paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9904129 | The proof uses the following property of valuations: MATH. Furthermore, when that minimum is attained in only one MATH, we have equality. Let MATH. Writing MATH one can pass to the valuation by: MATH . Subtracting, one obtains: MATH . This concludes the proof. |
math/9904129 | The NAME diagram of MATH at REF is MATH. (This latest set is convex, since the points lie on the curve MATH and this curve is convex). Therefore, there is a unique root MATH of MATH that minimizes MATH. Since MATH, where the sum ranges over all the roots, we have: MATH . Replacing by the actual values of the coefficien... |
math/9904129 | We can assume without loss of generality that the ordering of the MATH satisfies: MATH . For MATH we have: MATH . Hence, using MATH: MATH . By the same argument, for MATH: MATH . Hence, MATH . Set MATH for MATH. We know that the MATH are such that MATH. Hence MATH . Hence: MATH and hence MATH . By REF or by REF, MATH a... |
math/9904129 | We see from its NAME diagram that the polynomial MATH has distinct roots MATH with: MATH . So we have MATH, and MATH . Assume that there are MATH such that for each MATH, there is a machine MATH over MATH deciding MATH in time MATH. Its generic path is defined by a polynomial MATH of degree MATH. Let us fix MATH. In pa... |
math/9904130 | Let MATH be in MATH. Let MATH be the machine that recognizes MATH in polynomial time, where the input MATH is assumed to be in MATH. Although it is possible that a MATH is not in MATH, it is still possible to recognize MATH in polynomial time. Indeed, MATH is also in MATH. The machine MATH will check MATH and MATH. Now... |
math/9904130 | REF is trivial, refer to REF. CASE: Let MATH be the problem in REF . Since MATH is generic in MATH, all inputs in MATH should escape the canonical path. Hence, if MATH is the polynomial that defines the canonical path, MATH for MATH. But then it cannot be evaluated in time polylog(MATH), by REF . Hence, under REF , the... |
math/9904130 | Let MATH and assume that MATH. We have to show that MATH. For each MATH, one can embed MATH into some MATH as follows: Let MATH be the deterministic polynomial time machine to recognize MATH, and let MATH be the non-deterministic polynomial time machine to recognize MATH. We can assume without loss of generality that M... |
math/9904131 | This follows from REF and the definition of rigidity REF . |
math/9904131 | The proof follows from the identification of the MATH term in CITE or CITE. Also, note that MATH where MATH . Then MATH, MATH, and MATH. |
math/9904131 | If MATH for MATH, then from REF , the MATH term for the spectral sequence converging to MATH is zero. Thus, MATH for MATH. The result now follows from long exact sequence REF. |
math/9904131 | The theorem follows from REF . |
math/9904131 | We first compute MATH on the level of cochains. Consider the vector space basis MATH of MATH given by MATH . From CITE, the class of MATH in MATH is represented by the cochain MATH . The cochain map MATH inducing the isomorphism MATH satisfies MATH . Also, it is know that MATH generates MATH (as a MATH vector space). L... |
math/9904142 | Suppose that MATH is a cross product algebra. Since by REF is itself an algebra one obtains the first and second identity of REF by unitality of MATH and MATH. Using these relations yields the fourth identity of REF by application of MATH to REF. Again using the left unitality of MATH one concludes that MATH from which... |
math/9904142 | If MATH is an algebra isomorphism then define MATH and MATH. Using the particular definition of MATH in REF and the identities REF, it is verified immediately that MATH is an algebra morphism since MATH is an algebra morphism. Similarly one proves that MATH and MATH which is therefore an isomorphism. If on the other ha... |
math/9904142 | Define MATH. Then by REF it follows MATH, MATH, and MATH. Using consecutively that MATH is an algebra morphism, REF , and REF then yields MATH. Therefore MATH is an algebra morphism obeying the conditions of REF . Suppose that there is another MATH meeting the same stipulations as MATH. Then MATH. In the last equation ... |
math/9904142 | The proposition can be proven easily by REF and triviality of MATH. |
math/9904142 | The non-trivial part of the proposition can be verified easily with the help of the identity MATH, the associativity of MATH and MATH, and REF or REF respectively. |
math/9904142 | ``REF. REF.": For MATH we define MATH, MATH to be the morphisms splitting the idempotent MATH as MATH and MATH for some objects MATH. Then with the help of REF it follows MATH and MATH. Hence MATH. Now we define MATH, MATH, MATH, and MATH. Then using REF and the (co- )algebra property of MATH one verifies easily that M... |
