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math-ph/0005010
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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, the cohomology groups MATH of the NAME complex REF are isomorphic to the direct limit of the direct system MATH of NAME cohomology groups MATH of finite order jet manifolds MATH. The forthcoming REF completes the proof.
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math-ph/0005010
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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.
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math-ph/0005010
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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.
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math-ph/0005010
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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.
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math-ph/0005010
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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.
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math-ph/0005010
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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). It follows that a horizontal form MATH is MATH-closed (respectively, MATH-exact) if and only if MATH where MATH is a closed (respectively, exact) form. The decomposition REF complete the proof.
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math-ph/0005010
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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.
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math-ph/0005010
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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, MATH). REF states that, if MATH is a contractible fibre bundle and a MATH-exact form MATH on MATH is of finite jet order MATH (that is, MATH), there exists an exterior form MATH on MATH such that MATH. Moreover, a glance at the homotopy operators for MATH and MATH CITE shows that the jet order MATH of MATH is bounded for all exterior forms MATH of fixed jet order. Let us call this fact the finite exactness of the operator MATH. Given an arbitrary fibre bundle MATH, the finite exactness takes place on MATH over any open subset MATH of MATH which is homeomorphic to a convex open subset of MATH. Now, we show the following. CASE: Suppose that the finite exactness of the operator MATH takes place on MATH over open subsets MATH, MATH of MATH and their non-empty overlap MATH. Then, it is also true on MATH. CASE: Given a family MATH of disjoint open subsets of MATH, let us suppose that the finite exactness takes place on MATH over every subset MATH from this family. Then, it is true on MATH over the union MATH of these subsets. If REF hold, the finite exactness of MATH on MATH takes place since one can construct the corresponding covering of the manifold MATH REF . Proof of REF . Let MATH be a MATH-exact form on MATH. By assumption, it can be brought into the form MATH on MATH and MATH on MATH, where MATH and MATH are exterior forms of finite jet order. Due to the decompositions REF - REF , one can choose the forms MATH, MATH such that MATH on MATH and MATH on MATH are MATH-exact forms. Let us consider their difference MATH on MATH. It is a MATH-exact form of finite jet order which, by assumption, can be written as MATH where an exterior form MATH is also of finite jet order. REF below shows that MATH where MATH and MATH are exterior forms of finite jet order on MATH and MATH, respectively. Then, putting MATH we have the form MATH equal to MATH on MATH and MATH on MATH, respectively. Since the difference MATH on MATH vanishes, we obtain MATH on MATH where MATH is of finite jet order. Proof of REF . Let MATH be a MATH-exact form on MATH. The finite exactness on MATH holds since MATH on every MATH and, as was mentioned above, the jet order MATH is bounded on the set of exterior forms MATH of fixed jet order MATH.
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math-ph/0005010
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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 is zero on a neighborhood of MATH and, therefore, can be extended by REF to MATH. Let us denote it MATH. Accordingly, the exterior form MATH has an extension MATH by REF to MATH. Then, MATH is a desired decomposition because MATH and MATH are of finite jet order which does not exceed that of MATH.
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math-ph/0005010
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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 exact sequence REF falls into MATH and, similarly, the exact sequence REF does MATH . The equalities REF for the couples of numbers MATH, MATH, and the equality REF for MATH lead to the chain of isomorphisms MATH . The exact sequence REF for MATH contains the exact sequence MATH . Since MATH and MATH, the result follows from REF for MATH.
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math-ph/0005017
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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 calculation gives MATH . With MATH denoting the expectation with respect to the measure MATH, by NAME 's inequality and REF we therefore have the estimate MATH . From the factorization REF it follows that MATH . We will now make use of the fact that the MATH are i.i.d.random variables. For this purpose define the MATH matrices MATH recursively by MATH and MATH such that in particular MATH. Now it is easy to see that MATH . We now use the fact that the operator inequality MATH implies MATH and MATH for all MATH. In particular we have MATH from which we obtain the recursive estimates MATH and hence MATH . We remark that with the same arguments one proves the lower bound MATH . The relation REF, the estimate REF combined with REF and NAME 's lemma imply now MATH which proves REF . To establish the last claim of the theorem we recall the following well known estimates (see for example . CITE) MATH valid for all large MATH uniformly for all MATH in the (compact) support of MATH for fixed MATH. Using the estimate REF gives the estimate MATH for all large MATH. Since MATH, this concludes the proof of the theorem.
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math-ph/0005017
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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 improve on this estimate. As in CITE we set MATH with MATH. Moreover MATH only when MATH, which we recall can happen only if MATH. Therefore MATH . By means of the inequality MATH we immediately obtain MATH . Since MATH we can replace MATH either by MATH or by MATH. Now let us consider the operator MATH for finite MATH. Applying REF we obtain MATH . Repeating this procedure recursively we obtain MATH . From the existence of the spectral shift density REF by the NAME ergodic theorem it follows that MATH . From the obvious inequality MATH the bound REF follows. For large MATH we have the following asymptotics CITE uniformly in MATH on compact sets: MATH such that MATH and MATH. If the single-site potential MATH has MATH derivatives in MATH then MATH and MATH as MATH CITE. The estimate MATH is REF below.
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math-ph/0005017
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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.
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math-ph/0005032
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Since MATH is an identity, we have MATH . On the other hand, since MATH is an identity, we have MATH . Thus MATH.
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math-ph/0005032
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We know that MATH. Multiplying on the left by MATH gives MATH . By associativity, this gives MATH and so MATH .
