 | Assume that we want to replace MATH by a different lifting MATH, and choose maps MATH realizing MATH, MATH respectively; their respective extensions to MATH and MATH agree on MATH. We correspondingly have MATH and MATH in REF . Since MATH can be any fibrant GEM realizing MATH, we may assume it is a simplicial MATH-module, and thus MATH has a natural MATH-module structure. Set MATH. Then MATH induces a map MATH whose projection onto MATH is MATH. Moreover, because MATH and MATH agree with MATH when pulled back to MATH, we have MATH, and thus MATH factors through MATH, and this is a map of MATH-coalgebras because MATH is a coabelian MATH-comodule (actually, a MATH-comodule), and MATH is induced by group operations from the MATH-coalgebra maps MATH and MATH. See REF below. Moreover, in the abelian group structure on MATH we have MATH (see REF), so this is a coboundary, which proves independence of the choice of MATH. Now assume that there exists MATH REF with MATH. By the discussion in REF, we know that REF is a direct sum for MATH, and since MATH is cofree, we can choose MATH in REF to extend MATH by zero, so MATH in REF , and thus MATH. Conversely, if MATH, we can represent it by a coboundary MATH for some MATH-coalgebra map MATH, and thus get MATH, for MATH the projection. If we set MATH (we can subtract maps, because MATH is a graded MATH-module), we have MATH. We can therefore choose MATH realizing MATH, and then MATH, so that MATH so constructed yields MATH, as required. |
math/9912183 | First note that any simply-connected coalgebra over MATH is realizable by CITE and its Corollary, so in this case the theorem merely states that one always has a coherently vanishing sequence of characteristic classes. Given any connected space MATH, the cosimplicial space MATH defined by MATH (where MATH is the NAME monad MATH - compare CITE), is a cosimplicial resolution of MATH in the sense of REF - that is, MATH is a cosimplicial cofree resolution of MATH (see CITE). But then by REF , MATH has a coherently vanishing sequence of characteristic classes. Conversely, assume that MATH is a simply-connected unstable MATH-coalgebra with a coherently vanishing sequence of characteristic classes. By REF , any cosimplicial cofree resolution MATH may be realized by a cosimplicial space MATH. In particular, since MATH and MATH, we may assume that the same holds for each MATH, so that each MATH-GEM MATH is simply-connected. Because MATH is a resolution, MATH for MATH and MATH or MATH, and thus by CITE the homology spectral sequence for MATH converges strongly to MATH. Since the MATH-term of REF is concentrated along the MATH-line, we get MATH and this is an isomorphism of unstable coalgebras, since the edge homomorphism is induced by a topological map MATH. |
math/9912183 | If MATH, there is a map of MATH-comodules MATH such that MATH, and by the discussion in REF MATH can be lifted to a map MATH actually factoring through MATH. If we define a map of MATH-comodules: MATH, then MATH is just MATH in the following diagram: Note that because MATH is cofree, we can realize MATH by a map MATH in MATH, so MATH is realized by MATH. We may take the simplicial GEM MATH to be a simplicial MATH-module, with MATH the addition map, and define MATH to be the composite MATH. For every MATH we have MATH, so MATH induces a map MATH, with MATH and thus by REF the following diagram in MATH commutes: which yields a map of MATH-truncated objects MATH, or equivalently, a map MATH, which clearly induces an isomorphism in MATH for MATH. Note that for any choice of MATH we have MATH, by REF. Since MATH by REF , we have MATH, so by REF MATH, and since the following square commutes: and MATH is an isomorphism, so is MATH, by REF. Thus MATH is an isomorphism, too, so MATH is a weak equivalence by REF for MATH. |
