paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0107008 | The assumptions imply that MATH and MATH define the same partition of MATH into elliptic and hyperbolic elements. By REF it suffices to show that MATH and MATH have the same elliptic subgroups. Note that both trees are proper by REF . Exchanging names if necessary, assume that MATH is the tree referred to in REF or REF . If REF holds then MATH and MATH have the same elliptic subgroups by REF . If REF holds then REF together yield the same conclusion. |
math/0107008 | Let MATH, MATH, and MATH. Then MATH and MATH are MATH - trees (via MATH in the case of MATH), and they satisfy the hypotheses of REF . Hence there is an isomorphism of MATH - trees MATH. By CITE there is an isomorphism MATH. The element MATH and factorization MATH exist by CITE. |
math/0107008 | Since both trees have REF vertex stabilizers, they have the same elliptic subgroups. Then, as each tree is proper, the vertex stabilizers in each tree are precisely the maximal elliptic subgroups. Thus the trees have the same vertex stabilizers. By REF the trees are canonically isomorphic. It is here where strictness of both trees is required. |
math/0107008 | First note that MATH is torsion free: every hyperbolic element has infinite order and every elliptic element is contained in an infinite cyclic subgroup. Also note that for any MATH and MATH, the element MATH is hyperbolic if and only if MATH is, and when this occurs they have the same axis. We will say that two subgroups MATH are commensurable if MATH has finite index in both MATH and MATH. Commensurability is an equivalence relation on the subgroups of MATH. We say that two nontrivial elements MATH are commensurable if the infinite cyclic subgroups they generate are commensurable. This occurs if and only if MATH for some MATH. Thus, commensurable hyperbolic elements have the same axis. In a generalized NAME - NAME tree every edge stabilizer has finite index in its neighboring vertex stabilizers. This implies that all edge and vertex stabilizers are in a single commensurability class. Because these subgroups are infinite cyclic, this class contains all nontrivial subgroups of vertex stabilizers. Therefore every nontrivial elliptic element is commensurable with all of its conjugates. Now suppose that an element MATH is commensurable with its conjugates. If MATH is hyperbolic then let MATH be its axis. For every MATH, MATH is commensurable with MATH and hence its axis MATH is equal to MATH. But MATH, and so we have shown that MATH is MATH - invariant. Hence MATH contains any minimal subtree. These remarks imply that if a minimal subtree of a generalized NAME - NAME tree is not contained in a line, then the nontrivial elliptic elements are exactly those elements which are commensurable with all of their conjugates. Thus, MATH and MATH define the same partition of MATH into elliptic and hyperbolic elements. Then, as all vertex stabilizers have REF (by finite generation), REF implies that the two trees are related by a deformation. The other conclusions follow from REF . |
math/0107008 | We define MATH by choosing MATH to be any point of MATH with MATH. It is straightforward to check that MATH is a MATH - quasi-isometry, and a MATH - quasi-inverse of MATH. Next let MATH be any MATH - quasi-inverse to MATH, and take MATH and MATH. Then MATH . The middle term on the second line is equal to MATH and so the second line is bounded by MATH. Hence MATH as desired. |
math/0107008 | The implication MATH is given by REF . For REF , we claim that a collapse move is itself an equivariant quasi-isometry. Then REF implies that the relation defined by REF is an equivalence relation. Since the relation of REF is generated by collapse moves, and the implication MATH holds for these moves, it must hold in general. To prove the claim, we show that a collapse move is a MATH - quasi-isometry. The essential point is that if MATH can be collapsed, then the connected components of MATH have diameter at most MATH. Then, if MATH is a geodesic of length MATH, then at most MATH of its geometric edges are in MATH. Thus the distance MATH between the endpoints after the collapse satisfies MATH . Now consider a collapsible edge MATH (with MATH) and a translate MATH which is adjacent to MATH. The endpoints of MATH are in different MATH - orbits, so either MATH or MATH. In the former case we have that MATH, a contradiction. Hence the latter holds. It follows that every edge in the connected component of MATH containing MATH is incident to MATH. Hence this component has diameter at most MATH. For REF we will establish the fact that if MATH is any coarsely equivariant map of MATH - trees, then every vertex stabilizer of MATH fixes a vertex of MATH. We then apply this fact both to MATH and to a quasi-inverse MATH, which exists and is coarsely equivariant by REF . Then MATH and MATH have the same elliptic subgroups. Suppose MATH is MATH - equivariant. If MATH for some MATH, then MATH. Thus MATH, the subtree spanned by MATH, is a bounded subtree of MATH. Now every element MATH of MATH has a fixed point in MATH, since if it had positive translation length then the subset MATH would be unbounded. Note also that a fixed point of MATH lies on the path MATH and so every element of MATH has a fixed point in MATH. By REF either MATH has a fixed point in MATH or there is an end MATH fixed by MATH. Suppose the latter holds. Let MATH and MATH be vertices of MATH. The subtree spanned by MATH, MATH, and MATH meets MATH in a subtree containing MATH, so the intersection MATH is nonempty. As MATH is bounded, there is a nearest vertex MATH to MATH, which is also the nearest vertex of MATH to MATH. Fixing MATH, this shows that MATH for every MATH. Therefore MATH fixes MATH as each of its elements fixes both MATH and some vertex of MATH. |
math/0107008 | Let MATH be a vertex of MATH. We wish to show that MATH. For each MATH we have that MATH. Let MATH be the vertex defined by MATH . That is, MATH is the midpoint of the tripod spanned by MATH, MATH, and MATH. Note that MATH fixes the segment MATH. Suppose that MATH for every MATH. Then MATH fixes an interior point of MATH. However, as MATH is proper, the fixed point set of MATH is MATH, and therefore MATH for some MATH. Then MATH is the midpoint of MATH, a path of length less than MATH. |
math/0107008 | The first conclusion follows directly from REF . The second conclusion follows from REF . |
math/0107011 | We compute what at first seems to be a different functional, in two different ways. Call an edge on a puzzle piece right-side-up if its outward normal is parallel to an outward normal of the entire puzzle, upside-down if the outward normal is antiparallel. So on a right-side-up triangle, all three edges are right-side-up, and vice versa for an upside-down triangle. Whereas on a rhombus, two of the edges are right-side-up, two upside-down. Define the functional MATH by MATH where the sign is MATH if MATH right-side-up, MATH if MATH upside-down. We claim first that MATH. Consider the contribution a piece MATH makes to the sum in MATH: a MATH-triangle contributes nothing, a MATH-triangle contributes the three coordinates of a vertex (which sum to zero), and we leave the reader to confirm that a rhombus contributes the length of the corresponding edge in MATH. To show that MATH also matches the conclusion of the theorem, rewrite by switching the order of summation: MATH . For every edge MATH internal to the puzzle, this latter sum is MATH which cancels, whereas for every exterior MATH-edge it is MATH. The claim follows. And as stated before, this functional is a sum of honeycomb edge-lengths, so automatically nonnegative on MATH. |
math/0107011 | This is certainly an upper bound: by REF, the sum of all the constant coordinates of boundary edges is zero. Let MATH be a nondegenerate honeycomb, MATH a (possibly negative) real number such that MATH is smaller than the length of any of MATH's edges, and MATH two boundary edges. Then there exists a path MATH in the honeycomb tinkertoy MATH connecting MATH and MATH (which we can ask be non-self-intersecting). We can add MATH times the natural sign to the constant coordinates of MATH's edges along MATH and get a new honeycomb. This changes MATH's coordinate by MATH, and MATH's by MATH. By repeating this with other pairs, we can achieve arbitrary small perturbations of the boundary coordinates, subject to the sum staying zero. So MATH contains a MATH-dimensional neighborhood of MATH. |
math/0107011 | By rotating if need be, we can assume MATH looks like the left figure in REF , a simple degeneracy. Assume the path MATH comes from the Southwest, MATH going to the Southeast (the other three cases are similar). We can pull the edges in MATH at MATH down and create a vertical edge in the middle (as in REF example). To extend this change to the rest of the honeycomb, we apply the trading construction to MATH, but we must move edges a negative MATH times their natural signs (or else the vertical edge created will have negative length). In particular the first edge of MATH, whose natural sign is negative, has its constant coordinate increased. |
math/0107011 | By REF, MATH is simply degenerate and acyclic (this uses MATH regular and MATH a largest lift). Let MATH be the post-elision tinkertoy, of which MATH can be regarded as a nondegenerate configuration. We first claim that MATH is disconnected. For otherwise, we could use the trading construction to vary the boundary of MATH in arbitrary directions (subject to the sum of the coordinates being zero), and therefore MATH would not be on a facet of MATH. So we can write MATH, and MATH, where MATH are honeycombs with tinkertoys MATH respectively. The boundary MATH therefore satisfies the equation ``the sum of the boundary coordinates of MATH belonging to MATH is zero." (Likewise MATH.) This remains true if we deform MATH and MATH as individual honeycombs. By the genericity condition on MATH, MATH and MATH must each be connected, so varying them gives us a MATH-dimensional family of variations of MATH with MATH in the interior. By REF we have found the equation for the facet containing MATH. It remains to show that MATH turns clockwise to MATH at every intersection; actually we will only show that all the intersections are consistent (and switch the names of MATH and MATH if we chose them wrongly). For each intersection MATH, let MATH be the tinkertoy made from the honeycomb tinkertoy MATH by eliding all of MATH's simple degeneracies other than MATH. Since the fully elided MATH is acyclic with two components, each MATH is acyclic with one component. We can now attempt to trade MATH's boundary coordinates for MATH's. Fix a semiinfinite edge of MATH and one of MATH. Since MATH is acyclic there will be only one path MATH connecting them, necessarily going through MATH. By REF we can apply the trading construction to MATH, increasing the constant coordinate on our semiinfinite edge of MATH if MATH turns clockwise to MATH at MATH, decreasing it in the other case. If there exist vertices MATH such that MATH turns clockwise to MATH at MATH, but vice versa at MATH, then by trading we can move to either side of the hyperplane determining the facet. This contradiction shows that the intersections must be consistently all clockwise or all counterclockwise. |
math/0107011 | Plainly MATH satisfies this inequality with equality. We can deform MATH and MATH to nondegenerate honeycombs MATH and MATH; if we move the vertices of each little enough, they will not cross over edges of the other, and the result will again be a clockwise overlay MATH, satisfying the same equality. Build a puzzle MATH from MATH as follows: CASE: to each vertex in MATH, associate a REF-triangle CASE: to each vertex in MATH, associate a REF-triangle CASE: to each crossing vertex in MATH, associate a rhombus with the puzzle pieces glued together if the vertices in MATH share an edge. Then the fact that MATH turns clockwise to MATH means that the labels on the rhombi will match the labels on the triangles, so MATH will be a puzzle. An example is in REF . We can perturb MATH and MATH small amounts and vary the boundary coordinates of in arbitrary directions. This gives us a MATH-dimensional family of possible boundaries containing our original point MATH, all satisfying the stated inequality. By REF , MATH is in the interior of a facet determined by this inequality, which is the inequality MATH. |