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math-ph/0005032
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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 end, we will conclude that MATH, but we cannot assume that now.) Multiplying on the right by MATH and using associativity again gives MATH A similar argument shows that if MATH, then MATH.
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math-ph/0005032
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Exercise.
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math-ph/0005032
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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.
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math-ph/0005032
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Easy.
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math-ph/0005032
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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 component. Furthermore, MATH is a continuous path starting at MATH and ending at MATH, and so the inverse of any element of the identity component is again in the identity component. Thus the identity component is a subgroup.
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math-ph/0005032
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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. Every MATH complex matrix MATH can be written as MATH where MATH is the NAME canonical form. The only property of MATH we will need is that MATH is upper-triangular: MATH . If MATH is invertible, then all the MATH's must be non-zero, since MATH. Let MATH be obtained by multiplying the part of MATH above the diagonal by MATH, for MATH, and let MATH. Then MATH is a continuous path which starts at MATH and ends at MATH, where MATH is the diagonal matrix MATH . This path lies in MATH since MATH for all MATH. But now, as in the case MATH, we can define MATH which connects each MATH to REF in MATH, as MATH goes from REF to REF. Then we can define MATH . This is a continuous path which starts at MATH when MATH, and ends at MATH REF when MATH. Since the MATH's are always non-zero, MATH lies in MATH. We see, then, that every matrix MATH in MATH can be connected to the identity by a continuous path lying in MATH. Thus if MATH and MATH are two matrices in MATH, they can be connected by connecting each of them to the identity.
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math-ph/0005032
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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. (Note that since MATH, MATH.) This allows us to connect MATH to the identity while staying within MATH.
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math-ph/0005032
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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 MATH . As MATH ranges from REF to REF, this defines a continuous path in MATH joining MATH to MATH. This shows that MATH is connected. A slight modification of this argument, as in the proof of REF , shows that MATH is connected.
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math-ph/0005032
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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 connected. For let MATH be any matrix with negative determinant, and take MATH in MATH. Then MATH and MATH are in MATH, and can be joined by a continuous path MATH in MATH. But then MATH is a continuous path joining MATH and MATH in MATH.
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math-ph/0005032
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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.
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math-ph/0005032
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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 the sum is again continuous.
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math-ph/0005032
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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 because (and only because) MATH and MATH commute, MATH and so REF becomes MATH . To prove REF, simply note that MATH and so the two sides of REF are the same term by term. Point REF is evident from the proof of REF .
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math-ph/0005032
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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.)
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math-ph/0005032
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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 real MATH in the interval MATH. Now, MATH for MATH, and by analyticity this identity continues to hold on the whole set MATH. On the other hand, if MATH, then MATH so that MATH . Thus MATH makes sense for all such MATH. Since MATH for real MATH with MATH, it follows by analyticity that MATH for all complex numbers with MATH.
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math-ph/0005032
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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 is diagonalizable. Suppose that MATH, with MATH diagonal. Then MATH. It follows that MATH is of the form MATH where MATH are the eigenvalues of MATH. Now, if MATH, then certainly MATH for MATH. (Think about it.) Thus MATH and so by the Lemma MATH . CASE: MATH is not diagonalizable. If MATH is not diagonalizable, then, using the NAME canonical form, it is not difficult to construct a sequence MATH of diagonalizable matrices with MATH. (See REF .) If MATH, then MATH for all sufficiently large MATH. By REF , and so by the continuity of MATH and MATH, MATH. Thus we have shown that MATH for all MATH with MATH. Now, the same argument as in the complex case shows that if MATH, then MATH. But then the same two-case argument as above shows that MATH for all such MATH.
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math-ph/0005032
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Note that MATH so that MATH . This is what we want.
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math-ph/0005032
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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 letting MATH and using the continuity of the exponential.
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math-ph/0005032
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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 (using the continuity of the exponential) MATH which is the NAME product formula.
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math-ph/0005032
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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 polynomial must be zero. Thus the NAME canonical form of MATH will be strictly upper triangular. That is, MATH can be written as MATH . In that case (it is easy to see) MATH will be upper triangular, with ones on the diagonal: MATH . Thus if MATH is nilpotent, MATH, and MATH. CASE: MATH arbitrary. As pointed out in REF, every matrix MATH can be written as the sum of two commuting matrices MATH and MATH, with MATH diagonalizable (over MATH) and MATH nilpotent. Since MATH and MATH commute, MATH. So by the two previous cases MATH which is what we want.
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math-ph/0005032
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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 neighborhood of zero, with MATH and MATH. Now look at MATH . Making the change-of-variable MATH gives MATH . It follows that MATH is differentiable, since derivatives in the MATH variable go onto MATH, which is smooth. On the other hand, if we use the identity MATH in REF , we have MATH . Now, the conditions on the function MATH, together with the continuity of MATH, guarantee that MATH is close to MATH, and hence is invertible. Thus we may write MATH . Since MATH is smooth and MATH is just a constant matrix, this shows that MATH is smooth. Now that MATH is known to be differentiable, we may define MATH . Our goal is to show that MATH. Since MATH is smooth, a standard calculus result (extended trivially to handle matrix-valued functions) says MATH . It follows that for each fixed MATH, MATH . Then, since MATH is a one-parameter group MATH . Letting MATH and using REF from REF shows that MATH.
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math-ph/0005032
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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.
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math-ph/0005032
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This is immediate, since by REF , MATH and MATH.