math/9904142 | The identities MATH and MATH hold since MATH is an algebra morphism and MATH is a coalgebra morphism, whereas the identities MATH and MATH, as well as the equations MATH and MATH have been shown in the proof of REF . From the proofs of REF one can directly read off the structure of the morphisms MATH, MATH, MATH, and M... |
math/9904142 | If the conditions of REF hold, then by construction (see the proof of the theorem) the injections and projections of MATH and the morphisms MATH, MATH, MATH and MATH are related by MATH, MATH, MATH, and MATH, where MATH is the given bialgebra isomorphism. Thus MATH and MATH follow directly. |
math/9904142 | Obviously statement MATH and MATH are equivalent due to REF . Hence suppose statements MATH and MATH hold. Then by REF there exists a normalized cross product bialgebra MATH which is isomorphic to MATH via the isomorphism MATH. We define MATH . The morphisms MATH, MATH, MATH, MATH, MATH, and MATH are given in terms of ... |
math/9904142 | We prove REF . The remaining statements can be derived in a similar manner or follow directly by MATH-symmetric reasoning. Without loss of generality we may assume MATH and MATH. CASE:: Suppose that MATH is trivial. This means that MATH. Using REF and the unital identities of REF then yields MATH. On the other hand fro... |
math/9904142 | By definition cocycle cross product bialgebras are cocycle cross product algebras and cycle cross product coalgebras in particular. Hence MATH is an algebra and MATH is a coalgebra. Since cocycle cross product bialgebras are normalized one deduces with REF that MATH is an algebra morphism and MATH is a coalgebra morphi... |
math/9904142 | REF can be derived straightforwardly from REF and the relations REF. |
math/9904142 | The first identity in REF has been obtained from REF. With the help of REF the second identity will be derived. Application of MATH-symmetry completes the proof. |
math/9904142 | The relative associativity of MATH and MATH will be derived from the respective weak associativity of REF taking into account REF . |
math/9904142 | Using that MATH is a right module and applying REF yields the first identity of REF. |
math/9904142 | We use REF and the second identity of the comodule-algebra compatibility of REF to show that MATH is a left MATH-comodule algebra. In MATH-symmetric manner it will be proven that MATH is a right MATH-module coalgebra. |
math/9904142 | Since MATH is a is a left comodule algebra by REF the first identity in REF follows from the relative asociativity of MATH according to REF . The verification of REF needs a little more calculation. We prove the first identity of REF. The second one can be derived with similar techniques. We start with the second modul... |
math/9904142 | For the proof of the first identity of REF we use REF and the fourth equation of REF . To verify the second identity of REF, the module-algebra compatibility for NAME data, REF and the comodularity of MATH have to be applied successively. Using the module-algebra compatibility and the MATH-symmetric version of the four... |
math/9904142 | We use REF for MATH and the relative associativity of MATH REF to obtain the first identity of REF. The first identity of REF will be used explicitely in the proof of REF . Below we will give its detailed derivation. MATH where the first equation comes from the relative bialgebra property of MATH according to REF. The ... |
math/9904142 | The proof follows straightforwardly from the cocycle and cycle compatibilities of REF and the identities REF. |
math/9904142 | Because of MATH-symmetry we will only demonstrate the first identities of REF. Applying to the left hand side of REF the algebra-coalgebra compatibility (for MATH and MATH) and the entwining property of MATH REF yields the result. To obtain the first identity of REF we transform its left hand side consecutively using t... |
math/9904142 | The lemma follows straightforwardly from the (co- )associativity of MATH and MATH and the identities MATH. |
math/9904142 | We only consider the first identities of REF. The second identities in each row are MATH-symmetric analogues. The identities REF follow from REF by application of the mapping MATH. Relations REF are special cases of REF through composition with MATH. In a first step we derive REF for the case i=REF. For REF we obtain M... |
