paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0010245 | We have the NAME representation CITE MATH where MATH is a closed contour containing MATH in its interior. We write MATH . The entries of MATH have NAME series absolutely uniformly convergent in MATH. Hence it follows easily that MATH and MATH, as sums and products of the entries of MATH, have NAME series that are absol... |
math/0010245 | We have for any MATH that MATH . Here we have used that MATH, and hence MATH, commutes with all relevant time-frequency shifts. Thus the frame operator corresponding to MATH is given by MATH. By the spectral mapping theorem we have that MATH and this gives the first part of the result. Next we compute the tight frame g... |
math/0010245 | It is sufficient to consider the case MATH. We begin with the observation that for MATH satisfying MATH where MATH and MATH there holds MATH . REF is equivalent to the operator version of the NAME inequality CITE. For the reader's convenience we include a short proof. REF follows from MATH since MATH is a non-negative ... |
math/0010246 | The NAME scheme MATH of points in the projective plane exists as a special case of NAME 's construction in CITE, with a universal family MATH having the analogous universal property. We identify MATH as usual with an open subset of MATH, the complement of the projective line MATH ``at infinity." The projection of MATH ... |
math/0010246 | Let MATH be the universal family over MATH. The fiber MATH of MATH over a point MATH is the disjoint union of closed subschemes MATH and MATH in MATH of lengths MATH and MATH, respectively, with MATH and MATH. Hence over MATH, MATH is the disjoint union of flat families MATH, MATH of degrees MATH and MATH. By the unive... |
math/0010246 | Let MATH be the preimage in MATH of the open set MATH of points MATH where the MATH are all distinct. It follows from REF that MATH restricts to an isomorphism MATH, so MATH is irreducible. We are to show that MATH is dense in MATH. Let MATH be a closed point of MATH, which we want to show belongs to the closure MATH o... |
math/0010246 | It follows from the cell decomposition of CITE that the closed locus MATH in MATH consisting of points MATH with MATH supported on the MATH-axis MATH in MATH is the union of locally closed affine cells of dimension MATH. The subset MATH is equal to MATH and MATH is finite. |
math/0010246 | By symmetry among the points MATH of MATH we see that MATH, where MATH is the locus in MATH defined by the equations MATH. It follows from REF that MATH is isomorphic to an open set in MATH, where MATH is the closed subset MATH in MATH. As a reduced subscheme of MATH, the latter is isomorphic to MATH, where MATH is the... |
math/0010246 | By definition the blowup of MATH at MATH is MATH, where MATH is the NAME algebra. The ring MATH is a homogeneous subring of MATH in an obvious way, and since MATH generates MATH as a MATH-module, we have MATH, that is, MATH for some homogeneous ideal MATH. In geometric terms, using REF and the fact that MATH, this says... |
math/0010246 | The possible ideals MATH for the given MATH are the length MATH ideals in the local ring MATH, or equivalently, the one-dimensional subspaces of its socle. The fiber of MATH is therefore the projective space MATH, and we have MATH. First consider the maximum possible dimension of any fiber of MATH. Since both MATH and ... |
math/0010246 | REF gives us isomorphisms on the preimage of MATH between MATH and MATH, and between MATH and MATH. We can identify MATH with the closed subset of MATH consisting of points where MATH are the same in both factors, and MATH contains MATH. On the preimage of MATH, under the isomorphisms above, this corresponds to the clo... |
math/0010246 | We have the corresponding result for MATH in REF . We can assume by induction that the result for the nested scheme holds for smaller values of MATH for the base case note that MATH. Locally on a neighborhood of any point where MATH are not all equal, the result then follows from REF . The locus where all the MATH are ... |
math/0010246 | Let MATH. Let MATH be the generic locus, that is, the open set of ideals MATH for which MATH is a set of MATH distinct points. The NAME morphism MATH induces an isomorphism of MATH onto its image in MATH, and MATH is the locus where some two of the MATH have the same MATH-coordinate, and another two have the same MATH-... |
