paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9903147 | If MATH, we have MATH . Likewise, MATH. In both cases, equality holds only if MATH. The corollary now follows from REF . |
math/9903147 | There can only be a nonzero morphism MATH if MATH. It follows from REF that MATH . |
math/9903147 | We have MATH . REF symmetries of MATH force it to vanish, and the result follows. |
math/9903147 | We have MATH . |
math/9903147 | The NAME complex of MATH is bigraded, MATH, and since the differential MATH is homogeneous of bidegree MATH, the homology is also bigraded. In terms of this bigrading, we wish to calculate MATH; evidently, this vanishes unless MATH. The plethysm REF implies that MATH . We will derive a lower bound for the Laplacian MAT... |
math/9903147 | NAME 's formula shows that MATH . On these four summands, the operator MATH equals MATH, MATH, MATH and MATH, respectively. Thus, the only summands on which MATH vanishes are MATH, and MATH. |
math/9903153 | By (joint) infinite descent. Here, as in subsequent proofs, the infinite-descent ``boilerplating" is omitted. Note that none of the hypothetical MATH-games in REF can be the MATH-game, so all of these games MATH have options. Suppose REF holds; say MATH, MATH, MATH. Some option MATH or MATH must be a MATH-game. But the... |
math/9903153 | By infinite descent. A solution to REF yields an (earlier-created) solution to REF, which yields a solution to REF, which yields a solution to REF. |
math/9903153 | By contradiction. A solution to REF would yield a solution to REF. |
math/9903153 | By infinite descent (making use of earlier results as well). Suppose REF holds with MATH. Some option MATH or MATH must be a MATH-game. In the former event, we have MATH (since MATH), so that either MATH REF , MATH REF , or MATH REF ; in the latter event we have MATH REF . Suppose REF holds with MATH. Since MATH, it ha... |
math/9903153 | By contradiction. A solution to REF would yield a solution to REF. |
math/9903153 | By infinite descent. Suppose REF holds with MATH. Then some option of MATH must be a MATH-game; without loss of generality, we assume MATH. But MATH, and we have already ruled out MATH REF , MATH REF , and MATH REF , so we have MATH REF . Suppose REF holds with MATH. MATH must have a MATH-option or MATH-option MATH, bu... |
math/9903153 | By infinite descent. Suppose REF holds with MATH. For all MATH we have MATH, so that MATH must have some MATH-option; but this MATH-option cannot be of the form MATH, since MATH REF . Hence there must exist an option MATH of MATH such that MATH. This implies that MATH, since none of the cases MATH REF , MATH REF , MATH... |
math/9903153 | Suppose MATH with MATH. Let MATH be an option of MATH. Since MATH, MATH must have a MATH-option of the form MATH (for MATH some option of MATH) or of the form MATH (for MATH some option of MATH). In either case, we find that the MATH-game MATH, when added to some other game (MATH or MATH), yields a game of type MATH; t... |
math/9903153 | Suppose MATH with MATH. Notice that MATH for every option MATH of MATH. CASE: There exist options MATH, MATH of MATH (possibly the same option) for which MATH. Then its option MATH. Since MATH, REF gives MATH. But this contradicts REF , since MATH. CASE: There do not exist two such options of MATH. Let MATH be an optio... |
math/9903153 | Any two players can gang up on the third, by depleting neither heap until the victim has made his move, and then removing both heaps. |
math/9903153 | CASE: Suppose MATH. Then its options MATH and MATH are MATH-games. But since MATH is also an option of MATH, this is a contradiction. CASE: Suppose MATH. Then MATH, MATH, and MATH are all MATH-games, and in particular MATH must have a MATH-option. That MATH-option can be neither MATH nor MATH, so there must exist MATH,... |
math/9903153 | CASE: Take an arbitrary game MATH. We know that each of MATH, MATH is either of type MATH or type MATH (by REF above). If either of them is a MATH-game, then so is the other (by REF ), and if neither of them is a MATH-game, then both are MATH-games. Either way, MATH and MATH have the same type. CASE: The proof is simil... |
math/9903153 | Let MATH be the simplest game not identical to REF such that MATH. CASE: MATH. Then MATH. But REF , together with the fact that MATH is equivalent to every NAME MATH with MATH, tells us that this can't happen. CASE: MATH. The winning option of MATH can't be MATH, by REF , so it must be an option of the form MATH. But t... |
