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math/0003217
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Use REF for the two-form to straightforwardly take an exterior power and compute the total number of summands of one kind (with fixed MATH). Suppose the product contains some MATH; then it must come in pair with one of the adjoining edges - let it be MATH in the above notations, so there are four choices. Then MATH must come (if it is in the sum) with one of the two edges at its other end, and then the other edge at its other end must come with still another, and so on. Thus we have one factor of four, and many factors of two. Each new factor of four appears if we encounter an edge in MATH, so the total number of summands of one kind is at most MATH. Recalling the MATH in the two-form, MATH.
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math/0003217
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We can construct at most two paths around punctures ``going to the left" including the edge MATH - a path can be constructed by deciding at which end of MATH we start building it. Let one of these paths go through edges MATH, and the other - through MATH; let MATH and MATH be the sums of the simplicial coordinates of the edges in these paths. Then using the formula for MATH in terms of MATH-lengths of sectors REF we see that MATH . Since MATH in the domain MATH, we get MATH so that MATH. If there were only one path going through MATH, that is, if MATH, we would have MATH, and thus MATH, which is even better.
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math/0003217
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Assume for contradiction that MATH. Note that MATH by clearing the denominators and extracting full squares. Similarly MATH and both inequalities hold if MATH. Denote by MATH and MATH the edges at the other end of MATH. From above it then follows that MATH using REF . But this is just REF for edges MATH, MATH and MATH, and thus MATH. Suppose MATH; then it follows that MATH, and we can apply a similar argument to the edges MATH and MATH at the other end of MATH to obtain MATH. Continuing this process inductively, and assuming at each step that MATH, we end up with an infinite strictly increasing sequence of edges MATH, which is rather hard to achieve on a finite graph. Thus assuming that one triangle inequality among MATH-lengths fails, we have arrived at a contradiction.
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math/0003217
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For any vertex MATH denote by MATH, MATH and MATH the edges containing it, with MATH being the maximal among the three. Then using the triangle inequalities we have MATH .
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math/0003217
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We apply the Stoke's theorem multiply. Indeed, MATH .
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math/0003217
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Indeed, recall REF the definition of the MATH-lengths MATH in the usual notations. REF states that MATH is twice the sum of MATH-lengths of the sectors it traverses. What matters for us is that it is a sum of some MATH-lengths. For any MATH-length we have MATH . Thus for MATH we have MATH . Applying this trick to each of MATH yields the theorem.
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math/0003217
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Since MATH and MATH link, we have MATH. Thus MATH .
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math/0003217
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Start from the ends of the chain, and apply the lemma to eliminate edges one by one, coming from the ends towards MATH.
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math/0003217
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Split the integral into two parts depending on whether MATH or MATH. The computation for them is identical; if MATH, we have MATH .
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math/0003217
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The wheel is a collection of chains MATH. Keep the two edges of MATH in between which some chain MATH ends (existent by definition of a wheel), and integrate the other ones out using REF - by this we pick up a factor of MATH for each edge. Then use REF above to integrate out the last two remaining edges of the chain MATH - here we pick up a factor of MATH for two edges, that is, MATH per edge. Performing induction in MATH finishes the proof.
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math/0003217
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Indeed, recall that MATH by REF . Thus MATH . Since MATH for MATH, for the second summand we have MATH . For the first summand we compute MATH . Combining the above estimates, we get the lemma.
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math/0003217
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Construct a wheel MATH starting from edge MATH. If this wheel is not the whole graph, consider a chain MATH ending on MATH by at least one end. If it ends on MATH by the other end also, consider another such chain MATH and so on, until we either exhaust the graph, or get a chain MATH which has an end not on MATH. Then construct a wheel MATH at the other end of the chain MATH. If the union of these two wheels and the chains constructed is still not the whole graph, we repeat the process. As a result, we decompose the graph into a disjoint union of wheels MATH and chains MATH for MATH and MATH such that the chains MATH have both ends on MATH for MATH or MATH, and that MATH connects MATH and MATH for MATH. Then use REF to eliminate all edges of chains MATH except the terminal ones, which end at MATH. Using REF , we can then include these terminal edges of MATH's while implementing the proof of REF - integrating out the edges of MATH one by one. Doing this, we get a factor of MATH for eliminating two edges, instead of the smaller factor of MATH, which we were getting originally. At the last step of integrating over the edges of MATH we use the edge at the end of MATH for REF , and thus reduce the problem to MATH wheels. Induction in MATH then yields the desired result.
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math/0003217
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Combining the results of REF , we see that the integral in question is bounded above by MATH . Using REF , we can integrate out all variables except MATH, by acquiring an extra factor of MATH. Remembering the factor of MATH for choosing the minimal edge at each vertex, our final upper bound becomes MATH .
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math/0003217
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For MATH we construct a relation MATH in the following way. Consider two distinct punctures MATH and MATH connected by an edge MATH of a triangulation MATH - if such did not exist, that is, if all edges emanating from a puncture went back to the puncture itself, it would not be a triangulation of the surface. Shrinking MATH to a point, and collapsing triangles on both sides of MATH into arcs of a triangulations, thus identifying MATH and MATH, produces a new triangulation MATH. We define MATH to be the set of all pairs MATH obtained in such a way. Now consider some MATH. Any graph in MATH such that MATH can be reconstructed from MATH by ``blowing up" a pair of edges emanating from a vertex to triangles. Since there are MATH ways to choose a pair of edges of MATH, there are at most MATH triangulations MATH such that MATH. The argument works for all MATH, and thus MATH. Applying this argument until we decrease the number of punctures to one, and then utilizing NAME 's asymptotic computation for that case finishes the proof.
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math/0003219
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It is a standard fact that a vector in MATH is primitive if and only if it is the bottom row of some element of MATH. This implies MATH is surjective and that MATH. The rest is clear.
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math/0003219
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We have MATH by REF , and this equals MATH by REF .
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math/0003219
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Let MATH, MATH, MATH, and MATH be as in REF , with MATH assumed orientable. Assume MATH as well as MATH. Let MATH. Then MATH. By definition of MATH, MATH if and only if MATH. On the other hand, MATH, so MATH and MATH have the same orientation number if and only if MATH.
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math/0003219
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Apply REF twice.
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math/0003219
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We have MATH by the definition of MATH. Hence MATH. Thus the expression in REF is a square either of MATH or of MATH.
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math/0003219
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When MATH is orientable, this follows from REF , merely because MATH. Now assume MATH is non-orientable. Let MATH, with MATH. As we have said above, there is some MATH with MATH. By REF , MATH. The element MATH is in MATH by REF , so it is in MATH; clearly it carries MATH to itself while reversing orientation. Hence MATH acts non-trivially on MATH, meaning MATH.
