paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0012080 | Applying the NAME inequality one gets MATH . Combining REF with REF and taking the obvious inequality MATH into account one gets MATH . To complete the proof it remains to apply REF and note that MATH (see the proof of REF ). |
math/0012080 | CASE: By REF MATH . On the other hand by REF either MATH or MATH . Since maximum values of the formal deficiency indices are attained only simultaneosly, one gets MATH . REFThis assertion is a special case of REF . |
math/0012082 | The inclusion MATH is clear, as a relevant element MATH is also relevant after base change. So suppose that a homogeneous element MATH of degree MATH is relevant. Then we have a factorization MATH into homogeneous factors MATH of degree MATH such that the degrees MATH generate a subgroup of finite index. For MATH we can write MATH with suitable index sets MATH, homogeneous elements MATH of degree MATH and elements MATH. Therefore MATH . For every choice MATH, the element MATH is a product of elements of degrees MATH and so it is a relevant element in MATH. Therefore MATH and MATH belongs to its radical. |
math/0012083 | Let MATH be any point of MATH, and apply MATH to the exact sequence MATH to get MATH . Since the extension is non-split, MATH, and therefore MATH . By NAME duality, MATH. Since MATH we conclude that MATH. We read this as saying that, up to multiplication by a constant, there is a unique morphism of MATH into MATH. Assume MATH, and let MATH be any homomorphism. Compose MATH with the inclusion MATH to get a homomorphism MATH. Since there is a unique map (up to scalars) MATH, MATH can be restricted to a map MATH that fits in the diagram MATH . Since MATH is a line bundle, MATH can be either an isomorphism or zero. If MATH is an isomorphism, then the snake lemma shows that MATH, which is impossible because MATH is torsion-free. Therefore MATH must be zero, and then MATH is zero as well. We conclude that for MATH we have MATH . Note that MATH is locally free at MATH, so a local computation shows that MATH . Applying MATH to the exact sequence MATH gives MATH for all MATH. Since MATH is NAME with trivial dualizing sheaf, and MATH is locally free, NAME duality shows that this last group is REF for MATH, and is isomorphic to MATH for MATH. Any non-trivial homomorphism MATH would give by composition a non-trivial one MATH, contradicting our earlier result. We conclude that for MATH, MATH for all MATH. Now let MATH be any point of MATH, and apply MATH to the exact sequence MATH to get MATH . Since the first term is obviously non-trivial, we conclude that MATH . |
math/0012083 | The result follows at once from REF , the fact that the projective dimension of MATH is MATH, and the following lemma. |
math/0012083 | (An adaptation of the proof of CITE.) We have MATH by the adjunction of MATH and MATH. Furthermore, MATH by the projection formula. Since MATH is smooth at MATH, writing down the NAME resolution for MATH and pulling back via MATH we get a free resolution of MATH on MATH which can be used to compute MATH. This gives MATH where MATH. Now the hypercohomology spectral sequence MATH proves the result. |
math/0012083 | CITE. |
math/0012083 | Let MATH be the relative moduli space of semistable sheaves of rank REF, degree MATH on the fibers of MATH, computed with respect to a polarization MATH of fiber degree MATH. (For the existence of such a polarization see CITE.) This moduli problem is fine (this is apparently a known fact, CITE; or one can prove it directly, CITE), and using the results in CITE one concludes that MATH is smooth, and the universal sheaf on MATH induces an equivalence of derived categories MATH . It is easy to see that MATH is again a generic elliptic threefold: MATH and MATH are locally isomorphic over open sets in MATH, by an analysis similar to that of REF, simplified by the fact that there are no contractions caused by semistable sheaves. Thus MATH and MATH have the same singular fibers, and their discriminant loci are the same, therefore MATH is a generic elliptic threefold. The Jacobian of MATH can be identified in a natural way with MATH, and therefore we get an isomorphism MATH where MATH is the element of MATH that corresponds to MATH. A computation similar to the one in REF shows that the restrictions of MATH and MATH along MATH are the same, so by REF we must have MATH in MATH. Therefore, we have MATH . |
math/0012086 | We will not define the NAME or NAME representations, but we will use some of their basic properties, all of which can be found in CITE. The NAME representation of MATH can be decomposed into irreducible summands, one corresponding to each NAME diagram with MATH boxes. Let MATH denote the representation corresponding to the NAME diagram MATH. The NAME representation is the sum of those MATH for which MATH has one or two rows. The NAME diagrams with MATH boxes are MATH, MATH, MATH, MATH and MATH. The MATH which lie in the NAME representation are as follows. CASE: MATH is one-dimensional. CASE: MATH is the NAME representation. CASE: MATH can be defined by composing the NAME representation of MATH with the map from MATH to MATH given by MATH, MATH, and MATH. The NAME diagram MATH is a reflection of MATH. Reflection of the NAME diagram has the effect of substituting MATH in the corresponding representation. It is shown in CITE that the kernel of the NAME representation is invariant under this substitution. Thus the kernel of MATH is the same as that of the NAME representation. Finally, MATH is one-dimensional. If the NAME representation of MATH is faithful then so are the NAME and NAME representations. Conversely, suppose MATH is a non-trivial braid in the kernel of the NAME representation of MATH. Consider the commutator MATH. This lies in the kernel of MATH, and hence MATH. Since this is a commutator, it lies in the kernel of any one-dimensional representation. Since MATH is central in MATH, it also lies in the kernel of the representation corresponding to MATH. Thus it lies in the kernel of the NAME and NAME representations. It remains to show that MATH must be non-trivial. This is not difficult, but would take us too far afield. We therefore omit this part of the proof. |
math/0012086 | We can assume that the tine edges of MATH and MATH both intersect MATH transversely. Let MATH be the lift of MATH to MATH which contains MATH. Let MATH be the corresponding lift of MATH. Then MATH intersects MATH transversely for any MATH. Let MATH denote the algebraic intersection number of these two arcs. Then the following definition of MATH is equivalent to REF . MATH . Suppose MATH goes from MATH to MATH. Let MATH and MATH be disjoint small regular neighbourhoods of MATH and MATH respectively. Let MATH be a subarc of MATH which starts in MATH and ends in MATH. Let MATH be a loop in MATH based at MATH which passes counterclockwise around MATH. Similarly, let MATH be a loop in MATH based at MATH which passes counterclockwise around MATH. Let MATH be the ``figure eight" MATH . Let MATH be the lift of MATH which is equal to MATH outside a small neighbourhood of the puncture points. Then the following definition of MATH is equivalent to REF . MATH . Note that MATH is a closed loop in MATH. Since MATH is in the kernel of the NAME representation, the loops MATH and MATH represent the same element of MATH. They therefore have the same algebraic intersection number with any lift MATH of MATH. Thus REF will give the same result for MATH as for MATH. |
math/0012086 | Apply an isotopy to MATH so that MATH intersects MATH at a minimum number of points, which we denote MATH (in no particular order). Recall the definition given in REF . MATH . If MATH then clearly MATH. We now assume that MATH and prove that MATH. By applying a homeomorphism to our picture, we can take MATH to be a horizontal straight line through MATH with two puncture points above it and one puncture point below it. (The noodle has been pulled straight and the fork is twisted!) Let MATH and MATH be the upper and components of MATH respectively. Relabel the puncture points so that MATH contains MATH and MATH and MATH contains MATH. Consider the intersection of MATH with MATH. This consists of a disjoint collection of arcs which have both endpoints on MATH, and possibly one arc with an endpoint on MATH. An arc in MATH which has both endpoints on MATH must enclose MATH, since otherwise it could be slid off MATH to reduce the number of points of intersection between MATH and MATH. Thus MATH must consist of a collection of parallel arcs enclosing MATH, and possibly one arc with an endpoint on MATH. Similarly, each of the arcs in MATH either enclose one of the puncture points MATH or MATH, or have an endpoint on one of MATH or MATH. There can be no arc in MATH which encloses both MATH and MATH, since the outermost such arc together with the outermost arc in MATH would form a closed loop. An example of a noodle and a tine edge in MATH is shown in REF . We have omitted the handle of the fork, which plays no role in our argument. Let MATH and MATH be two points of intersection between MATH and MATH which are joined by an arc in MATH or MATH. This arc, together with a subarc of MATH, encloses one puncture point. Thus MATH . Also, MATH intersects MATH with opposite signs at MATH and MATH, so MATH . Thus MATH . Proceeding along MATH, we conclude that the values of MATH are the same for all MATH. Thus MATH evaluated at MATH is equal to MATH. Thus MATH is not equal to zero. |
math/0012086 | Let MATH be a homeomorphism which represents an element of the kernel of the NAME representation. We will show that MATH is isotopic relative to MATH to the identity map, and so represents the trivial braid. Let MATH be a noodle. As before, take MATH to be a horizontal line through MATH such that the puncture points MATH and MATH lie above MATH and MATH lies below MATH. Let MATH be a fork such that MATH is a straight line from MATH to MATH which does not intersect MATH. Then MATH. By the Basic Lemma, MATH. By the NAME Lemma, MATH is isotopic to an arc which is disjoint from MATH. By applying an isotopy to MATH relative to MATH, we can assume that MATH. By a similar argument using different noodles, we can assume that MATH fixes the triangle with vertices MATH, MATH and MATH. Thus MATH must be some power of MATH, the NAME twist about a curve parallel to MATH. It is easy to show that the NAME representation of MATH is the scalar matrix MATH. Thus the only power of MATH which lies in the kernel of the NAME representation is the trivial braid. |
math/0012086 | Let MATH be the tine edge in standard form as shown schematically in REF . The endpoints of MATH at MATH and MATH are shown. Segments of MATH are labelled with numbers to indicate the number of parallel copies required. A laborious computation or a short computer program can be used to check that MATH up to multiplication by a power of MATH. Both MATH and MATH are roots of this polynomial. |
math/0012086 | Put MATH and MATH in the standard form as in REF . Thus MATH is a horizontal straight line with MATH and MATH above it and MATH below it. Assume that MATH is the left endpoint of MATH. We show that the lowest power of MATH in MATH occurs with coefficient MATH. The highest power of MATH and the case where MATH is the right endpoint of MATH are handled similarly. Let MATH be the points of intersection between MATH and MATH. Recall REF , which states that MATH . Let MATH be such that MATH is minimal. We will show that there is only one such MATH. We proceed by induction MATH. The case MATH is trivial, so assume MATH. If MATH were to the right of MATH then there would be a subarc of MATH going from MATH around MATH in the clockwise (negative) sense to intersect MATH at a point MATH. Then MATH, which contradicts the minimality of MATH. Thus MATH must lie to the left of MATH. Let MATH be a vertical line from the top of the disk to a point on MATH between the puncture points MATH and MATH such that MATH does not intersect MATH. If MATH were to the left of MATH but to the right of MATH then there would be a subarc of MATH going from MATH around MATH in the clockwise sense, once again contradicting the minimality of MATH. Thus MATH lies to the left of MATH. Let MATH be the union of MATH with the portion of MATH which lies to the left of MATH. This is a noodle which intersects MATH at fewer than MATH points. The pairing MATH is the sum of those monomials MATH for which MATH lies to the left of MATH. Thus MATH is such that MATH is minimal in the calculation of MATH. By the induction hypothesis, there is only one such MATH, so we are done. |
