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hep-th/0104183 | If two cocycle representations MATH are given, then we have MATH, where MATH . By REF , the line bundles MATH and MATH are isomorphic. The bundle isomorphism MATH is defined by MATH where MATH and MATH is the function defined in the proof of REF . By REF we have MATH. This implies the formula in the proposition. |
hep-th/0104183 | Note that the CS action and the transition functions of the line bundle are expressed by summations of local terms. So REF are verified in the proof of REF . |
hep-th/0104183 | We should verify that the CS action of MATH satisfies the differential equation of the parallel transport determined by the connection MATH. For this purpose we construct a path in MATH by endowing MATH with a specific triangulation that is induced from the triangulation of MATH. Then explicit calculation of the action with respect to this triangulation shows the following MATH . This will complete the proof. |
hep-th/0104183 | First we take a map MATH. We can make a path of connections on MATH by setting MATH, where MATH is a MATH-valued REF-form on MATH and MATH if MATH. This is also a connection on the bundle over MATH, so we take MATH correspondingly. Using REF we have MATH and MATH is independent of MATH. We denote the curvature of MATH on the bundle over MATH by MATH. The curvature of MATH on the bundle over MATH is MATH. Finally we have MATH . Hence the connection MATH induced by MATH is a stationary point of MATH if and only if MATH. |
hep-th/0104216 | Clearly REF , together with the interchange sign factors implies REF after contraction with MATH. After substituting REF and rationalising, the identity is therefore quartic, and generically of NAME type. The only bilinear cases occur when MATH (in which case the condition MATH refers to the symmetry of MATH), or for MATH (for which MATH and MATH anticommute, and the charge conjugation of MATH determines whether the pseudoscalar identity holds). |
hep-th/0104216 | REF follows directly by contraction of (REF a) with MATH. REF uses REF together with anticommutativity of MATH and MATH (See (REF b)). |
math-ph/0104001 | In the case MATH, REF gives the following: MATH where MATH is the ``charge" variable. Since MATH and MATH, our principal specialization MATH is written in terms of MATH as follows: MATH . Thus, we obtain MATH . In order to compute the coefficient of MATH, we use the NAME triple product identity: MATH and also the following well-known identity (see REF): MATH . Replacing MATH by MATH in REF , MATH by MATH and MATH by MATH in REF , we rewrite the right side of REF by MATH and thus we have MATH . For convenience, we introduce the following functions: MATH . To prove this lemma, it is sufficient to prove the following two equations: CASE: MATH. CASE: MATH. First we shall prove REF . Since MATH, we have MATH . Hence we obtain MATH . We then have MATH . A straightforward computation replacing MATH by MATH yields MATH . By REF , we obtain MATH . It is known (see for example, REF) that the right side of REF has the following product expansion: MATH and hence REF follows. Next we shall prove REF . Since MATH, we have MATH . Hence we obtain MATH . Replacing MATH by MATH and MATH by MATH, we have MATH . Replacing MATH by MATH in REF and comparing it with REF , we get REF . |
math-ph/0104001 | By REF , we get the recurrence formula MATH . We now set for MATH, MATH , MATH and MATH . For the proof of this proposition, it is sufficient to show that MATH . We shall show REF by induction on MATH. Using these notation, REF is rewritten as MATH . Assume that REF is true for MATH. Then we obtain MATH proving the proposition. |
math-ph/0104001 | To prove REF , we let MATH in REF . Then, since MATH and MATH, we get MATH . Since MATH, we have MATH and hence MATH which is REF . To prove REF , we let MATH for MATH. First we consider in the case MATH, since MATH and by REF , it follows that MATH and hence we obtain MATH . In the case MATH, the same discussion as above yields MATH and hence we obtain MATH which is REF . |
math-ph/0104001 | This formula is shown just by the same argument as that in the proof of REF, by making use of a basis of MATH: MATH satisfying MATH and MATH where MATH. Since MATH and REF , we have MATH for MATH and MATH. Hence we have MATH proving the proposition. |
math-ph/0104003 | By REF defines a phase function on MATH. We have MATH and MATH is continuous on the subspace of restrictable distributions. Therefore we may use any standard OI regularization MATH and obtain MATH. Since the latter is an OI regularization in one dimension with phase function MATH and symbol MATH the assertion is proved. |
math-ph/0104003 | For the justification of REF we use the construction of the distributional product in REF via the pullback of the tensor product on MATH under the map MATH which embeds MATH as the diagonal into MATH. In doing so the original OIs may be approximated by smooth regularizations (for example, amplitude cut-offs in the integrands) the tensor products thereof being pulled back simply as smooth functions (meaning restriction to MATH in this case). It is easily seen then that the smooth functions obtained thereby converge weakly (as OI regularizations) to the OI given in REF . By continuity of the pullback (under the given wave front set conditions) this limit equals the pullback of the tensor product of the corresponding limits and therefore, in turn, is the distributional product MATH. Note that MATH is smooth in MATH (due to the cut-off MATH) and homogeneous of degree MATH outside the set MATH and is therefore a symbol of order MATH. The function MATH is smooth on MATH and homogeneous of degree MATH in MATH. If MATH then the gradient MATH for all MATH and hence MATH is a phase function. In case MATH the gradient vanishes exactly along one half-ray component of the set MATH (for example, along MATH if MATH). Although it is no longer a phase function in the strict sense, the distribution MATH is then defined as the sum of a classical integral, an OI, and a NAME transform of a MATH-function. We discuss this for the case MATH in detail, the other cases are completely analogous. Let MATH be a smooth function that is equal to MATH near MATH, has support in MATH, and satisfies MATH. Let MATH be smooth with compact support and MATH when MATH. We can split the integral defining MATH into three terms according to MATH. The first integral, then, is a classical one defining a smooth function, the second is an OI since the gradient of MATH does not vanish on the support of the integrand. In the third integral, we have MATH (insert MATH and use the fact that MATH and MATH on the support of the integrand) and hence the last term is equal to MATH which we interpret via the NAME transform of the MATH-function MATH on MATH as MATH in the sense of locally integrable functions - hence it is distribution on MATH. The weak continuity with respect to MATH follows from the smooth dependence of the phase function in the OI representation (compare CITE, before REF ) and the continuity of the NAME transform on MATH. |