math/9912183 | When MATH is of finite type, we can choose a cosimplicial resolution MATH, with a CW basis in which each MATH (and thus each MATH) is of finite type. Let MATH and MATH be cosimplicial spaces realizing MATH, which are resolutions (in the sense of REF) of MATH and MATH respectively, as in REF. By REF (respectively, REF ), we have MATH. Let MATH denote the diagonal of the bicosimplicial space MATH obtained from a given cosimplicial space MATH by applying the NAME MATH-resolution functor CITE dimensionwise to MATH. By REF there is a map of cosimplicial spaces MATH which is a weak equivalence in MATH, so induces an isomorphism in the MATH-terms of the homology spectral sequences for MATH and MATH. Since MATH and MATH are of finite type for each MATH, by CITE, MATH induces a homotopy equivalence MATH (and similarly MATH and MATH). However, for each MATH, the spaces MATH and MATH are MATH-GEMs, so they are MATH-complete (compare CITE), and thus MATH by CITE, and similarly for MATH, so we find that MATH and MATH are indeed MATH-equivalent. |
math/9912184 | CASE: First note that if MATH is a trivial cofibration in MATH, then MATH has a natural section MATH (with MATH) for any MATH: This is because by REF , MATH for MATH; since MATH is fibrant in MATH, we can choose a left inverse MATH for MATH, so MATH has a right inverse MATH, which is natural in MATH; so these maps MATH fit together to yield the required map MATH. This need not be true in general if MATH is not a weak equivalence, as the example of MATH shows. CASE: Given MATH represented by MATH with MATH (MATH), consider the diagram: in which MATH is a fibration by REF if MATH, so the lower left-hand square is in fact a homotopy pullback square (see CITE). By descending induction on MATH, (starting with MATH), we may assume MATH is nullhomotopic in MATH, as is MATH, so the induced pullback map MATH, is also nullhomotopic by the universal property. We conclude that MATH, and since MATH is a fibration by REF , we can choose MATH so that MATH. Thus MATH lifts to MATH, and MATH is surjective . CASE: Finally, the long exact sequence in homotopy for the fibration sequence MATH implies that MATH is monic, so MATH is, too. |
math/9912184 | To simplify the notation, we work here with topological spaces, rather than simplicial groups, changing back to MATH if necessary via the adjoint pairs of REF. Given a free simplicial MATH-algebra resolution MATH with CW basis MATH, where MATH for some MATH, and MATH is the free MATH-algebra generated by the graded set MATH, let MATH denote the cardinality of MATH, and set MATH. Define new ``spheres" and ``disks" of the form MATH and MATH. (This is to ensure that there will be at least MATH different representatives for each homotopy class in MATH or MATH.) By remark REF above, if we use the construction of REF in MATH (or in MATH, mutatis mutandis) with these ``spheres" and ``disks", and apply it to the space MATH, rather than to MATH, we obtain a resolution by spheres MATH. We define MATH by induction on the simplicial dimension; it suffices to produce for each MATH an embedding MATH commuting with MATH. If we denote MATH by MATH and set MATH, then we may assume by induction we have a monomorphism MATH (taking generators to generators, and commuting with face and degeneracy maps). For each MATH-algebra generator MATH in MATH, if MATH then MATH is represented by some MATH, and we can choose distinct (though perhaps homotopic) maps MATH for different generators MATH by our choice of MATH. Then by REF there is a wedge summand MATH in MATH (with no disks attached), and the corresponding free MATH-algebra coproduct summand MATH in MATH, generated by MATH, has MATH and MATH for MATH by REF, since MATH and thus MATH, and spheres indexed by nullhomotopic maps have disks attached to them. We see that MATH, so we may define MATH. If MATH, then all we need are enough distinct MATH-algebra generators in MATH: we cannot simply take MATH for nullhomotopic MATH, because of the attached disks; but we can proceed as follows: Since MATH and MATH, we have MATH distinct nonzero maps MATH with MATH. Define MATH, MATH; then MATH is, up to homotopy, a sphere wedge summand in MATH, and thus MATH is a MATH-algebra generator mapping to MATH under the augmentation. NAME, define MATH in MATH by MATH, MATH where MATH is a homoeomorphism onto the summand MATH in MATH. Then MATH and MATH but MATH; thus MATH is a MATH-algebra generator in MATH. By thus alternating the MATH and MATH we produce MATH distinct MATH-algebra generators in MATH for each MATH. |