math/9903154 | By REF , we have MATH . Therefore, MATH . Hence MATH . As a consequence, MATH . |
math/9903154 | For MATH, we have MATH . |
math/9903156 | Only the ``if" part is non-trivial. Assume that MATH is a projection band in MATH. Consider the complimentary band MATH. Then MATH. Since MATH is band preserving it follows that MATH leaves MATH invariant. But then the restriction of the operator MATH to MATH is a projection operator with the trivial kernel, and so MAT... |
math/9903156 | Let MATH be the maximal band of regularity of MATH. The existence of this band was established by NAME REF , and has been reproved in REF . The latter theorem asserts also that the complimentary band MATH is principally universally complete. An application of REF to the band MATH finishes the proof. |
math/9903156 | The implications MATH are obvious. To prove MATH assume, contrary to what we claim, that there is some MATH for which the collection MATH is not d-independent. Then there is a non-empty open subset MATH of MATH and some scalars MATH not all of which are zero such that MATH for all MATH from MATH. Note, however, that MA... |
math/9903156 | Let MATH be a maximal d-independent system in MATH. Assume, contrary to what we claim, that MATH is not a d-basis. Then there exists an element MATH that cannot be d-expanded with respect to MATH in the sense of REF . That is, we cannot find a full in MATH collection of bands MATH satisfying MATH. This implies that the... |
math/9903156 | To verify that MATH satisfies condition MATH take any band MATH in MATH and any element MATH that is not disjoint to MATH. Consider the principal ideal MATH generated by MATH. Then MATH for some compact NAME space MATH and MATH is a band in MATH. Since MATH satisfies the countable sup property, the compact space MATH s... |
math/9903156 | Suppose that MATH is a d-basis in MATH, where MATH. There are at least two elements MATH, MATH, which are not disjoint. Indeed, otherwise the element MATH would be a singleton d-basis in MATH. Let MATH and let MATH be an order isomorphism of MATH onto a dense subalgebra of some MATH space such that MATH. Since MATH and... |
math/9903156 | Suppose to the contrary that there is a finite maximal d-independent system MATH with MATH. Then after representing MATH on a compact space MATH, exactly as in the proof of the previous theorem, we can find at least one element in our system MATH, let it be MATH for definiteness, and a non-empty open subset MATH of MAT... |
math/9903158 | Since MATH, the map MATH extends uniquely to MATH by continuity. Thus, by the construction of MATH, the stratum MATH is projected homeomorphically to MATH. As the codimension of MATH in MATH is MATH, MATH. |
math/9903158 | The extensions can be defined by the following formulas: MATH . |
math/9903158 | Observe, that if MATH contains two elements of MATH, then it contains a pair of consecutive elements of the sequence MATH. If MATH or MATH, then MATH or MATH, respectively. If MATH or MATH, then MATH or MATH, respectively. |
math/9903158 | When MATH tends to infinity, all three vectors MATH, MATH, MATH lie in the plane containing MATH and the direction of the move of MATH. |
math/9903158 | All three vectors MATH, MATH, MATH lie in the plane containing the direction of the tangent vector to MATH at MATH. |
math/9903158 | NAME REF- REF prove the first statement of REF . Now we prove the rest. To evaluate the degree, we return to the arguments given in the first subsection of this section. Assume that our knot MATH is in general position with respect to the vertical projection. Calculate the degree by counting (with signs) points of the ... |
math/9903158 | Obviously, MATH . Now the result follows from REF. Appearance of MATH is related to the fact that closing a diagram long knot produces a new maximum point. |
math/9903158 | The local degree of MATH at a generic point MATH is MATH. Summing up the local degrees at all points of the preimage of some regular value, we obtain the right hand side of REF. |
math/9903165 | From the monotonicity assumptions on MATH and the assumption that dilations preserve the multiplication structure, it is easy to see that the multiplication law MATH on MATH must have the upper diagonal form MATH where MATH are polynomials. From REF we have MATH . Inserting this into REF and solving recursively for the... |