math-ph/0104003 | Let MATH then MATH and from REF we obtain MATH where in the last line we have made use of the symmetry properties of MATH. Changing coordinates in the inner integrals to MATH and again by the symmetry of MATH this reads MATH . Finally, since MATH and MATH for some polynomial in MATH, we may interchange the order of integration and arrive at MATH . Since MATH was arbitrary and the above upper bound for MATH shows that the inner integrand is in MATH with respect to MATH, the proposition is proved. |
math-ph/0104008 | The proof consists of a simple calculation. Consider MATH . |
math-ph/0104017 | The product first line REF is known to be associative. To assert that MATH-Char-MATH we need to check multiplicativity: now, for MATH: MATH where we used the multiplicativity of MATH due to the commutativity of MATH: indeed, one has, for MATH: MATH . Check that MATH: we have by the first line REF : MATH . Check of REF we have: MATH . The facts that product and inverse are continuous in the MATH-topology is clear. (ia) For the proof we write MATH . Char-MATH and MATH. Check of MATH: for MATH, MATH, we have: MATH . Check of MATH: MATH implies: MATH hence we have MATH with MATH. The NAME algebra property of MATH can be checked directly from REF , but will immediately result from (iia), which we now check. (iia) For the proof we write MATH-Char-MATH and MATH. Check of MATH: MATH implies for MATH: MATH . Check of MATH: for MATH, MATH one has, MATH hence we have MATH with MATH. Check of REF : it is clear that MATH is a NAME of the NAME algebra Lie-MATH. |
math-ph/0104017 | CASE: One has MATH, for MATH, MATH. For MATH one has MATH. CASE: We first notice the property MATH transposed of the known property MATH. Check of MATH: for MATH one has MATH, indeed one has: MATH . It follows that MATH. Now MATH turns the unit MATH into MATH. And, for MATH and MATH, one has using the known property MATH: MATH: MATH . REF reads MATH, transposed of the axiom MATH. Check of the cocommutativity REF : holds obviously on the generators MATH and MATH. Now REF propagates multiplicatively: if MATH fulfill MATH, MATH, MATH, owing to multiplicatively of MATH. |
math-ph/0104017 | MATH is a unital algebra with unit MATH by definition. Using the terminology of REF, we have that the coalgebra axioms MATH, MATH for MATH result by transposition from the algebra axioms MATH, respectively, MATH for MATH. NAME axioms for MATH: we already met MATH in REF , MATH is the known fact that MATH, we have met MATH and MATH in Definition-REF above. Finally the axiom MATH for MATH and MATH are transposed of each other. |
math-ph/0104017 | REF : REF: it suffices to take MATH and MATH, then MATH, then: MATH . MATH: for MATH and MATH, MATH thus MATH. MATH : MATH: it suffices to take MATH and MATH, then MATH . MATH: for MATH and MATH, MATH thus MATH. MATH : MATH,whilst MATH . CASE: MATH : MATH: it suffices to take MATH, then MATH . MATH: for MATH, MATH implying MATH. MATH : MATH: it suffices to take MATH, then: MATH . MATH: for MATH, MATH implying MATH as above. MATH : MATH, whilst MATH . CASE: Check of the first line REF : we check that MATH is a convolution inverse of MATH: MATH where we used REF , MATH, and MATH. Check of the second line REF : equate the derivations at REF of MATH. |
math-ph/0104017 | From MATH homogeneous, we conclude for the product MATH with coproduct MATH that MATH equals MATH . It is instructive to check instead REF , showing that MATH . |
math-ph/0104017 | We check that one has MATH for MATH. With MATH the decomposition of the unit associated to the direct sum MATH, we have MATH, thus for MATH and MATH, MATH hence MATH. The fact that REF are transposed of each other then implies that MATH is a derivation of the algebra MATH. The proof of MATH for MATH is analogous. Check of the two last claims: - for MATH ``MATH": MATH . - for MATH ``MATH" and MATH: MATH . |
math-ph/0104017 | CASE: Let MATH, MATH: by REF we have MATH . Hence: MATH implying MATH. We proved that MATH . Ker-MATH. (In fact we found that: MATH first subscript indicating the MATH-grade). Conversely let MATH . Ker-MATH, with MATH, MATH. By what precedes we have MATH, hence MATH, proving that MATH. CASE: For MATH, REF implies REF . However REF is linear with respect to MATH, hence holds for MATH by REF . |
math-ph/0104017 | REF follows from REF and the multiplicativity of MATH. REF assume that MATH with all MATH, MATH strictly positive, and MATH with all MATH, MATH strictly positive. It follows that MATH where all the tensor products but the two first have both factors of strictly positive degree. |
math-ph/0104017 | The requirements, for the MATH-linear MATH, respectively, MATH: MATH: MATH respectively, MATH determine MATH and MATH by induction with respect to the degree MATH as respective right- and left convolution inverses of MATH for the convolution product of End-MATH: but these then coincide because MATH: thus MATH is a bilateral convolution inverse of MATH, thus the (unique) antipode. The second consequence of progressiveness of MATH is that, given MATH, there is a drastic limitation (depending on MATH) of non-vanishing values MATH, MATH. We recall that (compare REF): - the algebra\ïc dual MATH of MATH becomes a topological algebra if equipped with the MATH-topology and the algebra-structure (MATH, MATH) stemming from the coalgebra-structure of MATH, - the subalgebra MATH generated in MATH by the unit MATH and the NAME MATH of infinitesimal characters becomes by transposition a NAME algebra MATH, MATH, MATH. |
math-ph/0104017 | CASE: Since by REF a general element of MATH is of the form MATH with MATH, and we have obviously MATH, it is no restriction for the proof of the implication REF to assume that MATH belongs to MATH projection MATH applied to MATH then suppresses all the products MATH containing at least one factor in MATH: the remaining products MATH stem from the successive NAME summations MATH in REF , thus have by progressiveness all their factors MATH such that MATH (observe, as noticed in footnote REF, that all MATH belong to MATH). The equality MATH requires MATH, otherwise MATH vanishes, whence REF for MATH, the extension to MATH being trivial. CASE: The expression of the n-fold convolution product of MATH: MATH yields for products of elements of MATH vanishing on MATH: REF thus follows from REF . Note that REF holds for MATH (obvious directly - embodying this into our proof would require the convention MATH). |
math-ph/0104017 | REF is the known device for proving metrizability of a topological vector-space whose topology is specified by a countable set of semi-norms. One checks all the properties stated in footnote REF. REF follows from the known fact that the closure of a topological algebra is an algebra. CASE: Check of MATH: let MATH, MATH with MATH, we have, for MATH: MATH hence MATH, hence MATH. |
math-ph/0104017 | REF we have from REF , with MATH: MATH whence our claim since MATH for MATH. (ia) the proof is the same up to notation. CASE: Particular case of what precedes. Now the exponential of an infinitesimal character, if it makes sense, is a character. |
