paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/0004014 | Suppose MATH is in the MATH-class and define MATH as in REF . Suppose MATH (for a proof of this statement, see REF ). The argument hinges on particular scaling properties of the functionals MATH and MATH, which enable us to convert REF into REF. Given MATH, let us for each MATH define MATH by MATH . Then we have MATH w... |
math-ph/0004014 | REF for MATH is well-known. Assume that MATH and observe that, due to the perfect scaling properties of both MATH and MATH, REF can alternatively be written as MATH . Let MATH be the principal eigenvalue respectively, an associated eigenvector of MATH in MATH with NAME boundary condition. Then MATH, which means that MA... |
math-ph/0004014 | Let MATH and consider MATH. Let MATH and define MATH. By a straightforward calculation, MATH. Let MATH. Then MATH, while MATH. This implies that MATH. Since MATH, the proof is finished. |
math-ph/0004014 | In the case MATH, use the shift-invariance of MATH, NAME 's inequality, and the monotonicity REF to obtain MATH . In the case MATH, instead of NAME 's inequality we apply MATH to deduce similarly as in REF that MATH . |
math-ph/0004014 | By combining REF and the left inequality in REF for MATH instead of MATH, we see that MATH. Since MATH, the left-hand side of REF, with MATH instead of ``MATH," is bounded below by MATH. By REF , MATH positive, finite and non-zero. |
math-ph/0004014 | Recall the notation of Subsection REF. By taking the expectation over MATH (and using that MATH is an i.i.d. field) and recalling REF, we have for any MATH that MATH . Consider the scaled version MATH of the local times MATH . Let MATH be the space of all non-negative NAME almost everywhere continuous functions in MATH... |
math-ph/0004014 | In the course of the proof, we use abbreviations MATH and MATH. Recall that MATH denotes an orthonormal basis in MATH (with inner product MATH) consisting of the eigenfunctions of MATH with NAME boundary condition. We first turn to the case MATH. Use the NAME expansion REF and the inequality MATH to obtain MATH . By NA... |
math-ph/0004014 | Let MATH. First, notice that the second term in REF can be estimated in terms of a sum: MATH . Thus, applying REF to MATH (that is, for MATH) with MATH replaced by MATH for some fixed MATH, raising both sides to the MATH-th power, and using REF we get MATH . Next we take the expectation with respect to MATH and note th... |
math-ph/0004014 | It is immediate from REF with MATH that MATH . According to REF, we have, for every MATH, MATH . Using this for MATH in REF, we see that, for sufficiently large MATH (depending only on MATH and MATH), the right-hand side of REF is no more than MATH. |
math-ph/0004014 | The idea is to construct a partition of unity MATH where MATH with MATH . Then we put MATH where MATH is the discrete gradient. Obviously, MATH is MATH-periodic in every component. The construction of MATH such that MATH satisfies REF is given at the end of this proof. Assuming the existence of the above partition of u... |
math-ph/0004014 | Having all the prerequisites, the proof is easily completed. First, MATH by REF . Therefore, combining REF with REF , we have that MATH whenever MATH is large enough. Invoking also the NAME expansion REF with respect to the eigenfunctions of MATH in MATH and the fact that MATH, we find that MATH . Now apply REF for MAT... |
math-ph/0004014 | From REF we immediately have MATH where MATH is as in REF. Here, for the upper bound we simply neglected MATH in REF, whereas for the lower bound we first wrote REF as a normalized sum of the right-hand side of REF with the walk starting and ending at all possible MATH, and then inserted MATH, applied REF, and then rec... |
math-ph/0004014 | Let MATH and let MATH. We want to apply REF with the random potential MATH and with MATH replaced by MATH for some fixed MATH. (Later we shall let MATH and pick MATH appropriately.) Recall the definition of MATH in REF and abbreviate MATH. Take logarithms in REF, multiply by MATH and use REF to obtain MATH almost surel... |
math-ph/0004014 | Let MATH and let MATH be twice continuously differentiable with MATH. If MATH, let MATH be a non-degenerate ball in MATH centered at MATH. Suppose that MATH does not belong to the exceptional null sets of the preceding assertions. In particular, there are unique infinite clusters MATH in MATH and MATH in MATH, and MATH... |
math-ph/0004014 | We begin with REF. Recall that MATH is also an eigenfunction for the transition densities of the random walk in MATH with potential MATH. Using this observation at time MATH, we can write MATH . Since MATH is nonpositive and MATH is bounded from below, we have MATH . Using the same strategy as in REF, we have MATH. Sin... |
