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math/0011254 | If MATH for MATH, then, by REF , MATH for all MATH. It is then clear that MATH by the MATH-continuity of MATH REF . |
math/0011254 | We proceed by contradiction. Assume there is MATH such that MATH and MATH. Therefore MATH by REF . Now by REF of MATH and REF we have MATH so that MATH . |
math/0011254 | For MATH we have MATH . |
math/0011254 | The first statement follows easily from the fact that the integral in REF is absolutely convergent. Changing variables in the first iterated integral below we get MATH . Now we split the outer integral on the righthand side as MATH. The first one easily evaluates to MATH taking into account REF . This term converges to zero in view of REF and an elementary error term in the prime number theorem. The second one is bounded by MATH where we have applied in succession NAME 's theorem, REF , and an elementary error tem for the prime number theorem of the form MATH. |
math/0011254 | That MATH is simply REF for MATH. For the second part we use REF : MATH . |
math/0011254 | The hypothesis on the zero of MATH implies, by REF , that MATH . Now assume by contradiction that MATH converges in MATH, so it must converge to MATH. On the other hand, noting that MATH when MATH and MATH, we get MATH . Then letting MATH we obtain MATH . Since the left-hand side goes to zero when MATH this contradicts REF . |
math/0011254 | For MATH REF translates into MATH . On the other hand if we apply MATH to MATH we get MATH . Hence MATH and MATH for some positive constant MATH. |
math/0011254 | The MATH defining REF , as well as REF give MATH in MATH. Assume by contradiction that MATH converges in MATH, then so does MATH, and therefore MATH which contradicts REF . Likewise the wholly analogous computation MATH in MATH yields the divergence of MATH in MATH. |
math/0011254 | Each MATH, so REF implies it is the right-continuous, step function MATH which is constant on every interval MATH . Therefore pointwise convergence trivially implies MATH . Now we proceed by induction. For MATH it is clear that REF gives MATH. Next assume for MATH that MATH for MATH, then the limit in REF yields MATH . But comparing this to the well-known MATH we obtain the desired MATH as MATH. |
math/0011254 | We have trivially MATH . Assume by contradiction that MATH. For the step functions involved this clearly implies pointwise convergence, then, from REF , we get MATH for each MATH, which, by the way, forces REF to hold for MATH too. But this immediately implies that MATH. However, MATH diverges in MATH by REF , so we have obtained a contradiction. |
math/0011255 | When MATH, the result follows from the formula MATH. Let MATH. Let MATH be an orthonormal basis for MATH. Let MATH be a MATH-basis for MATH. Then, by CITE, MATH is known to be a MATH-basis for the module MATH of MATH-derivations, where MATH . By the contraction MATH of a MATH-form and a derivation, the MATH-modules MATH and MATH are MATH-dual each other CITE CITE . Let MATH be dual to MATH. In other words, MATH (NAME 's delta). Then MATH is a MATH-basis for MATH. Then each MATH is obviously MATH-invariant and MATH . Therefore we have MATH . By CITE, the set MATH is a basis for MATH. In particular, MATH. Since MATH is MATH-invariant, MATH. This shows that MATH . Conversely let MATH. Then MATH. Thus MATH can be uniquely expressed as MATH . Act MATH on both sides, and we get MATH . Therefore, by the uniqueness of the expression, we have MATH and MATH. This implies that each MATH lies in MATH and that MATH . Thus we have shown the inclusion MATH . |
math/0011255 | Let MATH. We naturally interpret the ``empty exterior product" to be equal to the coefficient ring. Thus the result follows from the formula MATH. Let MATH. In the proof of REF , we have already shown that the both sides have the same MATH-basis MATH . |
math/0011263 | REF was proved for MATH in CITE. If MATH then for MATH . Therefore MATH . The NAME antipode MATH is computed from REF . We give the proof for MATH only. The general case will be treated elsewhere. Let MATH and MATH . Then we find together with REF of the main theorem: MATH . An action of MATH on tensors MATH and MATH is given as follows MATH . It would be desirable to study MATH-orbits on MATH and present full classification of all orbits in terms of invariants. However, this topic exceeds the scope of this paper and will be presented elsewhere. |
math/0011263 | MATH . |
math/0011265 | It suffices to distort the front MATH smoothly to a front MATH so that the resolution of MATH is the Lagrangian projection of the knot corresponding to MATH. We choose MATH to have the following properties; see REF for an illustration. Suppose that there are at most MATH points in MATH with any given MATH coordinate. Outside of arbitrarily small ``exceptional segments," MATH consists of straight line segments. These line segments each have slope equal to some integer between REF and MATH inclusive; outside of the exceptional segments, for any given MATH coordinate, the slopes of the line segments at points with that MATH coordinate are all distinct. The purpose of the exceptional segments is to allow the line segments to change slopes, by interpolating between two slopes. When two line segments exchange slopes via exceptional segments, the line segment with higher MATH coordinate has higher slope to the left of the exceptional segment, and lower slope to the right. It is always possible to construct such a distortion MATH. Build MATH starting from the left; a left cusp is simply two line segments of slope MATH and MATH for some MATH, smoothly joined together by appending an exceptional segment to one of the line segments. Whenever two segments need to cross, force them to do so by interchanging their slopes (again, with exceptional segments added to preserve smoothness). To create a right cusp between two segments, interchange their slopes so that they cross, and then append an exceptional segment just before the crossing to preserve smoothness. We obtain the Lagrangian projection of the knot corresponding to MATH by using the relation MATH. This projection consists of horizontal lines (parallel to the MATH axis), outside of a number of crossings arising from the exceptional segments. These crossings can be naturally identified with the crossings and right cusps of MATH or MATH. In particular, right cusps in MATH become the crossings associated to a simple loop. It follows that the Lagrangian projection corresponding to MATH is indeed the resolution of MATH, as desired. |
math/0011265 | For a term in MATH of the form MATH, where we exclude powers of MATH's, we wish to prove that MATH, MATH, and MATH for all MATH. Consider the boundary of the map which gives the term MATH. By definition, the portion of this boundary connecting MATH to MATH belongs to link component MATH on one hand, and MATH on the other. We similarly find that MATH and MATH. |
math/0011265 | Let MATH be a DGA with MATH. Consider an elementary automorphism of MATH sending MATH to MATH, where MATH does not involve MATH; since MATH is in MATH, it is easy to see that this automorphism descends to a map on characteristic algebras. We conclude that tamely isomorphic DGAs have tamely isomorphic characteristic algebras. On the other hand, equivalence of characteristic algebras is defined precisely to be preserved under stabilization of DGAs. |