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math-ph/0005032
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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 MATH are in MATH, as is MATH, since MATH is a group. But now because MATH is a matrix NAME group, the limit of things in MATH must be again in MATH, provided that the limit is invertible. Since MATH is automatically invertible, we conclude that it must be in MATH. This shows that MATH is in MATH. Now for point REF. Recall REF that MATH. It follows that MATH, and hence by the product rule REF MATH . But now, by REF , MATH is in MATH for all MATH. Since we have (by points REF) established that MATH is a real vector space, it follows that the derivative of any smooth curve lying in MATH must be again in MATH. Thus MATH is in MATH.
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math-ph/0005032
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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 the required properties. CASE: MATH. This follows from REF and our definition of MATH, by putting MATH. CASE: MATH for all MATH. This is immediate, since if MATH, then MATH. CASE: MATH. By REF , MATH . By the NAME product formula, and the fact that MATH is a continuous homomorphism: MATH . But then we have MATH . Differentiating this result at MATH gives the desired result. CASE: MATH. By REF , MATH . Using a property of the exponential and REF , this becomes MATH . Differentiating this at MATH gives the desired result. CASE: MATH. Recall from the proof of REF that MATH . Hence MATH where we have used the fact that a derivative commutes with a linear transformation. But then by REF , MATH REF : MATH. This follows from REF and our definition of MATH. CASE: MATH is the unique real-linear map such that MATH. Suppose that MATH is another such map. Then MATH so that MATH . Thus by REF , MATH coincides with MATH. CASE: MATH. For any MATH, MATH . Thus MATH.
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math-ph/0005032
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Easy. Note that REF guarantees that MATH is actually in MATH for all MATH.
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math-ph/0005032
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Recall that by REF , MATH can be computed as follows: MATH . Thus MATH which is what we wanted to prove. See also REF .
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math-ph/0005032
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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 all sufficiently large MATH. Suppose MATH. Then MATH. To show that MATH, we must show that MATH for all MATH. As MATH, MATH. Note that since MATH, MATH, and so MATH. Thus we can find integers MATH such that MATH. Then MATH. But MATH, and MATH is closed, so MATH. We think of MATH as MATH. Then MATH is a subspace of MATH. Let MATH denote the orthogonal complement of MATH with respect to the usual inner product on MATH. Consider the map MATH given by MATH . Of course, we can identify MATH with MATH. Moreover, MATH is an open subset of MATH. Thus we can regard MATH as a map from MATH to itself. Now, using the properties of the matrix exponential, we see that MATH . This shows that the derivative of MATH at the point MATH is the identity. (Recall that the derivative at a point of a function from MATH to itself is a linear map of MATH to itself, in this case the identity map.) In particular, the derivative of MATH at REF is invertible. Thus the inverse function theorem says that MATH has a continuous local inverse, defined in a neighborhood of MATH. Now let MATH be any neighborhood of zero in MATH. I want to show that MATH contains a neighborhood of MATH in MATH. Suppose not. Then we can find a sequence MATH with MATH such that no MATH is in MATH. Since MATH is locally invertible, we can write MATH (for large MATH) uniquely as MATH, with MATH and MATH. Since MATH and MATH is continuous, MATH and MATH tend to zero. Thus (for large MATH), MATH. So we must have (for large MATH) MATH, otherwise MATH would be in MATH. Let MATH. Note that MATH and MATH. Since the unit ball in MATH is compact, we can choose a subsequence of MATH (still called MATH) so that MATH converges to some MATH, with MATH. But then by the Lemma, MATH! This is a contradiction, because MATH is the orthogonal complement of MATH. So for every neighborhood MATH of zero in MATH, MATH contains a neighborhood of the identity in MATH. If we make MATH small enough, then the exponential will be one-to-one on MATH. (The existence of the matrix logarithm implies that the exponential is one-to-one near zero.) Let MATH denote the inverse map, defined on MATH. Since MATH is compact, and MATH is one-to-one and continuous on MATH, log will be continuous. (This is a standard topological result.) So take MATH to be a neighborhood of MATH contained in MATH, and let MATH. Then MATH is open and the exponential takes MATH homeomorphically onto MATH.
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math-ph/0005032
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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 . In light of the Proposition, MATH contains a neighborhood MATH of the identity. In particular, MATH is non-empty. We first claim that MATH is open. To see this, consider MATH. Then look at the set of matrices of the form MATH, with MATH. This will be a neighborhood of MATH. But every such MATH can be written as MATH and MATH can be written as MATH, so MATH. Now we claim that MATH is closed (in MATH). Suppose MATH, and there is a sequence MATH with MATH. Then MATH. Thus we can choose some MATH such that MATH. Then MATH and MATH. But by assumption, MATH, so MATH. Thus MATH, and MATH is closed. Thus MATH is both open and closed, so MATH.
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math-ph/0005032
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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.
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math-ph/0005032
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By REF , MATH is a real subalgebra of MATH complex matrices, and is thus a real NAME algebra.
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math-ph/0005032
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Observe that MATH whereas MATH . So we require that MATH or equivalently MATH which is exactly the NAME identity.
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math-ph/0005032
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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. The skew-symmetry means that if REF is complex linear in the first factor, it is also complex linear in the second factor. Thus we need only show that MATH . Well, the left side of REF is MATH whereas the right side of REF is MATH and indeed these are equal. It remains to check the NAME identity. Of course, the NAME identity holds if MATH and MATH are in MATH. But now observe that the expression on the left side of the NAME identity is (complex!) linear in MATH for fixed MATH and MATH. It follows that the NAME identity holds if MATH is in MATH, and MATH in MATH. The same argument then shows that we can extend to MATH in MATH, and then to MATH in MATH. Thus the NAME identity holds in MATH.