math-ph/0104017 | Equality MATH: Since MATH, compare REF , the inclusion MATH is obvious, the inclusion MATH following from the fact that the algebra MATH is generated in polynomial grade REF. The definition property MATH entails for MATH that MATH, and that MATH for MATH . Ker-MATH. Conversely for MATH as in REF one has MATH-Char-MATH indeed: - MATH whilst MATH, - for MATH: MATH whilst MATH, - for MATH: MATH hence MATH whilst MATH since MATH and MATH both vanish. |
math-ph/0104017 | CASE: Before proving REF for general MATH, we look for orientation at the cases MATH, and REF. REF amount to the second, respectively, third line REF . CASE: We first prove MATH . In order to simultaneously prepare the next steps we compute MATH for general MATH: computation indicated by the table (with the terms of MATH) in the first column, the terms of MATH in the first line, and the products at the intersection of the other lines and columns. Table MATH : Computation of MATH yields REF tensor products: - the MATH indicated by belong to MATH, thus vanish under MATH, - the MATH indicated by belong to MATH, thus also vanish under MATH, - the remaining REF: MATH and MATH, yield REF upon making MATH, MATH. We next prove MATH . Since MATH, MATH, is of the form MATH with MATH, we compute the relevant part of MATH whereby the -terms of MATH can be discarded since multiplications by MATH leave MATH invariant: we thus have the table: Table MATH: Computation of MATH yielding REF tensor products, all in MATH or MATH, thus vanishing under MATH. CASE: We first prove MATH . We need to classify the tensor products in REF: They consist of: - the MATH indicated by belong to MATH, thus vanish under MATH, - the MATH indicated by which belong to MATH, turned by MATH into MATH whose first factor vanishes under MATH, - the remaining REF: MATH, MATH in the set MATH of cyclic permutations of MATH, whose sum is turned by MATH into MATH, by REF equal modulo MATH to MATH in other terms MATH: we proved REF . We next prove MATH . Since MATH, MATH, is of the form MATH with MATH, we compute the relevant part of MATH whereby the -terms of MATH can be discarded since multiplications by MATH leave MATH invariant: we thus have the table: Table MATH : Computation of MATH yielding REF tensor products, all in MATH or MATH, thus vanishing under MATH, because MATH vanishes under MATH by REF . A more detailed analysis of this table registrates, amongst its tensor products: - the MATH indicated by which belong to MATH turned by MATH into MATH vanishing under MATH, - the MATH indicated by which belong to MATH turned by MATH into MATH vanishing under MATH because MATH vanishes under MATH: indeed MATH acting on the tensor products in MATH contained in MATH, and MATH vanishes under MATH by REF . - the remaining REF: MATH, MATH in the set MATH of cyclic permutations of MATH. We now reexamine REF . We have, MATH denoting equality up to negligible terms: MATH hence MATH, hence MATH . It should now be clear how things propagate recursively to yield a proof for general MATH. We assume that REF is as follows: its MATH tensor products consist of: - the MATH indicated by vanishing under MATH - the MATH indicated by vanishing under MATH - the remaining MATH with sum MATH, MATH in the set MATH of cyclic permutations of MATH, with REF holding for MATH. From what precedes it is clear that these features propagate from MATH to MATH. In particular we have: MATH and MATH . |
math-ph/0104017 | Each MATH is of the form MATH with MATH, MATH, MATH. We assume that MATH vanishes under all elements MATH and show that MATH: indeed, let MATH, MATH: then: - since by REF all the terms of h but the first lie in MATH-ker-MATH one has MATH: the requirement MATH thus entails the vanishing of the first term MATH. - next by REF one has MATH: asking this to vanish for all MATH thus entails the vanishing of the second term MATH. - assume that we have shown that MATH vanishes for MATH we have, by REF : MATH whose vanishing under all MATH entails MATH: we thus proved REF inductively. |
math-ph/0104017 | One has for MATH: MATH . |
math-ph/0104017 | CASE: Let MATH. First MATH indeed, for MATH of degree MATH: first MATH since MATH. Then: MATH . |
math-ph/0104017 | Performing a licit exchange of limits in: MATH we first show that we have MATH . Indeed, the horizontal limit comes from REF with MATH; the right vertical limit is REF . As for the left vertical limit, we have that MATH applied to: MATH yields MATH . Next the function MATH is holomorphic on the whole NAME sphere, thus must be constant: we have MATH, MATH, that is : MATH . Feeding REF into this yields: MATH equating coefficients of MATH then yields REF . Passage from REF to REF : MATH a is injective in restriction to the augmentation ideal MATH. Thus MATH thus also Res-MATH determines MATH, the key to the explicit dependence being REF , itself obtained as follows: integrating from REF to MATH, MATH yields MATH, it suffices now to inverse MATH in restriction to MATH. Applying REF to MATH yields REF for MATH. Inductive iteration then yields REF : assume it holds for MATH: applying MATH on both sides yields: MATH up to numbering of variables identical to REF . |
math-ph/0104017 | We need only check the additional cases of the NAME requirement. Now, for MATH, we have MATH and MATH by the fact that the algebra-derivation MATH of MATH is a NAME algebra-derivation of the NAME algebra Lie-MATH: for MATH we have MATH and MATH thus MATH. |
math-ph/0104017 | We apply the expansional formula MATH to MATH, MATH, MATH, yielding: MATH . We thus have MATH whence REF MATH . |
math-ph/0104017 | Progressiveness of MATH implies that REF specifies inductively a MATH-linear map MATH: MATH. Check of REF : obvious from REF and MATH. Check of REF : we have, by REF , for MATH, omitting parentheses: MATH whence REF owing to MATH. What is not obvious from REF is that MATH is multiplicative. We prove this by induction: for MATH, MATH we have, still omitting parentheses: MATH hence, by the induction hypothesis: MATH . On the other hand we have, using REF and the commutativity of MATH: MATH the two expressions containing the same terms. |
math-ph/0104017 | REF respectively, REF see below REF respectively, B. REF. CASE: MATH: by MATH, MATH and MATH: MATH . MATH: for MATH: MATH moreover, by MATH, MATH, MATH: MATH . MATH: by MATH for MATH: MATH hence MATH. |
math-ph/0104017 | REF one has MATH, indeed, for MATH: MATH . The other claims follow from known facts about split short exact sequences. |
math-ph/0104017 | We have MATH because MATH, MATH, MATH. Check of REF , we have: MATH whilst MATH . Check of REF , we have: MATH . Check of REF , MATH and MATH . |
math-ph/0104017 | CASE: Let MATH, MATH, then MATH. Furthermore: MATH . CASE: MATH entails MATH . CASE: MATH entails MATH . CASE: MATH entails MATH . Assume that MATH is a NAME algebra with antipode MATH. Check of REF : MATH entails MATH . Check of REF : MATH entails MATH whilst MATH entails MATH . |
math-ph/0104020 | Letting MATH, we have MATH . |
math-ph/0104020 | This is true since MATH . |