math/0011265 | For clarity, we first relabel the MATH as MATH. We may assume that the MATH are ordered so that MATH contains only terms involving MATH, MATH; see CITE. Let MATH be the smallest number so that MATH. We can write MATH, where MATH and the expression MATH does not involve MATH. After applying the elementary isomorphism MATH, we may assume that MATH and MATH. For any MATH such that MATH involves MATH, replace MATH by MATH. Then MATH does not involve MATH unless MATH; in addition, no MATH can involve MATH, since then MATH would involve MATH. Set MATH and MATH; then MATH and MATH does not involve MATH or MATH for any other MATH. Repeat this process with the next smallest MATH with MATH, and so forth. At the conclusion of this inductive process, we obtain MATH with MATH (and MATH), and the remaining MATH satisfy MATH; relabel these remaining generators with MATH's. |
math/0011265 | The statement for MATH is obvious. To calculate MATH, note that the image of MATH in MATH is generated by MATH, MATH, MATH, MATH, MATH, MATH, MATH, and MATH. |
math/0011265 | This follows immediately from the fact that MATH is generated by MATH. |
math/0011265 | We use REF . The first equality is obvious. For the second equality, we claim that, for fixed MATH and MATH, MATH only appears in conjunction with MATH in the expressions MATH and MATH, for arbitrary MATH. It then follows that MATH is the number of MATH of degree MATH, which is MATH. To prove the claim, suppose that MATH contains a term MATH. Since MATH and MATH, there must be another term in MATH which, when we apply MATH, gives MATH; but this term can only be MATH. The same argument obviously holds for MATH. |
math/0011265 | Note that MATH the characteristic algebra MATH and the choice of augmentation MATH determine the right hand side. On the other hand, the dimension of the degree MATH part of MATH is MATH if MATH, and MATH if MATH. It follows that we can calculate MATH and MATH from MATH, MATH, and MATH. Fix MATH. By REF , we can then calculate MATH and hence the NAME polynomial MATH . Letting MATH vary over all possible augmentations yields the proposition. |
math/0011265 | We perform a series of computations in MATH: MATH . Substituting for MATH and MATH in the relations MATH yields the relations in the statement of the lemma. Conversely, given the relations in the statement of the lemma, and setting MATH and MATH, we can recover the relations MATH. |
math/0011265 | Suppose otherwise, and consider the algebra MATH obtained from MATH by setting MATH. There is an obvious projection from MATH to MATH which is an algebra map; under this projection, MATH map to MATH, with MATH in MATH. But it is easy to see that MATH, with MATH and MATH, and it follows that there do not exist such MATH. |
math/0011265 | Let MATH be the characteristic algebra of the Legendrian mirror of MATH. Since the relations in MATH are precisely the relations in MATH reversed, REF implies that there do not exist MATH such that MATH. On the other hand, there certainly exist MATH such that MATH; for instance, take MATH and MATH. Hence MATH and MATH are not isomorphic. This argument still holds if some number of generators is added to MATH and MATH, and so MATH and MATH are not equivalent. The result follows from REF . |
math/0011265 | Similar to the proof of REF . |
math/0011265 | It is clear that the only expressions in MATH which are invertible from either side are products of some number of MATH, MATH, and MATH, with inverses of the same form. Since MATH all commute, the lemma follows. |
math/0011265 | Since MATH, MATH is certainly invertible from the right. Now consider adding to MATH the relations MATH, MATH, MATH, and MATH for all MATH not previously mentioned. The resulting algebra is isomorphic to MATH, in which MATH is not invertible from the left. We conclude that MATH is not invertible from the left in MATH either, as desired. |
math/0011265 | From REF , MATH and MATH are not equivalent. |
math/0011265 | In REF , we have already calculated the first-order NAME polynomials for MATH, once we allow the grading MATH to range in MATH. The polynomials for the link MATH and grading MATH are identical, except with the indices MATH and MATH reversed. It is easy to compute that there is no choice of MATH for which these polynomials coincide with the polynomials for MATH given in REF . The result now follows from REF . |
math/0011269 | The first and third assertions are clear. For the second assertion we need to check that MATH is coherent. The NAME formula implies that the maps defining the MATH's are MATH-linear. Therefore, we see by induction that MATH, being a kernel of a MATH-map of coherent sheaves and using CITE, is coherent. Let MATH. We show by induction on MATH that MATH is also torsion free. For MATH this is clear since MATH. Suppose that MATH is torsion free. From the short exact sequence MATH and the fact that MATH is locally free since MATH is smooth we find MATH . It follows that MATH and from the induction hypothesis we get that MATH is torsion free. The short exact sequence MATH and the induction hypothesis now show that MATH is an extension of torsion free modules hence also torsion free. In the course of the proof we showed that MATH is torsion free for each MATH. Consider now an open affinoid domain MATH. It follows by rank consideration that the sequence MATH must stabilize. Indeed, since both MATH and MATH are coherent, it follows that their restrictions are associated to the MATH-modules MATH and MATH. By NAME 's normalization lemma CITE the algebra MATH is finite over some NAME algebra MATH and therefore MATH is a product of fields. By tensoring with MATH and considering the dimensions over the various fields in the product it is clear that MATH for sufficiently large MATH, which implies, since MATH is torsion free, that MATH. It follows that MATH for a sufficiently large MATH and is thus associated with the module MATH. This proves that MATH is coherent and completes the proof of the lemma. |
math/0011269 | This is because on an affinoid space all extensions of MATH by itself are trivial. Indeed, MATH. |
math/0011269 | Immediate from REF . |
math/0011269 | This follows if we show that if MATH is the complement of MATH then MATH. This is true for the trivial isocrystal by purity CITE and then follows easily for all unipotent isocrystals by induction. |
math/0011269 | This follows since MATH and MATH is unipotent by REF . |
math/0011269 | We may assume that MATH is dense in MATH. Otherwise, since MATH is smooth we can extend MATH to the connected components with non-empty intersection with it and extend it as zero in the other components. We use induction on the rank of MATH. We have a short exact sequence MATH with MATH unipotent as well. Let MATH be the image in MATH of MATH. By the induction hypothesis it is the restriction to MATH of a subcrystal MATH of MATH. Pulling back REF via MATH we see that we may assume that MATH is surjective. Now by rank considerations, either MATH, in which case the result is clear, or MATH is an isomorphism. In this last case we find a splitting MATH over MATH and by REF this extends to MATH giving us our required subcrystal. |
math/0011269 | Any two NAME endomorphisms have a common power so that it is sufficient to prove the result for a single NAME endomorphism. We first show uniqueness. In fact we show this for a point of MATH in an arbitrary extension field MATH. Suppose MATH and MATH are both fixed by MATH with MATH. Then we have MATH . Therefore, MATH so MATH so the theorem implies MATH. We now show existence. For some possibly huge field MATH, which we may assume NAME by further extension, we have a point MATH. For the point MATH to be a fixed point we need MATH or MATH. Let MATH be the unique element such that MATH. By the theorem we may solve MATH and obtain our fixed point MATH. Since the fixed point MATH is unique, we have MATH for every NAME automorphism, which shows that MATH is already defined over MATH. The behavior of these points under composition is again clear from uniqueness. |