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math-ph/0005032
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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 uniquely in the form MATH, with MATH. This gives us a complex vector space isomorphism of MATH with MATH, and it is a triviality to check that this is a NAME algebra isomorphism. On the other hand, MATH is the space of all MATH complex skew-self-adjoint matrices. But if MATH is any MATH complex matrix, then MATH . Thus MATH can be written as a skew matrix plus MATH times a skew matrix, and it is easy to see that this decomposition is unique. Thus every MATH in MATH can be written uniquely as MATH, with MATH and MATH in MATH. It follows that MATH. The verification of the remaining isomorphisms is similar, and is left as an exercise to the reader.
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math-ph/0005032
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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 show that MATH and MATH satisfy the same differential equation, with the same initial conditions. We can see right away that MATH . On the other hand, differentiating MATH by means of the product rule gives MATH (NAME can verify that the last term on the right is correct by differentiating term-by-term.) Now, since MATH and MATH commute with MATH, they also commute with MATH. Thus the second term on the right in REF can be rewritten as MATH . The first term on the right in REF is more complicated, since MATH does not necessarily commute with MATH. However, MATH . But since MATH, MATH with all higher terms being zero. Using the fact that everything commutes with MATH gives MATH . Making these substitutions into REF gives MATH . Thus MATH and MATH satisfy the same differential equation. Moreover, MATH. Thus by standard uniqueness results for ordinary differential equations, MATH for all MATH.
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math-ph/0005032
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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-triangular matrices), then MATH is of the form MATH such a matrix commutes with both MATH and MATH. That is, MATH and MATH commute with their commutator. Since MATH is a NAME algebra homomorphism, MATH and MATH will also commute with their commutator: MATH . We want to show that MATH is a homomorphism, that is, that MATH. Well, MATH can be written as MATH for a unique MATH and MATH can be written as MATH for a unique MATH. Thus by REF MATH . Using the definition of MATH and the fact that MATH is a NAME algebra homomorphism: MATH . Finally, using REF again we have MATH . Thus MATH is a group homomorphism. It is easy to check that MATH is continuous (by checking that MATH, exp, and MATH are all continuous), and so MATH is a NAME group homomorphism. Moreover, MATH by definition has the right relationship to MATH. Furthermore, since the exponential mapping is one-to-one and onto, there can be at most one MATH with MATH. So we have uniqueness.
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math-ph/0005032
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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 remains only to verify REF . The idea is that if MATH is NAME algebra homomorphism, then it will take a big horrible looking expression involving `ad' and MATH and MATH, and turn it into the same expression with MATH and MATH replaced by MATH and MATH. More precisely, since MATH is a NAME algebra homomorphism, MATH or MATH . More generally, MATH . This being the case, MATH . Similarly, MATH . Assume now that MATH and MATH are small enough that B-C-H applies to MATH and MATH, and to MATH and MATH. Then, using the linearity of the integral and reasoning similar to the above, we have: MATH . This is what we wanted to show.
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math-ph/0005032
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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 it is legal to interchange limit and derivative. (We will consider this issue at the end.) Then we have MATH . We now apply the product rule (generalized to MATH factors) to obtain MATH . But MATH (where we have used the relationship between Ad and ad). So we have MATH . Observe now that MATH is a geometric series. Let us now reason for a moment at the purely formal level. Using the usual formula for geometric series, we get MATH . This is what we wanted to show! NAME this argument make sense at any rigorous level? In fact it does. As usual, let us consider first the diagonalizable case. That is, assume that MATH is diagonalizable as an operator on MATH, and assume that MATH is an eigenvector for MATH. This means that MATH, for some MATH. Now, there are two cases, MATH and MATH. The MATH case corresponds to the case in which MATH and MATH commute, and we have already observed that the Theorem holds trivially in that case. The interesting case, then, is the case MATH. Note that MATH, and so MATH . Thus the geometric series in REF becomes an ordinary complex-valued series, with ratio MATH. Since MATH, this ratio will be different from one for all sufficiently large MATH. Thus we get MATH . There is now no trouble in taking the limit as we did formally above to get MATH . We see then that the Theorem holds in the case that MATH is diagonalizable and MATH is an eigenvector of MATH. If MATH is diagonalizable but MATH is not an eigenvector, then MATH is a linear combination of eigenvectors and applying the above computation to each of those eigenvectors gives the desired result. We need, then, to consider the case where MATH is not diagonalizable. But REF , if MATH is a diagonalizable matrix, then MATH will be diagonalizable as an operator on MATH. Since, as we have already observed, every matrix is the limit of diagonalizable matrices, we are essentially done. For it is easy to see by differentiating the power series term-by-term that MATH exists and varies continuously with MATH. Thus once we have the Theorem for all diagonalizable MATH we have it for all MATH by passing to the limit. The only unresolved issue, then, is the interchange of limit and derivative which we performed at the very beginning of the argument. I do not want to spell this out in detail, but let us see what would be involved in justifying this. A standard theorem in elementary analysis says that if MATH pointwise, and in addition MATH converges uniformly to some function MATH, then MATH is differentiable and MATH. (For example, REF in NAME 's Principles of Mathematical Analysis.) The key requirement is that the derivatives converge uniformly. Uniform convergence of the MATH's themselves is definitely not sufficient. In our case, MATH. The NAME product formula says that this converges pointwise to MATH. We need, then, to show that MATH converges uniformly to some MATH, say on the interval MATH. This computation is similar to what we did above, with relatively minor modifications to account for the fact that we do not take MATH and to make sure the convergence is uniform. This part of the proof is left as an exercise to the reader.
