paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/0007007 | As for MATH in the proof of REF , with MATH in REF changed to MATH. |
math-ph/0007007 | We apply the variational principle, using MATH-fold products of MATH-particle states MATH as MATH-particle states MATH for all MATH with MATH. Setting MATH we see that the inequality holds, for each MATH and MATH. |
math-ph/0007007 | This results from combining REF with REF . |
math-ph/0007007 | We use the NAME inequality CITE MATH (MATH is normalized as MATH, the constant can be chosen to be MATH), which, together with NAME 's inequality MATH, implies that MATH where MATH is the one-particle reduced density matrix of MATH, and MATH is its density. Now if MATH, which we can of course assume, then MATH with MAT... |
math-ph/0007007 | For MATH consider the wave functions MATH. By REF there holds for each MATH the inequality MATH . Both sides of this inequality are NAME measurable functions of MATH. Observe MATH, MATH, and integrate both sides of REF with MATH. Divide by MATH, and use MATH in front of the kinetic energy term. |
math-ph/0007007 | From MATH we borrow a part of MATH to use REF , MATH sum over all pairs MATH , and multiply with MATH. This gives MATH . Inserting REF into REF completes the proof. |
math-ph/0007007 | This follows from the convergence of the energies, by analogous arguments as in CITE or CITE. |
math-ph/0007007 | Let MATH be the density of MATH. We have MATH where we used again the NAME inequality for the last step. By an analogous argument as in REF we see that the right side of REF converges to MATH as MATH. |
math-ph/0007007 | See REF . |
math-ph/0007007 | Since MATH is unique, MATH converges weakly to MATH by REF and the proof of REF . Apply REF to MATH and MATH, and note that MATH and MATH. |
math-ph/0007007 | We use the notations of the proof of REF . The first assertion is easily proved, using that MATH and MATH. To prove the second we denote MATH and MATH, and compute MATH where we have used MATH in the last step. Therefore MATH as MATH, which gives immediately the desired result. |
math-ph/0007009 | The symmetry and positive definiteness of MATH implies that MATH is also symmetric and positive definite. Hence there exists an orthogonal matrix MATH, MATH, such that MATH where MATH is a diagonal matrix with all positive eigenvalues MATH. Thus MATH which has the same MATH's as eigenvalues. Hence it is nonsingular. Su... |
math-ph/0007009 | The proof relies on a direct verification of the Gaussian function with REF as a solution to REF . This has been done many times by physicists CITE. Hence we will not repeat the lengthy computation. With the verification of the solution, and uniqueness of the fundamental solution to REF , the lemma is proven. |
math-ph/0007009 | Necessity: If the solution to REF is nonsweeping, then by REF and the sweeping theorem from REF, its fundamental solution has a stationary limit MATH. Since MATH has Gaussian distribution for MATH, the limit distribution is also NAME with finite variance. Therefore, MATH . By the general formula for MATH CITE and using... |
math-ph/0007009 | By REF with reversibility has symmetric MATH; and by REF its stationary process has covariance MATH. Therefore, MATH is symmetric, and MATH . Hence, the standard fluctuation-dissiplatin relation follows, and furthermore REF can be simplified into MATH. On the other hand, the standard fluctuation-dissipation relation is... |
math-ph/0007009 | The proof of this result is contained in the proof of REF , MATH. |
math-ph/0007013 | With REF in the hand, the proof goes along very much the same lines as in CITE. Indeed, let MATH be as in REF and set MATH in REF to be MATH, where MATH will be chosen later and MATH will tend to MATH. Let MATH be in the MATH-class and recall that MATH with MATH. Abbreviate MATH and MATH, and note that MATH. Then, usin... |
math-ph/0007013 | Let MATH be the embedded discrete-time simple random walk on MATH and let MATH be its local times defined by MATH. Let MATH denote the expectation with respect to the discrete-time walk, starting at MATH. Abbreviate MATH and MATH. Then, by REF, MATH where MATH, and MATH is a shorthand for MATH. Fix MATH and define MATH... |
math-ph/0007013 | Abbreviate MATH and let MATH. Then MATH . Using this estimate and the NAME inequality, we have for any MATH that MATH for any MATH. Set MATH and note that we have MATH as MATH, due to MATH. Consequently, the term with MATH is negligible and the right-hand side of REF is bounded by MATH. The claim is finished by the NAM... |