math-ph/0104020 | Let MATH and MATH be given. Since the probability of any MATH being the module MATH is MATH, the NAME of Large Numbers furnishes a MATH such that for all MATH, MATH where MATH is the number of the first MATH's that are modules in MATH. Now let MATH be given such that MATH. Then, there are at least MATH pairwise disjoint modules MATH in MATH, and hence MATH. The conclusion now follows. |
math-ph/0104020 | Without loss of generality, suppose that of the MATH bonds already specified, an odd number are negative MATH bonds. We can also without loss of generality assume that MATH is even. Then, the number of ways to get an odd number of negative MATH bonds is MATH . The number of ways to get an even number of negative MATH bonds is MATH . However, it is a well known fact that these two are equal. |
math-ph/0104020 | Suppose MATH is a ground state for MATH with any MATH such that MATH, and suppose that MATH is not a module in MATH. Then, if MATH is nonempty, it follows that MATH is not entropic relative to MATH. Define MATH. By REF we know that any site MATH can be involved in at most one unhappy bond in MATH; for if MATH were involved in more, then the set MATH would be entropic relative to MATH. Now consider the plaquette MATH. Since MATH is frustrated, we know that there is at least one bond in MATH that is also in MATH. We thus have four cases: CASE: MATH . By repeated application of REF to the plaquettes MATH, MATH, MATH, MATH, MATH, MATH, MATH, in that order, we deduce that the following bonds are in MATH: MATH, MATH, MATH, MATH, MATH, MATH, MATH, as seen in Picture REF. We now observe that the set MATH is entropic relative to MATH which is a contradiction. Therefore, REF is not possible. Following the structure of the reasoning in REF , we similarly find that none of REF MATH, REF MATH, or REF MATH is possible. But this contradicts the fact that at least one of these cases must hold. Therefore, MATH is a module in MATH. |
math-ph/0104020 | Let MATH be a ground state for MATH with any MATH such that MATH, and suppose that MATH is not a module in MATH. Define MATH. Then, any site in MATH can be involved in at most two unhappy bonds. Since plaquette MATH is frustrated, it has at least one unhappy bond. By symmetry, we can assume this bond to be MATH. We then see that plaquette MATH needs one more unhappy bond since it is unfrustrated. By symmetry, we can take this bond to be MATH. Next, since MATH is unfrustrated and MATH is frustrated, we see that the bonds MATH and MATH must be unhappy. We then see that plaquette MATH is unfrustrated. Since it already has one unhappy bond, it must have one more. We therefore have an entropic set consisting of either site MATH or MATH, which is a contradiction. Thus, there is indeed a non-trivial entropic set in MATH relative to MATH, and hence MATH is a module in MATH. |
math-ph/0104020 | Let MATH be a ground state for MATH with any MATH such that MATH, and suppose that MATH is not a module in MATH. Letting MATH, we first note that if MATH is any subset of MATH, then MATH since MATH is a ground state. Also, since each site MATH is involved in at most three bonds, it follows that at most one of the bonds containing MATH is in MATH. Now consider plaquette MATH. Since MATH is frustrated, at least one of the bonds in MATH must be in MATH. First consider the case that MATH. We now observe that if MATH, then MATH is entropic relative to MATH, while if MATH, then MATH is entropic relative to MATH. In view of this, and the fact that MATH is unfrustrated, we conclude that MATH. Using reasoning similar to that just used, together with the fact that MATH is frustrated, we see that either MATH or MATH. If MATH, then MATH is entropic relative to MATH. Hence, MATH. From this and the fact that MATH is frustrated, we either have MATH or MATH. If MATH, then MATH is entropic relative to MATH. Thus, MATH. Consequently, since MATH is frustrated, either MATH or MATH. If MATH, then defining MATH shows that MATH, which is a contradiction. Thus, MATH. However, we now observe that MATH is entropic relative to MATH, and so we conclude that MATH, which is a contradiction. Therefore, MATH, and hence, by symmetry, none of the bonds in MATH is in MATH, contradicting the fact that MATH is frustrated. Therefore, MATH is a module in MATH. |
math-ph/0104028 | Note that REF is equivalent to MATH . In turn, REF can be written as MATH . So the theorem is proven. |
math-ph/0104028 | Observe that MATH iff MATH and MATH. Then, by REF we get that MATH solve REF iff MATH is given by REF where MATH and REF is satisfied. |
math-ph/0104028 | Let MATH satisfies MATH and MATH . Assume that a pair MATH solve REF . Using MATH we can rewrite REF in the form MATH . A direct calculation yields MATH . Then MATH . Due to REF we come to the conclusion that MATH . Moreover from REF follows MATH and MATH . Conversely, assume MATH and MATH . Besides, let MATH and REF holds. Then we have MATH or equivalently MATH . In turn REF can be written as MATH . Putting MATH we come to REF . |
math-ph/0104028 | From REF we immediately get that MATH . Clearly, this expression is finite iff MATH . |
math-ph/0104028 | Note that MATH can be equivalently written as MATH . Then we have MATH-which is finite iff MATH . |
math-ph/0104028 | Let-MATH be invertible. First we shall show that MATH is invertible also and its inverse has the form REF . Let MATH and MATH . Then MATH that is, MATH . Then we get MATH . Since MATH we obtain MATH . Now let MATH. Then MATH . So, we have MATH . From REF follows MATH that is, MATH . Finally after inserting REF to REF we get MATH . This means that operator MATH is given by REF . Now we shall show that MATH . For this aim let us note that by a self-adjointness of MATH we have MATH, where MATH denotes orthogonal completion in MATH topology. Further let us note that MATH . In turn, since operator MATH has property MATH we get MATH . Using REF we can conclude that MATH is dense in MATH . Hence we get the density of MATH and MATH in MATH . Operator MATH belongs to MATH . Then by REF we get MATH . |
math-ph/0104028 | Let MATH have property N̂. Assume that MATH where MATH . Since MATH (see REF ) there exists MATH supported by the compact set MATH so that MATH . Clearly, MATH . Let MATH satisfies MATH . Expressing REF in the spherical system of coordinates one can show that MATH behaves like MATH as MATH . Thus MATH and MATH . Relying on the results of CITE we get that for all MATH the limits MATH exist in the uniform operator norm. Besides MATH . It known that MATH can be represented as the integral operator with the kernel given by MATH . NAME MATH can be written as MATH . Similarly as before expressing MATH in the spherical system of coordinates we get the following asymptotics MATH for MATH . Then the fact that MATH implies MATH . Thus MATH as well as MATH and REF is satisfied. |