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math-ph/0005032
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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 MATH. Thus MATH is in MATH. (This argument is precisely the one we used to show that the NAME algebra of a matrix NAME group is a closed under brackets, once we had established that it is a subspace.)
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Not written at this time.
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math-ph/0005032
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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 representation of MATH. The properties of MATH follow from the properties of MATH given in REF .
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This follows from REF.
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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 largest value of MATH for which MATH, and consider MATH . Since (by REF ) each application of MATH lowers the power of MATH by REF, MATH will kill all the terms in MATH whose power of MATH is less than MATH, that is, all except the MATH term. On the other hand, we compute easily that MATH . We see, then, that MATH is a non-zero multiple of MATH. Since MATH is assumed invariant, MATH must contain this multiple of MATH, and so also MATH itself. But now it follows from REF that MATH is a non-zero multiple of MATH. Therefore MATH must also contain MATH for all MATH. Since these elements form a basis for MATH, we see that in fact MATH, as desired.
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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, when considering complex representations we are interested only in complex invariant subspaces. Now, suppose that MATH is irreducible as a representation of MATH. If MATH is a (complex) subspace of MATH which is invariant under MATH, then certainly MATH is invariant under MATH. Therefore MATH or MATH. Thus MATH is irreducible as a representation of MATH. On the other hand, suppose that MATH is irreducible as a representation of MATH, and suppose that MATH is a (complex) subspace of MATH which is invariant under MATH. Then MATH will also be invariant under MATH, for all MATH. Since every element of MATH can be written as MATH, we conclude that in fact MATH is invariant under MATH. Thus MATH or MATH, so MATH is irreducible as a representation of MATH.
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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 have at least one eigenvector.
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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 ``raising operator," because it has the effect of raising the eigenvalue of MATH by REF, and we call MATH the ``lowering operator." We know that MATH. Thus MATH or MATH . Thus MATH . Similarly, MATH, and so MATH so that MATH . This is what we wanted to show. As we have observed, MATH must have at least one eigenvector MATH (MATH), with some eigenvalue MATH. By the lemma, MATH and more generally MATH . This means that either MATH, or else MATH is an eigenvector for MATH with eigenvalue MATH. Now, an operator on a finite-dimensional space can have only finitely many distinct eigenvalues. Thus the MATH's cannot all be different from zero. Thus there is some MATH such that MATH but MATH . Define MATH and MATH. Then MATH . Then define MATH for MATH. By the second part of the lemma, we have MATH . Since, again, MATH can have only finitely many eigenvalues, the MATH's cannot all be non-zero. With the above notation, MATH . We proceed by induction on MATH. In the case MATH we note that MATH. Using the commutation relation MATH we have MATH . But MATH, so we get MATH which is the lemma in the case MATH. Now, by REF. Using REF and induction we have MATH . Simplifying the last expression give the Lemma. Since MATH can have only finitely many eigenvalues, the MATH's cannot all be non-zero. There must therefore be an integer MATH such that MATH for all MATH, but MATH . Now if MATH, then certainly MATH. Then by REF , MATH . But MATH, and MATH (since MATH). Thus in order to have MATH equal to zero, we must have MATH. We have made considerable progress. Given a finite-dimensional irreducible representation MATH of MATH, acting on a space MATH, there exists an integer MATH and non-zero vectors MATH such that (putting MATH equal to MATH) MATH . The vectors MATH must be linearly independent, since they are eigenvectors of MATH with distinct eigenvalues. Moreover, the MATH-dimensional span of MATH is explicitly invariant under MATH, MATH, and MATH, and hence under MATH for all MATH. Since MATH is irreducible, this space must be all of MATH. We have now shown that every irreducible representation of MATH is of the form REF . It remains to show that everything of the form REF is a representation, and that it is irreducible. That is, if we define MATH, MATH, and MATH by REF (where the MATH's are basis elements for some MATH-dimensional vector space), then we want to show that they have the right commutation relations to form a representation of MATH, and that this representation is irreducible. The computation of the commutation relations of MATH, MATH, and MATH is straightforward, and is left as an exercise. Note that when dealing with MATH, you should treat separately the vectors MATH, MATH, and MATH. Irreducibility is also easy to check, by imitating the proof of REF . (See REF .) We have now shown that there is an irreducible representation of MATH in each dimension MATH, by explicitly writing down how MATH, MATH, and MATH should act REF in a basis. But we have shown more than this. We also have shown that any MATH-dimensional irreducible representation of MATH must be of the form REF . It follows that any two irreducible representations of MATH of dimension MATH must be equivalent. For if MATH and MATH are two irreducible representations of dimension MATH, acting on spaces MATH and MATH, then MATH has a basis MATH as in REF and MATH has a similar basis MATH. But then the map MATH which sends MATH to MATH will be an isomorphism of representations. (Think about it.) In particular, the MATH-dimensional representation MATH described in REF must be equivalent to REF . This can be seen explicitly by introducing the following basis for MATH: MATH . Then by REF (MATH), and it is clear that MATH. It is easy to see that MATH. The only thing left to check is the behavior of MATH. But direct computation shows that MATH as required. This completes the proof of REF .