math/0107011 | Consider a honeycomb MATH satisfying MATH for some MATH, but otherwise minimally degenerate; an example is in REF . (By MATH symmetry it is enough to consider the MATH case.) Note that there is only one nongeneric vertex, at the end of the multiplicity-two semiinfinite edge. It is straightforward to construct a similar such honeycomb for any MATH and MATH. We mimic the proof of REF , in using the trading construction to exhibit a MATH-dimensional family in MATH satisfying MATH. We can move the doubled edge wherever we like by translating the whole honeycomb. To trade the constant coordinates of any other two boundary edges, we use paths avoiding the bad vertex in MATH, which exist for MATH. This proves the first claim. For MATH we hit a snag - sometimes the only paths will go through the bad vertex. But we can check MATH directly. The regular inequalities are MATH . Sum the first two, and subtract the equality MATH to get MATH. The other two inequalities are proved in ways symmetric to this one. |
math/0107011 | It is enough to prove it for gentle paths of length two, and then string the MATH inequalities together. In REF we present all length two paths (up to rotation and dualization), and the corresponding pairs of edges in an overlay. In each case the angles around the associated region force the strict inequality. |
math/0107011 | If MATH arises from a clockwise overlay MATH, we can perturb MATH and MATH a bit to make them generic (as in the proof of REF ). Recall that a gentle loop is a gentle path whose first and last edges agree. Then by the proposition the corresponding edge in MATH is strictly longer than itself, contradiction. By the contrapositive of REF , MATH's inequality is inessential. |
math/0107011 | Of the two outer tines of the rake, one points toward the vertex, one points away. Every gentle path starting at the inward-pointing tine goes into the outward-pointing tine (to turn into the middle tine would not be gentle), and conversely every path from the outward-pointing can be extended; this is why they both have MATH gentle paths to them for some MATH. The gentle paths starting at the handle go either into the middle tine, or the outward-pointing outer tine. So if MATH gentle paths start from the middle tine, MATH start from the handle. |
math/0107011 | First note that for this purpose, it is enough to specify a honeycomb up to translation. To do that, it is enough to specify the lengths and multiplicities of the bounded edges, and say how to connect them. Not every specification works - the vertices have to satisfy the zero-tension condition of CITE, and the vector sum of the edges around a region must be zero. To specify a clockwise overlay, we must in addition two-color the edges MATH and MATH, and make such edges only meet at crossing vertices, such that the colors alternate MATH when read clockwise around each crossing vertex. We now build a clockwise overlay MATH as a sort of graph-theoretic dual of MATH - one vertex for each puzzle region of MATH, one bounded edge for each region edge of MATH (almost - two region edges on the same boundary of a region determine the same edge of MATH), one semiinfinite edge for each exterior edge of MATH. If MATH is a bounded edge of MATH corresponding to a region edge MATH of MATH, then MATH CASE: is perpendicular to MATH CASE: is labeled MATH or MATH depending on whether MATH is adjacent to a MATH or MATH region CASE: has multiplicity equal to the length of MATH CASE: has length equal to the number of gentle paths starting at MATH and ending on the boundary of MATH. This last number is finite exactly because there are no gentle loops. The vertices of MATH are zero-tension because the vector sum of the edges around a region in MATH is zero. The regions in MATH are all trapezoids, dual to the meeting of four regions in MATH at rakes, and close up by REF . For the second statement, note that no edge of MATH connects two distinct vertices in the same honeycomb, because the corresponding region edge in MATH does not bound two regions of the same type. |
math/0107011 | Let MATH be a witness to MATH produced during the proof of REF . By slight perturbation of MATH and MATH to nondegenerate MATH and MATH, small enough that no vertex of one crosses an edge of the other, we can create a MATH whose only degenerate edges correspond to the rhombi in MATH. (One can modify the construction in REF to give such a MATH directly, but it is not especially enlightening.) Puzzles are easily seen to be determined by the set of their rhombi, and by REF (to come) the number of rhombi in a puzzle is determinable from the boundary conditions. So if MATH is not rigid, so there exists another puzzle MATH with the same boundary, then this MATH has a rhombus that MATH doesn't. Then by its definition as a sum of edge-lengths, MATH. But by REF , MATH, contradiction. |
math/0107011 | Let MATH denote the radius-MATH neighborhood of MATH, that is, the set of pieces of MATH with a vertex on MATH. By the condition about nonconsecutive vertices, this neighborhood doesn't overlap itself, that is, every edge connected to MATH (but not in MATH) is connected to a unique vertex of MATH. Cut MATH up along its normal lines. It is easy to check that it falls into only four kinds of ``assemblages" up to rotation, listed in REF . Notice that for each assemblage, there is a unique other of the same shape, but rotated MATH. This pairs up the two triangular assemblages, the parallelograms each being self-matched. The crucial observation to make is that two matching assemblages have the same labels on the boundary (away from the normal lines). In particular, if we simultaneously replace each assemblage in MATH by the other one with the same shape, the new collection fits together (because all the normal lines have been reversed), fits into the rest of the original puzzle (because the labels on the boundary are the same), and gives a new gentle loop running in the opposite direction. |
math/0107011 | We will show that minimal gentle loops satisfy the condition of REF . First we claim that minimal gentle loops do not self-intersect. Let MATH be a gentle loop that does self-intersect, and let MATH be a vertex occurring twice on MATH such that the two routes through MATH are different. (If there is no such MATH, then MATH is just a repeated traversal of a loop that does not self-intersect.) There are MATH local possibilities, corresponding to choosing two of the rightmost three gentle paths in REF ; in each one we can break and reconnect the gentle loop to make a shorter one, contradicting minimality. Second (and this is the rest of the proof) we claim that minimal gentle loops do not have nonconsecutive vertices at distance MATH. If MATH is a counterexample, then there exists an edge MATH connecting two points on MATH. (This MATH is just an edge in the lattice, not necessarily in the puzzle MATH - it may bisect a rhombus of MATH or whatever.) Removing the endpoints of MATH from MATH separates MATH into two arcs; call the shorter one the minor arc and the longer the major arc. (If they are the same length make the choice arbitrarily.) Choose MATH such that the minor arc is of minimal length. For the remainder we assume (using puzzle duality if necessary) that the gentle loop is clockwise. We now analyze the local picture near MATH, which for purposes of discussion we rotate to horizontal so that the minor arc starts at the west vertex of MATH, and ends at the east vertex. This analysis proceeds by a series of reductions, pictured in REF . The first edge of the minor arc goes either west, northwest, or northeast (to have room for a gentle turn from the major arc); likewise, the last edge goes either southeast, southwest, or west. CASE: If the first edge of the minor arc went northeast, we could shift MATH to MATH connecting the second vertex of the minor arc to the last vertex, contradicting the assumption that the minor arc was minimal length. So in fact the first edge goes west or northwest, the last southwest or west (by the symmetric argument). CASE: We now involve the major arc. Its last edge goes northwest or northeast. If it goes northeast, then the first edge of the minor arc goes northwest (for gentleness). But then by REF MATH is oriented west; therefore we could shorten MATH to a loop that used MATH, contradicting MATH's assumed minimality. So the last edge of the major arc goes northwest. CASE: Therefore the first edge of the major arc goes southeast (since MATH doesn't intersect itself), so the last edge of the minor arc goes southwest. But then REF says that MATH is oriented east. At this point the west vertex of MATH has an oriented edge coming in from the southeast, one going out to the east, and another going out either west or northwest (at least). This matches none of the vertices (or their puzzle duals) in REF . The contradiction is complete; there was no such MATH. |
math/0107011 | If not, there exists a regular boundary MATH such that MATH. Let MATH be a largest lift of MATH; by REF MATH is simply degenerate. Since MATH is nonrigid, by REF it has a gentle loop; a minimal such loop MATH is breathable, by the proof of REF . Let MATH be the result of breathing MATH along MATH. We claim that some rhombus MATH of MATH overlaps some rhombus MATH of MATH in a triangle. To see this, divide MATH up along its normal lines into the assemblages of REF . At least one such assemblage must be a triangle, for otherwise the loop can not close up. (In fact there must be at least six triangles.) When we breathe the loop, the rhombus MATH in the original assemblage in MATH overlaps the rhombus MATH in the new assemblage in MATH in a triangle. Dually, the edges of the honeycomb tinkertoy MATH corresponding to those two rhombi meet at a vertex. Since MATH, and they are defined as the sum of certain edge-lengths of MATH, all those edges of MATH must be length zero. Therefore, the two adjacent edges in MATH corresponding to MATH and MATH are length zero. But then MATH is not simply degenerate, contradiction. |
math/0107011 | To see that MATH is a well-defined puzzle, we need to check that each new region is well-defined - that traversing its boundary we return to where we started. For MATH-regions there is nothing to show, for MATH-regions the whole region is inflated by the factor MATH, and for rhombus regions two opposite sides of the parallelogram are stretched. There is an evident correspondence between gentle loops in MATH and gentle loops in MATH, so by REF they are either both rigid or neither is. |
math/0107011 | The deflation of MATH is a MATH-puzzle consisting only of MATH-triangles, letting us count the number of MATH-triangles in the original puzzle (namely, MATH) and also the number of MATH-edges on the three sides (namely, MATH). By deflating the dual puzzle, one can count there to be MATH-triangles. The remaining area is MATH in units of triangle, and all that is available are rhombi which each use up REF. |
math/0107011 | CASE: Follow a path in the honeycomb, going Northwest whenever possible, Southwest when not, eventually coming out on an edge with constant coordinate MATH. Each Southwest sojourn increases the first coordinate, and each Northwest leaves it unchanged, so the original first coordinate must have been at most MATH. Replacing ``NAME with ``NAME gives the opposite inequality. The other two coordinates come from rotating this proof MATH. CASE: Since the sum of the three coordinates is zero by definition, the third one can be bounded in terms of the first two. CASE: This is just a special case of REF . |
