paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/0005010 | The proof is based on the fact that the NAME complex MATH of MATH is the direct limit of NAME complexes of exterior forms on finite order jet manifolds. Since the exterior differential MATH commutes with the pull-back maps MATH, these complexes form a direct system. Then, in accordance with the well-known theorem CITE,... |
math-ph/0005010 | Since every fibre bundle MATH is affine, MATH is a strong deformation retract of MATH, and so is MATH (see REF). Then, in accordance with the NAME - NAME theorem CITE, cohomology MATH of MATH with coefficients in the constant sheaf MATH coincides with that MATH of MATH. The well-known NAME theorem completes the proof. |
math-ph/0005010 | The complex REF is exact due to the NAME lemma, and is a resolution of the constant sheaf MATH on MATH since MATH are sheaves of MATH-modules. Then, by virtue of REF , we have the cohomology isomorphism MATH . REF below completes the proof. |
math-ph/0005010 | Since MATH is a strong deformation retract of MATH, the first isomorphism in REF follows from the above-mentioned NAME - NAME theorem CITE, while the second one is a consequence of the NAME theorem. |
math-ph/0005010 | The exact sequence REF is a resolution of the pull-back sheaf MATH on MATH. Then, by virtue of REF , we have a cohomology isomorphism MATH . The isomorphism REF follows from the facts that MATH is a strong deformation retract of MATH and that MATH is the pull-back onto MATH of the sheaf MATH on MATH CITE. |
math-ph/0005010 | Due to the relation MATH the horizontal projection MATH provides a homomorphism of the NAME complex REF to the complex MATH . Accordingly, there is a homomorphism MATH of cohomology groups of these complexes. REF show that, for MATH, the homomorphism REF is an isomorphism (see the relation REF below for the case MATH).... |
math-ph/0005010 | Being nilpotent, the vertical differential MATH defines a homomorphism of the complex REF to the complex MATH and, accordingly, a homomorphism of cohomology groups MATH of these complexes. Since MATH, the result follows. |
math-ph/0005010 | Let the common symbol MATH stand for the coboundary operators MATH and MATH of the variational bicomplex. Bearing in mind the decompositions REF - REF , it suffices to show that, if an element MATH is MATH-exact with respect to the algebra MATH (that is, MATH, MATH), then it is MATH-exact in the algebra MATH (that is, ... |
math-ph/0005010 | By taking a smooth partition of unity on MATH subordinate to the cover MATH and passing to the function with support in MATH, one gets a smooth real function MATH on MATH which is REF on a neighborhood of MATH and REF on a neighborhood of MATH in MATH. Let MATH be the pull-back of MATH onto MATH. The exterior form MATH... |
math-ph/0005010 | For MATH, the manifested isomorphism follows from the fact that MATH for any sheaf MATH on MATH. To prove other ones, let us replace the exact sequence REF with MATH and consider the short exact sequences MATH . They give the corresponding exact cohomology sequences MATH . Since sheaves MATH, MATH, are acyclic, the exa... |
math-ph/0005017 | Let MATH denote the MATH-matrix for the pair (MATH, MATH), which by the factorization property can be represented in the form MATH . In fact, this factorization property is a consequence of the multiplicativity property of the fundamental matrix (see CITE for a proof and for references to earlier work). A short calcula... |
math-ph/0005017 | In CITE we proved (see REF there and its proof) that for any two potentials MATH and MATH with (compact) disjoint supports one has MATH with MATH where MATH and MATH are the right and left reflection coefficients for the NAME equation with the potential MATH, MATH. Actually REF states that MATH for all MATH. Now we imp... |
math-ph/0005017 | As proved in CITE there is a constant MATH independent of MATH and MATH such that MATH where MATH, MATH, and MATH denotes the trace class norm (see for example, CITE). From the proof of REF it follows that MATH for all MATH. |
math-ph/0005032 | Since MATH is an identity, we have MATH . On the other hand, since MATH is an identity, we have MATH . Thus MATH. |
math-ph/0005032 | We know that MATH. Multiplying on the left by MATH gives MATH . By associativity, this gives MATH and so MATH . |
math-ph/0005032 | To show that MATH is the inverse of MATH, we must show both that MATH and MATH. Suppose we know, say, that MATH. Then our goal is to show that this implies that MATH. Since MATH, MATH . By associativity, we have MATH . Now, by the definition of a group, MATH has an inverse. Let MATH be that inverse. (Of course, in the ... |
