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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 MATH on each summand. Given a partition MATH, we calculate that MATH and MATH . On the summand MATH, we have MATH, MATH . Combining all of these ingredients, we see that MATH. If MATH, the right-hand side is bounded below by MATH; unless MATH and MATH, our summand does not contribute to MATH. Equivalently, MATH must lie in the interval MATH. It remains to consider the summands of MATH with MATH; these have the form MATH . On the summand MATH of MATH, the operator MATH equals MATH while on all other irreducible components of MATH, it is strictly less. It follows that the Laplacian can only vanish on the summand MATH, and only at that when MATH. |
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 then we have either MATH (every option MATH must be a MATH-game), which is REF, or MATH (every option MATH must be a MATH-game), which is REF. Suppose REF holds; say MATH, MATH, MATH. Then there exists MATH, which must satisfy MATH REF . Suppose REF holds; say MATH, MATH, MATH. Then there exists MATH, which must satisfy MATH REF . Suppose REF holds; say MATH, MATH, MATH. Then there exists MATH, which must satisfy MATH REF . Finally, suppose REF holds; say MATH, MATH, MATH. Then there exists MATH, which must satisfy MATH REF . |
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 has an option MATH of type MATH or type MATH (for if all options of MATH were MATH-games and MATH-games, MATH would be of type MATH or MATH). If MATH, then we have MATH REF , and if MATH, then we have MATH REF . Suppose REF holds with MATH. Then MATH REF . Suppose REF holds with MATH. Since MATH, it has an option MATH of type MATH or of type MATH (for if all options of MATH were MATH-games and MATH-games, MATH would be of type MATH or MATH). MATH yields MATH REF , and MATH yields MATH REF . Finally, suppose REF holds with MATH. Then there exists MATH, which must satisfy MATH REF . |
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, but if MATH then MATH REF , which can't happen; so MATH. Similarly, MATH has a MATH-option MATH. MATH, so MATH REF . (Note that the second half of this proof requires us to look two moves ahead, rather than just one move ahead as in the preceding proofs.) |
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 REF can occur. Similarly, every MATH has an option MATH such that MATH, MATH. Since MATH is a MATH-game, MATH and MATH are MATH-games and MATH is a MATH-game. One of the options of MATH must be a MATH-game; without loss of generality, say MATH. Since MATH and since none of the cases MATH REF , MATH REF , MATH REF can occur, MATH must be a MATH-game. But recall that MATH is a MATH-game, so that its option MATH is a MATH-game. This gives us MATH, which is an earlier-created solution to REF. |
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; this is impossible, by REF . |
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 option of MATH. Since MATH, and since there exists no MATH for which MATH, there must exist an option MATH of MATH such that MATH. MATH, by REF , but MATH cannot be of type MATH, since adding MATH yields a MATH-position. Hence MATH, and REF implies MATH. Since MATH, there must exist an option MATH with MATH. Everything we've proved so far about MATH applies equally well to MATH (since all we assumed about MATH was that it be some option of MATH). In particular, MATH must have an option MATH such that MATH. However, since MATH is an option of the MATH-position MATH, MATH. Hence MATH and MATH are two MATH-positions whose sum is a MATH-position, contradicting REF . |
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, contradicting REF . CASE: Assume MATH. Then either MATH or MATH (no other option of MATH can be of type MATH, by REF ), and in either case MATH. CASE: Suppose MATH. Then MATH, MATH, and MATH are all MATH-games. MATH must have a MATH-option, but MATH and no option MATH or MATH REF can be a MATH-game (by REF ), so MATH itself must be a MATH-game. Also, since MATH and MATH, there must exist MATH with MATH. Then MATH and MATH, which is inconsistent with MATH. CASE: Every option of MATH has a component heap of size REF or more, so MATH has no MATH-options, by REF . CASE: Suppose MATH. Then MATH can't be MATH (by REF ), so it must have an option MATH; MATH, contradicting REF . CASE: Suppose MATH. Then MATH can't be MATH (by REF ), so it must have an option MATH; MATH, contradicting REF . |
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 similar, except that one needs REF instead of REF . CASE: For all MATH, MATH and MATH. |
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 then MATH, which contradicts the assumed minimality of MATH. (MATH won't help us, since MATH, not MATH.) Case III: MATH. Letting MATH be any option of MATH, we have MATH. This contradicts the assumed minimality of MATH. |
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 MATH is the identity on MATH. |
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 MATH is an algebraic equation of degree not exceeding MATH. Therefore it cannot have more than MATH solutions, and this implies that the function MATH must be a constant on a non-empty open subset MATH of MATH, a contradiction. |
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 there exists a band MATH in MATH such that for any non-trivial band MATH, for any finite set of indices MATH, and for any scalars MATH the element MATH is not disjoint to MATH. Let us consider this band MATH. Since MATH satisfies MATH, we can find a non-zero MATH, a full in MATH system of pairwise disjoint bands MATH, and scalars MATH such that for each MATH we have MATH. It is easy to verify now that the system MATH is d-independent, a contradiction to the maximality of MATH. |