math/0012088 | The map MATH decomposes as MATH. Here MATH is the embedding. Set MATH. Then the support of MATH is contained in MATH, and MATH . Hence the corollary follows from the following lemma. |
math/0012088 | CASE: Since MATH, it is enough to show that MATH . Since MATH, we may reduce the assertion to MATH . Let MATH. Then MATH, and MATH. Hence REF is a consequence of MATH REF . CASE: Let us take a coordinate system MATH on MATH such that MATH is given by MATH. Assume that MATH. Set MATH. It is enough to show that MATH for MATH sufficiently close to MATH and MATH. For any MATH, MATH by the assumption. Hence MATH, which implies MATH. |
math/0012088 | Since the assertion is local on MATH we may assume that MATH and MATH is the projection MATH. Hence MATH. Let MATH denote the MATH-module of vector fields tangent to the fibers of MATH. With the above definition, a MATH-lattice for MATH is a coherent MATH-submodule MATH of MATH such that MATH generates MATH on MATH and such that MATH. Locally on MATH, there exists a coherent MATH-submodule MATH of MATH such that MATH. Let us prove that the coherent MATH-module MATH is a MATH-lattice in MATH. Since MATH generates MATH it is sufficient to prove that MATH is MATH-coherent. Locally on MATH, there exists a finite covering of MATH by MATH-conic open subsets MATH and MATH-modules MATH such that MATH is a MATH-lattice in MATH on MATH. Hence, for each MATH, there exists MATH such that MATH is contained in MATH on MATH. Consider the increasing sequence MATH of coherent MATH-submodules of MATH. For each MATH, the restriction of this sequence to MATH is contained in MATH, which is MATH-coherent, hence it is locally stationary. Therefore MATH is MATH-coherent. To summarize, we have constructed a MATH-module MATH, coherent over MATH-module and which generates MATH. We may now apply REF and consider the good filtration MATH associated to MATH. By the construction, MATH. Setting MATH for MATH, we find a coherent MATH-submodule MATH of MATH such that MATH and MATH. Hence, locally on MATH, there is a MATH-linear epimorphism MATH. Repeating the same construction with MATH replaced by MATH, we construct an exact sequence MATH . Since MATH there is a MATH-linear morphism MATH such that MATH. Set MATH. This is a coherent MATH-module and we have an isomorphism MATH . To conclude, we choose a finite resolution of MATH by finitely free MATH-modules, and tensorize over MATH by the flat MATH-module MATH. |
math/0012088 | CASE: The inclusion MATH is clear. Conversely, assume that there are sequences MATH and MATH such that MATH . Then the sequence MATH is contained in MATH. Therefore, MATH and MATH satisfy REF . CASE: Suppose that MATH. Then by MATH there exist sequences MATH in MATH, MATH in MATH, MATH such that MATH and MATH . Therefore MATH . Since MATH, we get MATH hence MATH. CASE: The condition is obviously necessary. Let us now assume that there exists MATH, MATH, such that MATH. Then there exist sequences MATH in MATH and MATH in MATH such that MATH and MATH. Taking suitable subsequences, we may assume that MATH converges to MATH. Suppose that MATH. Then MATH, a contradiction. If MATH is unbounded we get MATH hence MATH, a contradiction. In the other case, we may assume that MATH and setting MATH, we get MATH, a contradiction. |
math/0012088 | Since MATH, there exists a coordinate system MATH on MATH as a real analytic manifold and a coordinate system MATH of MATH such that MATH is defined by the equations MATH. We shall argue by induction on MATH. MATH . Let us prove the result for MATH. Assume that MATH is a real analytic hypersurface defined by the equation MATH, and set MATH . Assume that MATH. We shall show that the morphism MATH is an isomorphism. The morphism REF is given by MATH . Therefore to obtain REF it is enough to prove that MATH . Since MATH implies MATH by REF , which contradicts the assumption. Hence MATH . Denoting MATH then, by REF , MATH hence MATH. MATH . Let us set MATH and MATH. Since MATH in a neighborhood of MATH, the hypothesis of induction implies MATH . On the other hand, MATH. Hence by MATH . This ends the proof of REF . |
math/0012095 | Consider the normal form REF of MATH in the MATH-th components. The normal form for MATH is obtained as follows: MATH . Thus the operation on MATH induced by MATH is given by MATH . |
math/0012095 | We write the two sets of equations in REF as follows: MATH for each MATH. For generic values of MATH, use the first MATH equations the first set of MATH equations, we can solve for MATH. Similarly, we can solve for MATH from the first MATH equations of the second set of MATH equations. The remaining vectors MATH, MATH, have to satisfy another two equations obtained from the last equations in those two sets of MATH equations, respectively, by substituting MATH and MATH with their solutions in terms of MATH for MATH. It it then easy to check that these two equations are linearly independent when MATH. |
math/0012095 | The case of MATH can be checked directly. For MATH, we have MATH . |
math/0012095 | In MATH, let MATH be the sublattice generated by the translation vectors of partial conjugations. Then we have MATH . Let MATH be a non-zero vector perpendicular to MATH. We can choose such a MATH so that its coordinates are polynomials in MATH and the inner product MATH is identically zero. This can be achieved by considering generic values of MATH first. Then since MATH for generic values of MATH, it has to be zero identically. Let MATH. The inner product MATH is invariant under the translations by vectors in MATH. This is a desired link-homotopy invariant of ordered, oriented links since MATH for all MATH. The fact that the invariant MATH is of fine type is a direct consequence of the fact that the linking numbers and the triple linking numbers are all finite type invariants of string links CITE. If we have a singular link, we may put it into the form of the closure of a single string link. Since polynomials of finite type invariants are still of finite type, MATH vanishes on singular string links with sufficiently large number of double points. This implies that it is a finite type link invariant. |
math/0012095 | Using the fact that the rows of the cofactor matrix MATH of a given matrix MATH are perpendicular to different rows of MATH, we see that MATH is perpendicular to all the vectors of translation operation induced by partial conjugations as well as the vector MATH. Certainly, MATH. So MATH is a non-trivial link-homotopy invariant of ordered, oriented links with REF components. It is of finite type since it is a polynomial in MATH and MATH. Under the reversion of orientation, MATH changes to MATH. Since MATH, the invariant MATH is only changed by a sign under the reversion of orientation. |
math/0012098 | Let MATH be bounded, and let MATH. Since MATH we have that MATH. Thus, for each MATH, there exists a MATH-covering of MATH by MATH balls of equal diameter MATH. Hence MATH which gives MATH. Thus MATH, and so MATH, proving the lemma. |
math/0012098 | For MATH, let MATH be a grid of side length MATH covering MATH. Every square of side MATH is included in the ball of diameter MATH, which, in turn, is included in at most MATH squares of the grid MATH. Thus MATH proving the lemma. |
math/0012098 | It is enough to prove the reverse inequalities in relations REF . For MATH, let MATH be such that MATH. Since a square with a side length MATH is contained in at most four squares with a side length MATH, we have MATH. Similarly, MATH. Thus for each MATH, MATH . The conclusion now follows by REF . |
math/0012098 | Let MATH be a sequence of strictly increasing positive integers. For each MATH, let MATH and MATH be the terms under the limit signs in equality REF in the given order. It is enough to prove that MATH and MATH . Suppose that the set MATH is on a plane with the Cartesian coordinates MATH. For each integer MATH, let MATH be the grid of side length MATH intersecting the origin and with sides parallel to the coordinate axes. Thus MATH intersects with the MATH-axes at each point of the partition MATH. Let MATH. For MATH, let MATH be the number of squares of the grid MATH contained in the strip MATH and covering the set MATH defined by relation REF , where MATH. By the definition of MATH, MATH for each MATH. Summing over all indices MATH, it follows that the inequalities MATH hold for each MATH. By the second inequality in display REF , we have that REF . To prove REF , first suppose that MATH. Then by relation REF , MATH, and REF follow because MATH. Now suppose that MATH. In that case, REF follow from the first inequality in display REF and from the following two relations: MATH and MATH valid for any two sequences MATH and MATH of positive numbers such that MATH. |
math/0012098 | If MATH then MATH has bounded MATH-variation, and so REF holds by relation REF . If MATH, then REF holds because MATH by relation REF . Thus one can assume that MATH. Let MATH for some MATH. We claim that for each integer MATH, MATH . Indeed, if MATH then MATH. Suppose that MATH and MATH. Since for each MATH, MATH, by NAME 's inequality, we have MATH proving relation REF . If MATH, then by REF , MATH . Thus relation REF follows by the relation REF , proving the lemma. |
math/0012099 | It suffices to show that MATH is decomposable if MATH is decomposable. Suppose that MATH for some integral polytopes MATH and MATH in MATH each with at least two points. Then MATH. We need to show that both MATH and MATH have at least two points. Suppose otherwise, say MATH has only one point. Let MATH be any vertex of MATH such that MATH is a vertex of MATH. Since MATH, there are unique vertices MATH and MATH such that MATH. As MATH has at least two points, it has another vertex MATH such that MATH is one of its edges. Then, by REF , MATH has an edge MATH that starts at MATH and is parallel to MATH where MATH is a vertex of MATH different from MATH. The latter property implies that MATH for some real number MATH. Hence MATH as MATH has only one point and MATH. This means that MATH maps two vertices of MATH to one vertex of MATH, contradicting our assumption. |
math/0012099 | It follows directly from REF . |
math/0012099 | Suppose that MATH for some integral polytopes MATH and MATH. Then MATH or MATH is homothetic to MATH, say MATH. This means that there is a real number MATH and MATH such that MATH. Hence the vertices of MATH are MATH . Since MATH is integral, all the vertices MATH are integral and in particular MATH are integral. So MATH must be a rational number and the denominator of MATH divides MATH; hence MATH is an integer. As MATH, we have MATH or MATH. In either case, MATH is a trivial summand of MATH. Therefore MATH is integrally indecomposable. |
math/0012099 | Let MATH be the edge sequence of an integral summand MATH of MATH. By the final statement in REF , each edge of MATH occurs as the summand of some edge MATH of MATH where MATH is a primitive vector, and it is easily seen that its corresponding edge vector must be of the form MATH with MATH. The sum is zero simply because the boundary of MATH is a closed path. Conversely, any sequence of this form will determine a closed path. Since MATH is a polygonal sequence, MATH must define the boundary of a convex polygon. It will be a summand of MATH, with the other summand having edge sequence MATH. |
