paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0008220 | Without loss of generality MATH. Suppose MATH and MATH . Then MATH must be equal to one of MATH or MATH; but then MATH a contradiction. |
math/0008220 | We will show that the sum MATH converges to the desired integral REF , where the sum is over MATH. Similar arguments hold for MATH, MATH, and MATH. The value REF is then a weighted average of these sums, with weights MATH. Since MATH (see REF ), the weights are bounded in absolute value (less than MATH) and sum to MATH... |
math/0008220 | The graph MATH has MATH-edges. For MATH let MATH be the MATH-valued random variable indicating the presence of the MATH-th MATH-edge in a random matching. Then MATH, and so MATH . We have MATH and so MATH. In the case when one of MATH is greater than the sum of the others, we know from REF that MATH converges to MATH o... |
math/0008220 | Let MATH denote the coefficient of MATH in MATH. As we computed earlier, on the toroidal graph MATH the MATH-probability of a MATH-edge (respectively, MATH-, MATH-, MATH-edge) is given by MATH (respectively, MATH, MATH, MATH). The expected number of MATH-edges is MATH. Let MATH . Let MATH be the corresponding set of ma... |
math/0008220 | We show that the Hessian (matrix of second derivatives) is negative definite, that is, MATH, and MATH, for all MATH, except at the four points MATH or MATH. A computation using REF gives MATH . A second differentiation yields MATH and this quantity is strictly negative except at the points MATH. A similar calculation h... |
math/0008222 | The number MATH is an algebraic integer, so its MATH-adic absolute value is at most MATH. To determine how much smaller it is, first notice that MATH . In order for MATH to reduce to MATH modulo MATH, we must have MATH . However, this is impossible unless MATH, because MATH has order MATH in the residue field. Since MA... |
math/0008222 | In this proof, we will write MATH to indicate an unspecified power of MATH. Because the product in question is real and the only real power of MATH is MATH, we will in several cases be able to see that factors of MATH equal MATH without having to count the MATH's. Start by observing that MATH . (To prove the last line,... |
math/0008222 | The proof is based on the observation that for any non-zero numbers, the power sums of their reciprocals are minus the NAME coefficients of the logarithmic derivative of the polynomial with those numbers as roots, that is, MATH . To apply this fact to MATH, define MATH . Then MATH . This function is invariant under int... |
math/0008225 | The first step of the proof will be to show that the domain of MATH is indeed the whole space MATH. Using the positivity of the MATH operator we show that MATH which means that MATH. We need NAME 's inequality CITE MATH where MATH, MATH is the dimensionality of the space and for us MATH. Applying this inequality we obt... |
math/0008225 | For the type of spaces that we work with, a set is compact iff it is closed and we can build a MATH-net for any positive number MATH. The MATH-net is a finite set MATH such that for each MATH there is an element MATH verifying MATH. Thus compactness is equivalent to the possibility of building an arbitrarily good appro... |
math/0008229 | By construction the MATH - invariants defining MATH are decomposable (as sums of products of one - dimensional classes), hence they are elements in MATH which restrict to zero on the subgroup MATH. Thus the extension restricted to this subgroup defines an elementary abelian subgroup of rank equal to MATH which is the k... |
math/0008229 | Let MATH, we know that MATH is NAME and by a standard result in commutative algebra it follows that under that condition the cohomology will be a free module over any polynomial subring over which it is finitely generated (see CITE). Hence we only need to prove that MATH is a finitely generated MATH-module. To prove th... |
math/0008229 | Indeed MATH can be expressed as a central extension MATH which by construction yields a mod MATH LHS spectral sequence which collapses at MATH. |