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 satisfies the countable chain condition and consequently, as said earlier, the collection of essentially constant functions is dense in MATH in view of [HK,RR]. As shown in REF there is a non-zero essentially constant function MATH in MATH. It remains to note that MATH is a semi-component of MATH not only in MATH but also in MATH, and we are done. |
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 are d-independent, there exists a non-empty open subset MATH such that at least one of the functions MATH is not constant on each non-empty open subset MATH. Assume for definiteness that the function MATH is such. Then for any positive integer MATH the functions MATH belong to MATH and by REF they are d-independent on MATH. To conclude the proof we will verify that this contradicts the fact that the system MATH is a d-basis in MATH. Indeed, the latter implies that for each non-empty open set MATH and each MATH we can find a non-empty open subset of MATH on which the element MATH is a linear combination of the elements of our d-basis. Consequently we can find a non-empty open subset MATH of MATH on which each of the functions MATH is a linear combination of the functions of the d-basis MATH. But this is impossible if MATH since, as said above, the functions MATH are linearly independent on MATH. |
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 MATH such that for any open MATH and for any positive integer MATH the functions MATH, are linearly independent on MATH. Since MATH satisfies the countable sup property, the compact space MATH satisfies the countable chain condition. Fix a MATH. For each MATH-tuple of scalars MATH consider the function MATH . Observe that if MATH is another m-tuple. then by REF the functions MATH and MATH are d-dependent on MATH iff MATH for some scalar MATH. Let us fix an arbitrary uncountable collection of points MATH such that any MATH pairwise distinct points MATH in MATH are linearly independent. Then, by REF , the functions MATH are d-independent on MATH. Fix any MATH and consider the function MATH introduced above. Since MATH is a maximal d-independent system, the function MATH cannot be d-independent of this system on MATH. Therefore, there is a non-empty open subset MATH such that MATH coincides on MATH with a linear combination of the functions MATH. We claim next that whenever we take arbitrary MATH mutually distinct points MATH we necessarily have that MATH . Indeed, if an open set MATH is not empty, then on MATH each of the functions MATH is a linear combination of the functions from MATH. On the other hand, as said above, these functions MATH are linearly independent on MATH, a contradiction. We conclude the proof by showing that the existence of an uncountable family MATH of non-empty open sets satisfying REF , contradicts the countable chain condition in MATH. We will use induction on MATH. Assume that the above statement is true for some MATH, and let us prove it for MATH. Consider all sets of the form MATH where MATH and MATH are pairwise distinct points in MATH. Observe that by REF the sets MATH are pairwise disjoint. If each MATH, then we are done in view of the induction hypothesis. So some sets MATH are non-empty, and, since they are pairwise disjoint, there may be at most countably many of such sets. Let MATH be all these non-empty sets. Since each MATH depends on a finite number of indices in MATH, there are uncountably many MATH-s that have not been used. Consider all the sets MATH for which MATH depends on at least one of these unused MATH-s. Again by the induction hypothesis, all these MATH cannot be empty. Take an arbitrary such MATH that is not empty. However, this MATH must be disjoint from each MATH, and so we have a contradiction. |
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 preimage of a regular value MATH of MATH close to MATH. As we observed in REF, those of them which belong to each of the six copies of MATH contribute MATH. It remains to notice that the preimage does not intersect the copies of MATH. Indeed, each point of the preimage belonging to one of these copies would correspond to a configuration of MATH such that MATH is positioned in MATH almost strictly above MATH and MATH. The points MATH and MATH are not close to each other because they are separated on MATH by MATH. The projection of MATH is assumed to be generic. In particular, it does not have triple points. Therefore if MATH is sufficiently close to MATH, this configuration cannot appear. Note also, that although MATH divides MATH, and the regular value MATH may be chosen in any component of the complement of MATH, the above evaluation of the local degree does not depend on this choice. This completes the proof of REF . |
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 components of MATH we see that MATH for some polynomials MATH which depend on the MATH. The claim follows. |
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 trivially from the triangle inequality and REF. To show REF, It suffices to show that MATH for all balls MATH, where MATH and MATH is the mean of MATH on MATH. Fix MATH, and suppose that MATH has radius MATH. We divide into three cases, depending on the relative sizes of MATH, MATH and MATH. We first consider the case where MATH is larger than MATH. In this case MATH vanishes unless MATH is in MATH, in which case the oscillation is MATH. Since the MATH are disjoint and live in MATH the total contribution from these balls is acceptable. Next, we consider the case where MATH has size between MATH and MATH inclusive. For each scale MATH, there are at most MATH balls MATH of size MATH which give a non-zero contribution. Since each ball contributes at most MATH, we are done. Finally, we consider the case where MATH has size smaller than MATH. For each scale MATH, there are at most MATH balls MATH which contribute. But from the smoothness of MATH we see that each ball gives a contribution of MATH for some MATH. Summing in MATH we see that this contribution is also acceptable. |
