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math/0012118 | The statement follows from the following more general statement: Inside a regular neighborhood, MATH, of MATH, there are two claspers MATH and MATH as above, such that MATH is diffeomorphic rel boundary to MATH. Notice that since MATH avoids any caps that MATH may have, it in particular avoids the root leaf. We proceed by induction, the picture above serving as the base case. In the pictures that follow, the thicker lines denote a regular neighborhood of a clasper. To induct, we break the clasper MATH into a union of two simpler claspers as follows: The big box is a pictorial convenience to represent an arbitrary clasper. Inductively we get the following picture: Then using the base case on the left leaf of the right-hand clasper, we obtain By REF applied to the grey leaf on the right and a cancellation of the bottom NAME pair, we get Next we would like to cancel the grey-black NAME pair above. This requires some care because parts of MATH run parallel to the grey leaf MATH. However, in our construction, MATH avoids the caps of MATH. Thus we can split the regular neighborhood of MATH apart into the leaf, plus a parallel copy of that leaf through which other claspers wander: After that we apply a sequence of REF moves to obtain a clean NAME pair that can be cancelled. In the figure below we also push some black arcs into the grey area which after all only represents some neighborhood of the clasper: Thus we have finished the inductive step. |
math/0012118 | Push each intersection point of an edge with the given disk bounding MATH out toward the other leaves, using little fingers following the spine of the clasper MATH. Each such finger splits into two at a trivalent vertex of MATH, and stops right before a leaf (which is necessarily distinct from MATH). This describes a new disk MATH bounding MATH which has the property that on each edge MATH incident to a leaf MATH there are several parallel sheets of MATH being punctured by MATH (and there are no intersections of MATH with edges other than MATH). If the leaf MATH happens to be the root leaf, we push these sheets over the cap of MATH, introducing intersections with the knot, but eliminating the intersections with MATH. If MATH is not the root, we add a series of nested tubes that go around MATH, trading the intersections with MATH for genus on MATH. Thus MATH now bounds an embedded surface which intersects MATH but is disjoint from the clasper MATH. We perform the zip construction on MATH to segregate the knot intersections, where the first daughter MATH will inherit the half of MATH bounding a disk intersecting the knot. This first daughter is of type REF . The second daughter has the leaf coming from the half of MATH bounding a surface disjoint from the clasper MATH. Converting the clasper to a grope we get a grope of tree type MATH whose tip corresponding to MATH bounds a surface disjoint from the grope. Hence we really have a grope of increased class, but it has high genus at the tip MATH. REF now yields a sequence of cobordisms of type MATH as claimed. |
math/0012118 | By REF we may assume that all surface stages of the given grope are of genus one. Such a grope cobordism corresponds to a MATH-clasper surgery, which we proceed to simplify. CASE: First we make the leaves MATH-framed. This is accomplished using the following simple observation. Suppose MATH and MATH represent symplectic basis elements on a punctured genus one surface embedded in MATH. These have framings MATH, the diagonal terms of the NAME matrix. There is also the intersection pairing MATH. By REF. The formula MATH implies that if MATH and MATH, then MATH. By NAME twisting one can represent MATH by embedded curves meeting at a point. So MATH represent a MATH-framed basis of MATH. In particular, suppose MATH is a surface stage of the grope for which MATH is a tip, and MATH bounds a higher surface stage. Then MATH and we can let MATH be the tip in place of MATH. This takes care of all possibilities except the case when MATH and MATH are both tips of the grope which have nonzero framings. Here we perform some sleight-of-hand using claspers. Convert the grope to a clasper MATH. Then insert a NAME pair of leaves on the edge incident to MATH. This disconnects the clasper into two pieces MATH as in REF . The tips MATH and MATH each lie on exactly one of these claspers. The other leaf MATH of MATH bounds a grope MATH gotten from MATH, by considering MATH as the root leaf. MATH is the curve on the bottom stage of MATH which bounds the next surface stage, as pictured. By changing MATH to MATH as before, we convert the tip MATH of MATH to a zero-framed tip. Changing MATH to a clasper MATH by our procedure, we again have the clasper MATH with the leaf MATH . NAME the root MATH of MATH. Convert this back to an edge to achieve a clasper of the same type as MATH, but with one more tip zero framed. This clasper may be converted back to a grope if we wish. Notice that under our grope-clasper correspondence, the framings of tips (leaves) do not change. Do this until all tips are zero-framed. CASE: Next we make the leaves unknotted. It is an exercise to prove that there is a set of arcs from a knotted leaf to itself, such that cutting along these arcs yields a collection of unknots. Hence, given a knotted leaf, one can apply the zip construction to such a set of arcs, thereby reducing the number of knotted leaves in each resultant clasper. Repeat this procedure until you have a set of claspers with unknotted leaves. Note that we have now proved that any MATH-clasper surgery can be reduced to a sequence of MATH-clasper surgeries, each of which has MATH-framed leaves bounding disks. To continue, we need to clean up the intersection pattern of the disks. By pushing fingers of disks out to the boundary, one may assume each pair of disks intersects in clasp singularities; that is, the intersection pattern on each disk is a set of arcs from interior intersections with the clasper to the boundary of the disk. Secondly, we eliminate triple points. After we did the first step, there is a triple point which is connected by a double point arc to the boundary of one of the disks, such that there are no intevening triple points. Push a finger of the disk which is transverse to this arc along the arc and across the boundary. Repeat this until all triple points have been removed. This homotops the disks into a position such that the intersection pattern consists of disjoint clasp singularities. CASE: We now start with a clasper MATH which has MATH-framed leaves bounding disks MATH with only clasp intersections between each other. In addition, the disks MATH may have several types of intersections with MATH and the knot MATH, which we proceed to organize. Note that our theorem states that, modulo higher grope degree, we can reduce to only two types of singularities for the MATH: Either there is a single clasp (and no other intersections with MATH or MATH), or there is a single intersection with the knot MATH (and no intersections with MATH). We call such disks good for the purpose of this proof. The bad disks fall into several cases which we will distinguish by adding an index to the disk MATH which explains the failure from being good. The cases are as follows, where we list exactly the singularities of the disk, so unmentioned problems do not occur. If a disk MATH has intersections with edges of MATH, we call it MATH. more then one intersection with MATH, we call it MATH. has more than one clasp, we call it MATH. intersections with edges of MATH and with MATH, we call it MATH. intersections with edges of MATH or with MATH, and has clasps, we call it MATH. Just to be clear, the cases MATH, MATH and MATH above represent disks without clasps, whereas MATH has no intersections with edges of MATH or with MATH. It is clear that these cases represent all possibilities for a bad disk. Recall that a disk MATH was called a cap if it is embedded disjointly from MATH. In our notation, this means that a cap is either bad of type MATH (more than one intersection with MATH), or it is good (exactly one intersection with MATH). We ignore the case of a cap without intersections with MATH since then the surgery on the clasper has no effect on MATH. We now introduce a complexity function on claspers with given disks MATH as above. It is defined as a quintuplet MATH of integers MATH, ordered lexicographically. The MATH are defined as follows: CASE: MATH is minus the number of disks MATH which are caps. CASE: MATH is the total number of intersections of the knot with caps MATH. CASE: MATH is the total number of clasps. CASE: MATH is the number of bad disks of type MATH. CASE: MATH is the number of bad disks of types MATH. The proof proceeds by using the zip construction to split a bad disk of a clasper into two daughters. In each of the five cases given below we check that both daughter claspers have either smaller complexity or higher grope degree, so they are ``cleaned up". The five cases can be applied in an arbitrary order and they are performed as long as there is a bad disk on a daughter clasper (where we do not work