math/0008229 | We consider the NAME - NAME - NAME spectral sequence with mod MATH coefficients for the group extension MATH . Recall that MATH, where, if MATH denotes the usual NAME operator, MATH for MATH. If MATH is the natural projection, it induces an isomorphism between the subrings of the mod MATH cohomologies generated by REF ... |
math/0008229 | Let MATH denote the MATH - NAME subgroup of MATH, that is, the kernel of the natural projection MATH. This group fits into a commutative diagram of extensions: MATH . Note that MATH is MATH. As before, the MATH - invariants defining MATH are trivial mod MATH, hence MATH is also an exterior algebra on MATH one - dimensi... |
math/0008229 | By a result due to NAME and NAME (see CITE), the LHS spectral sequence associated to the defining extension above collapses at MATH if coefficients are taken in MATH for MATH sufficiently large. The improved lower bound MATH was obtained in CITE. This implies the collapse of the mod MATH spectral sequence, the statemen... |
math/0008229 | Indeed, the integral MATH term can only have MATH - torsion for finitely many primes MATH, the result follows from the mod MATH reduction sequence for MATH. |
math/0008230 | In the quotient we have MATH, while MATH. Also MATH while MATH, which is equal to MATH in the quotient. Thus MATH and these two copies of MATH commute with each other, giving a copy of MATH in the quotient. Next, note that MATH exchanges MATH, MATH, and also exchanges MATH, MATH, hence the two copies of MATH above, and... |
math/0008230 | We note that the lift of MATH is given as MATH . On the other hand, note that MATH commutes with MATH, MATH, consequently with each of the remaining four generators, while the first two generators commute with the third and fourth, and MATH, MATH, and all three have the central element MATH in common. The verification ... |
math/0008230 | The only thing that needs to be pointed out is that if there were a MATH which did not contain the central element MATH, (and hence a MATH obtained by adjoining MATH), then it would project non-trivially to one of the four MATH's in the quotient MATH. But we have identified the lifts of these groups as copies of groups... |
math/0008230 | We use the description of MATH as MATH where we write MATH . But we can replace this copy of MATH by MATH and MATH is defined as the identity on MATH and the correspondence above on MATH. |
math/0008230 | First we check the result for MATH. We have MATH and this is a presentation of MATH. The same calculations result for MATH using the automorphism above. Moreover, MATH commutes with MATH while MATH inverts it. So this change cancels out in the squares for the pair MATH, and we have verified that the groups asserted to ... |
math/0008230 | The normalizer of MATH is obtained by adjoining MATH, MATH. Thus there are three degree two extensions of MATH in MATH. The extension by MATH is MATH. Clearly, the extension by MATH does not give a MATH. Finally, consider the extension by MATH. Replace the second MATH by MATH . Then these two copies of MATH commute wit... |
math/0008230 | For any triple of indecomposable modules MATH, MATH, MATH appearing in the direct sum decompositions of MATH, MATH, and MATH above, we have MATH. Thus MATH. This shows that the map above is an isomorphism on the socle, hence it is injective. The NAME series of the two sides are equal, by a calculation, so we have an is... |
math/0008230 | Consider the long exact sequence in cohomology arising from the short exact sequence MATH. The maps MATH are just the restriction homomorphisms for MATH, which can be shown to be surjective in all degrees MATH. This implies that MATH is isomorphic to the kernel of the restriction map MATH for all MATH. This means that ... |
math/0008230 | MATH and MATH are interchanged by an outer automorphism of MATH. |
math/0008230 | There is an exact sequence MATH. |
math/0008230 | There is an exact sequence MATH. |
math/0008230 | There is an exact sequence MATH, and we consider the associated long exact sequence in cohomology. Since MATH, we see that MATH is one-dimensional if MATH. Furthermore, the induced maps MATH are surjective for all MATH. This, plus the fact that the socle of MATH is two-dimensional, determines the NAME series. For the i... |
math/0008230 | Use the exact sequence MATH. |