math-ph/0004014 | Suppose that MATH. This implies that MATH almost surely and, by the law of large numbers, MATH . Suppose that MATH does not belong to the exceptional sets of REF . For sufficiently large MATH, let MATH be such that REF holds. Let MATH. The strategy for the lower bound on MATH is that the random walk performs MATH steps... |
math-ph/0004014 | Since MATH is continuous, there is a ball MATH of radius of order MATH such that MATH. If MATH is so large that MATH, then MATH and the left-hand side of REF is bounded from above by MATH. The proof now proceeds in a different way depending whether MATH or MATH. In the following, the words ``percolation," ``infinite cl... |
math-ph/0004014 | Let MATH be the solution to MATH and let MATH be its scaled version: MATH. Define the tilted probability measure MATH . We denote expectation with respect to MATH by MATH. Consider the event MATH . Then MATH can be bounded as MATH . Applying the left inequality in REF, we obtain MATH . Since MATH and MATH is continuous... |
math-ph/0004014 | Suppose MATH has an atom at MATH with mass MATH. Then, noting that MATH are only the sites with MATH, we have MATH . Since MATH and MATH, we have MATH and MATH, whereby REF immediately follows. |
math-ph/0004014 | Suppose that MATH and MATH. Set MATH and consider the probability measure MATH with density MATH with respect to MATH. Invoking that MATH, we obtain MATH . Now use the Scaling Assumption and the fact that MATH as MATH to extract the term MATH from the exponential on the right-hand side (here we recalled that MATH). Mor... |
math-ph/0004014 | Fix MATH and MATH with MATH. Recall the notation REF. Let MATH be such that for all MATH and for all MATH . Such a MATH indeed exists, since MATH as MATH and since MATH is uniformly continuous on MATH. This implies that to prove REF it suffices to find an almost-surely finite MATH such that for each MATH there is a MAT... |
math-ph/0004024 | Local exactness of a vertical complex on a coordinate chart MATH follows from a version of the NAME lemma with parameters. We have the the corresponding homotopy operator MATH where MATH. Since MATH is a vector bundle, it is readily observed that the homotopy operator is globally defined, and so is the form MATH. |
math-ph/0004024 | Bearing in mind the monomorphism REF , let MATH be a closed form whose cohomology class belongs to MATH. We will show that, if MATH is MATH-exact, MATH is exact. The form MATH is represented by the sum REF , and we have MATH . Let MATH where MATH is an exterior form on some finite-order jet manifold MATH. Given a globa... |
math-ph/0004024 | Firstly, let us describe the cohomology group MATH of the first row in REF . In this complex, the cocycles are closed forms MATH, while the coboundaries are exact forms MATH where MATH. Because of the isomorphism REF , such a MATH-cocycle is a coboundary in the infinite-order NAME complex and is decomposed into the sum... |
math-ph/0004024 | Let MATH be a MATH-closed form, that is, MATH. It belongs to MATH if MATH. In accordance with the relation REF , it follows that MATH, that is, belongs to MATH. |
math-ph/0004024 | The obstruction for the cohomology group MATH to be trivial are the elements of the overlap MATH, that is, the forms MATH such that MATH, that is, MATH. By virtue of REF , such a form is given by the sum REF which reads MATH . Then, MATH is an exact form. |
math/0004002 | The assertion follows from REF , which is a part of REF (for reductive MATH). Indeed, reductivity of MATH implies reductivity of MATH CITE. If MATH is not finite, then it contains a non-trivial one-parameter subgroup MATH. For MATH, we have MATH whenever MATH. |
math/0004002 | As MATH does not contain MATH, there exists a non-trivial one-parameter subgroup MATH in MATH with finite intersection with MATH. The corresponding extension MATH is spherical iff MATH is spherical in MATH. |
math/0004002 | Under these assumptions, there exists a non-trivial one-parameter subgroup of MATH with finite intersection with MATH which normalizes (and even centralizes) MATH. |
math/0004002 | Suppose that there exists an affine embedding MATH with infinitely many MATH-orbits. Consider the morphism MATH. It determines an embedding MATH. Let MATH be the integral closure of the subalgebra MATH in the field of rational functions MATH. We have the following commutative diagrams: MATH . The affine variety MATH wi... |
math/0004002 | CASE: By NAME 's theorem, there exists a MATH-module MATH and a vector MATH having REF . Let us denote by MATH the character of MATH at MATH. Since MATH is observable in MATH, every finite-dimensional MATH-module can be embedded in a finite-dimensional MATH-module CITE. In particular, there exists a finite-dimensional ... |