math/0012099 | Certainly the language associated with NAME lies in NP as we may use a proper decomposition of MATH to verify membership of the language. We give a polynomial reduction of Partition to NAME which proves, since Partition is CITE, that NAME is NAME. Recall that the input to Partition is a sequence MATH of positive integers which we may take to be non-decreasing. Thus MATH. Let MATH. The question in Partition is whether there is a subsequence of MATH with sum MATH. Observe that we may assume that MATH is even, for otherwise the question is easily answered. Consider now the following instance of NAME: the edge sequence MATH where all MATH. Firstly, it is easy to check that this is indeed a polygonal sequence. Secondly, any polygon associated with the polygonal sequence has a proper decomposition if and only if the sequence MATH has a subsequence with sum MATH. Thus we have a polynomial reduction, which completes the proof. |
math/0012099 | (Note that by a vector operation we mean adding two vectors, multiplying a vector by a scalar, or adjoining a point to a set.) The running time is easy to see as each set MATH has size at most MATH. Also, the set IP can be computed in time MATH: since the edge sequence is already given one can enumerate points in MATH by scanning vertical line segments starting at MATH. One need only keep track of the top and bottom edges when moving the lines (that is, increasing MATH values) and the edges tell us the range of the MATH value for any given MATH value. (Note that MATH itself can be bound in terms of MATH, MATH and the area of the polygon using Pick's formula CITE.) To prove the correctness, observe that all the points in MATH are of the form MATH, MATH. REF ensures that MATH for some MATH and REF insists that MATH (note that MATH for all MATH). If one of the points in MATH is equal to MATH then MATH, and so the sequence MATH forms the edge sequence of a proper integral summand of MATH. On the other hand, for any proper integral summand MATH of MATH, MATH can be ``slid" into MATH at MATH, that is, MATH can be translated so that MATH is a vertex of MATH and MATH lies inside MATH. Hence all the vertices of MATH must lie in MATH and thus in IP. Consequently its edge sequence will be detected by our algorithm. |
math/0012099 | Supposing MATH, we may view the vector sum as a path from MATH to MATH, so the number of such paths is equal to the sum of the numbers of paths from MATH to MATH for MATH, using MATH. Hence the numbers of paths can be computed iteratively as described in the algorithm: the number MATH in MATH records the number of paths from MATH to MATH using MATH and the set MATH records all the pairs MATH, MATH, for which a path reaches MATH with its last edge being MATH with MATH. Thus the integer in cell MATH is the total number of closed paths MATH starting at MATH. By REF this is the number of integral summands of MATH. |
math/0012100 | Let MATH, and let MATH. Moreover, let MATH. The fact that a subspace MATH of MATH is NAME means that both MATH and MATH vanish on it identically, that is, MATH, MATH for all MATH, and MATH is maximal, that is, MATH dimensional, MATH, with this property. Note that MATH is non-degenerate on MATH, that is, it is a symplectic form on this vector space. Now, both MATH and MATH vanish on MATH since MATH. Let MATH denote the subspace of MATH which annihilates MATH, that is, MATH. Then any NAME subspace MATH of MATH satisfies MATH since MATH and MATH for all MATH. Note that MATH has codimension MATH in MATH (since MATH has codimension MATH in the NAME subspace MATH). Thus, MATH is a REF-dimensional vector space, and MATH descends to a symplectic form on it. The image MATH of a NAME subspace MATH in MATH is Lagrangian with respect to this form. There is a one-dimensional family of such Lagrangian subspaces; the image of MATH is one of them. Indeed, given any non-zero element MATH of MATH, there is a unique Lagrangian subspace of MATH which includes MATH, namely the span of MATH. This then determines a unique NAME subspace MATH of MATH with MATH. We claim that MATH is not a subspace of MATH. Indeed, suppose otherwise, that is, that MATH. The hypothesis on the full rank projection of MATH means that MATH are independent on MATH. The corresponding NAME vectors MATH under MATH in MATH are tangent to MATH, hence in MATH. Thus, they span a MATH-dimensional subspace MATH of this space. If MATH is a nonzero linear combination of the functions MATH, then MATH does not vanish on MATH, which implies that MATH does not vanish on MATH. Hence MATH, which means that MATH and MATH have trivial intersection. But this is a contradiction: by dimension counting, the codimension of MATH in MATH is MATH. Hence MATH is not a subspace of MATH. Thus MATH cannot all vanish identically on MATH. Since they all vanish identically on MATH, this is the only NAME subspace with this property. By the clean intersection assumption, MATH, that is, MATH and MATH are not the same. Hence the MATH do not all vanish on MATH, that is, the pull-back of MATH to MATH at MATH is non-zero for some MATH. By relabelling the coordinates, we may assume that MATH is non-zero. |
math/0012104 | The conclusion of this Lemma can fail only if there are MATH and MATH with MATH. This means that MATH . This can only happen if MATH is in the space spanned by MATH, or, equivalently, MATH is in the space spanned by MATH. This means that all the rows MATH of MATH satisfy MATH for some MATH. Interpreting a row of MATH as a vertex of MATH, this means that MATH is contained in the affine plane MATH. |
math/0012104 | Consider the mapping MATH . Since we assume MATH, we can apply REF and conclude that MATH is a diffeomorphism. The pull-back of the canonical symplectic structure in MATH by MATH is precisely MATH, because of NAME REF. NAME with that property are called symplectomorphisms. Since the volume form of a symplectic manifold depends only of the canonical REF - form, symplectomorphisms preserve volume. We compose with a scaling by MATH in the first MATH variables, that divides MATH by MATH, and we are done. |
math/0012104 | Let MATH be the set of covectors MATH with the property that MATH . Notice that MATH is a convex subset of MATH. Geometrically, the planes in MATH with MATH minimal correspond to the envelope of the graph of MATH. The set MATH contains MATH: For any given MATH, we set MATH. Since MATH is convex, MATH . For the converse, assume that there is MATH not in the closure of MATH. Then there is some MATH such that MATH . We define the following gradient vector field in MATH: MATH . Because the denominator is bounded below by MATH, this field is well-defined and NAME in all of MATH. Let us fix an arbitrary initial condition MATH, and let MATH denote a maximal solution of the vector field. Since the vector field has norm REF, MATH cannot diverge in finite time and therefore MATH is well-defined for all MATH. Now we look at the function MATH. Its derivative with respect to MATH is MATH. Therefore, MATH. We deduce from there that MATH . Hence, MATH, a contradiction. |
math/0012104 | The momentum map MATH maps MATH onto the interior of MATH. Indeed, let MATH be a row of MATH, associated to a vertex of MATH. Then there is a direction MATH such that MATH for some unique MATH. We claim that MATH. Indeed, let MATH, MATH a real parameter. If MATH is another row of MATH, MATH as MATH. We can then write MATH as: MATH . Since MATH is positive definite, MATH and MATH where MATH is the unit vector in MATH corresponding to the row MATH. It follows that MATH . When we set MATH, we have MATH on MATH, so we have a local diffeomorphism at each point MATH. Assume that MATH for MATH. Then, let MATH. The function MATH has the same value at MATH and at MATH, hence by NAME 's Theorem its derivative must vanish at some MATH. In that case, MATH and since MATH, MATH must vanish in some MATH. This contradicts REF . |
math/0012104 | By REF , MATH . Hence its MATH-th coordinate is: MATH . |
math/0012104 | We compute: MATH and hence MATH . Also, MATH . Therefore, MATH . We can now use REF to conclude the following: CASE: Determinant of the Condition Matrix MATH . We can now write the same formula as a determinant of a block matrix: MATH and replace the determinant by a wedge. The factor MATH comes from replacing MATH by MATH. |
math/0012104 | Given MATH, we define MATH as the space of MATH such that MATH vanishes. Using CITE (or REF p. REF below), we deduce that the average number of complex roots is: MATH . By REF , we can replace the inner integral by a MATH - form valued integral: MATH . Since the image of MATH is precisely MATH, one can add MATH extra variables corresponding to the directions MATH without changing the integral: we write MATH. Since MATH is equal to MATH, the average number of roots is indeed: MATH . In the integral above, all the terms that are multiple of MATH for some MATH will cancel out. Therefore, MATH . Now, we apply the integral formula: MATH to obtain: MATH . According to formul REF, the integrand is just MATH, and thus MATH . |
math/0012104 | Assume that MATH, with MATH colinear and MATH orthogonal to MATH. As the image of MATH is orthogonal to MATH, MATH is orthogonal to MATH, so MATH and hence MATH. We can therefore replace MATH by MATH without compromising equality REF . Since MATH was minimal, this implies MATH. |
math/0012104 | MATH . |
math/0012104 | The expected number of roots in MATH for a linear system is MATH . It is also MATH. This holds for all MATH, hence the volume forms are the same and MATH . |
math/0012104 | We set MATH, then MATH. |
math/0012104 | As in the complex REF , the expected number of roots can be computed by applying the co-area formula: MATH . Now there are three big diferences. The set MATH is in MATH instead of MATH, the space MATH contains only real polynomials (and therefore has half the dimension), and we are integrating the square root of MATH. Since we do not know in general how to integrate such a square root, we bound the inner integral as follows. We consider the real NAME space of functions integrable in MATH endowed with Gaussian probability measure. The inner product in this space is: MATH where MATH is NAME volume. If MATH denotes the constant function equal to MATH, we interpret MATH . Hence NAME inequality implies: MATH . By construction, MATH, and we are left with: MATH . As in the complex case, we add extra MATH variables: MATH and we interpret MATH in terms of a wedge. Since MATH we obtain: MATH . Now we would like to use NAME again. This time, the inner product is defined as: MATH . Hence, MATH . This time, MATH, so we bound: MATH . |
math/0012104 | Let MATH. As in the mixed case, we define: MATH where now MATH. Let MATH. We also define MATH to be the canonical projection mapping MATH to MATH and set MATH. Then, MATH . As before, we change coordinates in each fiber of MATH by MATH with MATH colinear to MATH, MATH in the range of MATH, and MATH othogonal to MATH and MATH. This coordinate system is dependent on MATH. In the new coordinate system, REF splits as follows: MATH . The integral MATH of MATH is the expected number of real roots on MATH, therefore MATH . In the new system of coordinates, MATH is defined by the equation: MATH . Since MATH, MATH . This implies: MATH . We can integrate the MATH variables MATH to obtain: MATH . This is MATH times the probability MATH for the linear case. |
math/0012104 | We parametrize: MATH . Then, MATH . We can choose an orthonormal basis MATH of MATH such that MATH. Then, MATH . Thus, MATH and hence MATH . |
math/0012104 | We will prove this Lemma locally, and this implies the full Lemma through a standard argument (partitions of unity in a compact neighborhood of the support of MATH). Let MATH be fixed. A small enough neighborhood of MATH admits a fibration over MATH by planes orthogonal to MATH. We parametrize: MATH where MATH is the solution of MATH in the fiber passing through MATH. Remark that MATH, and MATH. Therefore, MATH . Also, if one fixes MATH, then MATH is a parametrization MATH. We have: MATH . |
