paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905001 | We proceed by induction on MATH. For MATH, choose a system of parameters MATH and a neighbourhood MATH of MATH where MATH and MATH have regular representatives (which abusing notation we call MATH also) and MATH. Then the claim is clear, because MATH, and the virtual transform of MATH at MATH is MATH itself, so we can ... |
math/9905001 | By induction on MATH. The case MATH is clear, because then MATH is the closed subset of MATH determined by the ideal MATH. Suppose now MATH and the claim is true for MATH. Let MATH be a cluster and MATH the open neighbourhood given by REF . It will be enough to see that MATH is closed in MATH, because MATH can be cover... |
math/9905001 | Let MATH be a cluster and MATH the open neighbourhood given by REF . As in REF , it will be enough to see that MATH is closed in MATH. We also know from the proof of REF that MATH is defined by a finite number of polynomials MATH, MATH. Now for MATH, MATH for all MATH, so the polynomial MATH must be identically zero. T... |
math/9905001 | Choose MATH such that MATH for all MATH. Then MATH and the claim follows because the dimension of the fibers of the (non flat) family MATH is upper semicontinuous. |
math/9905001 | The morphism MATH defines a closed subset MATH, and we have to see that MATH. As MATH is closed and MATH it will be enough to see that MATH is the closure of MATH. Take MATH, with MATH. The family determined by MATH corresponds to an ideal sheaf MATH. By definition, MATH, so there is an open neigbourhood MATH of MATH i... |
math/9905001 | It is clear that MATH. On the other hand, as MATH is consistent, there are curves going sharply through MATH which miss MATH (compare CITE), so MATH. Abusing slightly notations, we will call MATH; as the length of MATH does not depend on MATH we have MATH and the schemes MATH form a flat family. |
math/9905001 | CASE: MATH is consistent in MATH, but not consistent in MATH, otherwise MATH. Therefore the proximity inequality at MATH must be satisfied for MATH but not for MATH. This means that MATH . Therefore, if MATH we must have MATH, and for any cluster MATH we have MATH. So MATH for all MATH, so MATH against the hypotheses. ... |
math/9905001 | We will proceed by induction on MATH. For MATH, there is nothing to prove. For MATH, we have either MATH or MATH so we can apply the induction hypothesis and MATH . On the other hand, a straightforward computation shows that MATH, therefore MATH. In these contitions, REF tells us that MATH thus finishing the proof. |
math/9905001 | By induction on MATH. The cases MATH are clear, so assume MATH and the claim to be true for MATH and MATH. This means that MATH, MATH are closed subvarieties and the morphism MATH is the restriction of MATH. So we have also a closed subvariety MATH and MATH so by the definitions, MATH is the strict transform of MATH un... |
math/9905001 | Write MATH with MATH. It is enough to define MATH and MATH . |
math/9905001 | It is clear that MATH; we will prove the claim by induction on MATH. For MATH the result is obvious. For MATH we have MATH. By semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. Now for MATH, unloading gives MATH, and the result follows from the i... |
math/9905001 | By semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. But for MATH, unloading gives MATH, and the result follows from REF . |
math/9905001 | The proof runs by induction on MATH, the case MATH being clear from REF . Assume MATH. For MATH general in MATH, consider the line MATH passing through MATH in the direction of MATH. If MATH we can assume (MATH being general) that MATH does not lie on MATH; for MATH this is automatic. For every curve MATH therefore MAT... |
math/9905001 | By REF we may assume MATH. But then MATH so MATH, and REF concludes. |
math/9905001 | By REF we may assume MATH. We will distinguish two cases according to the parity of MATH. CASE: Write MATH. As MATH is consistent in MATH, and MATH, we must have MATH, so by semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. It is clear that MATH ... |
math/9905001 | Consider MATH. Let MATH be the closed variety of MATH where MATH are proximate to MATH. It is easy to see that MATH is an irreducible variety and MATH. Then the claim may be equivalently stated as MATH where MATH. By semicontinuity it is enough to see that MATH and unloading gives that MATH is equivalent to MATH in MAT... |