math/9904142 | The morphisms on right hand side of the corresponding compatibility relations can be decomposed with the help of REF. Then the reduction REF will be applied to the particular tensor factors. And finally we use again REF to obtain the result. |
math/9904142 | The first identity of REF has been derived by successive application of the left module-algebra compatibility of REF and the relations REF. The second identity is its MATH-symmetric counterpart. To get the third identity we use REF, the first and the second equation of REF, the relations REF and the module-comodule com... |
math/9904142 | Both identities follow from the third identity of REF using REF. |
math/9904142 | We prove the first identity of REF graphically. MATH . The first equation in the graphic has been obtained with the help of the cycle-cocycle compatibility REF. In the second equation we use REF. The third equation is REF. Application of REF to the fourth diagram yields the fifth relation. The identities REF follow fro... |
math/9904142 | The identity follows straighforwardly from REF. |
math/9904142 | The identities REF follow from REF, whereas REF will be derived from REF and the entwining REF of MATH, MATH. |
math/9904142 | According to REF the identities REF have to be verified in order to prove that MATH is a cross product algebra. Similar MATH-symmetric procedures are needed to demonstrate that MATH is a cross product coalgebra. Without difficulties the unital identities of REF can be verified. The fifth relation of REF has been proven... |
math/9904142 | The proof will be splitted into several parts. At first we verify the following diagram. MATH where the identities REF in the first row are definitions and the equalities REF in the second row are special cases of REF. The morphisms in the third row will be obtained by applying MATH, MATH and MATH using the identities ... |
math/9904143 | Given any MATH, the sequence MATH is coherent. Conversely, given any coherent sequence MATH, we can define MATH by MATH where MATH. |
math/9904143 | As a vector space, MATH, where MATH consists of all functions supported on monomials of weight MATH. It follows that MATH as MATH vector spaces. Of course, there are exactly MATH monomials of weight MATH. Finally, if MATH then MATH, hence MATH. |
math/9904143 | Since MATH has a MATH-basis is given by all monomials of weight MATH, the two MATH-algebras are isomorphic as MATH-vector spaces. The multiplication in MATH is induced from the multiplication in MATH, with the extra condition that monomials of weight MATH are truncated. This is the same multiplication as in MATH. |
math/9904143 | We must show that if MATH, and MATH . , then MATH for MATH. We have that MATH. |
math/9904143 | The monomial MATH has bi-degree MATH. |
math/9904143 | We have that MATH, MATH for MATH. Hence REF implies that MATH . Putting MATH, MATH for MATH we can write this as MATH from which REF follows by taking logarithms. This implies REF as well. |
math/9904143 | CASE: Obvious. REF It suffices to show that for any subset MATH of cardinality REF or REF, there is a MATH with MATH. If MATH then there is an unique positive integer MATH such that MATH, and MATH is the desired generator. If MATH with MATH then we claim that there is a positive integer MATH such that MATH. Namely, cho... |
math/9904143 | The first and the last assertions are obvious. The second one follows from the proof of REF in the previous lemma. |
math/9904144 | By CITE, the line bundle on MATH associated with the divisor MATH is MATH. Let MATH be the canonical section of this line bundle; then MATH is an eigenvector of MATH, because its divisor MATH is MATH-stable. The closure in MATH of MATH is the divisor of a regular function on MATH, eigenvector of MATH of weight MATH, an... |
math/9904144 | Let MATH be the NAME module for MATH; it is a simple, self-dual MATH-module CITE. On the other hand, the line bundle MATH is MATH-linearized by construction of MATH, and the MATH-module MATH contains an eigenvector of MATH of weight MATH, unique up to scalar, by REF . Further, the image of this eigenvector under restri... |
math/9904144 | Let us prove REF . Since the divisor MATH is ample, this is a consequence of CITE when MATH. Moreover, since MATH, MATH are defined over MATH, it follows from the construction of MATH CITE that MATH, the boundary divisors MATH and the large NAME varieties MATH are all defined and flat over some open subset of MATH (in ... |
math/9904144 | By REF , the scheme-theoretic intersection MATH is reduced; this is equivalent to REF . For REF , we have to check that the image of each MATH in MATH is not a zero divisor. But the scheme-theoretic intersection MATH is reduced. Further, by CITE, each irreducible component of MATH has codimension MATH in MATH, and is n... |