math/9903165 | Without loss of generality we may assume that MATH is supported on the unit cube MATH. When MATH the lemma is clear. For MATH we write MATH for all MATH, where MATH and MATH . Clearly MATH, MATH have bounded MATH norm on the unit cube, and MATH has mean zero for each MATH. The lemma then follows from induction. |
math/9903165 | For any small MATH, we have the NAME approximations MATH . Combining all these estimates with MATH and letting MATH gives the result. |
math/9903165 | It suffices to show the two estimates MATH and MATH where BMO is defined with respect to the ball structure of the homogeneous group. The desired distributional estimate follows from REF thanks to the inequality MATH which follows immediately from the NAME inequality and the NAME decomposition. The estimate REF follows... |
math/9903165 | We first construct an auxilliary sequence MATH of integers and a sequence MATH of intervals of integers by the following iterative procedure. Let MATH be the interval MATH. For each MATH in turn, we choose MATH so that MATH has minimal radius among all the balls MATH. Removing the element MATH from MATH divides the rem... |
math/9903165 | From REF, and REF we have MATH . By REF, this becomes MATH . Since MATH, MATH is bounded. From this, REF and the observation that MATH, we thus have MATH . Inserting this into the previous estimate we thus obtain MATH which is the first part of REF. The MATH portion of REF then follows from REF. We now turn to REF. Fro... |
math/9903165 | Since there are at most MATH finitely overlapping MATH which need to be considered for REF, it suffices to show that MATH for each ball MATH. But this follows from REF after some re-arranging. |
math/9903165 | Suppose MATH is in the set in REF. Since MATH fails, we have MATH . We rewrite this as MATH where the vector MATH is defined by MATH . Since MATH, we may rewrite this as MATH where the MATH denotes that the MATH term is missing from the wedge product. Since MATH holds, we thus see that MATH . Also, from REF we see that... |
math/9903165 | From REF, it suffices to verify MATH . The first two estimates follow from REF, while the third is trivial. The fourth estimate follows from the chain rule and REF providing that MATH. |
math/9903174 | If MATH then MATH is nonempty by the (projective) dimension theorem (see for example, CITE) and the fact that MATH. Assume now that MATH. Identify the set of systems MATH with the vector space MATH. In analogy to the proof of CITE we compute the dimension of the coincidence set MATH . Using the same arguments as in CIT... |
math/9903174 | (Compare with CITE). If MATH then a simple dimension argument shows that MATH cannot be surjective. We therefore will assume that MATH. Consider once more the coincidence set MATH introduced in REF and consider the projection onto MATH. For a generic point inside MATH the fiber of the projection has dimension equal to ... |
math/9903174 | Let MATH be a set of real matrices whose fiber has dimension equal to MATH. (Since the set of real matrices inside MATH is not contained in an algebraic set, such real matrices exist.) Let MATH be the induced center. If the degree of MATH is odd then the finite morphism MATH is surjective over the real numbers. Indeed ... |
math/9903174 | Let MATH, MATH. There is a fixed MATH permutation matrix MATH such that the set MATH is equal to the cell MATH described in REF . The linear transformation MATH extends to a linear transformation in MATH and this linear transformation maps MATH isomorphically onto MATH. |
math/9903174 | MATH . The matrix to the right, an element of MATH, induces a linear transformation on the projective space MATH which maps MATH onto MATH. |
math/9903174 | The closure of MATH in the NAME variety MATH is the NAME variety MATH, and MATH is a product of NAME varieties. The degree formula of a product of projective varieties under the NAME embedding CITE is given by MATH . Combining these formulas gives the result. |
math/9903174 | Let MATH be a MATH-dimensional subspace where MATH represents a MATH matrix whose entries are given by linear polynomials in MATH. The NAME coordinates of MATH define a polynomial map from MATH to MATH. Homogenizing the map, we have a map MATH defined by the sublinear system MATH of MATH. The restriction of MATH to MAT... |
math/9903174 | Consider the characteristic map MATH introduced in REF . According to REF MATH is a finite morphism of degree MATH. By REF the degree of MATH is MATH. For a generic set of polynomials MATH the inverse image MATH contains MATH different solutions and all these solutions are contained in MATH. |