math/0008230 | A computer calculation using MAGMA for MATH shows that through degree REF the coefficients of the polynomial MATH agree with the ranks of the cohomology. Now according to the preceding tables, all algebra generators occur by this degree. Hence using the multiplicative structure of the spectral sequence we infer that it... |
math/0008230 | First it was shown that the elements of MATH are in the kernels of restriction by computing the restriction maps on the elements. Then for each degree MATH the restriction maps MATH for MATH in the list was written as a linear transformation and the intersection of the null spaces was computed to be zero . |
math/0008230 | From the lemma we see that MATH for MATH. But by the NAME series REF we have that MATH has the same dimension as MATH. On the other hand we know that MATH is generated by elements of degree at most MATH. So MATH, since all generators are in MATH. So MATH induces a surjective homomorphism MATH. But again because the NAM... |
math/0008240 | First suppose that the algebraic number MATH of negative complex points with respect to MATH is zero. Then, according to REF , the adjunction equality is fulfilled and, by REF, we can find an almost complex structure MATH homotopic to MATH and hence to MATH such that MATH is pseudoholomorphic with respect to MATH. If c... |
math/0008240 | For the sake of simplicity, let us assume that there is only one double point MATH. First consider the case that MATH is positive. Let MATH and choose an orientation preserving chart MATH around MATH such that CASE: MATH CASE: For small disks MATH in MATH around MATH, MATH,MATH and the restrictions MATH map the orienta... |
math/0008240 | We have a decomposition MATH, to which we can apply REF to obtain MATH . By REF , MATH. Substituting this into the last equation leads to the desired result. |
math/0008240 | As demonstrated in CITE, we can find an immersed surface in MATH having MATH positive double points, MATH negative double points and representing the class MATH. In fact, pick two generic lines in MATH and reverse the orientation of one of them to obtain two spheres MATH which intersect in one point with intersection n... |
math/0008240 | This follows immediately from REF . |
math/0008242 | Let MATH be a MATH-bridge link; let MATH be a rational-form diagram for MATH, and write MATH for MATH. The classification of MATH-bridge links implies that MATH is isotopic to any rational-form diagram associated to the fraction MATH. (If MATH is an integer, then it is easy to see that MATH is the trivial knot, which i... |
math/0008242 | None of MATH, MATH, and MATH contains negative powers of MATH; the lemma will be proved if we can show that MATH, where MATH . Define the auxiliary matrices MATH . Then MATH and MATH, and so MATH . But if we define a sequence of functions MATH, then an easy induction yields the recursion MATH with MATH and MATH. In par... |
math/0008242 | Let MATH be a Legendrian rational form for a MATH-bridge link MATH. The crossings of MATH are counted, with the same signs, by both the writhe of MATH and the NAME number of the Legendrian link MATH obtained from MATH; MATH, however, also subtracts half the number of cusps. Hence MATH by REF . Since MATH is ambient iso... |
math/0008243 | Let MATH where MATH and MATH are defined as in the statement of REF. To approximate the creation rate, we need to approximate MATH, which is the constant term of MATH. The constant term is given by the usual contour integral, which we will approximate using the saddle point method. Write MATH and MATH, so that MATH. No... |
math/0008243 | We have MATH where MATH and MATH. For fixed MATH, this function is clearly maximized at MATH. When MATH, it becomes a linear function maximized at MATH (for MATH). This yields MATH. Therefore, MATH if MATH is small enough compared to MATH. For MATH, we have MATH . It follows that MATH . Therefore, MATH . If we are not ... |
math/0008243 | From REF , we see that we can choose MATH to be the imaginary part MATH . If we substitute MATH for MATH and MATH for MATH and differentiate, then the first term of REF contributes the MATH-terms in the formula we are proving. To see this, recall that (up to an irrelevant multiple of MATH) MATH . After we express this ... |