math/0012104 | Every MATH admits an open neighborhood such that MATH restricted to that neighborhood is a diffeomorphism. This defines an open covering of MATH. Since MATH is locally compact, we can take a countable subcovering and define a partition of unity MATH subordinated to that subcovering. Also, if we fix a value of MATH, then MATH becomes a partition of unity for MATH. Therefore, MATH where the second equality uses REF with MATH. Since MATH, we are done. |
math/0012105 | The first assertion is a consequence of REF, which states that MATH (this fact may also be proved more directly). Hence the complete left MATH-projections on MATH are exactly the orthogonal projections MATH. If MATH is unital, this part of the argument would be easier. It is well known that MATH is an operator algebra with the NAME product. If MATH is a complete right MATH-ideal of MATH, then MATH is, by the first part, equal to a principal right ideal MATH. Here MATH is an orthogonal projection. Considered as subsets of MATH, we have MATH. But also MATH. So MATH by basic functional analysis. So MATH is a right ideal of MATH. Since MATH is unital, MATH, and MATH is a left identity for MATH. There exists a net in MATH which converges to MATH in the weak* topology. By a well known argument using the fact that the weak closure of a convex set equals its norm closure, one may replace the above net with a left contractive approximate identity for MATH (see for example, REF for details). Conversely, if MATH is a closed right ideal of MATH with a contractive left approximate identity, then MATH is a subalgebra of MATH with a left identity MATH of norm MATH by CITE Note that MATH is an orthogonal projection in MATH. Moreover MATH by a routine argument approximating elements in MATH by weak*-converging nets of elements in MATH (see for example, REF ). We have MATH . Thus MATH is a complete right MATH-summand of MATH by the first part, so that MATH is a complete right MATH-ideal of MATH. |
math/0012106 | Observe that if MATH . It follows that MATH where MATH . This says that MATH measures the deviation of MATH from being a MATH-module map. We must show that E is in the image of MATH . Recall that for MATH we have MATH where MATH denotes the algebra (wedge) product on MATH . Thus MATH . MATH . For MATH write MATH and MATH . In order to simplify notation we drop the summation symbol wherever the latter coproducts appear below. From our last calculation we have MATH and MATH . Because MATH is cocommutative, the full summations are equal: MATH . Thus MATH . It follows from coassociativity that MATH . Thus MATH is in the image of MATH and in fact MATH . To verify the NAME identity, apply MATH to the NAME expression in MATH . By the morphism condition just established, the result is the NAME identity valid in MATH . Assuming that MATH is injective, the NAME identity in MATH follows. |
math/0012106 | We need only evaluate MATH on elements of the form MATH and MATH. We begin with MATH where MATH, MATH and MATH are the evident unshuffles. This composition is equal to MATH which is equal to REF by the commutator relation. For the terms of the form MATH, the only unshuffles that we need to consider are those that result in terms of the form MATH and MATH . Recall that when MATH in the first term and when MATH in the second term, a coefficient of MATH must be introduced. So we have MATH which, after expanding the MATH terms of the unshuffles along with the signs mentioned above, is seen to equal the NAME relation, and hence is equal to REF. |
math/0012108 | Let MATH. Then MATH for all MATH, and thus MATH, from which MATH. The opposite relation holds by symmetry. |
math/0012108 | Choose for MATH a finite generating set MATH, and define the weight MATH by MATH for all MATH. Then MATH for all MATH, so MATH. |
math/0012108 | Clearly MATH for MATH, so MATH is a weight. Let MATH be a minimal form of MATH. The lemma asserts that one can suppose there are no consecutive letters in MATH in MATH; now two equal consecutive letters cancel, and the product of any two distinct letters in MATH equals the third one. For any arrangement MATH of MATH we have MATH, so the substitution of MATH for MATH will not increase the weight of MATH. |
math/0012108 | Let MATH be a minimal form of MATH. Thanks to REF we may suppose the number of MATH-s in MATH is at most the number of MATH-s plus one. Construct words MATH over MATH using MATH seen as a substitution on words; they represent MATH and MATH respectively. Note that MATH . As MATH and MATH, each MATH in MATH contributes MATH to the total weight of MATH and MATH; the same argument applies to MATH and MATH. Now, grouping together pairs of MATH-s in MATH and MATH-s, we see that MATH is a sum of left-hand terms, possibly MATH; while MATH is bounded by the total weight of the letters in MATH and MATH, which is precisely the sum of the corresponding right-hand terms. |
math/0012108 | We construct an injection MATH of MATH into the set of labelled binary rooted trees each of whose leaves is labelled by an element of MATH of weight bounded by MATH and each of whose interior vertices is labelled by an element of the subgroup MATH of MATH. For MATH, MATH is called its representation . It is constructed as follows: if MATH satisfies MATH, its representation is a tree with one vertex labelled by MATH. If MATH, let MATH be such that MATH, and write MATH. By REF , MATH, so we may construct inductively the representations of MATH and MATH. The representation of MATH is a tree with MATH at its root vertex and MATH and MATH attached to its two branches. We first claim that MATH is injective: let MATH be a tree in the image of MATH. If MATH has one node labelled by MATH, then MATH. If MATH has more than one vertex, let MATH be the label of the root vertex and MATH be the two subtrees connected to the root vertex. By induction on the number of vertices of MATH, we have MATH for unique MATH and MATH. Then as MATH is injective there is a unique MATH with MATH, and MATH. We next prove by induction on MATH that if MATH then its representation is a tree with at most MATH leaves. Indeed if MATH then MATH's representation has one leaf and MATH, while otherwise MATH's representation is made up of those of MATH and MATH. Say MATH and MATH; then by REF we have MATH. By induction these representations have at most MATH and MATH leaves. As MATH, we have MATH for all MATH; and by direct computation, MATH, so the number of leaves of MATH's representation is MATH as was claimed. We conclude that MATH is bounded by the number of representations with MATH leaves; there are MATH binary trees with MATH leaves, MATH choices of labelling for each of the MATH interior vertices, and MATH choices for each leaf; so REF follows. |
math/0012108 | The sequence MATH was shown to be of superpolynomial growth in CITE, so REF yields the claimed lower bound; an elementary proof of this lower bound appears also in CITE. For the upper bound, which is the main result of the present note, we invoke REF with MATH, noting that MATH and MATH, to obtain MATH. |
math/0012112 | First, we check that REF defines an action. Let MATH be two copies of MATH acting on MATH by MATH and MATH, respectively. Consider MATH as a MATH-equivariant principal MATH-bundle over MATH, and let MATH denote the pull-back to MATH. The action of MATH on MATH is free, and has MATH as a cross-section. Using MATH identify MATH. We claim that the induced action of MATH on MATH is the twisted MATH-action. Given MATH we compute MATH . The action of MATH takes this back to MATH, which proves the claim. Since the inversion map on MATH is anti-Poisson, MATH is a NAME map for the reversed NAME structure MATH on MATH. We check it is a moment map for the twisted action. Let MATH denote projection along MATH. Using the moment map condition for MATH, MATH which are the generating vector fields for the twisted action. |
math/0012112 | Since MATH preserves MATH and is anti-linear it takes MATH to its complex conjugate and so changes the sign of MATH. It follows that the involutions MATH and MATH are related by MATH. Therefore MATH changes the sign of the cocycle MATH dual to the bracket on MATH. |
math/0012112 | For each MATH let MATH be the inclusion MATH. Equip MATH with the MATH-action such that the maps MATH are equivariant, with respect to the MATH-action on MATH defined by MATH. Define MATH by MATH, and let MATH where MATH pull-back to MATH under MATH and vanish on MATH. Then REF is equivalent to MATH and REF is equivalent to the moment map condition MATH . These two equations also hold for the average of MATH under the MATH-action. Since MATH the averaging process changes only MATH, but not MATH. We may therefore assume that MATH is MATH-invariant. Let MATH be the unique vector field on MATH such that MATH and MATH. It is MATH-invariant, preserves MATH, and its flow MATH takes the slice at MATH to that at MATH. Let MATH be the isotopy of MATH defined by MATH. Then MATH implies MATH. Similarly, for MATH we have MATH . This shows MATH, or equivalently MATH. MATH-equivariance of the flow MATH implies that MATH intertwines the MATH-actions on MATH for the parameters MATH. Since the moment maps are determined up to a constant in MATH, this proves MATH. In the presence of a family of anti-symplectic involutions with MATH, REF-form MATH changes sign under the corresponding involution MATH of MATH. The vector field MATH, and therefore its flow, are MATH-invariant. Equivalently, MATH. |
math/0012112 | Recall that the definition of a MATH-valued moment map depends on the inner product MATH on MATH. For any MATH consider the rescaled inner product MATH, and let MATH. Replacing MATH with MATH replaces the map MATH by MATH and the form MATH by MATH . We obtain a family MATH of Hamiltonian MATH-spaces with MATH-valued moment map (relative to MATH), with MATH . Taking the linearizations of their products MATH we obtain a family of Hamiltonian MATH-spaces MATH where MATH . Consider the limit MATH. The family of moment maps MATH extends smoothly to MATH by MATH. Since the family of REF MATH extends smoothly to MATH by MATH, MATH extends smoothly to MATH by MATH. We thus have a family of compact Hamiltonian MATH-spaces, MATH connecting MATH with the linearization of MATH. The proof is completed by an application of REF , with MATH . To check REF for MATH, we note that the first term vanishes in our case since MATH is independent of MATH. Since MATH, also MATH is independent of MATH, and therefore MATH . |
math/0012112 | We proceed as in the proof of REF . Replacing MATH with MATH we obtain a family MATH of Hamiltonian MATH-manifolds with MATH-valued moment maps (relative to MATH) with MATH . Conjugating and linearizing we obtain a family MATH of Hamiltonian MATH-manifolds with MATH and MATH . These families extend smoothly to MATH by MATH and MATH, and connect the linearization of MATH with the space MATH. Therefore, the claim again follows from REF . |
math/0012112 | Given a solution MATH with product MATH, define MATH recursively as follows: put MATH, let MATH be the unique element with MATH, then let MATH the unique element with MATH, and so on. Let MATH be the image of MATH by the action of MATH. By construction MATH for MATH, and since the product is MATH we must have MATH. This shows that the map MATH is surjective. Starting the recursion with MATH rather than MATH replaces MATH by its image under the diagonal dressing action of MATH. This shows that the map is a bijection. |
math/0012112 | The equivalence of REF , first proved by NAME in CITE, follows from REF. The equivalence of REF follows from REF. The equivalence of REF follows from REF , since MATH acts trivially on the NAME in this case. It was proved independently by CITE. |
math/0012112 | The exterior differential of MATH is given by MATH . This shows MATH and together with MATH yields, MATH . To compute MATH, observe first that by the moment map condition, the contraction of MATH with any bivector field of the form MATH for MATH is given by MATH . The bivector field MATH is a linear combination of such terms. Using the defining property MATH of the cocycle, the first summand simplifies and we obtain MATH . By the structure equation MATH, the first terms in REF agree. By the following REF the second terms agree as well. |