math/9912184 | Start with MATH. For MATH, assume MATH. By REF , MATH (as in REF), so we can set MATH; but MATH need not vanish for MATH. However, given MATH, we may define MATH inductively, starting with MATH, by MATH (face and degeneracy maps taken in the external direction); we find that MATH is in MATH. If we define MATH by MATH, by the universal property of free MATH-algebras this extends to a map MATH, which together with the inclusion MATH yields a map MATH which is an isomorphism by the NAME Theorem (compare CITE). Thus we may set MATH, that is, the free MATH-algebra generated by MATH. Compare CITE. |
math/9912184 | By REF we may assume MATH has a (free) CW basis MATH, and that there is a resolution by spheres MATH (in MATH) and an embedding of simplicial MATH-algebras MATH. We may also assume that MATH is fibrant REF , with MATH a fibration. We shall actually realize MATH by a map of bisimplicial groups MATH. Note that once MATH has been defined through simplicial dimension MATH, for any MATH we have a commutative diagram (obtained by fitting together three of the long exact sequences of the fibrations REF). The vertical maps are induced by the inclusions MATH, and so on - see REF . The only difficulty in constructing MATH is that REF does not hold for MATH - that is, the maps MATH in the above diagram in general need not be isomorphisms - so we may have an element in MATH represented by MATH with MATH (but of course MATH). In this case we could not have MATH with MATH, so MATH would not be acyclic. It is in order to avoid this difficulty that we need the embedding MATH, since by definition this cannot happen for MATH: we know that MATH is surjective for each MATH, so MATH is, too, which implies that for each MATH: MATH which we shall call REF for MATH. (This implies in particular that MATH.) Note that given any fibrant MATH having REF for MATH for each MATH, if we consider the long exact sequence of the fibration MATH: MATH we may deduce that MATH for MATH . We now construct MATH by induction on the simplicial dimension : CASE: First, choose a fibration MATH realizing MATH. By REF, there is a map MATH realizing MATH, so MATH; since MATH is a fibration, we can change MATH to MATH with MATH . CASE: Let MATH denote the fiber of MATH. Since MATH is a surjection, we have MATH, and MATH maps MATH onto MATH, so MATH factors through MATH, and we can thus realize it by a map MATH. Set MATH (so MATH), with MATH equal to MATH, and change MATH to a fibration MATH. Again we can realize MATH by MATH with MATH, since MATH is a fibration; so we have defined MATH realizing MATH . CASE: Now assume we have MATH realizing MATH, with REF holding for MATH for MATH. For each MATH-algebra generator MATH (in degree MATH, say), REF implies that MATH, so by the exactness of REF we can choose MATH such that MATH. This allows us to define MATH so that MATH realizes MATH, as well as MATH realizing MATH. Because MATH is a free MATH-algebra, this implies the homotopy-commutativity of the outer rectangle in (as well as the lower square, by the induction hypothesis). Thus MATH, so MATH. By REF this implies MATH, so (since MATH is a free MATH-algebra) also MATH - which means that we can choose MATH so that MATH (since MATH is a fibration). Thus if we set MATH to be MATH, we have MATH. If MATH (in the notation of REF) we set MATH, and define MATH, and MATH respecyively by MATH and MATH. We see that MATH and MATH, and this will still hold if we change MATH into a fibration, and extend MATH to MATH. This defines MATH realizing MATH . CASE: It remains to verify that MATH so defined satisfies REF. However, REF implies that we have a map of short exact sequences: in which the left vertical map is an isomorphism and the right map is one-to-one, so MATH is one-to-one, too. Therefore, MATH, which implies that REF holds for MATH, too. This completes the inductive construction of MATH. |
math/9912184 | Choose free CW bases for MATH and MATH, and realize the resulting CW resolutions by MATH and MATH respectively, where (as in the proof of REF ) we may assume MATH is a fibration for each MATH. MATH will be defined by induction on MATH: MATH may be realized by a map MATH REF , and since MATH is a fibration and MATH, we can choose a realization MATH for MATH such that MATH. In general, MATH may be realized by a map MATH REF , and since MATH is a fibration, we may choose MATH so MATH. By induction this yields a map MATH such that MATH, so MATH is indeed a simplicial morphism (realizing MATH). |