math-ph/0104029 | If there are MATH and MATH such that REF - REF are satisfied then MATH and MATH defined by REF will satisfy NAME equation. It follows from our assumptions that MATH so the kernels of the NAME equation are bounded. They remain bounded after the division by MATH which turns REF into a NAME equation of the second kind. Since this equation has at most one solution in MATH we conclude that MATH. Then MATH from REF . |
math-ph/0104029 | Set MATH and MATH. Since MATH and both functions MATH are continuous they have the same sign in some neighbourhood of MATH. So the conditions of REF are satisfied in this case. Let MATH be the solution of REF then MATH, MATH and MATH. Therefore, MATH defined by REF lies in MATH and satisfies REF . Define MATH by REF then REF says that MATH and REF is satisfied. Finally, REF together with REF are equivalent to REF which means that REF is satisfied also. Thus, MATH and MATH solve the inverse problem on MATH. |
math-ph/0104029 | The argument is essentially the same as above with one difference: by REF MATH is now positive on the whole interval MATH and not only in a small neighborhood of MATH so that MATH. |
math-ph/0104029 | Recall that MATH is a solution to the parabolic problem with the initial data MATH. Under REF it is known that MATH (CITE, p. REF). Then for MATH the function MATH is in MATH. Similarly, MATH for a fixed MATH and, therefore, MATH. Thus, all of the conditions of REF are fulfiled. |
math-ph/0104029 | By REF it suffices to prove that MATH . The latter follows from the usual maximum principle for parabolic equations (CITE, p. REF). For the same reason MATH. Assume that MATH for some MATH. Since MATH one obtains MATH for some MATH. Then, by the strong maximum principle (CITE, p. REF), MATH in condradiction with our assumption. Thus, MATH on MATH as we claimed. |
math-ph/0104029 | It suffices to note that MATH and use REF to verify the necessary smoothness. The rest of argument is the same as in REF . |
math-ph/0104030 | This result is established in CITE so we only indicate the basic points of the proof. The assumption of REF implies the injectivity of MATH if MATH then the function MATH solves the NAME problem for the NAME operator in MATH and in MATH and satisfies the radiation condition at infinity. Thus MATH in MATH by the assumption of REF , and MATH in MATH by REF. Therefore MATH, where we have used the jump relation for the normal derivative of the single-layer potentials. The operator MATH is of NAME: it can be written as MATH where MATH is an isomorphism and MATH is compact as an operator from MATH into MATH. The injectivity of MATH together with its NAME property imply the conclusion of REF . |
math-ph/0104030 | The kernel of the operator MATH, defined by REF , and its first derivatives are continuous functions of MATH and MATH running through bounded sets, including the diagonal MATH. By REF the action of MATH is equivalent (up to the terms preserving smoothness) to taking the first order derivatives. Therefore the conclusion of REF follows. |
math-ph/0104030 | Let MATH be an arbitrary element of MATH. Denote MATH. Since MATH is a basis of MATH, one has MATH . If MATH then, applying the continuous operator MATH, one gets MATH, so MATH for all MATH since MATH is a basis of MATH. We have proved that MATH is a basis of MATH. Let us prove that if MATH is a NAME basis of MATH then MATH is a NAME basis of MATH, that is, there exists an isomorphism MATH of MATH onto MATH such that MATH, MATH . Let MATH be an orthonormal basis of MATH. Define a linear operator MATH by the formula: MATH in particular, MATH. Let us prove that MATH is an isomorphism of MATH onto MATH. If this is proved, then MATH, and REF is proved. Clearly MATH is linear, is defined on all of MATH, and is continuous. Only the continuity of MATH needs a proof. Let MATH in MATH. Then MATH, MATH. Thus: MATH where we have used the assumption that MATH a NAME basis of MATH. Thus MATH is a linear continuous, defined on all of MATH operator. Therefore MATH is bounded. Let us check that MATH is injective: if MATH, MATH, and MATH then MATH. Apply MATH and get MATH. Thus, MATH, since MATH is a basis. The injectivity of MATH is proved. To complete the proof one has to check that the range of MATH is the whole space MATH. Let us do this. Take an arbitrary MATH and define MATH. Let MATH, then MATH . Therefore MATH is an isomorphism of MATH onto MATH, and MATH is a NAME basis of MATH, as claimed. REF is proved. |
math-ph/0104040 | CASE: Is a direct consequence of the corresponding property for the bracket on MATH. CASE: Here we use that, for any MATH, and MATH and the statement follows from a straightforward computation. |
math-ph/0104040 | Just observe that the condition above kills the last term in REF . |
math-ph/0104040 | By NAME property, we only need to check that MATH when the MATH are of the form MATH (in fact, we will see that this expression vanishes). Now, we can distinguish three different cases. MATH case: there are at least two functions among the MATH (MATH). Rearrange the factors to get MATH and then compute, having in mind the previous remark, MATH case: there is exactly one function among the MATH (MATH). This time, rearrange to MATH case: there is no function among the MATH (MATH). Here, all we have are exact MATH-forms, and then MATH . |
math-ph/0104040 | REF is known from REF (see REF ). For REF to be understood, we only need to check it. First, note that if MATH, then clearly MATH. Next, consider any differential operator MATH and MATH; we have MATH thus, if we take MATH it results MATH, and so MATH . It is the first term on the last member what destroys skew-symmetry, but writing MATH we see that MATH thus, we have from REF , MATH . |
math-ph/0104040 | If MATH, we have MATH . |
math-ph/0104040 | As a consequence of MATH, we have that for all MATH, MATH, so MATH is MATH-linear and -by the localization lemma- it defines a MATH such that MATH . Next, we study what happens when MATH acts on MATH-forms. Just because MATH and MATH are so, MATH is a tensorial operator, and then MATH is MATH linear in all its arguments. Then, again by the localization lemma, MATH such that MATH . Now, acting on any MATH-form, with MATH, any of the previous operators gives MATH (note the degrees); thus, we have MATH . |
math-ph/0104040 | We will follow the ideas presented in CITE, where the authors consider the MATH case. If MATH, where MATH, we can write MATH, with MATH, and then, as MATH commutes with MATH, MATH so MATH . For the converse, consider an arbitrary MATH, neither necessarily verifying MATH, nor being tensorial. Let us determine its non tensorial part by taking brackets with a function MATH. We have MATH, so MATH such that MATH. Now, the mapping MATH is a derivation: if MATH, by the NAME property for the bracket MATH, MATH . Now, to each derivation from MATH to MATH, we can assign a MATH in the following manner. Let MATH and note that MATH as a consequence of MATH being a derivation. Also, we have MATH, so MATH is MATH- linear and there exists a MATH such that MATH . Let us study under what conditions this MATH is skew-symmetric. It suffices to consider only the case when MATH, and to observe that MATH . Now, from the proof of REF , we know that this is skew-symmetric if and only if MATH. So, under this condition we can take MATH . In the last step, we check that, for any MATH, MATH is a tensorial operator. As MATH and a differential operator is characterized by its action on forms of degree less or equal to its order, we only need to consider the case of a MATH-form. So, for MATH, we compute MATH . Thus, MATH is a tensorial operator, and applying the previous lemma, MATH such that MATH and then MATH, with MATH and MATH. If MATH, then must be MATH, so MATH and MATH. |