math/0010246 | The MATH-form MATH is MATH-invariant and thus defines a MATH-form on the smooth locus in MATH and therefore a rational MATH-form on MATH. If MATH is a point of MATH, then MATH is generated as an ideal in MATH by two polynomials, MATH and MATH where the complex numbers MATH are regular functions of MATH. This is so beca... |
math/0010246 | We have tautological sheaves MATH and MATH pulled back from MATH and MATH. The kernel MATH of the canonical surjection MATH is the line bundle with fiber MATH at the point MATH. From REF we have MATH. On the generic locus, the fiber MATH can be identified with the one-dimensional space of functions on MATH that vanish ... |
math/0010246 | Let MATH be the generic locus, the open set consisting of ideals MATH for which MATH consists of MATH distinct, reduced points. Note that MATH is clearly reduced over MATH, and since MATH is flat over MATH and MATH is dense, MATH is reduced everywhere. (For this argument we do not have to assume that MATH is irreducibl... |
math/0010246 | Since MATH and MATH are locally free sheaves, the sheaf homomorphism MATH in REF can be identified with a linear homomorphism of vector bundles over MATH. Let MATH denote the induced map on the fiber at MATH. The rank MATH of the fiber map is a lower semicontinuous function, that is, the set MATH is open for all MATH. ... |
math/0010246 | The set MATH of points MATH such that MATH is open and MATH-invariant. From the proof of REF we see that the MATH conjecture implies that MATH contains all the monomial ideals MATH. Since every MATH has a monomial ideal in the closure of its orbit, this implies MATH, so MATH is flat, and MATH is NAME. Let MATH, a local... |
math/0010246 | Let MATH. By REF , we have MATH, and REF implies that MATH is a free MATH-module. |
math/0010246 | By definition arithmetically normal means that MATH is a normal domain. Since MATH itself is a normal domain, this is equivalent to the ideals MATH being integrally closed ideals for all MATH. The powers of an ideal generated by a regular sequence are integrally closed, as is an intersection of integrally closed ideals... |
math/0010246 | The question is local on MATH, so without loss of generality we can assume MATH is affine. Then we are to show that MATH for MATH and that MATH is an isomorphism. REF implies that MATH for MATH. Let MATH and MATH. REF imply that MATH and MATH are both at least MATH. Hence the local cohomology modules MATH and MATH vani... |
math/0010246 | We have MATH and MATH. Obviously MATH is a free MATH-module, but there are also the ideals MATH to consider. Necessarily we must have MATH. Then MATH is the union of subspaces in MATH defined by the vanishing of at least MATH of the coordinates MATH. Its ideal MATH is generated by all square-free monomials MATH, where ... |
math/0010246 | The points of MATH are exactly the ideals MATH, where MATH is such that MATH. In particular, every such MATH belongs to MATH, for MATH. Since MATH for all such MATH, we have MATH. |
math/0010246 | Let MATH be a point of MATH, where MATH for some MATH. On MATH we have MATH, while on MATH we have MATH. Hence MATH, so MATH. |
math/0010246 | At a point MATH, MATH coincides locally with MATH for a unique MATH. Identifying MATH with MATH, the ideal of the subspace in REF is locally either MATH, if MATH, or MATH, otherwise. It follows that every MATH coincides locally with an ideal in MATH generated by square-free monomials in the variables MATH, and for such... |
math/0010246 | If MATH then for all MATH we have MATH, just as in the proof of REF . This implies REF . On MATH we have identically MATH, MATH for MATH, so the coordinate ring of MATH is generated by the remaining variables, namely MATH, MATH, MATH, MATH. This implies that the projection on these coordinates is an isomorphism of MATH... |
math/0010246 | Let MATH be the projection on the coordinates MATH. On MATH we have identically MATH, MATH for MATH. As in the proof of the preceding lemma, this implies that MATH induces an isomorphism of MATH onto its image, which in this case is MATH. Note that MATH is also the image under MATH of the whole of MATH. Let MATH be the... |
math/0010246 | We are to show that every MATH can be uniquely expressed as MATH with MATH and MATH. By hypothesis, this is true for the image of MATH in MATH, with MATH, for all MATH. The local ring MATH is a subring of MATH, and the unique coefficients MATH satisfying REF for any MATH also satisfy REF for MATH. Hence they do not dep... |