math/9912184 | If MATH is a loop space, it has a strictly associative MATH-multiplication MATH which induces MATH on MATH (compare CITE), so MATH extends to a cosimplicial space MATH by REF . Applying the functorial construction of CITE to MATH yields a (strict) cosimplicial augmented simplicial space MATH, and since we assumed MATH embeds in MATH for each MATH, MATH may also be rectified . Conversely, if MATH is a (strict) cosimplicial simplicial space realizing MATH, then we may apply the realization functor for simplicial spaces in each cosimplicial dimension MATH to obtain MATH (by REF). The realization of the codegeneracy map MATH induces MATH, so it corresponds to a MATH-space multiplication MATH (see CITE). The fact that MATH is a (strict) cosimplicial space means that all composite codegeneracy maps MATH are equal, and thus all possible composite multiplications MATH (that is, all possible bracketings in REF ) are homotopic, with homotopies between the homotopies, and so on - in other words, the MATH-space MATH is a MATH space (see CITE) - so that MATH is homotopy equivalent to loop space by CITE. Note that we only required that the codegeneracies of MATH be rectified; after the fact this ensures that the full cosimplicial simplicial space is rectifiable. |
math/9912184 | Choose a MATH-resolution of MATH which is (MATH)-connected in each simplicial dimension, and let MATH be as in REF. By definition of the cross-term MATH-algebras MATH in REF, they must involve NAME products of elements from all lower order cross-terms; but since MATH is a MATH-space by assumption, all obstructions of the form MATH vanish (see REF). Thus, the lowest dimensional obstruction possible is a third-order operation MATH (MATH), which involves a triple NAME product and thus takes value in MATH for MATH. If we apply the (MATH)-Postnikov approximation functor to MATH in each dimension, to obtain MATH, all obstructions to rectification vanish, and from the spectral sequence of REF we see that the obvious map MATH induces an isomorphism in MATH for MATH. Since MATH is a loop space by REF , so is its (MATH)-Postnikov approximation, namely MATH. |
math/9912191 | Suppose MATH is not contained in MATH, otherwise we are done. Fix MATH and let MATH be written in a canonical form MATH that is, MATH and MATH (with the convention that MATH or MATH may be trivial.) Note that this expression is not unique in general, but its length MATH only depends on MATH. Suppose first that all elements of MATH have length REF. If MATH contains two elements MATH and MATH then MATH is a word written in a reduced normal form, and has length REF, hence it is non torsion by REF , p. CASE: This contradicts that MATH is torsion. Thus, all MATH must belong to the same factor, and we are done. We can suppose that the first factor MATH of MATH is non trivial and that MATH is odd and MATH. Choose MATH such that MATH is the minimum over all lengths of elements in MATH of length MATH. If MATH then we obtain a contradiction as before. Thus MATH must be trivial and similarly MATH. Therefore, MATH has a normal decomposition of the form MATH for some MATH which we rename by MATH to simplify the notation. Similarly, all MATH must have odd length. Moreover, if MATH then MATH has a normal form starting with MATH, the first factor of MATH and ending with MATH. Of course, if MATH=REF then MATH. Consider then MATH. If MATH has length MATH, then MATH has length MATH, and its normal form begins and ends with elements in MATH. On the other hand, if MATH has length REF, then MATH. But, arguing as before, we see that such elements of length REF must be actually in MATH. Repeating this process we obtain a MATH such that MATH where MATH or MATH depending on MATH, if MATH is odd or even, respectively. If any of the elements in MATH has length MATH, then we can conclude that MATH which is a contradiction. Thus all elements in MATH must have length REF, and the first case applies, hence MATH is contained in some factor. This finishes the proof. |
