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math/0012168 | We shall use the following notation. MATH and MATH are contiguous intervals, MATH and MATH is the length of MATH . Let a constant MATH be given. One first shows that if MATH and MATH are contiguous with MATH and if MATH is sufficiently near the identity in the quasisymmetric topology, then MATH . Assume MATH is a sequence of elements of MATH converging in the right MATH-topology to MATH . This means that for sufficiently large MATH . Also, assume that for each MATH there is a function MATH approaching zero as MATH approaches MATH such that for all contiguous intervals MATH and MATH with MATH . Taking the product, we obtain MATH . Since there is a uniform bound on the quasisymmetric norm of MATH we may assume MATH and thus we may substitute in MATH and MATH . We obtain MATH . For given MATH we can pick MATH large enough so that MATH whenever MATH and MATH . Then MATH and this implies MATH has vanishing ratio distortion. |
math/0012168 | Because MATH by plugging in MATH and letting MATH approach MATH and MATH one sees that MATH must fix MATH and MATH . By postcomposition of MATH with a real dilation, we may assume MATH and this implies MATH . But any such map taking these values for arbitrarily large negative values of MATH cannot satisfy an inequality of the form MATH unless MATH . |
math/0012168 | We follow the proof given in CITE, page REF. To begin, assume there is a trigonometric polynomial MATH with MATH and MATH . Thus at some point MATH and we can assume that MATH . Since MATH is a maximum at MATH . Consider the trigonometric polynomial MATH of degree MATH . In the interval MATH there are MATH points where MATH takes the values MATH and between any two of these points the polynomial MATH takes values of opposite sign. Hence MATH has MATH different zeros in this interval, and so MATH also has MATH different zeros. One of these zeros is MATH since MATH . Also, MATH vanishes at MATH . Moreover, MATH has MATH zeroes between the zeros of MATH . Thus MATH has at least MATH zeros in this interval, and since it is a trigonometric polynomial of degree MATH it must be identically zero. Thus MATH is constant, but since MATH this implies MATH is constant. But this contradicts the statement that MATH changes sign and we conclude that the original assumption could not be correct, that is, we have MATH which means MATH . |
math/0012168 | We begin by proving that if such trigonometric approximations are possible for every MATH then MATH is in the NAME class. For each integer of the form MATH let MATH be a trigonometric polynomial of degree MATH such that MATH . Now select MATH so that MATH and write MATH in the form MATH where MATH and MATH . In general, define the difference operator MATH by MATH . Then MATH . From the hypothesis, MATH . Putting MATH we may rewrite MATH as a sum over scales: MATH . Each term MATH has norm bounded by MATH and is a trigonometric polynomial of degree less than or equal to MATH . So by REF , the second derivative of MATH is at most MATH . Thus, by the second mean value theorem MATH . By using REF , we obtain MATH . Putting REF together, we obtain MATH and this proves the first half of the theorem. To prove the other half, for every MATH we must construct a trigonometric polynomial MATH of degree MATH that approximates MATH in the sup-norm to within MATH . Let MATH be the NAME kernel defined by MATH where MATH is the NAME kernel and MATH . By convolution of MATH with the NAME kernel MATH one gets a trigonometric polynomial of degree MATH that approximates MATH to within MATH in the sup norm. For details of this proof we refer to CITE, pages REF, or to CITE. |
math/0012168 | Since a much easier proof of the same result is given in REF, here we only outline the argument given by NAME in CITE. Begin by using a result of CITE: if MATH then MATH . Then employ the NAME theorem. Let MATH be in MATH and MATH be a trigonometric polynomial of degree MATH with MATH . Let MATH and MATH . Then MATH and therefore MATH . Thus MATH is approximable in the sup norm to within MATH by a trigonometric polynomial of degree MATH . This implies that MATH is approximable to within MATH by a trigonometric polynomial of degree MATH therefore MATH is in the NAME class. The proof for the class MATH is similar. |
math/0012168 | A similar and much deeper result is true if MATH is replaced by any plane domain, CITE, CITE. Let MATH be any linear functional on the NAME space MATH that annihilates MATH . To show that MATH is dense in MATH it is sufficient to show that MATH annihilates MATH . By the NAME and NAME representation theorems, there exists a bounded measurable function MATH defined in MATH so that MATH . If we extend MATH and MATH to the lower half-plane by the rules MATH and MATH then we can write the formula for MATH as MATH . The assumption that MATH annihilates MATH implies that MATH whenever MATH is a real number. One shows that MATH has the following properties: CASE: MATH in the sense of distributions, CASE: MATH as MATH and CASE: MATH has a MATH-modulus of continuity, that is to say, given MATH there exists a MATH such that for every MATH and MATH with MATH and MATH and with MATH . Let MATH be a semi-disc in the upper half-plane with diameter of length MATH along the line MATH and with midpoint on the MATH-axis. The curved part of the boundary of MATH is parameterized by the curve MATH . Assume further that MATH is continuous on the real axis and MATH as MATH . Then the subspace of MATH comprising those MATH with these properties is dense in MATH . Since MATH is integrable and MATH is bounded, MATH where the limit is taken both as MATH and as MATH . On the other hand, from NAME 's formula, MATH . Because MATH is identically zero when MATH if we first take the limit in this line integral as MATH we obtain MATH . Because of the vanishing condition on MATH is dominated by a constant times MATH and thus vanishes as MATH . |
math/0012168 | From the preceding discussion we have seen how an element MATH of MATH determines by the correspondence an element MATH in MATH . Conversely, because of the residue REF , the extension formula and because MATH is dense in MATH, any element MATH of MATH determines by this correspondence an element of MATH . Note that MATH is equivalent to the condition that MATH . Since by definition elements MATH of MATH satisfy MATH as MATH they also satisfy MATH and MATH . One therefore sees that any quadratic polynomial vector field MATH annihilates all elements of MATH and so also annihilates MATH since MATH is dense in MATH . |
math/0012168 | Given MATH satisfying condition , the NAME REF yields a vector field MATH with the property that MATH is vanishing. Thus implies . It is easy to see that if MATH for MATH outside a sufficiently large compact set and if MATH is degenerating, then MATH and so implies . To see that implies , consider the following sequence of quadratic differentials MATH where MATH and MATH is arbitrary: MATH . Note that MATH which is a positive constant not depending on MATH and, for fixed MATH in the upper half-plane, MATH as MATH . By REF MATH and we know that this quantity approaches zero as MATH no matter which sequence MATH is selected. Thus implies . |
math/0012168 | We first observe that the pairing defined in REF is non-degenerate between MATH and MATH . Since it is a non-degenerate pairing between MATH and MATH and since MATH whenever MATH for all MATH implies MATH . Moreover, for MATH and MATH by the mean value property MATH where MATH . Because the pairing is non-degenerate, the mapping from MATH to MATH given by MATH is well-defined and injective. In order to show it is surjective it suffices to show the unit ball of MATH is compact with respect to the weak topology. To this end, assume MATH is a sequence in MATH with MATH and MATH is a linear functional of the form MATH where MATH for MATH outside sufficiently large compact subsets of MATH . By normal convergence MATH has a subsequence that converges uniformly on compact subsets to some MATH which (in order to avoid cumbersome notation) we denote also by MATH . Note that by NAME dominated convergence MATH and, since MATH . We divide the argument into three cases: either MATH or MATH or MATH . In the first case, MATH is degenerating and MATH . In the second case, since MATH if we put MATH then the denominator is bounded away from zero. Thus MATH is a degenerating sequence and MATH converges to zero, which implies MATH converges to MATH . In the third case MATH and so MATH converges to MATH . Thus, in all cases MATH converges to MATH . |
math/0012168 | Note that MATH and therefore the condition that MATH for all MATH in MATH implies MATH for all MATH in the lower half plane. On taking the third derivative with respect to MATH we find that MATH for all MATH in the lower half-plane. Since finite linear combinations of the form MATH where MATH are points in the lower half-plane, are dense in the space of integrable holomorphic quadratic differentials in the upper half plane (see CITE, CITE), we see that MATH for all MATH in MATH implies MATH for all MATH which implies MATH is in MATH . Conversely, if MATH is orthogonal to every MATH then MATH for every MATH in the lower half plane and, by integrating three times and normalizing so that MATH vanishes at MATH and MATH we find that MATH for all MATH in MATH . To see that MATH is surjective we apply the extension REF to the vector field MATH . That is, for given MATH representing an element in MATH we put MATH where MATH and MATH . Then it is a routine calculation (see CITE) to show that MATH . We leave it to the reader to show that the NAME map MATH is an isomorphism, and in particular that MATH if, and only if, MATH for every MATH in MATH . A detailed proof may be found in CITE, or in CITE, pp. REF. |