math/0011269 | This is just a reformulation of REF . |
math/0011269 | In the diagram above, the horizontal maps are induced by the elements MATH and MATH in the top and bottom row respectively. The vertical maps are given in REF For appropriate choice of NAME endomorphisms MATH and MATH on MATH and MATH respectively, inducing maps MATH and MATH on the corresponding principal spaces, we have MATH. It follows from this that MATH is MATH invariant and therefore must be equal to MATH. Spelling this out gives the result. |
math/0011269 | REF is CITE. REF. and REF. both follow from the fact, proved in the course of proving REF there, that there is a surjective morphism, NAME equivariant from its construction, MATH, and it is well known that MATH is mixed with positive weights. REF is stated in CITE and proved in the appendix to CITE. Finally, REF. is discussed in REF. |
math/0011269 | That MATH has the underlying MATH-vector space structure MATH is clear from the fact that MATH is a tensor functor. By REF we have MATH, hence MATH. Write the action of MATH on objects of MATH as MATH. Let MATH. Then we define a sequence MATH, where MATH. It is easy to see that the MATH define an element MATH and that MATH so defined is MATH-linear. Suppose now that MATH and MATH belong to MATH. The MATH-action factors through some MATH. For each MATH and MATH there are maps of MATH-modules, MATH and MATH, from MATH to MATH and MATH respectively. Their tensor product sends MATH to MATH hence MATH to MATH so MATH. This is the last assertion of the lemma from which the fact that MATH is a ring homomorphism follows easily. Indeed, we have MATH . |
math/0011269 | By REF the group MATH is the group corresponding to the NAME category MATH together with the fiber functor MATH sending a module to its underlying vector space. We recall that an element of MATH consists of a family of maps MATH, where MATH runs over the objects of MATH. For each MATH, MATH is an automorphism. Such a family should satisfy the following properties: CASE: The map MATH is the identity on MATH, where MATH is the unit object of MATH, that is, the MATH-module MATH. CASE: For any two objects MATH and MATH we have MATH. CASE: For every map MATH we have MATH. It is clear that MATH is a group. Let MATH. Clearly MATH acts on MATH. An element MATH therefore defines an automorphism MATH of MATH. It is immediately checked that the family MATH defines an element of MATH. This gives a map MATH. The inverse map is given as follows: Let MATH be an element of MATH. We define MATH. The elements MATH are compatible under the maps MATH since the identity elements are. They thus define an element MATH. The inverse map will map MATH to MATH and this map is well defined once we show that MATH. That MATH follows immediately from MATH. We next show that for any MATH the automorphism MATH is given by multiplication by MATH. Suppose MATH is in fact a MATH-module. Let MATH and define a map MATH by MATH. This is clearly a map of MATH-modules. Then MATH . By MATH-linearity this extends to give the claim. This shows that the composition MATH is the identity. It remains to check that MATH and for that it suffices to check that MATH. But this is now clear from REF of the family MATH and from the construction of MATH. |
math/0011269 | We need to show that MATH induces a bijection on MATH-points for any commutative MATH-algebra MATH. We first claim that for any MATH with MATH there is a unique MATH with MATH such that MATH. We show by induction the same for MATH. Then by uniqueness these solutions glue to give the unique MATH. For MATH there is nothing to prove. Suppose we already found MATH. Let MATH. This is an affine space for MATH where MATH. We have a map MATH given by MATH. Since MATH we find for MATH that MATH with MATH given by MATH. It follows from REF that MATH is invertible. This immediately implies that MATH is invertible and in particular there is a unique solution MATH to MATH. It remains to prove that if MATH, then the unique MATH so constructed is also in MATH. To do this we repeat the above argument with MATH replaced by MATH, MATH replaced by MATH and MATH replaced by MATH. The same argument works by using REF and the fact that the successive quotients MATH still have only negative weights. We need to prove that MATH. We have MATH . On the other hand MATH . We see that MATH and MATH are both solutions to the same equation which, since MATH has a unique solution, and they are therefore equal. |
math/0011269 | The only point which is not completely clear is the compatibility of the differentials on MATH and MATH. This follows, for example on functions, from the fact that the sections MATH are horizontal, hence for an isocrystal MATH and a section MATH we have MATH. |
math/0011269 | Suppose MATH. Let MATH. Since MATH is horizontal we have by REF that MATH. Applying REF to the inclusion MATH shows that MATH for any MATH. It follows that MATH is an abstract NAME function and the inclusion MATH defines a morphism of abstract NAME functions. On the other hand the MATH map provides a map from MATH to MATH showing that MATH. |
math/0011269 | The open subset has a non-zero open intersection with at least one MATH, it follows that MATH vanishes on MATH and by the proposition we have MATH. |
math/0011269 | If MATH and MATH then MATH is locally constant. Suppose its value on some MATH is MATH. Then MATH and MATH coincide on MATH hence MATH by REF . |
math/0011269 | Let MATH so MATH. Define the connection MATH as follows. As a MATH-module MATH and MATH. We let MATH and MATH be the projections on MATH and MATH respectively. Note that MATH is horizontal. The connection MATH is an extension of unipotent isocrystals and is therefore unipotent. However, it may not be integrable. We consider MATH. Choose some closed point MATH. Since MATH maps to MATH in MATH it follows that MATH. The NAME complex of MATH restricted to MATH is exact. Therefore we can choose MATH such that MATH. Then MATH and MATH. By REF we have MATH. We can analytically continue MATH to a collection MATH. We claim that the abstract NAME function MATH satisfies MATH. Indeed, MATH and MATH so MATH. It follows that the restriction of MATH to MATH induces a morphism MATH and our claim follows and with it the theorem. |
math/0011269 | If there is a subcrystal N such that MATH then MATH is a subobject of MATH. If MATH then the abstract NAME function MATH is a quotient object of MATH. Repeating this we get by rank considerations to a minimal subquotient. |
math/0011269 | Suppose that MATH, MATH, represent the same NAME function. The NAME function MATH is therefore MATH and it follows that for any MATH, MATH. Let MATH. Then by REF we have for any MATH that MATH. Let MATH be the projection from MATH to MATH. We claim that both MATH and MATH are isomorphisms. To show that MATH in injective we notice that MATH is injective and MATH is a subcrystal of MATH contained in MATH hence must be MATH by minimality. On the other hand, MATH is a subcrystal of MATH and MATH for any MATH hence again by minimality MATH. The same arguments apply by symmetry to MATH. It is now clear that the projections induce isomorphisms of abstract NAME functions MATH. But MATH on MATH and therefore MATH. Let MATH and MATH be two maps of MATH into MATH. Then MATH is a subcrystal of MATH containing MATH and therefore must be equal to MATH, hence MATH. |