math/9905001 | REF and the fact that all singularities of general curves sit at the proper base points of MATH are standard. For a proof of REF, compare CITE. Compare also NAME 's remark on REF. |
math/9905001 | This is an easy application of the residual exact sequence of the NAME method. There is a unique point MATH where the length of the component of MATH supported at MATH is bigger than that of MATH. Let MATH be a general straight line through MATH, and consider the residual exact sequence MATH . We have to prove that MAT... |
math/9905001 | Because of REF , MATH has no fixed part and MATH. By REF then, we only have to see that if MATH is composed of the curves of a pencil then MATH is a point with multiplicity MATH. Let MATH be composed of MATH curves of degree MATH in a pencil. Then MATH and MATH because MATH imposes independent conditions to curves of d... |
math/9905001 | Consider a scheme MATH where MATH is a tacnode scheme of order MATH supported at MATH and MATH is a cusp scheme of order MATH supported at MATH, all of them having their points in general position. We claim that the linear system MATH of curves of degree MATH containing MATH is nonempty and that a generic curve in it h... |
math/9905001 | Analogous to the proof of REF. |
math/9905001 | The cluster of infinitely near singular points of a MATH - singularity, with MATH even, is a weighted unibranched cluster with MATH free points with multiplicities MATH, so we are in a particular case of REF . If MATH is odd, then the cluster of infinitely near singular points of MATH is a weighted unibranched cluster ... |
math/9905007 | This follows from REF and CITE. |
math/9905007 | REF is in CITE, see also CITE. MATH follows from REF , and REF is trivial, by defining MATH from MATH. |
math/9905007 | We want to prove MATH . Multiplying to the left by MATH and to the right by MATH, this amounts to showing MATH as a map from MATH into MATH. To this end, pick a MATH such that MATH where MATH and MATH here refers to MATH-class in MATH. It suffices to verify the commutation on these MATH since they generate MATH. Pick p... |
math/9905007 | If MATH, MATH, choose again projections MATH and MATH such that MATH . Here we have identified MATH with a subgroup of MATH through MATH. Thus, for any real valued additive character MATH on MATH we have again MATH establishing the lemma. |
math/9905007 | By definition of MATH we have MATH . But MATH so MATH and hence MATH . Replacing MATH by MATH in the reasoning above, we obtain the other identity. |
math/9905007 | Suppose that MATH. Then as in the proof of REF we argue that MATH is properly outer where we use the fact that MATH for MATH. Thus we conclude that MATH is properly outer on MATH. |
math/9905007 | Without the condition MATH, this follows from CITE. To obtain this condition we consider each cycle MATH separately, and thus we may suppose that MATH is one cycle, that is, MATH except MATH. In this case we find MATH projections MATH, MATH, MATH in MATH such that MATH where MATH. Then we set MATH for MATH. It follows ... |
math/9905007 | For each MATH in REF we find a projection MATH such that MATH. However this only shows that MATH, and MATH will never be faithful. To overcome this difficulty we replace MATH by a smaller projection and we set MATH as in CITE. Note that we have not assumed that MATH. |
math/9905007 | First we take a sequence MATH of projections in MATH which represents MATH in REF and satisfies that MATH and for MATH we have that MATH . For any MATH with MATH it follows from the density of MATH in MATH that there exists a function MATH such that MATH . If MATH then MATH since MATH. Thus for a sufficiently large MAT... |
math/9905007 | By using the projections MATH in REF we obtain that MATH are almost mutually orthogonal when MATH and we must show that they are equivalent as projections in MATH. To show that they are equivalent, we regard MATH as homomorphisms of MATH into MATH and compute by REF - REF . REF MATH . With a similar computation for MAT... |
math/9905007 | By REF it suffices to show that there is a MATH such that MATH and MATH for MATH. (The existence of MATH is clearly necessary for the density. For the sufficiency, let us extract the apposite argument from CITE: Let MATH and put MATH. Then MATH. By NAME theorem, there exists a real polynomial MATH approximating MATH, a... |
math/9905007 | Note that MATH follows automatically. If MATH for an automorphism MATH of MATH, then there is an isomorphism MATH of MATH onto MATH such that MATH . By setting MATH and MATH, all the properties follow easily. |