math/0012112 | Since both sides are MATH-invariant, it suffices to verify the identity at points of MATH. Let MATH and MATH. Then MATH where MATH, and we have MATH . Let MATH be a slice at MATH for the coadjoint-action on MATH. There is a splitting, MATH where MATH (the orthogonal complement of the isotropy algebra) is embedded via the generating vector fields. Let MATH. By the NAME symplectic cross-section theorem, MATH is a symplectic submanifold, and the embedding MATH given by the generating vector fields defines a MATH-orthogonal splitting MATH where REF-form on MATH is given by the NAME formula, MATH . Let MATH be root vectors for the positive roots MATH, normalized by MATH. Then MATH form a basis of MATH, and MATH is the subspace corresponding to roots with MATH. By a short calculation, MATH . The splitting REF is also MATH-orthogonal. The pull-backs MATH and MATH to MATH differ by the pull-back by MATH of REF-form MATH. Since MATH is a co-isotropic subspace of MATH and MATH agree on that subspace, it follows that the top exterior powers of MATH and MATH are equal. Therefore, MATH . Since MATH this gives, MATH as required. |
math/0012112 | For all MATH, the number MATH is the trace of the operator MATH on MATH. For MATH, the operator MATH is nilpotent and therefore has zero trace. Suppose MATH, and let MATH. Since the pairing between MATH and MATH is given by MATH, we have MATH. On any root space MATH, MATH acts as a scalar MATH, hence has trace MATH. It follows that the trace of MATH on MATH is, MATH . |
math/0012112 | We claim that the statement of the Lemma is equivalent to the equation, MATH . Indeed, using the definition of MATH we have MATH . Since MATH the second term is MATH. To compute the first term, note that the dressing vector fields MATH, together with minus the right-invariant vector fields MATH, are the generators for the MATH-action on MATH. Therefore, using REF, and REF , MATH which identifies the first term with MATH. It remains to show REF. This condition is equivalent to the vanishing of the second order differential operator MATH on MATH, because MATH . Let MATH be the projection. Then MATH where MATH is the NAME operator on MATH corresponding to the invariant bilinear form MATH. Since MATH is MATH-invariant, we can replace the superscript ``R" by a superscript ``L". Hence MATH where we have used MATH. The vector fields MATH generate the right-MATH action and therefore vanish on right-MATH-invariant functions. It follows that MATH, so that MATH. |
math/0012112 | Using REF we compute, MATH . |
math/0012115 | Since MATH and MATH are independent, we may replace MATH by high positive powers of conjugates to ensure that the subgroup MATH of MATH generated by MATH is free with basis MATH, each nontrivial element of MATH is hyperbolic, and MATH is quasi-convex with respect to the action on MATH (see CITE). We will call such free subgroups NAME groups. Let MATH be the NAME graph of MATH with respect to the generating set MATH. Then MATH is a tree and each oriented edge has a label MATH. Choose a basepoint MATH and construct a MATH - equivariant map MATH that sends MATH to MATH and sends each edge to a geodesic arc. NAME implies that MATH is a MATH - quasi-isometric embedding for some MATH and in particular for every MATH the MATH - image of the axis of MATH in MATH is a MATH - quasi-axis of MATH. By MATH denote the axis of MATH in MATH, MATH. Choose positive constants MATH and define MATH for MATH. CASE: The key to the proof is the following observation. If MATH are fixed and the exponents MATH are chosen suitably large, then for any sufficiently long MATH - oriented segment MATH in the axis MATH of MATH and any orientation preserving MATH - qi embedding MATH (with respect to the MATH - orientation of MATH) there is a subsegment MATH of MATH containing a string of MATH edges labeled MATH whose image (pulled tight) is a segment in MATH consisting only of edges labeled MATH (this is true because MATH so the image will contain a whole fundamental domain for the action of MATH on MATH). REF illustrates the situation. Now assuming that MATH let MATH be a long arc, let MATH, and let MATH be such that MATH is MATH - close to the MATH - quasi-axis MATH of MATH, with matching orientations. Choose an arc MATH so that MATH is MATH - close to MATH. Then there is a MATH - quasi-isometry MATH obtained by composing MATH and MATH does not depend on the choices of MATH, only on MATH, MATH, and MATH. Combining this with the observation above, we conclude that MATH takes a long segment in an axis of a conjugate of MATH uniformly close to an axis of a conjugate of MATH with matching orientation, contradicting the assumption that MATH. Similarly, MATH for MATH. CASE: The proof is similar to the proof of REF , only now one uses MATH. Similarly, MATH for MATH. REF in addition MATH then MATH. If MATH, we obtain the situation pictured in REF where a long string of MATH's is close to a long string of MATH's with either the same or opposite orientation. Note that it is possible that all such pairs of strings have opposite orientation so the assumption MATH is necessary. Similarly, if MATH then MATH for all MATH. We now finish the proof. If MATH then the above claims conclude the argument. Otherwise, note that by REF we have MATH. Now replace MATH by MATH and repeat the construction. |
math/0012115 | It is clear that MATH is unbounded on MATH for any MATH: If we use MATH as a competitor path we have MATH . If MATH is large, then there are no copies of MATH in the MATH - neighborhood of an axis of MATH, which implies that MATH is zero on MATH. The proof of the other claim is similar. |
math/0012115 | Let MATH be the sequence from REF . We assume in addition (without loss of generality) that each MATH is cyclically reduced. Define MATH as MATH where MATH is chosen as in REF so that MATH is unbounded on MATH and so that it is REF on MATH for MATH (a high power of MATH cannot be translated into a MATH - neighborhood of an axis of MATH). It follows that MATH is not a linear combination of MATH, that is, the sequence MATH consists of linearly independent elements. We can easily arrange that MATH is contained in the commutator subgroup of MATH (this is automatic if MATH are in the commutator subgroup; otherwise, replace MATH by MATH with MATH - as in REF it follows that MATH). In that case any homomorphism MATH vanishes on MATH and it follows that the sequence MATH in MATH consists of linearly independent elements. |
math/0012115 | CASE: We will show that MATH has finite index in MATH. Let MATH be an infinite sequence of elements of MATH. Choose a MATH - quasi-axis MATH for MATH and let MATH. Since MATH and MATH commute, MATH is also a MATH - quasi-axis for MATH and the distance between MATH and MATH is uniformly bounded by MATH. Let MATH be such that MATH minimizes the distance between MATH and the MATH - orbit of MATH. Thus MATH is uniformly bounded (by MATH plus the diameter of the fundamental domain for the action of MATH on MATH); call such a bound MATH. Let MATH be from the definition of MATH. We note that MATH move both MATH and MATH by MATH. Therefore, the set of such elements is finite. From MATH we conclude that MATH and MATH represent the same MATH - coset and the claim is proved. CASE: Denote by MATH and MATH the corresponding conjugates of MATH. For notational simplicity, we will first assume that MATH is axial and that MATH is an axis of MATH. Without loss of generality we assume MATH. Choose MATH and let MATH be as in the definition of MATH for MATH. Let MATH be the size of the finite set from the definition of MATH. If MATH and MATH contain oriented MATH - close arcs MATH and MATH of length MATH (MATH is the translation length of MATH) then the elements MATH move each point of the terminal subarc of MATH of length MATH a distance MATH for MATH. It follows that MATH for distinct MATH so that MATH have equal positive powers. In general, when MATH is only a quasi-axis, one can generalize the above paragraph by replacing MATH etc. by larger constants that depend on MATH and MATH. Alternatively, one can modify MATH to make MATH axial: simply attach an infinite ladder (REF - skeleton of an infinite strip) along one of the two infinite lines to MATH; then attach such ladders equivariantly to obtain a MATH - space. Finally, subdivide each rung and each edge in MATH into a large number MATH of edges in order to arrange that the ``free" lines in the attached ladders are geodesics and axes for the corresponding conjugates of MATH. The group MATH continues to act on the new space MATH which is quasi-isometric to MATH. The statement for MATH implies the statement for MATH. CASE: Let MATH be a hyperbolic element. Again, without loss of generality, we will assume that MATH has an axis MATH. We aim to show that some translate of MATH, say MATH, has both ends distinct from MATH, since then MATH and MATH are independent hyperbolic elements. Suppose first that there is MATH such that MATH is not in the MATH - neighborhood of MATH, but it is asymptotic in one direction, that is, a ray in MATH is contained in the MATH - neighborhood of MATH. From REF we see that MATH and MATH cannot contain segments of length MATH that are oriented MATH - close to each other; in particular, one of MATH, MATH moves towards the common end, say MATH, and the other moves away from it. Now consider MATH for large MATH. This is a bi-infinite geodesic with one end MATH and the other end distinct from the ends of MATH, MATH. The translates MATH and MATH violate REF as they have oriented rays within MATH of each other. It remains to consider the case when every translate of MATH is within MATH of MATH. After passing to a subgroup of MATH of index REF if necessary, we may assume that MATH preserves the ends of MATH. Now proceed as in REF to show that MATH has finite index in MATH. CASE: This is similar to REF . We assume for simplicity that MATH are axial. By MATH denote the translation length of MATH, MATH. Let MATH be as in the definition of MATH for MATH with respect to some MATH in an axis MATH of MATH and MATH. Let MATH be the size of the corresponding finite set. Now assume that MATH has an axis that admits a segment of length MATH which is oriented MATH - close to a segment in MATH. Consider MATH, MATH. As before, we conclude that MATH for some MATH; thus MATH and MATH have common positive powers. CASE: Since the action of MATH on MATH is nonelementary, we can choose a NAME subgroup MATH. Let MATH. For notational simplicity we will assume that all nontrivial elements of MATH are axial and in fact that there is a MATH - invariant totally geodesic tree MATH (this can be arranged by modifying MATH as in the proof of REF except that now one attaches REF - skeleton of (tree)MATH instead of (line)MATH). Let MATH be an axis of MATH. Then MATH provides a segment MATH in MATH such that the set of MATH that move each point of MATH by MATH is finite. Now consider an infinite sequence MATH of elements of MATH with distinct (and hence non-parallel) axes MATH that overlap MATH in (oriented) finite intervals that contain MATH. If MATH then, according to REF , there is MATH such that MATH is MATH - close to MATH. Replacing MATH by MATH if necessary, we may assume that MATH moves each point of MATH by MATH. Thus MATH for some MATH, so that MATH and MATH are within MATH of each other, contradicting the choice of the sequence. Finally, note that we could have taken MATH for some MATH that does not commute with MATH, and that the argument shows that MATH for all but finitely many MATH. |
math/0012115 | This is a consequence of REF . |
