paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/9906019 | Since MATH has finite statistics, MATH and hence the centre of MATH are finite dimensional. Therefore if MATH is a covariant representation, MATH is trivial on such centre. Then, since MATH implements automorphisms of MATH, it implements an action of MATH by automorphisms of MATH, preserving any factorial component. Thus this action is implemented by a unitary representation MATH in MATH. Then MATH is a representation of MATH acting trivially on MATH. Clearly such a representation decomposes into representations of the irreducible components of MATH, so these are MATH-covariant. |
math-ph/9906019 | Of course we may restrict to the case MATH. Since MATH is the product of MATH with MATH, such reflections are implemented by the corresponding modular involutions MATH, MATH, and MATH is equivalent to MATH, both being conjugate endomorphisms of MATH, there exists a unitary MATH intertwining MATH and MATH. Since MATH, MATH implements the trivial action on MATH, hence, MATH being irreducibile, MATH is a constant and we may choose MATH selfadjoint. Then MATH is another representation of MATH realizing the covariance of MATH. Since MATH is irreducible, we get MATH, where MATH is a character of MATH. Applying this relation twice, we get MATH, namely MATH. Now observe that, since MATH is a NAME group representation, it is the exponential of a NAME algebra morphism MATH from the NAME algebra of MATH to MATH. Since MATH is simply connected, MATH exponentiates to a character, which we denote by MATH, whose square gives MATH, and we get MATH namely MATH and MATH yield the required representation of MATH. |
math-ph/9906019 | By property (MATH), MATH does not depend on MATH, as observed in REF . Concerning the relation between spin and statistics, we may define a function MATH, MATH, as in the proof of REF , which is indeed a local group representation namely, if MATH verify MATH, we have MATH. Setting MATH, we get MATH and MATH. Hence, for sufficiently small MATH, MATH . The proof now continues as in REF . |
math-ph/9906019 | Let MATH belong to MATH, MATH. By REF MATH because of the above property of MATH since the integrand is constantly equal to MATH on the set MATH. Then the localization of MATH and NAME duality imply that the range of MATH is contained in MATH. Setting MATH gives a conditional expectation of MATH onto the range of MATH that restricts to a conditional expectation MATH of MATH onto MATH if MATH. Since MATH is assumed to have finite index, MATH is automatically normal CITE. Therefore MATH is normal for MATH with MATH, hence for any MATH. Concerning the MATH-invariance of MATH we have, making use of the cocycle condition, MATH . |
math-ph/9906019 | We have MATH and MATH is locally normal because both MATH and MATH are locally normal. |
math-ph/9906019 | If MATH and MATH is localized in a double cone, the commutator function MATH vanishes on a right wedge, hence MATH. Since MATH is locally normal, MATH commutes with every MATH, thus with MATH; but MATH being irreducible, it is therefore a scalar equal to its vacuum expectation value: MATH as MATH is normal and MATH-invariant. |
math-ph/9906019 | One may repeat the proof of REF for each of the one-parameter light-like unitary translation groups. |
math-ph/9906019 | MATH is a simple MATH-algebra since it is the inductive limit of type III factors (that are simple MATH-algebras). Therefore MATH is one-to-one and the statement will follow if we show that MATH if cyclic for MATH. To this end we may use a classical NAME argument. If MATH is orthogonal to MATH, and MATH, then for all MATH we have MATH for MATH in a neighborhood of MATH, thus for all MATH by the spectrum condition shown by REF . Hence, setting MATH and MATH, MATH is orthogonal to MATH, thus MATH because MATH is irreducible since MATH by the local normality of MATH. |
math-ph/9906019 | We begin with the case where MATH is irreducible and assume for convenience that MATH. Notice then that MATH is finite-dimensional and, by covariance, globally MATH-invariant with MATH in the subgroup of boosts because these transformations preserve MATH. Therefore MATH is a finite-dimensional subspace of MATH globally invariant for MATH, MATH. By REF we thus have MATH for every element MATH, thus MATH because MATH is separating. It follows that if MATH and MATH namely MATH . Since the converse implication is obvious by wedge duality we have the equality of the two intertwiner spaces. Now if MATH is any endomorphism with finite index, MATH is finite-dimensional because MATH and MATH decomposes into a direct sum of irreducible endomorphisms of MATH which are covariant, therefore the preceding analysis shows that MATH in this case, too. Since MATH is translation invariant, we get MATH whenever MATH and, since MATH was arbitrary, the result follows. |
math/9906004 | Let MATH denote the NAME graph of MATH with respect to some finite generating set for MATH. Let MATH denote the almost invariant subset MATH of MATH and let MATH denote the almost invariant subset MATH of MATH. Recall from the start of this section, that if we identify MATH with the MATH - cochain on MATH whose support is MATH, then MATH is an almost invariant subset of MATH if and only if MATH is finite. Thus MATH is a finite collection of edges in MATH and similarly MATH is a finite collection of edges in MATH. Now let MATH denote a finite connected subgraph of MATH such that MATH contains MATH and the natural map MATH is onto, and let MATH denote a finite connected subgraph of MATH such that MATH contains MATH and the natural map MATH is onto. Thus the pre-image MATH of MATH in MATH is connected and contains MATH, and the pre-image MATH of MATH in MATH is connected and contains MATH . Let MATH denote a finite subgraph of MATH which projects onto MATH, and let MATH denote a finite subgraph of MATH which projects onto MATH. If MATH meets MATH, there must be elements MATH and MATH in MATH and MATH such that MATH meets MATH. Now MATH meets MATH is finite, as MATH acts freely on MATH. It follows that MATH meets MATH consists of a finite number of double cosets MATH . The result would now be trivial if MATH and MATH were each the vertex set of a connected subgraph of MATH. As this need not be the case, we need to make a careful argument as in the proof of REF. Consider MATH in MATH such that MATH and MATH are disjoint. We will show that MATH and MATH are nested. As MATH is connected, the vertex set of MATH must lie entirely in MATH or entirely in MATH . Suppose that the vertex set of MATH lies in MATH. For a set MATH of vertices of MATH, let MATH denote the maximal subgraph of MATH with vertex set equal to MATH. Each component MATH of MATH and MATH contains a vertex of MATH. Hence MATH contains a vertex of MATH and so must meet MATH. If MATH also meets MATH, then it must meet MATH. But as MATH is connected and disjoint from MATH, it lies in a single component MATH. It follows that there is exactly one component MATH of MATH and MATH which meets MATH, so that we must have MATH or MATH. Similarly, if MATH lies in MATH, we will find that MATH or MATH. It follows that in either case MATH and MATH are nested as required. |
math/9906004 | As MATH and MATH are translates of MATH or MATH, it suffices to prove that MATH is small if and only if it lies in a bounded neighbourhood of each of MATH, MATH, MATH, MATH. If MATH is small, it projects to a finite subset of MATH which therefore lies within a bounded neighbourhood of the image of MATH. By lifting paths, we see that each point of MATH lies in a bounded neighbourhood of MATH, and hence lies in a bounded neighbourhood of MATH and MATH. By reversing the roles of MATH and MATH, we also see that MATH lies in a bounded neighbourhood of each of MATH and MATH . For the converse, suppose that MATH lies in a bounded neighbourhood of each of MATH and MATH. Then it must lie in a bounded neighbourhood of MATH, so that its image in MATH must lie in a bounded neighbourhood of the image of MATH. As this image is finite, it follows that MATH must be small, as required. |