on claspers of higher grope degree). Since each MATH is bounded below, this cleaning up process must terminate. This can only happen if all disks are good (or the clasper has higher grope degree), which is the statement of our theorem. We now describe the five cases of the cleaning up process. In each case the label says which bad disk is being split, then we have to specify the splitting arc and the order of the daughter claspers. CASE: Suppose there is a bad disk of type MATH. By REF , this splits into a daughter clasper MATH of the same degree but with an extra cap, and into a sequence of claspers of higher grope degree. For MATH the number MATH is reduced. CASE: Suppose there is bad disk of type MATH. Split along an arc that divides the intersections with MATH into two smaller sets. Each daughter clasper inherits a cap with fewer intersections, so MATH goes down for both daughters (whereas MATH is unchanged). (Cl) Suppose there is bad disk of type MATH. Draw an arc along the disk separating the clasps into two smaller groups. The zip construction produces two daughter claspers MATH and MATH for which MATH are preserved. To calculate the change in MATH we need only consider the leaves of MATH and MATH as MATH does not see knot or edge intersections. The leaves of MATH differ from those of MATH, only by cutting off part of the leaf we are splitting along. By construction, this has fewer clasps, that is, MATH is reduced for both daughters MATH. MATH . Suppose there is bad disk of type MATH. Split along an arc separating the two types of intersections, such that MATH inherits the part of the leaf with just edge intersections. Since the intersection pattern for MATH is just a subpattern of the original, the entire complexity function cannot increase. But MATH clearly decreases for MATH because a new disk with only edge intersections has been created. On the other hand MATH has a new cap, so MATH decreases for it. (EK,Cl) Suppose there is bad disk of type MATH. Split along an arc which separates the clasps from the other types of intersections. Split in such a way that MATH inherits the part of the leaf which has the clasps. Now MATH is preserved in MATH. The cut leaf now has only clasp intersections, and since the intersections of the disks of MATH with everything are decreased, new disks with both clasp and other types of intersections are not created. Hence MATH decreases for MATH. Now we analyze MATH. Since MATH can only go down when we split, it suffices to show that MATH decreases. This follows by the same argument as case (Cl). We note that in the above five cases, when we split along a disk, the caps away from the split disk are preserved, as are the number of intersections of the knot with these caps. Furthermore, in the first daughter clasper MATH the four complexity functions MATH must each stay the same or go down, because the intersection pattern of MATH is just a subpattern of the one for the original clasper. The number MATH can only increase during a MATH-move, but then MATH goes down for the first daughter MATH (and MATH has higher grope degree). The intersection pattern for MATH changes in a more complicated way. The first problem is that it sits on a different knot: the knot modified by MATH, which adds intersections of the knot with the disks MATH. (We are applying MATH and MATH sequentially!) The second problem is that the edges of MATH wander around inside a neighborhood of MATH and add intersections as well. Therefore, the complexities MATH and MATH may increase from MATH to MATH in all moves above, except for MATH. We summarize the information of these moves in the following table. Observe that performing a move always implies a reduction of the relevant complexity MATH, which we have written first in its row. Other complexities may or may not increase, and in some cases they actually decrease. In that sense the table contains the worst case scenario for the complexities MATH of the two daughter claspers. The notation MATH means that MATH may increase (which is bad), whereas MATH is the good case where the complexity definitely decreases. Unmentioned complexities MATH either stay unchanged or decrease. We see from this worst scenario table that for all the five moves the total complexity goes down for both daughter claspers (or the grope degree increases). This completes our argument. |
math/0012122 | We restrict to the proof of the first inclusion the other one being similarly. By REF we may assume that MATH and MATH are reduced divisors. We need to show that MATH . The inclusion MATH follows from REF . To show the converse inclusion, let MATH be a resolution of singularities such that MATH, MATH, are SNC divisors. If MATH is a section of MATH defined over some open subset MATH of MATH, then MATH is a form in MATH that has at most logarithmic poles along the irreducible components of MATH and is holomorphic along the components of the proper transform of MATH. Hence MATH is a section of MATH so that MATH, as required. |
math/0012123 | If MATH, then by definition the restriction of MATH to the boundary of MATH lies in the image of the NAME projector MATH. In particular, if MATH, then the restriction of MATH to the boundary lies in the intersection of MATH and the image of MATH. The unique continuation property for MATH implies that this intersection is exactly the kernel of MATH, that is, the kernel of MATH is isomorphic to MATH. As a discrete self - adjoint operator, MATH is invertible if and only if MATH. Moreover, MATH is a self - adjoint projection satisfying the equation MATH. Thus MATH. |
math/0012123 | The first statement is a special case of CITE, where the homotopy groups of MATH and MATH are computed. The path connectedness of MATH was proved in REF. The path connectedness of MATH can be proved along the same lines: if MATH then MATH and hence MATH, where MATH, is a path in MATH connecting MATH and MATH (compare for example, CITE). |
math/0012123 | The first part was proved above. Since MATH, any element in MATH can be written in the form MATH for MATH. Since MATH it follows that MATH . Thus the restriction of MATH to the image of MATH is MATH . It follows that MATH is NAME (that is, MATH restricts to a NAME operator on the image of MATH) if and only if MATH is NAME, which occurs precisely when MATH is not in the essential spectrum of MATH. Similarly MATH is invertible (that is, the restriction of MATH to the image of MATH defines an isomorphism onto the image of MATH) if and only if MATH is not in the spectrum of MATH. The same argument also shows REF . Finally, since MATH is smoothing if and only if MATH is smoothing. Here we use that the projections MATH onto MATH are differential operators of order MATH. Since MATH are pseudo - differential and unitary the operator MATH is smoothing if and only if MATH is smoothing. |
math/0012123 | Suppose MATH is a subspace with MATH. Then it is easy to check that the orthogonal projections MATH are isomorphisms and hence MATH. Conversely, if MATH then let MATH be a unitary isomorphism. Then MATH is a subspace satisfying MATH. |
math/0012123 | We only sketch the proof, since this fact is well - known, at least when the MATH - function is regular at MATH, and the general case is proven by the same argument, because the pole of the MATH - function at MATH is determined by the asymptotics of the spectrum, whereas the spectral flow depends only on the small eigenvalues. Given MATH, choose a MATH so that MATH does not lie in the spectrum of MATH. Applying standard results from perturbation theory CITE we infer that MATH does not lie in the spectrum of MATH for MATH close enough to MATH, say MATH. Moreover the span of those eigenvectors of MATH whose eigenvalues lie in MATH varies continuously for MATH. Thus we can write MATH for MATH and MATH as a sum MATH . The sum MATH is finite, and so its analytic continuation to MATH is integer valued: MATH . Thus MATH . Notice that this equation depends on our choice of convention for defining the spectral flow. The function MATH varies smoothly in MATH since we have subtracted the eigenvalues that cross zero, and since no eigenvalues equal MATH in this interval. If we define MATH similarly to REF then MATH is smooth and MATH . Therefore, using REF we obtain MATH . Dividing by MATH proves the lemma over the interval MATH. The general case is obtained by covering the interval MATH by small subintervals and adding the results. For the last assertion, notice that if the dimension of MATH is independent of MATH, then MATH for all MATH. |
math/0012123 | Assume first that MATH is the NAME projector MATH and that the pair MATH is invertible. By REF this means that MATH and MATH are invertible. Putting REF together and taking into account that MATH is independent of MATH REF , we obtain MATH and thus MATH . Since MATH is of trace class we may choose a self - adjoint trace class operator MATH such that MATH. Then MATH where we have used the multiplicativity of the NAME determinant in the last line. Consequently MATH . Putting together REF we obtain REF for MATH and MATH such that MATH is an invertible pair. However, since both sides of REF depend continuously on MATH, REF remains valid for all MATH. Finally, if MATH are arbitrary then MATH . |