math-ph/0007013 | Pick any MATH. Let MATH be so large such that the sum in REF with MATH for all MATH exceeds MATH. Note that MATH. Combining the results of REF for MATH and REF , we derive, for sufficiently large MATH respectively,MATH, the bound MATH where MATH is the constant from REF. But MATH by our choice of MATH, which means that... |
math-ph/0007013 | The argument is based on the asymptotic sublinearity of MATH at infinity. However, in order to have sublinearity on the whole interval MATH, we first construct an auxiliary modification of MATH. Let MATH be such that MATH is positive, increasing, and concave on MATH. Let MATH to be the right derivative of MATH at MATH.... |
math-ph/0007013 | We begin by formalizing the event in REF; in order to later approximate continuous MATH by a discrete variable, we write MATH instead of MATH: MATH . Note that the probability of MATH does not depend on MATH and note that different MATH's are independent if the MATH's have distance larger than MATH from each other. The... |
math-ph/0007013 | Let MATH and fix MATH and MATH such that MATH. Let MATH, define MATH as in REF and let MATH be as in REF ; suppose MATH without loss of generality. Let MATH. As in CITE, the lower bound will be obtained by restricting the walk in REF to perform the following: The walk keeps jumping toward MATH, spending at most time MA... |
math-ph/0007014 | It suffices to prove that MATH for all MATH. There is a normalized vector MATH such that MATH. (MATH is antisymmetric according to the NAME principle.) We use the notation MATH to denote the inner product in NAME space and spin space. Then we can write MATH with MATH . As a (unnormalized) variational trial vector we ta... |
math-ph/0007014 | Let us first show that it suffices to prove that for any normalized sequence MATH, MATH, (not necessarily minimizing) tending weakly to zero MATH . To prove this let MATH be some minimizing sequence, that is, assume that MATH and that MATH . By the NAME Theorem we can assume that this sequence, as well as the sequence ... |
math-ph/0007014 | First, note that MATH if MATH, because MATH has this same monotonicity property. Therefore, for any sequence of MATH, MATH is monotonically decreasing and has a sequence-independent, finite limit, which we call MATH, and we note that MATH. To prove the opposite, namely MATH, we shall prove that MATH for every MATH. Let... |
math-ph/0007014 | Write MATH where MATH contains the annihilation operators and MATH contains the creation operators. We omit MATH and we omit the vector index of MATH for simplicity. Since MATH we learn from REF that MATH and MATH are in MATH. Since MATH strongly, MATH . The same holds if MATH is replaced by MATH and MATH by MATH. This... |
math-ph/0007014 | MATH . |
math-ph/0007014 | Each of the various energies converges as MATH by REF . |
math-ph/0007014 | Let MATH any smooth, bounded function on MATH with bounded first derivative. We easily compute MATH . We use the NAME REF to compute MATH . Now we choose MATH to be MATH and where MATH is a smooth cutoff function that is identically equal to REF outside the ball of radius REF, and identically zero inside the ball of ra... |
math-ph/0007014 | This proof is a slight modification of the one in CITE. The basic idea is to show that there is effectively no interaction between localized particles and low momentum photons. To make this idea explicit we write our Hamiltonian in a gauge different from the usual NAME gauge. To be precise, define MATH which is well de... |
math-ph/0007014 | We differentiate REF with respect to MATH and obtain MATH . The norms of the first and third terms can estimated precisely the same way as in REF , and yields a bound of the form MATH . For the second term, a straightforward calculation shows that MATH . The last term is dealt with in a similar fashion as the previous ... |
math-ph/0007014 | The Hamiltonian MATH has a normalized ground state MATH, by REF . Pick a sequence MATH tending to zero and denote the corresponding eigenvectors by MATH. This sequence is a minimizing sequence for MATH by REF . Since MATH is bounded there is a subsequence (call it again MATH) which has a weak limit MATH. Since MATH and... |
math-ph/0007014 | Our immediate aim is to compare the field energy MATH with the localized field energy MATH. For simplicity the various indices are supressed and MATH is replaced by MATH. The variable MATH plays no role here. Pick an orthonormal basis MATH of MATH in MATH. States of the form MATH where MATH is finite, form an orthonorm... |