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math/0003221
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Define the base algebra to be the target subalgebra of MATH, that is, MATH, and let MATH. Then, clearly, MATH, so that images of MATH and MATH commute. The comultiplication MATH regarded as a map to MATH is coassociative and compatible with the multiplication. It is a bimodule map since MATH for all MATH and MATH. We also have the compatibility condition: MATH . Next, let MATH, then we have MATH and MATH for all MATH, that is, MATH is a MATH module map, also MATH where we used the antipode axioms of a weak NAME algebra. Thus, MATH satisfies the counit axiom. The antipode MATH is an algebra anti-homomorphism satisfying MATH. The section MATH is given by MATH . Finally, we verify the antipode properties : MATH . Thus, MATH is a NAME algebroid. If MATH is a morphism between weak NAME algebras, then it is clear from our definitions that the pair MATH gives a morphism between the corresponding NAME algebroids.
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math/0003221
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Clearly, MATH is an algebra homomorphism. Its coassociativity follows from REF . Observe that the relations between MATH and MATH yield the following identities (recall that MATH is invertible when restricted on the counital subalgebras, since the latter are finite-dimensional) : MATH . Here and in what follows we write MATH and MATH. Using these properties and REF we check that MATH is still a counit for CASE : MATH . We proceed to verify the rest of the axioms of a weak NAME algebra, writing MATH : MATH for all MATH. The axioms involving MATH are checked using identities REF : MATH . We define a new antipode by MATH and compute (writing MATH for additional copies of MATH and MATH) : MATH where we used axioms of MATH and properties of the counital maps. Note that for MATH we get the identity MATH that is, MATH is the inverse of MATH. It can be proven similarly that MATH . Finally, let us write MATH and compute MATH whence MATH is a weak NAME algebra.
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math/0003221
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Clearly, we have MATH, MATH, and MATH. The relations between MATH and counit are straightforward. We also compute MATH . One verifies the identities REF in a similar way.
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math/0003221
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We check first two identities, leaving the rest as an exercise to the reader. Using the formulas for MATH and MATH we compute : MATH where we used the zero weight property of MATH.
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math/0003221
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The twist relations involving counit are obvious, for the rest we have, using identities from REF : MATH and one checks other identities similarly.
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math/0003221
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We need to show that the structure operations of MATH are obtained by transposing the corresponding operations of MATH. For the unit and counit this is obvious. The product of the elements MATH and MATH of MATH, where MATH, can be found by the evaluation against the elements of MATH using REF : MATH . On the other hand, we have MATH where MATH. Comparing the last two relations we conclude that the multiplication in MATH is opposite to the one induced by the comultiplication in MATH. Finally, the antipode defined by REF satisfies MATH for all MATH and MATH, while the transpose of MATH gives MATH which completes the proof.
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math/0003221
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Since MATH and MATH are finite dimensional, MATH is an isomorphism if and only if its image coincides with MATH, that is, MATH which happens precisely when MATH for all MATH. Since a NAME algebra MATH is a free MATH-module, we see that the rank of MATH has to be equal to MATH and, therefore, to MATH, by REF .
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math/0003221
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Let us write MATH, where MATH is the sum of all terms having the MATH-degree MATH in the first component. Using the structure of the MATH-matrix of MATH, we can write REF as a finite system of linear equations labeled by degree MATH: MATH where MATH stands for the terms involving MATH for MATH. Thus, REF can be solved recursively, starting with MATH, and the solution is unique provided that the operator MATH is invertible in MATH. Let us show that this operator is diagonalizable and compute its eigenvalues. For all MATH, MATH, and MATH we have MATH and therefore MATH whence the eigenvalues of MATH in MATH are the numbers MATH where MATH and MATH are characters of MATH. In particular, each MATH is a MATH-th root of unity. Clearly, the eigenvalue of the operator MATH in MATH is MATH. Putting these numbers together, we conclude that REF is invertible in MATH if and only if MATH for all MATH which is the case for generic MATH.
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math/0003221
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For all MATH we have MATH since MATH, whence MATH.
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math/0003221
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We have MATH hence MATH by the uniqueness of the solution of REF .
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math/0003221
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This statement amounts to showing that MATH . If we denote MATH and MATH then the above equality translates to MATH which follows from the quantum NAME equation after cancelling the first factor.
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math/0003221
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It is enough to show that such a solution MATH is unique for the equation MATH. Let us write MATH where MATH is the sum of all elements having MATH-degree MATH in the first component and MATH in the third one. Then the equation MATH transforms to the system MATH for all MATH and MATH such that MATH, where MATH. As in REF one can check that the operator MATH is invertible for generic MATH, therefore the above system can be solved recursively and the solution is unique.
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math/0003221
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Let us denote the left-hand side of REF by MATH and the right-hand side by MATH. We show that both MATH and MATH are solutions of the system MATH, then the result will follow from REF . We have : MATH where we used that MATH, properties of the MATH-matrix, and that MATH commutes with MATH. To establish that MATH note that both MATH and MATH are solutions of the equation MATH and therefore are determined uniquely by their parts of zero degree in the third component. Thus it suffices to compare these parts : MATH . The relations between MATH and MATH are obvious.
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math/0003221
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We directly compute : MATH . The identities MATH and MATH are clear.
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math/0003221
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It follows from the explicit REF for the universal MATH-matrix MATH of MATH, defining REF of MATH, and REF for MATH that MATH the ``coefficients" MATH are MATH-valued functions on MATH, and MATH run over MATH-tuples of non-negative integers. Note that the terms of MATH occurring in MATH are linearly independent from the rest and so are those occurring in MATH and MATH, where MATH is the MATH-tuple with MATH in the position corresponding to the single root MATH and MATH's elsewhere. Hence, the subspaces MATH belong to the image of MATH. In all of the three above cases we will show that MATH is invertible in MATH (and, therefore, MATH are non-zero scalars for all MATH) and that the generators of MATH lie in the algebra generated by the above subspaces MATH. Clearly, MATH is invertible, whence MATH is spanned by the elements MATH, that is, MATH for all MATH and MATH for MATH. Next, MATH is the coefficient with MATH in MATH. To determine it, note that by REF we have MATH where MATH stand for the terms of degree MATH in the first component. We use the recursive relation REF to find MATH : MATH from where we obtain MATH and consequently MATH which is invertible for all generic MATH. It follows that MATH is spanned by the elements MATH, whence MATH. Finally, MATH therefore, MATH is invertible and MATH.
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math/0003221
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We saw in REF that the image of MATH contains all the generators of the algebra MATH, therefore MATH. Since MATH is finite dimensional, MATH is an isomorphism.
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math/0003223
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Put MATH for MATH, MATH, and MATH for MATH, MATH. Our assumptions on MATH, MATH, and MATH imply that MATH for MATH if MATH and MATH for MATH otherwise; notice that MATH for MATH. Now it follows from the definition of MATH and the formulae in CITE that MATH. As MATH is a weight of a MATH-restricted MATH-module, we have MATH. This yields the lemma.