math/9912191 | The proof is similar to that of REF . |
math/9912191 | Let MATH be a torsion subgroup, and MATH such that MATH. If MATH we are done. Suppose then that MATH. Express MATH in a reduced normal form MATH that is, MATH and MATH, where MATH (MATH or MATH may be trivial). The relation MATH yields the following MATH . By the normal form theorem for free products with amalgamation CITE, this is only possible if MATH for all MATH. This concludes the proof. |
math/9912191 | Suppose that MATH with MATH and MATH such that MATH. By REF , we can suppose that MATH. Hence MATH implies MATH. In other words, MATH for all MATH. If MATH, then express MATH in reduced form and replace a first choice MATH by a different one if the first MATH-factor of MATH in REF is cancelled by MATH. Hence the normal form of the left hand in REF has non-trivial factors and the normal form theorem for free products CITE gives a contradiction, hence MATH. But MATH, hence MATH and thus MATH, as desired. Note that the lemma also follows from REF . |
math/9912191 | Let MATH and MATH. If MATH then MATH. Suppose that MATH. Note that MATH so we can consider the free product with amalgamation MATH. We shall show that MATH. Let MATH be a subgroup isomorphic to MATH and MATH such that MATH. By REF we can suppose that MATH or MATH. Suppose that MATH, the other case is easier. Let MATH be written in a reduced normal form. If MATH then MATH since MATH, and this is a contradiction. Hence MATH. As in REF we deduce that MATH, thus MATH since MATH is suitable REF . Hence MATH and we get a contradiction. If MATH, then we obtain MATH. So both MATH and MATH are in MATH, which is a contradiction. Similarly, if MATH we have that MATH and MATH are in MATH. This is again impossible. This concludes the proof. |
math/9912191 | Let MATH be a subgroup isomorphic to MATH and MATH such that MATH. By REF we can suppose that MATH already. Let MATH be written in a reduced form in MATH, where MATH and there is no subwords MATH with MATH or MATH with MATH (see CITE). If MATH then MATH, since MATH. This yields a contradiction. Thus MATH. We have MATH for every MATH. In other words, for a fixed MATH, the following holds MATH . By the normal form theorem for HNN extensions CITE, either MATH and MATH, or MATH and MATH. Suppose that MATH, the other case is analogous. Then MATH, thus MATH, and we can replace in REF the subword MATH by MATH. Repeating the same argument we obtain that MATH or MATH. Hence one of the two possibilities holds depending on MATH or MATH. So we have either MATH or MATH . In the first case we have an isomorphism MATH. Since MATH is complete there is MATH such that MATH. This yields MATH, that is, MATH and MATH are conjugate, and thus MATH and MATH are conjugate, but this is impossible by hypothesis. In the second case we have MATH by completeness. But, on the other hand MATH since MATH is written in a reduced form. We conclude that MATH is in MATH, as desired. |
math/9912191 | If MATH we take MATH. Suppose that MATH. If MATH and MATH are conjugate, we take MATH. Otherwise, we consider the HNN extension MATH where MATH is any isomorphism between MATH and MATH. By REF we know that MATH. Now, if MATH and MATH are conjugate in MATH, we take MATH. It follows automatically that MATH, since MATH is complete. If MATH and MATH are not conjugate in MATH, we consider a new HNN extension MATH where MATH is any isomorphism between MATH and MATH. Again MATH by REF . In that case we have an isomorphism MATH, which equals MATH for some MATH by completeness. Thus MATH. This shows that MATH, as desired. |
math/9912191 | Let MATH and view MATH and MATH as subgroups of MATH, respectively. Suppose that we have a subgroup MATH isomorphic to MATH and MATH such that MATH. By REF a conjugate of MATH is contained in MATH or MATH. Hence by hypothesis we can assume that MATH is already contained in MATH. Suppose that MATH. By REF it follows that either MATH or a conjugate of MATH is contained in MATH. In the first case MATH since MATH, so contradiction, and the second case is obviously impossible. Thus MATH as desired. |