math/0008243 | We will use the formula for MATH from REF. Let MATH be a small, simply-connected neighborhood in MATH of the smallest real interval containing MATH, such that the points MATH are not in MATH. (We checked at the end of the proof of REF that these points are not in MATH.) It follows from the definition of MATH that MATH ... |
math/0008243 | We assume that MATH, since otherwise MATH. As in the proof of REF, we will integrate MATH around a circle about the origin, where, as in REF , MATH . However, since we are looking only for an upper bound and not for an asymptotic estimate, we will not need the full saddle point method. We will only sketch the proof, be... |
math/0008243 | Since the discriminant of the polynomial is MATH, we see that it has two real roots whenever MATH. Clearly, MATH is a root iff MATH, and since the sum of the roots is MATH, it is the greater root iff also MATH. One can check the other claims similarly. |
math/0008243 | This simply amounts to showing that MATH is bounded as a function of MATH, MATH, and MATH, for MATH sufficiently small. If one computes MATH using the quadratic formula, and then differentiates it with respect to MATH, one finds that it equals a continuous function of MATH, MATH, and MATH (for MATH near MATH) divided b... |
math/0008243 | From REF , we see that MATH is approximated by MATH . (The error here is bounded by a constant multiple of MATH, which is MATH and hence MATH by REF.) Since MATH, we see that MATH is given to within MATH by the sum of MATH with MATH and its complex conjugate. REF will show that the latter two sums are MATH as MATH goes... |
math/0008243 | We need to show that MATH approximates MATH. Since REF implies that the creation rates are all non-negative, and MATH is the sum of a subset of the creation rates appearing in the sum giving MATH, the placement probability must be at least MATH. Also, given any point in the Aztec diamond, the north-going placement prob... |
math/0008243 | First suppose that MATH. The desired result will follow from the equation MATH together with the estimate given by REF. First, we show that REF applies to the creation rates appearing in the sum. Consider MATH as a function of MATH. Its first derivative at MATH is MATH which is greater than MATH since MATH. The only ro... |
math/0008243 | Since MATH is algebraic, it satisfies an equation MATH with MATH polynomials (not all identically zero). Let MATH. We will show that MATH has the desired property, using induction on MATH. We can choose the coefficients MATH so that they have no (non-constant) common factor. Fix MATH, and let MATH. Define MATH. Since t... |
math/0008243 | To prove this, we will apply the NAME REF says that MATH satisfies the conditions of REF, so there is an absolute upper bound for the number of roots that it can have as a function of MATH while MATH, MATH, and MATH are held fixed (unless it is identically zero for those values of MATH, MATH, and MATH). Before we apply... |
math/0008243 | Let MATH and MATH . For MATH, MATH (by REF) and MATH is bounded (by REF). Suppose MATH for all MATH. To bound the sum in the statement of the proposition, we will apply summation by parts. We have MATH . This sum is bounded in absolute value by MATH. The function MATH is monotonic on MATH and MATH (on the subintervals ... |
math/0008243 | We simply check that this formula satisfies the differential equations and boundary conditions. |
math/0008243 | We use induction on the size of MATH (holding MATH fixed and varying MATH). The case where this set is empty is trivial. Assume that the lemma is true whenever MATH, and suppose we have a situation in which MATH. It clearly suffices to consider the case in which MATH for some vertex MATH in MATH that is adjacent to at ... |
math/0008243 | Apply REF to the partial height functions MATH and MATH. |
math/0008243 | Let MATH be the highest extension of MATH to MATH, let MATH be the lowest extension of MATH to MATH, and define MATH and MATH similarly. (It is not hard to show that the complete height functions extending a given partial height function form a lattice under the usual partial ordering, so it makes sense to talk about t... |