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math/0003223
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Set MATH and let MATH. It is clear that MATH as MATH. Define subgroups MATH and MATH as follows. For MATH set MATH if MATH and MATH if MATH, MATH, MATH (we have MATH for MATH and MATH for MATH). For MATH, MATH, or MATH put MATH, MATH, or MATH, respectively, and MATH . Next, set MATH for MATH and MATH for MATH or MATH (here MATH for MATH and MATH). One easily observes that the sets of roots in brackets used to define MATH and MATH yield bases of the systems MATH and MATH, respectively. Denote these bases by MATH. In all cases MATH is conjugate to MATH in MATH. We have MATH, MATH, MATH, or MATH for MATH, MATH, MATH, or MATH, respectively. It is clear that the subgroups MATH and MATH commute. Set MATH. Let MATH, MATH, and MATH. It is not difficult to conclude that MATH is a maximal unipotent subgroup in MATH and MATH is such a subgroup in MATH. We can assume that MATH, MATH and MATH. We shall write a weight MATH in the form MATH where MATH is the restriction of MATH to MATH. Set MATH. It is clear that MATH for each MATH-module MATH. Taking this into account, it is not difficult to conclude the following. If MATH is a filtration of MATH, MATH, MATH, and MATH, then MATH . First suppose that MATH. Since passing to the dual representation does not influence the NAME form of MATH, one can assume that MATH for some MATH if MATH. As for MATH-large representations the estimates of CITE hold; we also assume that MATH is not MATH-large. Hence MATH for all MATH and long roots MATH (for all MATH if MATH or MATH). By the formulae for the maximal roots of the classical groups in CITE, this forces that MATH . Now we proceed to construct two composition factors MATH and MATH of the restriction MATH such that MATH and MATH. This will be done for almost all MATH. In exceptional cases we shall find one factor MATH such that MATH. By REF , this would yield the assertion of the theorem. Let MATH be a nonzero highest weight vector. Put MATH. The vector MATH generates an indecomposable MATH-module MATH with highest weight MATH. Using REF , one can deduce that MATH for all MATH. Here for MATH we take into account that MATH. Hence MATH is MATH-restricted. Now assume that either MATH, or MATH for some MATH. For such representations we construct another weight vector MATH that is fixed by MATH. Set MATH for MATH, MATH; otherwise take MATH as in REF . First suppose that MATH for some MATH REF . Choose maximal such MATH and put MATH. Now let MATH for all MATH REF . Our assumptions on MATH imply that MATH for some MATH; furthermore, one can take MATH for MATH and MATH for MATH. Choose minimal such MATH and set MATH if MATH or MATH and MATH for MATH and MATH. It follows from CITE that in all cases MATH. Using CITE and analyzing the roots in MATH and MATH and the weight system MATH, we get that MATH fixes MATH in all situations. Here it is essential that the case MATH with MATH is excluded. In the latter case we cannot assert that MATH fixes MATH. Set MATH, MATH. Now it is clear that MATH generates an indecomposable MATH-module MATH with highest weight MATH. We claim that MATH is MATH-restricted. Write down all the situations where MATH for some MATH. We have MATH in REF if MATH and MATH or MATH and MATH and in REF for MATH and all MATH; and MATH for MATH both in REF with MATH and in REF . In REF we also have MATH if MATH and MATH if MATH. In REF one gets MATH if MATH. In all other situations we have MATH. Now apply REF to conclude that MATH is MATH-restricted. Set MATH, MATH, MATH, and MATH, MATH. Obviously, MATH is a composition factor of MATH. It is well known that MATH. It is clear that MATH. Since MATH, we have MATH if MATH. So by REF , MATH . It follows from CITE that MATH. Hence MATH. One easily observes that MATH and MATH cannot both be trivial MATH-modules. Our assumptions on MATH and CITE imply that the dimension of a nontrivial irreducible MATH-module is at least MATH. In the exceptional case where MATH and MATH we need to evaluate MATH. First let MATH. As above, MATH. This implies that MATH contains a dominant weight MATH and MATH is greater than the size of the orbit of MATH under the action of the NAME group of MATH. The latter is equal to MATH for our values of MATH. By REF , this yields the assertion of the theorem for almost all MATH. It remains to consider the case where MATH and MATH. It is well known that then the restriction MATH is a direct sum of MATH-modules MATH. Since MATH, we get MATH and MATH. Now suppose that MATH. By the NAME tensor product theorem CITE, MATH can be represented in the form MATH where MATH is the NAME morphism of MATH associated with raising elements of MATH to the MATH-th power and all MATH. It is clear that the morphism MATH does not influence the NAME form of MATH. Hence one can assume that MATH where MATH for some MATH and both MATH and MATH are nontrivial. Set MATH, MATH, MATH and define by MATH the restriction of MATH to MATH. Now it follows from the definitions of MATH and MATH that MATH. By CITE, MATH and MATH. First suppose that MATH or MATH. Set MATH if MATH and MATH otherwise and denote by MATH the remaining representation from the pair MATH. Then MATH and CITE implies that MATH. Let MATH be the value of MATH if one formally sets MATH. Then by CITE, MATH which settles the case under consideration. Now assume that both MATH and MATH. Then MATH and MATH. Since MATH, we have MATH. Arguing as for MATH-restricted MATH, we can and shall suppose that MATH for some MATH if MATH. Put MATH and construct the composition factors MATH, MATH, of the restriction MATH as for MATH-restricted MATH before. Transfer the notation MATH, MATH, and MATH, MATH, to MATH. Again we have the exceptional case MATH and MATH where we do not construct MATH and consider MATH only. Obviously, MATH. As before, we deduce that MATH. By CITE, MATH and MATH. Let MATH be the number of NAME blocks of the maximal size in the canonical form of MATH as an element of MATH, MATH. Looking at the realizations of MATH as tensor products, one easily observes that MATH. Set MATH, MATH and consider MATH as MATH-modules in the natural way. In the general case the MATH-module MATH has a filtration two of whose quotients are isomorphic to MATH and MATH, respectively. In the exceptional case MATH is a quotient of a submodule in MATH. Observe that MATH. Using CITE that describes the canonical NAME form of a tensor product of unipotent blocks, we obtain that MATH and MATH. As for MATH-restricted MATH, we show that MATH if MATH and MATH and conclude that MATH in the general case and MATH in the exceptional cases with MATH. Now REF completes the proof.