math/9906004 | We need to show that MATH is transitive and that if MATH and MATH then MATH . Suppose first that MATH and MATH. The first inequality implies that MATH is small and the second implies that MATH is small, so that two of the four sets MATH are small. The assumption of REF implies that one of these two sets must be empty. As MATH, our definition of MATH implies that MATH is empty. Similarly, the fact that MATH tells us that MATH is empty. This implies that MATH as required. To prove transitivity, let MATH, MATH and MATH be elements of MATH such that MATH. We must show that MATH . Our first step is to show that MATH is small. As MATH and MATH are small, we let MATH be an upper bound for the distance of points of MATH from MATH and let MATH be an upper bound for the distance of points of MATH from MATH. Let MATH be a point of MATH. If MATH lies in MATH, then it lies in MATH and so has distance at most MATH from MATH. Otherwise, it must lie in MATH and so have distance at most MATH from some point MATH of MATH. If MATH lies in MATH, then MATH has distance at most MATH from MATH. Otherwise, MATH lies in MATH and so has distance at most MATH from MATH. In this case, MATH has distance at most MATH from MATH. It follows that in all cases, MATH has distance at most MATH from MATH, so that MATH lies in a bounded neighbourhood of MATH as required. As MATH is contained in MATH, it follows that it lies in bounded neighbourhoods of MATH and MATH, so that MATH is small as required. The definition of MATH now shows that MATH, except possibly when two of the four sets MATH are small. The only possibility is that MATH and MATH are both small. As one must be empty, either MATH or MATH. We conclude that if MATH, then either MATH or MATH. Now we consider two cases. First suppose that MATH, so that either MATH or MATH. If MATH, then MATH, so that MATH. As MATH and MATH, it follows from the first paragraph of the proof of this lemma that MATH. Hence, in either case, MATH . Now consider the general situation when MATH. Again either MATH or MATH. If MATH, then we have MATH. Now the preceding paragraph implies that MATH. Hence we again have MATH and MATH so that MATH. Hence MATH still holds. This completes the proof of the lemma. |
math/9906004 | REF are obvious from the definition of MATH and the hypotheses of REF . To prove REF , we observe that if MATH and MATH, then MATH and MATH must both be small. This implies that MATH itself is small, so that MATH or MATH must be small. But this contradicts the hypothesis that MATH is a non-trivial MATH - almost invariant subset of MATH . Finally we prove REF . Let MATH be an element of MATH such that MATH. Recall that, as MATH projects to a finite subset of MATH, we know that MATH lies in a MATH - neighbourhood of MATH, for some MATH. If MATH but MATH is not contained in MATH, then MATH and MATH are not nested. Now REF tells us that if MATH is such a set, then MATH belongs to one of only finitely many double cosets MATH. It follows that if we consider all elements MATH of MATH such that MATH, we will find either MATH, or MATH lies in a MATH - neighbourhood of MATH, for finitely many different values of MATH . Hence there is MATH such that if MATH then MATH lies in the MATH - neighbourhood of MATH. Similarly, there is MATH such that if MATH, then MATH lies in the MATH - neighbourhood of MATH . Let MATH denote the larger of MATH and MATH. Then for any elements MATH and MATH of MATH with MATH, the set MATH lies in the MATH - neighbourhood of each of MATH, MATH, MATH and MATH . Now suppose we are given MATH and wish to prove REF . Choose a point MATH in MATH whose distance from MATH is greater than MATH, choose a point MATH in MATH whose distance from MATH is greater than MATH and choose a path MATH in MATH joining MATH to MATH. If MATH, then MATH must lie in MATH and MATH must lie in MATH so that MATH must meet MATH. As MATH is compact, the proof of REF shows that the number of such MATH is finite. This completes the proof of REF of the lemma. |
math/9906004 | To prove the first part, we let MATH denote the set of all translates of MATH and MATH by elements of MATH, let MATH be the involution on MATH and let the relation MATH be defined on MATH by the condition that MATH if MATH is empty or the only small set of the four sets MATH. REF show that MATH is a partial order on MATH and satisfies all of NAME 's REF - REF . Hence we can construct a tree MATH from MATH. As MATH acts on MATH, we have a natural action of MATH on MATH . Clearly, MATH acts transitively on the edges of MATH. If MATH acts without inversions, then MATH has a single edge and gives MATH the structure of an amalgamated free product or HNN decomposition. The stabiliser of the edge of MATH which corresponds to MATH is the stabiliser MATH of MATH, so we obtain a splitting of MATH over MATH unless MATH fixes a vertex of MATH. Note that as MATH is finite, and MATH preserves MATH, it follows that MATH contains MATH with finite index as claimed in the theorem. If MATH acts on MATH with inversions, we simply subdivide each edge to obtain a new tree MATH on which MATH acts without inversions. In this case, the quotient MATH again has one edge, but it has distinct vertices. The edge group is MATH and one of the vertex groups contains MATH with index two. As MATH has infinite index in MATH, it follows that in this case also we obtain a splitting of MATH unless MATH fixes a vertex of MATH . Suppose that MATH fixes a vertex MATH of MATH. As MATH acts transitively on the edges of MATH, every edge of MATH must have one vertex at MATH, so that all edges of MATH are adjacent to each other. We will show that this cannot occur. The key hypothesis here is that MATH is non-trivial. Let MATH denote MATH or MATH, and note that REF of REF shows that MATH or MATH. Recall that there is MATH such that if MATH then MATH lies in the MATH - neighbourhood of MATH. If MATH denotes MATH, and MATH, it follows that MATH lies in the MATH - neighbourhood of MATH. Let MATH denote the distance of the identity of MATH from MATH. Then MATH must lie within the MATH - neighbourhood of MATH, for all MATH, so that MATH itself lies in the MATH - neighbourhood of MATH. Similarly, MATH lies in the MATH - neighbourhood of MATH. Now both MATH and MATH project to infinite subsets of MATH, so MATH cannot equal MATH or MATH . It follows that there are elements MATH-and MATH of MATH such that MATH, so that MATH-and MATH represent non-adjacent edges of MATH. This completes the proof that MATH cannot fix a vertex of MATH . To prove the last statement of the first part of REF , we will simplify notation by supposing that the stabiliser MATH of MATH is equal to MATH. One of the standard MATH - almost invariant sets associated to the splitting we have obtained from the action of MATH on the tree MATH is the set MATH in the preceding paragraph. We will show that MATH is MATH - almost equal to MATH. The preceding paragraph shows that MATH lies in the MATH - neighbourhood of MATH, and that MATH lies in the MATH - neighbourhood of MATH. It follows that MATH is MATH - almost contained in MATH and MATH is MATH - almost contained in MATH, so that MATH and MATH are MATH - almost equal as claimed. This completes the proof of the first part of REF . For the second part, we will simply comment on the modifications needed to the preceding proof. The statement of REF remains true though the proof needs a little modification. The statement and proof of REF apply unchanged. The statement of REF remains true, though the proof needs some minor modifications. Finally the proof of the first part of REF applies with minor modifications to show that MATH acts on a tree MATH with quotient consisting of MATH edges in the required way. This completes the proof of REF . |