math/0107011 | We prove the second statement: the first follows from the second, since MATH is MATH if MATH is the reversal of MATH, MATH otherwise. The structure constants for multiplication of NAME classes are well known to also be the structure constants for tensor products of polynomial representations of MATH (the first to observe this seems to be NAME; see CITE or CITE). The precise statement is as follows. Let MATH be the number of MATH-s after the MATH-th MATH in MATH, so MATH, and MATH is a partition of the number of inversions of MATH. Likewise construct MATH from MATH, and MATH from MATH. Then MATH . This latter can be calculated using honeycombs, as proved in CITE; it is the number of honeycombs with boundary coordinates MATH on the Northwest side, MATH on the Northeast, and MATH on the South. We now construct a map from our puzzles to these honeycombs. To create a honeycomb from a puzzle is a three-step process (follow along with the example in REF ): CASE: Place the puzzle in the plane MATH such that the bottom right corner is at the origin, and turn it MATH counterclockwise. CASE: At each boundary edge labeled MATH, attach a rhombus (outside the puzzle), then another (parallel to the first), and repeat forever. Fill in the rest of the plane with MATH-triangles. CASE: Deflate the extended puzzle, keeping the right corner at the origin. The honeycomb's vertices then come from the deflated MATH-regions, and the honeycomb's edges come from the deflated rhombus regions, with the multiplicity on the edge coming from the thickness of the original rhombus region. The resulting diagram obviously has finitely many vertices, all edges in triangular-coordinate directions, and semiinfinite edges only going NW/NE/S (coming from the MATH-edges on the boundary of the original puzzle). The remaining condition for it to be the diagram of a honeycomb is that each vertex have zero total tension; this is equivalent to the fact that the original MATH-region was a closed polygon. To see that this is a bijection, we construct the inverse map. Start with a honeycomb computing MATH. By REF , it fits inside the triangle with vertices MATH. Inflate each edge of the honeycomb intersected with the triangle to a rhombus region, the thickness given by the multiplicity of the edge, and each vertex to a polygon of MATH-triangles, the lengths of the edges of the polygon given by the multiplicities of the edges at the vertex. The result is a puzzle with boundary MATH. |
math/0107011 | These are REF combined with CASE: REF (the inequality is reversed because we are summing over MATH-s here instead of MATH-s) CASE: REF . |
math/0107011 | Let MATH denote the high weight of the determinant representation, and MATH denote the number of times MATH appears in MATH. Then using the equality MATH we can reduce to the case that MATH and MATH are nonnegative (so, high weights of polynomial representations). Therefore MATH is also nonnegative, for otherwise MATH would be zero for all MATH. From there, we can use REF to convert to a NAME problem, that is, counting puzzles rather than honeycombs. One then has to check that rescaling a honeycomb by the factor MATH corresponds to dual-inflation on puzzles. Since the original honeycomb MATH is rigid, so too is the corresponding puzzle MATH, therefore by REF so too is the dual inflation of MATH by the factor MATH, and therefore so is MATH. |
math/0107011 | All the proofs go through without modification, except for one: we need to check that when we breathe a gentle loop in a MATH-ary puzzle using the loop-breathing REF , we don't introduce any rhombi that cross from one constituent puzzle to the next, for that would remove a boundary edge from a puzzle. But the edges separating constituents are obviously on normal lines to the gentle path, and the loop-breathing construction does not remove these edges. |
math/0107011 | The machinery used here is geometric invariant theory, particularly the ``geometric invariant theory quotients are symplectic quotients" theorem CITE. To begin with, take MATH integral, and consider the graded ring MATH . By the NAME theorem, MATH is a product of three (partial) flag manifolds as an algebraic variety, and from its induced projective embedding inherits a symplectic structure. By NAME 's extension of NAME, it is symplectomorphic to the product of the MATH coadjoint orbits through MATH and MATH. We use the trace form on MATH to identify these with the corresponding isospectral sets of Hermitian matrices. Now consider the invariant subring MATH. By definition, MATH is the geometric invariant theory quotient of this product of three flag manifolds by the diagonal action of MATH. By REF, the NAME function of this variety is the NAME function of the polytope of honeycombs with boundary conditions MATH. In particular its leading coefficient (respectively, its degree), which as for any projective variety is the variety's symplectic volume (respectively, its complex dimension), is also the volume (respectively, real dimension) of the polytope. By the GIT/symplectic equivalence, this GIT quotient by MATH can alternately be constructed as a symplectic quotient by its maximal compact MATH. This construction takes the zero level set of the MATH moment map - here the moment map is the sum of the three matrices - and quotients it by MATH. Combining these results, we find that the symplectic volume (respectively, complex dimension) of the moduli space of zero-sum Hermitian triples is the volume (respectively, real dimension) of the polytope of honeycombs. The same holds for rational triples MATH because both sides behave the same way under rescaling, and then for arbitrary triples because both sides are continuous. |
math/0107013 | Let MATH be a local biholomorphism tangent to the identity at MATH, as in REF , and MATH the germ of a NAME real-analytic hypersurface which is preserved by MATH. By assumption, we have MATH. Thus, by REF , with MATH, applied to the local biholomorphisms MATH and MATH, both sending MATH into itself, we conclude that MATH. Hence each point of the complex hypersurface MATH is a fixed point of MATH. This completes the proof of REF . |
math/0107013 | We have to show that, if MATH and MATH are locally biholomorphic, then MATH and MATH. We introduce the following ordering of the pairs MATH. We write MATH if either MATH, or if MATH and MATH (or, equivalently, MATH). We prove the statement by contradiction. Suppose MATH. We differentiate REF by the chain rule MATH times in MATH and MATH times in MATH, evaluate the result at MATH and use the identities REF . On the right-hand side we obtain a sum of terms each of which has a factor of MATH with MATH. By the assumption MATH, all these derivatives are zero. Similarly, using the fact that MATH for MATH on the left-hand side, we conclude that MATH . Since MATH and MATH, we reach the desired contradiction. We must then have MATH. The opposite inequality follows by reversing the roles of MATH and MATH by considering the inverse mapping MATH. Hence, MATH as claimed. |
math/0107013 | As in the proof of REF , we differentiate REF by the chain rule MATH times in MATH and MATH times in MATH, evaluate the result at MATH and use the identities REF : MATH . By putting MATH and using REF we obtain MATH. Furthermore, the derivative MATH on the right-hand side of REF can be computed by differentiating REF in MATH at MATH: MATH . Note that for MATH, we obtain MATH, that is, MATH is real. After these observations, REF can be rewritten as MATH . We point out that MATH (and similarly MATH) if and only if MATH (in this case, MATH). It can be shown that MATH if and only if MATH is finitely nondegenerate at MATH (see for example . CITE). We will not use this fact in the present paper. It follows, that for MATH, REF reduces to the form MATH . If MATH, in which case MATH, we instead have MATH . Define MATH to be the order of vanishing at MATH of the function MATH. (It is not difficult to verify from REF that MATH. We will not use this property.) Together with REF , it follows from REF , respectively, that MATH. Let us write MATH where MATH and MATH are local biholomorphisms of MATH at MATH. Taking MATH-th roots from both sides of REF respectively we obtain MATH where MATH and the branch of the MATH-th root has to be chosen appropriately. We want to write MATH as a holomorphic function in MATH, MATH, MATH and MATH. For this, observe that there exists a constant MATH such that MATH where we used the principal branch of the MATH-th root near MATH (for MATH small). We substitute REF for MATH in REF and differentiate once in MATH at MATH to obtain MATH in both cases MATH and MATH. Solving REF for MATH and using REF we obtain MATH where MATH is a holomorphic function in its arguments, defined in a neighborhood of the subset MATH such that MATH for any MATH. In particular, we see from REF that the mapping MATH along the NAME variety MATH is completely determined by its first jet at MATH. In fact MATH depends only on the derivatives MATH, MATH, MATH. We shall now determine derivatives of MATH with respect to MATH along the NAME variety MATH in terms of jets of MATH at the origin. We begin with the derivatives MATH (or MATH), MATH. To get an expression involving them we use the same strategy as above, but now we differentiate REF MATH times in MATH and MATH times in MATH and then set MATH. We shall use the notation MATH etc. to denote strings of partial derivatives of positive order. We obtain MATH where the functions MATH and MATH are polynomials in MATH and MATH respectively with holomorphic coefficients in MATH. (More precisely, the coefficients are polynomials in MATH and MATH and in their derivatives.) We claim, however, that no term involving MATH occurs with a nontrivial coefficient in MATH when MATH. Indeed, in this case MATH and MATH. Thus, to obtain REF we differentiate once in MATH and MATH times in MATH. Consequently, no term of the form MATH can appear, as claimed. We observe that from the identity REF with MATH we obtain a polynomial expression for the derivative MATH in terms of MATH and MATH. (Recall that MATH.) After substituting in REF this expression for MATH, the right-hand side of REF for MATH and the right-hand side of REF for MATH, we obtain MATH where MATH is a polynomial in MATH with holomorphic coefficients in MATH. We also observe that the coefficient of MATH on the left-hand side does not vanish identically by the choice of MATH. As before, we get an expression for the derivatives MATH that occur on the right-hand side of REF by differentiating REF in MATH at MATH, this time MATH times: MATH where MATH is a polynomial in MATH with holomorphic coefficients in MATH. We solve REF for MATH in terms of MATH with MATH. Then, by induction on MATH and by REF , we obtain for any MATH an identity of the form MATH where MATH is a polynomial in MATH with holomorphic coefficients in MATH. Note that the term MATH only occurs in REF when MATH. We now substitute the right-hand side of REF for MATH and the right-hand side of REF for MATH in the identity REF to obtain MATH where MATH a polynomial in MATH with holomorphic coefficients in MATH. We claim that an identity of the form MATH holds, where MATH is a polynomial in MATH with holomorphic coefficients in MATH. We prove the claim by induction on MATH. For MATH, there is no occurrence of MATH on the right-hand side of REF . We now would like to divide both sides of REF by the first factor MATH on the left-hand side. We know from REF that this factor does not vanish identically. However, it may happen that it vanishes for MATH. In order to obtain a holomorphic function on the right-hand side after this division, we have to make sure that the vanishing order of the right-hand side with respect to MATH at the origin is not smaller that the vanishing order of MATH. Of course, by REF , this is true for those jet values of MATH that come from a local biholomorphism MATH sending MATH into MATH. On the other hand, we have no information about the vanishing order of MATH for other values of MATH. Hence we may not be able to divide. The idea for solving this problem is to extract the ``higher order part" of MATH that is divisible by MATH. Let MATH be the vanishing order of the function MATH at MATH. Then the vanishing order of the left-hand side of REF is at least MATH. By truncating the power series expansion in MATH of the coefficients of the polynomial MATH on the right-hand side of REF , we can write it in a unique way as a sum MATH such that both MATH and MATH are polynomials in MATH with holomorphic coefficients in MATH, each coefficient of MATH is a polynomial in MATH of order at most MATH with holomorphic coefficients in MATH and each coefficient of MATH is of vanishing order at least MATH with respect to MATH. We now remark that, whenever we set MATH for a mapping MATH satisfying REF , the first polynomial MATH must vanish identically in MATH. Therefore, the identity REF will still hold after we replace MATH by MATH. Now it follows from REF that MATH is divisible by MATH and, hence, we have proved the claim for MATH. The induction step for MATH is essentially a repetition of the above argument. Once REF is shown for MATH, we substitute it for MATH in the right-hand side of REF . Then we obtain a polynomial without MATH and the above argument can be used to obtain REF for MATH. The claim is proved. A formula for MATH similar to REF is obtained by substituting the result for MATH in REF (and also using REF ). The proof of REF is complete. |