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math/0003223
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Let MATH, MATH, MATH, and MATH be such as in the assertion of the proposition. Assume that MATH. Therefore we have MATH. Set MATH and denote by MATH the weight subspace of weight MATH in the MATH-module MATH. It is clear that the NAME group of MATH interchanges MATH and MATH; hence MATH. Put MATH, MATH, MATH, and MATH. Set MATH for MATH, MATH for MATH or MATH, and MATH for MATH. Let MATH be a nonzero highest weight vector and put MATH. By CITE, MATH. We need a subgroup MATH which can be defined as follows. Put MATH and MATH. The canonical NAME forms of MATH in the standard realizations of MATH and MATH are well known. We have MATH since the dimension of the first realization is at most MATH due to our assumptions. Taking into account these NAME forms, one easily obtains the values of MATH, MATH, and using REF , deduces the following facts: MATH for MATH and MATH, MATH for MATH and MATH, and MATH in all other cases where MATH is the subgroup defined in the proof of REF ; MATH, MATH; and MATH. Next, observe that MATH for MATH and MATH, MATH for MATH, and MATH for MATH. Our construction of the vector MATH shows that MATH fixes MATH if MATH. This forces that MATH generates an indecomposable MATH-module MATH with highest weight MATH. Then one immediately concludes that MATH. This yields that MATH for MATH and MATH, MATH for MATH, MATH for MATH, and MATH otherwise. It is clear that MATH. Denote by MATH the subset of weights of the form MATH and by MATH the irreducible MATH-module with highest weight MATH. By NAME 's theorem CITE, for each MATH the dimension of the weight subspace MATH coincides with that of the weight subspace in MATH whose weight differs from MATH by the same linear combination of the simple roots. Hence MATH does not depend upon MATH. Set MATH. Since MATH is an irreducible MATH-module and MATH, observe that MATH is a linear span of vectors of the form MATH. Now, analyzing the weight structure of MATH, we conclude that MATH and MATH. This implies that MATH REF and MATH do not depend upon MATH. It follows from CITE that MATH . Let MATH. Then MATH. Obviously, MATH if MATH and MATH for MATH. Thus REF implies that MATH in the first case and MATH in the second case. This forces that MATH in the first case and MATH in the second case. We have seen before that MATH with MATH, MATH, or MATH. Hence one can take MATH and MATH to complete the proof.
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math/0003232
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Here we must use some toric geometry. The ideal MATH defines a subset of the lattice MATH. The dual set of the ideal, MATH, defines a rational polytope MATH in the dual lattice MATH. To find a ``monomial log resolution" of MATH is to find a sequence of toric blowups which refine the polytope MATH in the appropriate sense. This can be done because MATH is rational. The blowups required are exactly those required torically to resolve the singularity of the space MATH. REF indicates how this process might be used to resolve the cusp. See CITE, for more information on toric resolutions.
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math/0003232
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The lemma hinges on three basic ideas. First, MATH is certainly a map from MATH to MATH, but because it is linear, MATH extends to all of MATH in a natural way. As a map of real vector spaces, MATH is continuous because it is linear. Second, for each of the effective toric divisors, or ``coordinate planes" MATH in MATH . The equality is strict because the blowups permitted are monomial. Third, we have the standard equation MATH, where MATH is the integral closure of MATH. We will prove the lemma by proving REF first for integral points MATH, then for rational points, and finally for real points. We will prove REF by using the strict positivity of MATH. Finally, we'll deduce REF by continuity. Let MATH be an integer point of MATH not in MATH. Then MATH so MATH. If instead MATH has rational coordinates, then we can clear denominators. Let MATH be integral. MATH, so MATH. (Here we have used the just-proved integer case, as well as the fact that the resolution MATH resolving MATH also resolves MATH.) Dividing by MATH gives MATH. If MATH has real coordinates, choose a rational MATH also not in MATH. MATH, so MATH. A standard convexity argument proves that if MATH then MATH. To prove REF, let MATH be in the interior of MATH. Choose MATH and MATH with MATH. Then MATH. MATH, and MATH is strictly positive in every coordinate, so MATH is in the interior of MATH. REF follows from the continuity of the map MATH.
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math/0003236
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The first three results are immediate from REF . For the fourth we apply the formula given in the theorem: MATH which gives the required result.
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math/0003236
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Recall that MATH. This determines MATH by naturality. First of all observe that a basis element MATH REF of MATH is primitive if and only if MATH. To complete the proof we evaluate the coproduct on the non-primitive height REF basis elements and on all the basis elements of greater height in this dimension. For MATH there are no basis elements of height greater than REF. For simplicity we use the reduced coproduct MATH so that MATH is primitive when MATH. Straightforward calculations give the following: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. The lemma follows immediately from these results. Similar calculations give the results for MATH and MATH.
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math/0003236
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These are immediate from the NAME squares in MATH REF and the NAME relations (see CITE).
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math/0003236
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By REF MATH is a linear combination of the elements given by REF . But since, by REF , it is annihiliated by MATH and MATH only elements of the given form can arise.
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math/0003236
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Suppose that MATH corresponds to MATH. Then MATH by REF . The result follows by REF .
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math/0003236
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Suppose that MATH. Notice that the height REF part of MATH is given by MATH (see REF) and so MATH using the obvious inclusion MATH as the elements of height REF. Furthermore, MATH is determined by the normal NAME - NAME numbers of MATH and so by the bordism class of MATH (see REF ). Suppose that MATH and MATH are two immersions of bordant manifolds MATH and MATH corresponding to MATH, MATH respectively. Since MATH and MATH are bordant manifolds MATH. It follows that MATH by REF . However MATH is not primitive (by REF ) and so not spherical. Hence MATH and so MATH as required. For MATH the above proof has to be modified to take account of height REF elements. This leads in this case to MATH which gives the same result.
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math/0003236
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Suppose that MATH and that MATH is an immersion corresponding to MATH. If MATH in REF then MATH. By REF , MATH if and only if MATH has coefficient MATH when MATH is written in terms of the basis MATH. This occurs if and only if the NAME - NAME number MATH. For, by REF , MATH is given by the NAME product MATH in MATH. By naturality, this product can be evaluated in MATH using MATH and the fact that MATH, the MATH-th elementary symmetric polynomial. In this case MATH and so the NAME product MATH is given by the coefficient of MATH. Hence MATH if and only if MATH. In the case MATH, by REF REF , there exists an embedding MATH of a manifold MATH bordant to MATH. Suppose that MATH corresponds to MATH. Then MATH since-MATH is an embedding with no double points. Hence, by REF , MATH and so MATH. For MATH put MATH where MATH. Then by NAME 's immersion theorem there are immersions MATH and MATH and the product of these gives an immersion MATH. Finally a standard verification shows that the normal NAME - NAME number MATH. The argument for MATH is almost identical. The manifold MATH either is a boundary or is bordant to MATH depending on the value of the normal NAME - NAME number MATH. The presence of the height REF term MATH in the NAME image shows that in this case any immersion of a non-boundary has an odd number of triple points.
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math/0003236
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For MATH, MATH and so there exists an immersion MATH by NAME 's immersion theorem CITE. Furthermore, for such MATH there is an immersion MATH with double point manifold of odd NAME characteristic REF . Taking the connected sum of this immersion and MATH gives an immersion of MATH with a double point self-intersection surface with NAME characteristic of opposite parity to that of the double point self-intersection surface of MATH. Hence both parities can arise.