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 remainder into two intervals MATH and MATH; we choose MATH to be the larger of the two intervals. We then increment MATH and iterate the above construction. One can easily show inductively that MATH for all MATH, so that all the MATH are well defined. Furthermore, one has MATH for all MATH between MATH and MATH. One of the sets MATH, MATH has a cardinality of at least MATH. If the former set is larger, we choose MATH to be the first MATH elements of this set and observe that MATH. Otherwise we choose MATH to be the first MATH elements of the latter set and observe that MATH. |
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. From REF, and REF, we have MATH where MATH is the quantity MATH . Since MATH, MATH is bounded, and so the first part of REF obtains. To show the MATH portion of REF, it suffices from REF and the chain rule to show that MATH. But by REF, and REF we have MATH and the claim follows from REF and the inequality MATH arising from the hypothesis MATH. We now show the MATH portion of REF. We consider the MATH derivatives for MATH and MATH separately. If MATH, then we see from REF that MATH so the claim follows from the first part of REF. If MATH, then from REF we have MATH . Since MATH is polynomial and MATH is bounded, it thus suffices by REF to show that MATH . But from REF, and REF, we have MATH and the claim follows from the first part of REF. We now turn to REF. It suffices to show that MATH . From REF we have MATH . By REF this is majorized by MATH . The claim then follows from REF. Finally, we show REF. We can rewrite the desired estimate as MATH . From the support assumptions on MATH and MATH we have MATH. Thus by the chain rule and REF, it suffices to show that MATH and MATH for all MATH. We may of course assume that MATH and MATH since the claims are trivial otherwise. But these estimates follow from REF, and REF, noting that MATH from the hypothesis MATH. |
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 . By REF, we thus have MATH . By REF, this simplifies to MATH . Since MATH, we have MATH. Applying this inequality, we obtain a telescoping product which simplifies to MATH . From REF we see that MATH is independent of MATH if MATH. If MATH, then MATH can vary with MATH. However, from REF we see that MATH unless MATH. In both cases we thus conclude that, up to an error of MATH, the quantities MATH can each take at most MATH values. From this, REF, and REF, we see that MATH lies in a union of MATH-neighbourhoods of hyperplanes. From REF and the fact that the frozen quantities MATH are comparable to REF, we thus see that MATH also lives in a union of MATH-neighbourhoods of hyperplanes. From REF and the boundedness of MATH, we thus see that MATH lives in a union of MATH-neighbourhoods of compact hypersurfaces. From REF we have MATH and so MATH also lives in the union of MATH-neighbourhoods of compact hypersurfaces. The desired cardinality bound on the possible MATH then follows from REF and a covering argument. |
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 CITE one computes MATH . Since MATH is projective the projection onto the second factor (namely MATH) is an algebraic set by the main theorem of elimination theory (see for example, CITE). This projection can result in an algebraic set of dimension at most MATH. The claim therefore follows. |
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. Let MATH be a system whose fiber has this dimension and let MATH be the corresponding center as defined in REF . By construction we have that MATH . In particular if MATH then MATH and the characteristic map MATH is a finite morphism. If MATH choose a subspace MATH having codimension MATH inside MATH and having the property that MATH . Such a subspace MATH exists by CITE. Let MATH be the central projection with center MATH and let MATH be the central projection with center MATH. Then MATH is a finite morphism which is surjective over MATH. MATH is a linear map, it is surjective as well and MATH . It follows that MATH is surjective as soon as MATH. |
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 over the complex numbers the inverse image MATH represents a finite set of complex conjugate points for every real point MATH. But then also MATH and MATH are surjective over the reals. |
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 MATH is the NAME embedding which is described in the beginning of REF. However, the map MATH is not well-defined on a subvariety of codimension REF, that is, MATH is not a ``morphism," but a ``rational map" with non-empty indeterminacy locus. To eliminate this indeterminacy locus we will construct a ``blow-up." Let MATH be the indeterminacy locus of pure codimension REF where MATH is homogeneous of degree MATH. We may assume that MATH is nonsingular and irreducible. If not, consider MATH the trivial flat family of MATH over MATH (or the projection onto MATH) where MATH is the affine line over MATH. Let MATH be the fiber over MATH, and consider the subvariety MATH of MATH given by MATH where MATH is homogeneous of degree MATH. Such MATH forms a flat family over MATH by varying MATH with MATH. Moreover, we can choose an appropriate MATH so that the generic MATH is nonsingular and irreducible. Let MATH be the ideal of MATH in MATH for MATH. Let MATH be the sublinear system of