math/9905007 | In view of REF it suffices to show this when MATH is a cocycle perturbation of MATH, that is, MATH with MATH a one-cocycle for MATH. We define an isomorphism MATH of MATH onto MATH by MATH . Since MATH, MATH, MATH naturally extends to an isomorphism MATH of MATH onto MATH. By setting MATH and MATH, all the properties f... |
math/9905007 | We have shown REF Suppose REF . Note that MATH and MATH are stable AF algebras and MATH are all isomorphisms such that MATH is commutative. Then by a standard intertwining argument we obtain isomorphisms MATH such that MATH is commutative and MATH, MATH, and MATH. Since MATH is purely infinite and MATH has only one tra... |
math/9905007 | We have shown REF Assume that REF holds. As in the proof of REF we obtain an isomorphism MATH of MATH onto MATH such that MATH for some unitary MATH. Since MATH, MATH is isomorphic to MATH, MATH, which is again isomorphic to MATH, we obtain an isomorphism MATH of MATH onto MATH such that MATH, extending MATH. The image... |
math/9905007 | By general theory we obtain that REF . From a part of the proof of REF we obtain REF . Assume that REF holds. Denoting by MATH the compact operators on MATH, we consider the systems MATH and MATH. Since MATH is a hereditary C*-subalgebra of MATH and MATH is a stable simple AF algebra, we have that MATH. In the spectral... |
math/9905010 | Immediate from the definition. |
math/9905010 | Notice all simple objects in one orbit give the same link invariant, so we could as well compute MATH by taking the sum only over a representative MATH of each orbit class, and replacing the factor MATH with MATH where MATH is the stabilizer of MATH and thus MATH is the number of elements in the orbit of MATH . Because... |
math/9905010 | See the end of REF. |
math/9905010 | It is well known that MATH is isomorphic to the group MATH by the map sending MATH to the homomorphism MATH on MATH such that MATH acts on the classical representation MATH by MATH (this is clearly a homomorphism descending to MATH and is injective by the faithfulness of the left regular representation. That the domain... |
math/9905010 | Suppose MATH is long, and MATH . Since MATH is the coefficient of MATH in the expansion of MATH in terms of simple roots, we have MATH for some positive integers MATH. To conclude the existence of a unique MATH as in the statement of the lemma, it suffices to show that MATH is a base Since the coefficient of MATH in th... |
math/9905010 | By REF , the weights occurring in the decomposition of the truncated tensor product of two weights lie in the product of their cosets in MATH and thus are annihilated by any homomorphism which annihilates the factors (since the reflections that generate the quantum NAME group preserve these cosets). Thus MATH is closed... |
math/9905010 | See end of REF. |
math/9905010 | See end of REF. |
math/9905010 | Of course since MATH on MATH can be identified with group multiplication on MATH the only closed subsets of MATH will be subgroups and hence of the form MATH . So assume MATH and that MATH is degenerate. We will show that MATH is in the range of MATH which suffices for the theorem. In that case if MATH then MATH so MAT... |
math/9905010 | Since we have exempted MATH is cyclic. |
math/9905010 | We note first that if MATH is in the NAME alcove, and MATH is an element of the quantum NAME group taking a weight MATH not in the NAME alcove into the NAME alcove, then the distance between MATH and MATH is strictly less than the distance between MATH and MATH . To see this, note that if MATH is in the NAME alcove, th... |
math/9905010 | We will make the argument for MATH noting parenthetically how it differs for MATH when not simply-laced. We will actually prove that MATH is nonzero, which is equivalent. Recall that since MATH is the weight MATH (respectively MATH the number of short simple roots). Thus in the sum REF , there is a contribution of MATH... |
math/9905010 | First, we note that MATH because they have the same inner product with the simple roots, and that MATH for all MATH . It suffices to check the second claim on simple roots. For MATH we have that MATH is another simple root of the same length, different from MATH and thus both sides of the equation are zero (the object ... |