math-ph/0005032 | Exercise. |
math-ph/0005032 | Let MATH be any element of MATH. Then MATH. Multiplying on the left by MATH gives MATH. Now consider MATH. Since MATH, we have MATH. In light of REF , we conclude that MATH is the inverse of MATH. |
math-ph/0005032 | Easy. |
math-ph/0005032 | Saying that MATH and MATH are both in the component containing the identity means that there exist continuous paths MATH and MATH with MATH, MATH, and MATH. But then MATH is a continuous path starting at MATH and ending at MATH. Thus the product of two elements of the identity component is again in the identity compone... |
math-ph/0005032 | Consider first the case MATH. A MATH invertible complex matrix MATH is of the form MATH with MATH, the set of non-zero complex numbers. But given any two non-zero complex numbers, we can easily find a continuous path which connects them and does not pass through zero. For the case MATH, we use the NAME canonical form. ... |
math-ph/0005032 | The proof is almost the same as for MATH, except that we must be careful to preserve the condition MATH. Let MATH be an arbitrary element of MATH. The case MATH is trivial, so we assume MATH. We can define MATH as above for MATH, with MATH, and MATH, since MATH. Now define MATH as before for MATH, and define MATH to be... |
math-ph/0005032 | By a standard result of linear algebra, every unitary matrix has an orthonormal basis of eigenvectors, with eigenvalues of the form MATH. It follows that every unitary matrix MATH can be written as MATH with MATH unitary and MATH. Conversely, as is easily checked, every matrix of the form REF is unitary. Now define MAT... |
math-ph/0005032 | MATH cannot be connected, for if MATH and MATH, then any continuous path connecting MATH to MATH would have to include a matrix with determinant zero, and hence pass outside of MATH. The proof that MATH is connected is given in REF . Once MATH is known to be connected, it is not difficult to see that MATH is also conne... |
math-ph/0005032 | REF shows that MATH may be thought of (topologically) as the three-dimensional sphere MATH sitting inside MATH. It is well-known that MATH is simply connected. |
math-ph/0005032 | In light of REF , we see that MATH and hence MATH . Thus the series REF converges absolutely, and so it converges. To show continuity, note that since MATH is a continuous function of MATH, the partial sums of REF are continuous. But it is easy to see that REF converges uniformly on each set of the form MATH, and so th... |
math-ph/0005032 | Point REF is obvious. Points REF are special cases of point REF. To verify point REF, we simply multiply power series term by term. (It is left to the reader to verify that this is legal.) Thus MATH . Multiplying this out and collecting terms where the power of MATH plus the power of MATH equals MATH, we get MATH . Now... |
math-ph/0005032 | Differentiate the power series for MATH term-by-term. (NAME might worry whether this is valid, but you shouldn't. For each MATH, MATH is given by a convergent power series in MATH, and it is a standard theorem that you can differentiate power series term-by-term.) |
math-ph/0005032 | The usual logarithm for real, positive numbers satisfies MATH for MATH. Integrating term-by-term and noting that MATH gives MATH . Taking MATH (so that MATH), we have MATH . This series has radius of convergence one, and defines a complex analytic function on the set MATH, which coincides with the usual logarithm for r... |
math-ph/0005032 | It is easy to see that the series REF converges absolutely whenever MATH. The proof of continuity is essentially the same as for the exponential. If MATH is real, then every term in the series REF is real, and so MATH is real. We will now show that MATH for all MATH with MATH. We do this by considering two cases. CASE:... |
math-ph/0005032 | Note that MATH so that MATH . This is what we want. |
math-ph/0005032 | The expression inside the brackets is clearly tending to MATH as MATH, and so is in the domain of the logarithm for all sufficiently large MATH. Now MATH where MATH is an error term which, by REF satisfies MATH. But then MATH and so MATH . Since both MATH and MATH are of order MATH, we obtain the desired result by lett... |