math/0011269 | Easy from REF . |
math/0011269 | We know that for any two opens MATH with MATH not empty the restriction MATH is injective by REF . To prove the sheaf property it therefore suffices to show that if MATH is an open covering of MATH and we have MATH such that MATH and MATH coincide on MATH then there is a MATH restricting to the MATH. Suppose MATH has a minimal representation MATH. For each pair MATH and MATH the restrictions of MATH and MATH to MATH are minimal by REF and represent the same function by assumption. It follows from REF that there is a unique isomorphism MATH carrying MATH to MATH and MATH to MATH. By the uniqueness these isomorphisms match well when MATH and MATH vary and therefore allow us to glue the MATH to an isocrystal MATH on MATH. It is also clear than that the MATH glue together to MATH and that the MATH taken together supply a well defined horizontal section MATH for any MATH. Thus we obtain our required NAME function. |
math/0011269 | If MATH is affine then there is a basis of strict neighborhoods of MATH in MATH which are affinoid compare the proof of CITE. The result therefore follows from REF . |
math/0011269 | Suppose MATH is a NAME function in a representation of degree at most MATH. Since MATH is affine the underlying MATH-module of MATH is free by REF . We may therefore write MATH with MATH and MATH. It thus suffices to prove the proposition for MATH, that is, prove that MATH. It follows from the proof of REF that a closed form in MATH has an integral in MATH and so it is clear by induction that MATH. For the other direction, suppose that MATH is a NAME function and that MATH can be written in a short exact sequence MATH where MATH is trivial and MATH has a filtration as in MATH but of length MATH. We may further find a splitting MATH, which need not be compatible with the connection. Since MATH can be written as MATH with MATH it suffices to prove the result with MATH for MATH. We first notice that MATH (since MATH is horizontal) so it even belongs to MATH by the induction hypothesis. Suppose that MATH. Then MATH vanishes on MATH and therefore equals MATH for some MATH. It follows that MATH. This belongs to MATH hence by the induction hypothesis to MATH. Note that in the MATH case we have MATH and the same argument implies simply that MATH hence that MATH is a constant. Since MATH is trivial a general MATH can be written as a combination of MATH's of the type that was already considered with coefficients in MATH. This completes the induction step and also shows the case MATH. |
math/0011269 | To define MATH we would like to extend the definition of the fiber functors MATH to MATH. This can be done as follows: Suppose MATH. Then it is MATH for some MATH defined on some strict neighborhood of MATH in MATH. For each MATH this neighborhood contains an annulus around MATH isomorphic to MATH for some MATH. Over MATH the underlying module to MATH trivializes by REF . Thus it is isomorphic to some MATH as in the proof of REF . Since MATH is surjective it follows easily that MATH has a full set of solutions on MATH and the functor that sends MATH to the set of solutions is the required fiber functor. Therefore, given a NAME function MATH, the collection MATH extends to give MATH and applying the section MATH we get an element in MATH. Thus, the definition of MATH extends immediately to provide the ring homomorphism MATH extending MATH into MATH and a similar map on differential forms. These maps are injective because they extends MATH which is already injective by REF . The maps MATH on functions and forms are clearly compatible with differentials. It therefore suffices to prove that the image of MATH is exactly MATH (a similar argument applies to differential forms). We claim that in fact MATH. By the way these spaces are constructed it is easy to see that using induction we only need to prove the following claim: If MATH and MATH, then MATH. This follows by the fact that MATH is uniquely determined by the condition of being MATH-equivariant and the fact that MATH is MATH-equivariant since MATH is an endomorphism of MATH. |
math/0011269 | Formal check. |
math/0011269 | First we claim that MATH depends only on MATH. Indeed, if we are given two short exact sequences for the same MATH, MATH: MATH and MATH: MATH, we may consider MATH which is the limit of the diagram MATH. There is then a map MATH which extends to a map of abstract NAME functions and a map of short exact sequences with both MATH and MATH. REF therefore proves the claim. We have seen REF that an abstract NAME function has a minimal subquotient. Since subquotients extend to maps of short exact sequences for an appropriate choice of filtration on the subquotient we now see that an abstract NAME function has the same MATH as its minimal representative, hence MATH depends only on the underlying NAME function. The linearity is now straightforward. |
math/0011269 | We can obtain the function MATH as follows: The isocrystal is MATH (as defined in REF ) with MATH, MATH and MATH. We use the obvious short exact sequence MATH. Then MATH. The pullback MATH is obtained by finding a preimage in MATH to MATH and applying to it the connection. Taking the preimage MATH we get that MATH is the NAME cohomology class of MATH. The section MATH is clearly given by MATH with MATH the isomorphism from MATH to MATH given by the projection on the second component. The result is now clear. |
math/0011269 | By covering a tight situation with affine ones it suffices to prove the second statement. Suppose MATH is affine. The surjectivity of MATH is clear from REF . To prove the exactness suppose MATH is in the kernel of MATH. It follows from REF that MATH can be written as MATH and by REF we have then MATH. We may assume that the MATH are independent over MATH. It now follows that MATH for each MATH, hence that MATH with MATH and therefore MATH for an appropriate choice of MATH. Thus, MATH. |
math/0011269 | We can take an element of MATH and represent it as a hyper-cocycle with respect to the covering MATH, MATH where MATH, MATH, the MATH form a one-cocycle, MATH and we have MATH on MATH. If MATH, then the image of MATH in MATH is given by the collection MATH. To compute the image under MATH we, following the procedure described after REF , choose lifts MATH and consider the cocycle resulting from the differences MATH. But MATH so each of the functions MATH is constant on MATH. Since the NAME cohomology of MATH with constant coefficients is trivial we can always arrange to fix the integrals MATH in such a way that MATH so that the application of MATH gives MATH. This is clearly the same as going along the diagram first down and then right. |
math/0011269 | The only if part is clear. For the if part we notice that we may modify MATH by elements from MATH so that we may assume MATH. But then it follows from the definition of MATH and the fact that MATH is a sheaf that MATH comes from an element MATH via restriction to the MATH and taking MATH and therefore MATH comes from MATH. |
