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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 MA... |
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 thu... |
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 subgro... |
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... |
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 ... |
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 ... |
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 (com... |
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 isomor... |
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... |
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-balanc... |
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 ... |
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 produc... |
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 th... |
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 Hermi... |
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 ... |
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,... |
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 ... |
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 f... |
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 f... |
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 co... |
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 PRE... |
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 h... |
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 ... |
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 MAT... |
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 ch... |
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 RE... |
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-modu... |
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 m... |
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 i... |
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 )... |
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... |
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/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, ... |
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 ... |
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 ... |
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 ... |
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 t... |
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 elem... |
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 ... |
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... |
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 b... |
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 ... |
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... |
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 tha... |
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 w... |
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 ob... |
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 gr... |
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 MAT... |
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 MAT... |
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.... |
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/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... |
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, whe... |
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/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. S... |
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 ident... |
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