math-ph/0005032 | Using the power series for the exponential and multiplying, we get MATH where (check!) MATH. Since MATH as MATH, MATH is in the domain of the logarithm for all sufficiently large MATH. But MATH where by REF MATH. Exponentiating the logarithm gives MATH and MATH . Since both MATH and MATH are of order MATH, we have (usi... |
math-ph/0005032 | There are three cases, as in REF MATH is diagonalizable. Suppose there is a complex invertible matrix MATH such that MATH . Then MATH . Thus MATH, and MATH. (Recall that MATH.) REF MATH is nilpotent. If MATH is nilpotent, then it cannot have any non-zero eigenvalues (check!), and so all the roots of the characteristic ... |
math-ph/0005032 | The uniqueness is immediate, since if there is such a MATH, then MATH. So we need only worry about existence. The first step is to show that MATH must be smooth. This follows from REF (which we did not prove), but we give a self-contained proof. Let MATH be a smooth real-valued function supported in a small neighborhoo... |
math-ph/0005032 | By definition of the NAME algebra, MATH lies in MATH for all real MATH. But as MATH varies from MATH to MATH, MATH is a continuous path connecting the identity to MATH. |
math-ph/0005032 | This is immediate, since by REF , MATH and MATH. |
math-ph/0005032 | Point REF is immediate, since MATH, which must be in MATH if MATH is in MATH. Point REF is easy to verify if MATH and MATH commute, since then MATH. If MATH and MATH do not commute, this argument does not work. However, the NAME product formula says that MATH . Because MATH and MATH are in the NAME algebra, MATH and MA... |
math-ph/0005032 | The proof is similar to the proof of REF . Since MATH is a continuous group homomorphism, MATH will be a one-parameter subgroup of MATH, for each MATH. Thus by REF , there is a unique MATH such that MATH for all MATH. This MATH must lie in MATH since MATH. We now define MATH, and check in several steps that MATH has th... |
math-ph/0005032 | Easy. Note that REF guarantees that MATH is actually in MATH for all MATH. |
math-ph/0005032 | Recall that by REF , MATH can be computed as follows: MATH . Thus MATH which is what we wanted to prove. See also REF . |
math-ph/0005032 | We follow the proof of REF in NAME and tom NAME. In view of what we have proved about the matrix logarithm, we know this result for the case of MATH. To prove the general case, we consider a matrix NAME group MATH, with NAME algebra MATH. Suppose MATH are elements of MATH, and that MATH. Let MATH, which is defined for ... |
math-ph/0005032 | Recall that for us, saying MATH is connected means that MATH is path-connected. This certainly means that MATH is connected in the usual topological sense, namely, the only non-empty subset of MATH that is both open and closed is MATH itself. So let MATH denote the set of all MATH that can be written in the form REF . ... |
math-ph/0005032 | The only non-trivial point is the NAME identity. The only way to prove this is to write everything out and see, and this is best left to the reader. Note that each triple bracket generates four terms, for a total of twelve. Each of the six orderings of MATH occurs twice, once with a plus sign and once with a minus sign... |
math-ph/0005032 | By REF , MATH is a real subalgebra of MATH complex matrices, and is thus a real NAME algebra. |
math-ph/0005032 | Observe that MATH whereas MATH . So we require that MATH or equivalently MATH which is exactly the NAME identity. |
math-ph/0005032 | The uniqueness of the extension is obvious, since if the bracket operation on MATH is to be bilinear, then it must be given by MATH . To show existence, we must now check that REF is really bilinear and skew-symmetric, and that it satisfies the NAME identity. It is clear that REF is real bilinear, and skew-symmetric. T... |
math-ph/0005032 | From the computations in the previous section we see easily that the specified NAME algebras are in fact complex subalgebras of MATH, and hence are complex NAME algebras. Now, MATH is the space of all MATH complex matrices, whereas MATH is the space of all MATH real matrices. Clearly, then, every MATH can be written un... |
math-ph/0005032 | Let MATH and MATH be as in the statement of the theorem. We will prove that in fact MATH which reduces to the desired result in the case MATH. Since by REF commutes with everything in sight, the above relation is equivalent to MATH . Let us call the left side of REF MATH and the right side MATH. Our strategy will be to... |