math/0011269 | We prove by induction on the number of differential forms the more precise statement saying that MATH is a NAME function of MATH and MATH and that its differential is indeed MATH as defined above. Suppose that we proved this for at most MATH differential forms. By the induction hypothesis it is clear that MATH is a NAME differential form and we easily compute that MATH . It follows that MATH can be integrated. Let MATH be its integral. It is immediate to see that the restriction of MATH to the diagonal MATH is MATH. By functoriality MATH is constant on the diagonal and we may therefore assume that MATH. But then for fixed MATH this last equality together with the fact that MATH implies that MATH. |
quant-ph/0011036 | CITE Suppose MATH. MATH is obviously linear, so to check that MATH is a quantum operation we need only prove that it is completely positive. Let MATH be any positive operator acting on the state space of an extended system, MATH, and let MATH be some state of that system. Defining MATH, we have MATH by the positivity of the operator MATH. It follows that MATH and thus for any positive operator MATH, the operator MATH is also positive, as required. This completes the first part of the proof. Suppose next that MATH is a quantum operation. Our aim will be to find an operator-sum representation for MATH. Suppose we introduce a system, MATH, with the same dimension as the original quantum system, MATH. Let MATH and MATH be orthonormal bases for MATH and MATH. It will be convenient to use the same index, MATH, for these two bases, and this can certainly be done as MATH and MATH have the same dimensionality. Define a joint state MATH of MATH by MATH . The state MATH is, up to a normalization factor, a maximally entangled state of the systems MATH and MATH. This interpretation of MATH as a maximally entangled state may help in understanding the following construction. Next, we define an operator MATH on the state space of MATH by MATH . We may think of this as the result of applying the quantum operation MATH to one half of a maximally entangled state of the system MATH. It is a truly remarkable fact, which we will now demonstrate, that the operator MATH completely specifies the quantum operation MATH. That is, to know how MATH acts on an arbitrary state of MATH, it is sufficient to know how it acts on a single maximally entangled state of MATH with another system. The trick which allows us to recover MATH from MATH is as follows. Let MATH be any state of system MATH. Define a corresponding state MATH of system MATH by the equation MATH . Notice that MATH . Let MATH be some decomposition of MATH, where the vectors MATH need not be normalized. Define a map MATH . A little thought shows that this map is a linear map, so MATH is a linear operator on the state space of MATH. Furthermore, we have MATH . Thus MATH for all pure states, MATH, of MATH. By linearity it follows that MATH in general. |
quant-ph/0011036 | Suppose MATH for some unitary MATH. Then MATH which shows that MATH and MATH generate the same operator. Conversely, suppose MATH . A little thought shows that for this equation to hold each MATH can be expressed as a linear combination of the MATH, MATH. Thus MATH from which we see that MATH is unitary, as required. |
quant-ph/0011036 | A very useful inequality in information theory is MATH, for all positive MATH, with equality if and only if MATH. Here we need to rearrange the result slightly, to MATH, and then note that MATH which is the desired inequality. The equality conditions are easily deduced by noting that equality occurs in the second line if and only if MATH for all MATH, that is, the distributions are identical. |
quant-ph/0011036 | CASE: Obvious from the relevant definitions. CASE: Since MATH we have MATH . Thus MATH. But MATH, so MATH with equality if and only if MATH is a deterministic function of MATH. CASE: Follows from the previous result. CASE: To prove subadditivity and, later, strong subadditivity we use the fact that MATH for all positive MATH, with equality if and only if MATH. This fact is easily proved using calculus. We find that MATH . NAME may easily be recovered by multiplying by a constant (to change the base of the logarithm to base MATH), and rearranging the expression. Notice that equality is achieved if and only if MATH for all MATH and MATH. That is, the subadditivity inequality is saturated if and only if MATH and MATH are independent. CASE: Follows from subadditivity and the relevant definitions. CASE: NAME subadditivity of NAME entropy follows from the same technique as used to prove subadditivity; the difficulty level is about the same as that proof. Interestingly, while carrying out the proof one notes that the equality conditions for strong subadditivity are that MATH forms a NAME chain. |
quant-ph/0011036 | Inserting the relevant definitions, the result is equivalent to MATH which is a rearranged version of the strong subadditivity inequality proved earlier. |
quant-ph/0011036 | We prove the result for MATH, and then induct on MATH. Using only the definitions and some simple algebra we have MATH which establishes the result for MATH. Now we assume the result for general MATH, and show the result holds for MATH. Using the already established MATH case, we have MATH . Applying the inductive hypothesis to the first term on the right hand side gives MATH so the induction goes through. |
quant-ph/0011036 | The first inequality was proved in REF on page REF. From the definitions we see that MATH is equivalent to MATH. From the fact that MATH is a NAME chain it is easy to prove that MATH is also a NAME chain, and thus MATH. The problem is thus reduced to proving that MATH. This is just the strong subadditivity inequality, which we already proved. Suppose MATH. Then it is not possible to reconstruct MATH from MATH, since if MATH is the attempted reconstruction based only on knowledge of MATH, then MATH must be a NAME process, and thus MATH by the data processing inequality. Thus MATH. On the other hand, if MATH, then we have MATH and thus whenever MATH we have MATH. That is, if MATH then we can infer with certainty that MATH was equal to MATH, allowing us to reconstruct MATH. |
quant-ph/0011036 | Let MATH and MATH be orthogonal decompositions for MATH and MATH. From the definition of the relative entropy we have MATH . We substitute into this equation the equations MATH and MATH where MATH. Notice that MATH satisfies the equations MATH and MATH (such matrices are called doubly stochastic). Substitution gives MATH . MATH is a strictly concave function, so MATH, where MATH, with equality of and only if there exists a value of MATH for which MATH. Thus MATH with equality if and only if MATH is a permutation matrix. This has the form of the classical relative entropy, from which we deduce that MATH with equality if and only if MATH for all MATH, and MATH is a permutation matrix. To simplify the equality conditions further, note that by relabeling the eigenstates of MATH if necessary, we can assume that MATH is the identity matrix, and thus that MATH and MATH are diagonal in the same basis. The condition MATH tells us that the corresponding eigenvalues of MATH and MATH are identical, and thus MATH are the equality conditions. |
quant-ph/0011036 | CASE: Clear from the definition. CASE: From the non-negativity of the relative entropy, MATH, from which the result follows. CASE: From the NAME decomposition, as discussed in REF, we know that the eigenvalues of systems MATH and system MATH are the same. The entropy is determined completely by the eigenvalues, so MATH. CASE: Let MATH and MATH be the eigenvalues and corresponding eigenvectors of MATH. Observe that MATH and MATH are the eigenvalues and eigenvectors of MATH, and thus MATH as required. CASE: Immediate from the preceding result. |