math/0107013 | We use the expansion of the function MATH (and similarly for the function MATH, using a MATH to denote corresponding objects associated to MATH) as follows MATH . Recall that MATH is of finite type at MATH if and only if MATH, that is, if MATH for some MATH. We define a positive integer associated to MATH by MATH and an integer MATH associated to MATH by the analogous formula. It follows easily, by setting MATH in REF , differentiating in MATH and then setting MATH as in the proof of REF , that MATH (that is, MATH is a biholomorphic invariant). Indeed, by using also the fact that MATH, we obtain in this way the identity MATH . By differentiating REF with respect to MATH and setting MATH, we also obtain MATH where we have used REF and the fact that MATH. By using the conjugate of REF to substitute for MATH, we obtain MATH where we have used the notation MATH. Observe, by differentiating REF with respect to MATH and setting MATH, that MATH . Since MATH by REF , we conclude from REF that MATH . Also, observe that MATH by the choice of MATH. It follows that MATH for all MATH in any sufficiently small punctured disc centered at REF. In such a punctured disc there exist MATH locally defined holomorphic functions MATH, differing by multiplication by a MATH-th root of unity, such that MATH and MATH for each choice of MATH. Moreover, each local function MATH extends as a multiple-valued holomorphic function (with at most MATH branches) to a neighborhood of MATH in MATH. If we choose MATH in a sufficiently small punctured disc centered at MATH, then also MATH and REF can be written MATH . MATH. Taking MATH-th roots from both sides of REF and substituting the conjugate of REF for MATH we obtain MATH with MATH and the branch of the MATH-th root has to be appropriately chosen. Analogously to the proof of REF , we write MATH where MATH is a constant and where we used the principal branch of the MATH-th root near MATH. We now use REF to rewrite REF as MATH where MATH is a holomorphic function as in REF . Here the problem arises that the values of MATH may differ from MATH and thus the values of the MATH-th root in REF are not uniquely determined. To solve this problem we use a trick to replace the function MATH by MATH . The new function (denote it by MATH instead of the old one) takes the right value MATH for MATH, so that the principal branch of its MATH-th root is defined, and the identity REF still holds for any local biholomorphism MATH of MATH sending MATH into MATH. We substitute REF for MATH in REF and differentiate once in MATH at MATH to obtain MATH . By REF , the coefficient on the right-hand side does not vanish identically. Hence we can solve MATH from REF as a multiple-valued holomorphic function MATH and thus ignore the fact that MATH is constant. We know from REF , however, that, whenever the arguments of MATH and MATH are jets of a local biholomorphism, the ``function" MATH is actually (locally) constant. The different values of MATH are due to the choice of different branches of MATH and MATH. Recall that the values of MATH and MATH may only differ by MATH-th roots of unity. If MATH and MATH are multiplied by the roots of unity MATH and MATH respectively, then MATH is multiplied by MATH. We claim that, for every fixed jet MATH the (multiple-valued) function MATH is uniformly bounded for MATH in some neighborhood of MATH. Indeed, it follows from the construction that the derivative MATH equals the MATH-th root of MATH. Furthermore, by REF , the functions MATH and MATH have the same vanishing order at MATH. Then, taking MATH-th powers of both sides in REF we obtain (single-valued) holomorphic functions of the same vanishing order at MATH (recall that both functions MATH and MATH do not vanish at MATH). Hence MATH is bounded as a ratio of two holomorphic functions having the same vanishing order. The claim is proved. We now substitute REF for MATH into REF and then use REF to obtain MATH . Since MATH for MATH near MATH, we may apply the implicit function theorem and solve for MATH in REF to conclude that MATH where MATH is an (a priori multiple-valued) holomorphic function defined in a domain MATH that contains all points MATH with MATH that are sufficiently close to the set MATH. A value of MATH depends on the values of MATH, MATH and MATH. We have seen that, if MATH and MATH are multiplied by MATH and MATH respectively, then MATH is multiplied by MATH. We now observe, that in fact the identity REF is invariant under this change. Hence we obtain exactly the same value for MATH for all possible values of MATH and MATH. We conclude that MATH is single-valued. A similar expression for MATH is obtained by substituting REF into REF . In summary, we obtain MATH where MATH is a (MATH-valued) holomorphic function in a domain MATH as in REF . Here we conjugated and switched the variables. To complete the proof of REF , we shall proceed as in CITE. We consider the equation MATH and try to solve it for MATH as a function of MATH in a neighborhood of a point MATH close to MATH. (We cannot apply the method of CITE at MATH, since the function MATH in REF may not be defined in a neighborhood of MATH.) We expand MATH in powers of MATH near MATH as follows MATH with MATH for all MATH, since MATH. Let MATH be the smallest integer MATH such that MATH with MATH fixed. The existence of MATH is guaranteed by the finite type condition. Moreover, if MATH is sufficiently small, MATH does not depend on MATH. After dividing REF by MATH and using REF we obtain MATH where MATH. Then, by the implicit function theorem, the equation MATH has MATH solutions of the form MATH, where MATH is a holomorphic function in a neighborhood of MATH with MATH, where MATH is a sufficiently small disc. Hence REF can be solved for MATH in the form MATH for MATH and MATH both sufficiently small. We now substitute REF for MATH in the identity REF to obtain MATH where MATH is a holomorphic function defined for all MATH with MATH in a neighborhood of MATH and MATH where MATH is the domain of definition of the function MATH in REF . It follows from the above description of MATH that MATH is in the domain of definition of MATH whenever MATH and MATH is in a sufficiently small neighborhood of MATH. Let us expand the function MATH in MATH and decompose it as MATH, where MATH . Since MATH is holomorphic in a neighborhood of MATH in MATH, the function of MATH on the right hand side of REF is independent of the value of the MATH-th root, independent of MATH and extends holomorphically to a neighborhood of MATH in MATH. Let us denote by MATH the value of MATH. Since the function MATH, in which we substitute MATH is single valued on MATH, for MATH sufficiently small (depending on MATH with MATH), we conclude that the function MATH is identically MATH. Hence, we must have MATH where MATH. Next, we decompose each MATH uniquely as follows MATH where MATH denotes the order of vanishing of MATH at REF. It is not difficult to see that we have MATH for some small MATH and constant MATH, where MATH denotes the maximum of MATH and MATH for MATH. Hence, the power series MATH defines a holomorphic function whose domain of definition contains any point MATH with MATH and MATH in a sufficiently small neighborhood of MATH. We now wish to show that MATH in REF can be replaced by MATH with MATH. For this, we decompose the function MATH uniquely as MATH, where MATH where MATH is the remainder polynomial in the division REF . Now, observe that MATH, with MATH, coincides with the function MATH. Since the right-hand side of REF is holomorphic in MATH near MATH, it is not difficult to see that MATH must be identically MATH, and that MATH . It remains to remark that the right-hand side of REF is holomorphic in MATH near MATH with any MATH sufficiently small and is independent of MATH. The proof of REF is complete. |
math/0107013 | The proof can be obtained by repeating the arguments from CITE. |
math/0107013 | We write MATH for the right-hand side of REF . For MATH, the proof is rather simple. Expand both sides of REF in powers of MATH and identify the coefficients of MATH. It is not difficult to see that one may solve the resulting equation for the coefficients MATH of MATH in terms of MATH, with MATH, unless MATH is an eigenvalue of MATH. The last possibility does not happen if MATH is sufficiently large and MATH is outside a countable union MATH of proper real-analytic subvarieties. For every MATH and MATH satisfying the hypotheses in the theorem, we conclude that MATH. The required statement follows by continuity and the fact that the complement of MATH is dense in a neighborhood of MATH in MATH. The details are left to the reader. For the rest of the proof we assume MATH. We shall write MATH for the MATH-th derivative of MATH in MATH evaluated at MATH and MATH for the column of the MATH derivatives MATH. We shall differentiate REF in MATH at MATH and apply the chain rule. Since MATH, the derivatives of the right-hand side MATH will be always evaluated at MATH. Hence they will be ratios of real-analytic functions where the denominator is some power of MATH. We consider the ring of all ratios of this kind and all polynomials (and rational functions) below will be understood over this ring (that is, a polynomial below will be a polynomial with coefficients in this ring). By taking the MATH-th derivative REF of the identity REF in MATH, evaluating at MATH, and using the chain rule, we obtain MATH for any MATH, where MATH, MATH is a MATH-valued matrix polynomial in MATH depending only on MATH and MATH is a MATH-valued polynomial in MATH depending on both MATH and MATH. In fact, MATH can be written as MATH with each MATH being a MATH matrix given by the formula MATH . We further fix integers MATH and MATH satisfying MATH, collect the identities REF in blocks for MATH and MATH and write them in the form MATH where MATH is the diagonal MATH matrix with eigenvalues MATH, each of multiplicity MATH, MATH are MATH-valued matrix polynomials in MATH and MATH is a MATH-valued polynomial in MATH depending on both MATH and MATH. Here we put all the terms containing MATH with MATH into MATH. We first try to solve the system REF with respect to the MATH highest derivatives MATH. We can do this provided the coefficient matrix MATH is invertible. In general, a solution can be obtained only modulo the kernel of MATH. Here the dimension of the kernel may change as MATH changes. To avoid this problem we consider only solutions MATH of REF with MATH (as we may by a priori assuming MATH). For these solutions, we obtain from REF an identity MATH where MATH is a MATH-valued polynomial in MATH and MATH. Here two cases are possible. If the kernel of MATH is trivial for some MATH, it is trivial for MATH outside a proper real-analytic subvariety. Then for such values of MATH, REF can be iterated to determine all derivatives MATH, for MATH, in terms of MATH. The proof is complete by continuity. If the kernel of MATH is nontrivial for all MATH, it has a constant dimension for MATH outside a proper real-analytic subvariety. Then we consider the system REF with MATH replaced by MATH. Here the MATH-tuple of new unknown derivatives MATH with the coefficient matrix MATH is involved. However, we can still extract some information when the image MATH is a proper subspace of MATH (which happens precisely when MATH), namely MATH . We now use the explicit form of MATH to conclude that, for every MATH sufficiently large, the matrix MATH on the left-hand side is invertible for MATH and for MATH outside a proper subvariety. By taking the union of these subvarieties for different MATH we see that, for each MATH outside a countable union of proper subvarieties, the matrices REF are invertible for all MATH. We write MATH . Thus MATH is a rational function in MATH. By applying MATH to both sides of REF , we obtain a rational expression for MATH in terms of lower order derivatives modulo the linear subspace MATH . Previously we fixed the first derivatives MATH. In this step, we further assume