math/9903178 | Observe that the MATH-module generated by MATH is the span of the elements MATH where MATH and where each MATH is a positive integer. To prove the first assertion, it is enough to check that this vector space is stable by multiplication by MATH where MATH. For this, write MATH. Then we have MATH . If MATH is such that ... |
math/9903178 | Observe that MATH is a MATH-module. Furthermore, it is spanned (as a vector space) by the images of MATH where MATH is a basis of MATH, and the MATH are positive integers. It follows that the MATH-module MATH is generated by the images MATH of the MATH (MATH). Observe that MATH is killed by MATH; thus, the MATH-module ... |
math/9903178 | We argue by induction on the number of elements in MATH. We may assume that MATH contains no proportional elements. Let MATH. Write MATH and consider MATH as a rational function on MATH. Observe that the poles of MATH are simple and along the hyperplanes MATH (MATH). Choose MATH among the poles of MATH. Choose a decomp... |
math/9903178 | First we claim that the space of functions which vanish at infinity is stable by the action of MATH. Indeed, let MATH vanish at infinity. Write MATH where MATH. For MATH, set MATH . The assumption that MATH vanishes at infinity means that MATH for all regular MATH and for all MATH in MATH. Let MATH; then, for all MATH ... |
math/9903178 | From REF we obtain MATH . So it is sufficient to check that MATH. For this, consider MATH where MATH and where MATH is linearly dependent. Choose MATH such that MATH for all MATH. We can find MATH such that MATH; then MATH . |
math/9903178 | First we consider the case where MATH. Then MATH because MATH is defined at MATH. So MATH maps MATH to MATH, and its trace is MATH. Now we assume that the formula holds for MATH, and we claim that it holds for MATH where MATH. Indeed, using the fact that MATH vanishes on derivatives, we obtain MATH which implies the cl... |
math/9903178 | Let MATH be a regular element of MATH. Then we have by the NAME formula: MATH . Now observe that MATH. Thus, we have MATH . |
math/9903178 | Let MATH and MATH a sequence of elements of MATH. Consider the element MATH of MATH. If MATH is contained in MATH, or if MATH, we are already in the desired set. If MATH is not in MATH and if MATH is not in MATH, then using the decomposition MATH we can strictly decrease the power of MATH in the expression of MATH. |
math/9903178 | Remark that MATH . Therefore, MATH maps MATH to MATH, and both members of REF vanish on MATH. Now consider an element MATH where MATH is generating. If the length of MATH is greater than MATH, both members of REF vanish. If MATH consists of one element, then MATH generates MATH and we obtain REF follows from the fact t... |
math/9903178 | Let MATH be the free vector space with basis the elements MATH, MATH. Let MATH be the kernel of the natural map from MATH to MATH. By definition, MATH is the space of linear relations between the MATH. We denote by MATH the subspace of MATH with basis MATH, and by MATH the subspace of MATH generated by the elements MAT... |
math/9903178 | REF is a consequence of REF ; a direct proof is a follows. Let MATH be a basis of MATH containing MATH, and let MATH be its image in MATH. Then MATH is a basis of MATH and all bases of MATH are obtained in this way. Moreover, MATH is a non-zero multiple of MATH. It follows that MATH is surjective. Clearly, the kernel o... |
math/9903178 | The formula for the NAME transform of MATH is straightforward. It implies surjectivity of MATH because this map is MATH-linear, and the MATH-module MATH is generated by the MATH where MATH is a basis of MATH such that MATH is contained in MATH (here we use the assumption that MATH is centrally symmetric). For injectivi... |
math/9903178 | It is enough to check this for MATH. Then, for any MATH, we have MATH where MATH denotes the boundary of the support of MATH; here the latter equality follows from NAME 's theorem. Because MATH is a union of polyhedral cones of smaller dimensions, the function MATH is in MATH. We thus have MATH which implies our formul... |
math/9903178 | It is sufficient to prove this formula for MATH. As we have for MATH: MATH we obtain MATH . So we see, from REF above, that the equation of REF is satisfied. |
math/9903178 | It is sufficient to prove this formula for MATH. (On MATH, both sides are equal to MATH, because MATH maps MATH to MATH). Thus it is sufficient to prove this formula for a derivative MATH of an element MATH, with MATH. Then MATH . If MATH is not a wall of MATH, then MATH has no jump along MATH. Thus the left-hand side ... |