quant-ph/0011036 | (Original?) The proof is to apply NAME 's inequality to MATH and MATH, MATH . The result will follow if we can prove that MATH. To do this, we apply the cyclic property of the trace and the completeness and orthogonality relations for the projectors to obtain MATH . A little thought shows that MATH commutes with MATH and thus with MATH, so MATH . This completes the proof. |
quant-ph/0011036 | We begin with the pure state case, MATH. Let MATH be a system with the same state space as the MATH, and introduce a system MATH with an orthonormal basis MATH corresponding to the index MATH on the probabilities MATH. Define MATH . Since MATH is a pure state we have MATH . Suppose we perform an orthonormal measurement on the system MATH in the MATH basis. After the measurement the state of system MATH is MATH . But orthogonal measurements never decrease entropy, so MATH. Observing that MATH for the pure state case, we have proved that MATH when the states MATH are pure states. Furthermore, equality holds if and only if MATH, which is easily seen to occur if and only if the states MATH are orthogonal. The mixed state case is now easy. Let MATH be orthonormal decompositions for the states MATH, so MATH. Applying the pure state result and the observation that MATH for each MATH, we have MATH which is the desired result. The equality conditions for the mixed state case follow immediately from the equality conditions for the pure state case. |
quant-ph/0011036 | (Adapted from CITE) We begin by proving the result for MATH, and then use this to establish the result for general MATH. Let MATH and MATH be any two vectors. Applying the NAME inequality twice and some straightforward manipulations, we have MATH . By hypothesis, MATH and MATH, so MATH . Let MATH be any unit vector. Then applying the previous result with MATH and MATH gives MATH . Thus MATH . Define MATH . Note that MATH is Hermitian, so by observation number REF on page REF, MATH where the last inequality is just REF . MATH is a positive operator, so by observation number REF on page REF and the previous inequality, MATH . Finally, by REF on page REF, and the commutativity of MATH and MATH, MATH which establishes that REF holds for MATH. Let MATH be the set of all MATH such that REF holds. By inspection, we see that MATH and MATH are elements of MATH. We now use the result for MATH to prove the result for general MATH. Suppose MATH and MATH are elements of MATH, so MATH . These inequalities are of the form REF for which the MATH case has already been proved. Using the MATH result we see that MATH . Using the commutativity assumptions MATH, we see that for MATH, MATH . Thus whenever MATH and MATH are in MATH, so is MATH. Since MATH and MATH are in MATH, it is easy to see that any number MATH between MATH and MATH with a finite binary expansion must be in MATH. Thus MATH is dense in MATH. The result now follows from the continuity in MATH of the conclusion, REF . |
quant-ph/0011036 | (NAME 's theorem) (Adapted from CITE) Define MATH . Observe that MATH and MATH commute, as do MATH and MATH, and MATH and MATH. Recall REF on page REF, that all these operators are positive with respect to the NAME inner product. By the lemma, MATH . Taking the MATH matrix element of the previous inequality gives MATH which is the desired statement of joint concavity. |
quant-ph/0011036 | CITE Define MATH . Note that the first term in this expression is concave in MATH, by NAME 's theorem, and the second term is linear in MATH. Thus, MATH is concave in MATH. Define MATH . Noting that MATH and using the concavity of MATH in MATH we have MATH . That is, MATH is a concave function of MATH. Finally, defining the block matrices MATH we can easily verify that MATH. The joint convexity of MATH now follows from the concavity of MATH in MATH. |
quant-ph/0011036 | Let MATH be the dimension of system MATH. Note that MATH . Thus MATH. The concavity of MATH follows from the joint convexity of the relative entropy. |
quant-ph/0011036 | CITE The two inequalities are, in fact, equivalent. We will use convexity of the conditional entropy to prove the first, and show that the second follows. Define a function of density operators on the system MATH, MATH . From the concavity of the relative entropy we see that MATH is a convex function of MATH. Let MATH, where MATH is a pure state of the system MATH and the MATH are probabilities. From the convexity of MATH, MATH. But for a pure state, MATH, as MATH and MATH for a pure state. It follows that MATH, and thus MATH which is the first inequality. Finally, to obtain the second inequality, introduce a fourth system, MATH, purifying the system MATH. Then MATH . Since MATH is a pure state, MATH and MATH, so the previous inequality becomes MATH as we set out to show. |
quant-ph/0011036 | CITE Introduce a system, MATH, with an orthonormal basis MATH of states with index MATH corresponding to the index of the states MATH. Suppose the initial state of MATH is MATH . Applying the joint entropy theorem on page REF and doing a little algebra, we see that MATH and MATH. The result now follows from the observation that MATH, which is a restatement of strong subadditivity. |
quant-ph/0011036 | The sufficiency of REF follows from the superdense coding technique discussed in REF. Sufficiency in the case MATH is obvious, so we suppose MATH. NAME prepares MATH maximally entangled pairs of qubits, and sends one qubit of each pair to NAME, who can use them in conjunction with sending MATH qubits to NAME to transmit MATH bits to NAME, using superdense coding. NAME uses her remaining allotment of MATH qubits to transmit the remaining MATH bits in the obvious way. The proof that REF are necessary follows from an application of NAME 's theorem and the non-increasing property of the NAME MATH under partial traces, as discussed in the previous section. The details are as follows. Let MATH be NAME 's MATH bits of information, which we assume is uniformly distributed over MATH. Without loss of generality, it can be assumed that the protocol between NAME and NAME is of the following form. For any value MATH of MATH, NAME begins with a set of qubits in state MATH and NAME begins with a set of qubits in a standard state MATH. The protocol consists of a sequence of steps, where at each step one of the following four processes takes place. CASE: NAME performs a unitary operation on the qubits in her possession. CASE: NAME performs a unitary operation on the qubits in his possession. CASE: NAME sends a qubit to NAME. CASE: NAME sends a qubit to NAME. After these steps, NAME performs a measurement on the qubits in his possession, which has outcome MATH. NAME 's goal is to maximize the mutual information between MATH and NAME 's input, MATH. NAME may be wondering about the possibility of a protocol in which NAME starts with a mixed state, or in which non-unitary operations are allowed. Note that using the techniques of REF any non-unitary operation can be simulated by a unitary operation, with the introduction of an appropriate environmental model, and by the purification procedure discussed in REF it is possible to simulate any protocol in which NAME starts with a mixed state by a protocol in which NAME starts with a pure state. Let MATH be the density operator of the set of qubits that are in NAME 's possession after MATH steps in the protocol have been executed, and MATH be the density operator of NAME 's system after MATH steps, averaged over all possible inputs, MATH. Due to NAME 's bound, it suffices to upper bound the final value of MATH. We consider the evolution of MATH and MATH. Initially, MATH, since