that MATH. We can then drop the dependence of MATH, MATH and MATH on the derivatives as we did for MATH. Taking also REF into account, we conclude that MATH is determined modulo MATH . If this space is zero-dimensional (for some MATH), the two REF determine MATH completely and yield a polynomial expression for these derivatives in terms of lower derivatives (for this value of MATH). We now observe that we may write MATH, where the matrix MATH tends to the identity (for MATH fixed) as MATH. Moreover the linear operator MATH, where MATH (and the integers MATH in MATH are replaced by a continuous variable MATH in the obvious way), is a ratio of analytic functions for MATH in a neighborhood of MATH. By NAME 's rule, there exist real-analytic functions MATH and MATH that represent bases for MATH and for MATH respectively for every MATH outside a proper subvariety. If we write MATH for the determinant of the matrix of MATH-vectors MATH then the intersection REF is positive dimensional if and only if MATH vanishes at MATH. Since MATH is real-analytic near MATH, we conclude that either MATH is identically REF or there exists MATH such that MATH for MATH and for MATH outside a proper subvariety MATH. In the second case the intersection REF is zero-dimensional for MATH and MATH and we obtain MATH as a polynomial expression in lower derivatives. By a simple inductive argument and the analyticity of MATH and MATH, it follows that MATH for all MATH and all MATH outside the union of MATH, MATH. The desired statement follows by continuity, since the union of MATH's is nowhere dense. Thus, we may assume that MATH. In this case, we consider REF with MATH replaced by MATH. By solving REF , with MATH in place of MATH, for MATH (modulo MATH), we conclude that MATH where MATH and MATH are defined as above. By applying MATH to both sides and substituting the right-hand side for MATH in REF , we deduce that MATH where MATH is as defined above and MATH is the invertible (for MATH large enough and MATH outside a proper subvariety MATH, provided that we have MATH) matrix MATH . As before we assume in this step that MATH and drop the dependence on these derivatives. Now observe that the assumption that MATH (which is equivalent to the intersections REF being nontrivial for all MATH sufficiently large) implies that the space MATH is of strictly lower dimension than MATH. This is a crucial observation. Let us write MATH for the inverse of MATH. It follows that we can solve REF for MATH modulo the subspace MATH . If the intersection MATH is zero-dimensional, we can find a polynomial expression for MATH in terms of lower derivatives by using REF . We claim again that this intersection will be either zero-dimensional for all sufficiently large MATH and MATH outside a proper subvariety MATH or positive-dimensional for all sufficiently large MATH and all MATH. The argument is as before. We can detect positive dimensionality of the intersection REF by the vanishing of suitable determinants formed by the vectors in REF with MATH replaced by MATH, where MATH (which also of course depends on MATH) is defined as follows. Let us factor the scalar MATH in MATH, writing MATH. Then we define MATH, where as before MATH. It is not difficult to see that MATH is analytic in MATH near MATH with MATH equal to the identity. Since any determinant formed by the vectors in REF with MATH replaced by MATH will be analytic, the claim now follows as above. As mentioned above, if the intersection REF is trivial for all sufficiently large MATH, we are done. If not, we must go iterate the procedure above, and start with REF with MATH replaced by MATH. In this way we will obtain a subspace MATH (in a way analogous to that yielding MATH). By the same argument as above, the fact that we are forced to go to the next iteration (that is, the intersection REF is nontrivial for all large MATH) implies that MATH has strictly lower dimension than MATH. The crucial observation is that, if we are forced to make another iteration, the dimension of the subspaces MATH drops. Hence, the process will terminate after at most MATH steps. The details of the iterations are left to the reader. Summarizing, we obtain a linear system for MATH (providing MATH is large enough) in terms of lower order derivatives and the matrix coefficient of MATH is polynomial in MATH and is invertible for MATH outside a countable union of proper subvarieties. Then the proof is completed by the analyticity of MATH and MATH and by continuity as before. |
math/0107013 | Let MATH be as in REF with MATH, for some MATH (which will be specified later). Observe that, for any local biholomorphism MATH sending MATH into itself, the restriction MATH is a local NAME between open neighborhoods of MATH and MATH in MATH. We shall write MATH, MATH. If we take MATH, MATH and MATH in REF , then MATH satisfies REF and hence REF . For a MATH as in REF , we set MATH . REF then imply that MATH for some real-analytic function MATH defined in a neighborhood of MATH. The assumption MATH and REF imply that MATH near MATH. Let us write, for MATH, MATH . We conclude, by the construction of MATH and REF , that MATH for MATH. To complete the proof of REF , consider the functions MATH, for MATH, which satisfy MATH. For MATH, we have MATH and hence both MATH and MATH satisfy the same system of differential equations MATH where MATH, as does any other MATH arising from a CR diffeomorphism MATH with MATH. Recall that MATH is one of the components of MATH. The existence of the integer MATH such that MATH (which is equivalent to MATH) if MATH now follows from REF , although the choice of the integer MATH appears to depend on the mapping MATH. However, we shall show that one can find a MATH that works for every MATH. Let MATH be the integer obtained by the above procedure applied to MATH where MATH is the identity mapping MATH. We conclude that if MATH sends MATH into itself and MATH, then MATH. We claim that the same number MATH satisfies the conclusion of REF for any MATH, MATH. Indeed, for any MATH, MATH as in REF , the mapping MATH sends MATH into itself and satisfies MATH. Hence, by the construction of MATH, we must have MATH which proves the claim. This completes the proof of REF . |
math/0107014 | We define the support MATH of a linear form MATH to be the set MATH. The cardinal number of MATH is called the length of MATH and is denoted by MATH. The length is invariant under the action of MATH, so that the length MATH of MATH is defined as that of a linear form contained in MATH. The bijections MATH preserve length. The proof will proceed by induction on the common length of the MATH in the hypothesis of Lemma. If MATH, then MATH consists of MATH. Applying REF to MATH we see that the statement of Corollary is true in this case. Suppose that the statement is true for all MATH of length less than MATH. If MATH denotes the set of all linear forms MATH, there is a natural injection MATH sending MATH to MATH. Then it is clear that MATH for any MATH. On the other hand, MATH has length less than that of MATH for any bijection MATH fixing elements of MATH unless MATH. In the latter case we have MATH . Let MATH be the totality of linear forms of the form MATH with MATH and MATH. Then, for each MATH and MATH, the linear form MATH either belongs to MATH or has length less than MATH. Moreover each MATH appears exactly once in this way. It is also easy to see that the totality of transforms of MATH with length less than MATH by permutations of MATH fills some orbits in MATH with multiplicities. If this set (with multiplicity) is denoted by MATH, then one has MATH . Since the first and second terms in the right hand side belong to MATH by REF and induction assumption, the left hand side does so. |
math/0107014 | We introduce another variable MATH with MATH and put MATH. Then, fixing MATH and MATH, we consider the function MATH . Note that MATH where MATH . Since MATH for MATH, we have MATH for MATH. We expand MATH with respect to MATH and MATH in the following form MATH . Then we see that there is a family MATH for each MATH such that MATH for MATH and MATH . We further expand each factor MATH in formal power series in the above expression. Then MATH is expanded in an (infinite) sum of expressions of the form MATH, where MATH is an element of MATH and it satisfies MATH. Hence MATH is written in a sum of expressions of the follwing form MATH . Since MATH for MATH, these REF belong to MATH by REF . Hence MATH also belongs to MATH. Therefore, if we specialize MATH to MATH (hence MATH to MATH) in MATH, we get an expansion MATH with MATH. Then MATH with MATH. But MATH does not have negative powers of MATH in its expansion as can be seen by taking a representative MATH of MATH which satisfies REF for each MATH in REF or REF . It follows that MATH belongs to MATH. This finishes the proof of REF for MATH. A similar easier argument shows that MATH belongs to MATH. |
math/0107014 | First we prove REF. The argument follows that of CITE. We repeat their argument for the sake of completeness. Recall that MATH. We regard the both sides of REF as meromorphic functions of MATH and consider the quotient MATH defined for MATH. It is doubly periodic in MATH with respect to the lattice MATH and has no poles. Its value at MATH is MATH. This implies the equality REF. We now prove REF . So suppose MATH and MATH. If MATH then MATH and the equality in REF is REF itself since MATH. Next we consider the case MATH. Since MATH, REF can not be applied. So we proceed in the following way. First observe that MATH . Hence, if we put MATH, we see easily that MATH . We then apply REF to the right hand side and get MATH . This proves REF . |
math/0107014 | In view of REF it is enough to show that MATH regardless of the sign of MATH. Suppose that MATH. Then MATH implies MATH. Since MATH and MATH, we have MATH. NAME that MATH. Then MATH implies MATH. Also MATH implies MATH. But MATH since MATH and MATH. Hence MATH. |
math/0107014 | We look at the coefficient of MATH for MATH and with MATH in REF which is equal to MATH. Noting that MATH is approaching to MATH when MATH approaches to MATH, we obtain REF. |
math/0107014 | Suppose MATH is of the form MATH . Then MATH is equal to MATH, and hence belongs to the image of MATH. Thus REF implies REF . Suppose that MATH and MATH is equal to MATH for any MATH. Then MATH, and hence MATH for MATH. Thus REF implies REF . Suppose MATH for any MATH. Then, by REF , MATH . Hence MATH is of the form MATH with MATH. Thus REF implies REF . |
math/0107014 | We put MATH and MATH. Then we have MATH . By REF for MATH. Hence MATH or MATH . Since MATH is a basis of the free module MATH, we see that MATH for MATH and MATH for MATH. |
math/0107014 | Write MATH. Let MATH be the core of MATH. Then MATH . Since MATH for MATH by the definition of the core and MATH for MATH does not depend on MATH in MATH by REF , MATH does not depend on the choice of MATH in MATH. |
math/0107014 | Let MATH be the core of MATH. Then MATH for MATH and MATH for MATH is independent of MATH in MATH by REF so that MATH for MATH and MATH for MATH is independent of MATH. Hence the sum MATH is a constant depending only on MATH which we shall denote by MATH. We put MATH. Then MATH. By REF MATH is of the form MATH for MATH, where MATH is independent of MATH. Therefore, if we write MATH, then MATH . If we compare this with MATH we see that MATH is of the form MATH and MATH. This shows that MATH depends only on MATH. |
math/0107014 | This follows readily from the fact that MATH is a basis of MATH for each MATH. |
math/0107014 | MATH by REF. But MATH which is independent of MATH. Hence we obtain MATH since MATH. |
math/0107014 | We postpone the proof of REF . As to REF , we take a generic vector MATH such that MATH, which is possible by REF . Since MATH is constant by REF for any MATH, MATH is also constant, which we denote by MATH. Since MATH by REF , MATH must be equal to MATH. |
math/0107014 | The proof is similar to that of REF . Put MATH. We expand MATH in the form MATH . Then MATH is a sum of expressions of the following form. MATH where MATH satisfying MATH. It follows that MATH belongs to MATH by REF . Noting that MATH, we see that MATH can be expanded in the form MATH with MATH. |
math/0107014 | REF is immediate. As was remarked in Note after REF , MATH equals MATH. Therefore, if MATH then MATH, and consequently MATH for all MATH, as is easily seen. Then REF follows. |