math-ph/0007014 | By the IMS localization formula we have that MATH where the second term goes to REF as MATH. With our assumption on MATH only the sets MATH with MATH contribute. From REF we get that MATH as first MATH then MATH. Certainly MATH and MATH from which the statement immediately follows. |
math-ph/0007014 | Since the energy of MATH is uniformly bounded we also know that MATH is uniformly bounded Let us describe the operator MATH in more detail. Recall that MATH where MATH denotes the vacuum vector in NAME space and MATH. Hence, using the definition of MATH, we find that MATH . The projection MATH annihilates the photons i... |
math-ph/0007014 | We write MATH with MATH. There are thus four terms in MATH . Using the NAME inequality, the MATH term can be bounded above by MATH. On the other hand, MATH, where MATH is the commutator MATH; the factor REF comes from the two polarizations MATH. Altogether, we obtain MATH . Finally, we use the NAME inequality again to ... |
math-ph/0007014 | In addition to REF , use the facts that for any MATH, MATH and MATH. |
math-ph/0007014 | We shall use the abreviation MATH . For our special choice of gauge MATH where we set MATH . Next, pick any MATH in MATH and calculate (recalling the definition of MATH in REF ) MATH with MATH. This extends, using an approximation argument, to all MATH and, in particular, to MATH. Here we note that, on account of REF a... |
math-ph/0007014 | First some notation: For any function MATH define MATH and MATH . Returning to REF with MATH replaced by MATH we have MATH which can be rewritten as MATH . Notice that without the first term on the right of the inequality sign, the structure of this inequality is the same as REF , except that, of course, MATH plays the... |
math-ph/0007020 | We first consider the case when MATH is self-adjoint, that is, MATH. Write MATH with MATH and MATH. Here MATH are the positive and negative parts of MATH, obtainable from the spectral representation of MATH or more explicitly as MATH. Obviously MATH. Since MATH, MATH is self-adjoint. For arbitrary MATH write MATH with ... |
math-ph/0007020 | We adapt a standard proof (see for example, CITE, p. REF) used in the context of positive maps on non-unital MATH-algebras. We first claim that it suffices to prove boundedness on MATH. Indeed, this follows from the decomposition REF and the related bounds. Assume that MATH is not bounded on MATH. Then there are MATH w... |
math-ph/0007020 | Assume that there is MATH with MATH, such that MATH for all MATH. If MATH is not positive definite, then we have a contradiction, so let MATH. We claim there is MATH which is not positive definite with MATH. In fact, we may take MATH to be a one-dimensional orthogonal projection onto an eigenvector of MATH times the co... |
math-ph/0007020 | Assume that MATH holds such that there is MATH with MATH. Then MATH holds for all selfadjoint MATH due to the bound MATH and therefore MATH for all MATH and all MATH. This in turn implies MATH for all MATH and all MATH. Since MATH by definition, there is MATH with MATH (see the decomposition REF ), so MATH is not ergod... |
math-ph/0007020 | It suffices to consider MATH, where MATH is any one-dimensional orthogonal projection (compare the proof of REF ). Let MATH be a unit vector, so we have to consider MATH and the claim follows. |
math-ph/0007020 | We use the same notation as in the proof of the previous lemma. In addition write MATH for MATH and set MATH. Then MATH and again the claim follows. |
math-ph/0007020 | The estimate REF for the case MATH is trivial, so let MATH. Since MATH is self-adjoint, there is a complete set orthonormal eigenvectors MATH with real eigenvalues MATH of MATH. With MATH we obtain MATH . Observe that MATH holds, since MATH and MATH. In the last estimate in REF we used the estimate REF . Furthermore MA... |
math-ph/0007020 | Here and in the proof of the next lemma we will mimic and extend REF, page REF in the present context. Set MATH such that MATH are all in MATH. Consider MATH in MATH, which defines an element in MATH. By REF Consider the following linear transformation MATH on MATH (equipped with the natural scalar product) given as th... |
math-ph/0007020 | Choosing MATH in REF we have the estimate MATH for all MATH. Thus REF follows trivially from REF in case MATH. So let MATH. Take MATH to be an orthonormal basis in MATH diagonalizing MATH with eigenvalues MATH. Then we obtain MATH . Here we have used REF for the first inequality. The second inequality is a consequence ... |
math-ph/0007020 | We want to prove that MATH is also an eigenvector with eigenvalue MATH, that is, MATH. Since MATH (such that also MATH and hence MATH by the uniqueness of MATH) it suffices to show that MATH holds for all MATH, where the MATH are as in the previous lemma. Also by the previous lemma we have MATH . Here we have used the ... |