NAME begins in a state independent of MATH. Now, consider how MATH and MATH change for each of the four processes above. CASE: NAME performing a unitary operation on her qubits does not affect MATH and hence has no effect on MATH or MATH. CASE: It is easy to verify that MATH and MATH are invariant under unitary transformations, so NAME performing a unitary on his qubits does not affect MATH and MATH, either. CASE: NAME sends a qubit to NAME. Let MATH denote NAME 's qubits after MATH steps and MATH denote the qubit that NAME sends to NAME at step number MATH. By the subadditivity inequality discussed in subsection REF and the fact that, for a single qubit MATH, MATH, MATH. Also, by the triangle inequality discussed in subsection REF, MATH. It follows that MATH and thus MATH . CASE: NAME sends a qubit to NAME. In this case, MATH is MATH with one qubit traced out. We saw in the previous section that MATH does not increase under partial trace, so MATH. Note also that MATH for this process, by the triangle inequality. Now, since MATH can only increase when NAME sends a qubit to NAME and by at most REF follows from the NAME bound. Also, since MATH can only increase when one party sends a qubit to the other and by at most REF follows from the observation that MATH is an upper bound on the NAME MATH, and the NAME bound. Finally, note that even if pre-shared entanglement is allowed, so MATH may start out greater than zero, MATH is still zero at the start of the protocol, and thus the reasoning leading to the constraint MATH still holds. This completes the proof. |
quant-ph/0011036 | CITE For completeness, we outline the construction of the projector MATH onto the typical subspace. Suppose MATH has orthogonal decomposition, MATH . Then MATH has orthogonal decomposition MATH where the sum is over all sequences MATH, MATH and MATH. We say a sequence MATH is MATH-typical if MATH . Intuitively, the sequence MATH can be thought of as the sequence of outputs produced by a classical source producing independent random variables, identically distributed according to MATH. Using the law of large numbers and taking the logarithm of the above definition, we see that in the limit as MATH goes to infinity, a typical sequence occurs with probability going to one. Furthermore, since the sum of a set of probabilities is at most one, and MATH we see that there at most MATH-typical sequences. We now define MATH to be the projector onto the MATH-typical subspace. Note that MATH so by the law of large numbers, for sufficiently large MATH, MATH, for any MATH. Setting MATH proves REF . Furthermore, by REF is diagonal in the same basis as MATH, and thus commutes with MATH, REF . The second property now follows from the identity MATH and the observations that MATH and MATH . |
quant-ph/0011036 | Proof of achievability The compression scheme used to prove achievability is exactly the same as that used by CITE, although the analysis is made slightly different by the use of dynamic fidelity as the reliability criterion. Let MATH be such that MATH. Define MATH to be the projector onto the MATH-typical subspace, and use MATH to denote the MATH-typical subspace. By the typical subspace theorem, for all MATH sufficiently large, MATH and MATH . Let MATH be any MATH dimensional NAME space containing MATH. The encoding is done in the following fashion. First a measurement is made, described by the complete set of orthogonal projectors MATH, with corresponding outcomes we will call MATH and MATH. If outcome MATH occurs nothing more is done and the state is left in the typical subspace. If outcome MATH occurs then we replace the state of the system with some standard state MATH chosen from the typical subspace. It follows that the encoding is a map MATH and has the operator-sum representation MATH where MATH and MATH is an orthonormal basis for the orthocomplement of the typical subspace. The decoding operation MATH is just the identity on MATH, MATH. With these definitions for the encoding and decoding it follows that MATH where the last line follows from the theorem of typical subspaces. But MATH can be made arbitrarily small and thus it follows that there exists a reliable compression scheme MATH of rate MATH whenever MATH. Proof of the weak converse In REF, on page REF, we prove the following result, known as the entropy-fidelity inequality. For any MATH and complete quantum operations MATH and MATH, MATH where MATH is the dimension of the NAME space of MATH and MATH is the coherent information defined by MATH and MATH is a non-negative quantity known as the entropy exchange. From the non-negativity of the entropy exchange it follows that MATH and thus from the entropy-fidelity inequality MATH . Applying REF to MATH and noting that MATH we see that MATH where MATH is the dimension of the source space MATH. Dividing by MATH and taking the limit as MATH we see that MATH (when the limit exists). Thus, for reliable transmission we obtain MATH . It follows that if MATH then all compression schemes must be weakly unreliable. Proof of the strong converse Suppose MATH purifies MATH. Then taking MATH copies of MATH, MATH purifies MATH. Define MATH where MATH is the identity operation on MATH. From REF of the NAME number enumerated on page REF we see that MATH can be written in the form MATH where MATH and MATH . By definition we have MATH and thus MATH . To bound the dynamic fidelity we examine the individual terms in this equation. Note that MATH where MATH. Let MATH be the projector onto the support of MATH. Notice that MATH as an operator inequality, and thus MATH . But MATH and by the second part of the typical subspace theorem it follows that for sufficiently large MATH, MATH . Putting it all together we have MATH for sufficiently large MATH. Note, incidentally, that how large MATH needs to be depends only on MATH and MATH, and not on MATH. Inserting this equation into REF gives the result MATH for all MATH, for sufficiently large MATH. It follows that if MATH then MATH as MATH, which is what we set out to prove. |
quant-ph/0011036 | NAME proof of the weak converse: We use the ``generalized entropy-fidelity" lemma from REF. This result states that for any set of operations MATH such that MATH and MATH are trace-preserving, MATH . Applying this result with MATH, MATH, MATH, and MATH gives the result by the same reasoning used earlier in the proof of the weak converse without a classical side channel. NAME proof of the strong converse: Define MATH . The proof of the strong converse in the presence of a classical side channel now proceeds exactly as that for the strong converse without a classical side channel. |
quant-ph/0011036 | Since the set of density operators is compact, there exists a finite MATH-net MATH. Suppose two or more of these density operators have the same entropy. For example, suppose MATH. Then we set MATH and perturb MATH by a small amount, for example MATH where MATH is chosen in such a way that MATH, but MATH. (Note that if MATH then it may be necessary to perturb MATH using a different state, say a pure state). It is easy to see that by perturbing all the density operators MATH by a distance less than MATH it is possible to ensure that the resulting perturbed set MATH is entropy distinct. Moreover, the resulting set is a MATH-net, since MATH where MATH has been chosen such that MATH. This completes the proof of the lemma. |