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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. Suppose, then, that the Proposition holds for all representations with dimension strictly less than MATH, and that MATH. If MATH is irreducible, then again we have a direct sum with only one summand, and we are done. If MATH is not irreducible, then there exists a non-trivial invariant subspace MATH. Taking MATH in the definition of complete reducibility, we see that there is another invariant subspace MATH with MATH and MATH. That is, MATH as a vector space. But since MATH and MATH are invariant, they can be viewed as representations in their own right. (That is, the action of our group or NAME algebra on MATH or MATH is a representation.) It is easy to see that in fact MATH is isomorphic to MATH as a representation. Furthermore, it is easy to see that both MATH and MATH are completely reducible representations, since every invariant subspace MATH of, say, MATH is also an invariant subspace of MATH. But since MATH is non-trivial (that is, MATH and MATH), we have MATH and MATH. Thus by induction MATH (as representations), with the MATH's irreducible, and MATH, with the MATH's irreducible, so that MATH.
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math-ph/0005032
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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 suppose that MATH. Since MATH is assumed invariant, MATH will be in MATH for any MATH. We need to show that MATH is perpendicular to MATH. Well, since MATH is unitary, then for any MATH . But MATH is assumed invariant, and so MATH. But then since MATH, MATH. This means that MATH for all MATH, that is, MATH. Thus MATH is invariant, and we are done.
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math-ph/0005032
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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 in fact MATH . That is, MATH is a unitary representation with respect to the inner product MATH. Thus MATH is completely reducible by REF .
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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 invariant. NAME invariance means that MATH for all MATH and for all NAME sets MATH, where MATH . It is a fact which we cannot prove here that every matrix NAME group has a left NAME measure, and that this measure is unique up to multiplication by a constant. (One can analogously define right NAME measure, and a similar theorem holds for it. Left NAME measure and right NAME measure may or may not coincide; a group for which they do is called unimodular.) Now, the key fact for our purpose is that left NAME measure is finite if and only if the group MATH is compact. So if MATH is a finite-dimensional representation of a compact group MATH acting on a space MATH, then let MATH be an arbitrary inner product on MATH, and define a new inner product MATH on MATH by MATH where MATH is left NAME measure. Again, it is easy to check that MATH is an inner product. Furthermore, if MATH, then by the left-invariance of MATH . So MATH is a unitary representation with respect to MATH, and thus completely reducible. Note that MATH is well-defined only because MATH is finite.
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REF .
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math-ph/0005032
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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 MATH and MATH are operators on MATH, then compute that MATH . This shows that MATH are equal on elements of the form MATH. Since every element of MATH can be written as a linear combination of things of the form MATH (in fact of MATH), MATH and MATH must be equal on the whole space.
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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.
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math-ph/0005032
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Using the product rule, MATH . This is what we wanted to show.
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math-ph/0005032
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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.
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math-ph/0005032
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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 for it not to have any non-trivial subspaces. This means that MATH must be one-dimensional. (Recall that we do not allow MATH to be zero-dimensional.)
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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 irreducible, we must have MATH or MATH. Thus MATH is either one-to-one or zero. Suppose MATH is one-to-one. Then the image of MATH is a non-zero subspace of MATH. On the other hand, the image of MATH is invariant, for if MATH is of the form MATH for some MATH, then MATH . Since MATH is irreducible and image-MATH is non-zero and invariant, we must have image-MATH. Thus MATH is either zero or one-to-one and onto. Proof of REF. Suppose now that MATH is an irreducible complex representation, and that MATH is a morphism of MATH to itself. This means that MATH for all MATH, that is, that MATH commutes with all of the MATH's. Now, since we are over an algebraically complete field, MATH must have at least one eigenvalue MATH. Let MATH denote the eigenspace for MATH associated to the eigenvalue MATH, and let MATH. Then for each MATH . Thus applying MATH to an eigenvector of MATH with eigenvalue MATH yields another eigenvector of MATH with eigenvalue MATH. That is, MATH is invariant. Since MATH is an eigenvalue, MATH, and so we must have MATH. But this means that MATH for all MATH, that is, that MATH. Proof of REF. If MATH, then by REF MATH is an isomorphism. Now look at MATH. As is easily checked, the composition of two morphisms is a morphism, so MATH is a morphism of MATH with itself. Thus by REF , MATH, whence MATH.
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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 REF we see that MATH . Thus on the one hand MATH, while on the other hand MATH. Let us compute MATH. By definition, MATH. But, MATH, where as usual MATH . We know that there is a basis MATH for MATH such that MATH is an eigenvector for MATH with eigenvalue MATH. This means that MATH is also an eigenvector for MATH, with eigenvalue MATH. Thus in the basis MATH we have MATH . But we are assuming the MATH is odd! This means that MATH is an odd integer. Thus MATH, and in the basis MATH . Thus on the one hand, MATH, while on the other hand MATH. This is a contradiction, so there can be no such group representation MATH. REF m is even. We will use the following: There exists a NAME group homomorphism MATH such that REF MATH maps MATH onto MATH REF MATH and REF the associated NAME algebra homomorphism MATH is an isomorphism which takes MATH to MATH. That is, MATH. CASE: Now consider the representations MATH of MATH. I claim that if MATH is even, then MATH. To see this, note that MATH . Thus MATH. But as in REF . Only, this time, MATH is even, and so MATH is an integer, so that MATH. Since MATH, MATH for all MATH. According to REF , for each MATH, there is a unique pair of elements MATH such that MATH. Since MATH, it makes sense to define MATH . It is easy to see that MATH is a NAME group homomorphism (hence, a representation). By construction, we have MATH . Now, if MATH denotes the NAME algebra representation associated to MATH, then it follows from REF that MATH . But the NAME algebra homomorphism MATH takes MATH to MATH, that is, MATH. So MATH, or MATH. Thus MATH, which is what we want to show.
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math-ph/0005032
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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.