math/9906004 | The given splitting of MATH over MATH corresponds to an action of MATH on a tree MATH such that MATH-has a single edge, and some edge of MATH-has stabiliser MATH. Let MATH denote the fixed set of MATH, that is, the set of all points fixed by MATH. Then MATH is a (non-empty) subtree of MATH. As MATH normalises MATH, it must preserve MATH, that is, MATH. Suppose that MATH. As MATH has finite index in MATH, we let MATH denote a set of coset representatives for MATH in MATH, where MATH. As MATH acts transitively on MATH, we have MATH. Edges of MATH all have stabiliser MATH, and so edges of MATH all have stabiliser MATH. As MATH does not lie in MATH, these stabilisers are distinct so the intersection MATH contains no edges. The intersection of two subtrees of a tree must be empty or a tree, so it follows that MATH is empty or a single vertex MATH, for each MATH. Now MATH preserves MATH and permutes the translates MATH, so MATH preserves the collection of all the MATH's. As this collection is finite, MATH has a subgroup MATH of finite index such that MATH fixes a vertex MATH of MATH. As MATH has finite index in MATH, it follows that MATH itself fixes some vertex of MATH, which contradicts our assumption that our action of MATH on MATH corresponds to a splitting of MATH. This contradiction shows that MATH must equal MATH, so that MATH is normal in MATH-as claimed. |
math/9906004 | Our first step will be to show that MATH and MATH must be commensurable. Without loss of generality, we can suppose that MATH is small. The other small set can only be MATH, as otherwise MATH or MATH would be small which is impossible. It follows that for each edge of MATH, either it is also an edge of MATH or it has (at least) one end in one of the two small sets. As the images in MATH of MATH and of each small set is finite, and as the graph MATH is locally finite, it follows that the image of MATH in MATH must be finite. This implies that MATH has finite index in the stabiliser MATH of MATH. By reversing the roles of MATH and MATH, it follows that MATH has finite index in MATH, so that MATH and MATH must be commensurable, as claimed. Now let MATH denote MATH, so that MATH stabilises both MATH and MATH, and consider the images MATH and MATH of MATH and MATH in MATH. As MATH has finite index in MATH and MATH, it follows that MATH and MATH are each finite, so that MATH and MATH are almost invariant subsets of MATH. Further, two of the four sets MATH have finite image in MATH, so we can assume that MATH and MATH are almost equal, by replacing one of MATH or MATH by its complement in MATH, if needed. Let MATH denote the intersection of the conjugates of MATH in MATH, so that MATH is normal in MATH, though it need not be normal in MATH. We do not have MATH, but because MATH has finite index in MATH, we know that MATH has finite index in MATH and hence also in MATH, which is all we need. Let MATH and MATH denote the images of MATH and MATH respectively in MATH, and consider the action of an element MATH of MATH on MATH . Trivially MATH. As MATH and MATH are almost equal, MATH must be almost equal to MATH. Now we use the key fact that MATH is associated to a splitting of MATH so that its translates by MATH are nested. Thus for any element MATH of MATH, one of the following four inclusions holds: MATH, MATH, MATH, MATH. As MATH is almost equal to MATH, we must have MATH or MATH . But MATH has a power which lies in MATH and hence stabilises MATH. It follows that MATH, so that MATH lies in MATH. Thus MATH is a subgroup of MATH . Similarly, MATH must be a subgroup of MATH, so that MATH. This completes the proof of REF. Note that it follows that MATH, that MATH and MATH and that MATH and MATH are almost equal or almost complementary. In order to prove REF, we apply the preceding work to the case when the second splitting is obtained from the first by conjugating by some element MATH of MATH. Thus MATH and MATH which is MATH - almost equal to MATH by REF . Hence if two of the four sets MATH are small, then so are two of the four sets MATH small. Now the above shows that MATH, so that MATH normalises MATH. This completes the proof of the lemma. |
math/9906004 | The preceding lemma showed that the hypotheses imply that MATH equals MATH and also that the images MATH and MATH of MATH and MATH in MATH are almost equal or almost complementary. By replacing one of MATH or MATH by its complement if needed, we can arrange that MATH and MATH are almost equal. We will show that in most cases, the two given splittings over MATH and MATH must be equivalent, and that the exceptional cases can be analysed separately to show that the splittings are conjugate. Recall that by applying REF , we can use information about MATH and its translates to construct a MATH - tree MATH and hence the original splitting of MATH over MATH. Similarly, we can use information about MATH and its translates to construct a MATH - tree MATH and hence the original splitting of MATH over MATH. We will compare these two constructions in order to prove our result. As MATH and MATH are almost equal subsets of MATH, it follows that there is MATH such that, in the NAME graph MATH of MATH, we have MATH lies in a MATH - neighbourhood of MATH and MATH lies in a MATH - neighbourhood of MATH. Now let MATH denote one of MATH or MATH, let MATH denote one of MATH or MATH and let MATH and MATH denote the corresponding sets obtained by replacing MATH with MATH. Recall that MATH is small if and only if its image in MATH is finite. Clearly this occurs if and only if MATH lies in a MATH - neighbourhood of MATH, for some MATH. It follows that MATH is small if and only if MATH is small. As MATH and MATH are associated to splittings, we know that for each MATH, at least one of the four sets MATH is empty and at least one of the four sets MATH is empty. Further the information about which of the four sets is empty completely determines the trees MATH and MATH. Thus we would like to show that when we compare the four sets MATH with the four sets MATH, then corresponding sets are empty. Note that when MATH lies in MATH, we have MATH, so that two of the four sets MATH are empty. First we consider the case when, for each MATH, only one of the sets MATH is small and hence empty. Then only the corresponding one of the four sets MATH is small and hence empty. Now the correspondence MATH gives a MATH - isomorphism of MATH with MATH and thus the splittings are equivalent. Next we consider the case when two of the sets MATH are small, for some MATH. REF implies that MATH normalises MATH. Further if MATH, then MATH-is almost equal to MATH or MATH. Let MATH denote the normaliser of MATH in MATH, so that MATH acts on the left on the graph MATH and we have MATH. Let MATH denote the subgroup of MATH consisting of elements MATH such that MATH is almost equal to MATH or MATH. Now we apply REF from CITE to the action of MATH on the left on the graph MATH . This result tells us that if MATH is infinite, then it has an infinite cyclic subgroup of finite index. Further the proof of this result in CITE shows that the quotient of MATH by MATH must be finite. This implies that MATH has two ends and that MATH has finite index in MATH. To summarise, either MATH is finite, or it has two ends and MATH has finite index in MATH. Let MATH be an element of MATH whose image in MATH has finite order such that MATH. As MATH is associated to a splitting of MATH, we must have MATH or MATH. As MATH has finite order in MATH, we have MATH, for some positive integer MATH, which implies that MATH so that MATH itself lies in MATH. It follows that the group MATH must be trivial, MATH, MATH or MATH. In the first case, the two trees MATH and MATH will be MATH - isomorphic, showing that the given splittings are equivalent. In the other three cases, MATH is non-empty and we know that, for any MATH, two of the four sets MATH are small. Thus in these cases, it seems possible that MATH and MATH will not be MATH - isomorphic, so we need some special arguments. We start with the case when MATH is MATH. In this case, the given splitting must be an amalgamated free product of the form MATH, for some group MATH. If MATH denotes an element of MATH, then MATH. Thus MATH acts on MATH and MATH with inversions. Recall that either the two partial orders on the translates of MATH and MATH are the same under the bijection MATH, or they differ only in that MATH but MATH, for all MATH. If they differ, we replace the second splitting by its conjugate by some element MATH, so that MATH is replaced by MATH and we replace MATH by MATH . As MATH is MATH - almost equal to MATH, the partial orders on the translates of MATH and MATH respectively are the same under the bijection MATH except possibly when one compares MATH, MATH and MATH, MATH, where MATH . In this case, the inclusion MATH tells us that MATH, and the inclusion MATH tells us that MATH. We conclude that the partial orders on the translates of MATH and MATH respectively are exactly the same, so that MATH and MATH are MATH - isomorphic, and the two given splittings are conjugate by an element of MATH. Now we turn to the two cases where MATH is infinite, so that MATH has finite index in MATH and MATH has two ends. As MATH normalises MATH, REF shows that MATH is normal in MATH. As MATH has two ends, it follows that MATH, so that MATH is MATH or MATH. It is easy to check that there is only one splitting of MATH over the trivial group and that all splittings of MATH over the trivial group are conjugate. It follows that, in either case, all splittings of MATH over MATH are conjugate. This completes the proof of REF . |