math-ph/0104040 | If MATH for some MATH, it is clear that MATH. For the converse, we have in the proof of the preceding proposition MATH, with MATH and MATH, so MATH, MATH and MATH. |
math/0104003 | First suppose that MATH. As indicated earlier in this section, each length MATH word MATH on the symbols MATH defines a permutation MATH. From this construction, it is easy to see that the recording tableau of MATH under the BRKV variation of the RSK algorithm is equal to the recording tableau of MATH under the RSK algorithm. Thus it is enough to prove that the probability that the word MATH has BRKV recording tableau MATH is MATH. This is immediate from REF . Now the case MATH can be handled by introducing MATH extra symbols between MATH and MATH - call them MATH and choosing each with probability MATH. Thus the random word is on MATH and these extra symbols. Each word defines exactly one permutation - the symbols MATH are treated as positive. By the previous paragraph, the probabiilty of obtaining recording tableau MATH is equal to MATH where the associated MATH are defined by MATH . As MATH, this distribution on permutations converges to that of a MATH shuffle, and the generating function of the MATH converges to MATH . |
math/0104003 | Let MATH be the MATH matrix MATH (MATH). Let MATH be the MATH matrix MATH (MATH). We evaluate in two ways MATH. One on hand, MATH . Clearly this is a NAME determinant. The symbol is MATH . On the other hand, the NAME formula gives that MATH where MATH (respectively, MATH) is the MATH matrix formed by using the columns (respectively, rows) indexed by MATH. Writing MATH associates the subsets MATH with partitions with at most MATH parts. Then MATH as desired. |
math/0104003 | The coefficient of MATH on the left hand side is one over the number of permutations on MATH symbols which commute with a permutation with MATH cycles of length MATH. This is MATH which agrees with the coefficient of MATH on the right hand side. |
math/0104003 | Let MATH in both sides of the statement of REF . Writing MATH and applying the strong NAME theorem gives that the determinant converges to MATH . However we know that MATH because this equation is simply the NAME cycle index REF with MATH replaced by MATH. |
math/0104003 | We associate to a realization of the BR REF point process a random permutation MATH as follows. First take the deck size to be the number of points which has distribution NAME(MATH). Rank the MATH coordinates of the points in increasing order, where one breaks ties for negative MATH coordiantes by defining the point with the larger MATH coordinate to be smaller and breaks ties for positive MATH coordinates by defining the point with larger MATH coordinate to be larger. Then MATH is defined as the rank of the MATH coordinate of the point with the MATH-th smallest MATH coordinate (with probability one there is no repetition among MATH coordinates). For example, if the BR point process yields the points MATH then the resulting permutation would be (in REF-line form) MATH . It is easy to see that this distribution on permutations is the same as that arising from a MATH shuffle. The second part follows from the first part and REF . |
math/0104003 | By REF , the sought probability is MATH . Writing MATH (page REF) and MATH the result now follows by an argument as in REF . |
math/0104003 | This follows easily from REF together with the fact that if MATH goes to the pair MATH under the RSK correspondence, the the descent set of MATH is equal to the descent set of MATH and the descent set of MATH is equal to the descent set of MATH REF . |
math/0104003 | For each MATH, let MATH be a random MATH matrix formed by letting each entry equal MATH with probability MATH, MATH with probability MATH, and MATH with probability MATH. Let MATH be the first time that all rows of MATH containing no zeros are distinct; from the inverse description of MATH shuffles this is a strong uniform time in the sense of REFB-REFD of NAME MATH, since if all cards are cut in piles of size one the permutation resulting after riffling them together is random. The separation distance after MATH applications of a MATH shuffle is upper bounded by the probability that MATH CITE. Let MATH be the event that rows MATH and MATH of MATH are the same and contain no zeros. The probability that MATH occurs is MATH. The result follows because MATH . |
math/0104003 | Given the results of REF, the proof of the first part runs along exactly the same lines as in the proof of REF. The second assertion follows from the observation that a MATH shuffle followed by reversing the order of the cards is conjugate (by the longest length element in the symmetric group) to a MATH shuffle. Alternatively, arguing as in the proof of REF, one sees that the effect of reversing the cards on the cycle index of a MATH shuffle is to get MATH . |
math/0104004 | MATH : If all MATH are centered, that is, MATH, and alternating, that is, MATH, then the assertion follows directly by the definition of freeness and by the relation MATH because at least one factor of MATH for MATH is of the form MATH and thus the assertion follows by induction. The essential part of the proof consists in showing that on the level of cumulants the assumption `centered' is not needed and `alternating' can be weakened to `mixed'. Let us start with getting rid of the assumption `centered'. For this we will need the following lemma - which is of course a special case of our theorem. CASE: Let MATH und MATH. Then we have: MATH . To simplify notation we consider the case MATH, that is, we want to show MATH . We will prove this by induction on MATH. MATH : the assertion is true, since MATH . MATH: Assume we have proved the assertion for all MATH. Then we have MATH . According to our induction hypothesis only such MATH contribute to the above sum which have the property that MATH is a one-element block of MATH, that is, which have the form MATH . Then we have MATH hence MATH . Since MATH we obtain MATH . Let MATH. Then this lemma implies that we have for arbitrary MATH the relation MATH that is, we can center the arguments of our cumulants MATH REF without changing the value of the cumulants. Thus we have proved the following statement: Consider MATH and MATH REF with MATH. Then we have MATH . It remains to weaken the assumption `alternating' to `mixed'. For this we will need the following lemma. CASE: Consider MATH, MATH and MATH. Then we have MATH . Examples: MATH . For MATH we denote by MATH that partition which is obtained by identifying MATH and MATH, that is, for MATH we have MATH (If MATH and MATH