math/0004002 | Let MATH be the couple from REF . Denote by MATH the stabilizer MATH of the vector MATH. By REF - REF and since MATH is isomorphic to MATH, MATH is an overgroup of MATH with MATH. By REF, the closure of MATH in MATH is a cone, so by REF does not hold for MATH. REF completes the proof. |
math/0004002 | For completeness, we give the proof in the case, where MATH is affine (the only case we need below). It suffices to prove that any MATH-invariant element of MATH is the residue class of a MATH-invariant rational function on MATH. For any MATH, we shall write MATH if MATH. Such ``congruences" are MATH-stable and may be ... |
math/0004002 | If MATH is as above, then Knop has shown that the algebra MATH admits a non-trivial MATH-invariant grading, whose homogeneous components are sums of isotypic components of the MATH-module MATH, see CITE and its proof. This grading is constructed as follows. Under the above assumptions, there is a central valuation MATH... |
math/0004002 | It remains to prove that REF holds for MATH whenever any extension of MATH by a one-dimensional torus is spherical. As MATH is reductive, MATH is reductive, too. If there exists no one-parameter extension of MATH at all, then MATH is finite and MATH is affinely closed by REF . Otherwise MATH. As the spherical case is c... |
math/0004002 | The reductive group MATH acts on MATH, which is the field of rational functions on a projective line. If the kernel of this action has positive dimension, then it contains a one-dimensional torus extending MATH to a non-spherical subgroup. Otherwise, either MATH is finite or MATH and each subtorus of MATH has a dense o... |
math/0004002 | We follow the proof of REF . Put MATH; then MATH. We modify the proof of REF to obtain a MATH-module MATH and a MATH-eigenvector MATH such that MATH, MATH is a finite extension of MATH, and MATH contains infinitely many MATH- (not MATH-) orbits. Arguing as in the proof of REF , we see that the closure MATH of MATH is M... |
math/0004002 | The NAME algebra of MATH equals MATH, where MATH is the centralizer of MATH in MATH. We have MATH, and MATH. If MATH is semisimple, then MATH, and MATH is finite. REF implies the assertion for this case. Now suppose that MATH is not semisimple. If there exists a non-spherical extension of MATH by a one-dimensional toru... |
math/0004002 | We use exactly the same arguments as in the proof of REF replacing an embedding of MATH with infinitely many orbits from CITE by an embedding of MATH-modality MATH constructed in CITE. |
math/0004002 | Clearly, MATH. Taking an affine cone over the projective embedding constructed in REF , one obtains an affine embedding of MATH of modality MATH, where MATH is a finite extension of MATH. Using the construction from the proof of REF , we get an affine embedding of MATH of modality MATH. The obvious inequality MATH comp... |
math/0004002 | REF follows from REF . To prove REF , we can use REF . If there exists a one-dimensional torus in MATH such that the extension MATH is non-trivial and MATH, then there exists an affine embedding of MATH of modality MATH. Conversely, suppose that MATH is an affine embedding of modality MATH. We need to find a one-dimens... |
math/0004002 | If MATH is not semisimple, then for a central one-dimensional subtorus MATH one has MATH. If MATH is semisimple, then for any one-dimensional subtorus MATH there exists a NAME subgroup MATH which does not contain MATH, and there is a MATH-orbit on MATH of dimension MATH. This implies MATH. |
math/0004003 | Given any morphism MATH we construct a morphism MATH as follows: For all MATH, define a map MATH in the obvious way: send MATH with MATH to the morphism MATH that sends MATH to MATH. Now define MATH to be the composite MATH where MATH is the product operation of MATH. This map is easily see to preserve the MATH-action.... |
math/0004003 | This is due to REF and the requirement that every morphism preserve the unit map. |
math/0004003 | Here MATH is the standard graded module-pullback MATH. It is clear that this kernel is an operad, and a glance at the exact sequence in homology induced by MATH (and the fact that MATH and MATH induce homology isomorphisms in all dimensions and for all degrees - see REF) shows that it is also MATH. |
math/0004003 | Suppose MATH is some element. Then we have a commutative diagram MATH which we can expand to a diagram MATH which gives us an element of MATH. To conclude that this defines a morphism of operads, we must show that it respects compositions as in REF, or that the kernel of the map MATH of graded MATH-modules is an ideal ... |