math/0010246 | REF implies that any free MATH-module basis of MATH is also a basis of MATH, so MATH. Similarly MATH. |
math/0010246 | Let MATH be an element of MATH. Since MATH is free with basis MATH we can write MATH with MATH and MATH. Of course this is also the unique expression for MATH in terms of the basis MATH of the MATH-vector space MATH. Hence MATH for MATH, so we have MATH. |
math/0010246 | Let MATH denote the subspace MATH, and let MATH be its ideal and MATH its coordinate ring. By definition, MATH, where MATH is the intersection of the ideals MATH for all MATH. Since the MATH coordinates are independent on each MATH, the coordinate rings MATH are free, and hence torsion-free, MATH-modules. The ring MATH... |
math/0010246 | The requirement that MATH for all MATH, for a given MATH and some MATH satisfying REF, is equivalent to MATH. |
math/0010246 | We will show that MATH spans MATH as a MATH-vector space. Since the MATH-degree enumerator of MATH is clearly MATH, this implies the result by REF . Let MATH be the subset of MATH containing only the elements defined in REF. By REF we have MATH . It suffices to show that MATH spans MATH. For this it suffices in turn to... |
math/0010246 | First we verify that the specified elements do belong to MATH. Observe (glancing ahead) that for each MATH, MATH, MATH, the ideal MATH displayed on the right-hand side in REF is generated by polynomials which vanish on MATH, so we have MATH. It is mostly routine now to check case-by-case that the relevant elements MATH... |
math/0010246 | In each case the ideal listed on the right-hand side is clearly contained in MATH. Conversely, in the proof of REF we showed that MATH is generated (and even spanned as a MATH-module) by elements belonging to the ideal on the right-hand side. |
math/0010246 | Let MATH be the coordinates in indices MATH, so MATH. Obviously MATH is a free MATH-module with basis MATH the set of all monomials in the variables MATH, and each ideal MATH is spanned by a subset of MATH. By REF , MATH is a free MATH-module with a basis MATH which is a common basis for the ideals MATH. It follows tha... |
math/0010246 | We already have the result on MATH by REF . Let MATH be a point of MATH. By REF , MATH coincides locally at MATH with a subspace arrangement isomorphic to MATH, for a set MATH with two elements. Unraveling the definitions of MATH and the local isomorphism, we find that MATH coincides locally, as a reduced closed subsch... |
math/0010246 | A manifestation of translation invariance is that the endomorphism MATH of MATH defined by the substitutions MATH carries the ideals MATH into themselves, as is clear from their definition in REF. We therefore have MATH, where MATH is the defining ideal of MATH. Since MATH, we have an induced ring homomorphism MATH. We... |
math/0010246 | By REF , we have MATH, where MATH. By REF , we have MATH where MATH. Replacing MATH with MATH, this reduces the problem to showing that MATH in MATH for MATH. For this MATH we have set-theoretically MATH . By REF , the ideal MATH is equal to its radical, and its zero set is MATH. Applying REF with MATH to the union of ... |
math/0010246 | Note that by REF , MATH is reduced and equal to the union over all MATH of MATH. Thus MATH belongs to the lattice of ideals in REF , as does MATH. Therefore MATH is reduced. For MATH, MATH coincides locally with the arrangement MATH from REF , and we may as well replace MATH with MATH. First suppose MATH. Fixing some M... |
math/0010246 | Define an equivalence relation MATH on MATH by the rule MATH if and only if MATH. It induces an equivalence relation MATH on functions MATH by the rule MATH if and only if MATH for all MATH (this differs from the equivalence relation in REF in that it depends on both the MATH and the MATH coordinates). It is easy to se... |
math/0010246 | It is clear that MATH contains the union in REF. Given a point MATH of MATH, let MATH. By the definitions of MATH and MATH, we have MATH for some MATH such that MATH, and MATH for some MATH such that MATH. A priori, MATH and MATH might be different, but REF allows us to take them both to be the maximum MATH for which M... |
math/0010246 | Note that MATH is the union of the subspaces MATH, over all MATH such that MATH. Let MATH be the closed subset MATH . Thus MATH is the union of the components of MATH that are not components of MATH. For any subset MATH, it follows that we have MATH . For MATH we show that MATH. Fix a point MATH and let MATH be the max... |