math/9906004 | Recall that MATH is associated to a splitting of MATH over MATH. It follows that MATH is associated to the conjugate of this splitting by MATH. Thus MATH and MATH are associated to splittings of MATH which are each conjugate to one of the two given splittings. If MATH and MATH are each translates of MATH or MATH, the nestedness of the translates of MATH shows that one of the two small sets must be empty as claimed. Similarly if both are translates of MATH or MATH, then one of the two small sets must be empty. If MATH is a translate of MATH or MATH and MATH is a translate of MATH or MATH, we apply REF to show that the splittings to which MATH and MATH are associated are conjugate. It follows that the two original splittings were conjugate as required. |
math/9906004 | Let the MATH splittings MATH of MATH be over subgroups MATH with associated MATH - almost invariant subsets MATH of MATH, and let MATH. We will start by supposing that no two of the MATH's are conjugate. We will handle the general case at the end of this proof. We will apply the second part of REF to MATH . Recall that our assumption that the MATH's have intersection number zero implies that no translate of MATH can cross any translate of MATH, for MATH. As each MATH is associated to a splitting, it is also true that no translate of MATH can cross any translate of MATH. This means that the set MATH is almost nested. In order to apply REF , we will also need to show that for any pair of elements MATH and MATH of MATH, if two of the four sets MATH are small then one is empty. Now REF shows that if two of these four sets are small, then either one is empty or there are distinct MATH and MATH such that MATH and MATH are conjugate. As we are assuming that no two of these splittings are conjugate, it follows that if two of the four sets MATH are small then one is empty, as required. REF now implies that MATH can be expressed as the fundamental group of a graph MATH of groups whose MATH-th edge corresponds to a conjugate of a splitting of MATH over the stabilizer MATH of MATH. As MATH is associated to a splitting of MATH over MATH, its stabiliser MATH must equal MATH. Further, it is clear from the construction that collapsing all but the MATH-th edge of MATH yields a conjugate of MATH, as the corresponding MATH - tree has edges which correspond precisely to the translates of MATH . Now suppose that we have a graph of groups structure MATH for MATH such that, for each MATH, MATH, collapsing all edges but the MATH-th yields a conjugate of the splitting MATH of MATH. This determines an action of MATH on a tree MATH without inversions. We want to show that MATH and MATH are MATH - isomorphic. For this implies that MATH and MATH have the same underlying graph, and that corresponding edge and vertex groups are conjugate, as required. Let MATH denote an edge of MATH, and let MATH denote MATH or MATH. There are edges MATH of MATH, MATH, such that the set MATH of all translates of MATH and MATH is nested and NAME 's construction applied to MATH yields the MATH - tree MATH again. We will denote MATH by MATH. The hypotheses imply that there is MATH such that the stabiliser MATH of MATH equals MATH, and that MATH is MATH - almost equal to MATH, where MATH is one of the standard MATH - almost invariant subsets of MATH associated to the splitting MATH. Let MATH denote MATH so that MATH is MATH - almost equal to MATH. Now REF shows that MATH is MATH - almost equal to MATH, so that MATH is MATH - almost equal to MATH. Now consider the MATH - equivariant bijection MATH determined by sending MATH to MATH. The above argument shows that if MATH is any element of MATH, and MATH is the corresponding element of MATH, then MATH-and MATH are MATH - almost equal. We will show that in most cases, this bijection automatically preserves the partial orders on MATH and MATH, implying that MATH and MATH are MATH - isomorphic, as required. We compare the partial orders on MATH and MATH rather as in the proof of REF . For any elements MATH and MATH of MATH, let MATH and MATH denote the corresponding elements of MATH. Thus MATH is small if and only if MATH is small. We would like to show that when we compare the four sets MATH with the four sets MATH, then corresponding sets are empty, so that the partial orders are preserved by our bijection. Otherwise, there must be MATH and MATH in MATH such that two of the sets MATH are small. If MATH-and MATH are translates of MATH and MATH, then REF tells us that the splittings MATH and MATH are conjugate. As we are assuming that distinct splittings are not conjugate, it follows that MATH. Now the arguments in the proof of REF show that either the splitting MATH is an amalgamated free product of the form MATH, with MATH, or MATH is normal in MATH and MATH is MATH or MATH. If the second case occurs, then there can be only one splitting in the given family, so it is immediate that MATH and MATH have the same underlying graph, and that corresponding edge and vertex groups are conjugate. If the first case occurs and the partial orders on translates of MATH and MATH do not match, we must have MATH but MATH, for all MATH. We now pick MATH and alter our bijection from MATH to MATH so that MATH maps to MATH and extend MATH - equivariantly to the translates of MATH and MATH. This ensures that the partial orders on MATH and MATH match for translates of MATH. By repeating this for other values of MATH as necessary, we can arrange that the partial orders match completely, and can then conclude that MATH and MATH are MATH - isomorphic as required. We end by discussing the case when some of the given MATH splittings are conjugate. We divide the splittings into conjugacy classes and discard all except one splitting from each conjugacy class, to obtain MATH splittings. Now we apply the preceding argument to express MATH uniquely as the fundamental group of a graph MATH of groups with MATH edges. If an edge of MATH corresponds to a splitting over a subgroup MATH which is conjugate to MATH other splittings, we simply subdivide this edge into MATH sub-edges, and label all the sub-edges and the MATH new vertices by MATH. This shows the existence of the required graph of groups structure MATH corresponding to the original MATH splittings. The uniqueness of MATH follows from the uniqueness of MATH, and the fact that the collection of all the edges of MATH which correspond to a given splitting of MATH must form an interval in MATH in which all the interior vertices have valence MATH. This completes the proof of REF . |