math/0107014 | Suppose that MATH is divisible by MATH. Then, by REF MATH considered as a polynomial in MATH has roots at all MATH-th roots of unity other than MATH. Hence it must be divisible by MATH. On the other hand it is a polynomial of degree MATH with constant term MATH by REF . Therefore we must have MATH. Suppose that MATH. Then the same reasoning as above proves REF . If MATH, then MATH is divisible by MATH. Since the constant term and the coefficient (as a polynomial of MATH) of the highest term do not vanish by assumption and REF , MATH must be of the form REF . |
math/0107014 | The equality REF with MATH implies that MATH for all MATH with MATH by REF . This implies that MATH by REF . Then, using REF we see that MATH. Similarly the equality REF with MATH implies that MATH for MATH and MATH for MATH. This implies that MATH by REF , and yields, together with REF , the equalities MATH and MATH in case MATH. |
math/0107014 | By REF the fan associated with MATH has MATH-dimensional cones and MATH-dimensional cones. Such a fan is unique (up to automorphisms of the lattice MATH) and coincides with the fan associated with MATH. Since a toric variety is determined by its fan, MATH must be MATH. |
math/0107014 | The tautological line bundle MATH is a subbundle of MATH, and the tangent bundle along the fibers MATH of MATH is isomorphic to MATH. Hence MATH . Since the tangent bundle MATH is isomorphic to MATH, and MATH, we have MATH . |
math/0107014 | Let MATH be the primitive edge vectors of the MATH-dimensional cones. In view of REF it suffices to show that, under a suitable numbering, they satisfy the relations MATH or MATH . We first deal with the case MATH. From the completeness we see that each MATH-dimensional cone is a face of at least MATH-dimensional cones, and it is a face of at most MATH-dimensional cones because the number of MATH-dimensional cones are MATH. Since the number of MATH-dimensional cones is MATH, we conclude that there are two edge vectors, say MATH and MATH such that MATH spans MATH-dimensional cones with each of remaining vectors MATH, and each MATH spans MATH-dimensional cones with MATH. Thus the projected multi-fan MATH has exactly MATH-dimensional cones. It is complete and non-singular as a projected multi-fan of a complete non-singular multi-fan MATH. It follows that MATH is equivalent to the fan of MATH, and the projected edge vectors MATH satisfy the relation MATH . This implies the relation MATH . Similarly we have MATH . If MATH then MATH since MATH, and MATH lie on a hyperplane MATH. Since the multi-fan MATH is complete and non-singular, the primitive vectors MATH and MATH lie on the different sides of that hyperplane and must satisfy a relation as described in REF . If MATH then MATH and MATH and MATH are linearly dependent primitive vectors. Therefore we must have MATH. Thus REF holds. This proves REF in the case MATH. The case MATH is similar and easier. We see that there are four primitive edge vectors MATH in MATH-dimensional vector space MATH such that MATH . By the same reasoning as above we derive MATH . |
math/0107014 | By REF the fan MATH associated with MATH has MATH-dimensionl cones and MATH-dimensional cones. Moreover the number of MATH-dimensional cones is MATH in case MATH. By REF MATH is equivalent to that of a MATH-bundle over MATH or MATH- bundle over MATH. Among such manifolds those with MATH divisible by MATH are of the form given in REF . |
math/0107014 | Let MATH be a reduced orbifold chart such that MATH is invariant under the action of the torus MATH. The covering group MATH of MATH acts effectively on MATH where MATH was introduced in REF. The isotropy group MATH is isomorphic to the kernel MATH of the natural homomorphism MATH as was shown in CITE. On the other hand MATH and MATH are identified with MATH and MATH. Therefore the kernel MATH is isomorphic to MATH. Hence MATH is isomorphic to MATH. |
math/0107027 | By comparing the stabilizer subgroups of the closed orbits determined by corresponding points under the étale isomorphism it follows that MATH and clearly MATH. Because the equations of MATH are homogeneous there is a MATH-action on MATH (multiplying all matrices by MATH). The limit point MATH of any representation is the trivial representation. Starting from a simple representation MATH, any neighborhood of MATH contains a point determined by MATH for suitable MATH proving MATH. To prove that MATH observe that the set of all points of representation type MATH form an open subset of MATH (follows from the étale local description), whence if MATH is irreducible this set is dense. |
math/0107027 | Assume MATH is a counterexample with a minimal number of vertices. There are at most two directed arrows between two vertices (MATH is not contained) so we can define the graph MATH replacing a pair of directed arrows by a solid edge. Then, MATH is a tree (MATH is not contained). We claim that the component of MATH for every internal (not a leaf) vertex is at least two. Assume MATH in internal and has dimension one, then any non-zero trace MATH along a circuit in MATH passing through MATH (which must be the case by minimality of the counterexample) can be decomposed as MATH where MATH is part of the circuit along a subtree rooted at MATH. But then MATH when evaluated at representations of the preprojective algebra of the corresponding subtree, contradicting minimality of the counterexample. Hence, MATH is a binary tree (MATH is not contained) and even a star with at most three arms (MATH is not contained). If MATH does not contain MATH for MATH as subgraph, then MATH is a NAME quiver and one knows that in this case there are no nontrivial invariants, a contradiction. If MATH is the vertex-simple concentrated in vertex MATH, we claim that MATH for every vertex MATH. Indeed, it follows from CITE that for any non-isomorphic simple MATH-representations MATH and MATH of dimension vectors MATH and MATH we have MATH . Therefore, twice the dimension of MATH at MATH is smaller or equal to the sum of the dimensions of MATH in the two (maximum three) neighboring vertices. Fill up the arm of MATH corresponding to the longest arm of MATH with dimensions starting with MATH at the leaf and proceeding by the rule that twice the dimension is equal to the sum of the neighboring dimensions, then we obtain a dimension vector MATH such that MATH where MATH is the imaginary root of MATH, a contradiction. |
math/0107027 | CASE: There is a point MATH determined by a semi-simple representation MATH of representation type MATH. A neighborhood of MATH is étale isomorphic to a neighborhood of MATH in MATH. It is well known that MATH contains points of representation type MATH whence MATH is a dimension vector of a simple representation of MATH (take a simple of MATH and add zero matrices for the remaining arrows). By REF it follows that MATH. CASE: Let MATH and take a decomposition (representation type) MATH with all MATH, MATH and MATH minimal. Note that we can take all MATH whenever MATH (as then there are infinitely many non-isomorphic simples of dimension vector MATH). As a consequence MATH only has loops at vertices where MATH is equal to one and MATH is a simple root for MATH here we used irreducibility of MATH in order to apply REF . By REF there is a non-loop tame subsetting MATH contained in MATH and if MATH then we have a decomposition MATH which has strictly smaller total number of multiplicities unless MATH . Induction on the total dimension finishes the proof. |
math/0107027 | MATH : We claim that MATH is the unique maximal representation type in the ordering of inclusion in NAME. Assume not and let MATH be another maximal type, then MATH and by REF there is a tame setting contained in MATH but then there are non-loop polynomial invariants, whence MATH is not maximal. MATH : Follows from REF . |
math/0107028 | MATH : Consider the complex moment map MATH where MATH is the subspace of MATH-tuples MATH such that MATH. For MATH such that MATH we consider the element MATH in MATH. The inverse image MATH. By a result of CITE one knows that the geometric points of the quotient scheme MATH are the isomorphism classes of MATH-dimensional semi-simple representations of MATH. Because MATH is a minimal element of MATH all MATH-dimensional representations of MATH must be simple (consider the dimension vectors of NAME components) so each fiber of the quotient map MATH is isomorphic to MATH. The fact that MATH is smooth if MATH is a minimal non-zero element of MATH follows from computing the differential of the complex moment map, see also CITE. Because MATH is a principal MATH-fibration, MATH is an NAME algebra and as the total space MATH is smooth it follows that also the base space MATH is smooth. MATH : If MATH is an NAME algebra, it follows that MATH is a principal MATH-fibration. If in addition the basespace is smooth, so is the top space MATH. The assertion follows from NAME 's characterization of MATH-smoothness, REF . |
math/0107028 | With MATH we will denote the locally closed subvariety of MATH consisting of all geometric points MATH of representation type MATH. Observe that there is a natural ordering on the set of representation types MATH where the closure is with respect to the NAME topology. Clearly, if we can prove that all points of MATH are singular then so are those of MATH. Let MATH be a point outside of the NAME locus of representation type MATH then by REF is suffices to prove that MATH is singular in MATH. Assume that MATH has MATH loops in the vertex MATH where MATH. This means that there are infinitely many nonisomorphic simple MATH-representations of dimension vector MATH, but then MATH where MATH and by the above remark it suffices to prove singularity for MATH. That is, we may assume that the quiver setting MATH is such that the symmetric quiver MATH has loops only at vertices MATH where the dimension MATH. Assume moreover that MATH is indivisible (which by the above can be arranged once we start from a type MATH such that MATH has loops). Recall from CITE that invariants of quivers are generated by traces along oriented cycles in the quiver. As a consequence we have algebra generators of the coordinate ring MATH which is a graded algebra by homogeneity of the defining relations of the preprojective algebra MATH. To prove singularity it therefore suffices to prove that the coordinate ring is not a polynomial ring. Let MATH be the quiver obtained from MATH by removing all loops (which by the above reduction exist only at vertices where the dimension is one). Because the relations of the preprojective algebra are irrelevant for loops in such vertices we have that MATH is a polynomial ring (in the variables corresponding to the loops) over MATH. By CITE we know that MATH is singular, finishing the proof in this case. The remaining case is when MATH contains no loops (that is, all MATH are real roots for MATH) and when MATH is divisible. Because MATH is the dimension vector of a simple representation of MATH we know from CITE that the quiver setting MATH is such that MATH contains a subquiver say on the vertices MATH which is the double of a tame quiver such that MATH where MATH is the imaginary root of this tame subquiver. Consider the representation type of MATH for MATH . If we can show that a points is the MATH-stratum of MATH is singular, then the quotient scheme is singular in the trivial representation and we are done. Consider the quiver MATH, then it has loops in the vertex corresponding to MATH. Moreover, MATH is indivisible so we can repeat the argument above. The fact that MATH was assumed to be divisible asserts that MATH is not the NAME type MATH, finishing the proof. |