math/9903178 | Observe that MATH extends to a continuous function on MATH if and only if it has no jumps along walls. This amounts to MATH for any wall MATH (because MATH maps MATH to MATH, and MATH is injective on the latter). Equivalently, MATH for all regular MATH and for all MATH. Because MATH vanishes at infinity, this means tha... |
math/9903178 | Because of the relation MATH modulo MATH we see that the image of the element MATH in MATH changes sign, if we flip one of the elements MATH in MATH to MATH. We thus may assume that the relation is MATH for some MATH. Then the cones MATH REF are the maximal cones in a polyhedral subdivision of MATH, and we conclude by ... |
math/9903178 | Clearly, each MATH is contained in MATH. Conversely, let MATH. If MATH lies in no MATH then the segment MATH has a non-empty interior in MATH and is contained in the union of all facets of the MATH. It follows that this segment is contained in a facet of some MATH. Thus, MATH is in the hyperplane generated by this face... |
math/9903178 | Let MATH, then MATH or MATH is in MATH. Using the relation MATH, we see that MATH has a representative with support in MATH, for any linearly independent MATH. This shows existence. For uniqueness, it is enough to check that any MATH with support in some acute cone MATH must be zero. This is shown in the proof of REF f... |
math/9903178 | Observe first that the formula makes sense: because MATH is regular, MATH is regular for MATH sufficiently small and MATH. If the formula holds for MATH then it holds for MATH where MATH, because both MATH and MATH are MATH-linear. Thus it suffices to check the formula for MATH where MATH. Then MATH whereas MATH is res... |
math/9903179 | We proceed by induction on MATH. For MATH there is nothing to show (MATH is just one point). Let MATH and MATH. CASE: The subfunctor MATH of MATH given by MATH is representable by a locally closed subspace MATH. This can be seen as follows: Consider the description of MATH as an algebraic subset of the NAME of codim MA... |
math/9903179 | Again, we proceed by induction on MATH. With the notations introduced in the proof of REF , we can assume that the first MATH triples MATH are pairwise different and occur precisely MATH-times among all such triples (in particular, MATH). Recall that we assumed MATH precisely for MATH. (Note that MATH if MATH). For any... |
math/9903179 | Note that REF implies MATH. Hence, due to Nori's theorem (compare CITE), MATH for all curves MATH. We show that there are (at least) two different components of MATH: by REF , MATH and REF , gives the existence of a nonempty component MATH of MATH having the expected dimension MATH (the expected dimension in the constr... |
math/9903180 | Define an operator MATH by MATH . We first check that MATH commutes with the action of differential operators with constant coefficients. Using the equation MATH and the main REF of MATH, we obtain MATH . It remains to see that MATH and MATH coincide on MATH. For this, we will use the following formula. If MATH is a po... |
math/9903180 | We have MATH . In computing MATH, the variable MATH is fixed to a non-zero value. The result MATH is a meromorphic function of MATH. It is thus sufficient to prove that MATH belongs to the space MATH. We check this for MATH where MATH and MATH is a generating sequence. Let MATH and MATH . As MATH is generating, the set... |
math/9903180 | Indeed, if MATH is a smooth function on MATH with compact support, consider the series MATH . The coefficient MATH is rapidly decreasing in MATH, as the function MATH is smooth and compactly supported. Thus MATH is also a rapidly decreasing function of MATH. Furthermore MATH depends holomorphically on MATH. So the resu... |
math/9903180 | Consider first the one variable case. The set MATH is MATH. Let MATH be the integral part of MATH. Fix MATH. Consider the locally constant function of MATH defined by MATH . We extend this function as a locally MATH-function on MATH (defined except on the set MATH of measure MATH). We have the equality of generalized f... |
math/9903180 | By a method entirely similar to the proof of REF , we see that the operator MATH satisfies the relation MATH for MATH, MATH. Thus to prove that MATH on MATH, it is sufficient to prove that they coincide for MATH. In this case, we obtain MATH . |