MATH consisting of the same polynomials of degree MATH as for MATH, except that MATH is replaced by MATH . For each MATH, we take the blowing-up MATH of MATH with the exceptional divisor MATH, satisfying the following diagram: MATH with MATH and MATH where MATH is the pulled-back sublinear system on MATH . On the other hand, we take the blowing-up of MATH, the ideal of the subscheme MATH in MATH: MATH with the exceptional divisor MATH. Since MATH is a smooth curve, the composition MATH is flat (compare CITE). The fiber of MATH over MATH is MATH, that is, the blowing up of MATH, the ideal of MATH in MATH. Note that the restriction of the exceptional divisor MATH to a fiber, MATH, is MATH. The flatness of MATH assures that certain numerical invariants remain constant in the family (see CITE). In particular, the self-intersection MATH is independent of MATH. This is why we can assume that MATH is irreducible and nonsingular. Let us proceed with the proof of the theorem. We write MATH. We have the following diagram: MATH where MATH is the blowing-up of MATH. MATH. By using the NAME relation MATH inductively as in the previous cases, we get the following: MATH which yields MATH . |
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, then the corresponding term in the right-hand side is in the MATH-module generated by MATH: indeed, if MATH then MATH is a basis of MATH. On the other hand, if MATH then our term is the inverse of MATH with MATH and MATH. So the assertion follows by induction on MATH. Similarly, any element of MATH is a linear combination of elements MATH where MATH, MATH is linearly independent and the MATH are positive integers. If moreover MATH is not generating, then MATH is in MATH. If MATH is generating, then we can express MATH as a polynomial in the variables MATH, and we obtain MATH. |
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 is a non zero quotient of MATH. The latter is a simple MATH-module, isomorphic to MATH; therefore, MATH is isomorphic to MATH, too. Iterating this argument, we construct an ascending filtration of the MATH-module MATH, each submodule being generated by certain MATH's, with successive quotients isomorphic to MATH. |
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 decomposition MATH. Then MATH (the fraction field of MATH) is identified with the field of rational functions in the variable MATH, with coefficients in MATH. Therefore, we have a restriction map MATH. Consider the image MATH of MATH in MATH. The restriction map extends to an homomorphism MATH by restriction to generic points. We have also a residue map MATH with respect to the variable MATH, defined by the formula MATH for any integer MATH such that MATH. As MATH is a simple pole of MATH, we have simply MATH where MATH denotes the image of MATH in MATH. If MATH is a basis of MATH which contains MATH, then MATH is a basis of MATH. Therefore, MATH is in MATH. Consider a generator MATH of MATH, with MATH and MATH non generating. Write MATH . If MATH, then MATH. If MATH, the set MATH contains MATH and is non generating. Thus, its restriction MATH is non generating. We see that MATH can be written as MATH for some MATH, so that MATH. If MATH, it follows from the above discussion that MATH. Therefore, by the induction hypothesis, we have MATH: thus, MATH has no pole along MATH. By the beginning of the proof, MATH has no pole at all, so that MATH. |
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 such that MATH is regular, we also have MATH and therefore, the function MATH vanishes at infinity. Now MATH which implies our claim. Assume now that there exists a non-zero MATH which vanishes at infinity. As in the proof of REF , we can write MATH where the sum is over all linearly independent subsets MATH which are not bases, and where each MATH is in MATH. Furthermore, we may assume that the number of MATH such that MATH is non-zero is minimal (among all possible decompositions of all non-zero MATH which vanish at infinity). Choose MATH such that MATH, and choose a non-zero MATH such that MATH for all MATH. Then MATH for large MATH. But all successive derivatives of MATH vanish at infinity and are in MATH. Moreover, MATH is a decomposition with fewer terms than MATH. Thus, MATH for some positive MATH. Choose MATH minimal with this property, and set MATH. Then MATH is a non-zero element of MATH which vanishes at infinity, and MATH. But then the function MATH is constant for any MATH, a contradiction. |
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 claim. It follows that the formula holds for any MATH. If MATH then the left-hand side vanishes. On the other hand, the function MATH is in MATH; thus, MATH is in MATH and the right-hand side vanishes, too. |
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 that MATH is homogeneous of degree -REF. |
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 MATH. Let us show that MATH. If MATH is an ordered basis of MATH, we set MATH. If MATH is not in MATH, then there exists a MATH such that the set MATH is linearly dependent. Using the relation MATH, we replace MATH by a linear combination of elements MATH where MATH is obtained from MATH by replacing one of the elements MATH with MATH by MATH. It follows that the numbers MATH are strictly smaller than MATH , so that by induction, we obtain MATH. This shows that the set MATH generates MATH, and that MATH is spanned by MATH and by MATH. We now show that MATH. We need to check that if MATH as a rational function, then all MATH are equal to MATH. We prove this by induction on the number of elements in MATH. Remark that all elements of MATH contain MATH. For a set MATH of MATH linearly independent vectors, let MATH be the hyperplane generated by MATH. For a hyperplane MATH, set MATH. We write MATH with MATH . Choose a hyperplane MATH generated by MATH, for some MATH. Then the residue operator MATH kills all elements MATH except MATH, which is mapped to MATH. Thus, MATH. But remark that if we consider the ordered set MATH, the set MATH consists exactly of the elements MATH such that MATH. We conclude by applying the induction hypothesis to the vector space MATH and the system MATH, for all MATH. |