math/9906004 | The idea of the proof is much as before. We let MATH denote the almost invariant subset MATH of MATH, and let MATH denote MATH. We want to apply the first part of REF . As before, the assumption that MATH implies that MATH is almost nested. However, in order to apply REF , we also need to know that for any pair of elements MATH and MATH of MATH, if two of the four sets MATH are small then one is empty. In the proof of REF , we simply applied REF . However, here the situation is somewhat more complicated. REF below shows that if MATH and MATH are both small, then MATH must lie in a certain subgroup MATH of MATH. Thus it would suffice to arrange that MATH is nested with respect to MATH, that is, that MATH and MATH are nested so long as MATH lies in MATH. Now REF below tells us that there is a subgroup MATH commensurable with MATH and a MATH - almost invariant set MATH equivalent to MATH such that MATH is nested with respect to MATH. It follows that if MATH-and MATH are any elements of MATH and if MATH and MATH are both small, then one of them is empty. We also claim that, like MATH, the set MATH is almost nested. This means that if we let MATH denote MATH, we are claiming that MATH. Let MATH denote MATH. The fact that MATH is equivalent to MATH means that the pre-images in MATH of MATH and of MATH are almost equal almost invariant sets which we denote by MATH and MATH. If MATH denotes the index of MATH in MATH, then MATH and similarly MATH is an integral multiple of MATH. As MATH and MATH are almost equal, it follows that MATH, and hence that MATH as claimed. This now allows us to apply REF to the set MATH. We conclude that MATH-splits over the stabiliser MATH of MATH, that MATH contains MATH with finite index and that one of the MATH - almost invariant sets associated to the splitting is equivalent to MATH. It follows that MATH is commensurable with MATH and that one of the MATH - almost invariant sets determined by the splitting is equivalent to MATH . This completes the proof of REF apart from the proofs of REF . |
math/9906004 | The first inclusion is clear. The second is proved in essentially the same way as the proof of the first part of REF . Let MATH be an element of MATH, and consider the case when MATH ( the other case is similar). Recall that this means that the sets MATH and MATH are both small. Now for each edge of MATH, either it is also an edge of MATH or it has (at least) one end in one of the two small sets. As the images in MATH of MATH and of each small set is finite, and as the graph MATH is locally finite, it follows that the image of MATH in MATH must be finite. This implies that MATH has finite index in the stabiliser MATH of MATH. By reversing the roles of MATH and MATH, it follows that MATH has finite index in MATH, so that MATH and MATH must be commensurable, as claimed. It follows that MATH, as required. |
math/9906004 | REF tells us that the given set is contained in the union of a finite number of double cosets MATH. If MATH, we claim that the double coset MATH is itself the union of only finitely many cosets MATH, which proves the required result. To prove our claim, recall that MATH is commensurable with MATH. Thus MATH can be expressed as the union of cosets MATH, for MATH . Hence MATH so that MATH is the union of finitely many cosets MATH as claimed. |
math/9906004 | Consider an element MATH in MATH. REF tells us that MATH and MATH are commensurable subgroups of MATH. Let MATH denote their intersection and let MATH denote the intersection of the conjugates of MATH in MATH. Thus MATH is of finite index in MATH and MATH and is normal in MATH. Now consider the quotient MATH. Let MATH and MATH denote the images of MATH and MATH respectively in MATH. As before, MATH and MATH are almost invariant subsets of MATH which are almost equal or almost complementary. Now consider the action of MATH on the left on MATH. If MATH is in MATH, then MATH, so that MATH . If MATH and MATH are nested, there are four possible inclusions, but the fact that MATH excludes two of them. Thus we must have MATH or MATH. This implies that MATH as some power of MATH lies in MATH and so acts trivially on MATH . We conclude that if MATH is an element of MATH such that MATH and MATH are nested, then MATH stabilises MATH and so lies in MATH. Hence MATH lies in MATH. It follows that for each element MATH of MATH, the sets MATH and MATH are not nested. Recall from REF that MATH and MATH are not nested-MATH consists of a finite number of cosets MATH of MATH in MATH. It will be convenient to denote this number by MATH. Thus, for MATH, the set MATH and MATH are not nested-MATH consists of MATH cosets MATH of MATH in MATH. It follows that MATH lies in the union of MATH cosets MATH of MATH in MATH. As MATH, it follows that MATH lies in the union of MATH cosets MATH of MATH in MATH and hence that MATH has index at most MATH in MATH . A similar argument shows also that MATH has index at most MATH in MATH. Of course, the same bound applies to the index of MATH in MATH, for each MATH. Now we define MATH . Clearly MATH is a subgroup of MATH which is normalised by MATH. Now each intersection MATH has index at most MATH in MATH, and so MATH is an intersection of subgroups of MATH of index at most MATH. If MATH has MATH subgroups of index at most MATH, it follows that MATH has index at most MATH in MATH. Hence each element of MATH normalises a subgroup of MATH of index at most MATH in MATH. As MATH has only finitely many such subgroups, we have proved that there are a finite number of finite index subgroups MATH of MATH, such that MATH is contained in the union of the groups MATH, MATH, as required. |
math/9906004 | We will consider how MATH can intersect the normaliser of a subgroup of finite index in MATH. Let MATH denote a subgroup of MATH of finite index. We denote the image of MATH in MATH by MATH. Then MATH is an almost invariant subset of MATH. We consider the group MATH, which we will denote by MATH. Then MATH acts on the left on MATH, and we have MATH or MATH, for every element MATH of MATH, because every element of MATH satisfies MATH or MATH. Now we apply REF from CITE to the action of MATH on the left on the graph MATH. This result tells us that if MATH is infinite, then it has an infinite cyclic subgroup of finite index. Further the proof of this result in CITE shows that the quotient of MATH by MATH must be finite. This implies that MATH has two ends and that MATH has finite index in MATH. Hence either MATH is finite, or it has two ends and MATH has finite index in MATH . Recall that there are a finite number of finite index subgroups MATH of MATH, such that MATH is contained in the union of the groups MATH, MATH. The above discussion shows that, for each MATH, if MATH denotes MATH, either MATH is finite, or it has two ends and MATH has finite index in MATH. We consider two cases depending on whether or not every MATH is finite. Suppose first that each MATH is finite. We claim that MATH contains MATH with finite index. To see this, let MATH, so that MATH is a subgroup of MATH of finite index, and note that MATH is the union of a finite collection of groups MATH each of which contains MATH with finite index, so that MATH is the union of finitely many cosets of MATH. It follows that MATH also contains MATH with finite index and hence contains MATH with finite index as claimed. If we let MATH denote the intersection of the conjugates of MATH in MATH, then MATH is the required subgroup of MATH which is normalised by MATH . Now we turn to the case when MATH is infinite and so MATH has two ends and MATH has finite index in MATH . Define MATH to be MATH. As MATH contains MATH with finite index, MATH is the intersection of only finitely many conjugates of MATH. As MATH is contained in MATH, each of these conjugates of MATH is commensurable with MATH. It follows that MATH is a subgroup of MATH of finite index in MATH which is normalised by MATH. This completes the proof of the lemma. |