math/0107028 | Assume that MATH and of representation type MATH. Then, MATH is smooth in a neighborhood of the closed MATH-orbit of the semisimple representation MATH (closedness follows from CITE). By a result of CITE we know that the normalspace MATH to the orbit in MATH is equal to the space of self-extensions MATH . As a consequence we can identify this space with the representation space MATH where MATH is the quiver on MATH vertices MATH having CASE: MATH loops in vertex MATH, and CASE: MATH directed arrows from MATH to MATH . Moreover, the action of the stabilizer subgroup of MATH (which is MATH) on the normal; space is the basechange action of this group on MATH. By the Luna slice theorem CITE we have a MATH-equivariant étale isomorphism between CASE: a neighborhood of the orbit of MATH in MATH, and CASE: a neighborhood of the orbit of MATH in the principal fiber bundle MATH . Combining this étale description with the one from REF we deduce an étale MATH-isomorphism between the representation scheme MATH (in a neighborhood of the trivial representation) and the representation space MATH (in a neighborhood of the trivial representation). But then MATH and in CITE it was shown that for a preprojective algebra the smooth locus coincides with the NAME algebra. The only way the trivial representation can be a simple representation of MATH (or indeed, even of MATH) is when MATH has only one vertex and the dimension vector is MATH. But then the representation type of MATH is MATH finishing the proof. |
math/0107030 | Since the MATH - action on MATH is induced from the action on the image of the curve in MATH (and which is discarded by the map MATH), this is obvious. |
math/0107030 | If MATH then the codimension of MATH is at least one. Therefore MATH since MATH. |
math/0107030 | First of all notice that MATH factors through the projection map MATH to the factor corresponding to the vertex MATH. Hence it suffices to consider the behaviour of cotangent line bundles under MATH, forgetting the last MATH points. This map MATH again factors into MATH . We will prove the statement for the map MATH - the Theorem then follows by induction and NAME 's Theorem REF . So we want to show that (in the case MATH) MATH . It is well - known (see for example, CITE) that MATH where MATH is the divisor in MATH where the MATH - bubble contains the marked points MATH and MATH. Now note that MATH is constant for any MATH, hence MATH. This yields REF . |
math/0107030 | By NAME 's REF , it suffices to consider MATH, MATH. In this case, by REF , the left hand side of REF equals MATH . We therefore have to prove that MATH . For the case that MATH is different from MATH, we have shown in the proof of REF (see CITE) that MATH . Hence we only have to consider the case when MATH; since it is very similar to the previous case, we will only outline its proof. As in CITE, let MATH. Let MATH, MATH and MATH. We will also write MATH instead of MATH. Note that for the vertex MATH, we always have MATH. Therefore MATH which finishes the proof. |
math/0107030 | Suppose that for given MATH, MATH and MATH, the invariant REF is non - zero. Since MATH, we must have MATH. However, the real dimension of MATH is equal to MATH, hence MATH, and therefore MATH. Now suppose MATH is zero and therefore MATH as well, that is MATH is a multiple of the trivial class MATH. If MATH as well, then the invariant is zero since it just equals the cup product of the cohomology classes MATH. So suppose MATH. Then MATH where the MATH run over a basis of MATH. Since MATH, at least one of MATH has to be positive. On the other hand, we have MATH . So as soon as MATH the corresponding three - point invariant in the sum above is zero. This proves the lemma. |
math/0107030 | Suppose that for given MATH, MATH and MATH, the invariant REF is non - zero. Then, by the previous Lemma, MATH. Hence we have to consider homology classes MATH, and MATH being a multiple of the class of top degree, say MATH. The MATH - skeleton of the moment polytope of MATH has the following edges: CASE: for MATH, edges between MATH and MATH, and between MATH and MATH, all having homology class MATH; CASE: for MATH or MATH, an edge between MATH and MATH of homology class MATH; CASE: for MATH, an edge between between MATH and MATH of homology class MATH. The reader will now easily convince himself that there is no simple graph MATH in this class such that MATH, MATH, MATH, MATH and MATH all have non - zero equivariant NAME class on MATH, unless MATH. Finally note that MATH as cohomology classes, which finishes the proof. |
math/0107030 | The only simple graph MATH such that MATH, MATH, MATH and MATH all have non - zero equivariant NAME class on MATH is the one with one edge between MATH and MATH where all but the last marked point are at MATH. Applying the formula derived in this note yields the following for the NAME - NAME invariant MATH: MATH where we have used MATH for MATH by a similar argument as in CITE, that is by applying CITE. |
math/0107030 | The relation MATH follows directly from the previous three Lemmata. Now let us consider the other multiplicative relation: MATH . Hence we obtain MATH and by comparing the coefficients of the MATH, the relation follows. |
math/0107031 | Let MATH and MATH, MATH. Choose a basis MATH for MATH so that MATH is a basis for MATH. The MATH-matrix MATH has the following block structure in this basis: MATH . Here MATH is a skew-symmetric matrix of order MATH and MATH is a rectangular MATH-matrix. Consider the block matrix of order MATH that depends on a parameter MATH: MATH . We look at the rank of MATH in two different ways, which gives us the required inequality. CASE: Since MATH, we have MATH. Also, up to a permutation of columns, the rectangular matrix MATH is nothing but MATH. Thus, MATH. CASE: If MATH, then a chain of elementary transformations brings MATH in the following form: MATH . Therefore if MATH, then MATH . Thus, MATH . From the semi-continuity of the rank, it follows that MATH for MATH, which yields the required inequality. |
math/0107031 | Take MATH and then MATH or MATH. |
math/0107031 | Fix a vector space direct sum MATH. Choose MATH so that MATH is maximal. For any MATH, one has MATH. If MATH, then dimension of the Right-hand side is equal to MATH. Consequently, for all but finitely many MATH's, we obtain MATH. |
math/0107031 | Apply the previous Proposition to MATH. Here MATH. |
math/0107031 | Since MATH is regular, MATH. Hence MATH. |
math/0107031 | CASE: Let MATH be the simple ideals in MATH and MATH, where MATH and MATH. Since MATH, it suffices to prove that MATH whenever MATH is simple. CASE: Consider MATH as MATH-module: (MATH-REF) MATH. Here MATH is the multiplicity in MATH of the respective irreducible MATH-module. Restricting further to MATH, we obtain: MATH . A highest weight vector in MATH relative to MATH is nothing but a homogeneous element of MATH. Since each MATH lies in MATH, any highest weight vector in MATH relative to MATH is also a highest weight vector relative to MATH and hence relative to MATH. This means that the above two decompositions have ``the same" irreducible constituents. In other words, MATH is an irreducible MATH-module for any MATH-tuple MATH occurring in REF . This clearly implies that any such MATH-tuple contains at most one nonzero coordinate. Let MATH REF denote the sum of all nontrivial MATH-submodules, that is, the sum of all summands in REF , where MATH. Since each MATH-tuple contains at most one nonzero coordinate, MATH, the sum of vector spaces. Note that MATH (MATH), because any nonzero element would generate a MATH-submodule with a nontrivial action of both MATH and MATH. Denoting MATH, we see that MATH is a reductive NAME subalgebra. If MATH is a reductive subalgebra of a simple NAME algebra MATH, then the subalgebra MATH generated by MATH equals MATH. Indeed, MATH is stable relative to both MATH and MATH, so that it is an ideal in MATH, see CITE. Apply this to MATH, for arbitrary MATH. Here MATH, that is, it is nonzero, if MATH. Then the relations of the previous paragraph show that the subalgebra generated by MATH lies in MATH, that is, it does not contain MATH. Thus, MATH cannot be simple, if MATH. |
math/0107031 | This is proved, although not stated explicitly, in CITE. |
math/0107031 | Let MATH be a NAME subalgebra of MATH. Consider the reductive NAME subalgebra MATH. It is clear that MATH is the centre of MATH, that is, MATH. By the construction of MATH, we have MATH. Furthermore, MATH, since MATH. Finally, the very definition of MATH implies that MATH is distinguished in MATH and therefore MATH is concentrated in positive even degrees. |
math/0107031 | Without loss of generality, we assume that MATH is simple. Let MATH be a MATH-triple. Recall from REF the non-negative grading on MATH determined by MATH: MATH . Take any MATH. By REF , MATH commutes with MATH and, by the assumption, with MATH. From the MATH-theory it follows that the NAME subalgebra generated by MATH and MATH is MATH. Hence MATH lies in the centre of MATH, that is, MATH. This means that MATH is distinguished and therefore is even. By REF , we have MATH for all MATH. Then invoking REF , we conclude that (MATH-REF) MATH. This condition for a distinguished nilpotent element MATH was considered by NAME in CITE. Let MATH and let MATH be a nonzero element in MATH. NAME proved that MATH is a regular semisimple element in MATH whenever REF is satisfied, see REF in CITE Claim For every MATH there exists a unique MATH such that MATH. As MATH is distinguished, MATH is nilpotent and therefore MATH. Therefore MATH, if MATH. The uniqueness of MATH follows from the fact that MATH contains no nonzero nilpotent elements. So, it remains to prove that such a MATH exists. Consider the equation on MATH . (Note that MATH in view of our assumption on MATH and MATH.) Since MATH is Abelian, MATH is orthogonal to MATH with respect to the Killing form on MATH. Therefore MATH, that is, the above equation has a solution. Because MATH and MATH, we also deduce that there exists a solution MATH lying in MATH, as required. It follows from the Claim that MATH, which completes the proof of REF . |
math/0107031 | From REF , we conclude that MATH. Since MATH is even, we are done. |
math/0107031 | Consider the MATH-grading of MATH associated with a MATH-triple MATH. We have MATH and MATH, where MATH is a reductive subalgebra of MATH. Consider MATH as element of MATH, using the Killing form, and compute its stabiliser MATH. It readily follows from REF that MATH. Combining REF yields MATH where MATH is a regular element in the MATH-module MATH. Thus, it suffices to prove that the last expression is equal to MATH. Note that MATH is the isotropy representation for the affine homogeneous space MATH. This implies that MATH as MATH-modules and there exists an open subset MATH such that the stabilisers MATH REF are reductive and conjugate in MATH, see CITE. Therefore we may assume that MATH is such a reductive generic stabiliser. By CITE, MATH is a `symmetric' subalgebra of MATH whenever MATH. In particular, MATH is a spherical homogeneous space. In this case, one knows that MATH (see CITE). Here MATH. Since MATH is an orthogonal MATH-module, the field of invariants MATH is the quotient field of MATH. Therefore MATH and, since MATH, we conclude that MATH. |
math/0107031 | Choose a basis MATH for MATH so that MATH is a basis for MATH, MATH. Then MATH is a basis for MATH. Write the matrix MATH in this basis. It has the following block structure: MATH, where MATH (respectively, MATH) is a square matrix of order MATH (respectively, MATH). Notice that some fragments of MATH are MATH-matrices in its own right. Indeed, MATH and MATH. It follows that MATH and MATH, in view of REF . REF a The block structure of MATH shows that MATH which yields the first inequality. The second inequality stems from REF b Applying REF to MATH and MATH, we obtain (MATH-REF) MATH. Combining this with REF yields the desired inequality. CASE: The hypothesis means that MATH, that is, MATH is non-singular. It also implies that MATH. Hence MATH which together with REF gives the second equality. To obtain the first equality, notice that MATH and therefore MATH. Thus, MATH which completes the proof. |
math/0107031 | Set MATH. It is the uneffectivity kernel for the MATH-action on MATH. Letting MATH, we conclude from REF that MATH, where MATH is a REF-dimensional torus and MATH is the unipotent radical of MATH. Of course, we choose MATH so that all MATH's are eigenvectors of it. Considering the MATH-action on MATH, we may identify MATH, the NAME algebra of MATH, with the space MATH. Letting MATH, we obtain a MATH-stable complete flag MATH. CASE: Suppose conditions MATH are satisfied. Obviously, these conditions mean that MATH for all MATH. In other words, MATH is dense in MATH. More precisely, we have MATH and MATH. Thus, MATH, and finiteness follows. CASE: Conversely, suppose MATH has finitely many orbits in MATH. Then each MATH must contain a dense orbit. This already implies that all MATH's to be different. Indeed, assume that MATH for some MATH. Let MATH be a generic element of MATH. Then MATH is a MATH-invariant rational function on MATH, and therefore a dense orbit cannot exist. Thus MATH for all MATH. Next, let MATH be such that MATH. Clearly MATH whenever MATH. Writing the condition that the vectors MATH REF are linearly independent, we obtain exactly conditions MATH with fixed MATH. |