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 of MATH contains MATH. Conversely, if MATH is mapped to REF by MATH, then MATH is defined on MATH and thus, MATH. We can write MATH where MATH and MATH. Then MATH is in MATH, too, whence MATH and MATH. CASE: Let MATH be another ordered basis of MATH. Consider the element MATH . If MATH is non-zero, then we must have MATH for some non-zero MATH and some index MATH. If moreover MATH is non-zero, then we must have MATH for some non-zero MATH and some MATH (because MATH is not a multiple of MATH). Continuing in this way, we see that either MATH or there exists a permutation MATH of MATH and non-zero MATH such that MATH for all MATH. Then we have MATH. On the other hand, set MATH . If MATH, then there exist a unique index MATH and a unique non-zero MATH such that MATH . Further, MATH for all MATH. If moreover MATH, then MATH for uniquely defined MATH and MATH. Further, MATH for all MATH. Continuing, we obtain if MATH: MATH for a permutation MATH and non-zero MATH; then we have MATH. This is equivalent to the set of conditions of the first part of the proof, with MATH and MATH. CASE: Let MATH such that MATH. Let MATH and MATH be as above; then MATH, MATH and MATH are linearly dependent. Write MATH, MATH and MATH, then MATH and MATH. If MATH then MATH, MATH and MATH are linearly dependent, which contradicts the hypothesis MATH. Similarly, we cannot have MATH. Thus, MATH, that is, MATH. In this way we obtain MATH for all MATH; because MATH and MATH are in MATH, it follows that MATH. |
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 injectivity of MATH, we observe that any function MATH is supported in the acute cone MATH, and that MATH if and only if MATH vanishes outside a set of measure zero. Moreover, the set of functions MATH (where MATH) is dense in the space of smooth, rapidly decreasing functions on MATH. |
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 formula. |
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 of the equality in REF is equal to MATH. The right-hand side is also MATH, as there are at least MATH vectors in MATH which are not in MATH. If MATH is a wall of MATH, there exists MATH such that MATH where MATH is a basis of MATH. Write MATH. Write an element MATH as MATH with MATH and MATH. Then the left-hand side is the function MATH. If MATH is divisible by MATH, then MATH. Thus, both sides vanish. If MATH only depends on MATH, then MATH whence the right-hand side is MATH. |
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 that MATH vanishes at order REF there. This proves the equivalence of REF in the case where MATH. The general case follows by induction on MATH. Indeed, recall that MATH for any MATH. Thus, using the induction hypothesis for MATH, REF is equivalent to: MATH and MATH vanish at order MATH at infinity. Then MATH (because MATH vanishes at order REF at infinity) and REF is equivalent to: MATH vanishes at order MATH at infinity. The proof of equivalence of REF ' and REF ' is similar. |
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 the lemma below. |
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 facet, a contradiction. So MATH for some MATH. Assume that MATH for some MATH. Then MATH and MATH are disjoint segments with non-empty interiors in MATH. Moreover, because MATH, the closures of both segments contain MATH, a contradiction. |
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 for MATH; this proof adapts with minor changes, as follows. Embed MATH into the vector space MATH of functions on MATH. The additive group of MATH acts on MATH by translations; we denote by MATH this action. For a polyhedral cone MATH which contains a line MATH, we have MATH for all MATH. Thus, for MATH, we have MATH whenever MATH. Because MATH, it follows that there exist MATH (non necessarily distinct) such that MATH for all MATH. Moreover, we can find MATH such that MATH on MATH and that MATH for all MATH. Replacing MATH by MATH, we may assume that MATH for all MATH. Let MATH. We can choose MATH such that MATH for any non-empty subset MATH of MATH and for MATH. We have MATH . By assumption, MATH is identically zero on the open half-space MATH. It follows that MATH. |
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 restriction of MATH to MATH. But MATH as follows from the definition of MATH. |
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 MATH vector spaces of MATH given by NAME. In the local coordinates MATH associated to given stairs (compare CITE, II REF) the subspace MATH is defined by the vanishing of all MATH with MATH and the condition that not all MATH, MATH vanish. CASE: We introduce the following notations: CASE: MATH denotes the universal family, MATH, respectively MATH the blowing-up of MATH (respectively of the trivial section MATH in MATH), and MATH the exceptional divisor of MATH. CASE: MATH (MATH an oriented coloured tree with set of vertices MATH), MATH, denote the cluster graphs obtained from MATH by removing the root MATH (compare REF ). Set MATH . Without restriction, we can assume that the roots of MATH, MATH, are vertices in MATH (corresponding precisely to the infinitely near points MATH of level REF in MATH), while the roots of MATH, MATH, are not in MATH. We introduce the subsets MATH which (clearly) satisfy REF