math-ph/0007020 | Assume there are two eigenvectors MATH and MATH in MATH with eigenvalue MATH, normalized to MATH. By REF , its proof and by REF we have that MATH is an eigenvector with eigenvalue MATH. In particular both MATH and MATH are unitaries. Since MATH is also an eigenvector for the same eigenvalue for all MATH by the same arg... |
math-ph/0007020 | We use familiar arguments. Let the MATH form an orthonormal basis of eigenvectors for MATH arranged such that MATH and MATH. Here the MATH's are the eigenvalues, that is, MATH. Since MATH is a simple eigenvalue, MATH. By NAME 's theorem we have MATH with MATH and MATH. Obviously MATH holds such that MATH and the claim ... |
math-ph/0007020 | We recall that MATH is the norm limit of MATH as MATH. Since the space of compact operators is closed with respect to the norm topology it suffices to prove that all MATH are compact. By assumption to each MATH there is a sequence MATH of finite rank operators in MATH such that MATH . Since MATH is of finite rank MATH ... |
math-ph/0007020 | Take the MATH's to form an orthonormal basis in which MATH is diagonal, that is, such that MATH takes the form REF and choose MATH with MATH as above with MATH. Then we have MATH for all MATH, such that MATH is idempotent. Also the kernel of MATH consists of all elements which may be written in the form MATH. Moreover ... |
math-ph/0007022 | Given that the argument is not that different from the derivation of REF , we permit ourselves to present here just a brief summary. We start by introducing a convenient extension of the space MATH over which the random process MATH is defined. Let MATH and let MATH be a probability measure on MATH having MATH-marginal... |
math-ph/0007022 | It suffices to consider the point-charge processes, since any point process may be regarded as a special case with unit charges. Under the assumption of tight fluctuations, REF says that there is a measurable function MATH of the charge configuration such that MATH . Since the charge MATH assumes only values which are ... |
math-ph/0007022 | The argument in the discrete case is similar to the above, with the only notable difference being that the charge MATH now includes MATH in addition to an integer multiple of MATH. Thus in this case the ``antiderivative functional" MATH provided by REF satisfies MATH and hence MATH defines a cyclic factor as described ... |
math-ph/0007026 | If MATH is linear and MATH in REF is not zero, then MATH which is inconsistent. |
math-ph/0007029 | Consider the NAME operator defined on MATH by MATH . Since MATH is compact, the spectrum of MATH is discrete and its first (simple) eigenvalue and the corresponding (normalized) eigenfunction are analytic functions of the (real) parameter MATH CITE. We thus expand MATH and the corresponding eigenfunction MATH as a powe... |
math-ph/0007029 | The spectral problem corresponding to MATH is MATH which has discrete spectrum MATH. We will prove that if MATH then MATH for all MATH. To this end rewrite REF as MATH . It is not difficult to see that MATH is an eigenfunction if and only if MATH for some real number MATH different from zero. For MATH the operator MATH... |
math-ph/0007029 | It only remains to show the result for MATH larger than MATH. Clearly in this case MATH. Consider now the family of potentials given by MATH . Note that although MATH is not continuous on the circle, it can be approximated by continuous functions without affecting our results. For this family of potentials we obtain th... |
math-ph/0007029 | The first inequality follows directly by writing the eigenvalue problem as MATH and applying the previous corollary with MATH. The second part is a consequence of the fact that for MATH larger than MATH the principal eigenvalue must be larger than MATH. |
math-ph/0007029 | Fix a point MATH in MATH and denote by MATH the geodesic ball centred at MATH with radius MATH. Let now MATH and define the potential MATH (As before, this is discontinuous but can be approximated by continuous functions without changing the results.) By subtracting MATH on both sides of the equation for the eigenvalue... |
math-ph/0007030 | The linearity and antisymmetric properties are obvious. Two other properties are secured because CASE: MATH commutes with convolutions REF and sends zero function to itself REF ; CASE: The commutator MATH satisfies both to NAME and NAME identity. For example the NAME identity could be verified as follows: MATH where RE... |