math/0012169 | Let MATH be a triangulation of MATH. It is an easy application of NAME 's formulas for REF-ball and REF-sphere that the number of tetrahedra in a triangulation of any REF-ball without interior vertices equals the number of vertices plus interior edges minus three (such formula appears for instance in CITE). Hence our task is to prove that MATH has at most MATH interior edges. For this, we classify the interior edges according to how many vertices of MATH they are incident to. There are only four edges not incident to any vertex of MATH (the edges MATH, MATH). Moreover, MATH contains at most MATH edges incident to two vertices of MATH (that is, diagonals of MATH), since in any family of more than MATH such edges there are pairs which cross each other. Thus, it suffices to prove that MATH contains at most MATH edges incident to just one vertex of MATH, that is, of the form MATH or MATH with MATH. Let MATH be any vertex of MATH. If MATH equals MATH or MATH then the edges MATH and MATH are both in the boundary of MATH; for any other MATH, exactly one of MATH and MATH is on the boundary and the other one is interior. Moreover, we claim that if MATH is an interior edge in a triangulation MATH, then the triangle MATH appears in MATH. This is so because there is a plane containing MATH and having MATH as the unique vertex on one side. At the same time the link of MATH is a cycle going around the edge. Hence, MATH must appear in the link of MATH. It follows from the above claim that the number of interior edges of the form MATH in MATH equals the number of vertices of MATH other than MATH and MATH in the link of MATH. In a similar way, the number of interior edges of the form MATH in MATH equals the number of vertices of MATH other than MATH and MATH in the link of MATH. In other words, if we call MATH and MATH (the MATH, MATH in the index and of the vertices are reversed, because in this way MATH is monotone with respect to MATH, and MATH with respect to MATH), then the number of interior edges in MATH incident to exactly one vertex of MATH equals MATH. Our goal is to bound this number. As an example, REF shows the intersection of MATH with a certain triangulation of MATH (MATH). The link of MATH in this triangulation is the chain of vertices and edges MATH (the star of MATH is marked in thick and grey in the figure). MATH consists of the chains MATH and MATH and the isolated vertex MATH. In turn, the link of MATH is the chain MATH and MATH consists of the isolated vertices MATH and MATH. Observe that MATH has at most three connected components, because it is obtained by removing from MATH (a path) the parts of it incident to MATH and MATH, if any. Each component is monotone in the direction of MATH and the projections of any two components to a line parallel to MATH do not overlap. The sequence of vertices of MATH ordered in the direction of MATH, can have a pair of consecutive vertices contained in MATH only where there is a horizontal edge in MATH or in the at most two discontinuities of MATH. This is true because MATH is a regular MATH-gon. We denote MATH the number of horizontal edges in MATH and MATH this number plus the number of discontinuities in MATH (hence MATH). Every non-horizontal edge of MATH produces a jump of at least two in the MATH-ordering of the vertices of MATH, hence we have MATH . Analogously, and with the obvious similar meaning for MATH and MATH, MATH . Since MATH can be completed to a triangulation of MATH, and exactly four non-interior edges of MATH are horizontal or vertical, we have MATH, that is . MATH. Hence, MATH . Thus, there are at most MATH interior edges in MATH of the form MATH or MATH and at most MATH interior edges in total, as desired. |
math/0012169 | The polytopes constructed are quite similar to MATH constructed earlier except that MATH is non-regular (in REF ) and the segments MATH and MATH are longer and are not orthogonal, thus ending with different polytopes. The polytopes are shown in REF describes a maximal dissection of each of them, in five parallel slices. Observe that both polytopes have four vertices in the plane MATH and another four in the plane MATH. Hence, the first and last slices in REF completely describe the polytope. CASE: The vertices in the planes MATH and MATH form convex quadrangles whose only integer points are the four vertices. This proves that the eight points are in convex position and that the polytope MATH contains no integer point other than its vertices. Let us now prove the assertions on maximal dissections and triangulations of MATH: CASE: Consider the paths of length three MATH and MATH, which are monotone respectively in the directions orthogonal to MATH and MATH. Using them, we can construct two triangulations of size five of the polytopes MATH and MATH, respectively. But they do not fill MATH completely. There is space left for the tetrahedra MATH and MATH. This gives a dissection of MATH with twelve tetrahedra. All the tetrahedra are unimodular, so no bigger dissection is possible. REF A triangulation of size REF can be obtained using the same idea as above, but with paths MATH and MATH of lengths three and two respectively, which can be taken from the same triangulation of the square MATH. To prove that no triangulation has bigger size, it suffices to show that MATH does not have any unimodular triangulation. This means all tetrahedra have volume MATH. We start by recalling a well-known fact (see REF). A lattice tetrahedron has volume MATH if and only if each of its vertices MATH lies in a consecutive lattice plane parallel to the supporting plane of the opposite facet to MATH. Two parallel planes are said to be consecutive if their equations are MATH and MATH. Suppose that MATH is a unimodular triangulation of MATH. We will first prove that the triangle MATH is in MATH. The triangular facet MATH of MATH, lying in the hyperplane MATH, has to be joined to a vertex in the plane MATH. The two possibilities are MATH and MATH. With the same argument, if the tetrahedron MATH is in MATH, its facet MATH, which lies in the hyperplane MATH, will be joined to a vertex in MATH, and the only one is MATH. This finishes the proof that MATH is a triangle in MATH. Now, MATH is in the plane MATH and must be joined to a vertex in MATH, that is, to MATH. Hence MATH is in MATH and, in particular, MATH uses the edge MATH. MATH is symmetric under the rotation of order two on the axis MATH. Applying this symmetry to the previous arguments we conclude that MATH uses the edge MATH too. But this is impossible since the edges MATH and MATH cross each other. CASE: This polytope almost fits the description of MATH, except for the fact that the edges MATH intersect the boundary and not the interior of the planar quadrangle MATH. With the general techniques we have described, it is easy to construct halving dissections of this polytope with sizes from REF to REF. Combinatorially, the polytope is a REF-antiprism. Hence, REF shows that its minimal triangulation has REF tetrahedra. The rest of the assertions in the statement were proved using the integer programming approach proposed in CITE, which we describe in REF . We have also verified them by enumerating all triangulations CITE. It is interesting to observe that if we perturb the coordinates a little so that the planar quadrilateral MATH becomes a tetrahedron with the right orientation and without changing the face lattice of the polytope, then the following becomes a triangulation with ten tetrahedra: MATH. |
math/0012169 | Let MATH be a mismatched region and MATH the plane containing it. Since a mismatched region is a union of overlapping triangles, it is a polygon in MATH with a connected interior. If two triangles forming the mismatched region have interior points in common, they should be facets of tetrahedra in different sides of MATH. Otherwise, the two tetrahedra would have interior points in common, contradicting the definition of dissection. Triangles which are facets of tetrahedra in one side of MATH cover MATH. Triangles coming from the other side of MATH also cover MATH. Now, take triangles coming from one side. As mentioned above, they have no interior points in common. Their vertices are among the vertices of the tetrahedra in the dissection, thus among the vertices of the polytope MATH. Hence, the vertices of the triangles are in convex position, thus the triangles are forming a triangulation of a convex polygon in MATH whose vertices are among the vertices of MATH. For the second claim, suppose there were distinct mismatched regions having an interior point in common. Then their intersection should be an interior segment for each. Let MATH be one of the mismatched regions. It is triangulated in two different ways each coming from the tetrahedra in one side of the hyperplane. The triangles in either triangulation cannot intersect improperly with the interior segment. Thus the two triangulations of MATH have an interior diagonal edge in common. This means the triangles in MATH consists of more than one connected components of the auxiliary graph, contradicting the definition of mismatched region. |