math/0010246 | Let MATH be a point of MATH. The NAME MATH-algebra MATH is a quotient of the polynomial ring MATH, and therefore has a MATH-basis MATH, where MATH. Equivalently, MATH minimally generates MATH as a MATH-module, so with MATH as in the remarks following REF , we have MATH, MATH. This implies MATH, MATH, that is, MATH, MAT... |
math/0010246 | Let MATH, MATH, and MATH, where MATH. Then MATH, so the defining ideal of MATH is MATH. Since MATH is flat, we have MATH. From the definition it then follows that MATH, which is the algebraic equivalent of REF. |
math/0010246 | Let MATH be the leading coefficient of an element MATH with leading form MATH, as in REF. Similarly let MATH be the leading coefficient of MATH with leading form MATH. Then we have MATH, and the leading form of MATH is MATH where MATH. |
math/0010246 | Consider the filtration MATH of MATH, where MATH. Without loss of generality we can assume that MATH, since the hypotheses and conclusion of the Lemma are unaltered if we replace MATH by a larger integer MATH and define the extra subsets MATH for MATH to be empty. We have MATH and therefore, by REF , MATH for all MATH.... |
math/0010246 | Let MATH be a point of MATH, and let MATH be a point of the fiber. Note that this implies MATH. By REF there is a unique MATH containing MATH and a unique MATH containing MATH. Clearly we must have MATH. By the definition of MATH, we have MATH, and MATH vanishes at MATH and hence at MATH. Note that MATH determines MATH... |
math/0010246 | Let MATH be the projection. If MATH is a point of MATH such that the fiber MATH has at least MATH points then in the notation of REF we have MATH. By REF it follows that MATH. Since MATH is reduced, this implies MATH. |
math/0010246 | By REF we have MATH in MATH. This containment implies equality trivially if MATH or MATH. For MATH we have MATH. For MATH, MATH is empty, so MATH for MATH. Hence we may assume MATH. For MATH, the definition yields that MATH is the kernel of the ring homomorphism MATH corresponding to the projection MATH, that is, the i... |
math/0010246 | First consider a point MATH. By REF , MATH lies on a unique component MATH, for some MATH, and MATH coincides locally with MATH. Since MATH maps MATH into MATH, every point of MATH lying over MATH belongs to a component MATH with MATH and hence MATH. We can replace MATH with the union MATH of only these components, wit... |
math/0010246 | By REF , the coordinate ring MATH of MATH is a torsion-free MATH-module. By REF we have MATH for all MATH and all MATH. Thus the hypotheses of REF are satisfied and we have REF immediately. For REF also to yield us the last conclusion we need to have MATH for MATH and MATH, where MATH is the scheme-theoretic preimage M... |
math/0010246 | REF implies that MATH is a free MATH-module and therefore a subring of MATH. Let MATH be the ideal of MATH as a subscheme of MATH, so MATH. Both MATH and MATH belong to the lattice in REF . Moreover MATH for all MATH, as follows easily from REF . Hence REF implies that MATH is equal to its radical for all MATH, and by ... |
math/0010246 | For each basis element MATH, let MATH be the unique non-negative integer MATH such that MATH belongs to MATH but not to MATH. Similarly, let MATH be the unique MATH such that MATH belongs to MATH but not to MATH. Thus by REF , the set of elements MATH is a free MATH-module basis of MATH and the set MATH is a free MATH-... |
math/0010246 | We can assume we are given a common ideal basis of MATH, represented by a set of homogeneous polynomials MATH not involving the variables MATH, MATH. We will show that MATH is the required basis of MATH. Let MATH. From the definition it is immediate that MATH, where MATH. This implies that MATH is a free MATH-module wi... |
math/0010246 | Choosing an arbitrary representative MATH of each MATH-coset belonging to MATH, we can assume that MATH is given as a subset of MATH. Then MATH is a free MATH-module with basis MATH. This basis need not be a common basis of the ideals MATH; to construct MATH we have to modify it. Given MATH, let MATH be the set of indi... |