math-ph/0005032 | Recall that the NAME group has the very special property that its exponential mapping is one-to-one and onto. Let ``log" denote the inverse of this map. Define MATH by the formula MATH . We will show that MATH is a NAME group homomorphism. If MATH and MATH are in the NAME algebra of the NAME group (MATH strictly upper-... |
math-ph/0005032 | We begin by proving that the corollary follows from the integral form of the NAME formula. The proof is conceptually similar to the reasoning in REF . Note that if MATH and MATH lie in some NAME algebra MATH then MATH and MATH will preserve MATH, and so also will MATH. Thus whenever REF holds, MATH will lie in MATH. It... |
math-ph/0005032 | It is possible to prove this Theorem by expanding everything in a power series and differentiating term-by-term; we will not take that approach. We will prove only form REF of the derivative formula, but the form REF follows by the chain rule. Let us use the NAME product formula, and let us assume for the moment that i... |
math-ph/0005032 | Since by REF is a subspace of MATH, it remains only to show that MATH is closed under brackets. So assume MATH. Then MATH and MATH are in MATH, and so (since MATH is a subgroup) is the element MATH . This shows that MATH is in MATH for all MATH. But MATH is a subspace of MATH, which is necessarily a closed subset of MA... |
math-ph/0005032 | Not written at this time. |
math-ph/0005032 | REF states that for each NAME group homomorphism MATH there is an associated NAME algebra homomorphism MATH. Take MATH and MATH. Since the NAME algebra of MATH is MATH (since the exponential of any operator is invertible), the associated NAME algebra homomorphism MATH maps from MATH to MATH, and so constitutes a repres... |
math-ph/0005032 | This follows from REF. |
math-ph/0005032 | It suffices to show that every non-zero invariant subspace of MATH is in fact equal to MATH. So let MATH be such a space. Since MATH is assumed non-zero, there is at least one non-zero element MATH in MATH. Then MATH can be written uniquely in the form MATH with at least one of the MATH's non-zero. Let MATH be the larg... |
math-ph/0005032 | Let us make sure we are clear about what this means. Suppose that MATH is a complex representation of the (real) NAME algebra MATH, acting on the complex space MATH. Then saying that MATH is irreducible means that there is no non-trivial invariant complex subspace MATH. That is, even though MATH is a real NAME algebra,... |
math-ph/0005032 | Let MATH be an irreducible representation of MATH acting on a (finite-dimensional complex) space MATH. Our strategy is to diagonalize the operator MATH. Of course, a priori, we don't know that MATH is diagonalizable. However, because we are working over the (algebraically closed) field of complex numbers, MATH must hav... |
math-ph/0005032 | The following lemma is the key to the entire proof. Let MATH be an eigenvector of MATH with eigenvalue MATH. Then MATH . Thus either MATH, or else MATH is an eigenvector for MATH with eigenvalue MATH. Similarly, MATH so that either MATH, or else MATH is an eigenvector for MATH with eigenvalue MATH. We call MATH the ``r... |
math-ph/0005032 | The proof is by induction on the dimension of the space MATH. If MATH, then automatically the representation is irreducible, since then MATH is has no non-trivial subspaces, let alone non-trivial invariant subspaces. Thus MATH is a direct sum of irreducible representations, with just one summand, namely MATH itself. Su... |
math-ph/0005032 | So, we are assuming that our space MATH is equipped with an inner product, and that MATH is unitary for each MATH. Suppose that MATH is invariant, and that MATH is also invariant. Define MATH . Then of course MATH, and standard NAME space theory implies that MATH. It remains only to show that MATH is invariant. So supp... |
math-ph/0005032 | Suppose that MATH is a representation of MATH, acting on a space MATH. Choose an arbitrary inner product MATH on MATH. Then define a new inner product MATH on MATH by MATH . It is very easy to check that indeed MATH is an inner product. Furthermore, if MATH, then MATH . But as MATH ranges over MATH, so does MATH. Thus ... |
math-ph/0005032 | This proof requires the notion of NAME measure. A left NAME measure on a matrix NAME group MATH is a non-zero measure MATH on the NAME MATH-algebra in MATH with the following two properties: REF it is locally finite, that is, every point in MATH has a neighborhood with finite measure, and REF it is left-translation inv... |
math-ph/0005032 | REF . |