math/0012169 | MATH . Do an inflation of each mismatched region. This produces as many holes as mismatched regions, say MATH of them. Each hole is bounded by two triangulations of a polygon. This is guaranteed by the previous lemma. Denote by MATH the number of vertices of the polygon associated to the MATH-th mismatched region. In each of the holes introduce an auxiliary interior point. The point can be used to triangulate the interior of the holes by filling in the holes with the coning of the vertex with the triangles it sees. We now have a triangulated ball. Denote by MATH the size of the original dissection. The triangulated ball has then MATH tetrahedra in total. The number of interior edges of this triangulation is the number of interior edges in the dissection, denoted by MATH, plus the new additions, for each hole of length MATH we added MATH interior edges. In a triangulation MATH of a MATH-ball with MATH boundary vertices and MATH interior vertices, the number of tetrahedra MATH is related to the number of interior edges MATH of MATH by the formula: MATH. The proof is a simple application of NAME 's formula for triangulated REF-spheres and REF-balls and we omit the easy details. Thus, we have the following equation: MATH . This can be rewritten as MATH. Taking into account that MATH (because diagonals in a polygon are interior edges of the dissection), we get an inequality MATH . Finally note that in a mismatching dissection we have MATH and MATH. This gives the desired lower bound. MATH . Now we look at the proof of the upper bound on dissections. Given a MATH-dissection, we add tetrahedra of volume zero to complete to a triangulation with flat simplices that has the same number of vertices. One can also think we are filling in the holes created by an inflation with (deformed) tetrahedra. The lemma states that mismatched regions were of the shape of convex polygons. The MATH-simplices forming a mismatched region were divided into two groups (those becoming apart by an inflation). The two groups formed different triangulations of a convex polygon, and they had no interior edges in common. In this situation, we can make a sequence of flips (see CITE) between the two triangulations with the property that any edge once disappeared does not appear again (see REF ). We add one abstract, volume zero tetrahedron for each flip, and obtain an abstract triangulation of a MATH-ball. The triangulation with flat simplices we created is a triangulated MATH-ball with MATH vertices. By adding a new point in a fourth dimension, and coning from the boundary MATH-simplices to the point, we obtain a triangulated MATH-sphere containing the original MATH-ball in its boundary. From the upper bound theorem for spheres (for an introduction to this topic see CITE) its size is bounded from above by the number of facets of a cyclic MATH-polytope minus MATH, the number of MATH-simplices in the boundary of MATH. REF-dimensional cyclic polytope with MATH vertices is well-known to have MATH facets (see CITE), which completes the proof after a trivial algebraic calculation. |
math/0012169 | The key idea is as follows: suppose we have a REF-dimensional convex polytope MATH and two triangulations MATH and MATH of it with the following properties: removing from MATH the tetrahedra that MATH and MATH have in common, the rest is a non-convex polyhedron MATH such that the triangulations MATH and MATH of it obtained from MATH and MATH do not have any interior REF-simplex in common (actually, something weaker would suffice: that their common interior triangles, if any, do not divide the interior of the polytope). In these conditions, we can construct the dissection we want as a bipyramid over MATH, coning MATH to one of the apices and MATH to the other one. The bipyramid over the non-convex polyhedron MATH will be a mismatched region of the dissection. For a concrete example, start with NAME 's polyhedron whose vertices are labeled MATH in the lower face and MATH in the top face. This is a non-convex polyhedron made, for example, by twisting the three vertices on the top of a triangular prism. Add two antipodal points MATH and MATH close to the ``top" triangular facets (those not breaking the quadrilaterals see REF ). For example, take as coordinates for the points MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH . Let MATH be this non-convex polyhedron and let MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH. MATH cones vertex REF to the rest of the boundary of MATH, and MATH vertex REF. Any common interior triangle of MATH and MATH would use the edge REF. But the link of REF in MATH contains only the points REF, and the link in MATH contains only REF. Let MATH be the convex hull of the eight points, and let MATH and MATH be obtained from MATH and MATH by adding the three tetrahedra MATH, MATH and MATH. |
math/0012169 | In what follows we use the word cap to refer to the MATH-gon facets appearing in a prism or antiprism. We begin our discussion proving that any triangulation of the prism or antiprism has at least the size we state, and then we will construct triangulations with exactly that size. We first prove that every triangulation of the MATH-prism requires at least MATH tetrahedra. We call a tetrahedron of the MATH-prism mixed if it has two vertices on the top cap and two vertices on the bottom cap of the prism, otherwise we say that the tetrahedron is top-supported when it has three vertices on the top (respectively bottom-supported). For example, REF shows a triangulation of the regular REF-prism, in three slices. REF represent, respectively, the bottom and top caps. REF is the intersection of the prism with the parallel plane at equal distance to both caps. In this intermediate slice, bottom or top supported tetrahedra appear as triangles, while mixed tetrahedra appear as quadrilaterals. Because all triangulations of a MATH-gon have MATH triangles there are always exactly MATH tetrahedra that are bottom or top supported. In the rest, we show there are at least MATH mixed tetrahedra. Each mixed tetrahedra marks an edge of the top, namely the edge it uses from the top cap. Of course, several mixed tetrahedra could mark the same top edge. Group together top-supported tetrahedra that have the same bottom vertex. This grouping breaks the triangulated top MATH-gon into polygonal regions. Note that every edge between two of these regions must be marked. For example, in REF the top cap is divided into REF regions by REF marked edges (the thick edges in the Figure). Let MATH equal the number of regions under the equivalence relation we set. There are MATH interior edges separating the MATH regions, and all of them are marked. Some boundary edges of the top cap may be marked too (none of them is marked in the example of REF ). We can estimate the marked edges in another way: There are MATH edges on the boundary of the top, which appear partitioned among some of the regions (it could be the case some region does not contain any boundary edge of the MATH-gon). We claim that no more than two boundary edges per region will be unmarked MATH. This follows because a boundary edge is not marked only when the top supported tetrahedron that contains it has the point in the bottom cap that is directly under one of the vertices of the edge. In a region, at most two boundary edges can satisfy this. Hence we get at least MATH marked edges on the boundary of the top and at least MATH marked edges in total. Thus the number of mixed tetrahedra is at least the maximum of MATH and MATH. In conclusion, we get that, indeed, the number of mixed tetrahedra is bounded below by MATH. Note that we only use the combinatorics and convexity of the prism in our arguments. We will show that minimal triangulations achieve this lower bound, but then, observe that if MATH is even, in a minimal triangulation we must have MATH and no boundary edge can be marked, as is the case in REF . If MATH is odd, then we must have MATH and at most one boundary edge can be marked. The proof that any triangulation of a MATH-antiprism includes at least MATH tetrahedra is similar. There are MATH top-supported and bottom-supported tetrahedra in any triangulation and there are MATH marked edges between the regions in the top. The only difference is that, instead of REF, one has at most one unmarked boundary edge per region. Thus there are at least MATH marked edges in the boundary of the top, and in total at least MATH marked edges in the top. Hence there exist at least MATH tetrahedra in any triangulation. For a MATH-antiprism we can easily create a triangulation of size MATH by choosing any triangulation of the bottom MATH-gon and then coning a chosen vertex MATH of the top MATH-gon to the MATH triangles in that triangulation and to the MATH triangular facets of the MATH-antiprism which do not contain MATH. This construction is exhibited in REF show the bottom and top caps triangulated (each with its MATH marked edges) and REF an intermediate slice with REF mixed tetrahedra appearing as quadrilaterals. For a MATH-prism, let MATH and MATH, MATH denote the top and bottom vertices respectively, so that the vertices of each cap are labeled consecutively and MATH is always an edge of the prism. If MATH is even we can chop off the vertices MATH for odd MATH and MATH for even MATH, so that the prism is decomposed into MATH tetrahedra and a MATH-antiprism. The antiprism can be triangulated into MATH tetrahedra, which gives a triangulation of the prism into MATH tetrahedra, as desired. Actually, this is how the triangulation of REF can be obtained from that of REF . If MATH is odd we do the same, except that we chop off only the vertices MATH and MATH (no vertex is chopped in the edge MATH). This produces MATH tetrahedra and a MATH-antiprism. We triangulate the antiprism into MATH tetrahedra and this gives a triangulation of the MATH-prism into MATH tetrahedra. |
math/0012169 | Let the vertices of the prism be labeled MATH and MATH so that the MATH's and the MATH's form the two caps, vertices in each cap are labeled consecutively and MATH is always a side edge. For the upper bound in REF , we have to prove that a triangulation of MATH has at most MATH interior diagonals. The possible diagonals are the edges MATH where MATH is not in MATH modulo MATH. This gives exactly twice the number we want. But for any MATH and MATH the diagonals MATH and MATH intersect, so only one of them can appear in each triangulation. We now prove that the upper bound is achieved if MATH is in the conditions of REF . In fact, the condition on MATH that we will need is that for any MATH, the point MATH sees the triangle MATH from the same side as MATH and MATH (that is, ``from above" if we call top cap the one containing the MATH's). With this we can construct a triangulation with MATH tetrahedra, as follows: First cone the vertex MATH to any triangulation of the bottom cap (this gives MATH tetrahedra). The MATH upper boundary facets of this cone are visible from MATH, and we cone them to it (again MATH tetrahedra). The new MATH upper facets are visible from MATH and we cone them to it (MATH tetrahedra more). Now, one of the upper facets of the triangulation is MATH, part of the upper cap, but the other MATH are visible from MATH, so we cone them and introduce MATH tetrahedra. Continuing the process, we will introduce MATH, MATH tetrahedra when coning the vertices MATH, which gives a total of MATH tetrahedra, as desired. The triangulation we have constructed is the placing triangulation CITE associated to any ordering of the vertices finishing with MATH. A different description of the same triangulation is that it cones the bottom cap to MATH, the top cap to MATH, and its mixed tetrahedra are all the possible MATH for MATH. This gives MATH mixed tetrahedra, and MATH tetrahedra in total. We finally prove the lower bound stated in REF . Without loss of generality, we can assume that our prism has its two caps parallel (if not, do a projective transformation keeping the side edges parallel). Then, MATH can be divided into two prisms in the conditions of REF of sizes MATH and MATH with MATH: take any two side edges of MATH which posses parallel supporting planes and cut MATH along the plane containing both edges. By REF , we can triangulate the two subprisms with MATH and MATH tetrahedra respectively, taking care that the two triangulations use the same diagonal in the dividing plane. This gives a triangulation of MATH with MATH tetrahedra. This expression achieves its minimum when MATH and MATH are as similar as possible, that is, MATH and MATH. Plugging these values in the expression gives a triangulation of size MATH. |