math/9906004 | The previous lemma tells us that there is a subgroup MATH of finite index in MATH such that MATH normalises MATH. Let MATH denote the almost invariant subset MATH of MATH, and let MATH denote the almost invariant subset MATH of MATH . Suppose that the index of MATH in MATH is infinite. Recall from the proof of the preceding lemma that MATH has two ends and that MATH has finite index in MATH. We construct a new non-trivial MATH - almost invariant set MATH as follows. Since the quotient group MATH has two ends, MATH splits over a subgroup MATH which contains MATH with finite index. Thus there is a MATH - almost invariant set MATH in MATH which is nested with respect to MATH. Further, MATH is normal in MATH and the quotient group must be isomorphic to MATH or MATH. Let MATH be coset representatives of MATH in MATH so that MATH. We take MATH. It is easy to check that MATH is MATH - almost invariant and that MATH is nested with respect to MATH . Now suppose that the index of MATH in MATH is finite. We will define the subgroup MATH of MATH. The index of MATH in MATH is at most two. First we consider the case when MATH. We define MATH to be the intersection of the translates of MATH under the action of MATH. Thus MATH is invariant under the action of MATH. As all the translates of MATH by elements of MATH are almost equal to MATH, it follows that MATH so that MATH is also an almost invariant subset of MATH. Let MATH denote the inverse image of MATH in MATH, so that MATH is invariant under the action of MATH. In particular, MATH is nested with respect to MATH, as required. Now we consider the general case when MATH. We can apply the above arguments using MATH in place of MATH to obtain a subgroup MATH of MATH and a MATH - almost invariant subset MATH of MATH which is equivalent to MATH, and whose translates are nested with respect to MATH. We also know that MATH is MATH - invariant. Let MATH denote the image of MATH in MATH, let MATH denote an element of MATH and consider the involution of MATH induced by MATH. Then MATH is a non-trivial almost invariant subset of MATH and MATH. Define MATH, so that MATH and let MATH denote the pre-image of MATH in MATH. We claim that the translates of MATH and MATH are nested with respect to MATH. First we show that they are nested with respect to MATH, by showing that MATH is MATH - invariant. For MATH, we have MATH as MATH must be normal in MATH. It follows that MATH. As MATH, we see that MATH is MATH - invariant as required. In order to show that the translates of MATH and MATH are nested with respect to MATH, we will also show that MATH is empty. This follows from the fact that MATH which is clearly empty. This completes the proof of REF . |
math/9906004 | Suppose that there exists a subset MATH of MATH which is MATH - almost equal to MATH, such that MATH. We have MATH as MATH and MATH are MATH - almost equal. So, it is enough to show that for every MATH, either MATH or MATH is MATH - finite. Suppose that MATH. Consider MATH which is a union of a finite number of right cosets MATH, MATH. Since MATH, MATH. For any MATH, MATH. Thus MATH is at a bounded distance from MATH and hence MATH has finite image in MATH. Similarly, if MATH, MATH projects to a finite set in MATH. For the converse, suppose that MATH and let MATH denote the projection from MATH to MATH. By hypothesis, MATH or MATH is finite. The proof of REF tells us that there is a positive number MATH such that, for every MATH, the set MATH is contained in a MATH - neighbourhood of MATH or MATH. Let MATH, the MATH - neighbourhood of MATH and let MATH. If MATH and MATH, then MATH and thus MATH. If MATH and MATH, then MATH and thus MATH. It only remains to show that MATH is MATH - almost equal to MATH . This is essentially shown in the third and fourth paragraphs of the proof of REF . |
math/9906004 | Let MATH be the NAME graph of MATH with respect to a finite system of generators and let MATH. As in the proof of REF , for a set MATH of vertices in a graph, we let MATH denote the maximal subgraph with vertex set equal to MATH. We will show that exactly one component of MATH has infinite image in MATH. Note that MATH has exactly one infinite component as MATH has only two ends. Let MATH denote the set of vertices of the infinite component of MATH and let MATH denote the inverse image of MATH in MATH. If MATH has components with vertex set MATH, then we have MATH. Let MATH denote the vertex set of a component of MATH, and let MATH be the stabilizer in MATH of MATH. Since MATH is finite, we see that MATH is finite. Hence MATH has more than one end. Now our hypothesis that MATH is of surface type implies that MATH has finite index in MATH and thus MATH is finite. If MATH, we see that MATH divides MATH into at least three infinite components. Thus MATH and so MATH is connected. The other components of MATH have finite image in MATH. Similarly, exactly one component of MATH has infinite image in MATH. The same argument shows that for any finite subset MATH of MATH containing MATH, the two infinite components of MATH and-MATH have connected inverse images in MATH. Recall that if MATH crosses MATH strongly, then MATH crosses MATH. We will next show that if MATH does not cross MATH strongly, then MATH does not cross MATH. Suppose that MATH projects to a finite set in MATH. Take a compact set MATH in MATH large enough to contain MATH and MATH. By the argument above, if MATH is the infinite component of MATH, then its inverse image MATH is connected and is contained in MATH. Any two points in MATH can be connected by a path in MATH and thus the path does not intersect MATH. Thus MATH is contained in MATH or MATH. Hence MATH or MATH is empty. Suppose that MATH is empty. Then MATH. Since MATH projects to a finite set, we see that MATH projects to a finite set. Similarly, if MATH is empty, then MATH projects to a finite set in MATH. Thus, we have shown that if MATH projects to a finite set, then either MATH or MATH projects to finite set. Thus MATH does not cross MATH. |
math/9906004 | Observe that REF shows that, by changing MATH up to commensurability, and changing MATH, we may assume that the translates of MATH are almost nested with respect to MATH and nested with respect to MATH or MATH. If we do not have almost nesting for all translates of MATH, then there is MATH outside MATH such that none of MATH is MATH - finite. In particular, none of these sets is MATH - finite. But these four sets are each invariant under MATH and the fact that the strong intersection number vanishes shows that at least one of them has boundary which is MATH - finite. Since MATH is not in MATH, we have a contradiction to our hypothesis that MATH with MATH. This completes the proof. |
math/9906004 | Let MATH be a non-trivial MATH - almost invariant subset of MATH, let MATH be an element of MATH and let MATH, so that MATH-has stabiliser MATH. Let MATH denote the intersection MATH which has finite index in both MATH and in MATH because MATH lies in MATH. Thus MATH and MATH are both almost invariant subsets of MATH. As MATH is of surface type, the pair MATH has two ends so that MATH and MATH are almost equal or almost complementary. It follows that MATH is MATH - almost equal to MATH or MATH, that is, MATH or MATH. Recall from REF , that if MATH denotes MATH or MATH, then MATH. It follows that in our present situation MATH must equal MATH. By REF , we see that there are a finite number of subgroups MATH of finite index in MATH such that MATH is contained in the union of the normalizers MATH. As MATH has infinite index in MATH, one of the MATH, say MATH, has infinite index in its normalizer MATH. As MATH is of surface type, the pair MATH has two ends, so we can apply REF from CITE to the action of MATH on the left on the graph MATH. This result tells us that MATH is virtually infinite cyclic. Further the proof of this result in CITE shows that the quotient of MATH by MATH must be finite so that MATH has finite index in MATH. |
math/9906004 | The proof of REF goes through because of our assumption that MATH. The mistake in CITE occurs in the proof of REF where it is implicitly assumed that if MATH crosses MATH, then it crosses MATH strongly. Since we have assumed that the two intersection numbers are equal, the argument is now valid. |
math/9906007 | If MATH is a relative equilibrium then MATH for some MATH. Since MATH is quadratic homogeneous, we have MATH for all MATH. Hence MATH. On the other hand, since MATH is definite, the ray MATH is transverse to the level set MATH, hence MATH. Contradiction. |