and which correspond to clusters MATH with origin MATH, MATH, given by CASE: those points in MATH which are infinitely near to MATH, MATH, CASE: the intersection points MATH, MATH, of the strict transform of MATH with the exceptional divisor of MATH, the blowup of MATH in MATH (where MATH), MATH. Note that the points MATH are already fixed by MATH, while MATH can be chosen arbitrarily in MATH, such that all the MATH are pairwise distinct. CASE: Let MATH be such that MATH and such that the infinitely near points corresponding to the vertices in MATH are in the prescribed position given by MATH. We show that there exists a cartesian diagram of germs MATH obviously implying the statement of REF . MATH . We consider the strict transform MATH of the universal family, given by the ideal (sheaf) MATH associated to MATH. Here, MATH denotes the total transform of MATH under MATH, and MATH the ideal of the exceptional divisor in MATH. By semicontinuity of the fibre dimension of the finite morphism MATH, it follows that there is a locally closed subset MATH such that for any MATH we have MATH. In particular, the restriction of MATH to the preimage of MATH defines a flat morphism, whence, by the universal property of MATH there exists a morphism MATH . MATH . There is an isomorphism of germs MATH (compare , for example, CITE), and we can consider REF morphism of germs MATH . The preimages under MATH of the (germs at MATH of the) locally closed subsets MATH are (locally) isomorphic to MATH (if MATH), respectively to MATH (if MATH). MATH . Finally, locally at MATH, MATH is the preimage under MATH of MATH which, by the induction hypothesis, is a locally closed subset. |
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, let MATH be the union of those connected components of the strict transform MATH of the universal family which satisfy CASE: MATH, CASE: the infinitely near points of MATH corresponding to the vertices in MATH are in the prescribed position given by MATH, CASE: the infinitely near points of MATH corresponding to the vertices in MATH are on MATH (respectively on its strict transform) for all MATH, MATH. In particular, MATH and the fibres of the restriction of MATH, MATH have constant (vector space) dimension (MATH). Hence the MATH are flat and, by the universal property of MATH, we obtain morphisms MATH . We complete the proof by showing that the composed morphism MATH is dominant with irreducible and equidimensional fibres on the irreducible set MATH. Here, MATH if MATH (MATH being the infinitely near point in MATH corresponding to the root of MATH), and MATH if MATH. Let MATH be any MATH-tuple of pairwise different points MATH, MATH if MATH, (MATH, MATH). Then there is a curve germ MATH, topologically equivalent to MATH, having tangent directions MATH. Moreover, we can choose MATH such that the local branches of MATH and MATH with tangent direction MATH, MATH, coincide. By chosing the subtree MATH corresponding to MATH, we obtain a zero-dimensional scheme MATH with associated cluster graph MATH. By construction, MATH corresponds to a point in the fibre MATH. On the other hand, any point in the image is of this form and MATH . Hence, by the induction hypothesis, the fibres are irreducible and equidimensional. In the same manner, the dimension statement follows from the induction hypothesis, since the dimension of the image of MATH equals the number of free points of level REF in MATH. |
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 construction of CITE follows from the MATH-transversality in REF ). On the other hand, we construct a family MATH of bigger dimension: let MATH, MATH, MATH, MATH be generic curves of degrees MATH, MATH, MATH, respectively. The curve MATH has degree MATH and MATH ordinary cusps as its only singularities, one at each intersection point in MATH. Varying MATH in the spaces of curves of degrees MATH, respectively, we obtain a subfamily MATH in MATH. Note that the equality MATH with slightly deformed curves MATH implies MATH . Indeed, if MATH then they have MATH common cuspidal points belonging to MATH and MATH. Hence, by NAME 's theorem, MATH. The tangent line to MATH at each cusp is tangent to both, MATH and MATH, that means, the intersection number of MATH and MATH is at least MATH, whence MATH. We can conclude that MATH and, due to MATH, that MATH, MATH. Therefore, by REF , MATH . |
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 polynomial and MATH a simple fraction, then MATH . To see this, recall that the function MATH is homogeneous of degree MATH. As MATH, MATH is a sum of homogeneous terms of positive degree. Thus, for homogeneity reasons, MATH. Let MATH be regular and let MATH. As the function MATH is an element of MATH, we obtain by REF , MATH . Thus MATH . |
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 is not empty. We fix MATH. We have MATH and MATH. For MATH, we set MATH, with MATH. We consider the NAME expansion at MATH of the holomorphic function of MATH: MATH . This is of the form MATH where MATH is homogeneous of degree MATH in MATH. Let MATH, then MATH. We see that the function MATH has a NAME expansion of the form MATH where MATH is homogeneous of degree MATH in MATH. Thus MATH . Via the NAME series at MATH, the function MATH can be expressed as an infinite sum of homogeneous elements with finitely many negative degrees. As the iterated residue MATH vanishes on elements of degree not equal to MATH, and MATH is homogeneous of degree MATH, we see that the sum is finite and that MATH is in the space MATH as claimed. |
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 result of the summation MATH exists and is a holomorphic function of MATH. |