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math/0003236
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Evaluating the dual NAME operations on the elements of REF we obtain CASE: MATH; CASE: MATH; CASE: MATH. Hence MATH is in the submodule spanned by the set MATH . For MATH and MATH it follows that MATH and so, by NAME 's embedding theorem REF , MATH is bordant to a manifold MATH which has an embedding MATH. Let MATH correspond to MATH. Then MATH since MATH is an embedding. Furthermore, MATH since these are determined by the bordism class. It follows, by REF , that MATH since this is necessarily a primitive class in MATH. Hence MATH where, by REF , MATH gives the parity of the NAME characteristic of the double point self-intersection surface of the immersion. For MATH, given any immersion MATH the manifold MATH is necessarily a boundary since this is true of all REF - manifolds and so, if MATH is the corresponding element, MATH. The formula for MATH then follows from the value for the NAME image coming from REF . The presence of the height REF term MATH when MATH has odd NAME characteristic indicates that such immersions have an odd number of quadruple points.
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math/0003236
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This is immediate from the preceding proposition since, by REF , the bordism class of MATH is determined by MATH.
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math/0003236
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By NAME 's embedding theorem REF , MATH is bordant to a manifold MATH which has an embedding MATH. The double point self-intersection surface of MATH is empty. Hence, by REF , the double point self-intersection surface of MATH is a boundary.
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math/0003236
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Suppose that MATH and MATH are two immersions of bordant manifolds corresponding to MATH, MATH respectively. As in the proof of REF , since MATH and MATH are bordant manifolds MATH. It follows that MATH and so lies in the submodule of MATH spanned by the set given by REF . We consider the non-zero elements of this submodule in turn. First of all, by REF , neither MATH nor MATH is primitive and so neither of these elements can be spherical. This leaves the element MATH, which is primitive. The element MATH is not spherical. Accepting this for the moment, it follows that MATH and so MATH as required.
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math/0003236
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This proposition can be thought of as describing how detection by a certain NAME - NAME invariant corresponds to detection by a NAME - NAME invariant. The proposition follows from work of NAME and NAME REF . The special case of MATH appears as REF and the justification given there extends to the general case.
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math/0003236
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Suppose, for contradiction, that there exists an element MATH such that MATH . NAME gives a natural isomorphism MATH. Since the corresponding homology suspension MATH kills NAME products and commutes with the Kudo - NAME operations, MATH . Hence, by REF , MATH where MATH is an element such that MATH. We now show that this is impossible. First of all notice that, since MATH, MATH for MATH. This means that away from dimension MATH we can do calculations in MATH by calculating in MATH. In particular, MATH is generated by MATH and so MATH. Here MATH is the universal NAME - NAME class. Secondly, we can evaluate the action of the NAME squaring operations on MATH using the fact that the operations commute with suspension and using their action on MATH which is determined by the NAME formula (see REF ). In evaluating the operations we use the fact that MATH which means that we can write MATH for some positive integer MATH. Then, MATH . This contradicts REF and so proves the lemma, completing the proof of REF and so the proof of REF in the case of MATH.
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math/0003236
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To prove that decomposable manifolds of dimension MATH embed up to bordism it is sufficient to prove that each product MATH, with MATH, MATH and MATH, embeds up to bordism in MATH. To do this we make use of the following result which follows easily from the NAME embedding theorem. If the manifold MATH immerses in MATH, the manifold MATH embeds in MATH and MATH then MATH embeds in MATH. Since MATH and MATH, MATH it follows that MATH or MATH. Suppose without loss of generality that MATH. Then, by NAME 's embedding theorem CITE, MATH is bordant to a manifold which embeds in MATH. In addition, by the NAME immersion theorem, MATH immerses in MATH. Hence, by the lemma, MATH is bordant to a manifold which embeds in MATH. The numerical condition of the lemma is automatically satisfied: MATH since MATH. The final part of the proposition is immediate from REF as in the proof of REF .
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math/0003236
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We make use of particular indecomposable manifolds constructed by NAME CITE. In dimension MATH this manifold MATH is formed from the product MATH by identifying MATH with MATH. In NAME 's notation this is MATH. He shows that the cohomology ring of MATH is given by MATH where MATH and MATH, and the total tangent NAME - NAME class of MATH is given by MATH . This implies that the total normal NAME - NAME class of MATH is given by MATH from which it follows that MATH, MATH and MATH. Hence MATH is an orientable manifold such that the normal NAME - NAME number MATH. Let MATH be an immersion corresponding to MATH. Using the argument in the proof if REF it follows from the non-vanishing of this NAME - NAME number that MATH has coefficient REF when MATH is written in terms of the basis MATH. Hence from REF it follows that either MATH or MATH . However, since MATH is orientable, MATH lies in the image of MATH in MATH. Since MATH the element MATH does not come from MATH which eliminates the first of the above possibilities. This proves the formula in the proposition for the manifold MATH. However, since any other indecomposable manifold MATH is bordant to MATH modulo a decomposable manifold the same formula holds for an immersion of MATH by REF . The NAME characteristic of the double point self-intersection surface is now given by REF since, by REF , MATH.
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math/0003236
|
Almost everything is given by REF . The final observation about the NAME - NAME number is clear since, by REF , MATH occurs in MATH if and only if MATH does and by the argument in the proof of REF this is equivalent to MATH.
|
nlin/0003005
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The second-order system obtained by differentiating REF with respect to MATH is MATH . On the other hand, the NAME - NAME equations associated to the Lagrangian REF are MATH . If MATH is closed then MATH so that REF are indeed identical.
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nlin/0003034
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To establish the identity REF it suffices to show that the operation MATH satisfies the Jacoby identity. In accordance with REF , any element MATH can be uniquely represented as MATH, where MATH, MATH. Let MATH . Using the formula MATH and the fact that MATH is a NAME algebra we conclude that MATH where the index MATH on the left hand side means the projection onto MATH parallel to MATH. If we apply REF to project the Jacoby identity for MATH onto MATH, we make certain that the NAME identity for the operation MATH holds. In order to prove REF we need the following relation MATH where the left hand side is defined by REF . It can be checked by a direct calculation. Rewriting REF in terms of MATH-bracket with the help of REF we see that REF follows from the NAME identity for MATH. It remains to prove the second part of REF . Let MATH be a MATH-algebra. Put MATH where MATH is a NAME algebra generated by all operators of left multiplication in MATH. Recall that the left multiplication operator MATH is defined as follows MATH. The vector space MATH becomes a NAME algebra if we define MATH . Obviously, the bracket REF is skew-symmetric. One can easily show that the identities REF are equivalent to the NAME identity for REF . It follows from REF that the decomposition REF is a MATH-gradation. To define a decomposition REF we take for MATH the set MATH and MATH for MATH. As we see from REF , MATH is a subalgebra in MATH. For MATH and MATH thus defined, REF is of the form MATH. This relation is fulfilled according to REF .