math/9904144 | We will use the duality between line bundles and curves: each closed curve MATH in MATH defines an additive map MATH where MATH is the degree of the restriction of MATH to MATH. In fact, MATH only depends on the classes of MATH and MATH up to rational equivalence. Further, MATH is rationally equivalent to a positive in... |
math/9904144 | F rom the exact sequence of sheaves on MATH: MATH we see that MATH injects into MATH. The latter is equal to MATH . We have indeed MATH because MATH is generated by the regular sequence MATH by REF . We now need the following For a weight MATH, the following conditions are equivalent: CASE: MATH is dominant. CASE: MATH... |
math/9904144 | The canonical filtrations of the MATH fit together into a filtration MATH of MATH. REF implies that MATH (the MATH-th power of the ideal generated by MATH) and that MATH . Thus, the associated graded ring is isomorphic to the polynomial ring MATH. By CITE MATH form a regular sequence in MATH. Further, by CITE, the grad... |
math/9904144 | Set MATH. For MATH, let MATH be the sum of MATH-weight spaces in MATH (for the right MATH-action) over all weights in the coset MATH. Then each MATH is a MATH-submodule of MATH, and we have MATH . Further, the MATH-module MATH is freely generated by any MATH-eigenvector. Choose MATH and a dominant weight MATH in the co... |
math/9904144 | With notation as in REF , we have for any MATH and any index MATH: MATH by REF and the argument thereafter. Since MATH, it follows that MATH for all MATH. Recall now that each NAME variety MATH is NAME. Denote by MATH its canonical sheaf, a MATH-linearized sheaf. By CITE, we have an isomorphism of MATH-linearized sheav... |
math/9904144 | We begin by proving that MATH is NAME. For this, we use the notation introduced in the proof of REF . We check by decreasing induction on MATH that each MATH is NAME. If MATH then MATH, a non-singular variety. For arbitrary MATH, we have an exact sequence MATH . Further, we know that MATH is NAME; and, by the induction... |
math/9904144 | By CITE, MATH is equidimensional. Further, MATH is NAME, as MATH is. Because MATH is smooth, it follows that MATH is flat. For MATH, the scheme-theoretic fiber MATH identifies to MATH . Set MATH. Then MATH is stable under right multiplication by MATH, and left multiplication by MATH (the isotropy group of MATH in MATH)... |
math/9904144 | As the map MATH is flat and MATH-invariant, and MATH is isomorphic to affine line, the fibers MATH and MATH have the same class in MATH. Thus, the direct images of these fibers under MATH are equal in MATH. This proves the first equality. For the second one, we use the notation introduced in the proof of REF . Set MATH... |
math/9904144 | Observe that MATH where the second equality follows from REF . In particular, we have for MATH: MATH . As the MATH-linear forms MATH are linearly independent, we obtain MATH . |
math/9904144 | Let MATH be the class of the MATH-linearized line bundle MATH in MATH. As the restriction of this line bundle to the diagonal is MATH, we have MATH . By REF , the latter is equal to MATH . To complete the proof of the first equality, it suffices to check that MATH . For this, using NAME duality as in the proof of REF ,... |
math/9904144 | Because MATH and MATH are NAME, the same holds for MATH. And because MATH is birational and MATH is normal, we have MATH. We now show that MATH for MATH. For this, it suffices, by REF (see for example, CITE), to show that MATH for MATH and for any regular dominant weight MATH. Consider the line bundle MATH and its high... |
math/9904144 | Since MATH is the complete intersection in MATH of the NAME divisors MATH, by REF , it follows that MATH is NAME. Similarly, MATH is NAME and its canonical sheaf is the restriction to MATH of MATH . The latter is equal to MATH, as we saw in the proof of REF . This proves REF . The multiplication by MATH defines exact s... |
math/9904144 | Recall that MATH. By REF , the latter is isomorphic to MATH . Using REF , we obtain similarly that MATH. We prove that MATH by descending induction on MATH. If MATH then MATH and MATH. In this case, the assertion follows from CITE. In the general case, let MATH be a simple reflection such that MATH; let MATH be the par... |
math/9904148 | This lemma is well known, but in view of the difficulty we have to locate an explicit reference, we present a proof here. For simplicity, we only present the proof for the case MATH. The arguments in general case are exactly the same, but the notation is more complicated. We also identify the one dimensional torus with... |