math/0008243 | Let MATH be a lattice-path connecting a point MATH on the boundary of MATH to the point MATH. Let MATH be the partition of the space of possible height functions in which two height functions are regarded as equivalent if they agree at MATH. Let MATH be the conditional expectation MATH, the function from the set of hei... |
math/0008245 | We show first that MATH is incompressible. Of course this follows by standard techniques, by thinking of MATH as having a polyhedral metric of non-positive curvature and using the NAME - NAME Theorem to identify the universal covering with MATH (compare CITE). Since MATH is totally geodesic and geodesics diverge in the... |
math/0008245 | The proof is extremely similar to that for REF so we only remark on the ideas. First of all the conditions in the statement of REF play the same role as the link conditions in the definition of a cubing of non-positive curvature. So we can homotop disks which have boundary on MATH to reduce the graph of intersection of... |
math/0008245 | This follows immediately from REF , by observing that since the boundary of every meridian disk meets the double curves at least four times, there are no non-trivial MATH - gons in the complement of MATH in MATH for MATH and no MATH - gons. Hence MATH is almost cubed, as MATH in MATH has similar properties to MATH in M... |
math/0008245 | The argument is very similar to those for REF above, so we outline the modifications needed. Suppose there is a compressing or boundary compressing disk MATH for one of the surfaces MATH. We may assume that all the previous MATH are incompressible and boundary incompressible by induction. Consider MATH, the graph of in... |
math/0008245 | Note that the cases where either MATH has incompressible boundary or is non-orientable, are not so interesting, as then MATH and MATH are NAME and the result follows by NAME 's theorem CITE. So we restrict attention to the case where MATH and MATH are closed and orientable. Our method is a mixture of those of CITE and ... |
quant-ph/0008030 | Let MATH be a normal state of MATH. Then, by hypothesis, MATH is a normal state of MATH. Define a state MATH on MATH by MATH . Since MATH is normal, MATH. Define a state MATH on MATH by MATH . Since MATH is normal, MATH. Now, REF entail that MATH for any MATH, and thus MATH for any MATH. However, a state of the NAME al... |
quant-ph/0008030 | By hypothesis, the bijective mapping MATH must map the self-adjoint part of MATH onto that of MATH. Extend MATH to all of MATH by defining MATH . Clearly, then, MATH preserves adjoints. Recall that a family of states MATH on a MATH-algebra is called full just in case MATH is convex, and for any MATH, MATH for all MATH ... |
quant-ph/0008030 | For clarity, we suppress the representation map MATH. Suppose that MATH is a bounded function. We show that if MATH, then MATH for all MATH. The NAME operators on MATH satisfy the commutation relation (BR CITE , REF ): MATH . Using REF , we find MATH and from this, MATH. Now let MATH be in the domain of MATH. Then a st... |
quant-ph/0008030 | By NAME REF , MATH is a ``type III" NAME algebra which, in particular, contains no atomic projections. Since MATH is irreducible and MATH factorial, either MATH and MATH are disjoint, or they are quasi-equivalent. However, since MATH, the weak closure of the NAME representation clearly contains atomic projections. More... |
quant-ph/0008030 | Again, we use the fact that MATH REF does not contain atomic projections, whereas MATH REF does. Suppose, for reductio ad absurdum, that MATH and MATH are not disjoint. Since both these states are pure, they induce irreducible representations, which therefore must be unitarily equivalent. Thus, there is a weakly contin... |
quant-ph/0008030 | We shall prove the contrapositive. Suppose, then, that there is some extension MATH of MATH to MATH that is dispersion-free on all bounded functions of MATH. Then MATH is multiplicative for the product of the bounded operator MATH with any other element of MATH (KR CITE , REF ). Hence, by REF , MATH for all MATH and MA... |
quant-ph/0008030 | MATH may be thought of as a real NAME space relative to either of the inner products MATH defined by MATH . We shall use NAME and NAME 's REF : MATH are unitarily equivalent if and only if the positive operator MATH on MATH is a trace-class relative to MATH. (Since unitary equivalence is symmetric, the same ``if and on... |