belong to different blocks, then MATH has one block less than MATH; if MATH and MATH belong to the same block, then the number of blocks does not change; of course, we identify partitions of the set MATH with partitions from MATH; the property `non-crossing' is preserved under the transition from MATH to MATH.) Example: Consider MATH . Then we have MATH and MATH . With the help of this definition we can state our assertion more generally for MATH for arbitrary MATH: Assume that our assertion is true for all MATH, that is, MATH . Then it is quite easy to see that we have for arbitrary MATH with MATH: MATH . We will now prove the assertion of our lemma by induction MATH. MATH: The assertion is true because MATH . MATH: Let the assertion be proven for all MATH, which implies, as indicated above, that we have also for all MATH with MATH . Then we have MATH . By using this lemma we can now prove our theorem in full generality: Consider MATH and MATH (MATH). Assume that there exist MATH with MATH. We have to show MATH . This follows so: If MATH, then the assertion is already proved. Thus we can assume that there exists a MATH with MATH, implying MATH. In that case we can use the above lemma to obtain MATH . To show that this vanishes we will again use induction on MATH: The first term MATH vanishes by induction hypothesis, since two of its arguments are lying in the different algebras MATH and MATH. Consider now the summand MATH for MATH with MATH . Then we have MATH, and by induction hypothesis this can be different from zero only in the case where all arguments in each of the two factors are coming from the same algebra; but this would impy that in the first factor all arguments are in MATH and in the second factor all arguments are in MATH. Because of MATH this would imply MATH, yielding a contradiction with MATH. Thus all terms of the right hand side have to vanish and we obtain MATH . MATH : REF gives an inductive way to calculate uniquely all mixed moments; according to what we have proved above this mixed moments must calculate in the same way as for free subalgebras; but this means of course that these subalgebras are free. |
math/0104004 | We have MATH because cumulants which have both MATH and MATH as arguments vanish by REF . |
math/0104004 | We rewrite the sum MATH in the way that we fix the first block MATH of MATH (that is, that block which contains the element REF) and sum over all possibilities for the other blocks; in the end we sum over MATH: MATH . If MATH then MATH can only connect elements lying between some MATH and MATH, that is, MATH such that we have for all MATH: there exists a MATH with MATH. There we put MATH . Hence such a MATH decomposes as MATH where MATH . For such MATH we have MATH and thus we obtain MATH . This yields the implication MATH . We can now rewrite REF in terms of the corresponding formal power series in the following way (where we put MATH): MATH . This yields REF . Since REF describes uniquely a fixed relation between the numbers MATH and the numbers MATH, this has to be the relation REF . |
math/0104004 | We just have to note that the formal power series MATH and MATH from REF and MATH, MATH, and MATH are related by: MATH and MATH . This gives MATH thus MATH and hence also MATH . |
math/0104004 | Each MATH can be decomposed as MATH . In such a case we have of course MATH . Thus we obtain (omitting the arguments) MATH . |
math/0104004 | By bilinearity, it is enough to prove the assertion for processes MATH . Then the left hand side is MATH . Note that by linearity it suffices to consider the cases where the two time intervals are either the same or disjoint. In the first case we have MATH (because MATH is free from the increment MATH), whereas in the second case one of the increments is free from the rest and thus, because of the vanishing mean of the increment, the expression vanishes. But his gives exactly the assertion. |
math/0104004 | Let us just give a sketch of the proof. We restrict here to biprocesses of the form MATH. The extension to sums of such biprocesses follows the same ideas. Put MATH . We want to obtain an operator norm estimate for MATH by using MATH . This means we must estimate the MATH-th moment of our integral for MATH. This is much harder than the case MATH, but nevertheless it can be done by using again the crucial property MATH if MATH is free from the increment MATH. By using also NAME inequality for non-commutative MATH-spaces one can finally derive an inequality of the form MATH . Note that the structure of this inequality resembles the recursion formula for the NAME numbers MATH, MATH . By induction, one derives now from the above implicit inequality the explicit one MATH . We take now the MATH-th root and note that MATH . Thus we obtain MATH . MATH gives the assertion. |
math/0104004 | Let MATH be an interval and consider decompositions into disjoint subintervals MATH, MATH. For an interval MATH we denote by MATH the corresponding increment of the free Brownian motion, that is, MATH . The main point is now to show that (with MATH denoting NAME measure) MATH where we take the usual limit with width MATH of our decomposition going to zero. As said above we want to see that this convergence even holds in operator norm. We will sketch two proofs of this fact, one using the abstract properties of freeness, whereas the other works in a concrete representation on full NAME space. CASE: The assertion follows from the following two facts about freeness: CASE: let MATH be a semicircular family, that is, each MATH is semicircular and MATH are free; then, for a random variable MATH which is free from MATH we have CITE: MATH are free CASE: let MATH be free random variables with MATH for all MATH; then we have CITE MATH REF We realize MATH on the full NAME space as MATH; then we have to estimate the four terms MATH, MATH, MATH, and MATH. NAME of these terms tend to zero by simple norm estimates, the only problematic case is MATH . (This corresponds of course to the NAME formulas for MATH and MATH, namely the only non-zero term is MATH, see CITE.) To prove this later statement, one can model MATH by the sum of creation and annihilation operators on the full NAME space as MATH for MATH with MATH being orthogonal to MATH, hence MATH free from all MATH. Then one has to expand this representation of MATH and bring it, by using the NAME relations MATH, into a normal ordered form MATH. Finally, note that, again by the NAME relations, only the term for MATH contributes to the sum of our statement. |
math/0104004 | One has to check the statement for polynomials by using the product form of the NAME formula. All expressions in the statement make also sense for nice functions and the statement extends by continuity. For a polynomial MATH we have MATH . The first term gives directly the first term in our assertion (this is just a non-commutative first derivative), whereas the second yields MATH which can be identified with the second term in our statement. One should note: the fact that the trace and not the identity acts on the expression between two differentials results finally in the unusual form of the second order term in the functional form of the NAME formula; it is not a non-commutative version of the second derivative, but a mixture of derivative and difference expression. |