math-ph/0005032 | Define a map MATH from MATH into MATH by MATH . Since MATH and MATH are linear, and since MATH is bilinear, MATH will be a bilinear map of MATH into MATH. But then the universal property says that there is an associated linear map MATH such that MATH . Then MATH is the desired map MATH. Now, if MATH are operators on MA... |
math-ph/0005032 | Suppose that MATH is a smooth curve in MATH and MATH is a smooth curve in MATH. Then we verify the product rule in the usual way: MATH . Thus MATH . This being the case, we can compute MATH: MATH . This shows that MATH on elements of the form MATH, and therefore on the whole space MATH. |
math-ph/0005032 | Using the product rule, MATH . This is what we wanted to show. |
math-ph/0005032 | We prove the group case; the proof of the NAME algebra case is the same. If MATH is in the center of MATH, then for all MATH, MATH . But this says exactly that MATH is a morphism of MATH with itself. So by Point REF of NAME 's lemma, MATH is a multiple of the identity. |
math-ph/0005032 | Again, we prove only the group case. If MATH is commutative, then the center of MATH is all of MATH, so by the previous corollary MATH is a multiple of the identity for each MATH. But this means that every subspace of MATH is invariant! Thus the only way that MATH can fail to have a non-trivial invariant subspace is fo... |
math-ph/0005032 | As usual, we will prove just the group case; the proof of the NAME algebra case requires only the obvious notational changes. Proof of REF. Saying that MATH is a morphism means MATH for all MATH and all MATH. Now suppose that MATH. Then MATH . This shows that MATH is an invariant subspace of MATH. Since MATH is irreduc... |
math-ph/0005032 | REF m odd. In this case, we want to prove that there is no representation MATH such that MATH and MATH are related as in REF . (We have already considered the case MATH in REF .) Suppose, to the contrary, that there is such a MATH. Then REF says that MATH for all MATH. In particular, take MATH. Then, computing as in RE... |
math-ph/0005032 | Let MATH be a NAME algebra isomorphism. By REF , there exists an associated NAME group homomorphism MATH. Since MATH is also a NAME algebra homomorphism, there is a corresponding NAME group homomorphism MATH. We want to show that MATH and MATH are inverses of each other. Well, MATH, so by the Point REF, MATH. Similarly... |
math-ph/0005032 | For REF , let MATH act on MATH and MATH on MATH. We assume that the associated NAME algebra representations are equivalent, that is, that there exists an invertible linear map MATH such that MATH for all MATH and all MATH. This is the same as saying that MATH, or equivalently that MATH (for all MATH). Now define a map ... |
math-ph/0005032 | CASE: Verify Point REF. Since MATH is connected, REF tells us that every element MATH of MATH is a finite product of the form MATH, with MATH. But then if MATH, we have MATH . So we now need only prove Point REF . CASE: Define MATH in a neighborhood of the identity. REF says that the exponential mapping for MATH has a ... |
math-ph/0005032 | REF . |
math-ph/0005032 | MATH is simply connected. |
math-ph/0005032 | REF . |
math-ph/0005032 | Since we are working over the complex numbers, MATH has at least one eigenvalue MATH. Let MATH be the eigenspace for MATH with eigenvalue MATH. I assert that MATH is invariant under MATH. To see this consider MATH, and compute MATH . This shows that MATH is either zero or an eigenvector for MATH with eigenvalue MATH; t... |
math-ph/0005032 | By REF , MATH decomposes as a direct sum of irreducible invariant subspaces MATH. Each MATH must be one of the irreducible representations of MATH, which we have classified. In particular, in each MATH, MATH can be diagonalized, and the eigenvalues of MATH are integers. Thus MATH can be diagonalized on the whole space ... |
math-ph/0005032 | Apply REF to the restriction of MATH to MATH, and to the restriction of MATH to MATH. |
math-ph/0005032 | The definition of a root tells us that we have the commutation relation MATH. Thus MATH A similar argument allows us to compute MATH. |
math-ph/0005032 | Let MATH be the direct sum of the weight spaces in MATH. Equivalently, MATH is the space of all vectors MATH such that MATH can be written as a linear combination of simultaneous eigenvectors for MATH and MATH. Since REF MATH always has at least one weight, MATH. On the other hand, REF tells us that if MATH is a root v... |