math/0004003 | Its inverse sends a morphism MATH to MATH. |
math/0004003 | Its inverse sends a morphism MATH to MATH. |
math/0004003 | The map MATH fits into an exact sequence MATH where MATH. The first equivariance condition in the definition of an operad (see REF) implies that MATH maps into MATH, where MATH is induced by the composite MATH . It follows that MATH so that the image of the kernel of MATH lies in the kernel of MATH and the MATH is well... |
math/0004003 | By REF our free algebra is given by MATH REF implies the existence of chain maps MATH for all MATH and MATH-tuples MATH of positive integers. The map on the left is just transposition of the factors. This induces a morphism of NAME MATH . The associativity conditions imply that compositions are preserved by MATH, and t... |
math/0004003 | In this case, we set MATH . As before, we make crucial use of REF . We must proceed more delicately than in REF, though. We will use the approach to operads in REF and define composition operations. A composition MATH (where MATH) implies the existence of a chain-map MATH - this is just the transposition and adjoint. D... |
math/0004003 | This result is due to NAME. The uniqueness of MATH follows by induction and: CASE: MATH is determined by its values on MATH CASE: the image of the contracting chain-homotopy, MATH, lies in MATH. CASE: the boundary map of MATH is injective on MATH (which implies that there is a unique lift of MATH into the next higher d... |
math/0004003 | Existence and uniqueness follow from REF, setting MATH and MATH and the fact that, in dimension MATH, so any chain maps between them that commute with augmentation must be equal. In dimension MATH for all MATH. In higher dimensions, set MATH for MATH, MATH and for all MATH. The boundaries of MATH and MATH can be arbitr... |
math/0004003 | We have already proved that MATH is an operad - see REF. To see that the coproduct is an operad morphism, observe that the composition operations of MATH are coalgebra morphisms. This follows from the NAME Theory of Constructions once again, observing that the MATH can be defined by this theory - as the unique map MATH... |
math/0004003 | For all MATH, let MATH denote the standard MATH-simplex, whose vertices are MATH and whose MATH-faces are MATH, with MATH, MATH. We define MATH to be a free functor on models that are simplices and use the NAME Theory of Constructions to define maps: MATH where MATH is a MATH-simplex with chain complex MATH, and MATH a... |
math/0004003 | We form the push out of coalgebras over MATH: MATH which is possible because of REF . This gives us the diagram MATH where MATH and MATH are induced by the inclusions. They are clearly m-coalgebra morphisms. |
math/0004003 | We will prove this for MATH - the arguments for MATH and MATH are identical. The claim follows by induction on the number of columns. We show that the following equation (in MATH) is satisfied at column MATH of REF for all MATH: MATH . Every commutative subdiagram of the form: MATH results in an equation MATH and any c... |
math/0004003 | Again, we will prove the result for MATH - the arguments for MATH and MATH are identical. We use the description of localized categories in CITE: they define it via ``generators and relations". The only relations that exist in MATH are: |
math/0004003 | Again, we only prove the statement for MATH. This is an immediate consequence of REF. The statement that MATH is an elementary equivalence follows from the fact that, as a chain-complex, MATH, where MATH is acyclic and MATH-torsion free. The equation in MATH follows from the commutativity of REF, which implies that MAT... |
math/0004003 | The first statement follows immediately from REF, which implies that we can permute leftward elementary equivalences with the rightward maps and compose - thus simplifying the morphism. The second statement follows from the naturality of the push-outs used in the simplification: we can simplify the upper and lower rows... |
math/0004003 | This is an immediate consequence of REF. |
math/0004003 | The first statement follows from the fact that MATH is full. The second follows from the fact that, in an abelian category, we can always form push-outs. We, consequently, take a diagram of elementary homotopies (like REF) and, whenever we encounter a subdiagram of the form MATH and we form the push out of MATH . It is... |
math/0004003 | If MATH is a morphism in MATH, MATH, or MATH, this is clear: the inclusion is a direct summand (as a chain-complex) and a homology equivalence. In the general case (that is, in MATH, MATH, or MATH) MATH is a composite of two elementary equivalences (in opposite directions). |