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 functions of MATH: MATH . We compute the derivative in MATH of the left hand side. It is equal to MATH where MATH is the delta function of the set of integers. We compute the derivative in MATH of the right hand side. This function of MATH is constant on each interval MATH. The jump at the integer MATH is MATH . It follows that the derivative in MATH of the right hand side is also equal to MATH. Thus MATH where MATH is a constant. We verify that MATH is equal to MATH by using periodicity properties in MATH. It is clear that MATH is a periodic function of MATH as is MATH . It follows that MATH is also a periodic function of MATH. This implies MATH. Consider now, for MATH, MATH . We just saw that MATH . To determine MATH for MATH, we use the differential equation in MATH . Using decreasing induction over MATH, we see that MATH is a MATH-function of MATH, equal to a polynomial function of MATH of degree MATH on each interval MATH and with coefficients rational functions of MATH. For example, we obtain the value of the convergent series MATH . When MATH, we use the differential equation MATH so that, as we have already used, MATH . More generally, MATH is supported on MATH, in particular is identically MATH on MATH. We return to the proof of REF . For a simple fraction MATH, consider the function MATH . We first prove that it is a locally MATH-function, which is constant when MATH varies in an alcove. Let MATH be a basis of MATH. Let MATH. If MATH is the decomposition of MATH on the basis MATH, set MATH. The function MATH is constant when MATH varies in an alcove. Consider the sublattice MATH . We say that MATH is a MATH-basis, if MATH. In general, the quotient MATH is a finite set; let MATH be a set of representatives of this quotient. We can choose MATH in the following standard way. We consider the box MATH . Then we can take MATH . Define MATH . The set MATH is also a set of representatives of MATH. If MATH is a MATH-basis of MATH, this set is reduced to the single element MATH. Remark that the set MATH is constant when MATH varies in an alcove MATH. We denote it by MATH. If MATH is an alcove, and MATH a basis of MATH, we set MATH . Thus an alcove MATH together with a basis MATH produces a particular element MATH of MATH. Consider on the set MATH the locally constant function of MATH defined by MATH when MATH is in the alcove MATH. This defines a locally MATH-function of MATH, still denoted by MATH, defined except on the set MATH of measure MATH. This locally MATH-function of MATH defines a generalized function of MATH which depends holomorphically on MATH. We have the equality of generalized functions of MATH: MATH . If MATH is a MATH-basis of MATH, this follows from the formula in dimension MATH. In general, we consider MATH and the dual lattice MATH. Then MATH. We set MATH . For any set of representatives MATH of MATH, we have: MATH if MATH is not in MATH, while this sum equal MATH if MATH. Thus, MATH . This holds as an equality of generalized functions of MATH. Further, we have by the one-dimensional case: MATH . It follows that MATH is a locally MATH-function of MATH, as is MATH. It remains to determine the value of this function when MATH is in an alcove. For MATH, we have MATH so that the sum MATH is independent of the choice of the system of representatives MATH of MATH. We choose MATH. Then MATH because MATH for all MATH. Every function MATH, homogeneous of degree MATH, is obtained from an element of MATH by the action of a differential operator with polynomial coefficients. This operator is of degree MATH, if multiplication by MATH is given degree MATH, while derivation MATH is given degree MATH. Using REF , we see that REF follows from the fact that the function MATH is constant on each alcove. |
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 set of all MATH such that MATH for all MATH and MATH. The intersection of this set with a small neighborhood of MATH is contained in the complement of the union of the complex hyperplanes MATH, for MATH. The function MATH is analytic in MATH and MATH. Furthermore, when MATH, the function MATH is holomorphic at MATH. We prove this by induction on the codimension of MATH. If MATH, this follows from the explicit formula for MATH. Let MATH be an indivisible element of MATH such that MATH is contained in the real hyperplane MATH . We assume first that MATH is an element of MATH. We number it the first vector MATH of the basis MATH. We set MATH, MATH, etc; then MATH. Our subspace MATH is contained in MATH. Thus, we have MATH . By induction, MATH is analytic in MATH for MATH, except if there exist MATH such that MATH is in a hyperplane generated by MATH (the orthogonal of MATH in MATH) and some vectors of MATH. As MATH, we see that the singular set of MATH is contained in MATH. Furthermore, the function MATH is holomorphic in MATH near MATH. Assume now that MATH is not an element of MATH. We add it to the system MATH if MATH is not an element of MATH. Writing MATH, we obtain one of the NAME relations of the system MATH where MATH. A MATH-wall is a hyperplane of MATH generated by MATH and MATH vectors of MATH; then these vectors are distinct from MATH, because MATH. Thus, all MATH-walls for the basis MATH are also MATH-walls for the basis MATH. By our first calculation, it follows that MATH is analytic when MATH is not on a translate of a MATH-wall. Moreover, we have MATH so that the function MATH is holomorphic in MATH in a neighborhood of MATH. By the induction hypothesis applied to MATH, the function MATH is holomorphic on a non-empty open subset of MATH. So this function, considered as a function of MATH, has no pole along MATH. This proves REF , and hence REF when MATH is a simple fraction. The operator MATH satisfies also the commutation relation of REF . Thus, using differential operators with polynomial coefficients, we obtain the statement of REF when MATH is any element in MATH. |