math-ph/0005032 | Let MATH be as in the definition. Consider the subspace MATH of MATH spanned by elements of the form MATH with each MATH, and MATH. (If MATH, it is understood that MATH-in REF is equal to MATH.) I assert that MATH is invariant. To see this, it suffices to check that MATH is invariant under each of the basis elements. B... |
math-ph/0005032 | Uniqueness is immediate, since by the previous Proposition, MATH is the highest weight, and two distinct weights cannot both be highest. We have already shown that every irreducible representation is the direct sum of its weight spaces. Since the representation is finite-dimensional, there can be only finitely many wei... |
math-ph/0005032 | Let MATH be a highest weight cyclic representation with highest weight MATH and cyclic vector MATH. By complete reducibility REF , MATH decomposes as a direct sum of irreducible representations MATH . By REF , each of the MATH's is the direct sum of its weight spaces. Thus since the weight MATH occurs in MATH, it must ... |
math-ph/0005032 | We now know that a representation is irreducible if and only if it is highest weight cyclic. Suppose that MATH and MATH are two such representations with the same highest weight MATH. Let MATH and MATH be the cyclic vectors for MATH and MATH, respectively. Now consider the representation MATH, and let MATH be smallest ... |
math-ph/0005032 | We already know that all of the weights of MATH are of the form MATH, with MATH and MATH integers. We must show that if MATH is the highest weight, then MATH and MATH are both non-negative. For this, we again use what we know about the representations of MATH. The following result can be obtained from the proof of the ... |
math-ph/0005032 | Note that the trivial representation is an irreducible representation with highest weight MATH. So we need only construct representations with at least one of MATH and MATH positive. First, we construct two irreducible representations with highest weights MATH and MATH. (These are the so-called fundamental representati... |
math-ph/0005032 | We need a map MATH with the property that MATH for all MATH. This is the same as saying that MATH, or equivalently that MATH. But in light of REF , we can take MATH. |
math-ph/0005032 | Equivalent representations must have the same weights and the same multiplicities. |
math-ph/0005032 | Same as for MATH. |
math/0005001 | The first claim follows by replacing everything by absolute values in REF. The second claim follows by applying REF with MATH replaced by MATH. |
math/0005001 | The estimate REF follows from REF, NAME 's theorem, and the identity MATH . From REF we see that the left-hand side of REF is less than or equal to the right-hand side. To prove the reverse inequality, apply REF for MATH and for functions MATH which split as tensor products of functions on MATH and functions on MATH. |
math/0005001 | For MATH, let MATH be the MATH-linear operator defined by MATH . From duality we have MATH . By NAME we thus have MATH . This gives REF. This (together with REF ) implies that the left-hand side of REF is less than or equal to the right-hand side. To prove the reverse inequality, apply REF with MATH for MATH and use du... |
math/0005001 | By REF we may take MATH. We can rewrite the left-hand side of REF as MATH . From NAME 's theorem we have MATH . By NAME, the left-hand side of REF is thus less than or equal to MATH . The claim follows. |
math/0005001 | The right-hand side of REF (and thus REF) follows immediately from REF . The left-hand side follows from REF and setting MATH, MATH, MATH. |
math/0005001 | By adding dummy elements to MATH, MATH if necessary we may assume that MATH. Consider the quantity MATH . By REF applied to each summand, this is less than or equal to MATH where MATH is the tensor product of all the MATH for MATH. By REF it thus suffices to show that MATH . From the overlap of the MATH we have MATH . ... |
math/0005001 | The lower bound for REF follows from REF , so it suffices to show the upper bound. Write MATH, where MATH . From the support properties of MATH we see that for fixed MATH there are at most MATH values of MATH for which MATH does not vanish, and similarly with the roles of MATH and MATH reversed. Restricting MATH to tho... |
math/0005001 | By a limiting argument we may assume that MATH is finite. From REF we have MATH . Applying REF we obtain the MATH side of REF. To obtain the MATH side, apply REF with MATH, MATH, and MATH. Finally, REF comes from REF and testing REF with MATH and MATH being large characteristic functions. |
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