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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 by the formula MATH . It is trivial to check that MATH is a homomorphism. Furthermore, differentiation shows that the associated NAME algebra homomorphism is MATH for all MATH. Then by REF in the Theorem, we must also have MATH, that is, MATH for all MATH. But this shows that MATH and MATH are equivalent. Point REF follows immediately from Point REF, by taking MATH. MATH .
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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 local inverse which maps a neighborhood MATH of the identity into the NAME algebra MATH. On this neighborhood MATH we can define MATH by MATH . That is MATH . (Note that if there is to be a homomorphism MATH as in REF, then on MATH, MATH must be MATH.) It follows from REF to the NAME formula that this MATH is a ``local homomorphism." That is, if MATH and MATH are in MATH, and if MATH happens to be in MATH as well, then MATH. (See the discussion at the beginning of REF.) REF : Define MATH along a path. Recall that when we say MATH is connected, we really mean that MATH is path-connected. Thus for any MATH, there exists a path MATH with MATH and MATH. A compactness argument shows that there exists numbers MATH such that MATH for all MATH between MATH and MATH. In particular, for MATH, we have MATH for MATH. Thus we can define MATH by REF for MATH. Now, for MATH we have by REF MATH. Moving the MATH to the other side, this means that for MATH we can write MATH with MATH. If MATH is to be a homomorphism, we must have MATH . But MATH has already been defined, and we can define MATH by REF . In this way we can use REF to define MATH for MATH. Proceeding on in the same way, we can define MATH successively on each interval MATH until eventually we have defined MATH on the whole time interval MATH. This in particular serves to define MATH. CASE: Prove independence of path. In REF , we ``defined" MATH by defining MATH along a path joining the identity to MATH. For this to make sense as a definition of MATH we have to prove that the answer is independent of the choice of path, and also, for a particular path, independent of the choice of partition MATH. To establish independence of partition, we first show that passing from a particular partition to a refinement of that partition doesn't change the answer. (A refinement of a partition is one which contains all the points of the original partition, plus some other ones.) This is proved by means of the NAME formula. For example, suppose we insert an extra partition point MATH between MATH and MATH. Under the old partition we have MATH . Under the new partition we write MATH so that MATH . But (as noted in REF ), REF of the NAME formula REF implies that for MATH and MATH sufficiently near the identity MATH . Thus the right sides of REF are equal. Once we know that passing to a refinement doesn't change the answer, we have independence of partition. For any two partitions of MATH have a common refinement, namely, the union of the two. Once we know independence of partition, we need to prove independence of path. It is at this point that we use the fact that MATH is simply connected. In particular, because of simple connectedness, any two paths MATH and MATH joining the identity to MATH will be homotopic with endpoints fixed. (This is a standard topological fact.) Using this, we want to prove that REF gives the same answer for MATH and MATH. Our strategy is to deform MATH into MATH in a series of steps, where during each step we only change the path in a small time interval MATH, keeping everything fixed on MATH and on MATH. Since we have independence of partition, we can take MATH and MATH to be partition points. Since the time interval is small, we can assume there are no partition points between MATH and MATH. Then we have MATH where MATH is defined as in REF . But notice that our value for MATH depends only on MATH and MATH, not on how we get from MATH to MATH! Thus the value MATH doesn't change as we deform the path. But if MATH doesn't change as we deform the path, neither does MATH, since the path isn't changing on MATH. Since MATH and MATH are homotopic with endpoints fixed, it is possible (by a standard topological argument) to deform MATH into MATH in a series of small steps as above. CASE: Prove that MATH is a homomorphism, and is properly related to MATH. Now that we have independence of path (and partition), we can give a simpler description of how to compute MATH. Given any group element MATH, MATH can be written in the form MATH with each MATH in MATH. (This follows from the (path-)connectedness of MATH.) We can then choose a path MATH which starts at the identity, then goes to MATH, then to MATH, and so on to MATH. We can choose a partition so that MATH. By the way we have defined things MATH . But MATH so MATH . Now suppose that MATH and MATH are two elements of MATH and we wish to compute MATH. Well, write MATH . Then MATH . We see then that MATH is a homomorphism. It remains only to verify that MATH has the proper relationship to MATH. But since MATH is defined near the identity to be MATH, we see that MATH . Thus MATH is the NAME algebra homomorphism associated to the NAME group homomorphism MATH. This completes the proof of REF .
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REF .
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math-ph/0005032
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MATH is simply connected.
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REF .
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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; thus MATH is invariant. Thus MATH can be viewed as an operator on MATH. Again, since we are over MATH, the restriction of MATH to MATH must have at least one eigenvector MATH with eigenvalue MATH. But then MATH is a simultaneous eigenvector for MATH and MATH with eigenvalues MATH and MATH.
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math-ph/0005032
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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, and all of the eigenvalues are integers.
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math-ph/0005032
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Apply REF to the restriction of MATH to MATH, and to the restriction of MATH to MATH.
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math-ph/0005032
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The definition of a root tells us that we have the commutation relation MATH. Thus MATH A similar argument allows us to compute MATH.
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math-ph/0005032
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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 vector corresponding to the root MATH, then MATH maps the weight space corresponding to MATH into the weight space corresponding to MATH. Thus MATH is invariant under the action of all of the root vectors, namely, under the action MATH and MATH. Since MATH is certainly invariant under the action of MATH and MATH, MATH is invariant. Thus by irreducibility, MATH.