math/9912191 | Suppose that MATH and that MATH. Let MATH and MATH be two isomorphic copies of MATH. Choose a non trivial element MATH and let MATH and MATH be its copies in MATH and MATH respectively. Now define MATH where MATH is the normal subgroup generated by MATH. Then MATH by REF and moreover MATH. If MATH we first embed MATH where MATH and MATH is the normal closure of MATH. Then, by the previous lemma MATH. Now, since MATH, we can apply the first case. |
math/9912191 | Let MATH be a subgroup isomorphic to MATH and let MATH be such that MATH. Since MATH and MATH is regular we have MATH for some index MATH. We can also suppose that MATH. Now MATH, hence MATH. |
math/9912191 | By REF we may assume that MATH is infinite, hence the last cardinal condition becomes simply MATH. Applying REF MATH-times and REF once we can find an equipotent group MATH, such that MATH and MATH, inductively we find MATH such that MATH for each natural MATH. The union MATH of this chain belongs to MATH by REF and obviously satisfies MATH. |
math/9912191 | Let MATH. We have to construct a group MATH in MATH such that MATH with MATH. This can be obtained easily by using alternatively REF a countable number of times. The limit group will still lie in MATH by REF . Moreover it is in MATH by construction. |
math/9912191 | Let MATH be a subgroup isomorphic to MATH, and MATH such that MATH. As in previous results we can assume that MATH and that MATH, written in a reduced form, has length bigger than two. Then we have MATH both of them inside MATH. Since MATH there exists a MATH such that MATH. We can also suppose that the automorphism group MATH is already in MATH, by REF . Hence, the composition MATH is an automorphism, which is inner by completeness. Thus, MATH and thus MATH. This is a contradiction, since MATH was written in a reduced form. |
math/9912191 | Let MATH be the isomorphism mapping MATH to MATH. By REF. As in REF consider the HNN extension MATH. We must show that MATH. Clearly MATH and consider any MATH with MATH and any MATH with MATH. By the argument in REF we may assume that MATH and MATH with MATH. Now we apply REF as we did in REF . |
math/9912191 | Since MATH satisfies all conditions for groups in MATH except that MATH, we know that MATH and MATH lies in some subgroup which is in MATH hence MATH is conjugation by some element in MATH. Hence we can suppose without loss of generality that MATH. Now let MATH be another copy of MATH. As before there is an element MATH such that MATH. The subgroup MATH and MATH generate a subgroup MATH which is generated by MATH elements, and it follows from properties of MATH that we may also assume MATH. By hypothesis on MATH there is an element MATH such that MATH. Therefore, for all MATH and MATH, we will have MATH . In particular, if MATH, then MATH for all MATH. Since MATH it follows that MATH. Thus MATH for all MATH. By the same reason MATH. We have shown that MATH for all MATH. Since MATH, we conclude MATH. |
math/9912191 | Since elements of MATH and MATH have length REF in MATH, we have that MATH and MATH embed in MATH, by REF . It also follows that MATH and MATH are malnormal subgroups of MATH; see CITE or CITE. Hence MATH satisfies REF . Let us prove that MATH is in MATH: Let MATH be a subgroup isomorphic to MATH and MATH such that MATH. We now apply a version of the Torsion Theorem, see CITE for a fixed element MATH (of finite order) in MATH. Hence, there exists an element MATH such that the conjugate MATH belongs to MATH or MATH; we may assume MATH. From MATH follows MATH, hence MATH and MATH . NAME of MATH in MATH implies that MATH also belongs to MATH. By REF, hence MATH and thus MATH. Finally we show that REF holds. Let MATH be a given monomorphism which extends to a monomorphism MATH with domain MATH. Suppose that MATH is conjugation by some element MATH. We shall assume for simplicity that MATH (a similar argument will prove the general case). Since MATH extends MATH we have MATH for some MATH. In other words, we have the following relation in MATH . Note that MATH has length REF in MATH and that MATH cannot cancel with MATH in MATH. Hence, if we view MATH as a new element, then the new word is written in a weakly cyclically reduced form in MATH. REF tells us that this word must contain MATH of some word of MATH. This forces MATH, that is, MATH. But this contradicts the assumption made on MATH. Hence MATH as desired. |