math/0012173 | In fact, from REF it follows that there is MATH such that MATH. Then, MATH is a monomorphism. Since by REF it is also an strict epimorphism, it follows that it is an isomorphism, and consequently so is MATH. |
math/0012173 | REF follows immediately from REF. Let now MATH, and assume MATH is any arrow such that MATH. Since MATH is an strict epimorphism see REF, to prove REF it will be enough to show that given any two arrows MATH , the implication ``MATH " holds. By REF we can assume MATH and MATH invertible. Let MATH, then MATH, thus also MATH. Thus MATH. |
math/0012173 | By REF the functor MATH lifts into the category of transitive MATH-sets, for REF . REF then essentially means that this lifting is full and faithful. Since every transitive MATH-set is a quotient of the MATH-set MATH, it follows by the comparison lemma CITE, Expose III, REF that the topos of sheaves for the canonical topology on MATH is equivalent to the topos of MATH-sets. It is immediate to check see REF that the data in REF is a connected atomic site with a representable point, and any connected atomic topos with a representable point can be presented in this way (see CITE). This finishes the proof. |
math/0012173 | It follows immediately from REF. The coverings above force a relation to be, in turn, univalued, injective, everywhere defined, and surjective. |
math/0012173 | It follows immediately from REF. |
math/0012173 | For the first assertion it suffices to show that this map sends covers into covers on the respective sites of definition. But this is clear. The second assertion follows by the diagram in REF. |
math/0012173 | REF is clear, REF follows easily by induction. REF follows for each MATH by induction on MATH; on MATH by associativity, and on limit ordinals it is straightforward. |
math/0012173 | It is clearly a covering system by REF. It remains to see it is closed under composition. Consider MATH, with MATH, and MATH. Take an ordinal MATH such that MATH and use REF. |
math/0012173 | By definition of prorepresentable functor, MATH is a cofiltered category. Since all maps MATH in MATH are epimorphisms, for any object MATH the transition morphisms corresponding to an arrow MATH in MATH, MATH are all injective functions. By construction of filtered colimits in MATH it follows that the canonical maps of the colimit MATH are injective natural transformations (thus monomorphisms in the category MATH). This implies that MATH is a poset. |
math/0012173 | Let MATH be such that MATH is an isomorphism. We shall see that MATH is a monomorphism (and thus, by REF, an isomorphism). Let MATH be such that MATH. Clearly it follows that MATH. Take any MATH use REF and let MATH. In this way MATH and MATH define arrows MATH in MATH. It follows from REF that we must have MATH. Observe that within this argument we have also shown that MATH is faithful. |
math/0012173 | Let MATH. Then, there are arrows MATH and MATH such that MATH, MATH. It follows from REF that MATH, thus MATH, contrary with the assumption. In the second case we do in the same way. |
math/0012173 | We prove first that under the assumption in the proposition, for arbitrary MATH, the following implication holds: MATH . Take MATH and MATH, MATH such that MATH, MATH (recall MATH is cofiltered). It follows that MATH, thus by REF MATH. By REF take MATH such that MATH. Let MATH. Clearly, MATH and MATH. We have also MATH . By REF. Thus, we have MATH . It follows then from REF that we must have MATH. This finishes the proof of REF. It follows from REF that MATH. Since F is faithful REF this implies MATH. The proof finishes by definition of strict epimorphism REF . |
math/0012173 | By induction on the generation of covers (see REF. MATH) Let REF be the object in the two basic empty covers: MATH or MATH, By REF a pullback of the form MATH can not be, since it implies that MATH would have non empty content. Consider now MATH, MATH, MATH, the basic cover MATH, and the pullback (see REF: MATH . Let MATH, and take MATH, MATH in MATH. Since the function MATH is surjective REF , we can take MATH such that MATH, and let MATH be the image of MATH in MATH, MATH. We have then MATH. This shows that MATH (corresponding to the index MATH in the cover) has non empty content. The same argument applies to the remaining basic covers MATH. MATH) Consider now the cover MATH, with MATH and MATH. Take MATH such that MATH has non empty content, and for this MATH take MATH such that MATH has non empty content. limit ordinal MATH . In this case the proof is even more immediate. |
math/0012173 | Clear, since for the empty cover can not exist any index. |
math/0012173 | Consider first that by the corollary above a cover of MATH can not be empty. Let now MATH be one of the other basic covers, and consider the MATH cover determined by following pull-back: MATH . Take MATH, MATH in MATH. Consider all the MATH such that MATH, and let MATH be the images of these MATH in MATH, MATH. This defines, for each such MATH . We start now the induction on the generation of covers see REF. We deal simultaneously with the case MATH and the case MATH. Consider the MATH cover determined by a MATH cover as above, and for each MATH, a MATH cover MATH (the case MATH is included considering all these covers to be the identity). For each MATH (and MATH, MATH) as above, consider the following diagram, defined as a pullback in MATH . By assumption, for each MATH, there is MATH. Composing we have MATH. Finally, by REF and the inductive hypothesis we have MATH . Since all these arrows (one for each MATH) send MATH, by REF they all correspond to a same single arrow MATH in MATH. In conclusion, we have two arrows MATH, MATH such that for MATH, if MATH, then MATH. It follows by REF that there exist MATH such that MATH. Since the composite MATH is the arrow MATH, it is also the case that MATH. Thus we have MATH. The same argument applies to the other remaining basic covers MATH. The case of a limit ordinal is evident. This finishes the proof . |
math/0012173 | Consider the following chain of equivalences (or bijections) justified, in turn, by definition of MATH, (NAME and) construction of MATH, and REF respectively: MATH . |
math/0012173 | Notice that the NAME group MATH for the action of MATH on MATH is given by MATH (see REF). Thus, clearly, this statement is the particular case of REF, when MATH and MATH. |
math/0012173 | We shall see that MATH. Take any MATH and let MATH. Then: MATH . Taking the infimum against MATH, MATH . The terms in the supremum are equal to MATH except if MATH. This finishes the proof. |
math/0012173 | Assume MATH and MATH. The multiplication MATH of MATH is given by MATH . Since MATH is a morphism of groups as well as of locales, MATH . It follows that MATH. Thus MATH. The second assertion is obvious. |
math/0012173 | CASE: Let MATH, and let MATH be any element in MATH. Since MATH it follows that MATH. Thus it must be MATH REF Take the connected component of MATH in the product MATH and the two projections (the action in the product is given by MATH). |
math/0012173 | Let MATH be such that MATH (see REF. We shall see first that MATH taken in MATH. Let MATH, MATH be such that MATH. Take MATH, MATH, MATH and MATH as in REF). Then MATH, and thus MATH. By assumption it follows MATH, thus MATH. From this, since MATH is surjective REF , it follows there exists a function MATH such that MATH. It remains to see that MATH is a morphism of MATH-sets. We do this now. Let MATH be any two points in Y. Take MATH, MATH. We have MATH, which equals MATH unless MATH. With this: MATH . |
math/0012173 | Take MATH, MATH, MATH and MATH, MATH, REF . To prove the statement it is enough to show that REF . First we prove the following implication: MATH . Assume MATH. Then, MATH, and, MATH. Thus MATH. It follows MATH. With this, now we prove MATH. Let MATH be in MATH, that is, MATH. Take MATH such that REF . Then, for any MATH, MATH the last inequality justified by REF. It follows that MATH, thus MATH. This finishes the proof. |
math/0012173 | Let MATH be the underline set functor MATH. Then the map MATH given by MATH creates (thus also reflects) isomorphisms (compare with REF). |
math/0012173 | The first assertion is given by REF. The second follows from REF. |
math/0012173 | REF just says that MATH lands into MATH. We shall prove: CASE: The functor MATH which is faithful since MATH is, see REF is also full. CASE: Given any transitive MATH-set MATH, there exists MATH and an strict epimorphism MATH in MATH. proof of REF. This is just the meaning of REF . Given a morphism of MATH-sets MATH, choose any MATH, and let MATH. By REF in MATH. proof of REF. Choose any MATH. Consider MATH. Clearly MATH. Then, by REF, and the construction of REF , since MATH is cofiltered, it follows that there is MATH and MATH, such that MATH. The proof finishes then by REF. The theorem follows from REF by the comparison lemma CITE, Expose III, REF. |
math/0012173 | From REF it follows that the easy direction on this equivalence is given by REF, and the hard direction by REF. |