math/0004003 | The only-if part of this is clear. The if part follows immediately from the existence of algebraic mapping cylinders. |
math/0004003 | This follows from results in CITE, which characterizes the homotopy category of simplicial sets as a localization of the category of simplicial sets by weak equivalences. REF on page REF shows that a morphism MATH is an isomorphism in the homotopy category if and only if: CASE: MATH is an isomorphism for any set MATH C... |
math/0004003 | Consider an elementary homotopy MATH . It induces MATH where the map MATH is an equivalence since MATH in MATH. This factors through to the algebraic mapping cones. |
math/0004003 | By abuse of notation, we will denote the algebraic mapping cylinder by MATH (even though MATH is not necessarily a morphism). Let the chain complex of MATH (the unit interval) be given by: |
math/0004003 | We construct MATH to have an underlying chain complex that is the algebraic mapping cylinder of MATH. Since MATH is a natural transformation, and since MATH is free on the MATH, it follows that MATH is constructed from MATH. Consequently, REF implies that we can extend the m-structures of MATH and MATH to an m-structur... |
math/0004003 | It is only necessary to note that these are functors from the category of ordered pairs of simplicial sets and that the respective functors are free and acyclic on models composed of pairs of simplices. |
math/0004003 | The key fact here is that the m-structure of MATH is trivial, thereby eliminating any obstruction to the existence of MATH. The statements about co-commutativity and co-associativity follow immediately from the definition of MATH and the remaining statements are clear. |
math/0004003 | This follows from the Duality REF , which states the existence of an operad morphism: MATH . Now we compose the structure map of MATH with this to get an operad morphism MATH . The conclusion follows by noting that, as a non-MATH operad MATH. |
math/0004003 | See CITE for a proof. This involves considering MATH and applying the identities that modules and comodules must satisfy. |
math/0004003 | REF follow by an inductive application of the boundary formula in REF. The formula for MATH follows by a straightforward but tedious computation. |
math/0004003 | Most of the statements follow from the functoriality of MATH. The last statement follows by considering the MATH and passing to the limits: it is not hard to see that MATH for finite MATH. When we pass to the limits, there still exists a map, but it is not necessarily an isomorphism. |
math/0004003 | Recall that, in the notation of REF, MATH. The boundary of MATH is MATH, where MATH is given, on MATH by (see REF ) MATH and MATH is given by MATH . The diagram MATH commutes because the composite MATH is of degree MATH, and the diagram MATH is also seen to commute, after we permute MATH with MATH. he conclusion follow... |
math/0004003 | We must verify that the identities in REF are satisfied. The statement that MATH is a free group-object implies that MATH by a recursive application of REF. The identity in REF gives us: MATH . |
math/0004003 | The proof is very similar to REF (we take the duals of all of the maps there). |
math/0004003 | This is very similar to the proof of REF - we have formed the mapping sequence representing the tensor product MATH using REF (regarding MATH as a mapping sequence concentrated in the MATH term) and added a term representing the twisting cochain. |
math/0004003 | A lengthy computation (see CITE) shows that equivalence of MATH-structures implies an equation like REF in MATH. The corresponding equation in MATH will hold if there exists a coalgebra, MATH, over MATH such that the structure map MATH is injective. But this follows from REF. |
math/0004003 | This follows from REF and the acyclicity of a MATH-operad's components. |
math/0004003 | We appeal to REF to get a MATH-module structure on MATH. The conclusion follows from REF, and REF. |
math/0004003 | The only thing to be proved is that the maps MATH pull back to the upper row as maps of the MATH. But this is an immediate consequence of: |
math/0004003 | In this case, we translate the formula in REF into compositions in MATH: MATH with MATH terms, where MATH for MATH and and, for MATH where MATH represents the inclusion induced by raising all indices The expression for MATH is clearly a direct translation of the equation for MATH in REF into compositions in MATH. The t... |
math/0004003 | We know the following two things about the upper rows of REF: CASE: They are independent of MATH CASE: The composites of the maps MATH and MATH, respectively with the corresponding vertical maps satisfy the defining identities (see REF) of a mapping sequence (because every square of the diagrams commutes and because th... |