math/9904148 | It follows from REF , that there exists a formal power series MATH with non-negative coefficients, such that MATH . Our goal is to show that MATH. It follows from REF, that both the equivariant NAME counting series and the equivariant NAME series are additive with respect to MATH. More precisely, if MATH denotes the di... |
math/9904159 | The isomorphism on the left-hand side has been discussed in REF . The one in the middle is the special case MATH, MATH of the isomorphism of graded MATH-algebras MATH induced from the NAME class homomorphism MATH that associates to a character MATH the first NAME class of the line bundle MATH. Here MATH denotes the one... |
math/9904159 | We proceed by induction on the number of cones in the fan MATH. For MATH, the assertion is obvious. For the induction step, we choose a maximal cone MATH and consider the NAME sequences associated to MATH and MATH, both for MATH and for MATH. It suffices to prove that in the commutative diagram MATH obtained from REF ,... |
math/9904159 | For a cone MATH, let MATH be the character which coincides with MATH on MATH. Since the map MATH is an isomorphism of sheaves, it suffices to show that the ``local" equivariant NAME class MATH is mapped onto MATH. Observing that the inclusion MATH of the fixed point MATH induces an isomorphism MATH, we may further rest... |
math/9904159 | Let MATH be a simplicial refinement of the defining fan MATH for MATH, and denote with MATH the corresponding equivariant MATH-resolution of singularities. By REF , the assertion holds for MATH, since then MATH. By the equivariant version of REF, NAME, NAME, and NAME as stated in CITE (see also CITE or CITE), we may in... |
math/9904159 | Since there are only finitely many open subsets in MATH, it suffices to verify the sheaf axioms for two open subsets MATH. We thus have to prove the exactness of the sequence MATH . That follows from REF : The exactness is obvious if MATH is odd; for even MATH, the vector space MATH vanishes and thus, the sequence is p... |
math/9904159 | We define the sheaf MATH inductively on the MATH-skeleton subfans MATH starting with MATH for MATH. Suppose that for some MATH, the sheaf MATH has already been constructed on MATH, so in particular, for each MATH-dimensional cone MATH, the module MATH is given. It thus suffices to define MATH together with a restrictio... |
math/9904159 | Since the fan space MATH is covered by finitely many affine fans, it suffices to prove that for each cone MATH, the restriction map MATH is surjective. Using the results on graded modules recalled in the ``NAME, that is a consequence of condition (LME). - The vanishing of MATH follows immediately from the same conditio... |
math/9904159 | The Normalization property is obviously satisfied, since we have MATH . The Pointwise Freeness condition will be verified in REF , and the Local Minimal Extension requirement, in REF . |
math/9904159 | Since REF says that for intersection cohomology, MATH behaves like the product MATH, the implication ``i REF" is obvious, and ``ii REF" follows immediately from the Vanishing REF . For the implication ``iii REF", we observe that the assumption implies the degeneration of the intersection cohomology NAME spectral sequen... |
math/9904159 | Let MATH denote a toric MATH-resolution as in the proof of the Vanishing REF . By the ``classical" (that is, non-equivariant) Decomposition Theorem of NAME, NAME, NAME and NAME, we know that MATH is isomorphic to a direct summand of MATH, and, according to REF and NAME, that module vanishes in odd degrees. |
math/9904159 | By the ``relative affine orbit splitting" MATH and REF as well as its analogue MATH, we see that it is sufficient to consider the case of a MATH-dimensional cone MATH. We use the same notations as in the previous section; in particular, we write MATH and MATH. Furthermore, by the arguments of the preceding section, we ... |
math/9904159 | REF ` REF ': Since MATH is a sheaf, we see that we have a natural inclusion MATH of MATH-modules. By REF , each MATH is a free MATH-module. For MATH, we have MATH, so the right hand side is a free MATH-module. Moreover, every submodule of a torsion-free module is again torsion-free. ` REF ': If MATH is a maximal cone o... |
math/9904159 | We intend to prove that the intersection cohomology NAME number MATH is non-zero. With MATH and MATH, we consider the following part of the exact NAME sequence: MATH . The zeroes at both ends are due to the fact that the toric varieties MATH and MATH are equivariantly formal. The ``affine orbit splitting" REF provides ... |