math/0104007 | REF : Is clear from MATH in MATH as MATH and the convolution formula. CASE: Although this involves only marginal changes in the proof of REF , we recall it here to make the presentation more self-contained. Let MATH with MATH REF then with MATH where the expression within brackets on the right-hand side is bounded by some constant MATH, dependent on MATH and MATH only but independent of MATH, as soon as MATH with MATH chosen appropriately (and dependent on MATH, MATH, MATH, and MATH). Therefore the assertion is proved by putting MATH and MATH. REF : is proved by similar reasoning . |
math/0104007 | The necessary changes in the proof of REF , are minimal once we established the following If MATH and MATH with MATH then MATH. Using the short-hand notation MATH and MATH we have MATH . Hence we need to estimate terms of the form MATH when MATH. Let MATH be a closed set satisfying MATH and put MATH. Since MATH is a temperate distribution there is MATH and MATH such that MATH . MATH implies that each term in the sum on the right-hand side can be estimated for arbitrary MATH by MATH if MATH varies in MATH. Since MATH we obtain MATH with a constant MATH depending on MATH, MATH, MATH, MATH, and MATH but MATH still arbitrary. Choosing MATH, for example, we conclude that MATH has a uniform MATH-growth over all orders of derivatives. Hence it is a MATH-regular NAME function. Referring to the proof (and the notation) of REF , we may now finish the proof of the theorem simply by carrying out the following slight changes in the two steps of that proof. Ad REF : Choose MATH such that MATH in a neighborhood of MATH and write MATH . The first term on the right can be estimated by the same methods as in CITE and the second term is MATH-regular by the lemma above. Ad REF : Rewrite MATH and observe that the reasoning of CITE is applicable since MATH by the above lemma. |
math/0104007 | Let MATH then REF implies MATH and integration with respect to MATH from MATH to MATH yields MATH . |
math/0104007 | Let MATH then MATH which proves the first assertion. The second assertion follows from MATH with the short-hand notation MATH. |
math/0104007 | We use the characterizations in REF and the remarks on p. REF following those; choosing a smooth compactly supported wavelet MATH of order MATH we may therefore state the following: MATH belongs to MATH if and only if there is MATH such that MATH . Now the proof is straightforward. First let MATH. If MATH then MATH by NAME 's inequality. If MATH we use REF and set MATH to obtain MATH where MATH is a wavelet of order at least MATH. Hence REF gives an upper bound MATH uniformly in MATH. Hence REF follows. Finally, if we know that MATH and MATH then combination of REF gives if MATH uniformly in MATH. Hence another application of REF proves the assertion. |
math/0104008 | Exercise in homological algebra. |
math/0104008 | CASE: The map MATH is surjective, the MATH-module MATH being generated by MATH (compare CITE or CITE) REF Multiplying the monad of MATH by MATH, respectively by MATH and taking the cohomology, one obtains the following commutative and exact diagram : MATH which shows that the map MATH is surjective. |
math/0104019 | We assume with no loss of generality that MATH and MATH. Let MATH be a separability element in MATH. Let MATH. Then MATH satisfies MATH for all MATH. Since MATH, it follows that MATH is idempotent. Similarly, MATH satisfies MATH, MATH and is idempotent. Then MATH . Then MATH and MATH is central. Then MATH, whence MATH is idempotent. |
math/0104019 | CASE: Let MATH be a separability element for MATH. Let MATH be its image in MATH induced by MATH. If MATH, then: MATH . We easily conclude that MATH is a separability element for MATH. CASE: Since MATH is an ideal in both MATH and MATH, the canonical epimorphism MATH sends the separable extension MATH onto a separable extension MATH CITE. |
math/0104019 | CASE: Assume MATH is QF and MATH is injective. By the NAME theorem, it will suffice to show that MATH is projective. Since MATH is projective, then flat, we note that MATH is injective, then projective. Since MATH is projective, we note that MATH is projective. But the evaluation mapping MATH is a split epi, for if MATH is a MATH-separability element, then MATH defines a splitting MATH-monic, where MATH. Hence MATH is isomorphic to a direct summand in MATH and is projective. Assuming that MATH is QF, and MATH is injective, we argue similarly that MATH is injective-projective and that MATH is isomorphic to direct summand in the projective MATH-module MATH. (SEMISIMPLICITY). Suppose MATH is semisimple and MATH is a module. It suffices to note that MATH is projective. Since MATH is projective and the map MATH defined as above is a split MATH-epimorphism, it follows that MATH is isomorphic to a direct summand of the projective module MATH. Similarly, we argue that given MATH semisimple and module MATH, MATH is projective. (WEAK GLOBAL DIMENSION.) If MATH is a projective resolution of MATH, then MATH is a projective resolution as well since MATH is flat and MATH is projective. Recall that MATH is the MATH'th homology group of the chain complex MATH for each non-negative integer MATH, so MATH is the MATH'th homology group of MATH. At the level of chain complex, there is a split epi MATH which implies that MATH is isomorphic to a direct summand in MATH for each MATH. This shows that MATH. A similar argument with left modules shows that MATH. |
math/0104019 | The first statement follows from the fact that a separable extension is both right and left semisimple extension and properties of these CITE. For the second statement, we first establish an interesting isomorphism below involving MATH and its dual MATH. On the one hand, since MATH is a separable extension, MATH is a split extension of MATH, for if MATH is a separability element we define a bimodule projection by MATH by MATH (compare CITE). Then as MATH-bimodules, MATH for some MATH: moreover, by restriction this is true as MATH-bimodules. On the other hand, since MATH is split, it follows that for some MATH-bimodule MATH, which is left and right projective MATH-module, MATH as MATH-bimodules; whence MATH as MATH-bimodules. Putting together the two isomorphisms for MATH, we obtain MATH . Since MATH is f.g. projective, there is a module MATH such that MATH. Then applying MATH to this: MATH . Combining this with REF , we obtain MATH . This establishes that MATH. We similarly conclude MATH by combining the split extension MATH with the MATH-bimodule isomorphism MATH and the existence of MATH such that MATH. |
math/0104019 | Since there is a bimodule MATH such that MATH, we combine this with MATH to see that REF is a MATH-isomorphism, whence MATH is a right QF extension CITE. Similarly, we show MATH to be a left QF extension. |
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