math/0012115 | The details of this proof are similar to the proof of REF and we only give a sketch. We will use the following principle in this proof. If MATH is a rank REF free group and MATH a homomorphism then there is a rank REF free subgroup MATH such that either MATH contains no hyperbolic elements or else MATH is injective on MATH and MATH is NAME (it follows from MATH that either MATH contains two independent hyperbolic elements, in which case the latter possibility can be arranged, or MATH contains no hyperbolic elements, or MATH is virtually cyclic, and then the first possibility holds). Say the first projection of MATH induces an action which is MATH. Therefore there is a free group MATH such that the first projection of MATH is NAME. Now apply the above principle with respect to each coordinate to replace MATH by a subgroup so that each coordinate action is either NAME or has no hyperbolic elements. For concreteness, we assume that coordinates MATH are NAME and NAME have no hyperbolic elements (MATH). We still call MATH and MATH the basis elements of MATH. We will adopt the convention in this proof that for MATH the MATH projection of MATH is denoted by MATH. The proof of REF (see the last sentence) shows that after replacing MATH by MATH for some MATH if necessary, we may assume that MATH for MATH. Next, elements MATH and MATH for MATH will have the property that MATH for MATH (see REF in the proof of REF ). We could then construct a sequence MATH as in the proof of REF (in the same manner as in the previous sentence) so that MATH for MATH and MATH. In addition, we want to arrange that MATH for MATH (note that we cannot hope to arrange MATH since MATH might have the same MATH and MATH projections). This can be done by modifying the expression for MATH so that it reads (for example) MATH with MATH. The idea is that MATH would force the situation where a long string of MATH's is close to both a long string of MATH's and a long string of MATH's, implying MATH. Of course, it can be arranged that this is false by replacing MATH with MATH from the previous paragraph. We now define quasi-homomorphisms MATH by the formula MATH for large MATH. These maps clearly extend to quasi-homomorphisms on MATH. The first summand in the above formula is unbounded on the cyclic subgroup MATH and it is positive on large positive powers of MATH. The second through MATH summands are nonnegative on large powers of MATH thanks to the fact that MATH for MATH. Finally, the other summands are bounded on MATH since MATH is not hyperbolic for MATH. Thus MATH is unbounded on MATH. A similar argument shows that MATH is bounded on MATH for MATH, so that the elements of MATH induced by MATH are linearly independent. By choosing the MATH's to lie in the commutator subgroup of MATH as before, we obtain an infinite linearly independent set in MATH and hence in MATH. |
math/0012115 | Let MATH be in the stabilizer of both MATH and MATH. Then there is an isotopy of MATH so that MATH and MATH. It follows that for some MATH depending only on the complexity of the graph MATH we have that MATH is isotopic to the identity. Therefore MATH consists of elements of finite order and is consequently finite (every torsion subgroup of a finitely generated virtually torsion-free group is finite). |
math/0012115 | The first two bullets in the definition of MATH are clear. Our proof of the remaining property is motivated by NAME 's proof (as explained in CITE) that the curve complex has infinite diameter. We recall the construction and REF 's space of projective measured foliations on MATH (see CITE and CITE). Let MATH be the set of all curve systems in MATH and by MATH denote the intersection pairing, that is, MATH is the smallest number of intersection points between MATH and MATH after a possible isotopy. Let MATH and by MATH denote the space of formal products MATH for MATH and MATH where we identify MATH with the subset MATH. Extend MATH to MATH by MATH . Consider the associated function MATH defined by MATH . Then MATH is injective and we let MATH denote the closure of the image of MATH. An element of MATH can be viewed as a measured foliation on MATH. The pairing MATH extends to a continuous function MATH . There is a natural action of MATH on MATH given by scaling. The orbit space MATH is NAME 's space of projective measured foliations and it is homeomorphic to the sphere of dimension MATH (assuming this number is positive). The intersection pairing is not defined on MATH but note that the statement MATH makes sense for MATH. The mapping class group MATH of MATH acts on MATH by MATH and there is an induced action on MATH, MATH, and MATH. Let MATH be a pseudo-Anosov mapping class. Then MATH fixes exactly two points in MATH and one point MATH is attracting while the other MATH is repelling. All other points converge to MATH under forward iteration and to MATH under backward iteration. It is known that MATH implies MATH and similarly for MATH. Continuity of MATH implies the following fact: If MATH is a neighborhood of MATH then there is a neighborhood MATH of MATH such that if MATH, MATH and MATH then MATH. We will use the terminology that MATH is adequate for MATH if the above sentence holds. A similar fact (and terminology) holds for neighborhoods of MATH. Given MATH, choose closed neighborhoods MATH of MATH and MATH of MATH with MATH so that CASE: MATH is adequate for MATH and MATH is adequate for MATH, and CASE: if MATH and MATH then MATH. Assume now that two curve systems MATH and MATH belong to a quasi-axis MATH of a pseudo-Anosov mapping class MATH and that they are sufficiently far away from each other, so that after applying a power of MATH and possibly interchanging MATH and MATH we may assume that MATH and MATH. Assume, by way of contradiction, that MATH is an infinite sequence in MATH and MATH, MATH for all MATH. Note that if MATH is a curve system disjoint from MATH then MATH, and inductively if MATH then MATH. We therefore conclude that MATH and MATH. After passing to a subsequence, we may assume that the sequence MATH converges to MATH and MATH. Note that MATH by the choice of MATH and MATH. First suppose that the curve systems MATH are all different. To obtain convergence in MATH one is required to first rescale by some MATH, that is, MATH where MATH can be taken to be the length of MATH in some fixed hyperbolic structure on MATH. Under the assumption that MATH are all distinct, we see that MATH and this implies that MATH that is, that MATH, contradiction. The case when MATH are all distinct is similar. Finally, if MATH and MATH take only finitely many values, we may assume by passing to a subsequence that both MATH and MATH are constant. But then MATH and REF implies that the sequence MATH is finite. |
math/0012115 | Let MATH be a homomorphism. By the NAME - NAME theorem CITE either the image of MATH is finite or the kernel of MATH is contained in the center. When MATH is a nonuniform lattice, the proof is easier and was known to NAME before the work of NAME - NAME (see NAME 's comments to REF on NAME 's list). Since the rank is MATH the lattice MATH then contains a solvable subgroup MATH which does not become abelian after quotienting out a finite normal subgroup. If the kernel is finite, then MATH is a solvable subgroup of MATH which is not virtually abelian, contradicting CITE. Now assume that MATH is a uniform lattice. If the kernel MATH is finite then there is an unbounded quasi-homomorphism MATH by REF . But then MATH is an unbounded quasi-homomorphism contradicting the NAME - NAME result that every quasi-homomorphism MATH is bounded. |
math/0012117 | Define the projection operators on the circle MATH and MATH with MATH. Thus in particular, we have MATH. Let MATH denote the multiplication operator MATH . Direct calculation shows that MATH . First, we show that MATH, MATH and MATH are trace class. Indeed MATH, where MATH. MATH acts on the basis MATH for MATH, as follows : MATH. We find MATH where MATH denotes the characteristic function of the set MATH. But MATH, and hence by REF, we have the trace norm estimate MATH . Now write MATH where MATH and MATH with matrix elements MATH . Write MATH where MATH, MATH denotes convolution on MATH by MATH, MATH and MATH, MATH are the projections onto MATH and MATH respectively. From REF, it is clear that MATH is bounded from MATH with norm estimate MATH . On the other hand, a similar calculation to REF shows that MATH is trace class from MATH and MATH which implies MATH . Similarly, we have MATH . Thus MATH, MATH and MATH are trace class. Moreover, if we set MATH, MATH, then from (the proofs of) REF, it is clear that as MATH, MATH, MATH, MATH in trace norm, and hence the NAME determinants converge to the corresponding determinants. Also for MATH sufficiently large, the winding number of MATH is the same as the winding number of MATH, and so we see that to prove REF, it is enough to consider MATH's which are non-zero and analytic in a neighborhood of MATH. Henceforth we will assume that MATH is analytic : this analyticity assumption is not necessary and is used only to give a particularly simple proof of REF below. In the below, we only present the proof of REF. The proof of REF is similar. Formally, we proceed as follows. Suppose MATH is finite rank so that MATH is also finite rank. We have MATH . Using MATH and MATH, the right-hand-side reduces to MATH . The first term in both cases is equal to MATH. Using REF and MATH for the last determinant, REF becomes MATH which is the desired result, up to the winding number MATH. For the case in hand, however, MATH is not a trace class operator and the above ``proof" breaks down. We circumvent the difficulty by approximating the operator MATH by finite rank operators, and the missing factor MATH will appear along the way. Let MATH be the projection MATH . Note that MATH is a trace class operator since it has finite rank. Clearly MATH strongly, and hence by REF, MATH . Now since MATH is trace class, proceeding as above in REF, we have for MATH, MATH . Thus we have MATH where for MATH, MATH . We observe that CASE: MATH and MATH are trace class. CASE: MATH strongly as MATH. CASE: MATH in trace norm as MATH. CASE: MATH and MATH are uniformly bounded in operator norm as MATH when MATH is small enough. The third property follows using REF as MATH is trace class and MATH strongly. For a moment, we assume that MATH is small so that REF is satisfied. Now we rewrite the right hand side of REF as MATH . From REF above, we have MATH in trace norm. Using REF , we now have MATH . Rewrite MATH as MATH . Then as MATH, MATH and hence MATH in trace norm as MATH strongly and MATH is in trace class. Also MATH strongly, and MATH, MATH are uniformly bounded as MATH for MATH small enough. Thus by similar arguments leading to REF, we have MATH . Now by REF, we note that MATH where MATH is MATH with MATH replaced by MATH, and MATH is the operator of multiplication by MATH. Thus we have MATH by REF . Since MATH, and MATH, we have MATH strongly. Since MATH is a trace class, we obtain MATH . Therefore from REF, MATH . Since MATH does not depend on MATH, we obtain the value of this determinant by letting MATH in both sides of REF. But by REF below, for small MATH, MATH as MATH. On the other hand, from REF, MATH converges to MATH. Therefore MATH, and we obtain, for small MATH, MATH as desired. The result for all MATH now follows by analytic continuation. |
math/0012117 | In REF, we obtained the result REF for MATH and MATH. For general analytic MATH and MATH small, the proof remains the same until REF . The second component in the asymptotics of MATH is now MATH, and hence REF is changed to MATH . The calculation for REF is similar and we skip the details. |
math/0012117 | The formal procedure (without considering the winding number) is as follows. For MATH, let MATH be the projection operator on MATH, and let MATH be the projection operator on MATH. Since we have from REF MATH the determinant on the left-hand-side in REF, denoted by MATH, is equal to MATH . First we pull out the term MATH, then use REF to obtain MATH . Now note that (recall MATH) MATH . Using REF and then multiplying two determinants, we have MATH . Finally, using MATH in the determinant on the right-hand-side of REF, we obtain MATH which is precisely REF from REF. The rigorous proof is also similar to the proof of REF . Let MATH be the projection on MATH as in REF. We take MATH large so that MATH. The analogue of REF is now MATH where MATH which becomes, by the same argument leading to REF, MATH with MATH in REF where MATH is replaced by MATH, This then leads to the desired result as in the single interval case. |