math/0012176 | We will prove REF is proved in the same way. We have to show that MATH satisfies the identities VREF and VREF, since VREF and VREF hold by assumption. These identities are linear combinations of ``vertex monomials" of the form MATH (with some order of parentheses) where MATH's are either equal to MATH or are formal variables. We have to show that for any specification MATH the identity vanishes. Note that these identities are multilinear. Let MATH be the graded closure of MATH, that is, the minimal graded subspace of MATH containing MATH. Since all homogeneous components of the vertex operators from MATH are pairwise local, the space MATH satisfies all the assumptions of the theorem, so we can assume that MATH is graded. Consider all the vertex operators MATH such that MATH for some MATH. Every two such vertex operators are local, so since the theorem is known to be true in the case of ordinary vertex operators, these MATH's generate a vertex algebra MATH. Let MATH be an identity which we have to check. Since MATH is multilinear, it is enough to check it for homogeneous MATH, where MATH. Sometimes we must set MATH and MATH. When we substitute these expressions for MATH into MATH and apply REF, we get a linear combination of vertex monomials of the form MATH where MATH is a polynomial in MATH is a permutation of MATH and the products MATH's are applied according to some order of parentheses. We can cancel the common factor MATH. Now we observe that the only remaining factor in these monomials that depends on MATH's is MATH, therefore, for fixed MATH the map MATH is a polynomial map from MATH to MATH. But when all MATH this map is equal to REF, because then MATH generate a vertex algebra which satisfies the identity MATH, therefore, since MATH, MATH must be identically REF. |
math/0012176 | First of all we note that MATH is a derivation of all products: MATH because so are both MATH and MATH. Therefore if MATH then MATH and the first statement follows. For the second statement, it is enough to assume that all the generators from MATH are homogeneous. A pair MATH can be written as MATH for some MATH and MATH. Now the statement follows from REF . |
math/0012176 | We will show that the homomorphism MATH constructed above has the same universality property as MATH. Then the fact that MATH is an isomorphism follows from uniqueness of MATH. Let MATH be a homogeneous homomorphism of MATH into a space of generalized series with coefficients in some NAME algebra MATH. This induces a homomorphism MATH defined by MATH for any MATH and MATH. It is easy to see that MATH for every MATH and MATH. Hence there is a homomorphism MATH such that all the corresponding diagrams commute. |
math/0012176 | CASE: Let MATH. Then MATH and MATH commutes with all the rest of the factors in REF. Also, MATH commutes with all the factors in REF except MATH, whose commutators are given by REF, so REF follows. CASE: It follows that MATH for every MATH. Hence we have for MATH . CASE: Using that MATH, we get MATH . On the other hand, set MATH. The locality of MATH and MATH is REF, hence, using REF, MATH . CASE: Let us calculate MATH by the associativity REF . We have, using REF: MATH . The other relation is proved in the same way. CASE: Let us first calculate MATH. Since MATH we have MATH . Denote MATH and let MATH. Then MATH . Using the elementary fraction decomposition MATH we get that MATH so that MATH . The rest of non-trivial commutation relations between the factors in REF are MATH so we finally get, using that MATH, MATH . |
math/0012176 | Fix a root MATH satisfying REF, and let MATH. Let MATH be the representation map. Then MATH is a twisted MATH-module if and only if MATH. It is enough to require that MATH where MATH is (the image under MATH of) the eigenvector of MATH given by REF. Moreover, it is enough to require this only for a finite set of generators of MATH, for example for all the MATH's corresponding to the MATH-orbits of an integer basis of MATH. Set MATH so that (see REF) MATH . Set also MATH and then MATH . It follows that the field MATH is homogeneous if and only if MATH is homogeneous and the values MATH are the same modulo MATH for all MATH. Assume that MATH is homogeneous. Since MATH, we have MATH for some MATH. Let MATH be the automorphism of MATH such that MATH. It is easy to see that MATH. Take some MATH. Denote by MATH the coefficient of MATH in MATH. It follows that MATH. If MATH, then the coefficient of MATH in MATH is equal to REF. This gives us the following system of linear equations: MATH . Since MATH=REF, the solution of this system is MATH. Taking MATH instead of MATH we get exactly REF . REF follows from the fact that in order to have MATH we must have MATH for all MATH. |
math/0012176 | Set MATH. If MATH, then MATH. Yet by REF , we have MATH. Therefore, MATH acts by REF on every twisted MATH-module, hence by REF , so does MATH. |
math/0012176 | Let MATH be the decomposition of MATH into a disjoint union of MATH-orbits. Choose some MATH for every MATH. Let MATH. Then the set MATH is a set of generators of MATH. We note that MATH for MATH, where MATH is the commutator map given by REF. Let MATH. It is easy to see that MATH if and only if MATH. We can assume that MATH for some MATH and MATH. Clearly, MATH's span MATH over MATH. Assume also that our choice of MATH yields the minimal possible value of MATH. We claim that in this case MATH is a MATH-basis of MATH. Indeed, otherwise there is an invertible matrix MATH such that the MATH-th column of MATH is REF. Let MATH be the MATH-th column of MATH for MATH, and let MATH be the closure of MATH with respect to the action of MATH. Then MATH is a generating set of MATH, closed under MATH and, since MATH the number of MATH-orbits in MATH with non-zero degree is at most MATH. Suppose there is a linear relation between elements of MATH of the form MATH . In MATH this relation becomes MATH, where MATH is a product of some values of the cocycle MATH. In MATH this relation becomes MATH for some other constant MATH. For MATH we have MATH, hence in MATH we get the relation MATH. By the change of variables MATH we can make MATH. A look at REF for the commutator map MATH shows that if MATH for some MATH, then MATH for any other MATH. Summing up, we get that MATH is a homomorphic image of the algebra MATH where MATH for MATH and MATH if either MATH or MATH. The MATH-grading on MATH is defined by MATH for MATH and MATH for MATH where MATH is a MATH-basis of MATH. It is enough to show that MATH is MATH-graded semisimple. Let MATH, be an arbitrary vector in MATH. Denote by MATH the homogeneous component of MATH of degree MATH. Note that the element MATH is invertible. For a graded subspace MATH denote by MATH the homogeneous component of MATH of degree MATH. The component MATH is the homomorphic image of the group algebra of finite group MATH . By NAME 's theorem (see for example, CITE), MATH and hence MATH is a finite dimensional semisimple algebra. We claim that if MATH are two graded left ideals such that MATH, then MATH. Indeed, if MATH, then MATH, hence MATH. It follows that MATH is graded (left) NAME. Let MATH be a graded left ideal. Then MATH is graded maximal if and only if MATH is a maximal ideal of MATH. Therefore, since MATH, we must have MATH, and that finishes the proof. |
math/0012176 | Let MATH. By REF we have MATH, where MATH is MATH-graded module over MATH. Hence it follows from REF that MATH is decomposed into a direct sum of graded irreducible MATH-modules, and REF implies that MATH is decomposed into a direct sum of irreducible MATH-modules. A MATH-module MATH is simple if and only if MATH is a simple MATH-graded MATH-module. Such MATH must be isomorphic to a simple homogeneous ideal of MATH up to a shift of weights. The weights of MATH are restricted by REF . We claim that a simple object of MATH is determined up to an isomorphism by a choice of roots MATH, given by REF, for each generating MATH-orbit MATH of MATH, a choice of simple homogeneous ideal MATH of MATH and an equivalence class MATH. Indeed, assume all these choices are made. Since the extension MATH is fixed, the choice of MATH's determines the MATH-graded semisimple algebra MATH. The set MATH is an equivalence class in MATH. Let MATH be such that MATH for MATH. By REF we have that MATH, so now we further specify MATH. It follows that there are at most finitely many isomorphism classes of simple objects in MATH. |
math/0012178 | First we claim that for any integer MATH we have MATH where the equality holds if and only if MATH or MATH. If MATH or MATH we see immediately that the equality holds. If MATH then we have MATH . This proves our assertion above. Since MATH, we have MATH . Thus MATH. Now suppose the equality holds in REF . Then MATH or MATH. Because MATH, REF is immediate. Since MATH . REF follows. Note that MATH. By REF , we know that MATH is equal to the sum of those MATH many nonzero MATH, thus each of the nonzero MATH has to be equal to MATH. This proves REF. |
math/0012178 | CASE: By REF we have MATH . CASE: For any natural number MATH write MATH for the sum of the entries in the MATH-th nonzero column (from the left) of MATH. Let MATH. Suppose MATH with MATH. Then MATH consists of only MATH or MATH entries and each MATH by REF . Since MATH, we have MATH. It follows that MATH has at most MATH digits, of which MATH digits are equal to REF, and at most one digit is equal to REF. Moreover, this REF can not be at the rightest position. So the rightest MATH columns of MATH are zero. By removing those MATH zero columns of MATH we obtain MATH for MATH. It is clear that MATH. Explicitly we have MATH for every MATH. By REF we then have MATH . This proves REF. CASE: Now let MATH with MATH. Arguing as above, we observe that MATH consists of only MATH or MATH digits and each MATH has MATH many REF's with at most one REF in between. REF can only occur if MATH. Indeed, if there is a REF in between, then either MATH or MATH has to contribute a REF at the corresponding position of MATH. This situation can only happen if for MATH or MATH we have MATH . However, MATH can only occur if MATH. Suppose MATH. Then MATH for MATH, and in between any two consecutive nonzero columns there are MATH zero columns. It is then clear that MATH is defined by MATH and MATH. By MATH we have MATH . This proves REF. Suppose that MATH. We argued above that either MATH or MATH. If MATH then we remove the leftest non-zero column from MATH. This way, one obtains MATH for MATH with MATH. The relation between MATH and MATH is given by MATH for MATH and MATH for MATH. This gives a MATH correspondence between the two sets MATH and MATH . If MATH then we define MATH with MATH by removing the two leftest non-zero columns of MATH. In this case MATH for MATH and MATH for MATH. This gives a MATH correspondence between the two sets MATH and MATH . From MATH it now easily follows that MATH . This proves REF. |