math/9912191 | The proof is the same as that of REF , but we must indicate how to assign ordinals conveniently to the constructions described in REF. In particular we next show that MATH is not empty. As MATH, we identify the automorphism group MATH of MATH with the first MATH ordinals in MATH; the identity element MATH goes to MATH. Similarly, for every MATH we consider an isomorphic copy MATH of MATH, and we identify MATH with the first MATH elements in MATH. Let MATH. Hence MATH satisfies REF . If MATH then MATH has a unique reduced normal form MATH, where each MATH belongs to some factor MATH and successive factors are different. Let MATH be the maximum of the norms MATH for MATH. Now we assign to MATH any ordinal in MATH. There is enough room in this segment, since MATH is uncountable and MATH, we simply choose any free place. Hence MATH is a MATH-group. This yields a new version of REF . The assignments for REF are clear by continuity and note that we can do this because the length of the chains is strictly smaller than MATH. Suppose that MATH and MATH have been already embedded in MATH, and let MATH for some common subgroup MATH. If MATH consider all possible reduced words MATH with MATH in one of the factors of MATH. Assign MATH to MATH such that MATH is the minimum of all norms of such words. The same idea works for the rest of results in REF. In particular, for HNN extensions we assign the new element MATH which conjugates MATH and an isomorphic copy MATH to a new ordinal MATH such that MATH. |
math/9912191 | The proof is straightforward. |
math/9912191 | For a fixed MATH with MATH we have exactly MATH different subsets in MATH of size MATH, and MATH by GCH. Now the number of group structures supported by some set MATH is MATH. By GCH again and MATH we have MATH. We conclude MATH. |
math/9912191 | We have to check the conditions listed in REF . Let MATH such that MATH. Then there are groups MATH coded by MATH and MATH respectively, such that MATH is a subgroup of MATH. This yields MATH, or equivalently MATH, as desired. CASE: Let MATH such that MATH. Let MATH, MATH and MATH groups in MATH coded by MATH, MATH and MATH, respectively, such that MATH, MATH. Consider MATH the subgroup of MATH generated by MATH and MATH. It is clear that MATH and moreover MATH. Using the fact that MATH is already in MATH, we can add to MATH the elements of MATH, with norm in the domain of MATH, in order to obtain a new group MATH such that MATH. Finally, define MATH. The proofs of REF to REF are straightforward. REF. (Indiscernibility) Suppose that MATH and MATH is an order-isomorphism in MATH. Let MATH be a group coded by MATH. Using MATH we can define a strong isomorphism MATH as follows: if MATH for some MATH and MATH, define MATH. Clearly, the image MATH is a group in MATH which has code MATH. Hence MATH. The fact that MATH implies MATH is also clear. REF. (Amalgamation property) Let MATH and MATH coded by MATH and MATH respectively. Let MATH coded by MATH, such that MATH and MATH. Consider MATH the free product of MATH and MATH amalgamating the common subgroup MATH. Then MATH by REF . Now by REF there is a group MATH such that MATH. Take MATH. |
math/9912191 | Using REF , the desired group is the free group MATH modulo the small cancellation MATH. More precisely MATH is the quotient of MATH by the normal subgroup MATH generated by the symmetrized closure MATH of MATH. Hence REF hold. |
math/9912191 | It is clear that MATH is closed upwards in MATH. Now we have to show that MATH is dense in MATH. In the case when MATH then MATH and density is obvious. Suppose that MATH and let MATH be any element in MATH. Let MATH be a MATH-group which is coded by MATH. As in the proof of REF we can find a group MATH, such that MATH, which contains a subgroup MATH isomorphic to MATH and MATH. In particular, we have MATH. Moreover, the group MATH can be constructed in such a way that MATH. Therefore, MATH and MATH as desired. |
math/9912191 | It is clear that every MATH is closed upwards. The fact that MATH is dense in MATH is shown in REF , where we can find a small cancellation group MATH with the required properties. If necessary, use also REF to obtain a group MATH in MATH such that MATH. |