math/9905010 | If MATH is invertible, NAME shows it is degenerate for MATH if and only if MATH for every MATH which by REF is true if and only if MATH annihilates all MATH . To determine whether MATH is odd or even, we need to check whether MATH which is to say whether MATH is an even or odd multiple of MATH . Notice MATH (Using the ... |
math/9905010 | Since MATH the image of MATH under MATH annihilates MATH and hence MATH . Since MATH so the image of MATH under MATH annihilates MATH or equivalently the image of MATH annihilate MATH . Thus the image of MATH annihilates MATH . By REF , this means MATH (which of course is contained in MATH) will consist of degenerate u... |
math/9905010 | A little thought will convince the reader that this result is almost immediate assuming that the intersection (a group of degenerate invertibles) acts freely on MATH . This is in fact the case, though one would like a more direct argument than the one below. Let MATH be the group MATH . Recall that, if MATH is the vect... |
math/9905010 | Of course it suffices to see that, with MATH as defined in REF, MATH is MATH . Let MATH be a generator of MATH then MATH is even degenerate and no smaller power of MATH is degenerate. To say that MATH for some MATH is degenerate for MATH is to say that MATH since it suffices to check the degeneracy condition against a ... |
math/9905010 | Of course if MATH is modular then MATH and it is easy to see that the category is a product of the two corresponding subcategories. The central observation is that if MATH is any abstract graph and MATH is a choice of an object MATH for each edge MATH and MATH for each vertex MATH where MATH is constructed out of MATH ... |
math/9905016 | Both sets represent MATH. |
math/9905016 | Because MATH there is a MATH-cofinal family MATH in MATH on which our map constant, say with value MATH. Because the map is monotone this MATH is the ordinal that we are looking for. |
math/9905016 | Let MATH be a coherent family of functions, with values in MATH. For MATH and MATH let MATH and MATH. The sets MATH generate a MATH-ideal on the countable set MATH and one readily checks that MATH whenever MATH and MATH. Now apply REF to find MATH such that MATH for all MATH and MATH. This defines MATH by MATH iff MATH... |
math/9905016 | Take, for each MATH, the monotone enumeration MATH of MATH and apply the equalities to find MATH such that MATH. Use these MATH to define open sets MATH; observe that MATH. It follows that MATH is infinite. We let MATH be the closed set MATH; its image MATH meets every MATH in an infinite set. For every MATH let MATH b... |
math/9905016 | We fix MATH and show how to find MATH and MATH for each MATH. We transfer the almost disjoint family MATH to MATH by setting MATH and MATH. It is fairly straightforward to show that MATH is neat; one stretches the bijection MATH to find an injection MATH from MATH to MATH that maps every MATH onto a branch of MATH and ... |
math/9905016 | Let MATH and consider the lifting MATH. We define a lifting MATH in a fairly obvious way: first fix MATH such that MATH for MATH and, if need be, redefine, for the duration of this proof, the values MATH for MATH and MATH so as to get MATH whenever this is needed. Then, given MATH put MATH and put MATH . We define, for... |
math/9905016 | Consider a potential sequence MATH of points in MATH where MATH and MATH disagree. By the disjointness condition and because the symmetric difference of MATH and MATH is always finite we can assume that MATH does not meet MATH when MATH. Let MATH and MATH. Observe that MATH and MATH determine the same open subset of MA... |
math/9905016 | If MATH then MATH and if MATH then MATH; now let MATH code witnesses: if MATH then MATH and if MATH then MATH. |
math/9905016 | Take MATH and fix MATH and MATH such that MATH for all MATH. Now simply let MATH and MATH; clearly MATH for all MATH. We see that MATH. |
math/9905016 | The first equality is clear: by construction MATH and MATH, so that MATH. To prove the second equality let MATH be an infinite subset of MATH such that MATH for all MATH; because MATH and MATH are finite-to-one we can assume that MATH. But then we would have a contradiction because on the one hand MATH while on the oth... |
math/9905016 | Let MATH be the set of MATH in MATH for which MATH is infinite. Thin out MATH to get MATH infinite for all MATH in MATH. Choose MATH as per REF for MATH and MATH, so that MATH and MATH. Because the sets MATH are pairwise disjoint we can a one-to-one choice function MATH for the family MATH such that MATH for all MATH. ... |