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math-ph/0005032
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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. By definition, MATH is invariant under MATH and MATH. It is thus also invariant under MATH. Now, REF tells us that applying a root vector MATH to a weight vector MATH with weight MATH gives either zero, or else a new weight vector with weight MATH. Now, by assumption, MATH is a weight vector with weight MATH. Furthermore, MATH and MATH are root vectors with roots MATH and MATH, respectively. (See REF .) Thus each application of MATH or MATH subtracts MATH or MATH from the weight. In particular, each non-zero element of the form REF is a simultaneous eigenvector for MATH and MATH. Thus MATH is invariant under MATH and MATH. To show that MATH is invariant under MATH and MATH, we argue by induction on MATH. For MATH, we have MATH. Now consider applying MATH or MATH to a vector of the form REF . Recall the commutation relations involving a MATH or MATH and a MATH or MATH: MATH . Thus (for MATH and MATH equal to REF or REF) MATH, where MATH is either MATH or MATH or zero. Hence (for MATH equal to REF or REF) MATH . But MATH is in MATH by induction, and MATH is in MATH since MATH is invariant under MATH and MATH. Finally, MATH is invariant under MATH since MATH. Thus MATH is invariant. Since by REF contains MATH, we must have MATH. Since MATH is a root vector with root MATH and MATH is a root vector with root MATH, REF tells us that each element of the form REF is either zero or a weight vector with weight MATH. Thus MATH is spanned by MATH together with weight vectors with weights lower than MATH. Thus MATH is the highest weight for MATH. Furthermore,every element of MATH can be written as a multiple of MATH plus a linear combination of weight vectors with weights lower than MATH. Thus the weight space corresponding to MATH is spanned by MATH; that is, the weight space corresponding to MATH is one-dimensional.
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math-ph/0005032
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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 weights. It follows that there must exist a weight MATH such that there is no weight MATH with MATH. This says that there is no weight higher than MATH (which is not the same as saying the MATH is highest). But if there is no weight higher than MATH, then for any non-zero weight vector MATH with weight MATH, we must have MATH . (For otherwise, say, MATH will be a weight vector with weight MATH.) Since MATH is assumed irreducible, the smallest invariant subspace containing MATH must be the whole space; therefore the representation is highest weight cyclic. MATH .
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math-ph/0005032
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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 occur in some MATH. On the other hand, REF says that the weight space corresponding to MATH is one-dimensional, that is, MATH is (up to a constant) the only vector in MATH with weight MATH. Thus MATH must contain MATH. But then that MATH is an invariant subspace containing MATH, so MATH. Thus there is only one term in the sum REF , and MATH is irreducible.
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math-ph/0005032
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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 invariant subspace of MATH which contains the vector MATH. By definition, MATH is a highest weight cyclic representation, therefore irreducible by Proposition. CASE: Consider the two ``projection" maps MATH, MATH and MATH, MATH. It is easy to check that MATH and MATH are morphisms of representations. Therefore the restrictions of MATH and MATH to MATH will also be morphisms. Clearly neither MATH nor MATH is the zero map (since both are non-zero on MATH). Moreover, MATH, MATH, and MATH are all irreducible. Therefore, by NAME 's Lemma, MATH is an isomorphism of MATH with MATH, and MATH is an isomorphism of MATH with MATH. Thus MATH.
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math-ph/0005032
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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 classification of the irreducible representations of MATH. Let MATH be any finite-dimensional representation of MATH. Let MATH be an eigenvector for MATH with eigenvalue MATH. If MATH, then MATH is a non-negative integer. Now, if MATH is an irreducible representation of MATH with highest weight MATH, and if MATH is a weight vector with weight MATH, then we must have MATH. (Otherwise, MATH wouldn't be highest.) Thus applying the above result to the restrictions of MATH to MATH and to MATH shows that MATH and MATH must be non-negative.
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math-ph/0005032
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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 representations.) The standard representation of MATH is an irreducible representation with highest weight MATH, as is easily checked. To construct an irreducible representation with weight MATH we modify the standard representation. Specifically, we define MATH for all MATH. Using the fact that MATH, it is easy to check that MATH so that MATH is really a representation. (This is isomorphic to the dual of the standard representation, as defined in REF.) It is easy to see that MATH is an irreducible representation with highest weight MATH. Let MATH denote MATH acted on by the standard representation, and let MATH denote a weight vector corresponding to the highest weight MATH. (So, MATH.) Let MATH denote MATH acted on by the representation REF , and let MATH denote a weight vector for the highest weight MATH. (So, MATH.) Now consider the representation MATH where MATH occurs MATH times, and MATH occurs MATH times. Note that the action of MATH on this space is MATH . Let MATH denote this representation. Consider the vector MATH . Then applying REF shows that MATH . Now, the representation MATH is not irreducible (unless MATH or MATH). However, if we let MATH denote the smallest invariant subspace containing the vector MATH, then in light of REF , MATH will be highest weight cyclic with highest weight MATH. Therefore by REF , MATH is irreducible with highest weight MATH. Thus MATH is the representation we want.
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math-ph/0005032
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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.
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math-ph/0005032
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Equivalent representations must have the same weights and the same multiplicities.
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math-ph/0005032
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Same as for MATH.
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math/0005001
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The first claim follows by replacing everything by absolute values in REF. The second claim follows by applying REF with MATH replaced by MATH.
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math/0005001
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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.
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math/0005001
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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 duality.
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math/0005001
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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.
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math/0005001
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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.
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math/0005001
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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 . Similarly with the roles of REF reversed. The former claim then follows by interpolation. The latter claim then follows from the former claim and the Comparison principle.
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math/0005001
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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 those pairs for which MATH does not vanish and applying REF , the claim follows.
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math/0005001
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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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