math/9906007 | Since the action of MATH preserves the one-form MATH, the action of MATH on MATH is Hamiltonian. The action of MATH on MATH is also Hamiltonian: MATH is a corresponding moment map. Consequently MATH is a symplectic orbifold diffeomorphic to MATH; from now on we identify MATH and MATH. The Hamiltonian action of MATH on MATH descends to Hamiltonian action on MATH. Since MATH is compact and the action of MATH is Hamiltonian, the set of fixed points MATH is nonempty. In fact MATH is a disjoint union of connected symplectic suborbifolds of MATH (see for example CITE for more details on Hamiltonian group actions on orbifolds). For any point MATH, the MATH orbit MATH is contained in the MATH orbit MATH. Since MATH acts on MATH without fixed points we in fact have that MATH for any MATH. Consequently the union of manifolds MATH consists of periodic manifolds of the Hamiltonian MATH. It remains to check that for a connected component MATH of MATH, the manifold MATH is a nondegenerate periodic manifold of MATH. Now the time MATH map of the flow of the Hamiltonian vector field of MATH on MATH is given by MATH where MATH is the exponential map. So let MATH be a point in MATH. Then MATH is a relative MATH equilibrium of MATH. Hence the differential MATH at MATH is proportional to the differential of the MATH moment map, which is MATH. Hence MATH. Therefore it's enough to compute the differential at MATH of the ``NAME map" MATH, MATH, where MATH is the smallest positive number with MATH. Since the MATH orbit of MATH is a circle, the isotropy group of MATH is of the form MATH, where MATH is a finite abelian subgroup of MATH and MATH is a subtorus of MATH of codimension one. Moreover, we can split MATH as MATH where MATH is isomorphic to MATH and contains MATH. Let us next assume, to make the exposition simpler, that MATH is trivial. Then it follows from the slice theorem that a neighborhood of MATH in MATH is MATH equivariantly diffeomorphic to a neighborhood of MATH in MATH where MATH. Here MATH acts on MATH by MATH where MATH are nontrivial characters of MATH. Let MATH, MATH denote the projections. Then MATH. We claim that for all MATH between MATH and MATH, MATH is of the form MATH where MATH are irrational numbers. Note that the claim implies immediately that the algebraic multiplicity of the eigenvalue REF of the differential of the ``NAME map" MATH is MATH, hence that MATH is a nondegenerate periodic manifold of MATH. The claim holds because the one-parameter subgroup MATH is dense in MATH. More specifically, let MATH be a basis of the integral lattice of the torus MATH which is compatible with the splitting MATH (so that MATH and MATH is a basis of the integral lattice of MATH). Then MATH for some MATH. Moreover, since the one-parameter subgroup defined by MATH is dense in MATH, the sum MATH is not a rational number for any MATH tuple of rational numbers MATH. Since MATH, MATH. Consequently MATH and the claim follows (note that MATH are integers). If the group MATH is not trivial, then a neighborhood of MATH in MATH is modeled by the quotient MATH, where MATH acts on MATH by multiplication and on MATH linearly by MATH characters, so that the actions of MATH and MATH commute. The same argument as above still works: for the NAME map on the quotient to have an eigenvector in MATH with eigenvalue REF it is necessary for MATH to be a root of unity. But this is impossible as we have seen. This proves REF . |
math/9906007 | We use open sets in our definition of the category. Let MATH be a generic component of the moment map for the action of MATH on MATH so that MATH is precisely the set of critical points of MATH. The function MATH is NAME. Therefore MATH decomposes as a disjoint union of the unstable orbifolds MATH of the gradient flow of MATH. Clearly MATH. Now ``thicken" MATH's by replacing them with their tubular neighborhoods MATH inside MATH. Then MATH, the sets MATH's are open and MATH. Hence MATH. |
math/9906007 | The proof is a standard computation that uses the local normal form of the moment map of NAME and of CITE. Similar computations are carried out in CITE and in CITE. We use the version of the normal form theorem described in CITE which we now recall without proofs: Let the symbols MATH, MATH, MATH, MATH, MATH, MATH and MATH have the same meaning as above. The null directions of the restriction MATH is MATH. Hence MATH is naturally a symplectic vector space. Denote the corresponding symplectic form by MATH. Moreover, the linear action of the isotropy group MATH on MATH descends to a linear symplectic action on MATH. Denote the corresponding homogeneous moment map by MATH. The MATH invariant inner product on MATH chosen above defines MATH equivariant embeddings: MATH and MATH. Note that by construction of MATH and MATH we have that MATH for any MATH and any MATH. There exists a closed two-from MATH on the associated bundle MATH and an open MATH equivariant embedding MATH of a neighborhood the zero section of MATH into MATH with the following properties. CASE: MATH. CASE: MATH. CASE: MATH for all MATH. CASE: The embedding MATH, MATH is symplectic. Consequently for a small enough neighborhood MATH of REF in MATH, MATH is a symplectic slice through MATH. We now prove that MATH is the desired symplectic slice. Since MATH, the Hessian MATH is well-defined and behaves well under restrictions. In particular, MATH. Since MATH is MATH invariant and since for any MATH we have MATH. Therefore MATH is constant and hence MATH. Since the null directions of MATH is MATH, it follows that for any symplectic slice MATH through MATH which is tangent to MATH, we have MATH . Combining this with the previous computation we see that the rank of MATH is at most MATH. Thus by REF is positive definite for any symplectic slice which is tangent to MATH. It is easy to check that the manifold MATH is such a slice. We finally show that MATH . Now for any MATH we have MATH (since MATH by construction of MATH and MATH). Therefore MATH. |
math/9906008 | There exists a constant MATH such that MATH for all conjugacy classes MATH in MATH and circuits MATH in MATH representing MATH. Choose MATH such that MATH. We conclude that MATH for all conjugacy classes MATH. |
math/9906008 | Since MATH is indivisible, its initial edge and its terminal edge are contained in MATH. Suppose that both endpoints of MATH are contained in MATH. By REF , one of them is contained in a contractible component of MATH. Let MATH denote this endpoint. By REF , MATH is necessarily a zero stratum, and we have MATH. This implies that MATH. Since MATH is the collection of contractible components of MATH, we conclude that MATH, which implies that MATH. This contradicts our assumption that MATH is a NAME path. |
math/9906008 | Let MATH be a homotopy inverse of MATH. Since the restriction of MATH to MATH is hyperbolic, there exist numbers MATH such that MATH for all circuits MATH in MATH. There exists a path MATH such that MATH is an immersed circuit in MATH, satisfying MATH and MATH. We can find some constant MATH, depending only on MATH and MATH, such that MATH, MATH and MATH. We distinguish two cases that are not mutually exclusive. CASE: MATH. In this case, the bounded cancellation lemma tells us that MATH where MATH is the bounded cancellation constant of MATH. CASE: MATH. The same reasoning as in the previous case shows that MATH, where MATH is the bounded cancellation constant of MATH. Since MATH and MATH are homotopy inverses of each other, we can find some constant MATH such that MATH holds for all paths MATH and preimages MATH (MATH is a preimage of MATH under MATH). We conclude that MATH where MATH is some subpath of MATH satisfying MATH. Let MATH. Choose MATH large enough to satisfy MATH and let MATH. Clearly, MATH. If the length of MATH is at least MATH, we conclude that MATH . |
math/9906008 | We will show by induction that MATH . For MATH, the bounded cancellation lemma implies that MATH so the claim holds for MATH. Assume that the claim is true for some MATH. Again, the bounded cancellation lemma tells us that MATH by induction. Hence, we conclude that MATH . |