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 MATH: MATH . We thus need to prove that, for MATH, MATH . But MATH is given by a sum over the full lattice MATH, while MATH is only over the regular elements of MATH. Thus, we can write (in many ways) MATH as a linear combination of operators MATH where each MATH is an admissible subset of MATH contained in the real hyperplane MATH. Now NAME formula will follow from Let MATH be an admissible subset of MATH contained in the real hyperplane MATH. Then, for MATH . It suffices to prove that MATH for some diagonal basis MATH. A total order on MATH provides us with a special diagonal basis MATH of MATH (see for example REF .) We choose this order such that MATH is minimal. In this case every element of MATH is of the form MATH with MATH. We claim that for each MATH, MATH . Indeed, we use the notation of REF and write MATH. Then our set MATH is contained in MATH. Thus MATH . We see that for MATH fixed and regular, MATH where MATH and MATH has poles at most on the complex hyperplanes MATH for MATH. Thus the claim follows from REF . |
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 a finite measure, MATH . Moreover, the fact that MATH is uniformly bounded with respect to MATH yields that MATH is purely absolutely continuous, MATH where MATH . In order to prove REF we first observe that MATH takes on boundary values MATH for a.e. MATH in MATH-norm by REF. Next, choosing an orthonormal system MATH we recall that the quadratic form MATH exists for all MATH where MATH has NAME measure zero. Thus one observes, MATH . Let MATH then MATH (MATH denoting the NAME measure on MATH) and hence MATH . Since MATH is dense in MATH and MATH one infers MATH for a.e. MATH completing the proof. |
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 follows MATH . For MATH the operator MATH is invertible for MATH sufficiently large and therefore, for such MATH one can go to the limit MATH in REF to arrive at MATH . Since for MATH the operator MATH is dissipative, one concludes as in REF that MATH . Similarly one concludes that MATH is reduced by the subspace MATH and MATH . By REF one obtains for MATH . Combining REF and taking into account the fact that MATH one concludes that the subspace MATH reduces MATH and that MATH . Moreover, for MATH . Changing variables MATH using the fact that MATH is invertible for MATH then yields MATH where in obvious notation MATH denotes a bounded operator in MATH whose norm is of order MATH as MATH . Combining REF we get the asymptotic representation MATH . Together with REF this proves REF. |
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 of the spectral measures MATH associated with the NAME representation of MATH in REF is a finite measure on MATH one obtains from MATH for some MATH . Next we will show that MATH . While REF - REF are clear from REF, and REF, we need to prove the asymptotic relation REF. The difference of the first and the third resolvent as well as the difference of the second and fourth resolvent under the trace in REF is clearly of MATH in norm using the resolvent equation and the fact that MATH is a bounded operator. On the other hand, the operator under the trace in REF is the difference of two rank-one operators by REF and hence at most of rank two. Hence the trace norm of the operator under the trace in REF is also of order MATH as MATH . Integrating REF taking into account REF then proves that MATH . Next we abbreviate MATH and compute (compare REF) MATH where MATH by taking into account MATH and MATH which in turn follows from REF, the fact that MATH is a bounded operator, and MATH as MATH . Thus, MATH by REF. This implies MATH by REF and hence proves REF. |
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 MATH . Next, using MATH (compare REF), one obtains MATH since MATH . Relation REF is shown as follows. Since MATH and MATH (being distinct from the NAME MATH-function) belongs to some measure MATH in REF with MATH one concludes from CITE that also MATH in the corresponding exponential NAME representation of MATH (compare REF) and hence MATH . Next, we note MATH by REF which proves REF. To prove REF one estimates MATH using REF. Similarly, MATH by REF. This completes the proof. |
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 that we have to replace REF by REF . |
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 at level MATH). The proof of REF is the same as the proof of assertion REF. (Note that with the notation of CITE MATH is contained in MATH.) REF is a direct consequence of (MATH), taking MATH and using the injectivity of MATH. It remains to prove REF . Because of REF , we may assume that MATH and MATH are affine. Let MATH be a section of the projection MATH such that the restriction of MATH to MATH is a piecewise morphism. The existence of such a section has been shown in the proof of REF. Since MATH is cylindrical at level MATH, MATH is contained in MATH. Let MATH be the mapping MATH . We will prove the following assertion: CASE: The map MATH is a piecewise morphism. Using REF , the proof of REF is the same as in the proof of REF, except that we have to replace the assertion that MATH in CITE is a piecewise morphism by the slightly weaker REF above. It only remains to prove REF . Let MATH be in MATH and MATH in MATH. Using REF we see that MATH if and only if there exists MATH in MATH such that MATH and MATH. Thus, the graph of the map MATH is constructible and REF follows from the next lemma. |
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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