math/0012123 | We assume first that MATH is invertible. MATH is invertible by REF . In view of REF the space of those MATH so that MATH is invertible is path connected. Choose a smooth path MATH in MATH starting at MATH and ending at MATH so that MATH is invertible for all MATH. The spectral flow of MATH equals zero since the kernel is zero along the path and so REF shows that MATH is smooth. Hence MATH is smooth. Also, the map MATH is smooth since MATH for all MATH and hence MATH is holomorphic on MATH. REF states that the two smooth functions of REF are the lifts to MATH of the same function to MATH, and they both start at MATH. Hence they coincide for all MATH. Now let MATH be arbitrary. We may choose a path MATH in MATH such that MATH is invertible for MATH, MATH, and such that at MATH exactly MATH eigenvalues of MATH cross MATH from the upper half plane to the lower half plane and no eigenvalues cross from the lower half plane to the upper half plane. To see this let MATH be the orthogonal projection onto MATH. The projection MATH is a pseudo - differential operator. Now put MATH . By our choice of MATH we then have MATH . Moreover, for MATH we have from the first part of this proof MATH . From REF one infers MATH and since MATH at MATH exactly MATH eigenvalues of MATH cross MATH from MATH to MATH and no eigenvalues cross from MATH to MATH. Hence MATH completing the proof. |
math/0012123 | This follows from the fact that MATH and MATH preserve this splitting. Hence, MATH. Note that the splitting REF induces the splitting REF by restricting to the boundary. |
math/0012123 | We know from REF that MATH is path connected. This assures the existence of a path MATH. Furthermore, notice that MATH since MATH and MATH preserve the splitting of MATH. REF imply that MATH and MATH . Note that in REF MATH is taken with respect to MATH and in REF MATH is taken with respect to MATH (compare REF. of the warning above). In view of REF we find MATH and consequently, MATH . Adding REF and using REF gives the desired formula. |
math/0012123 | Fix a Lagrangian subspace MATH and let MATH. The only part which is not straightforward is the claim that MATH, MATH is a NAME pair. We use the following criterion (compare CITE). Two orthogonal projections MATH in a NAME space form a NAME pair (invertible pair) if and only if the operator MATH is NAME (invertible). One calculates MATH . Hence if MATH then the pair MATH is NAME (invertible) if the pair MATH is NAME (invertible). |
math/0012123 | Note again that in view of REF the space MATH is path connected. Using MATH one obtains a map MATH: MATH . Since MATH is the constant map at MATH, one sees that the path MATH is homotopic to the composite of the paths MATH and MATH . The claim now follows from the homotopy invariance and additivity of the spectral flow. |
math/0012123 | By REF the space MATH is isomorphic to MATH. Hence, if we can show that MATH is independent of MATH, then MATH . Consider MATH. In view of REF this means MATH . Since MATH we infer MATH. Conversely, given MATH put MATH. Then REF implies that MATH. REF is an immediate consequence of REF . |
math/0012123 | We note that it was proved in CITE that MATH is smooth. Consider the two parameter family of operators on MATH . By REF for fixed MATH the dimension of the kernel of MATH is independent of MATH. By the homotopy invariance of the spectral flow this implies MATH . Since MATH the right hand side equals MATH. The left hand side equals MATH and since MATH is invertible its spectral flow vanishes and we reach the desired conclusion. |
math/0012123 | Fix a Lagrangian subspace MATH and choose a smooth path MATH from MATH to the NAME - NAME - NAME projection MATH. Set MATH. By REF we have MATH . Using REF we obtain MATH . This proves the first line of REF. The second line of REF and the last assertion follow from REF . |
math/0012123 | It was remarked after REF that MATH and MATH. Consequently, MATH and hence MATH is trace class. REF implies that MATH is equal to MATH . Applying REF to the three summands in REF and taking REF into account gives the assertion. |
math/0012123 | CASE: Let MATH be the NAME algebra. Then the quotient map MATH sends MATH onto MATH. Moreover, MATH acts freely (from left and right) on the fibers. Thus we obtain a fibration MATH. The claim now follows since MATH is contractible. To see the latter we note that for any MATH - algebra MATH the set MATH is contractible. The contraction is given by MATH, MATH. This is well - defined since MATH. CASE: Let MATH. Then MATH splits into MATH and MATH. Now put MATH. |
math/0012123 | First note that if MATH and MATH then since MATH is compact one has MATH, in particular MATH. If MATH are different paths with MATH then consider the closed paths MATH and MATH, where MATH denotes the path MATH traversed in the opposite direction. Since the pointwise product of closed paths MATH is homotopic to MATH we find MATH . This shows that MATH is well - defined. The homotopy invariance is straightforward from the definition and the homotopy invariance of the winding number. CASE: This assertion is a consequence of REF. CASE: That MATH follows immediately from the definition. For MATH the third identity follows from REF . (note the normalization REF of MATH). If MATH is arbitrary we apply REF. and choose a continuous path MATH such that MATH and such that MATH is independent of MATH. The claim now follows from the homotopy invariance REF. |
math/0012123 | We apply REF. with MATH and obtain using REF. MATH . |
math/0012123 | In all four cases the diffeomorphism is given by MATH . By NAME 's Theorem CITE the space MATH is contractible and hence we obtain the claimed homotopy equivalences. |
math/0012123 | In view of REF the right hand side of REF induces a group isomorphism MATH. It remains therefore to check the sign convention and the convention for paths with endpoints not in MATH. Let MATH. Then, by definition, MATH for MATH small enough. A straightforward calculation shows MATH and thus for MATH small enough we have, in view of REF, MATH . To check the convention for paths with endpoints not in MATH we consider the paths MATH, respectively, MATH. By definition we have for MATH small enough MATH and, analogously, MATH . According to our convention for the winding number we have, on the other hand, MATH . In view of REF the proof is complete. |
math/0012123 | CASE: In view of REF and the previous theorem we find MATH . CASE: Using the previous Theorem and REF we obtain (we write MATH instead of MATH) MATH . |
math/0012123 | By REF we have MATH . Now the claim follows immediately from the homotopy invariance of the NAME index. |
math/0012123 | We prove REF first. From REF and the definition of MATH we infer MATH . To prove REF we assume, without loss of generality, that MATH is compact. Let MATH. Then we obtain from REF MATH . |
math/0012123 | Certainly MATH is a NAME space and MATH leaves MATH invariant. If we can prove that MATH is Lagrangian in MATH then from REF we infer MATH. What remains, therefore, is to prove the second part of REF without using the fact that MATH. We note first that we have an orthogonal direct sum decomposition MATH . Also, MATH is an isotropic subspace of MATH. In fact, if MATH then, since MATH is Lagrangian, MATH. Writing MATH then MATH. Next consider MATH such that MATH. Thus for all MATH we have MATH. Hence MATH and consequently, since MATH is closed, MATH. We may write MATH. From MATH we infer MATH and hence MATH. Summing up we have proved MATH. Since MATH is isotropic this implies MATH. Thus MATH is a Lagrangian (in particular closed) subspace of MATH. From REF it is now clear that MATH is isomorphic to MATH. |
math/0012123 | The proof is exactly the same as the proof of CITE. Namely, the isometry MATH maps the domain of MATH onto the domain of MATH and it anticommutes with MATH. Hence MATH and we are done. |
math/0012123 | Denote by MATH the manifold with boundary obtained by removing MATH from MATH. As in REF for MATH we denote by MATH the MATH - invariant of the operator on MATH obtained from MATH by imposing the boundary condition MATH at MATH and the boundary condition MATH at MATH. The MATH gluing formula for the MATH - invariant REF then implies MATH for MATH the NAME - NAME - NAME projection with respect to a Lagrangian subspace MATH. One easily checks that MATH hence REF implies MATH . Also by REF MATH . Plugging this into REF and applying again the MATH splitting formula for the MATH - invariant we get MATH . |
math/0012123 | We freely use the notations of REF and its proof. By REF we have for all MATH . Hence MATH and the assertion follows from REF . |
math/0012123 | The proof is exactly the same as the one of REF . One only has to invoke REF instead of REF . |