math/0012178 | From REF it follows that MATH for all MATH. By NAME REF we have MATH. This proves REF . (See also CITE.) From now on we assume MATH. For any integer MATH define MATH . Consider MATH as a function in MATH, it is clear that MATH is monotonically decreasing and converges to MATH when MATH approaches MATH. Choose MATH such that MATH and such that MATH is a multiple of MATH and such that MATH. For all MATH we have MATH; that is, MATH . Therefore, for all MATH and MATH one has MATH . On the other hand, it follows that MATH . Hence, for all MATH one has MATH . Thus REF are satisfied. But MATH, so we have MATH . Now we apply our assumption that MATH to REF and get MATH . Hence MATH by combining REF . Thus MATH. This proves REF . Assume MATH. Define MATH . Consider MATH as a function in MATH, it is monotonically decreasing to MATH as MATH approaches MATH. Recall from above that MATH is a multiple of MATH and MATH. Now we may assume MATH above is chosen large enough so that MATH and MATH. Since for all MATH we have MATH, we get MATH . It follows that for all MATH and MATH one has MATH . On the other hand, MATH . Therefore, for all MATH one has MATH . Hence REF are satisfied. But MATH implies that MATH . By REF, this implies MATH . Recall from REF we have MATH . Put these together we get for MATH that MATH . Now recall REF, for MATH, we have MATH . Assume MATH or MATH. By applying REF recursively for MATH, and also by using REF one gets for all MATH that MATH . When MATH we have MATH it follows that MATH. Apply this to the congruence of MATH one get MATH hence MATH. This proves REF . |
math/0012178 | Every MATH-rank zero hyperelliptic curve of genus MATH has an equation of the form MATH where MATH is a polynomial of degree MATH. This follows from the NAME formula (see CITE) as an easy exercise. By applying an obvious isomorphism of MATH we can assume MATH is monic and odd. From now on we write MATH with MATH. Any isomorphism between curves with an equation of the form MATH is of the form MATH for some polynomial MATH of degree MATH, some MATH and some MATH-th root of unity MATH. (This polynomial MATH has nothing to do with the integer MATH defined in this paper.) We want to show that there exists a triple MATH such that MATH sends MATH to MATH, where the polynomial on the right-hand-side is odd in MATH and has its MATH-coefficient vanishing. Also, we want to show that there are only finitely many such MATH. Let MATH be a variable. Write out MATH for polynomials MATH. Let MATH for some MATH such that MATH is odd in MATH. Let MATH for MATH. It suffices to show that the MATH-coefficient MATH of MATH is a non-constant function in MATH so that it vanishes after specializing MATH to one of finitely many MATH. Since MATH is odd, for MATH we have MATH, and for MATH we have recursively MATH. For MATH even, MATH is odd with a term MATH. So by the hypothesis on MATH, there exists a MATH such that MATH. Let MATH be the biggest MATH with MATH. Then we have MATH for all MATH. We consider the cases MATH and MATH separately. If MATH then MATH. Recall that MATH is odd, hence MATH. For any MATH one has that MATH. By applying this MATH times one observes that MATH. If MATH then MATH is even for all MATH. Hence MATH is odd, MATH and MATH. In both cases MATH is not constant. This finishes the proof. |
math/0012178 | By REF , any supersingular hyperelliptic curve of genus MATH can be given by an equation of the form MATH with MATH. If MATH then we also have MATH. For MATH (respectively, MATH) there are only MATH (respectively, MATH) non-zero coefficients left. |
math/0012178 | The given curve is of the form in REF hence supersingular. By REF every genus-MATH hyperelliptic curve of MATH-rank zero has an equation MATH for some MATH, which is not supersingular if MATH by REF . |
math/0012190 | In the following, when we write a condition concerning the MATH-th component of a MATH vector (for example, the case MATH for MATH in the next paragraph or MATH in the proof of REF ), we mean that the condition is void. First we prove that if MATH the condition MATH follows from REF . Suppose otherwise, there exists MATH such that MATH and MATH. (Here, MATH.) Then, we have MATH which is a contradiction. Similarly, if MATH the condition MATH follows from REF . Now we will prove that for all MATH, the conditions MATH for MATH follow from MATH. (The proof is similar for MATH.) Suppose otherwise, there exists MATH and MATH such that MATH, MATH, MATH and MATH (and thereby MATH) for MATH. Set MATH so that we have MATH. Then we have MATH which is a contradiction. Here we used the notation MATH . |
math/0012190 | Using REF with MATH, we see that these two sets of numbers satisfy the same recursion with the same initial condition. |
math/0012190 | First we show that for any nonzero MATH there exists MATH such that MATH. Consider the lexicographic ordering of monomials MATH. Namely, the monomial MATH is higher than MATH if MATH, or if MATH and MATH, and so on. Let MATH be the highest monomial present in MATH of MATH in REF . Then, taking MATH we have MATH. Next we show that for any nonzero MATH there exists MATH such that MATH. For MATH let MATH be the set of indices MATH such that MATH, MATH and MATH, with MATH . By the PBW theorem the monomials MATH span MATH. Set MATH . Take MATH and MATH. Using the definition of the coupling, we have MATH . The assertion follows from this. |
math/0012190 | We use induction on MATH. The identity REF for MATH takes the form MATH . The case MATH is just the NAME relation: MATH. We obtain REF by induction on MATH. Assume MATH, MATH. Then MATH follows from the identity MATH. Indeed on the Right-hand side of REF for MATH, the only term left is exactly MATH. The MATH case of identity REF is proved similarly. Now assume REF are proved for MATH and let us prove them for MATH. It is enough to prove REF, then REF is done by the same argument switching the roles of MATH and MATH. We use induction on MATH. The case MATH follows from the identity MATH . Suppose we have the statement for MATH, then the case MATH follows from the identity MATH . |
math/0012190 | We use the induction on MATH. Assume the statement is proved for MATH. (We assume nothing if MATH.) We will prove it for MATH. To do that we prove by the inverse induction on MATH the identity MATH where MATH. The case MATH is exactly REF for MATH. The identity REF for MATH follows directly from REF. Assume we have REF for MATH. Let us prove it for MATH. For that we prove the identity MATH for MATH by induction on MATH. For MATH we have exactly REF for MATH. If the statement is proved for MATH then the statement for MATH follows from the relation MATH and REF. For MATH we obtain REF for MATH and the proof is finished. |
math/0012190 | The fact that the map MATH is well defined follows directly from the defintion. Note that the vectors MATH of the form MATH, where MATH is the highest weight vector, span MATH. Indeed, as shown in the proof of REF , the orthogonal complement of the span of such vectors is trivial. Therefore it is enough to check REF for MATH. For such vectors REF is clear from REF . |
math/0012190 | Consider the case MATH with MATH, with MATH fixed. Denote the variables MATH such that MATH by MATH REF in some ordering. We can carry out the evaluation in two steps: MATH, where MATH is the evaluation of all the variables except MATH REF and MATH is the evaluation of the variables MATH. Let MATH. Since MATH and MATH, we have MATH . Differentiating the left hand side of this equality by MATH and using the symmetry of MATH with respect to MATH, we can deduce that MATH . Therefore, MATH has a zero of order at least two at MATH for each MATH. After evaluation, MATH is divisible by MATH. |
math/0012190 | Without loss of generality, we can assume that MATH. If MATH the assertion follows immediately. Suppose MATH. Set MATH, where MATH is the polynomial function of REF . It is enough to show that MATH is divisible by MATH, because the evaluation of the prefactor in REF only contains a pole of order MATH at this point. Let MATH. We obtain MATH in two steps: MATH, where MATH is the evaluation of all the variables except those in MATH. Using the fact that MATH if MATH, MATH . Therefore, MATH is divisible by MATH, and hence MATH is divisible by MATH. |
math/0012190 | The assertion follows by a similar argument as in the proof of REF from the restriction on MATH that it is zero if MATH or MATH. |
math/0012190 | If MATH then there is nothing to prove due to REF . Therefore, without loss of generality we assume MATH. Let MATH . Note that for MATH, we have MATH because MATH if MATH. Therefore, MATH is a polynomial. From REF we obtain MATH . Now, it follows by induction on MATH that for MATH the polynomial MATH is of the form MATH where MATH is a polynomial and MATH are polynomials independent on MATH of degree at least MATH in MATH. Indeed, if we have the statement for MATH, then the case MATH follows from REF with MATH. Therefore MATH is of degree at least MATH in MATH and the lemma follows. |
math/0012190 | This follows from REF . |
math/0012190 | For a rigged partition MATH we denote by MATH the monomial symmetric polynomial corresponding to the monomial MATH. The space of rational functions MATH can be parameterized by the set of rigged partitions MATH by associating MATH to MATH where MATH. The statement follows from the injectivity of MATH and the equality of dimensions REF . |
math/0012190 | Note that MATH. If MATH we have MATH, and we label the complement as MATH. If MATH we have MATH, and we label the complement as MATH. |