math/0012117 | We have MATH since MATH. Fix a number MATH. We split the integral into two pieces : MATH . In the first REF , using a standard argument and the convergence in distribution REF above, the limit becomes as MATH . For the second REF , the region is a union of two (not necessarily disjoint) pieces : MATH . Note that since MATH, MATH is either MATH or MATH. Over region REF , MATH . Similarly, MATH . Now from the tail estimates in REF below, the moment REF as MATH is equal to REF plus a term which can be made arbitrarily small if we take MATH large enough. However, from REF , for MATH large, REF is arbitrarily close to MATH. Thus we have proved the theorem. |
math/0012117 | We need to show that MATH . Fix MATH. We split the integral into two parts as in REF : CASE: MATH, and REF MATH. In REF , the integral is finite. In REF , the argument yielding REF implies that MATH . (Note that the additional terms corresponding to MATH in REF and MATH in REF are not necessary here as MATH is a smooth measure.) We will prove the finiteness of the last two expected values for MATH. First, we prove that MATH for any MATH. Note that by REF below, for MATH, MATH for some MATH. In particular, we have for any MATH, MATH . Thus, integrating by parts, MATH . Using REF, NAME 's lemma and REF, we have MATH . The proof of the finiteness of the second expected value in REF is similar using REF. |
math/0012117 | This is similar to REF (again note that MATH in CITE satisfies MATH.) Indeed, the proof of REF only requires the fact that MATH. In our case, MATH, which is clearly between MATH and MATH. The monotonicity can be found in REF . |
math/0012117 | Without any loss we can assume MATH. Note that since MATH, MATH . If MATH, then the expected value in REF is zero, and the bound is trivial. Thus we assume that MATH. Integrating by parts and using REF , MATH for large MATH, where MATH. Note that since MATH, MATH. We distinguish two cases : MATH where MATH is a fixed constant satisfying MATH, where MATH appears in REF . CASE: For all MATH, MATH. Note that for MATH, MATH . Hence from the estimate REF, we have MATH for MATH with a new constant MATH. Therefore, from REF, we obtain MATH . CASE: There is MATH such that MATH. We write REF as MATH using REF, where MATH is defined by REF with MATH and MATH. As in REF , for MATH, MATH and hence, the last integral is less than MATH. For the other terms, since MATH we have for MATH, MATH as MATH. Noting that MATH for some constant MATH, we have MATH for MATH with some constant MATH. Hence MATH . CASE: Bound REF : Recalling REF, if MATH, the expected value in REF is zero and the bound is trivial. Thus we assume that MATH. Integrating by parts and using REF , we have for some constants MATH, MATH for large MATH, where MATH as before. Given MATH, we take MATH so that for MATH, MATH. We distinguish two cases : MATH where MATH is a fixed constant as above. CASE: For all MATH, MATH. From the estimate REF, using MATH, MATH . CASE: There is MATH such that MATH. We write REF as MATH using REF, where MATH is defined by MATH in REF with MATH and MATH. The last integral is less than MATH with new constants MATH. For the other terms, note that since MATH and MATH, we have for MATH, MATH . Thus, we obtain MATH . |
math/0012117 | We need to prove REF. CASE: Estimate REF : It is enough to prove REF for MATH where MATH with MATH in REF . As noted above, since MATH is positive and MATH, its eigenvalues MATH satisfies MATH. Hence for MATH, by REF , MATH for any MATH. Therefore in order to prove REF, it is enough to show that MATH for some constants MATH where the summation is over the set (note MATH in REF) MATH with MATH. We will show that for MATH in MATH, MATH . Since MATH, REF follows from MATH . In order to show REF, since MATH it is enough to checking that MATH which is equivalent to check that MATH . But since MATH, MATH and hence REF is proved. Now using REF, the sum in REF satisfies MATH where the second inequality is due to the monotonicity of the function MATH. Since MATH, we obtain REF. CASE: Estimate REF : By a similar argument as in REF , it is enough to show that MATH for some constants MATH where MATH is the set MATH . For MATH in MATH, we have MATH . Now using REF, the sum in REF satisfies MATH . |
math/0012118 | Symmetry holds by definition. An annulus can be used to produce a grope of arbitrary class by gluing a trivial standard model into a puncture. Thus annuli can be used to demonstrate reflexivity in all cases. Transitivity should follow from gluing two grope cobordisms together. This can be done ambiently in MATH but extra care has to be taken to keep the glued grope embedded. For REF it can be seen as follows. One proves by induction on the number of surface stages that a grope cobordism MATH can be isotoped arbitrary close to a REF-dimensional complex MATH. One may assume that this spine MATH contains all the tips of the grope and one boundary circle MATH. Since we may use a strong deformation retraction of MATH onto MATH, the spine MATH (and in particular, MATH) is not moved during the isotopy. However, the other boundary circle MATH then undergoes quite a complicated motion and ends up running parallel to all of MATH. This means that in the following we have to be careful about introducing crossing changes on MATH because that might change the knot types of both boundaries of MATH. To prove transitivity of grope cobordism, we assume that two grope cobordisms MATH of class MATH are given with the knots MATH and MATH being isotopic. After pushing long enough towards the spines MATH, we may assume that MATH and MATH are disjointly embedded. This isotopy does not change the knot types on the boundaries of the gropes, even though it may change REF component link type of these boundaries (but that's irrelevant for our purposes). We now use our assumption and start moving the knots MATH and MATH closer to each other until they are parallel. At this point, we have to be careful not to change the knot types of MATH and MATH, for example, we can't just arbitrarily push MATH around in MATH. In fact, as pointed out above, MATH may not cross MATH at all, whereas it can cross MATH without changing any knot types. The same applies vice versa to MATH. To avoid changing our knot types, we first embed an isotopy between MATH and MATH into an ambient isotopy and run it until these knots are parallel, but with possibly parts of MATH still sitting in between them. Then we push MATH across MATH and MATH across MATH until MATH and MATH are honestly parallel in MATH. Finally, we consider tiny thickenings of the newly positioned MATH and MATH to gropes and glue them together using our pararellism. This may require twisting the annular region around, say MATH, so that the gluing in fact produces an embedded grope of class MATH as desired. Notice that the twisting does not affect the isotopy class of MATH or MATH. Now we turn to REF , that is, transitivity of capped grope cobordism. We use the same notation as in the previous case. In addition, we denote by MATH the caps of the grope MATH. By definition, the boundaries MATH are the tips of the grope and hence contained in the spine MATH. Hence the isotopy which pushes MATH towards the spine MATH can be done relative to MATH and we decide to do this isotopy with all of MATH fixed. This implies that the relevant data are the disjointly embedded caps MATH (except for the usual intersection points on MATH), together with the spine MATH which intersects the interiors of the caps. Next we implement the assumption that the caps only intersect the bottom stage of the original grope MATH. Since MATH this will still be true for a tiny neighborhood of MATH in MATH, which we now proceed to call MATH. By general position, the intersections of this thin grope MATH with the interiors of the caps are thus given by short arcs which run either from MATH to MATH, or from MATH to MATH. The case MATH to MATH does not occur because we chose MATH to be part of the spine MATH. Before proceeding with the argument, we devote a paragraph to what happens if a cap were allowed to intersect higher stages of the original grope. Then the intersections with the thin grope MATH would not be short arcs but rather certain unitrivalent trees which represent a normal slice through a grope. For example, for each intersection with the second stage one would see a small NAME tree in the cap, and the four univalent vertices of the H would lie on MATH. This can be illustrated in REF (which is not capped). The second surface the big evident NAME surface with two dual bands. An intersection of some disk through one of these bands also picks up intersections with the grope's bottom stage, which has one band which traces around the boundary of the second stage. Similarly, for each intersection of a cap with the MATH-th stage of the grope one would see a small tree with MATH trivalent vertices and MATH univalent vertices (which would lie on MATH). Thus the topology of these intersections distinguishes the different stages of the grope. In the following, we shall refer to all such intersections with stages of the grope above the bottom as ``NAME. Only the bottom stage produces arcs of intersections, and only in this case can the preferred boundary MATH appear in the interior of a cap. Now consider two capped gropes of class MATH with caps MATH respectively MATH and grope bodies MATH which we already assume to be pushed close to the spines MATH (keeping the caps constant). We then do the same move as in the uncapped case, making MATH and MATH parallel in MATH. This can be done keeping the caps constant because MATH and MATH are disjoint parts of MATH, and similarly for MATH. After twisting an annulus as before, we may glue the grope bodies along the common boundary MATH to obtain an embedded grope MATH of class MATH. The intersection arcs of the caps with the glued up annular regions (around MATH and MATH) now all run from MATH to MATH, hitting the intersection MATH once on the way. These are intersections of the caps with the new grope's bottom stage, and hence are allowed. We need to clean up the intersections of the caps which intersect each other and also the higher stages of the new grope. These intersections are totally arbitrary, except the two sets of caps are disjoint and the MATH caps avoid the higher stages of MATH and the MATH caps avoid the higher stages of MATH. A consequence of the first fact is that there are no triple points of intersection among the caps. After pushing little fingers across the boundary of the caps, there are no circles of double points, but we gain some new intersections of caps MATH with a top stage of MATH and vice versa. Now consider one cap MATH and recall that near its boundary a normal slice of MATH is NAME, with MATH on the univalent vertices. This implies that we may push every intersection that does not contain this knot MATH off MATH and across the normal slice. In particular, all intersections with MATH can be removed this way: Every ribbon and clasp intersection can be pushed across the boundary of MATH because only crossing changes between MATH and MATH are introduced (and all knot types stay the same). Doing this clean up procedure with each of the caps MATH, we end up with disjointly embedded caps for MATH, but possibly intersecting all stages of this grope. The next step, now that all the caps are disjoint, is to remove intersections of the caps MATH with higher stages of the grope MATH, and vice versa. Suppose that a cap MATH intersects higher stages of MATH. It will do so along some unitrivalent graph, but any univalent vertices are part of MATH. Thus we may push all of these intersection out of the cap MATH and across the normal slice, introducing crossings of MATH and MATH, which do not change the isotopy class of either. Similarly, higher stages of MATH will only intersect caps MATH so that they can be pushed off again without changing the knot types. This leads to a capped grope cobordism of class MATH between MATH and MATH and thus transitivity is proven. |