math/0012123 | We first consider the case MATH. Since MATH is invertible, its spectral flow vanishes. We apply REF , and REF to calculate MATH . Now suppose that MATH is arbitrary. Choose smooth paths MATH in MATH and MATH such that MATH and such that MATH are independent of MATH. The existence of MATH follows from REF by considering MATH. In view of REF the dimension of the kernels of MATH and MATH are constant and hence the spectral flow of MATH and MATH vanishes. We may therefore compose the paths MATH without changing the spectral flow. Also MATH in view of REF. In sum, without loss of generality we may assume that the family MATH satisfies MATH. Now consider the path MATH in MATH. In view of REF this path is homotopic rel endpoints to a path MATH. Putting MATH we see that MATH is homotopic rel endpoints to the path MATH. Since homotopies with fixed endpoints neither change the spectral flow nor the NAME index we find MATH . |
math/0012123 | REF states that MATH . Applying REF to the right hand side of REF and using REF yields MATH finishing the proof. |
math/0012123 | We note again that MATH is trace class (compare the proof of REF ). Thus MATH is compact and hence the triple index MATH is well - defined for any MATH. Let MATH be a smooth path in MATH from MATH to the NAME MATH. Notice that MATH since MATH and MATH are constant paths. From REF , and REF we infer MATH . |
math/0012123 | If MATH, then MATH, and similarly MATH. This proves the first assertion and the first part of the fourth assertion. If MATH, then MATH is orthogonal to MATH, since the decomposition REF is orthogonal. Since the decomposition REF is also orthogonal, MATH has the orthogonal decomposition MATH. Write MATH, and similarly MATH. Then MATH . Since MATH, the first assertion implies that MATH, so that MATH. The second assertion follows from this and the consequence MATH of the DeRham theorem. The third assertion is proved similarly, as is the second part of the fourth assertion. The last assertion follows from the identity MATH and the fact that MATH equals MATH up to a non - zero constant. |
math/0012123 | CASE: Notice that MATH preserves MATH. Let MATH denote the MATH eigenspaces of MATH acting on MATH. It is easy to check that since MATH is orthogonal to MATH, the projections MATH restrict to isomorphisms on MATH. Thus the MATH eigenspaces of MATH on MATH have the same dimension (or are both infinite). This shows that MATH is a symplectic subspace of MATH. Clearly MATH is a Lagrangian subspace of MATH. CASE: For MATH, take MATH and apply the first assertion. For MATH, take MATH. For MATH, take MATH; then MATH. That MATH is symplectic was discussed above; hence the direct sum MATH is symplectic. CASE: We apply REF with MATH. We have MATH and MATH. Since MATH form a NAME pair and MATH is finite - dimensional, also MATH is NAME. Consequently MATH is closed and we reach the desired conclusion using REF . |
math/0012123 | Naturality of the exterior derivative implies that MATH for any MATH. It suffices, therefore, to show that MATH and MATH, since MATH and hence MATH . Following CITE, since MATH and MATH, MATH has a NAME expansion on the collar MATH of the form MATH where MATH, MATH, and MATH. REF can be used to extend MATH to a bounded form on MATH so that the extension still satisfies MATH. Notice that MATH since MATH and MATH is independent of the collar parameter. Thus MATH decays exponentially on the infinite collar MATH. Write MATH. Then MATH . The last step follows from NAME 's theorem. As MATH increases to infinity, the last integral converges to zero since MATH is bounded on MATH and MATH exponentially decays. It follows that MATH and MATH are orthogonal in MATH. Now MATH with this sum expressed as a sum of homogeneous components. Thus MATH and MATH both vanish for each MATH, and therefore MATH and MATH both vanish. |
math/0012123 | REF shows that if MATH satisfies MATH and MATH, then MATH and MATH are closed forms. Thus they represent classes in MATH. Since MATH, it follows that MATH is a closed form on MATH representing a class in MATH . The identification of cohomology with harmonic forms takes MATH to MATH and so MATH . The space MATH is a Lagrangian subspace, as is MATH by a standard argument using NAME duality. Since any two Lagrangian subspaces of a finite - dimensional symplectic vector space have the same dimension, MATH. |
math/0012123 | REF shows that the finite - dimensional vector space MATH is a symplectic subspace of MATH. Let MATH be the symplectic reduction of MATH with respect to the isotropic subspace MATH as in REF : MATH . Then MATH is a Lagrangian subspace of MATH. NAME 's theorem CITE says MATH (The sign in the exponent MATH differs from CITE because in that paper the collar of MATH is parameterized as MATH.) Thus we need only to identify the limit of MATH. To help with the rest of the argument the reader should observe that the dynamics of MATH favor the vectors with a non - zero component in eigenspaces corresponding to positive eigenvalues. Let MATH denote the complete list of eigenvalues of MATH in the range MATH. Thus MATH . Given MATH, we use this decomposition to write MATH . Let MATH denote the largest MATH so that MATH is non - zero (and hence MATH). Then MATH . This shows that MATH where MATH . Write MATH and MATH so that MATH . Set MATH . CASE: MATH. CASE: The spaces CASE: MATH, CASE: the image of MATH, CASE: MATH, and CASE: the MATH solutions of MATH on MATH are all isomorphic. CASE: MATH. Assuming these three facts, the rest of the proof of REF is completed as follows. Note that MATH. We define MATH to be the orthogonal complement of MATH in MATH. Since MATH is a Lagrangian subspace (obtained by modding out MATH and MATH), it follows from REF that MATH completing the proof of REF . |
math/0012123 | For each integer MATH let (twisted coefficients in MATH are to be understood for all cohomology groups) MATH and let MATH . Consider the map MATH in the NAME - NAME sequence for MATH. Then there is a short exact sequence MATH where MATH. Moreover, the NAME - NAME sequence gives a short exact sequence MATH . Thus MATH . Also MATH . Combining REF and summing up over MATH even one obtains MATH . The symplectic space MATH decomposes as a symplectic sum MATH (one way to see this is to notice that MATH and hence MATH preserves the parity of a harmonic form since MATH is MATH - dimensional). The Lagrangian subspace MATH decomposes accordingly into a sum of Lagrangian subspaces MATH. Hence MATH. Similarly MATH. Thus the last three terms in REF cancel. Since MATH and MATH, MATH completing the proof of REF . |
math/0012123 | CASE: For MATH, the intersection MATH is isomorphic to the kernel of MATH acting on the closed manifold MATH obtained by stretching the collar of MATH. But this kernel is a homotopy invariant, isomorphic to MATH, and in particular its dimension is independent of MATH. To compute MATH, we use REF , or, more conveniently, its consequences REF. These show that MATH, which by REF is also isomorphic to MATH. Notice that the full conclusion of REF is used here. REF. and REF. are proven by the same argument. We prove REF. From the definition of MATH REF there is an exact sequence MATH . It follows easily that for any subspace MATH there is an exact sequence MATH REF identifies MATH with the image MATH and with MATH. Thus the dimension of MATH is independent of the length of the collar of MATH. REF identifies MATH with the image of MATH, hence its intersection with MATH is independent of the length of the collar as well. Thus the middle term in the exact sequence REF is isomorphic to MATH and in particular its dimension is independent of the length of the collar; this shows that MATH is independent of MATH for MATH. Now consider the case when MATH. In the decomposition REF of MATH, MATH and so using REF shows that MATH equals MATH . Therefore, MATH equals to MATH. CASE: It follows immediately from the definition that the triple index is additive in the following sense: let MATH be Hermitian symplectic NAME spaces and MATH (respectively, MATH) be projections in MATH (respectively, MATH) such that the triple indices MATH are well - defined. Then the triple index of MATH is well - defined in the Hermitian symplectic NAME space MATH and we have MATH . In the decomposition REF we have MATH . Using the additivity of MATH we see that MATH equals MATH which equals MATH by REF . |
math/0012123 | Combining REF to the paths of NAME data spaces obtained by stretching the collars of MATH to their adiabatic limit, we see that (switching from projection to Lagrangian notation) MATH . It then follows from REF (and in particular REF) that MATH . REF shows that MATH if MATH or MATH. |
math/0012123 | REF implies that MATH and that MATH . Subtracting REF from REF yields MATH . The difference MATH is equal to MATH by REF . Moreover, it follows easily from the definitions that MATH and since MATH, that MATH . Hence REF reduces to MATH . We will show that MATH . Assuming REF, the proof of REF is completed by combining REF , taking MATH. It remains, therefore, to prove REF. The proof is similar to the proof of REF implies that as the collar of MATH is stretched to its adiabatic limit, the dimension of the intersection of MATH with MATH is independent of MATH, as is the dimension of the intersection of MATH with MATH. REF then implies that MATH is equal to MATH. Using additivity of the triple index with respect to the decomposition REF, the calculation of MATH REF, and REF , we conclude that MATH . REF then implies that MATH . |