math/0004003 | We begin with an observation: Any MATH-structure on MATH induces one on MATH. It follows that we can form MATH and MATH and that there exist canonical morphisms MATH and MATH . It is also not hard to see that these morphisms will be injective if the m-structure morphism of MATH is. CASE: The morphisms MATH and MATH are... |
math/0004003 | See REF. |
math/0004003 | Our first claim is proved in REF. Our second claim follows by a variation on the construction in REF: we replace each term of MATH by MATH in all the subscripts in REF. The conclusion follows from the definition of the MATH - particularly the fact that they were defined exactly like the MATH with a shift in indices. |
math/0004003 | The first claim follows from REF. To prove the second, let MATH be the adjoint of MATH in REF and let MATH be the adjoint of MATH defined in REF. We form the tensor product MATH . Now we compose this on the right with the map MATH defining the action of MATH on MATH and with MATH on the left to get a map MATH . The ten... |
math/0004003 | That MATH is an operad follows from the fact that MATH is an operad for any comodule over a MATH-operad (by the discussion preceding REF) and REF. That these composition operations commute with the coproduct of MATH follows from: CASE: the definition of the coproduct MATH CASE: the fact that MATH (by REF), CASE: for ea... |
math/0004003 | This is a straightforward consequence of the fact that the MATH are acyclic in above dimension MATH if MATH is MATH. We can use the identity that a twisting cochain must satisfy - REF - to compute MATH inductively. We set MATH, where MATH is a sequence of nonnegative integers and the direct sum is taken over all such s... |
math/0004003 | To conclude that the pullbacks can be taken, we invoke REF. |
math/0004003 | We must show that it maps equivalent objects to equivalent objects. It is easy to see that MATH maps split injections (of MATH) into multiplicative split injections (of MATH) and a simple spectral sequence argument shows that maps induced by homology equivalences are homology equivalences. |
math/0004003 | Let MATH be a sequence of nonnegative integers and let MATH be a direct summand see REF. Note that MATH contains a summand MATH where MATH. REF implies that the composition operations of MATH coincide with those of MATH. Since the twisted differential of MATH consists of terms MATH with at least one of the MATH see REF... |
math/0004003 | There are several cases to consider: CASE: Replace MATH by MATH equivalent to it MATH where MATH and MATH are both elementary equivalences. If we combine this equivalence with the morphism above, we get MATH and we put this into canonical form (see REF) by taking the push out MATH where MATH and MATH are elementary equ... |
math/0004003 | First replace REF by canonical representations in terms of morphisms in the nonlocalized category MATH. We get the somewhat gory diagram MATH where CASE: all maps are morphisms in MATH. CASE: the maps MATH, MATH, MATH, and MATH are elementary equivalences in MATH. In order to put the composites MATH and MATH into canon... |
math/0004003 | Let MATH be inclusion of the fiber and consider the commutative diagram of spaces MATH where: CASE: both rows are homotopy equivalences CASE: MATH is the map in REF CASE: MATH is the projection of the fibration to its base This gives a commutative diagram of m-coalgebras, where all maps are defined in the category MATH... |
math/0004003 | This follows from REF by making MATH equal the basepoint of MATH. |
math/0004003 | This follows immediately from the equivalence of homotopy theories of pointed, REF-reduced, simplicial sets and pointed simply connected spaces - by the adjoint functors MATH (REF-reduced singular complex) and MATH (topological realization). See CITE for details. |
math/0004003 | The hypothesis implies that the chain-complexes are chain-homotopy equivalent, hence that the MATH, MATH, have the same homology. This implies that the lowest-dimensional nonvanishing homology groups - say MATH in dimension MATH - are isomorphic. We get a diagram of morphisms in MATH . Where CASE: The maps MATH, MATH, ... |
math/0004003 | We prove this result by an inductive argument somewhat different from that used in REF . We build a sequence of fibrations MATH over MATH in such a way that CASE: the morphism MATH lifts to MATH - that is, we have commutative diagrams MATH . For all MATH, MATH will be a fibration over MATH with fiber a suitable NAME sp... |
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