math/9903180 | Let MATH be a basis of MATH. Although we are not able to give a nice formula for the function MATH, we can still obtain an inductive expression that suffices to give some informations on it. Consider the set MATH, that is, the complement of the union of MATH-walls together with their translates by MATH. Let MATH be the... |
math/9903180 | From the definitions of MATH and MATH, we obtain for any MATH: MATH . Thus, it is enough to prove that MATH for MATH, because MATH generates MATH as a MATH-module by REF . For MATH in an alcove MATH, we can define the operator MATH by MATH . The kernel formula holds for the operator MATH. In particular, we obtain for M... |
math/9903186 | For convenience of the reader we reproduce here the proof first presented in CITE. It suffices to consider MATH and MATH . Since MATH by the NAME hypothesis on MATH and the fact MATH as MATH, the scalar NAME function MATH satisfies MATH . By standard results (see, for example, CITE, CITE), REF yields MATH where MATH is... |
math/9903186 | Let MATH. By REF we infer MATH . Adding REF, differentiating REF with respect to MATH proves REF for MATH. The result extends to all MATH by continuity of MATH in MATH-norm. |
math/9903186 | An explicit computation shows MATH for all MATH . Since MATH for MATH as a result of analytic continuation, one obtains MATH proving REF. (Here the interchange of the MATH and MATH integrals follows from NAME 's theorem considering REF in the weak sense.) |
math/9903186 | Given MATH one infers MATH by NAME 's theorem. Combining REF one concludes MATH . Since MATH and MATH are both bounded operators and by REF their quadratic forms coincide, one obtains REF. |
math/9903186 | NAME 's result CITE, applied to the NAME space of MATH-operators yields MATH (with MATH denoting convergence in MATH-norm). Combining REF, and REF then yields REF. |
math/9903186 | The a.e. existence of the norm limit in REF and the invertibility of MATH in MATH is a consequence of REF . By REF of logarithms of dissipative operators one infers MATH . By REF, MATH is reduced by the subspace MATH and MATH . The operator MATH restricted to the invariant subspace MATH then can be represented as follo... |
math/9903186 | REF is clear from the identities MATH and REF follows since MATH maps MATH into MATH and hence MATH . REF is then clear from REF. |
math/9903186 | Using REF, MATH the resolvent equation MATH and MATH one verifies REF then follows upon integrating REF. The integration constant MATH can be determined by a somewhat lengthy perturbation argument as follows. For brevity we will temporarily use the following short-hand notations, MATH . First we claim MATH . Since none... |
math/9903186 | Differentiating REF with respect to MATH taking into account REF - REF yields MATH . In order to verify REF we need to estimate various terms. For brevity we will again temporarily use the short-hand notations introduced in REF. Thus, REF becomes MATH and we need to verify the last line in REF . By REF, one concludes M... |
math/9903187 | REF is a direct consequence of REF . The proof of REF is similar to the proof of REF. |
math/9903187 | The proof is literally the same as the one of REF, noticing that REF also holds for a closed subscheme MATH of MATH with MATH. |
math/9903187 | The proof of Definition-REF generalizes to the present case because REF holds also for MATH a scheme of finite type over MATH (replacing ``dimension" by ``relative dimension"), and because REF holds for ``semi-algebraic" replaced by ``MATH-semi-algebraic" (compare REF below), both with identically the same proofs. Note... |
math/9903187 | Direct verification. |
math/9903187 | By resolution of singularities we may assume that MATH is smooth. If MATH is induced by a proper birational morphism from MATH to MATH, then REF is a direct consequence of REF. In the general case it is a direct consequence of REF and the following REF . |
math/9903187 | Let MATH in MATH be large enough. We may assume that MATH is cylindrical at level MATH. That MATH is cylindrical at level MATH is an easy consequence of the following assertion: CASE: For all MATH in MATH and MATH in MATH, with MATH, there exists MATH in MATH with MATH and MATH (whence MATH, since MATH is cylindrical a... |
math/9903187 | Well known. |
math/9903187 | It follows from REF that a point of MATH which projects to a point in MATH is in the MATH-orbit of a point of the form REF . To conclude observe that, in the basis MATH, MATH-invariant polynomials are sums of monomials of the form MATH, with MATH dividing MATH. |
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