math/9904159 | It is known (see, for example, CITE) that the natural homomorphism MATH is surjective (and even an isomorphism, though we do not need this stronger result). To investigate the target, let MATH be a completion of the fan. The set of cones MATH defines a closed invariant subvariety MATH of the compact toric variety MATH ... |
math/9904159 | We have to prove that the toric variety MATH has vanishing intersection cohomology in each odd degree MATH if MATH has. To that end, we look at the exact sequence MATH where the final term vanishes since MATH is odd. It clearly suffices to prove that MATH vanishes. We may identify this relative group with MATH by excis... |
math/9904159 | As in the proof of REF , we have to show that MATH vanishes in odd degrees. There is an isomorphism MATH where MATH is a finite subgroup of MATH acting diagonally on MATH, such that MATH. As passing to the quotient by MATH does not influence the rational (intersection) cohomology (see CITE), we obtain MATH thus proving... |
math/9904159 | We choose a vector MATH such that either MATH lies in the interior of the support REF , or that MATH lies in the interior of the complement REF , and complete the given fan MATH by adding the new ray MATH together with all cones of the form MATH, where MATH is a cone in the boundary of the support. The ``old" fan then ... |
math/9904163 | Choose a set MATH such that CASE: there is MATH with MATH and MATH, but CASE: for each MATH and MATH, if MATH then MATH. Clearly it is possible; necessarily MATH (remember that MATH is typical). Fix MATH. Applying bigness to MATH we get MATH such that MATH and MATH. On the other hand, by the choice of the set MATH (and... |
math/9904163 | Let MATH and let MATH witness that MATH is densely representable by MATH. Let MATH (for MATH). Then the sets MATH are dense in MATH by REF . We claim that they witness the failure of MATH. So suppose that MATH is a directed set which meets each MATH. For every MATH choose MATH such that MATH. Look at the set MATH - it ... |
math/9904163 | Let MATH, MATH, MATH a basis of a topology on MATH, MATH. Assume that some condition in MATH forces that MATH is not a MATH - family. Then we find MATH, MATH, MATH and MATH - names MATH for elements of MATH (for MATH) such that MATH . For each MATH choose a sequence MATH and a condition MATH such that MATH". (So necess... |
math/9904163 | In REF use REF; in other cases use directly the assumption that MATH is typical. |
math/9904163 | For a condition MATH we may find a stronger condition MATH with the following property CASE: for each MATH, MATH there are MATH, MATH, MATH, and MATH, MATH, MATH (objects, not names) such that the condition MATH forces the following: CASE: MATH, MATH, MATH, CASE: MATH, MATH for each MATH, CASE: for each MATH such that ... |
math/9904163 | Like REF. |
math/9904164 | All REF follow easily from the properties of quasi-Hopf algebras as stated in REF. We will give a rather detailed proof such that the unexperienced reader may get used to the techniques used when handeling formulae involving iterated non coassociative coproducts. We will in the following denote MATH. Equality REF follo... |
math/9904164 | The definitions in REF imply MATH, which by REF further implies MATH . Thus we are left to show the identity MATH (denoting MATH): MATH where the last equality follows from REF. |
math/9904164 | Clearly, REF implies MATH. Conversely, if MATH then REF implies MATH, hence MATH. |
math/9904164 | Using REF and MATH we compute MATH . Conversely MATH by REF . Thus MATH is indeed an isomorphism of vector spaces. We are left to show that MATH also respects the quasi-Hopf MATH-bimodule structures. By definition we have MATH and therefore MATH . Here we have used REF in the second and third line, respectively. Thus M... |
math/9904164 | If MATH then by REF MATH, whence MATH. The inverse implication follows from MATH and MATH. |
math/9904164 | Denote MATH and MATH the projections onto the corresponding coinvariants. To prove that MATH is bijective we claim MATH where MATH is given by MATH . By REF MATH is well defined and we have MATH thus proving REF. To prove REF let MATH and MATH. Then REF MATH where in the second line we have used REF , in the third line... |
math/9904164 | If MATH is a morphism of quasi-Hopf MATH-bimodules, then MATH, MATH is MATH-linear and MATH . Thus, MATH is uniquely determined by its restriction MATH, and by REF MATH provides an equivalence of categories with reverse functor given by MATH and MATH. By REF these functors preserve the monoidal structures provided we s... |
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