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nlin/0003034
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We do not give a complete proof of REF . Its first part can be proved in the same manner as the first part of REF . We explain only how to construct MATH for a given MATH-algebra. We take for MATH the NAME algebra MATH of all linear endomorphisms of MATH. The vector space MATH becomes a MATH-graded NAME algebra if we define MATH . It is not difficult to show that REF implies that REF the vector space MATH generated by all elements of the form MATH is a NAME subalgebra in MATH and REF the vector space MATH and defined above subalgebras MATH and MATH satisfy REF .
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quant-ph/0003074
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Let MATH be a pure state of MATH, and let MATH be the corresponding ultrafilter. CASE: Suppose that MATH for all compact subsets MATH of MATH. For each MATH, let MATH. Then, MATH and since MATH is an ultrafilter, MATH. However, MATH and therefore MATH is not closed under countable meets. CASE: If MATH for some compact set MATH in MATH, then MATH contains MATH where MATH is the family of open neighborhoods of some point MATH. For each MATH, let MATH . Then, MATH for each MATH. However, MATH since MATH has NAME measure zero. Thus, MATH and MATH is again not closed under countable meets.
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quant-ph/0003074
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Let MATH be given as above. For each MATH, let MATH . From the Ultrafilter Extension Theorem, MATH is contained in a Boolean ultrafilter MATH if and only if the meet of any finite collection of elements in MATH is nonzero. Let MATH. If MATH for each MATH, then MATH is an open set that contains MATH. If MATH, and MATH for MATH, then MATH is an open set that contains MATH, and since MATH, MATH is a nonempty open set. In either case, the open set MATH is nonempty. Since NAME measure does not vanish on any nonempty open set, we have MATH . Thus, there is an ultrafilter MATH that contains MATH. Since MATH when MATH, it follows that MATH when MATH.
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quant-ph/0003074
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We will consider the case where MATH. In light of the lemma, it will suffice to construct a family MATH of open subsets of MATH such that MATH when MATH and such that MATH for all MATH. For each MATH, let MATH. For each MATH, let MATH, and let MATH . Note that MATH for all MATH, and in fact MATH has nonempty interior. Thus, if we set MATH then MATH is a nonempty open set. We must show that MATH when MATH and that MATH for all MATH. Suppose that MATH. We may assume that MATH, in which case MATH, and MATH . It is easy to verify that the sequence MATH is in MATH for each MATH. But MATH and since MATH, it follows that MATH.
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quant-ph/0003074
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Let WUF (weak-ultrafilter principle) denote the statement that there is a free ultrafilter on MATH, and let PS denote the statement that there is a pure state on MATH. We must show that ZF+DC+MATH-PS is consistent. Since ZF+DC+MATH-WUF is consistent CITE, it will suffice to show that ZF+MATH(MATH-WUF). Thus, suppose that PS holds; that is, there is an ultrafilter MATH on MATH. By REF , MATH is not closed under countable meets; that is, there is a family MATH of elements of MATH such that MATH for all MATH, but MATH. (It's important to note that REF uses no choice axiom stronger than DC.) Define a family MATH of subsets of MATH by MATH . It is easily verified that MATH is an ultrafilter in MATH that contains all subsets of MATH whose complements are finite.
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quant-ph/0003074
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See REF of Ref. REF.
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quant-ph/0003074
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Suppose first that MATH. Then, MATH and so MATH. It then follows easily by induction that MATH when MATH for some MATH. Suppose now that MATH is a diadic rational; that is, MATH for some MATH, where MATH. Note then that MATH and so MATH . Finally, let MATH be an arbitrary element in MATH. Then, MATH for some strictly decreasing sequence MATH of diadic rationals. Now, for every MATH, there is some MATH such that MATH for all MATH, and therefore MATH for all MATH. We may also assume that MATH when MATH. Thus, MATH . Since this is true for all MATH, it follows that MATH.
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quant-ph/0003074
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Recall that the NAME compactification MATH of MATH is the unique compact NAME space such that every bounded continuous function MATH can be extended uniquely to a continuous function MATH. Let MATH denote the set of bounded, uniformly continuous functions from MATH into MATH. Since a uniform limit of elements in MATH is again in MATH, it follows that there is a unique compact NAME space MATH such that every MATH has a unique extension to a continuous function MATH. We may refer to MATH as the uniform compactification of MATH. Let MATH denote the set of continuous functions from MATH into MATH. Thus, there is an algebraic isomorphism from MATH onto MATH and we may identify MATH with the family of functions in MATH with range MATH. We show that every (pure) state MATH of MATH has a unique extension to a (pure) state MATH of the MATH-algebra MATH. We may then appeal to the result that any pure state MATH on MATH is of the form MATH for some MATH CITE. Let MATH be a state of MATH. Let MATH denote the set of functions from MATH into MATH. If MATH, then MATH and we may define MATH . Using REF , it is easy to see that MATH is homogenous with respect to positive real numbers, and it is straightforward to verify that MATH is additive over positive functions. Thus, MATH is an additive function from MATH into MATH. It follows then that MATH extends uniquely to a linear mapping from MATH into MATH CITE. Suppose now that MATH is a pure state of MATH, and let MATH be the unique extension of MATH to MATH as defined above. Suppose that MATH, where MATH are states of MATH and MATH. Then MATH and MATH are states of MATH, and MATH. However, since MATH is a pure state of MATH, we have MATH and since extensions are unique, it follows that MATH. Thus, MATH is a pure state of MATH. Thus, there is some MATH such that MATH for each MATH. We show that if MATH, then the state MATH does not converge. Indeed, since MATH is dense in MATH, there is a net MATH in MATH such that MATH. Let MATH be a function that vanishes at infinity. We show that MATH. Let MATH be given. Then, there is a MATH such that MATH for all MATH with MATH. However, since the set MATH is compact in MATH and MATH, there is some MATH such that MATH for all MATH. Since MATH, MATH. Since this is true for any MATH, it follows that MATH and MATH.
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quant-ph/0003101
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It is easily verified that applying each MATH with probability MATH to a qubit puts that qubit in the totally mixed state MATH (no matter if it is entangled with other qubits). Operator MATH just applies this treatment to each of the MATH qubits, hence MATH for every MATH.
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quant-ph/0003101
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This is easily verified: applying MATH and MATH, each with probability REF/REF, puts any qubit from MATH in the totally mixed state. Operator MATH does this to each of the MATH qubits individually.
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quant-ph/0003101
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If MATH can be written as a mixture of MATH-states, then MATH .
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quant-ph/0003101
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Let MATH, MATH be the superoperator of REF , and let MATH. Since MATH for all MATH-qubit states MATH, we have that MATH and MATH are unitarily related in the way mentioned in REF: there exists a unitary MATH matrix MATH such that for all MATH we have MATH . We can view the set of all MATH matrices as a MATH-dimensional vector space, with inner product MATH and induced matrix norm MATH (as done in CITE). Note that MATH if MATH is unitary. The set of all MATH forms an orthonormal basis for this vector space, so we get: MATH . Hence MATH and MATH.