math/0012123 | The first assertion follows immediately from the definition of MATH. The second assertion is clear. For the third, notice that MATH is independent of MATH, and that the MATH eigenspace of MATH is isomorphic to MATH. In particular the MATH eigenspace of MATH is constant dimensional, and so MATH is continuous. These facts imply that MATH is continuous in MATH. |
math/0012123 | We explained above why the MATH - invariant depends on the flat connection MATH only through the conjugacy class of its holonomy representation. By taking the double of MATH we obtain a closed Riemannian manifold MATH over which the connections MATH and MATH extend flatly. Letting MATH denote the extension to MATH, we know from CITE that the difference MATH is independent of the metric on MATH and depends only on the conjugacy class of the holonomy representation of MATH (see the paragraph following this proof). REF shows that MATH . Notice that by REF the subspaces MATH and MATH are independent of the Riemannian metric on MATH. Hence solving for MATH in REF yields an expression which is unchanged when the Riemannian metric is altered on the interior of MATH. |
math/0012123 | This follows by applying REF to MATH and MATH and subtracting. |
math/0012123 | The first and second assertions follow from the first and second assertions of REF . For the third assertion, we first claim that the group MATH of symplectic automorphisms of MATH is path connected. To see this, fix a Lagrangian subspace MATH of MATH. The map MATH taking MATH to MATH is a fibration with fiber the subgroup MATH consisting of those symplectic automorphisms which leave MATH invariant. Next, MATH fibers over MATH by mapping MATH to the restriction MATH. The fiber MATH of this map consists of those symplectic transformations MATH so that MATH restricts to the identity on MATH. Writing MATH it is easy to see that MATH consists of all matrices of the form MATH with MATH an arbitrary real matrix. Thus MATH is contractible, and since MATH is path connected, MATH is also path connected. Finally, since MATH fibers over the path connected space MATH with path connected fiber MATH, it is itself path connected. Choose a path MATH from the identity matrix to MATH. The third assertion of REF shows that MATH varies continuously in MATH. Thus the same is true of the integer-valued function MATH. Therefore this function is constant, completing the proof of the third assertion. To prove the last fact, Notice that MATH, where MATH is the span of MATH and MATH is the span of MATH. It is easy to calculate that with respect to these bases, MATH . Thus MATH . |
math/0012123 | We prove this for MATH, the general case is obtained by reparameterizing. Let MATH denote the path of projections to the NAME data space MATH, where MATH. Applying REF to the path MATH we see that MATH equals MATH, which by REF equals MATH . Switching to Lagrangian notation and parameterizing by MATH instead of MATH we can rewrite REF as MATH . We use the homotopy invariance of the NAME index to simplify these terms. Consider first MATH. We will show this term vanishes. The path MATH is homotopic to the composite of MATH and the constant map at MATH, and the constant path at MATH is homotopic to the composite of MATH and its inverse. Since MATH for all MATH, MATH and so by path additivity of the NAME index, MATH . REF says that MATH for MATH, but by reparameterizing we see that the intersection is zero for all MATH; obviously MATH. Hence MATH and so MATH . Consider now the term MATH . The constant path at MATH is homotopic to the composite of MATH and its inverse, and the path MATH is homotopic to its composite with the constant path at MATH. Therefore, MATH . Substituting REF into REF finishes the proof. |
math/0012124 | This is just a direct consequence of the fact that MATH in the polytope described in REF , see CITE. It is also a straightforward computation by using the fact that if MATH, then MATH . |
math/0012124 | Note that as a consequence of REF , we have MATH and MATH. Using REF, we conclude that there exist MATH and MATH all simple enjoying the properties stated above. |
math/0012124 | The proof given here is just a generalization of the proof of REF. Among the MATH choose a minimal one, which we denote by MATH. Let us reindex the MATH so that MATH for MATH. Note that since MATH is minimal we have MATH for MATH. Since MATH is standard, we can choose MATH such that MATH, the class of MATH in MATH is MATH and the class of MATH in MATH is MATH. Let MATH and restrict REF to MATH. Then MATH is standard on MATH for MATH and MATH for MATH. By the linear independence of standard monomials, REF restricted to MATH is not zero. Hence MATH. This implies that MATH. According to REF (or REF) we have MATH; note that MATH (or MATH) can not be equal to MATH, because MATH is non standard. From this argument we deduce that MATH for all MATH. Let MATH and MATH. Now MATH and MATH. By weight consideration, we have MATH. Furthermore MATH implies that MATH, or equivalently MATH since both belong to MATH. Therefore MATH, which implies that MATH. Therefore MATH. In other words MATH. |
math/0012124 | From REF , we know that for any pair MATH on the right-hand side, MATH and MATH. Moreover if MATH, then due to weight considerations, that is, MATH, we see that MATH. |
math/0012124 | Let MATH, MATH, MATH and MATH. A necessary condition for MATH to appear on the right-hand side of REF is MATH. Here, we shall prove that this condition immediately implies the assertion, that is, MATH. The proof is by induction on MATH. The fact that MATH implies (see the proof of REF ) MATH . Note that MATH and that MATH . Then due to the equality in REF , we must have MATH. There are four cases to consider. MATH . CASE: MATH. This implies that MATH. Hence denoting MATH and MATH, REF implies that MATH. By induction we are done. MATH . CASE: MATH. Let MATH be the smallest number such that MATH (if such a MATH less than MATH does not exist, let MATH). Note that MATH. Set CASE: MATH, CASE: MATH, CASE: MATH if MATH and CASE: MATH if MATH. Since MATH, we have MATH for MATH. Therefore MATH. Using the fact that for MATH, we have MATH, then MATH . From REF , we can conclude that MATH. The rest follows by induction. MATH . CASE: MATH is similar to REF . MATH . CASE: MATH is similar to REF . |
math/0012124 | As before, denote MATH, MATH. Note that MATH (that is there exist liftings MATH in MATH such that MATH). REF implies that the restriction of REF to the NAME variety MATH is MATH, where MATH. Since standard monomial basis is characteristic free, this holds in any characteristics. Hence MATH. |
math/0012124 | Let MATH. Denote by MATH. Then we have MATH. Now if MATH, then we proceed by induction on the length of MATH. Otherwise, let MATH be the largest integer such that MATH. We have MATH, thus by REF , we have that MATH commute with MATH for MATH. Thus MATH . By induction, MATH. Therefore the right-hand side is MATH and we have proved that the definition of MATH does not depend on the choice of the reduced expression. We are left to show that these elements form a basis for MATH. We claim that MATH. It is clear that MATH. Now, since MATH satisfies the condition of REF , we have MATH, thus MATH . We have therefore our claim. Now by the definition of MATH, we have MATH where MATH and for each MATH, we have MATH in MATH, MATH in MATH and MATH. It is now clear that the MATH's are independent. Further, one deduces from the expression for MATH above that the MATH-submodule generated by the MATH's is a direct summand of the tensor product MATH. Finally, by standard monomial theory, the cardinal of MATH is the rank of MATH. So the result follows. |
math/0012124 | Recall from the proof of REF that MATH where MATH and for each MATH, we have MATH in MATH, MATH in MATH and MATH. Let us apply MATH to MATH. Then from the explicite expression of MATH above, this is either MATH or MATH depending if MATH appears in the right hand side or not. On the other hand, if we replace MATH by the right hand side of the straightening relation, then it is clear from REF that the same evaluation yields MATH. But this is non zero from REF . So it must be MATH. |
math/0012124 | By REF , there exists a flat deformation whose general fiber is MATH and whose special fiber is a variety defined by a binomial ideal associated to a distributive lattice. This latter is toric as shown in CITE. |
math/0012129 | Let MATH be a point of MATH and MATH be a point of MATH, that is, MATH consists of a collection of pairwise distinct points MATH of MATH, to each of which there is an assigned element MATH. We define the corresponding point of MATH as follows: In the triple MATH, MATH is the MATH-bundle induced from MATH. Let now MATH be the MATH-bundle induced from MATH. For each MATH-dominant weight MATH orthogonal to MATH, MATH, let us denote by MATH the corresponding associated line bundle. By construction, we have the injective bundle maps MATH (here MATH is the NAME module corresponding to MATH), which satisfy NAME relations. We set MATH. The corresponding line bundles MATH are then MATH. Thus, by composing we obtain the maps MATH which are easily seen to satisfy the NAME relations, as required. |