math/0012190 | Observe MATH when MATH varies from MATH to MATH. For MATH, MATH increases from MATH to MATH by MATH if MATH REF or stays constant otherwise. In the case MATH, for MATH, MATH decreases from MATH to MATH by MATH if MATH REF or stays constant otherwise. In the case MATH, for MATH, MATH decreases by MATH at MATH REF or stays constant otherwise. In particular, we have MATH. For MATH, MATH. The equality REF follows from these observations with the convention REF for MATH. |
math/0012190 | We will prove that the inverse map MATH is given by MATH . Let us prove that the composition MATH is the identity map. Consider MATH and MATH. Set MATH. Since MATH, the smallest MATH elements in MATH are MATH. Therefore, MATH. Note that MATH . Therefore, we have MATH . Let us prove that the pair MATH given by REF is MATH-admissible. We define MATH and MATH as before from MATH and MATH. It is clear that MATH (MATH). The number MATH satisfies MATH. Since MATH, we have MATH. Let the smallest MATH elements of the set MATH be MATH (MATH). We will show that MATH. Then, it follows that MATH (MATH). Since MATH, we have MATH. Then, we have MATH . Since MATH, we have MATH. |
math/0012190 | Set MATH (MATH). We have obviously MATH (MATH). For MATH we have further MATH. Therefore, the minimal element is given by REF . |
math/0012190 | The minimal set MATH among MATH satisfying REF is given by MATH. Then, MATH is given by REF . |
math/0012190 | Assume that MATH and MATH REF for some MATH. As we noted at the beginning of this section we have MATH (MATH). This implies MATH (MATH). Therefore, we have MATH . We have MATH . Sub REF MATH. From MATH we have MATH. Using REF we have MATH . Therefore we have MATH and MATH. This implies MATH, and therefore MATH. Using MATH we have MATH. This is a contradiction because MATH . Sub REF MATH. We have MATH because MATH. From MATH follows MATH and using REF we have MATH . Therefore, we have MATH, MATH and MATH again. If MATH, because of REF we have MATH. Using MATH we have MATH. This is a contradiction because MATH . If MATH, we proceed as follows. If MATH, it leads to a contradiction as above. If MATH, because of REF we have MATH. It implies MATH. However, this is prohibited by REF . |
math/0012190 | We lead to a contradiction assuming that for some MATH and MATH satisfying MATH we have MATH . We set MATH so that MATH. A simple calculation as REF shows MATH . Here we used MATH for MATH. Note that the last two terms in the Right-hand side of REF is non-negative. We consider three cases MATH, MATH and MATH, separately. CASE: MATH. Because of MATH we have MATH. From this follows MATH . Using REF we have MATH, which is a contradiction. CASE: MATH. Sub REF MATH. Using MATH and REF we have MATH. Then, we have MATH . This is a contradiction. Sub REF MATH. Because of REF we have MATH. Therefore, noting that MATH, we have again MATH . This is a contradiction. Sub REF MATH. We have MATH because otherwise MATH and and using MATH we have MATH, which is a contradiction. We will prove by induction the following statements for MATH: MATH . Then, MATH leads to REF , which is a contradiction. We first note that MATH and MATH are valid. These are the basis for the induction. From MATH follows MATH. Using MATH we have MATH, and therefore MATH. Because of MATH from MATH follows MATH. Finally, we show that for MATH from MATH follows MATH. Unless MATH we have again MATH . CASE: MATH. Sub REF MATH. We will prove by induction the following statements for MATH. MATH . Then, from MATH, we have MATH. Using this we have MATH . This is a contradiction. As we have noted above we have MATH. Because of MATH and REF , we have MATH. This is MATH. Assume that MATH and MATH are valid for some MATH. From MATH follows MATH. Since MATH by MATH, we have MATH. If MATH, because of REF we have MATH. If MATH, we use REF . Note that MATH and MATH. If MATH, we have MATH, which contradicts REF . Otherwise, we have MATH. Thus we have derived MATH from MATH and MATH. Using REF we can derive MATH from MATH . Suppose that we have MATH for some MATH. Unless MATH we have MATH . This is a contradiction. We have derived MATH from MATH. Sub REF MATH. We will prove by induction the following statements for MATH. MATH . Then, from MATH we have MATH. Therefore we have MATH which is a contradiction. We have MATH and MATH. It is obvious that from MATH and MATH follows MATH. Suppose that MATH is valid for some MATH. In particular, we have MATH. If MATH, using REF or REF we have MATH unless we have MATH and MATH, which contradicts REF . Therefore, by using REF (if MATH) or REF (if MATH) we have MATH. Suppose that MATH is valid for some MATH. Unless MATH from REF we have MATH . This is a contradiction. Thus, we have proved MATH. Sub REF MATH. Because of MATH we have MATH. Because of REF , this is a contradiction. |
math/0012190 | Assume that MATH for some MATH and MATH (MATH). We have MATH (MATH). This implies MATH (MATH). Therefore, we have MATH. We have MATH . Sub REF MATH. From REF we have MATH. Using REF we have MATH . This is a contradiction because we assumed MATH. Sub REF MATH. We have MATH . This is a contradiction. |
math/0012190 | Suppose that for some MATH and MATH such that MATH we have MATH, MATH REF and MATH. We set MATH. We have MATH . CASE: MATH. We have MATH . Using REF we have MATH, which is a contradiction. CASE: MATH. From REF we have MATH and MATH. Therefore, we have MATH which is a contradiction. |
math/0012190 | If MATH, using REF we have MATH . If MATH, set MATH. We have MATH and MATH. Therefore, using REF we have MATH . |
math/0012190 | The proof is completely parallel to REF (we use REF ) except that REF for MATH follow directly from REF . |
math/0012190 | It is enough to show the bijectivity of MATH between the subset of MATH with a fixed MATH and the subset of MATH with MATH given by REF . Because of REF , and REF , these two subsets are both empty or REF are both valid. In both cases, the bijectivity is clear. |
math/0012190 | We use induction on MATH. If MATH, the statement is obvious because the union REF is for a single element MATH. We reduce the proof for MATH to MATH. Fix MATH such that MATH, and denote MATH. We take the union of MATH over a maximal string MATH REF of color MATH: MATH . This is maximal in the sense that there is no arrow of color MATH pointing to MATH or from MATH. Each arrow of color MATH belongs to one and only one maximal string of color MATH. If MATH, MATH, MATH and MATH for MATH. If MATH, there exists a sequence MATH such that MATH and MATH where MATH. Note that in the case MATH, the situation is the same if we set MATH. Consider the restriction MATH in MATH. Unless MATH, MATH and MATH are independent of MATH. If MATH, we have MATH . From these observations follows that MATH REF are disjoint, and the union is characterized by the conditions that MATH . Since MATH for MATH, there is no restriction on MATH for MATH. In particular, there is no restriction for MATH. Now, we modify the graph. We discard MATH REF from MATH and replace the set MATH by the union MATH characterized by REF . Carrying out this process for all the maximal strings of color MATH, we obtain a new graph MATH and the sets MATH (MATH). Observe that MATH satisfies the restriction MATH and there is no arrow of color MATH in MATH. We see that the graph MATH is isomorphic to MATH where MATH. The isomorphism maps MATH to MATH and identifies the color MATH in the former with the color MATH in the latter. We have MATH for MATH, and MATH if and only if MATH. Therefore, the condition for MATH in MATH is exactly the same as the condition for MATH in the subset MATH at the level MATH. Thus we have proved REF . |
math/0012190 | If MATH, we have MATH since MATH. From this follows that MATH, and therefore, MATH. Therefore, there is no restriction on MATH in MATH. We take the union of the riggings MATH subject to the restriction on MATH. This is equivalent to the special case of REF where MATH are replaced by MATH, respectively. Therefore, the left hand side of REF is disjoint and the equality holds. |
math/0012190 | We will prove this by induction on MATH with respect to the ordering defined in REF. We see that the statement is true for the maximal element MATH. In fact, if MATH the statement MATH follows from MATH and MATH if and only if MATH. This is the base of the induction. Now assume that the statement is true for all MATH of the first kind such that MATH. We will show that there exists a subset MATH satisfying MATH . This will close the induction steps. We fix MATH and define MATH . Namely, we take the disjoint union over MATH such that MATH with MATH. Note that the element MATH is fixed, MATH move around the interval MATH and new elements MATH are added in the interval MATH. We have REF obviously. By the same argument as in the proof of REF we obtain MATH where MATH. We have the following values of MATH and MATH. If MATH, then MATH . If MATH, then MATH . Now we will prove REF . We have MATH. Therefore, MATH . We take MATH in REF to be MATH. We have MATH and MATH if and only if MATH . By case checking one can prove that the set of integers consisting of the values of MATH where MATH runs over REF , contains MATH appearing in REF . For example, if MATH, we have MATH and MATH. Therefore, we obtain MATH . Other cases are similar. |
math/0012190 | If MATH we have MATH . First we show that if MATH is an interval of the first kind REF we have MATH . From this follows that MATH. If MATH then MATH. Therefore, in order to show REF one can forget the restriction on MATH. Then, it follows from REF . To finish the proof, we show that MATH. Consider MATH where MATH and MATH. Set MATH . We will show that MATH . Note that MATH and MATH. Therefore, the condition for the riggings MATH in MATH is stronger than MATH. Namely, we have MATH. For any MATH we can find MATH such that MATH . Then we have MATH . REF follows from this. |
math/0012190 | We show that REF is equal to REF by case checking for each case of the ordering of MATH. There are REF cases. Here we give the details for the case MATH. Other cases are similar. If MATH the intervals MATH which appear in REF satisfying MATH are of the form MATH where MATH. For such MATH we have MATH . Therefore, the pair of integers MATH runs over the set MATH. On the other hand we have MATH . This completes the proof. |