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quant-ph/0003101
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For ease of notation we assume without loss of generality that MATH uses no ancilla, so we assume MATH is a MATH-qubit state and omit MATH (this does not affect the proof in any way). We first show that MATH whenever MATH and MATH (MATH is well-defined but somewhat of an abuse of notation, since the matrix MATH is not a density matrix). This is implied by the following REF equalities: MATH . The first and second equality imply MATH, the first and third equality imply MATH. Hence MATH. We now define MATH and show that it is a PQC. Intuitively, MATH will map every state from MATH to a tensor product of MATH . NAME states by mapping pairs of bits to one of the four NAME states. The second bits of the pairs are then moved to the second half of the state and randomized by applying MATH to them. Because of the entanglement between the two halves of each NAME state, the resulting MATH-qubit density matrix will be MATH. More specifically, define MATH with MATH as in REF . Also define MATH. It remains to show that MATH for all MATH: MATH . In the step marked by MATH we used that MATH if MATH.
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quant-ph/0003101
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Diagonalize the ancilla as MATH, so MATH. First note that REF entropy REF imply: MATH . Secondly, note that MATH . Combining these two inequalities gives the theorem.
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quant-ph/0003101
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Sufficiency follows from REF and necessity from REF .
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quant-ph/0003101
|
Sufficiency follows from REF and necessity from REF .
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quant-ph/0003144
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It is instructive to start with the special case in which for some MATH, there exist two or more distinct values of MATH for which MATH. For this case let the set MATH be an orthonormal basis of a separable NAME space. Define a subset MATH of models satisfying REF as all models of the form MATH, where MATH with MATH and MATH arbitrary real-valued functions. By invoking REF , one checks that any such model has the claimed perfect fit; yet the set contains many unitarily inequivalent models, which predict conflicting statistics for some possible quantum measurement. This proves the special case. For the general case, modify the definitions above to MATH where MATH, for all MATH the projection MATH has dimension greater than REF, and MATH ranges over all unit vectors of the eigenspace defined by MATH. In particular, for any MATH, dim REF can be as large as one pleases. Then even if there is only one outcome that is ever recorded, there are still many unitarily inequivalent models that perfectly fit the data.
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quant-ph/0003144
|
For any set of measured outcomes, there exists a perfectly fitting model MATH of the form in the proof of REF for the general case, for which MATH, and a corresponding perfectly fitting model MATH such that MATH. For these two models, MATH .
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quant-ph/0003144
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The models MATH and MATH under the stated condition are unitarily equivalent to a pair of models that differ only in MATH with MATH. The proposition then follows from REF .
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cond-mat/0004005
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CASE: First, let us note that MATH is a smooth function with MATH . This implies that there is at least one solution MATH. CASE: For MATH we have MATH at least as long as MATH, that is, as long as MATH. CASE: For MATH, we note that MATH . Thus there are no solution in MATH. CASE: Finally, for MATH, we have MATH therefore there is a unique MATH such that MATH.
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gr-qc/0004057
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Given a smoothing kernel MATH, the embedding of the delta function is given by MATH . So MATH which is clearly undefined in the limit as MATH.
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gr-qc/0004057
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While this may seem trivial, it is possible for zero to have a nowhere zero representative. However as the representative is negligible, it decays faster than any power of MATH. So its inverse cannot be moderate. We make these statements formal. Suppose that MATH such that MATH. Then this must also be true for their representatives, so MATH, where MATH. Now, MATH is moderate so MATH such that MATH . Now, MATH is negligible, so MATH with MATH . Choice of MATH large enough guarantees that REF cannot both be true, thereby contradicting our assumption that z was invertible.
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gr-qc/0004057
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Suppose that MATH, for some MATH. Then substituting the point value MATH in this equation implies that the generalised number REF has an inverse, MATH. But this is a contradiction, as the generalised number zero is not invertible by REF , thus proving our result.
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gr-qc/0004057
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Let MATH be given. Then MATH as MATH is REF outside of MATH, and as MATH is bounded in this region, this limit clearly converges to zero. So we have MATH. Now, MATH which is also clearly zero, thus providing our result.
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gr-qc/0004057
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For simplicity we assume that MATH. Then, MATH . Thus we have shown that MATH. A similar proof applies for MATH.
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gr-qc/0004057
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Suppose that MATH such that MATH. Let MATH. Then we have MATH . Then we clearly have that MATH and MATH. As MATH is smooth, by the Intermediate Value Theorem, we have that MATH such that MATH. Now, let REF be represented by MATH. Then, in particular, MATH will satisfy MATH with MATH . So we can choose some MATH such that MATH whenever MATH. But we have that MATH, and in particular in MATH we have MATH. This is a contradiction, and thus MATH has no natural inverse.
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gr-qc/0004057
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We have MATH . Now, MATH has compact support so the integrand is well-behaved as MATH, and as the integration region vanishes, the limit gives zero, thus the result is immediate.
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gr-qc/0004057
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We have MATH . Choice of MATH arbitrary with MATH implies that the first term must vanish, similarly if MATH then the second term vanishes, which implies that MATH (using the standard definition of the support of a distribution).
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hep-th/0004029
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The uniqueness of such a function follows immediately from the NAME - NAME Theorem. To show the existence we shall consider the following function MATH . The set of REF and the constraint MATH (the latter means that MATH is an elliptic function) form the system of MATH equations on the functions MATH . For generic data this system is non-degenerate. Then it has the only solution (up to the permutations) and therefore defines the function MATH uniquely.
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hep-th/0004029
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Consider a function MATH. It's straightforward to check that the function MATH satisfy all defining properties of the function MATH. The uniqueness of MATH implies that MATH.
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hep-th/0004029
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To obtain these equations one has to divide REF by MATH and compare the residues of the both sides of the obtained equation at the points MATH .
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math-ph/0004004
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MATH . The fact that MATH follows from the fact that the MATH- and MATH-mode are independent of each other. For a calculation of the explicit expression for MATH, see CITE. MATH . This is simply the two-point function of the state MATH, see CITE.
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math-ph/0004004
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The first statement follows again from the general theory on normal fluctuation operators CITE. Also, it is an easy calculation to show that MATH . Because of the presence of the energy gap MATH, this can be written as: MATH . Therefore, using the fact that MATH is an equilibrium (KMS) state, one gets MATH . The explicit expression for the NAME two-point function MATH then follows from MATH .
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math-ph/0004004
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This follows immediately from the relations MATH .
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math-ph/0004004
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This follows from the time invariance of MATH, that is, MATH: MATH and the equations of motion REF .
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math-ph/0004004
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We compute this quantity using the correlation REF , rewritten in the form MATH . We take for MATH the operator MATH and then let MATH, and use REF . One gets MATH and by the virial theorem REF : MATH . Analogously MATH . On the other hand, MATH and hence MATH . After substitution in REF one gets MATH or alternatively MATH .