quant-ph/0011036 | (Main theorem) Returning to the main theorem, for each MATH, let MATH be a finite entropy distinct MATH-net. We will use these nets to construct an increasing sequence MATH of MATH-nets. Set MATH. Given MATH we construct MATH as follows. By perturbing each element in MATH by a distance at most MATH to create a perturbed set MATH it is possible to ensure that the resulting union MATH is entropy distinct. Observing that MATH is a MATH-net we see that MATH is a finite, entropy distinct MATH-net such that MATH . Define MATH . It is now easy to see that MATH is a countable, entropy distinct subset of the set of density operators. Furthermore, MATH is dense, since given MATH and MATH, choose MATH such that MATH, and then MATH such that MATH . This completes the proof of the theorem. |
quant-ph/0011036 | The symmetry property is obvious from the definition. As already noted, the entropy-boundedness is an immediate consequence of the definition of MATH, and the concavity of the entropy. Also as noted, the entropy-entanglement inequality MATH follows from the concavity of the conditional entropy. We give two proofs that more systems means more entanglement. The first proof is from the operational definition of entanglement. If we have enough singlet pairs to make MATH good copies of MATH then by throwing away all MATH copies of MATH we obtain MATH good copies of MATH. The result follows. The result also follows easily from the entropic definition of entanglement. Suppose MATH . Define MATH and let MATH be any ensemble for MATH. From the concavity of entropy it follows that MATH . This establishes the result. To prove that adding uncorrelated systems does not change the entanglement, note first that MATH. Suppose MATH is an ensemble for MATH such that MATH . Suppose MATH is an eigenensemble decomposition for system MATH. Then it is clear that MATH is an ensemble for MATH, and thus MATH . Thus MATH. |
quant-ph/0011036 | (convexity theorem for coherent information) The theorem follows from the concavity of the conditional entropy, REF , on page REF. By REF . The theorem follows immediately from the concavity of the conditional entropy. |
quant-ph/0011036 | To prove the lemma, notice that by the second part of the data processing inequality, REF , MATH . Applying REF gives MATH and combining the previous two inequalities gives MATH where the second step follows from the quantum NAME inequality, REF . But the binary NAME entropy MATH is bounded above by MATH and MATH, so MATH . This completes the proof. |
quant-ph/0011036 | Define MATH . This always exists and is finite, since MATH for some MATH. Fix MATH and choose MATH sufficiently large that MATH . Suppose MATH is any integer strictly greater than MATH. Then by REF , MATH . Using the fact that MATH (an immediate consequence of REF ) with MATH gives MATH where MATH is the integer immediately below MATH. Plugging the last inequality into REF gives MATH . But MATH and MATH, so MATH . This equation holds for all sufficiently large MATH, and thus MATH . But MATH was an arbitrary number greater than MATH, so letting MATH we see that MATH . It follows that MATH exists, as claimed. |
quant-ph/0011036 | What this theorem tells us is that we cannot reliably transmit more than MATH qubits of information per use of the channel. When the source entropy exists, it tells us we cannot transmit sources with entropy greater than MATH; when the entropy of the source is not defined, it still rules out transmission of sources for which the limsup in the expression (which is always defined) is too large. For unitary MATH we have MATH and thus MATH . By REF with MATH, and the fact that MATH it now follows that MATH . (Note that MATH here is the dimension of a single copy of the source NAME space, so that we have inserted MATH for the overall dimension MATH of REF ). Taking MATH-s on both sides of the equation completes the proof of the theorem. |
quant-ph/0011036 | Again, this result places an upper bound on the rate at which information can be reliably transmitted through a noisy quantum channel. The proof is very similar to the earlier proof of a bound for unitary encodings. One simply applies REF with MATH and MATH again invoking the fact that MATH, and chooses MATH to be the coding that maximizes this expression, to give: MATH . Taking MATH-s on both sides of the equation completes the proof. |
quant-ph/0011036 | By the second step of the data processing inequality, REF , MATH for each MATH, and noting also that by the completeness of MATH, MATH, we obtain MATH . Applying now the convexity theorem for coherent information, MATH we obtain MATH . But MATH is trace-preserving since MATH is trace-preserving and MATH is trace-preserving, and thus by REF , MATH . Finally, an application of the quantum NAME REF along with the observations that the entropy function MATH appearing in that inequality is bounded above by one, and MATH gives MATH as we set out to prove. |
quant-ph/0011036 | Let MATH be the state of system MATH when system MATH is traced out, MATH. Let MATH be an orthonormal decomposition for system MATH. Then there exist vectors MATH in the state space of system MATH such that MATH . But we know that MATH, from which we deduce MATH. Thus, we can find orthonormal MATH such that MATH, and thus MATH . Setting MATH and noting that MATH completes the proof. |
quant-ph/0011046 | We define a NAME machine MATH that will output a sequence MATH where MATH is a positive rational number. At any time MATH, let MATH be defined as follows. If there is no MATH for which some MATH has been outputted then MATH; otherwise, MATH is the maximum of those MATH. The machine MATH will have the following property for all MATH: MATH . To define MATH, take a universal NAME machine MATH. Let MATH simulate MATH simultaneously on all inputs. If at any stage of the simulation, some MATH has been found, then MATH checks whether it can interpret MATH as a positive rational number MATH, and whether it can output the triple MATH while keeping REF . If yes, the triple is outputted, otherwise it is not, and the simulation continues. Define MATH. Then it is easy to check that MATH satisfies the conditions of the proposition. To show MATH, note that the random variable whose distribution is MATH can be represented as a function of the coin-tossing infinite sequence. It is not difficult to check that the function in question now can be implemented by a prefix NAME machine. |
quant-ph/0011046 | Both these statements are proved via standard approximations. |
quant-ph/0011046 | The proof of the existence of MATH is completely analogous to the proof of REF . To prove MATH, note first that the form of its definition guarantees that MATH is a lower semicomputable semi-density, and therefore MATH. It remains to prove MATH. Since MATH is lower semicomputable, there is a nondecreasing sequence MATH of elementary semi-density matrices such that MATH, with MATH. For MATH, let MATH. Each of the nonnegative self-adjoint operators MATH can be represented as a sum MATH . Thus, MATH, with a computable sequence MATH, where MATH. The vectors MATH and the values MATH can be chosen elementary. Noting MATH finishes the proof. The statement of sum representations using projections and elementary density matrices is weaker than the statement about MATH. |
quant-ph/0011046 | Easy, see CITE. |
quant-ph/0011046 | Use REF. |
quant-ph/0011046 | The function MATH is lower semicomputable with MATH, hence it is dominated by MATH. This shows MATH. On the other hand, the semi-density matrix MATH is lower semicomputable, so MATH, MATH, hence MATH . |
quant-ph/0011046 | Let MATH, then MATH, hence MATH. |
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