math/0012190 | By REF we have shown REF for MATH. Let us reduce the proof to the case MATH. Suppose that MATH. Then, we have MATH. We will reduce this case to the case where MATH and MATH replaced by MATH and MATH, respectively. Note that the union is taken over MATH such that MATH, that is, MATH. Therefore, we refer to the structure of colored graph in this set. Recall the definition of MATH given by REF . If MATH varies on a maximal string of color MATH, then only MATH changes. However, we see that the MATH is independent of MATH because for MATH we have MATH. It follows that the vector MATH is constant on the maximal string. Therefore, we can take the union over MATH on maximal strings of color MATH only on the riggings MATH forgetting MATH. Taking unions over all of the maximal strings of color MATH, we can rewrite the left hand side of REF as the union of the resulting subsets over such MATH that satisfies MATH, that is, of the form MATH. The subgraph of MATH consisting of such MATH is isomorphic to MATH by mapping MATH to MATH and identifying the color MATH in the former with the color MATH in the latter. We have MATH . Therefore, we have MATH . Note also that MATH . Thus, we have reduced the case MATH, MATH, MATH, MATH to MATH. |
math/0012190 | The proof of REF is parallel to that of REF . In REF the restriction on MATH in MATH is independent of MATH and the restriction on MATH is of the form MATH . Here MATH is a MATH-vector independent of MATH. In REF the restriction on MATH in MATH is independent of MATH and the restriction on MATH is of the form MATH . Here MATH is a MATH-vector independent of MATH. In REF the union is taken over MATH such that MATH where MATH is given by REF , and in REF the union is taken over MATH such that MATH where MATH is given by REF . Recall that in the proof of REF we take the union over the maximal strings of color MATH as the first inductive step. Similarly, in the setting of REF we take the union over strings MATH REF of the form MATH REF and MATH. Here MATH . The difference between two cases is that MATH is fixed in REF , while MATH varies in REF . However, if we consider MATH instead of MATH, MATH is fixed and two cases are completely parallel. Therefore, the union is obtained by substituting MATH by MATH and make the restriction on MATH unmarked. |
math/0012190 | Suppose that MATH and MATH where MATH. Then, we have MATH, and MATH and MATH are uniquely determined from MATH. We have MATH . Note that MATH and MATH. Therefore, we have MATH for MATH. Since MATH, we have MATH . From these observations follows that MATH . The set MATH satisfies MATH and MATH if and only if MATH where MATH. By a similar argument as the proof of REF , taking the union of MATH over MATH, we obtain MATH. |
math/0012190 | Let us observe that there is a correspondence between MATH and MATH. We have REF . Moreover, because of MATH if and only if MATH. Finally, note that MATH . In this way, REF follows from REF . |
math/0012190 | The proof of this lemma is parallel to the proof of REF . Note that MATH (see REF ) does not depend on MATH. The summation extends to all MATH satisfying MATH where MATH . The restriction on MATH in MATH is given by MATH. It is marked if and only if MATH. The restriction on MATH in the union over MATH is given by MATH. The marking will change as follows. Because of MATH the restriction on MATH is unmarked if MATH. In the interval MATH, the marking is unchanged because MATH is fixed. Decompose the interval MATH into subintervals MATH constituting MATH and MATH constituting MATH in such a way that MATH and MATH. Note that MATH and MATH may be empty. The restriction on MATH is marked if and only if MATH for some MATH. This is equivalent to say that it is marked if and only if MATH for some MATH except for MATH. Therefore, we have the marking given in REF . |
math/0012190 | Fix MATH. First, we can rewrite each subset MATH in terms of subsets which have the same restriction on MATH as MATH but a different unmarked restriction on MATH. If the restriction is MATH, we rewrite it as the difference of two unmarked restrictions MATH and MATH. We extend this argument to all MATH with marked conditions by using the inclusion-exclusion principle, which will be explained below. Let MATH be the decomposition considered in the proof of REF . Denote by MATH the set of integers MATH for MATH except when it is equal to MATH. This is exactly the set of MATH such that the restriction on MATH is marked in MATH. For each MATH we denote by MATH the subset obtained from MATH by replacing all the elements MATH with MATH. Because of REF we have MATH . We set MATH . For a subset MATH we denote by MATH its support function, that is, MATH if MATH and MATH if MATH. The inclusion-exclusion principle tells us that MATH . Thus we obtain MATH . Denote by MATH the set of integers MATH for MATH except when it is equal to MATH. This is exactly the set of MATH such that the restriction on MATH is marked in MATH. For each MATH we denote by MATH the subset obtained from MATH by replacing all the elements MATH with MATH. Because of REF we have MATH . Denote by MATH the set of MATH satisfying REF . Note that MATH . Therefore, we have MATH . The inclusion-exclusion principle again tells us that MATH . Therefore, we obtain REF . |
math/0012190 | Suppose that MATH is given by REF for MATH. If MATH, then MATH implies MATH, and REF is obvious. If MATH, the union in the left hand side is over MATH such that MATH where MATH . Since MATH, by a similar argument as the proof of REF , we take the union of MATH and obtain MATH. |
math/0012190 | This is a consequence of REF (with MATH replaced by MATH) and REF . |
math/0012191 | The fact that the map REF is an isomorphism between MATH and MATH implies that MATH does not vanish for MATH . Clearly the functions MATH belong to the kernel of the operator in the right-hand side of REF. It has leading term REF and the nonvanishing of MATH implies REF. |
math/0012191 | The relation REF for MATH holds because MATH . To check the one for MATH we conjugate MATH by MATH (the factor in front of the right-hand side of REF). The result is MATH . This is the difference operator from the l.h.s. of REF with MATH and MATH which gives the proof of REF for MATH . The cases of MATH and MATH are handled in a similar fashion. |
math/0012191 | In the case MATH the functions MATH and MATH are well defined. From the remark in the beginning of this subsection it follows that REF and the definitions of MATH imply REF. The case MATH follows by continuity on MATH . The inclusion MATH in REF clearly follows from REF - REF. Because MATH is a regular difference operator with support MATH to prove REF it suffices to show that the functions MATH are linearly independent. Assume that MATH for some complex numbers MATH such that MATH or MATH . Applying MATH to this equality and using REF, we get MATH that is, MATH . For MATH this gives MATH . Dividing the two sides of REF for two consecutive values of MATH we get MATH . This gives MATH which is a contradiction. REF is proved analogously. |
math/0012191 | We compare the coefficients of MATH and MATH in REF and use REF for the operator MATH . This gives the formulas MATH which are equivalent to REF . Similarly comparing the coefficients of MATH in REF gives MATH which implies REF, taking into account REF. |
math/0012191 | Consider first the case MATH . Let MATH . REF implies MATH . Let MATH and MATH denote the MATH and MATH objects associated with the functions MATH (see the beginning of REF for the appropriate definitions). Obviously MATH and MATH . In this notation, the statement of the theorem is equivalent to MATH . To prove that the l.h.s. of REF is contained in the right-hand side, let us fix an operator MATH . There exists a difference operator MATH and a function MATH such that MATH . We will prove that all operators from MATH are MATH-invariant. This in particular shows that all functions from MATH are MATH-invariant and so are all operators from MATH . If MATH then there exists a differential operator MATH for which MATH . The fact that the hypergeometric function MATH is symmetric with respect to MATH and MATH and REF for MATH imply MATH . Combining REF, we conclude that MATH . This is only possible if MATH . The harder part of the proof of REF is to show that any MATH-invariant difference operator from MATH belongs to MATH . It is sufficient to prove that for any MATH there exists MATH and MATH such that MATH . First let us write REF in terms of MATH . REF remains unchanged: MATH while REF becomes MATH with MATH . The algebra MATH has a natural MATH filtration where MATH consists of all operators from MATH with support MATH . Denote MATH . We will prove that any difference operator MATH can be decomposed as a sum MATH where MATH . Since MATH any MATH-invariant rational function in MATH is a rational function in MATH by induction on MATH REF implies that MATH . A straightforward computation yields MATH and thus MATH . So MATH for some MATH . Here we use the MATH-invariance of MATH . Denote for simplicity MATH and let MATH . Using REF we obtain MATH for some other MATH . There exists a polynomial MATH for which MATH because the polynomial in the l.h.s. is clearly MATH-invariant. Denote by MATH the difference operator in REF. The l.h.s. of REF implies that MATH belongs to MATH and the right-hand side implies MATH which completes the proof of REF. |
math/0012191 | First note that MATH and similarly MATH are MATH-invariant polynomials in MATH . To prove the first statement of the lemma we use a similar computation. Restricting to the case MATH . The proof of the second statement is analogous. Assuming MATH and MATH we obtain MATH with MATH Since MATH the first condition is equivalent to MATH and the second one to MATH . To finish the proof of REF we just observe that MATH . The remaining cases for MATH are treated analogously. The identity REF follows from the analogous formula MATH and REF . Throughout this proof, for a real number MATH by MATH we denote its integer part. |
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