math/9912191 | Let MATH represent the group MATH and let MATH be such that MATH. If MATH, then MATH and density is obvious. Suppose that MATH. We have to construct a MATH-group MATH represented by MATH extending MATH such that MATH. We consider the three possible cases: CASE: MATH. In this case we use REF as follows. Define MATH, where MATH is a new element in MATH with MATH. Of course, MATH, so we can construct MATH with MATH and MATH, as explained in the proof of REF . Hence MATH and MATH are conjugate in MATH. CASE: MATH and MATH. Here, we use REF (REF of its proof). We can find an element MATH of infinite order such that MATH has infinite order as well. From REF we note that MATH are conjugate in MATH. Hence MATH. CASE: If MATH has finite order, we argue similarly. Take a conjugate MATH of MATH in some MATH such that MATH has infinite order. Hence MATH contains all elements of infinite order and REF apply. We conclude in any case that MATH is a density system. |
math/9912191 | We apply REF and let player I choose the density systems above. REF follows immediately from REF follows from REF follows from REF . |
math/9912192 | Consider MATH. We have to check that MATH commutes with MATH and MATH. We shall consider MATH (the case of MATH is similar, but simpler). Denote MATH. It is sufficient to give proof for MATH, then the general case will follow. Consider the diagram MATH . Take MATH. Apply MATH. We get MATH, where MATH. Here MATH, MATH, MATH, MATH. Apply MATH. We obtain MATH where in the last expression the argument of MATH and MATH is MATH and we denote MATH, MATH. Now let us apply first MATH, then MATH. Calculate: MATH applying MATH we obtain MATH where the argument of MATH and MATH in the last expression is MATH. Multiplying through, we obtain exactly the same terms as in REF with the opposite sign. Notice that MATH as a form is skew-symmetric in even columns. Thus we can swap MATH and MATH, cancelling the minus sign, and obtain MATH as desired. Stability of MATH is proved in the same way, and we omit the calculation. Let us turn to the relation with the isomorphisms REF. Consider the following diagram. The claim is that it is commutative. To check this, take MATH. We have: MATH now, MATH as desired. (Here MATH stands for MATH without the row MATH.) In a similar way the equality MATH is checked. |
math/9912192 | To find relations between MATH and MATH, for MATH, it is sufficient to consider the case MATH. (The general case is formally reduced to it by considering dual forms on extended space MATH and by setting MATH.) Then for MATH we have MATH where MATH. Notice that the range of MATH in the first line of REF contains MATH. Simultaneously interchanging MATH and MATH and the indices MATH and MATH, we obtain MATH where MATH. Now we see that all terms except for the last one in MATH would cancel the similar terms in MATH. Notice that MATH. It follows that MATH which equals zero by REF . Consider now MATH and MATH. For MATH we readily have MATH . Similarly, for MATH we obtain MATH again by REF . Finally, let us find the relation between operators MATH and MATH. Notice that MATH. For MATH by a direct calculation similar to REF using REF , we obtain the equality MATH . Apply now the transformation MATH. That means setting MATH, MATH, MATH, MATH. We arrive at MATH from where REF follows. Notice that by this calculation we showed that the operator in the right-hand side of REF gives another expression for the isomorphism MATH. |
math/9912192 | Since MATH is a differential graded algebra, generated by elements MATH over MATH (locally), it is sufficient to check REF for two cases: MATH and MATH, where MATH is a function. The first case was considered above. Consider MATH. Then, by definition, MATH . Apply MATH. We obtain MATH as desired. |
math/9912192 | Let MATH be in MATH. Consider MATH. Recall that the action of this operator consists in setting MATH, MATH, MATH, MATH in the argument. We shall find MATH and MATH. Directly from REF: MATH now, MATH . Comparing with REF, we immediately conclude that MATH . Applying MATH to both sides of REF, we obtain the desired identity REF. (Notice that MATH and MATH commute.) |