math/0012118 | The way to push genus down the grope is shown in REF . It shows how to trade genus of a stage with the previous stage. NAME run an arc from the previous stage across the current such a way as to separate the genus. Then run a small tube along the arc, increasing the genus of the previous stage. The dual depicted by MATH in the picture. In order to make the tree type of the grope behave as on the left of REF , we push off a parallel copy of MATH. (In the capped situation, MATH will have caps, which should be included when pushing off a parallel copy. The new caps will also only intersect the bottom stage.) The parallel copies of MATH may intersect, a fact we have depicted in REF . (In REF dimensions, however, they do not intersect if the grope is framed, so there is no further problem.) However, we can still iteratively apply this procedure, despite the self intersections until all the genus is at the bottom stage. But we can further subdivide the resulting grope cobordism with genus MATH at the bottom to a sequence of MATH cobordisms with genus one at the bottom stage, as on the right of REF . We claim that each genus one grope cobordism MATH is embedded. This can be seen schematically in REF which is supposed to show that the only intersections that arise come from parallel copies MATH and MATH which will eventually belong to distinct gropes MATH and MATH. This follows from the fact that the tree type of the gropes only changes as in REF which implies that at each step parallel copies correspond to distinct branches emanating out of a box. In the last step of the pushing down procedure, these different branches actually become distinct gropes MATH. If there are caps, note that they will still only intersect the bottom stage of the grope MATH they are attached to, even though they may intersect higher stages of MATH. |
math/0012118 | Construction of the unframed clasper Assume the grope is augmented with pushing annuli. Then each surface stage of the grope has two surfaces which attach to it, and these are either pushing annuli or higher surface stages of the grope. In order to simplify terminology, refer to both these types of surface as higher surfaces. Let MATH be a surface stage of the embedded grope, with higher surfaces MATH and MATH attaching to it. Then MATH is a point MATH, and in a neighborhood of this point MATH, MATH divides MATH into four quadrants. We distinguish two of these as follows. Let MATH be an ordered basis of the tangent space MATH constructed as follows. Let MATH be transverse to MATH and pointing into MATH. Choose MATH tangent to MATH. Choose MATH tangent to MATH in such a way that MATH is a positive orientation of MATH. The two quadrants lying between MATH and MATH and between MATH and MATH are called positive quadrants, see REF . There were two choices in selecting MATH, namely which surface is called MATH (MATH versus MATH) and which direction of MATH the vector MATH points along (MATH versus MATH.) Changing MATH to MATH will also change MATH to MATH in order to preserve the orientation MATH. Therefore, the positive quadrants do not change. If one changes MATH to MATH, then the role of MATH and MATH is reversed. But MATH is still positive, and hence the positive quadrants are those between MATH and MATH, as before. We are now ready to define the unframed clasper MATH in MATH. The leaves include those ends of the pushing annuli which are not attached to anything. (These are the tip leaves.) There is one more leaf which is a meridian to MATH.(This is the root leaf.) This leaf punctures the bottom stage of the grope in a single point. Every surface stage contains a node of MATH where the higher surfaces intersect. Hence each pushing annulus has a node on its boundary. This is connected by an embedded arc in the annulus to the tip leaf at the other end. Each surface stage except the bottom stage contains two nodes: one on the boundary and one in the interior. Connect these by an embedded arc in the surface stage whose interior misses the attaching regions for the higher surfaces, and such that it emanates from the interior node in a positive quadrant. Finally connect the node on the bottom stage to the intersection of the root leaf with the stage by an embedded arc whose interior avoids the attaching regions for the higher surfaces, and which emanates from the node in a positive quadrant. REF shows the construction for a grope of class MATH. Figuring out the framing The tip leaves of the clasper have obvious framings along the annuli they are contained in. Similarly each edge has an obvious framing as a subset of a surface. Framing a node is depicted in REF . Notice that the edge on the surface approaching via a positive quadrant. We glue together the perpendicular framings of the two edges associated to the higher surfaces with two triangles inside the positive quadrants. The framing of the approaching edge is naturally glued to one of these triangles. We can frame the root leaf using the meridional disk it bounds. This needs to be glued to the perpendicular framing of the incident edge. This is shown in REF , where we again use two triangles to glue up different parts of the clasper. Notice that this is the only place at which the clasper is not a subset of the grope but the triangles are defined as in the discussion of positive quadrants. Proving that this works We proceed by induction on the number of surface stages, the base case being a surface of genus one. Let MATH be the base surface, MATH the augmented surface and MATH the clasper we just constructed. The pair MATH in MATH can be realized as the restriction of an orientation preserving embedding into MATH of the genus two handlebody, which is a regular neighborhood of the standard picture given in REF . By REF is an embedding of the given picture, ignoring the clasper MATH. We precompose this embedding with a suitable orientation preserving automorphism of the regular neighborhood which fixes MATH pointwise and MATH setwise. Clearly the edges on the pushing annuli can be straightened out by twists supported in the annuli's interiors, and these twists extend to the regular neighborhood. Hence it suffices to straighten out the edge MATH which runs along MATH. Let the annuli be called MATH and MATH. The interior of MATH lies in the (open) annulus MATH. It can therefore be straightened via NAME twists. It also can approach MATH in two ways: by the two positive quadrants. There is an automorphism of MATH rel MATH taking one quadrant to the other. This is depicted in REF . Because of this lemma, it suffices to check that MATH in the standard model of REF . (We need the embedding to preserve orientations because an orientation is required to associate a well-defined link to the clasper.) The standard model is redrawn in REF , with heavy lines deleted from the ambient REF-ball to make it a regular neighborhood of MATH. The clasper is cleaned up a little bit in the second frame, and then the second NAME cancellation from REF is used to produce MATH in the third frame. Finally, an isotopy moves MATH to the knot MATH as shown in the remaining frames. Now for the inductive step. This follows from REF . Pictured is a top stage of the grope and part of the clasper MATH we constructed. In frame REF we have broken the edge of the clasper that lies on the top surface into two claspers MATH and MATH. This is the first NAME cancellation from REF and gives MATH. By induction we know that the clasper surgery MATH has the pictured effect on MATH since the indicated section, MATH of the leaf of MATH cobounds the surface stage corresponding to the clasper MATH with the pictured arc MATH. This gives rise to a new clasper MATH which corresponds to the grope which is gotten by forgetting about the indicated surface stage. MATH and MATH still bound this new grope, and by induction MATH which we saw is equal to MATH. |
math/0012118 | We have drawn MATH in REF , and replaced the clasper by MATH-framed surgery on the associated link. The curve MATH bounds the genus one surface MATH. Note that part of MATH travels over an attached MATH handle. Two dual curves on MATH each cobound an annulus with a parallel copy of the two lower leaves. These annuli are denoted MATH and MATH, and each also runs over an attached MATH handle. |
math/0012118 | REF is proven similarly to REF, using a sort of inverse to NAME 's move REF, which is the identity in REF . Now consider REF . One can plug either of the two pairs of arcs (clasped respectively unclasped) on the right of REF into the shaded region. After applying NAME 's version of the zip construction (using claspers with boxes) as shown in the figure, one obtains a (disconnected) clasper with boxes MATH, and two claspers MATH and MATH, containing the MATH-twists. Whether one gets MATH or MATH depends on what one plugs into the shaded region. There is an important subtlety here. Surgery along a rooted clasper (without boxes by definition) only affects the pair MATH inside a regular neighborhood of the clasper and its root disk, and is fixed outside of this neighborhood. On the other hand, for claspers with boxes, one may have to choose many roots, modifying the pair MATH inside a regular neighborhood of the clasper and its root disks. In REF , these added roots must include some of the little ``lassoes" coming out of the boxes. Hence the clasper MATH actually modifies MATH and MATH to two claspers MATH for MATH. Note that since MATH differ by a finger move, so do MATH. By the above move MATH . On the other hand by NAME 's move REF, MATH. Thus we have found a knot MATH in MATH and two claspers MATH which differ by a finger move in MATH and satisfy the desired identities: MATH and MATH. REF is REF and the proof is esentially the reverse of the above argument. |
math/0012118 | We only prove the cases when the tree MATH has at least REF edges. The other case is similar. By REF a clasper surgery on MATH along MATH can be thought of as changing the clasper surgery on some knot MATH from MATH to MATH as in REF . Now apply REF twice as follows: This implies our claim MATH. |
math/0012118 | CASE: Any tree of class MATH can be changed into a sequence of MATH-trees using geometric IHX. This can be proved by introducing the following function on rooted class MATH trees MATH: MATH is the maximum length of a chain of edges. Given MATH, consider a chain of maximal length MATH, and suppose it misses some internal vertices. Let MATH be an internal vertex of distance MATH from MATH. Then, by geometric IHX, this tree can be realized as a sequence of two trees with higher MATH: Hence we can keep applying IHX until we have a sequence of trees with maximal MATH, which as we have seen means that that a maximal chain hits every internal vertex. This is just a rooted MATH-tree. CASE: Let MATH denote the set of knots related to the unknot by capped tree clasper surgeries of NAME degree MATH. Similarly let MATH denote those knots which are related to the unknot by degree MATH capped tree claspers whose tree type is that of the half grope. Define MATH to be those knots related to the unknot by degree MATH rooted tree clasper surgeries, and let MATH bethe analogous object, restricting to half grope trees. (By REF , MATH.) We have the following map of short exact sequences: MATH and, by REF , the middle map is an isomorphism. By CITE, MATH is a group, a fact which implies that everything in the above diagram is a group (under connected sum). By the MATH lemma, the right hand map MATH is an isomorphism, as desired. Recall that all of the above quotients are defined as in the introduction, and are in particular not just quotient monoids. |
math/0012118 | Write out the left hand side clasper surgery as a surgery on the usual REF component link corresponding to the NAME. Then slide the visible part of the knot or clasper over one component of the Borromean rings. |
math/0012118 | Break MATH into a union of NAME and inductively apply REF . |
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