quant-ph/0008030 | Suppose that MATH and MATH are disjoint; that is, MATH. First, we show that MATH, where MATH is the quadratic form on MATH which, if densely defined, would correspond to the total MATH-quanta number operator. Suppose, for reductio ad absurdum, that MATH contains some unit vector MATH. Let MATH be the state of MATH defi... |
quant-ph/0008031 | Clearly a pure tensor satisfies the exchange property. To prove the converse, we proceed by induction over MATH. We pick a basis MATH of MATH and write MATH where MATH. If MATH for some MATH, the exchange property for the case MATH implies that MATH satisfies the exchange property, so is a pure tensor by the inductive ... |
quant-ph/0008031 | It is useful to associate to MATH a linear map MATH. If MATH has rank MATH then MATH and we are in REF . So we may assume MATH has rank MATH. We will consider MATH as a MATH by MATH matrix. Consider the homogeneous degree MATH polynomial MATH. There are MATH cases to consider: REF there are exactly two points in MATH w... |
quant-ph/0008031 | The four cases of the statement correspond to the four cases of REF . REF is obvious. REF follows as MATH is locally equivalent to MATH for some MATH; by NAME 's theorem MATH is locally equivalent to MATH. In REF , we have MATH where MATH and MATH are linearly independent for each MATH. There is no harm in assuming tha... |
quant-ph/0008031 | Let MATH be the image of an algebraic mapping MATH, where MATH is the locally closed algebraic subvariety comprised of MATH-uples MATH where MATH are distinct, MATH and MATH belong to some MATH-plane, and MATH. It is easy to see that MATH is irreducible; thus standard results in algebraic geometry say that MATH is cons... |
quant-ph/0008031 | Let MATH be the rank of MATH and MATH the rank of the range of MATH. Thus MATH is a linear combination of pure tensors MATH. Write MATH where MATH and MATH. Then we have MATH, so that MATH is a linear combination of the pure tensors MATH and MATH. In the other direction, assume that the range of MATH is contained in th... |
quant-ph/0008031 | The NAME group MATH acts naturally on MATH and preserves each MATH. Let MATH be the subspace spanned by MATH, MATH and MATH. Clearly the closure of MATH is the closure of the MATH-saturation MATH. We can compute its dimension as follows. We consider the infinitesimal equation of the NAME algebra MATH on MATH. For MATH,... |
quant-ph/0008031 | We associate to MATH as before a linear map MATH and compute the rank of its image. If MATH has rank MATH it is clear that MATH has rank MATH, so we can assume MATH is injective. We can think of MATH as parameterizing a line MATH in MATH. If this line is not contained in the hypersurface MATH, then MATH of its points h... |
quant-ph/0008047 | Let MATH be the operation maximizing MATH in the definition of MATH. Clearly, if we compose MATH with any operator of the form MATH, this leaves MATH unchanged. The same must then be true after averaging over MATH (``twirling" CITE). We may thus assume MATH, where MATH is the twirling superoperator. We find MATH must h... |
quant-ph/0008047 | Let MATH be primal optimal for MATH. Then MATH . Similarly, let MATH and MATH be primal optimal for MATH and MATH respectively. Then MATH is primal feasible for MATH, thus giving the second inequality. |
quant-ph/0008047 | Let MATH be an operator satisfying the constraints above. Then for any operators MATH, MATH, MATH, we have: MATH . If MATH, MATH, MATH, and MATH then the last four terms are all nonnegative, and we have MATH and thus MATH . In fact, by the theory of duality for SDPs, this inequality can be made tight, to wit: MATH mini... |
quant-ph/0008047 | If MATH and MATH are dual optimal for MATH and MATH, then MATH . Similarly, if MATH is dual optimal for MATH, then MATH . |
quant-ph/0008047 | For MATH, take MATH, MATH. For MATH, take MATH, MATH. |
quant-ph/0008047 | For the first inequality, let MATH and MATH be primal optimal for MATH and MATH; then MATH is primal feasible for MATH, giving the inequality. For the second inequality, let MATH be dual optimal for MATH. Then, taking MATH, we have MATH . |