math-ph/0007030 | The proof is a straightforward calculation using REF. We will carry them separately for cases of MATH and MATH. Let MATH, MATH. Then: MATH where the line REF follows from the first case in REF is exactly the first case in REF. The second case MATH (symbolically corresponding toMATH) is also not difficult but somehow lo... |
math-ph/0007030 | The properties follows from the corresponding properties of MATH-mechanical brackets REF and conservation of algebraic identities by representations REF . |
math-ph/0007030 | It is known (see CITE) that the above four properties are a direct consequence of those from REF . Again the properties are very well known for the quantum commutator and the NAME brackets. |
math-ph/0007030 | To establish first statement one verifies images of MATH under representations MATH REF and MATH REF by a direct calculation. We proceed with a derivation of REF . Let CITE MATH be the left and the right invariant vector fields on MATH correspondingly. They generate the right MATH and the left MATH shifts on MATH corre... |
math-ph/0007035 | Proof can be found in CITE. |
math-ph/0007035 | It is obvious that for MATH the operator MATH is a contraction. The second inequality will be proven in a more general context in REF. Since MATH are mutually orthogonal invariant spaces of MATH from REF follows that MATH . Proof of the fact that MATH and MATH are adjoint can be found in CITE. Proofs of the other two e... |
math-ph/0007035 | For any polynomial MATH we have MATH therefore approximating the square root by polynomials we obtain MATH. Note that for MATH we have MATH therefore MATH . If in the preceding calculations we replace MATH by MATH (and vice versa) and replace MATH by MATH we obtain the opposite inequality. |
math-ph/0007035 | Proof is straightforward, we shall omit it. |
math-ph/0007035 | Note that an expression is in the mentioned form if and only if it does not contain any subexpression being the left hand side of one of REF - REF . If it does not hold by replacing the left hand side of appropriate equation by the right hand side we obtain an expression (or a sum of expressions) which is either shorte... |
math-ph/0007035 | Let MATH be an isometry such that MATH limited to MATH is equal to identity. Such isometry exists because MATH and MATH have the same dimension. We define now an second quantization of MATH, that is, an isometry MATH by the formula MATH. The wanted MATH-isomorphism is MATH . The uniqueness of such isomorphism follows f... |
math-ph/0007035 | Proof of the first two inequalities can be found in CITE. Now we show the third one. MATH . The last inequality can be proven similarly. |
math-ph/0007035 | The last inequality holds since for each MATH we have MATH . |
math-ph/0007035 | Let MATH where MATH are disjoint intervals. Since MATH is left - adapted for any MATH we have MATH therefore MATH what proves claimed theorem. |
math-ph/0007035 | Proof of an analogue fact can be found in the paper of CITE. |
math-ph/0007035 | Let us consider a NAME space MATH such that there exists an operator MATH which restricted to MATH is a unitary operator MATH and restricted to MATH is equal to MATH. We introduce a process MATH, where MATH denotes the orthogonal projection on the subspace MATH for a set MATH. For any operator MATH we define MATH. Of c... |
math-ph/0007035 | It is enough to prove that for each MATH and all vectors MATH . Let MATH. We have MATH and the operator norm of the second summand does not exceed MATH. We shall use the notation introduced in proof of REF . It follows from the fact that MATH has measure MATH that MATH for any fixed vector MATH. If we rewrite REF repla... |
math-ph/0007035 | MATH because MATH and MATH. |
math-ph/0007035 | It is enough to notice that MATH and recall the definition of the norm MATH and REF . |
math-ph/0007035 | For a unital vector MATH orthogonal to MATH we have (see REF ) MATH and therefore for some constants MATH which depend only on MATH and MATH we have MATH . The second part of the lemma follows from the majorized convergence theorem. |
math-ph/0007035 | For MATH functions MATH uniformly tend to MATH. Therefore by preceding lemmas appropriate norms of biprocesses MATH tend to MATH what proves that the limit in the first equation holds in the operator norm. For MATH and MATH for each MATH functions MATH by REF uniformly tend to MATH and REF shows that the limit in the f... |