math/0107031 | Let MATH be a nilpotent matrix and let MATH be a MATH-triple. Then, denoting by MATH the usual MATH matrix power, one easily proves by induction that MATH . Then MATH . This equality is an incarnation of REF MATH for classical NAME algebras. Indeed, it is well known (see for example, CITE) that, for MATH, MATH is the span of all nonzero powers MATH. For MATH and MATH, the same holds with the odd powers of MATH. This proves REF . For MATH, it is known that MATH is the span of all odd powers if and only if the partition of MATH has at least three parts, whence REF . REF is obvious. The conclusions REF follows by REF. |
math/0107031 | CASE: The element MATH is being determined by the relation MATH . This already shows that MATH. Letting MATH and MATH, we obtain MATH. On the other hand, MATH, where MATH are defined by the equality MATH. Hence all MATH, Since MATH is regular, we have all MATH. It follows that all MATH, too. Thus, MATH is regular as well. REF applied in our situation gives the following: For any MATH there exists a unique MATH such that MATH. Since MATH, this construction gives the whole centraliser of MATH. Thus, each MATH has only `positive' and `negative' parts; MATH, where MATH and MATH. The totality of all positive parts forms MATH. Because MATH is Abelian, the totality of all negative parts forms an Abelian subalgebra of dimension MATH. Since MATH is one of such negative parts and it is regular, the totality of all negative parts is precisely MATH. |
math/0107031 | Take any MATH such that MATH. Writing MATH (in the obvious notation), we obtain MATH. Set MATH. Then MATH. Since MATH, the space MATH is the complexification of a space in the compact real form MATH. Thus, MATH is nondegenerate on MATH and hence on MATH. |
math/0107031 | Since both MATH and MATH are regular nilpotent elements in MATH, there exists a unique MATH such that MATH. To order to define MATH, we used the relation MATH. Because MATH, we have MATH as well, and therefore MATH. Since MATH, we conclude by REF . |
math/0107031 | Set MATH and MATH. We begin with proving that MATH has the required dimension, that is, MATH. Since MATH is regular semisimple, MATH. Hence MATH. Because MATH is Abelian, we see that MATH is orthogonal to MATH with respect to the Killing form. In other words, MATH and hence MATH. Consequently, MATH, as MATH. Using the fact the NAME subalgebra MATH is ``diagonally" embedded in MATH, we obtain MATH . It follows that MATH. Given MATH, we have MATH. Since MATH, each nonzero element of MATH is a sum of a nonzero element in MATH and an element in MATH. Thus, the validity of REF follows from that of REF . |
math/0107031 | Using the Killing form, each MATH can be considered as element of the dual space MATH. Let us describe a convenient model for MATH. In our situation, MATH. It follows that MATH. Since MATH (which actually holds for any distinguished element), we may identify MATH with MATH. The MATH-module structure on MATH is given by the usual NAME bracket in MATH coupled with the subsequent projection to MATH, with kernel MATH. Using the notation of the proof of REF , we have MATH. It is already proved therein that MATH, MATH, and MATH. Hence MATH. Thus, we only need to show that MATH. As was noted in the proof of REF , a stronger condition holds for-MATH; namely, MATH. Next, we can restate REF as follows: MATH induces a nondegenerate pairing between MATH and MATH. Using the invariance of MATH, this also yields a nondegenerate pairing between MATH and MATH. It follows that MATH. Since MATH, we see that MATH. This means that the image of MATH in MATH is of dimension MATH, which completes the proof. |
math/0107031 | The symmetricity of MATH is obvious. By REF , MATH is a basis for MATH. The matrix MATH in this basis is MATH. This proves REF . By REF , MATH remains nonsingular after having been evaluated against MATH. Hence MATH is nonsingular. Furthermore, MATH vanishes precisely on the complement of the open MATH-orbit in MATH. In our situation, MATH is the unique weight vector of minimal weight with respect to MATH. Let MATH be the MATH-stable complementary hyperplane. Since MATH is a semi-direct product of a REF-dimensional torus and a unipotent group, it is easily seen that MATH (compare proof of REF ). Because MATH is the annihilator of MATH, we are done. |
math/0107033 | We show that for all MATH and MATH it holds that MATH is independent of MATH. This would imply the claim of the Lemma. We denote by MATH the children of MATH, and by MATH the children of MATH. We then have MATH . Note that REF implies that if MATH is set of nonzero elements and MATH, then the uniform distribution measure on MATH satisfies for all MATH that MATH . Thus the probability at the right hand side of REF : MATH does not depend on MATH. Now the claim follows. |
math/0107033 | We note that if MATH, and MATH, then MATH. Similarly, if MATH, then MATH. Thus, for all MATH we have MATH and MATH. Relations REF now follows. Looking at REF one sees that the elements of MATH which satisfy MATH are exactly those elements which satisfy the equation MATH where MATH are some none zero constants. Thus the elements MATH of MATH which satisfy MATH and MATH are exactly those elements of MATH which satisfy the equations MATH and MATH . Note that by REF the number of solutions of these equations, MATH, in MATH is the same as the number of solutions of these equation over MATH in the variables MATH. It now follows that for every permutation MATH and MATH: MATH . Moreover, when MATH, MATH . Now, MATH as needed. |
math/0107033 | We will write MATH to denote that MATH and MATH are indistinguishable, and MATH when there exist MATH such that MATH. It is clear that if MATH, then MATH. On the other hand, suppose that MATH. We will show that there exist a MATH such that MATH . This would imply that indeed MATH (see REF ). We may write REF equivalently, MATH . Writing MATH, we may write the left hand side of REF as: MATH . Summing over all MATH with MATH in REF we obtain: MATH . (by REF we may write MATH for MATH when MATH). Since MATH there exists MATH such that MATH. Thus: MATH where we have used the fact that by REF MATH. It now follows that there exists a MATH for which REF holds. |
math/0107033 | If follows from the NAME theorem that MATH where MATH is a normal variable which does not depend on MATH. The claim follows, as for all MATH we can couple both MATH and MATH with the same normal variable MATH in such a way that MATH for both MATH and MATH. |
math/0107033 | Since MATH is irreducible and aperiodic it follows that there exists a MATH such that the matrix MATH is strictly positive. Thus, for every two states MATH and MATH we may construct a MATH level tree such that the root of the tree is labeled by MATH and all of the leaves are labeled by MATH. Let MATH and MATH. Suppose that MATH. Then there exists a labeling of the tree of MATH levels which has MATH at the root and MATH at level MATH, and such the census of MATH is MATH. We will prove the lemma by constructing a labeling where the root is labeled by MATH, level MATH is labeled by MATH, and the census of MATH is MATH. We denote by MATH the vertices at level MATH. For MATH we define MATH to be the census of the subtree rooted at MATH for the labeling MATH. Let MATH be a non-negative vector which satisfies MATH. Define MATH. Note that if MATH then there exists a labeling of the MATH levels tree, denoted MATH, which has MATH as its census and such that the label of the root for this labeling is MATH. Since MATH it follows that MATH for some integer valued vector MATH. Note that if for root value MATH, we could label MATH of the vertices of level MATH by labels MATH with multiplicity MATH, then we are done. Since using these vertices, it is possible to build the MATH part of the census. Now using the other vertices at level MATH it is possible to build the MATH part of the census (whatever labels these vertices have). Note that MATH . Therefore by assigning MATH of the vertices at level MATH the task of producing the prescribed labels at level MATH we obtain the required result. |
math/0107035 | MATH is REF . MATH follows from the NAME - NAME theorem CITE which says that MATH is quasi-isometric to MATH, together with the fact that MATH is quasi-isometric to MATH (see REF ). MATH is immediate. |
math/0107035 | The intuition behind the proof is that the geodesic flows on MATH and MATH are topologically conjugate; to put it another way, for each geodesic MATH on MATH, the connection on MATH moves MATH to a MATH - bilipschitz path in MATH and hence that path is close to a geodesic in MATH. In order to carry this out uniformly up in the geodesic foliation bundle MATH we shall apply structural stability tools from the theory of hyperbolic dynamical systems, as encapsulated in the NAME REF. A MATH - dimensional foliation of a subset of MATH has uniformly smooth leaves if the leaves are defined locally by immersions from open subsets of MATH into MATH such that for each MATH the partial derivatives up to order MATH are uniformly bounded away from zero and from infinity. A foliation of a smooth manifold MATH has locally uniformly smooth leaves if MATH is covered by coordinate charts in each of which the leaves are uniformly smooth. The property of (locally) uniformly MATH leaves is similarly defined by omitting the words ``for each MATH". Let MATH be the dimension of MATH. There exists a unique, MATH dimensional foliation MATH of MATH with locally uniformly smooth leaves such that MATH is transverse to the fibers of MATH, and the foliation of MATH obtained by intersecting MATH with the fibers of MATH is identical to the foliation by geodesic flow lines. Before proving the claim we apply it to prove REF . Since the conclusion of the lemma is local, it suffices to prove it when MATH is a MATH - lipschitz, MATH - cobounded affine arc MATH. Choose a compact set MATH such that any MATH - lipschitz, MATH - cobounded affine arc MATH may be translated by MATH to lie in MATH. Let MATH be the space of all MATH - lipschitz affine arcs MATH. The conclusion of the lemma is invariant under the action of MATH and so we may assume MATH. By enlarging MATH we may assume that MATH is a smooth codimension REF submanifold of MATH. By restricting MATH to MATH we obtain a hyperbolic surface bundle MATH and its geodesic flow bundle MATH; fix a smooth Riemannian metric on MATH. Let MATH be the restriction of the foliation MATH to MATH. For each MATH, the foliation MATH restricts to a REF - dimensional foliation MATH of MATH with uniformly smooth leaves, transverse to the fibers of MATH. Also, the Riemannian metric on MATH restricts to a smooth Riemannian metric on MATH. There is a unique vector field MATH on MATH which is tangent to MATH and is perpendicular to the geodesic flow lines, such that each MATH projects to a positive unit tangent vector in MATH. By uniqueness the foliation MATH is invariant under the antipodal map on MATH, implying that MATH is antipode invariant on MATH. Assuming as we may that the Riemannian metric on MATH is also antipode-invariant, it follows that MATH is antipode-invariant, and so descends to a vector field on MATH. This vector field spans the desired connection on MATH. Note that the connection on MATH is uniformly smooth along leaves of MATH, and as MATH varies over MATH the connection varies continuously; this follows from leafwise uniform smoothness of MATH. Let MATH be the connection flow on MATH, which has connection maps MATH, that is, MATH for MATH. To prove that MATH is MATH bilipschitz, it suffices by a standard result in differential equations to prove that MATH . Note that MATH is continuous as a function of MATH. Since MATH is compact, MATH has a finite upper bound MATH, and so MATH is the desired bilipschitz constant. Further compactness arguments show that the connection lines project to MATH - quasihorizontal lines in MATH for uniform MATH, and that the uniform continuity clause holds. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.