math/0012129 | Let MATH be a MATH-point of MATH. To construct a map of functors, MATH we must exhibit the maps MATH for all MATH-modules MATH and not just for those of the form MATH. However, we can do that because any MATH can be tensored with a MATH-dimensional representation of MATH corresponding to a MATH-dominant weight MATH, so that the new representation will be of the form MATH. By construction, the above map MATH is a closed embedding. The fact that MATH is a scheme (and not an ind-scheme) follows from the fact that we can choose MATH such that MATH is faithful as a representation of MATH. Let MATH be a MATH-point of MATH. Then if MATH is a MATH-module (whose weights, we can suppose, are MATH for some MATH-dominant weight MATH), the maps MATH are regular, since MATH, by the definition of MATH. Hence, MATH is contained in MATH. To show that this inclusion is an isomorphism on the level of reduced schemes, one has to check that the fact that MATH implies that MATH is a sum of positive coroots of MATH, which is obvious. |
math/0012129 | Let MATH be a MATH-point of MATH. By definition, this means that the divisors MATH and MATH in MATH do not intersect. Let MATH be a MATH-point of MATH which projects under MATH to the corresponding point of MATH. Set MATH, MATH, MATH. By assumption, MATH. We define a new MATH-bundle MATH as follows: over MATH, MATH is by definition the trivial bundle MATH; over MATH, MATH is identified with MATH; the data of MATH, being a trivialization of MATH over MATH defines a patching data for MATH. By construction, MATH is trivialized off MATH, let us denote this trivialization by MATH. We introduce the second MATH-bundle MATH in a similar fashion: MATH and MATH. From the construction, MATH acquires a trivialization MATH. In a similar way, from MATH we obtain two pairs MATH and MATH, which project under MATH to MATH and MATH, respectively. Thus, from the MATH-point MATH we obtain two MATH-points MATH and MATH of MATH and MATH, respectively. The map in the opposite direction is constructed in the same way. |
math/0012129 | By construction, a MATH-point of MATH is a data of a MATH-bundle MATH on MATH, with given reductions MATH and MATH to MATH and MATH, respectively, such that these reductions are mutually transversal on MATH, the MATH-bundle induced from MATH is trivialized, and the maps MATH are isomorphisms for MATH-dominant characters MATH of MATH. Therefore, over MATH all the three principal bundles MATH, MATH and MATH are trivialized in a compatible way, and the MATH-bundle induced from MATH is exactly MATH. Thus, MATH indeed defines a point of MATH. The map in the opposite direction, that is, MATH, is constructed in exactly the same way. The second assertion of the proposition, that is, the fact that MATH becomes identified with MATH is obvious from the construction. |
math/0012129 | Let us analyze what it means to have a MATH-point of the Cartesian product MATH . By definition, we have a MATH-bundle, a pair of MATH-bundles MATH and MATH and two systems of maps MATH and MATH for every MATH-module MATH: MATH which satisfy the NAME relations, with the condition that the MATH's are surjective, and the MATH's are injective over every geometric point of MATH. We define the open sub-stack MATH by the following condition: for every geometric point MATH, the MATH- and MATH-structures defined on MATH by means of MATH and MATH at the generic point of MATH are mutually transversal. Let MATH be a MATH-point of MATH. It is clear that the maps MATH as in the definition of MATH indeed define a MATH-point of MATH. Conversely, given a MATH-point of MATH, we define a MATH-point of MATH as follows: Firstly, we set the ``background" MATH-bundle to be MATH. Let MATH denote the induced MATH-bundle under MATH. Secondly, by construction, there exists an open dense subset MATH, such that MATH admits reductions simultaneously to MATH and MATH, which are, moreover, transversal. Hence, over MATH, we have identifications MATH and MATH. Therefore, it remains to show that MATH is such that the maps MATH, which are defined on MATH, extend as regular maps to the entire MATH, provided that MATH is of the form MATH for a MATH-module MATH. However, note that for MATH of the above form the composition MATH is an isomorphism. (As a remark, let us observe that in char REF, the maps MATH are isomorphisms for any MATH-module MATH, which is not necessarily the case over MATH.) Hence, by composing MATH and MATH we do obtain a regular map MATH which is what we had to show. |
math/0012129 | Note that a MATH-point of MATH belongs to MATH if and only if the following condition holds: for every MATH-dominant weight MATH, orthogonal to MATH for MATH, the corresponding map MATH is such that there exists a short exact sequence MATH such that MATH is a vector bundle on MATH, the support of MATH is MATH-finite and over any geometric point MATH, the length of MATH is exactly MATH. Given a MATH-point of MATH, we can compose the above embedding of sheaves with MATH and we obtain a map between line bundles, such that over every geometric point MATH its total amount of zeroes is MATH. This readily implies the first assertion of the lemma. To prove the second assertion, observe that MATH . However, the condition on the degree forces that MATH . Hence, MATH and the fact that its embedding into MATH coincides with MATH follows from the construction. |
math/0012129 | Since both MATH and MATH are smooth, it is enough to check the surjectivity on the level of tangent spaces. Thus, let MATH be a MATH-bundle and let MATH be the induced MATH-bundle. We must show that MATH is surjective if MATH. In general, the cokernel of this map is MATH. However, the irreducible subquotients of MATH as a MATH-module are all MATH-modules, which appear in the NAME series of MATH. Hence, the assertion of the proposition follows from the definition of MATH. |
math/0012129 | Consider the corresponding non-symmetrized power of the curve MATH. Over it we can consider the scheme MATH and we have a natural proper map MATH which covers the usual symmetrization map MATH. Let us denote temporarily by MATH the direct image MATH. The fact that the map which defines convolution of perverse sheaves on the usual affine NAME MATH is semi-small implies that the above map MATH is small. Hence, MATH is the NAME extension of its restriction to the open sub-scheme MATH. In particular, it carries a canonical action of the product of symmetric groups MATH, because this is obviously so over MATH, and MATH. By construction, the *-restriction of MATH to MATH can be identified with MATH . Therefore, it remains to see that the MATH-action on MATH corresponds to the natural action of the group on MATH. We prove the latter fact as follows: Since taking the global cohomology is a fiber functor for the category of spherical perverse sheaves on MATH, it suffices to analyze the MATH-action on the direct image of MATH under MATH, in which case the assertion becomes obvious. |
math/0012129 | To simplify the notation, we will take the element MATH corresponding to the decomposition which consists of one element: MATH. We need to calculate the *-restriction of MATH to MATH . By definition, our embedding MATH factors through MATH. Note that MATH is exactly the preimage of the main diagonal MATH. Therefore, the sought-for complex is, according to REF , the direct sum over MATH of MATH . Using REF, we obtain that MATH corresponding to MATH with MATH equals MATH . However, MATH which is what we had to show. |
math/0012129 | According to CITE, MATH has weights MATH, since MATH is pure of weight MATH. At the same time, CITE implies that MATH has weights MATH. Hence, we obtain that MATH is pure of weight MATH. Hence, the assertion of the corollary follows from the decomposition theorem. |
math/0012129 | Let us identify MATH with the quotient MATH. Since we know already that MATH is a scheme of finite type, for MATH deep enough in the dominant chamber, the preimage of MATH under the projection MATH is contained inside the subgroup MATH . Let us now consider MATH, which via the action of MATH can be identified with MATH. We can view MATH as a quotient MATH, via the action of MATH on MATH, viewed as an element of MATH. We obtain that the preimage of MATH in MATH is contained in MATH. Hence, MATH . |
math/0012129 | The initial observation is that each MATH is a scheme of finite type. We know this fact for MATH, since the above intersection is a locally closed sub-scheme in the NAME space MATH. In general, this assertion can be proven either by introducing the corresponding analog of the NAME space over a global curve, or by a straightforward local argument. Let us view MATH as a sub-scheme of MATH, such that MATH. As in the case MATH, we obtain that that the preimage of MATH under MATH is contained in a sub-scheme of the form MATH. Hence, the action of MATH maps MATH inside MATH . |