math/0010246 | Let MATH. By definition, MATH, so MATH. By REF , MATH. Since MATH does not vanish identically on any MATH, it is a non-zero-divisor in MATH. Hence multiplication by MATH is an isomorphism of MATH onto MATH. We may assume we are given a homogeneous common ideal basis MATH of MATH. Suppose that in a given MATH-degree MAT... |
math/0010246 | We caution the reader immediately that the intersections in REF are not scheme-theoretically reduced in general, so it is not enough to check the result set-theoretically. Instead, we must use our knowledge of the local picture on MATH to write down equations. If MATH is a point of the intersection on either side of RE... |
math/0010246 | To simplify notation let MATH and let MATH be the ideal MATH in MATH. We first prove that MATH is a free MATH-module. The ideal MATH is isomorphic to MATH. Since MATH in MATH, we have MATH, and the latter is a free MATH-module by hypothesis. The locus MATH is isomorphic to the preimage MATH in the special arrangement M... |
math/0010246 | We prove the theorem by induction on MATH and MATH. The base case MATH is given by REF . The base case for MATH is MATH. Note that for MATH we have MATH, for all MATH, and that the only non-trivial MATH is MATH, which already appears in MATH. Thus the case MATH is included in the case MATH. For MATH and MATH we can ass... |
math/0010246 | Set MATH, and let MATH be the polygraph over MATH, a subspace arrangement in MATH. Let MATH be the Cartesian product of MATH copies of the symmetric group MATH, acting on MATH by permuting the factors in MATH in MATH consecutive blocks of length MATH. In other words each MATH fixes the coordinates MATH on MATH, and for... |
math/0010246 | By the preceding remarks, the Theorem holds for MATH. NAME the exact sequence MATH over MATH with MATH, we get an exact sequence MATH with MATH a free MATH-module. Since MATH for all MATH, and MATH by REF, we have MATH. Let MATH. Note that MATH is the affine open set MATH, so inverting MATH is the same thing as localiz... |
math/0010247 | If MATH, this was already pointed out by CITE. Now suppose MATH. Let MATH be a MATH-tree of degree MATH. If we choose any univalent vertex to be the root of MATH, then we can associate to this rooted MATH-tree an element in MATH. For example, if MATH is the MATH-graph in REF , with the counter-clockwise orientation at ... |
math/0010247 | We construct a manifold MATH from MATH by adjoining handles to MATH along the framed link MATH and by removing tubular neighborhoods of properly imbedded disjoint disks MATH in MATH obtained by pushing MATH into the interior of MATH. Then MATH, so it only remains to observe that the pair MATH is acyclic. Let MATH. It i... |
math/0010247 | Recall that for each edge of MATH there are two components of MATH, one at each end of the edge (see REF ). We will call them companion components. We will construct MATH and MATH by assigning, for each edge of MATH, one of the associated companions to MATH and the other to MATH. This choice can be represented by an or... |
math/0010247 | First note that it follows from REF that the composition MATH factors through MATH. This yields the commutative REF , assuming for the moment that MATH. To see that MATH for MATH, and, at the same time, identify the composition of the maps in the bottom line of the diagram, we first consider a MATH-clasper MATH in MATH... |
math/0010247 | REF are proved above. To prove REF first consider MATH. Suppose that MATH and MATH. Then MATH and so MATH. For MATH we proceed by induction, using the following commutative diagram. MATH . The columns are exact and, by induction, the middle row is exact. It then follows that the top row is exact. |
math/0010247 | By REF MATH and so MATH. |
math/0010247 | Let MATH denote the subset (it is not a subgroup) of MATH consisting of all MATH such that MATH is an isomorphism, where MATH is the inclusion and MATH is the projection with MATH. Note that MATH. We will define a ``retraction" MATH, which is not a homomorphism, but will satisfy MATH and so prove the theorem. We identi... |
math/0010247 | CASE: Suppose MATH. Then MATH and, therefore, by NAME 's lemma, if MATH is the meridian disk of MATH corresponding to MATH, MATH bounds a disk MATH . By standard cut and paste techniques we can assume that the MATH are disjoint. We can now extend MATH over each MATH by mapping it onto MATH. Since the complement of MATH... |