math/9905016 | Fix MATH such that for all MATH and MATH in MATH: if MATH and MATH then MATH. Now note that MATH, so that MATH is finite. |
math/9905016 | Assume we have MATH with MATH infinite for all MATH and MATH whenever MATH. Choose MATH as per REF for the sets MATH and MATH and choose MATH such that MATH for all MATH. We obtain a contradiction as before: MATH and MATH are disjoint, hence MATH and MATH are almost disjoint. On the other hand MATH and MATH; the inters... |
math/9905018 | CASE: Every circle centre defined by three different sites from MATH is a NAME vertex in some MATH-th and MATH-th order NAME diagram. As there are MATH different circles, the first claim follows. CASE: Consider the arrangement of bisectors MATH. Fix one bisector MATH. Without loss of generality we can assume that bisec... |
math/9905018 | We show that MATH is a rank function for MATH. A rank function maps an element MATH from a poset to a unique level in such a way that the level corresponds with the length of any maximal chain from MATH to MATH. Let MATH, with MATH. Then every point MATH has the MATH points from MATH as its MATH nearest neighbors. Now ... |
math/9905018 | We apply REF to MATH and MATH. MATH . We join the two sums by applying the Symmetry REF .and evaluate the result by using REF . MATH . The lemma follows from combining the two equations. |
math/9905018 | Using REF we can write MATH in terms of numbers of faces. MATH . |
math/9905018 | We prove the theorem by induction. We use inversion. Inversion changes the point-inside-circle relation in REF-dimensional space in a point-below-plane relation in REF-dimensional space. See CITE, page REF for more details and further references. The inversion map MATH is defined by MATH . It lifts points in the plane ... |
math/9905018 | This follows directly from REF . |
math/9905018 | We prove the property by induction. MATH is zero by definition. The number of vertices MATH in the first order NAME diagram equals the number of circles of order zero, MATH. Thus, MATH [induction step] Assume we have proved that MATH . We can rewrite this, by using induction again, as MATH . Now we evaluate MATH. MATH ... |
math/9905018 | Applying REF we get MATH . The claim now follows from evaluating REF using this expression. |
math/9905018 | Write MATH. Then we get from the definition that MATH where MATH, as MATH counts the empty set. MATH is the number of zero dimensional faces plus the number of MATH dimensional faces, so MATH. Applying REF gives MATH . Straightforward calculations show that MATH . So it follows that MATH . |
math/9905032 | Another identity due to NAME (see REF, or REF ) reads MATH . Substituting MATH we get REF. Substituting MATH yields MATH . Let MATH the difference of the left-hand side and the right-hand side in REF. Using REF and the recurrence relation MATH we find that MATH. Hence for any MATH it is a periodic function of MATH and ... |
math/9905032 | It is convenient to set MATH. Since the operator MATH is invertible we have to check that MATH . This is clearly true for MATH; therefore, it suffices to check that MATH where MATH and MATH. Using the formulas MATH one computes MATH where MATH. Similarly, MATH . Now the verification of REF becomes a straightforward app... |
math/9905032 | Using the relation MATH and the definition of MATH one computes MATH . Clearly, the relation REF remains valid for MATH. It remains to consider the case MATH. In this case we have to show that MATH . Rewrite it as MATH . By REF this is equivalent to MATH . Examine the right - hand side. The terms with MATH vanish becau... |
math/9905032 | Straightforward computation using a formula due to NAME, see REF or REF . |
math/9905032 | Follows from a computation done in the proof of REF MATH and the following corollary of REF MATH . |
math/9905032 | Our argument is similar to an argument due to NAME and NAME, see the proof of REF. The recurrence relation REF implies that MATH . Consequently, the difference between the left-hand side and the right-hand side of REF is a function which depends only on MATH. Let MATH and MATH go to infinity in such a way that MATH rem... |
math/9905032 | By REF , the restriction of MATH on MATH is the square of the kernel MATH. Since the latter kernel is real and symmetric, the kernel MATH is nonnegative. Hence, it remains to prove that its trace is finite. Again, by REF , this trace is equal to MATH . This sum is clearly finite by REF. |