math/0012129 | First, from REF , it is easy to see that each MATH has the following form: it is the intersection cohomology sheaf of MATH tensored with a complex over MATH. There is a finite map MATH, defined as in REF, which normalizes MATH. Hence, it suffices to analyze the *-restriction to MATH of the direct image of MATH under this map. However, the preimage of MATH in MATH is MATH and we can assume that the *-restriction of MATH to this sub-stack is known by induction. In particular, all its direct summands are supported on sub-stacks of the form MATH . Since, MATH and MATH, none of these sub-stacks maps onto the main diagonal in MATH. |
math/0012161 | Because MATH is regular, the MATH defined above do not depend on MATH and are all equal to MATH thus REF reads MATH . Now writing MATH and MATH completes the proof. |
math/0012161 | Let MATH be the NAME graph of MATH relative to MATH defined above. Fix the endpoints MATH, the coset of MATH, and give MATH the length labelling. Let MATH and MATH be the circuit and proper circuit series of MATH. In this setting, expressing MATH and MATH, we see that MATH is the growth rate MATH of circuits in MATH, and MATH the growth rate MATH of proper circuits in MATH. As both MATH and MATH are power series with non-negative coefficients, MATH is the radius of convergence of MATH and MATH the radius of convergence of MATH. Let MATH and consider the function MATH . This function is strictly increasing for MATH, has a maximum at MATH with MATH, and is strictly decreasing for MATH. First we suppose that MATH, so MATH is monotonously increasing on MATH. We set MATH in REF and note that, for MATH, it says that MATH has a singularity at MATH if and only if MATH has a singularity at MATH. Now as MATH is the singularity of MATH closest to MATH, we conclude by monotonicity of MATH that the singularity of MATH closest to MATH is at MATH; thus MATH . Suppose now that MATH. If MATH, the right-hand side of REF would be bounded for all MATH while the left-hand side diverges at MATH. If MATH, there would be a MATH with MATH; and MATH would have a singularity at MATH. The only case left is MATH. |
math/0012161 | Present MATH as MATH, with MATH a quasi-free group and MATH the normal subgroup of MATH generated by the relators in MATH, and apply REF . |
math/0012161 | Recall that MATH means that both terms are sums of paths, say MATH and MATH, such that the minimal length of paths in the symmetric difference MATH tends to infinity. Now the difference between MATH and MATH consists only of paths in MATH that exit MATH, and thus have length at least MATH. The same argument holds for MATH. |
math/0012161 | If MATH is finite, then MATH, the degree of MATH; after a large even number of steps, a random walk starting at MATH will be uniformly distributed over MATH (or over the vertices at even distance of MATH, in case all circuits have even length). A long enough walk then has probability MATH (or MATH if all circuits have even length) of being a circuit. If MATH is infinite, we consider two cases. If MATH, that is, MATH is MATH-transient, the general term MATH of the series MATH tends to MATH. If MATH is MATH-recurrent, then, as MATH is quasi-transitive, MATH by CITE. We then approximate MATH by the sequence of its balls of radius MATH: MATH where we expand the circuit series of MATH as MATH. The same proof holds for the MATH. Its particular case where MATH is a NAME graph appears in CITE. |
math/0012161 | By REF implies the other two. By REF , and a computation on trees done in REF to deal with the case MATH, REF implies REF. It remains to show that REF implies REF. Assume that MATH is rational, not equal to MATH. As the MATH are positive, MATH has a pole, of multiplicity MATH, at MATH. There is then a constant MATH such that MATH for infinitely many values of MATH CITE. It follows by REF that MATH and the graph MATH is finite, of cardinality at most MATH. |
math/0012161 | For any language there exists a unique minimal (possibly infinite) automaton recognising it (CITE is a good reference). Let MATH be the minimal automaton recognising MATH. Recall that this is a graph with an initial vertex MATH, a set of terminal vertices MATH and a labelling MATH of the graph's edges such that the number of paths labelled MATH, starting at MATH and ending at a MATH is MATH if MATH and MATH otherwise. Extend the labelling MATH to a labelling MATH by MATH . Because MATH is saturated, and MATH is minimal, MATH; then MATH is the set of labels on proper paths from MATH to some MATH. Choosing in turn all MATH as MATH, we obtain growth series MATH counting the formal sum of paths and proper paths from MATH to MATH. It then suffices to write MATH . |
math/0012161 | Say MATH has MATH bumps, at indices MATH so that MATH. We will show the evaluation at MATH of the left-hand side of REF yields MATH. We claim there is a bijection between the subsets MATH of MATH and the pairs MATH where MATH is a path and MATH is a bump scheme for MATH with MATH; and further MATH. First, take a MATH and a MATH such that MATH. The path MATH is obtained by shuffling together the edges of MATH and MATH, and this partitions the edges of MATH in two classes, namely REF those coming from MATH and REF those coming from MATH and MATH. Let MATH be the indices of the bumps MATH in MATH coming from MATH, that is, such that MATH and MATH belong to the class REF . One direction of the bijection is then MATH. Conversely, given a subset MATH consider the set MATH. Split it in maximal-length runs of consecutive integers MATH. For all runs MATH we do the following: to MATH of even cardinality we associate a squiggle MATH of length MATH along MATH; to MATH of odd cardinality we associate a squiggle MATH of length MATH at MATH; then we delete in MATH the edges MATH. This process constructs a bump scheme MATH while pruning edges of MATH, giving a path MATH with MATH. These two constructions are inverses, proving the claimed bijection. It now follows that MATH . |
math/0012161 | The function MATH is monotonously increasing between MATH and MATH, where it reaches its maximum. The same argument applies as that given in the proof of REF . |
math/0012161 | Sums and products of algebraic series are algebraic. If MATH satisfies the algebraic relation MATH, then its formal inverse satisfies the relation MATH so is also algebraic. |
math/0012161 | Clearly MATH; if MATH converges over MATH then MATH converges over MATH where MATH; then MATH. The second assertion follows because in this case MATH. |
math/0012161 | This follows from MATH . |
math/0012164 | The equivalence MATH is proven in CITE. MATH. Let MATH be a MATH-bialgebroid in the sense of REF , and define (compare CITE) MATH . The map MATH is an algebra morphism since MATH. The fact that MATH is MATH-bilinear follows by an elementary calculation. Explicitly, for any MATH, MATH, MATH we have MATH-where we used REF to derive the last equality, thus proving that MATH is left MATH-linear. Similar calculation that uses REF , proves the right MATH-linearity of MATH. Next we prove that REF hold for MATH. Using REF we have MATH. Now using the first part of the counit REF for MATH, we obtain MATH, that is, REF for MATH. The condition (AREF ) follows from MATH and the second part of the counit REF together with REF . This shows that MATH is a MATH-bialgebroid with an anchor in the sense of REF , that is, MATH. In fact there is more, and this is MATH. Since MATH and MATH are MATH-bimodule maps, they are left MATH-module maps. Furthermore, both MATH and the corestriction MATH of MATH to MATH, are MATH-algebra maps. Therefore, by the observation at the end of REF, MATH and MATH are maps of MATH-rings, and hence also maps of MATH-bimodules. REF then implies that MATH is a MATH-bialgebra. MATH. Let MATH be a bialgebroid with an anchor and define MATH, MATH. CITE shows that MATH is a map in MATH, and a counit for MATH. This also implies that MATH, hence, in particular that MATH. Furthermore by CITE, for all MATH, MATH. Therefore MATH, and similarly for the target map MATH. This proves REF , and we conclude that MATH is a bialgebroid in the sense of REF . The above result gives MATH in terms of MATH. Looking at REF one can apply MATH once again. In this way one obtains that for a bialgebroid with an anchor MATH the structure maps MATH (the corestriction of MATH to MATH) and MATH are in fact maps of MATH-rings. Following REF the implication MATH is then obvious. |
math/0012164 | Clearly, MATH is an algebra map. We prove now that MATH is an anti-algebra map if and only if MATH is a right MATH-comodule algebra and the braided commutativity relation REF holds. Take any MATH, then MATH, and MATH. Suppose MATH is an anti-algebra map. Then applying MATH to the above equality one obtains REF . It follows then that MATH, that is, MATH is an algebra map, hence MATH is a MATH-comodule algebra as required. Conversely, suppose that MATH is a right MATH-comodule algebra and REF holds. Then MATH . Assume now that MATH is an anti-algebra map. We prove that MATH if and only if MATH. MATH is a right MATH-module via REF , that is, using REF we have MATH . MATH if and only if for all MATH, MATH we have that MATH or, equivalently, MATH . Since the tensor product is defined over MATH and equality REF holds we have MATH . Thus we conclude that MATH if and only if MATH, that is, if and only if MATH. Therefore we have proven that MATH is an anti-algebra map and MATH if and only if MATH is a braided commutative algebra in MATH. It is then straightforward to check that all the remaining conditions in REF hold. Finally we prove that MATH defined in the theorem is the antipode of MATH. The canonical projection MATH has a well defined section MATH, MATH. Since for all MATH, MATH, MATH and MATH, we have MATH that is, MATH is an anti-algebra map. REF follows from the definition of MATH, MATH while REF can be established by the following computation MATH . It remains to prove REF . The left hand side of REF equals MATH . REF , evaluated at MATH implies that MATH for all MATH and MATH. Now the right hand side of REF reads MATH that is exactly the left hand side of REF . Hence, MATH is an antipode of MATH. |