math/0010247 | MATH is defined by MATH (see REF ). To see that MATH is one-one we only need note that MATH, which follows from REF and that MATH. To prove that MATH is onto we need: If MATH, then MATH. The idea is given by the schematic pictures below. The first picture shows the complement of MATH. MATH is the complement of the stri... |
math/0010247 | If MATH, then we can lift MATH to an endomorphism MATH of MATH and we can realize MATH by an imbedding MATH. If MATH denotes the imbedded copy of MATH, then define MATH, attaching MATH to MATH. MATH is a homology bordism between MATH and MATH and clearly maps to MATH under MATH. |
math/0010247 | CASE: Write MATH, where MATH represents the MATH disjoint handles. Then the imbedding MATH which defines MATH can be regarded as the restriction of an identification of MATH with the upper hemisphere MATH of MATH - the holes in MATH which are removed to obtain MATH are the disks along which one end of each handle of MA... |
math/0010247 | CASE: Since MATH for all MATH, this follows immediately from the fact that MATH is onto CITE. CASE: Note that MATH iff. MATH for all MATH, where MATH, the normal closure of MATH. To see this first note that MATH iff. MATH for every MATH. Secondly note that MATH iff. MATH for all MATH - but this is equivalent to MATH fo... |
math/0010247 | CASE: If MATH then MATH. We will abuse notation and allow ourselves to denote the induced bases of MATH and MATH by MATH and MATH, respectively. If MATH, then MATH in MATH and so MATH in MATH , since MATH. CASE: Let MATH be a lift of MATH. Then we can write MATH for some MATH and MATH, where the latter inclusion is som... |
math/0010247 | Recall the definition of MATH from REF . It is well-known that MATH, the MATH-th term of the lower central series, if and only if every MATH - see for example, CITE. We therefore have MATH and the induced MATH is injective. That MATH follows from REF . It also follows directly from the definitions that the composite MA... |
math/0010247 | The first assertion is clear from the definitions. The rest of the theorem follows from the observation that the following diagram is commutative MATH and the fact that MATH is onto. |
math/0010247 | We need to show that MATH. First note that MATH - this follows from REF . For the ontoness we can apply the ontoness argument of REF with the extra fact that MATH. This follows from the observation that the following diagram is commutative MATH where MATH is defined by MATH. |
math/0010249 | From the isomorphism given by REF and recalling that the transform MATH preserves the MATH groups of WIT sheaves, it is enough to show that the same claim holds for the morphism REF . By the very definition of the NAME map, the tangent map to MATH may be identified with the map MATH obtained in the following way. Let M... |
math/0010249 | Since MATH is smooth, and MATH is injective and is an immersion, it is also an embedding. Now, let MATH. It is enough to show that the NAME product MATH vanishes when applied to pairs MATH of elements in MATH where MATH and MATH are represented, respectively, by the sequences MATH with MATH. It is enough to remark that... |
math/0010249 | Since MATH is prime, every sheaf in MATH is properly stable. Let MATH and assume that the subsheaf MATH destabilizes MATH. Let MATH. Standard computations show that if MATH is not MATH-stable then MATH . Setting MATH we have MATH, with MATH and MATH. This is impossible whenever MATH is prime. The statement about local ... |
math/0010251 | Let MATH be MATH-stable and assume that MATH is a proper sub MATH-module of dimension vector MATH. Restricting MATH to a representation of MATH we see that MATH and is a subrepresentation of MATH. Because MATH we have MATH, which is impossible as MATH is MATH-stable. Let MATH be a simple MATH representation and assume ... |
math/0010251 | Let MATH be a MATH-stable representation of MATH of dimension vector MATH and consider the MATH-dimensional representation MATH . It is clear from the definition that MATH is the local quiver setting corresponding to MATH. As before, there is a semi-invariant MATH such that MATH and MATH is a semi-simple representation... |
math/0010251 | Let MATH such that MATH and MATH a subrepresentation of dimension vector MATH. Then, MATH for if MATH then MATH and the composed map MATH must have a kernel contradicting the fact that MATH is an isomorphism. If MATH, then MATH and the restriction of MATH to MATH gives an linear isomorphism MATH. Finally, the decomposi... |