math/9905032 | Since MATH is the square of the operator with the kernel MATH, it suffices to check that the latter operator commutes with MATH, with the above choice of MATH and MATH. But this is readily checked using REF. |
math/9905032 | By replacing MATH by MATH, we may assume that MATH. By NAME and NAME formulas, we have MATH . Choose some large MATH and split the circle MATH into REF parts as follows: MATH . REF and the equality MATH imply that the integral MATH decays exponentially provided MATH is large enough. On MATH, REF applies for sufficientl... |
math/9905032 | We shall use Debye's asymptotic formulas for NAME functions of complex order and large complex argument, see, for example, REF. Introduce the following function MATH . REF can be rewritten as follows MATH . The asymptotic formulas for NAME functions imply that MATH where MATH provided that MATH in such a way that MATH ... |
math/9905032 | First, we check REF . In the case MATH this was done in the previous lemma. Suppose, therefore, that MATH is a regular sequence in MATH and consider the asymptotics of MATH. Because the function MATH is an entire function of MATH and MATH we have MATH where MATH is arbitrary; we shall take MATH to be some small but fix... |
math/9905032 | Let MATH be a regular sequence and let the numbers MATH and MATH be defined by REF. We shall assume that MATH for all MATH. The validity of the theorem in the case when MATH for some MATH will be obvious form the results of the next section. We have MATH where the first line is the definition of MATH and the second is ... |
math/9905032 | Consider the left-hand side of REF and choose for each MATH a pair of functions MATH such that MATH and MATH as MATH. Then, on the one hand, the probability in left-hand side of REF lies between the corresponding probabilities for MATH and MATH. On the other hand, the probabilities for MATH and MATH can be expressed in... |
math/9905032 | Assume first that MATH. We denote MATH . It will be convenient to use the following notation MATH . REF reads MATH where MATH. We have the following estimates as MATH . Substituting this into REF, we obtain REF for MATH. Assume now that MATH. Denote MATH . Introduce the notation MATH . REF reads MATH where MATH. Again ... |
math/9905032 | First suppose that MATH. Set MATH. We shall use REF with MATH. Provided MATH is chosen small enough and MATH is sufficiently large, MATH will be close to MATH and we will be able to use NAME expansions. For MATH we have MATH and, similarly, MATH . Since the function MATH is concave, we have MATH . The constant here is ... |
math/9905032 | From REF we have MATH . Let us split the sum in REF into two parts MATH that is, one sum for MATH and the other for MATH, and apply REF to these two sums. Note that MATH here corresponds to MATH in REF ; this produces factors of MATH and does not affect the estimate. Let the MATH's stand for some positive constants not... |
math/9905032 | As shown in CITE, the Airy kernel has the following integral representation MATH . REF implies that for any integer MATH . Let us fix MATH and pick MATH according to REF . Since, by assumption, MATH and MATH lie in compact set of MATH, we can fix MATH such that MATH. Set MATH . Then the inequalities MATH are satisfied ... |
math/9905032 | It is clear that REF implies the convergence of MATH to MATH in the weak operator topology. Therefore, by REF , it remains to prove that MATH as MATH. We have MATH where the MATH correction comes from the fact that MATH may not be a number of the form MATH, MATH. By REF we have MATH . Similarly, MATH . Since we already... |
math/9905032 | For MATH, set MATH . As is well known (for example, CITE), MATH is a continuous map from MATH to MATH. Next, MATH evidently is a continuous function on MATH. Consequently, REF is well defined and is a continuous function on MATH. When MATH, REF agrees with the conventional definition, because then MATH . This concludes... |
math/9905032 | Indeed, MATH approximates MATH in the topology of MATH. |
math/9905032 | Indeed, this is true for finite - dimensional MATH, and then we use the continuity argument. |
math/9905032 | Given a finite subset MATH, we assign to it, in the natural way, a projection operator MATH. Then, by elementary linear algebra, MATH . Assume MATH becomes larger and larger, so that in the limit it covers the whole MATH. Then the left-hand side tends to the left-hand side of REF. On the other hand the right-hand side ... |
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