math/0012164 | This can be proven by dualising the proof of CITE. Although not elementary this is quite straightforward and we leave it to the reader. |
math/0012166 | It suffices to consider the case MATH. Let MATH denote the graph whose vertices consist of the sets MATH and MATH, and whose edges consist of all pairs MATH and MATH for MATH. The assumption that the diagram be cocartesian is equivalent to the connectedness of the graph MATH. Every loop of this graph gives rise to a contraction of type REF and hence introduces a factor MATH. The number of such loops is MATH. |
math/0012166 | Treating each orbit separately, we may assume that MATH, that is .that the group MATH acts transitively on MATH. We will identify the defect MATH as the genus of an oriented closed compact surface MATH so that the claim becomes obvious. Let MATH. Let MATH be constructed as follows: Take MATH as the set of vertices. For all MATH and MATH add an oriented edge MATH from MATH to MATH. For each MATH glue in a (black) triangle along the edges MATH, MATH, and MATH. Finally, for each MATH and each orbit MATH of MATH glue in a (white) MATH-gon along the edges MATH, MATH,, MATH for some element MATH. (If MATH is MATH or MATH, then the MATH-gon is rather degenerate: a disc with MATH respectively MATH edges.) Every edge bounds one white and one black polygon. The resulting NAME is a connected compact oriented surface MATH with NAME characteristic MATH where MATH, MATH, and MATH are the numbers of black polygons corresponding to the orbits of MATH, MATH, and MATH, respectively. Hence MATH. |
math/0012166 | The map MATH is surjective and invariant under the MATH-action, so that its restriction to MATH is also surjective. Moreover, an invariant vector MATH in MATH is determined by its components MATH in MATH, where MATH runs through a system of representatives for all possible cycle types, that is, all partitions of MATH, and each component MATH is symmetric with respect to exchanging the values corresponding to orbits of the same length, and conversely. This shows that both vector spaces have the same dimensions. |
math/0012166 | For any MATH, MATH, MATH consider the following diagram of orbit spaces: MATH . The arrows of type MATH and MATH will correspond to ring homomorphisms whereas arrows of type MATH will contravariantly induce module homomorphisms. By definition, we have MATH with MATH and MATH . REF applies to the central diamond MATH in the diagram above: let MATH be defined by MATH . Then by REF: MATH . We apply the projection formula and get MATH with MATH . In this expression MATH is the function MATH . Summing up, we have MATH . The same symmetric expression arises if we compute MATH in the same way. This proves associativity. The following terms contribute to the degree of MATH: the degrees of MATH, MATH and MATH (given by REF ), the degrees of MATH, MATH and MATH, and the degrees of the two multiplications in the definition of MATH. The total balance is: MATH . |
math/0012166 | Let MATH. The following diagram of orbit spaces MATH commutes. Hence MATH. Moreover, MATH. It follows that MATH . |
math/0012166 | This is a consequence of REF . |
math/0012166 | The integral induces a MATH-invariant bilinear form on MATH. The only non-trivial pairings are of type MATH . As MATH and MATH have the same orbit spaces, this map is the composition MATH . But MATH, the integral on MATH. This shows that MATH induces a non-degenerate pairing on MATH. Since the pairing is invariant, its restriction to MATH is also non-degenerate. |
math/0012166 | By definition, multiplication by MATH is the same as applying the operator MATH. Now move the operator MATH as far to the right as possible thus introducing commutators. The claim of the proposition then follows from the following facts: CASE: MATH, CASE: MATH, CASE: MATH. Here REF holds degree reasons, and REF follow directly from REF. |
math/0012166 | Recall that the number of permutations MATH of a given cycle type MATH is given by MATH and that its signature is given by MATH . Therefore MATH where the index MATH in the last line means that we take the component of weight MATH only. By CITE, the last expression equals the total NAME class of the tautological sheaf MATH. |
math/0012166 | This proposition is an adaptation of a result of CITE to our situation. Without loss of generality we may assume that MATH is of the form MATH. Moreover, MATH. Therefore, we may fix a permutation with a disjoint cycle decomposition MATH with a certain ordering and assume that MATH. (Of course, then MATH is no longer contained in MATH, but recall that MATH is defined on the larger ring MATH.) Then by definition, MATH . Now let MATH be a single transposition, say MATH. We distinguish two cases according to whether MATH and MATH are contained in the same MATH-orbit or not. We analyse the affect of multiplying MATH and MATH, and take the sum over all transpositions afterwards. case: MATH and MATH are contained in different cycles of MATH, say MATH . Then MATH, that is, the multiplication with MATH merges the two orbits. The genus defect MATH vanishes. Hence the multiplication map MATH is essentially given by multiplying the coefficients corresponding to the two orbits MATH and MATH. Assuming that MATH and MATH are contained in the MATH's and MATH's orbit of MATH, this consideration shows MATH where MATH is the sign arising from the permutation of the MATH's. If MATH runs through all transpositions, there are MATH possibilities to hit the orbits MATH and MATH. Thus the right hand side in REF occurs with multiplicity MATH. Up to the sign, this yields the first term on the right hand side of REF . case: MATH and MATH are contained in the same cycle of MATH, say MATH . Then MATH, that is, the given cycle is split into two smaller cycles. Again the genus defect vanishes. The multiplication REF is essentially given by the comultiplication MATH, where the two factors on the right hand side correspond to the two new orbits. Hence if the cycle MATH is, say, MATH, then MATH where again MATH is the sign arising from the permutation of the MATH's. There are precisely MATH choices of (ordered!) pairs MATH from the cycle MATH such that the cycle splits into two cycles of lengths MATH and MATH. Again up to the sign, this corresponds to the second term on the right hand side of REF . Note that the factor MATH arises since transpositions are unordered. Summing up we see, that multiplication by MATH has the same effect - via MATH - as multiplication by MATH as described by REF . |
math/0012166 | The embedding MATH preserves products provided that MATH. Therefore, there are identities MATH . The proposition follows from this by a simple calculation see the proof of CITE. |
math/0012166 | The assertion follows from REF by induction on cohomological degree and weight. The calculation itself is identical to the proof of CITE. |
math/0012166 | Let MATH be given. We compute the terms MATH and MATH: MATH . If in this sum the index MATH takes values MATH, then the transposition MATH permutes the final REF's in MATH and thus has no effect. This part of the sum therefore equals MATH . This is the second half of the commutator. Adding up, we find MATH . There is no need to evaluate the right hand side. It suffices to note that conjugation by MATH has the effect of making the point MATH part of the orbits which support the diagonally embedded class MATH. By the definition of the multiplication in MATH, this leads to a contraction of MATH and MATH. But the result of the multiplication does not change if we replace the pair MATH by MATH or MATH. This is all we need. The second claim follows from the first when combined with REF . |
math/0012168 | (See CITE, page REF). Pre and postcomposition of MATH by affine maps yields a map MATH with the same constant MATH of quasisymmetry. To show MATH is NAME continuous at a point MATH, it suffices to show MATH is NAME continuous at MATH and to assume the intervals MATH and MATH contain MATH and that MATH and MATH . By plugging in MATH REF yields MATH . Repeated applications of REF with MATH and MATH yield MATH . Since MATH is increasing, this implies that for MATH . But since MATH the previous inequality implies MATH . To finish the proof we note that MATH and MATH have the same NAME exponent MATH . |
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