math/0010251 | Assume that MATH is the dimension vector of a MATH-stable representation of MATH, then the set of irreducible representations of MATH is an open subset of MATH. Assume that MATH and consider the vertex spaces (the eigenspaces) MATH and MATH as subspaces of the MATH-dimensional representation MATH (via the identificatio... |
math/0010252 | As in the proof of REF , we divide the diagram MATH into rectangular blocks MATH, MATH , and compute the weight within each block MATH. Clearly choosing a pair MATH in MATH is equivalent to, for each MATH, first choose the MATH left-most (respectively, the rest MATH) columns of length MATH (respectively, MATH) for MATH... |
math/0010252 | Recall that MATH. Suppose there exists a pair MATH in MATH, then there should be an integer MATH such that MATH and MATH for any MATH. In view of the definition of MATH we must have MATH and MATH, for MATH is a horizontal strip. It follows that MATH and MATH. Furthermore, if MATH is also in MATH, we must have MATH. Sum... |
math/0010254 | Distinguish one point MATH. Draw the segments MATH. Some of them may be obstructing, but by splitting these into smaller ones, one gets a planar graph connecting all points in MATH and whose edges are non obstructing. The result then follows from the main theorem in CITE and its reformulation in the appendix of CITE. |
math/0010254 | This is a consequence of one of the NAME relations (see CITE or CITE, théorème A. REF) namely that MATH . The proof of this relation is by induction on MATH. The case MATH is checked by a direct computation, and for other MATH we have MATH where the first equality is by induction and the second by the case MATH. |
math/0010254 | The convex hulls of MATH and MATH cannot intersect, thus the generators corresponding to their edges commute pairwise. |
math/0010254 | We prove REF by induction on the cardinality of MATH. With the notations of REF, let MATH, MATH so that MATH, and let MATH, so that MATH. Then by REF applied to MATH we have MATH. By induction hypothesis, we have MATH. As MATH, and as MATH commutes to MATH (since MATH), we get the result by induction. REF : denote by M... |
math/0010254 | REF is easy. REF comes from REF and the following remark: when MATH and MATH are in the same orbit, multiplying by MATH splits this orbit into two orbits, thus increasing by MATH the number of orbits; when MATH and MATH are in different orbits, multiplying by MATH merges their orbits, thus decreasing by MATH the number... |
math/0010254 | CASE: Let MATH. As the converse implication is trivial, we only have to check that MATH . Let MATH such that MATH. Consider the image in MATH of this identity: MATH. By REF , we have MATH and MATH, and, using REF , we have MATH and thus MATH. Consequently, when concatenating reduced decompositions for MATH and MATH, on... |
math/0010254 | By REF can be transformed into MATH by a finite rewriting process MATH in which, at each step, the elementary transformation MATH is CASE: either of the type MATH CASE: or of the type MATH where MATH are such that their product is defined in MATH and equal to MATH. Suppose the length MATH of the rewriting process is mi... |
math/0010254 | The statement follows CITE, and the proof is exactly the same. |
math/0010254 | We can follow the proof of CITE, replacing REF by REF . |
math/0010254 | We apply REF to the set MATH of MATH such that MATH and MATH. To check the assumptions of that lemma, we need that if MATH and if MATH then MATH exists and MATH (then we have MATH). Since MATH and MATH divide MATH, by cancellability and REF MATH and MATH have a common multiple in MATH, so by REF MATH exists and by REF ... |
math/0010254 | The statement is CITE and the proof is the same. |
math/0010254 | By REF the products of both sides with MATH are equal. By REF we will be done if we show that MATH is in MATH. By REF there exists MATH such that MATH and MATH. As MATH and MATH we have MATH for some MATH (by definition of MATH and cancellability in MATH). Hence MATH and MATH. So MATH. The result follows since any divi... |
math/0010254 | Assume that all elements of the family MATH divide MATH. If we can apply REF to the set MATH of elements of MATH which divide all common multiples of the MATH, it will give the result. Let us check the assumption of REF. This set MATH is finite as it is included in the set of divisors of MATH. The first assumption of R... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.