math-ph/0007035 | For MATH let MATH be simple adapted biprocesses. We assume that intervals MATH form a partition, that is, that they are disjoint. Note that we can replace partition MATH by a refined partition MATH so that MATH. MATH . The second and the third summands tend by REF to the second and the third summands of the right hand ... |
math-ph/0007042 | It follows easily from the relations REF , taking into account REF . Indeed the matrix NAME REF is linearizable via the position MATH which yields for MATH the linear ODE MATH . Hence, if REF matrices MATH are known, the matrix MATH can be evaluated, via REF , by solving the linear (generally nonautonomous) matrix ODE ... |
math-ph/0007042 | The second Proposition of the Lemma entails that the solutions of REF can be obtained by first solving the ODEs REF to determine the MATH matrices MATH and by then obtaining from these, according to the first Proposition of the Lemma, the MATH matrices MATH. |
math/0007005 | First, recall from the NAME theorem that the decomposition MATH is given by MATH . Now fix MATH and endow a vector space MATH with the flag module structure defined by MATH . If we replace MATH by MATH for some MATH, the expression for MATH is just multiplied by MATH, so up to isomorphism (of graded modules), the flag ... |
math/0007005 | It remains to show that the above construction can be reversed, so assume that MATH is a collection of points in MATH satisfying REF . Choose a (nonzero) representative MATH for each MATH, and endow a vector space MATH with the flag module structure defined by REF . By REF , this rule is compatible with relations of ty... |
math/0007005 | First, let MATH be monogressive and MATH-effective. If a point MATH has homogeneous coordinates MATH (see REF ), then by MATH-effectiveness, the coordinates of MATH are MATH. Using REF, we therefore see that MATH sends MATH to the following point in MATH: MATH . Using the change of ``variables" MATH (symmetric differen... |
math/0007005 | Consider a monogressive MATH-effective cell MATH. For each MATH, write again MATH, MATH (hence MATH by monogressivity). Reorder the MATH in such a way that MATH. By MATH-effectiveness, each MATH must appear in the subarray MATH, and each MATH in the subarray MATH. Now let MATH, and note that for each MATH, replacing MA... |
math/0007005 | The vector MATH is a highest weight vector, of weight MATH. Now apply REF , noting that the dimension of the simple module of highest weight MATH is precisely MATH. |
math/0007005 | This is immediate from the action of the generators of MATH on the basis elements of MATH and MATH, as described in REF: the formulas obtained there are symmetric in MATH and MATH. |
math/0007005 | The first statement amounts to the commutativity of REF , which immediately follows from REF. For the second statement, consider the composite map MATH corresponding to the line bundle MATH, and denote by MATH the linear span of the image of this map. Claim A. The subspace MATH is contained in the unique simple MATH-su... |
math/0007006 | For all MATH, let MATH and MATH. We need to show that MATH. By induction over MATH, we may assume that the left hand side is equal to MATH, so we must show that MATH. We decompose MATH, where MATH and MATH. Orthogonality of MATH already implies that for each MATH, MATH commutes with all elements of MATH and of MATH. Si... |
math/0007006 | Use REF and the fact that the MATH are complete. |
math/0007006 | Let MATH be another nonincreasing numbering of the elements of MATH. For any incomparable roots MATH, the commutator MATH is trivial, so it is enough to show that the sequence MATH can be rearranged into MATH by successively swapping adjacent pairs of incomparable roots. We have MATH for some MATH, and by assumption, t... |
math/0007006 | First, note that if MATH is MATH-stable, then for every MATH, MATH is also MATH-stable; we may therefore restrict the proof to the case where MATH. Since for each root MATH, the quotient MATH is a projective line, the last statement follows from the first one and from REF . Now choose a representative MATH for the refl... |
math/0007006 | The last part follows from MATH-stability (see REF ). For the first part, recall again CITE that if MATH are orthogonal roots, then MATH; since MATH, we then also have MATH. Now use this fact in the definition of MATH. |
math/0007006 | For each MATH, choose an isomorphism MATH such that MATH for all MATH and all MATH CITE. Consider first the case where MATH (that is, MATH). The set of all MATH, MATH, is an open dense subset of MATH (compare the proof of REF ), and we have MATH . By orthogonality, the MATH are linearly independent, hence the morphism ... |
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