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math-ph/0004004
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As shown in CITE, two fluctuation operators MATH are equal in the algebra of fluctuation operators whenever MATH that is, whenever the variance of the difference MATH of the operators vanishes. This is expressing in a mathematical rigorous setting, the phenomenon of coarse graining on the level of fluctuations. Therefore we calculate MATH . One finds, using the explicit knowledge of the state MATH REF : MATH hence leading to the following equality, as operators: MATH . From REF , it follows that MATH and from REF , MATH . Therefore MATH . This implies necessarily MATH and again the equality of the operators MATH in the ground state, as a result of coarse graining.
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math-ph/0004007
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In principle we adopt the method presented in CITE, and modify it appropriately where necessary. Let us first recall that if MATH is suitably chosen, with MATH on a neighbourhood of the interval MATH, the operator MATH has the same spectrum in MATH as MATH itself. Furthermore, the symbol MATH of MATH has an asymptotic expansion that coincides on MATH with that of MATH. Since below we are localising in energy to the interval MATH so that we can consider MATH instead of MATH, from now on we simply suppose that MATH. Let now MATH be given with MATH on MATH and such that the spectrum of MATH in supp MATH is discrete. This might require to choose MATH small enough. Then consider the (energy localised) quantum mechanical time evolution operator MATH . Up to an error of order MATH in trace norm, this operator can be approximated by a semiclassical NAME integral operator with kernel MATH if MATH is small enough (see, for example, CITE). We remark that here the amplitude MATH takes values in MATH, thus it also represents the internal (that is, spin) degrees of freedom. As explained in CITE, the phase MATH in REF is then given as the solution of the NAME equation MATH . In leading semiclassical order the transport equation for MATH can be solved by a separation of the internal degrees of freedom from the translational ones. The latter lead to the expression known from the scalar case (see, for example, CITE), whereas the modifications required by the matrix character of MATH are provided by the solution MATH of REF , see CITE. For the present purpose, however, one only needs the initial condition MATH. In a next step we consider MATH with NAME transform MATH such that MATH where MATH denotes the operator trace on MATH. Approximating MATH with the help of REF , in leading semiclassical order one then has to calculate MATH with the method of stationary phase. The stationary points of the phase MATH are given by MATH such that MATH is a periodic point of the Hamiltonian flow MATH with period MATH. Since we have assumed that MATH is not a critical value for MATH, the periods of MATH on MATH do not accumulate at zero, see CITE. Hence, if the support of MATH is chosen small enough, the manifold of critical points contributing to REF is given by MATH. In analogy to CITE we thus obtain MATH . Moreover, since we require MATH to be such that no MATH is a critical value and all MATH are compact, REF holds true with MATH replaced by MATH uniformly in MATH. In a last step we now apply the NAME Lemma of CITE, which takes non-negative weights into account. To this end we have to ensure that MATH is non-negative. However, since MATH is bounded, this can always be achieved by adding a suitable constant. Moreover, according to our above choice of MATH we find that MATH and MATH for all MATH. Therefore MATH . Repeating the above reasoning with the identity instead of MATH, one can express the number MATH of eigenvalues in MATH semiclassically, MATH . Thus, REF together finally yield the NAME limit REF .
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math-ph/0004007
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An application of the NAME inequality on the right-hand side of REF yields an upper bound for the quantity REF that reads MATH . In order to determine the limit as MATH of REF one can now apply REF , which gives MATH . Quantum ergodicity then follows, if the bound on the right-hand side can be shown to vanish. This will indeed be possible in the limit MATH. In order to achieve this we first employ the ergodicity of MATH in that we choose MATH in the relation REF , and thus find MATH for MATH-almost all MATH and MATH-almost all MATH. In terms of the principal symbol REF of the auxiliary operator MATH this means MATH . Our next goal is to calculate the right-hand side of REF . To this end we represent the hermitian MATH matrix MATH as a linear combination of MATH and the NAME matrices MATH, that is, MATH, where MATH. We then recall that for every MATH the adjoint map MATH is defined as MATH. Thus, MATH can be expanded in terms of the NAME matrices MATH such that MATH. The map MATH that results in this way can be identified as the universal (twofold) covering of MATH by MATH. Therefore MATH . We now show that the second term on the right-hand side of REF vanishes. To this end we multiply the integral over MATH, which yields a MATH matrix, with an arbitrary orthogonal matrix MATH from the left. We then exploit the fact that there exists MATH such that MATH, together with the left invariance of the NAME measure, in order to conclude that MATH . Since hence the MATH matrix represented by the integral over MATH is invariant under left multiplication by an arbitrary element of MATH, it must be the zero matrix. Therefore, the right-hand side of REF vanishes. We now choose some MATH such that REF holds and therefore obtain that MATH holds for MATH-almost all MATH. Thus, after an integration over MATH, this together with REF implies that MATH . According to a standard argument in the proof of quantum ergodicity for scalar Hamiltonians, see CITE, the vanishing of MATH in the semiclassical limit implies the existence of a density-one subsequence MATH such that REF holds. Another standard, diagonal construction then ensures that a subsequence can be chosen that is independent of the observable, see CITE.
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math-ph/0004014
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Let MATH be the function given by MATH . By our Scaling Assumption, MATH on MATH. Note that both MATH and MATH are convex, non-positive and not identically vanishing with value REF at zero. Consequently, MATH and MATH are continuous and strictly negative in MATH. Moreover, by applying NAME 's inequality to the definition of MATH, we have that MATH and MATH are both non-decreasing functions. Next we shall show that MATH tends to a finite non-zero limit for all MATH. Let us pick a MATH and a MATH and consider the identity MATH which results by comparing REF with the ``time" parameter interpreted once as MATH and next time as MATH. Invoking the monotonicity of MATH, it follows that MATH . This implies that MATH is bounded away from zero, because we have MATH where MATH stands for minimum. Since MATH was arbitrary, MATH is also uniformly bounded, by replacing MATH with MATH. Let MATH be defined for each MATH as a subsequential limit of MATH, that is, MATH with some (MATH-dependent) MATH. By our previous reasoning MATH is non-zero, finite and, for all MATH, it solves for MATH in the equation MATH . Here we were allowed to pass to the limiting function MATH on the left-hand side of REF because MATH is continuous and the scaling limit REF is uniform on compact sets in MATH. But MATH is non-decreasing while MATH is strictly decreasing, so the solution to REF is unique. Hence, the limit MATH exists in MATH for all MATH. It is easily seen that MATH is multiplicative on MATH, that is, MATH. Since MATH for MATH, by the same token we also have that MATH is non-decreasing. These two properties imply that MATH and that MATH for any MATH, and MATH, MATH integer such that MATH. Consequently, MATH with MATH. By plugging this back into REF and setting MATH we get that MATH . REF are thus established by putting MATH, which is REF. Clearly, MATH, in order to have the correct monotonicity properties of MATH and MATH. To prove also the second statement in REF, we first write MATH which, after taking the logarithm, dividing by MATH, and noting that MATH as MATH, allows us to conclude that MATH . The limit for general MATH is then proved again by sandwiching MATH between MATH and MATH and invoking the monotonicity of MATH.
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