paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003217 | 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 mus... |
math/0003217 | 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 t... |
math/0003217 | 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,... |
math/0003217 | 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 . |
math/0003217 | We apply the Stoke's theorem multiply. Indeed, MATH . |
math/0003217 | 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 ... |
math/0003217 | Since MATH and MATH link, we have MATH. Thus MATH . |
math/0003217 | Start from the ends of the chain, and apply the lemma to eliminate edges one by one, coming from the ends towards MATH. |
math/0003217 | Split the integral into two parts depending on whether MATH or MATH. The computation for them is identical; if MATH, we have MATH . |
math/0003217 | 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 MA... |
math/0003217 | 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. |
math/0003217 | 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 ... |
math/0003217 | 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 . |
math/0003217 | 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/0003219 | 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. |
math/0003219 | We have MATH by REF , and this equals MATH by REF . |
math/0003219 | 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. |
math/0003219 | Apply REF twice. |
math/0003219 | We have MATH by the definition of MATH. Hence MATH. Thus the expression in REF is a square either of MATH or of MATH. |
math/0003219 | 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 MA... |
math/0003221 | 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 a... |
math/0003221 | 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 writ... |
math/0003221 | 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. |
math/0003221 | 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. |
math/0003221 | The twist relations involving counit are obvious, for the rest we have, using identities from REF : MATH and one checks other identities similarly. |
math/0003221 | 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... |
math/0003221 | 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 . |
math/0003221 | 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 ... |
math/0003221 | For all MATH we have MATH since MATH, whence MATH. |
math/0003221 | We have MATH hence MATH by the uniqueness of the solution of REF . |
math/0003221 | 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. |
math/0003221 | 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 R... |
math/0003221 | 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 t... |
math/0003221 | We directly compute : MATH . The identities MATH and MATH are clear. |
math/0003221 | 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 th... |
math/0003221 | 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. |
math/0003223 | 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/0003223 | 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 (... |
math/0003223 | 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.... |
math/0003232 | 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 sen... |
math/0003232 | 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... |
math/0003236 | 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. |
math/0003236 | 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 th... |
math/0003236 | These are immediate from the NAME squares in MATH REF and the NAME relations (see CITE). |
math/0003236 | 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. |
math/0003236 | Suppose that MATH corresponds to MATH. Then MATH by REF . The result follows by REF . |
math/0003236 | 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 i... |
math/0003236 | 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. B... |
math/0003236 | 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 s... |
math/0003236 | 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 cor... |
math/0003236 | This is immediate from the preceding proposition since, by REF , the bordism class of MATH is determined by MATH. |
math/0003236 | 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. |
math/0003236 | 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 subm... |
math/0003236 | 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. |
math/0003236 | 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 t... |
math/0003236 | 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,... |
math/0003236 | 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 - ... |
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 | 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. |
nlin/0003034 | 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 ... |
nlin/0003034 | 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 d... |
quant-ph/0003074 | 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 ... |
quant-ph/0003074 | 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 f... |
quant-ph/0003074 | 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. ... |
quant-ph/0003074 | 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 th... |
quant-ph/0003074 | See REF of Ref. REF. |
quant-ph/0003074 | 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 d... |
quant-ph/0003074 | 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 i... |
quant-ph/0003101 | 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. |
quant-ph/0003101 | 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. |
quant-ph/0003101 | If MATH can be written as a mixture of MATH-states, then MATH . |
quant-ph/0003101 | 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... |
quant-ph/0003101 | 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 ... |
quant-ph/0003101 | 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. |
quant-ph/0003101 | Sufficiency follows from REF and necessity from REF . |
quant-ph/0003101 | Sufficiency follows from REF and necessity from REF . |
quant-ph/0003144 | 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 an... |
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 . |
quant-ph/0003144 | 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 . |
cond-mat/0004005 | 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 therefo... |
gr-qc/0004057 | 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. |
gr-qc/0004057 | 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 re... |
gr-qc/0004057 | 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. |
gr-qc/0004057 | 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. |
gr-qc/0004057 | For simplicity we assume that MATH. Then, MATH . Thus we have shown that MATH. A similar proof applies for MATH. |
gr-qc/0004057 | 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 th... |
gr-qc/0004057 | 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. |
gr-qc/0004057 | 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). |
hep-th/0004029 | 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... |
hep-th/0004029 | 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. |
hep-th/0004029 | 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 . |
math-ph/0004004 | 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. |
math-ph/0004004 | 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 explic... |
math-ph/0004004 | This follows immediately from the relations MATH . |
math-ph/0004004 | This follows from the time invariance of MATH, that is, MATH: MATH and the equations of motion REF . |
math-ph/0004004 | 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-ph/0004004 | 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. Therefo... |
math-ph/0004007 | 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 ... |
math-ph/0004007 | 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 wil... |
math-ph/0004014 | 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 definiti... |
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