quant-ph/0008047 | By the theorem, we have, writing MATH: MATH . |
quant-ph/0008047 | (First proof) Let MATH be primal optimal for MATH. Then MATH is primal feasible for MATH, so MATH . (Second proof) Let MATH be dual optimal for MATH. Then MATH . Here we used the facts that for a positive superoperator MATH and an arbitrary operator MATH, MATH . |
quant-ph/0008047 | We first consider MATH. Writing MATH, we have MATH with MATH subject to the constraints MATH . Since increasing MATH increases the feasible set, the maximum cannot decrease. Dually, MATH which is nondecreasing in MATH for any choice of MATH. For MATH, we proceed similarly; taking MATH, we have: MATH with MATH subject t... |
quant-ph/0008047 | For the first claim, take MATH, MATH, at which point MATH, so MATH. For the second claim, take MATH . Finally, for the third claim, take MATH . In each case, the lower bound coming from MATH agrees with the upper bound coming from MATH, and thus both MATH and MATH are optimal. |
quant-ph/0008047 | We need to show that for any MATH, MATH . Choose MATH between MATH and MATH, and consider the dual SDP bound with MATH . Then MATH is p.p.t., so MATH; the first term is bounded below REF by the following lemma. |
quant-ph/0008047 | Let MATH be the projection onto the positive part of MATH then we need to show that MATH is bounded below REF. Fix MATH, and consider the statement MATH. For this to be true, we must certainly have MATH . Letting MATH be the largest value of MATH such that these inequalities simultaneously hold for infinitely many MATH... |
quant-ph/0008047 | Indeed, this is true for each of the functions MATH and MATH individually, so must be true for their sum. |
quant-ph/0008047 | Let MATH be the superoperator MATH integrating with respect to the uniform probability measure on MATH. This is trace-preserving, MATH-local (thus p.p.t), and satisfies MATH. The first claim thus follows from the theorem. Similarly, if MATH is the superoperator MATH then the theorem applies to MATH. |
quant-ph/0008047 | Let MATH be primal optimal for MATH such that MATH is invariant under MATH and MATH. The representations of this group are in one-to-one correspondence with the integers MATH, with MATH and MATH. Writing MATH we have MATH . We next observe that MATH iff MATH and MATH iff MATH . Similarly, the partial transpose MATH is ... |
quant-ph/0008047 | We observe that MATH. Thus if we show that MATH, the proof of REF will apply to give REF ; taking MATH gives REF , and the equations for MATH follow immediately. It thus remains to show MATH (since the other inequality is immediate). Taking MATH we find MATH . |
quant-ph/0008047 | By the above argument, we may assume MATH. Now, MATH . We find that the optimal MATH satisfies MATH . Plugging in, we obtain the stated bound. |
quant-ph/0008047 | Fix an integer MATH, and consider the set MATH consisting of tensor products MATH with each MATH; note, in particular, that MATH is a set of mutually orthogonal projections. Since MATH we have MATH where we define MATH to be the number of factors equal to MATH. Let us then define an operator MATH . We observe that MATH... |
quant-ph/0008047 | That this is an upper bound was shown in CITE, so it suffices to prove the lower bound. We construct a protocol in two steps. First, suppose MATH possesses a transitive group of symmetries; that is, a transitive group MATH of permutations such that MATH . (For instance, the operator MATH is symmetric under the transiti... |
quant-ph/0008047 | To any subset MATH, we associate a projection MATH which satisfies MATH . For each MATH and each integer MATH, let MATH be the minimum over MATH of the largest eigenvalue of MATH subject to the constraint MATH. Then MATH (take MATH). Since MATH the theorem follows by the classical analogue of REF . |
quant-ph/0008047 | Since nonnegative linear combinations and limits of positive operators are positive, it suffices to prove the result for MATH. In that case, MATH factors as a tensor product of the following operators: MATH . The first two are clearly positive; that the third is positive is a special case of the following lemma. |
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