math/9906008 | Given MATH, choose an exponent MATH according to REF , for MATH. Express MATH as a concatenation of paths MATH such that MATH and MATH. There exist preimages MATH such that MATH is their concatenation. We claim that MATH for all MATH. Suppose otherwise, i. e., MATH for some MATH. Because of our choice of MATH, REF imply that MATH can be written as a concatenation MATH, where MATH splits as a concatenation of two pre-Nielsen paths with one illegal turn each. This implies that for some exponent MATH, MATH contains the concatenation of two NAME paths, which is impossible because of REF . Hence, MATH, and the lemma follows by induction. |
math/9906008 | Fix some MATH. Let MATH be a nontrivial circuit in MATH. We will distinguish several cases, and in each case we will show that there exist numbers MATH, MATH and MATH such that there exists a collection MATH of subpaths of MATH with the following properties: CASE: For every integer MATH and for every MATH, we have MATH where MATH is a subpath of MATH such that MATH. We say that MATH has the desired growth. CASE: There is no overlap between distinct paths in MATH. CASE: The sum of the lengths of the paths in MATH is at least MATH. If the numbers MATH, MATH and MATH depend only on the case in question, but not on MATH, then the theorem follows immediately because the growth of the subpaths in MATH provides a lower bound for the growth of MATH. We distinguish the following cases. CASE: MATH or MATH. If MATH, then MATH is legal, so it has the desired growth under forward iteration. Otherwise, let MATH be the collection of maximal legal subpaths of MATH of length at least MATH. The choice of MATH and REF guarantee that the subpaths in MATH have the desired growth under forward iteration, so we only have to show that they account for a definite fraction of the length of MATH. An elementary computation will verify this. Let MATH be the length of the longest path whose endpoints are vertices and whose length is strictly less than MATH. If MATH denotes the sum of the lengths of the segments in MATH, we have MATH and MATH. This implies that MATH independently of MATH. CASE: MATH. There are two subcases to consider. CASE: MATH. In this case, we define MATH to be the set of subpaths left after removing from MATH the maximal legal segments of length greater than MATH. Then we obtain MATH by removing from MATH the subpaths with fewer than four illegal turns. REF (with MATH) show that the elements of MATH have the desired growth under backward iteration, so we only have to show that MATH accounts for a definite positive fraction of the length of MATH. This fraction is minimal if MATH contains only one subpath (MATH cannot be empty) and if all the paths in MATH have exactly three illegal turns. We first find a lower bound for the number of legal segments of the path in MATH: Let MATH be the number of legal segments of MATH of length greater than MATH, and let MATH be the number of legal segments in MATH. Then MATH and MATH, which implies MATH. Moreover, the number of edges in a path provides a lower bound for the length of that path, so MATH is also a lower bound for MATH, and we conclude that MATH . CASE: MATH. In this case, the length of MATH is bounded by MATH, so there are only finitely many circuits to consider. Since MATH is atoroidal, the length of all circuits tends to infinity under forward iteration, and we can easily find an exponent MATH with the property that, say, MATH for all circuits of length at most MATH and for all MATH. The cases considered above account for every circuit MATH. This completes the proof. |
math/9906008 | By CITE, there exists some number MATH such that for all MATH there exist numbers MATH and MATH such that MATH. Now MATH has the desired property. |
math/9906008 | Fix some MATH such that MATH. Although there may be infinitely many paths MATH whose endpoints are vertices and whose MATH-length is at most MATH, there is only a finite number MATH of intersections MATH of such paths with MATH. Choose MATH according to REF , with MATH and MATH as above. We will show that MATH is the desired exponent. Let MATH be a path in MATH with MATH. Suppose that the first two statements do not hold for MATH. We want to show that the third statement is satisfied. In order to avoid case distinctions, we assume that MATH; the proof in the case MATH is a straight-forward modification of the following argument. Let MATH denote the MATH-legal segments of MATH for MATH. By assumption, the MATH-length of each MATH is bounded by MATH. Moreover, as the turn between two consecutive subpaths is MATH-illegal, the last edge in MATH is contained in MATH if MATH, and the first edge in MATH is contained in MATH if MATH. Fix some MATH. By REF , there exist numbers MATH and MATH such that MATH, MATH, and MATH. An elementary topological argument shows that there exist subpaths MATH of MATH with the property MATH and MATH. Let MATH be the shortest such subpath. We will show that MATH can be expressed as a concatenation MATH, where MATH are pre-Nielsen paths, and MATH is a path that is constant or contained in MATH. There exists a unique shortest subpath MATH of MATH such that MATH contains the first illegal turn of MATH and MATH. Similarly, let MATH be the shortest subpath of MATH such that MATH contains the second illegal turn of MATH and MATH. Note that the extremal (that is, initial and terminal) edges of MATH are (possibly partial) edges in MATH. We have MATH for some path MATH. If MATH were a path of positive MATH-length, this would imply that MATH, contradicting our choice of MATH and MATH. We conclude that MATH is contained in MATH or constant. We claim that MATH. In order to see this, we need to understand the cancellation that occurs between the two maximal MATH-legal subpaths of MATH. In the tightening process, the terminal edge of the first subpath cancels with the initial edge of the second subpath until the last edge of the first subpath forms a nondegenerate turn with the first edge of the second subpath. Since MATH contains a MATH-illegal turn, the resulting turn is necessarily MATH-illegal; in particular, it is contained in MATH. This shows that the part of MATH that is cancelled is completely determined by MATH. Similarly, the part of MATH that is cancelled is completely determined by MATH, etc. Since MATH, this shows that MATH. We conclude that MATH. Let MATH denote the first (possibly partial) edge of MATH. The map MATH expands the MATH-length of MATH by MATH, and it maps vertices to vertices, so MATH contains at least one entire edge in MATH, which implies that MATH has MATH-length at least MATH. The same argument applies to the last edge of MATH, which implies that MATH is completely determined by the extremal edges of MATH. Applying this argument to MATH and MATH, we conclude that MATH, hence MATH is a NAME path. The same argument shows that MATH is a NAME path. Repeating this argument for all indices MATH, we conclude that MATH splits as a concatenation MATH, where MATH and MATH are as in the third statement, the paths MATH are pre-Nielsen, and the paths MATH are contained in MATH or constant. This completes the proof. |
math/9906008 | Given MATH, choose an exponent MATH according to REF , with MATH. Express MATH as a concatenation of paths MATH such that MATH and MATH. There exist preimages of MATH such that MATH is their concatenation. We claim that MATH for all MATH. Suppose otherwise, i. e., MATH for some MATH. Because of our choice of MATH, REF implies that MATH can be written as a concatenation MATH, and MATH splits as a concatenation of three pre-Nielsen paths (with one illegal turn each) with (possibly empty) segments in MATH in between. This implies that for some exponent MATH, MATH contains three NAME paths with segments in MATH in between, which is impossible because of REF . Hence, MATH, and the lemma follows by induction. |
math/9906008 | We will proceed by induction up through the filtration of MATH (as in the previous sections, we equip MATH with the metric constructed in REF). The restriction of MATH to MATH is a homotopy equivalence, and we claim that MATH is of exponential growth. If it were a zero stratum, this would imply that MATH, but MATH. If it were of polynomial growth, it would give rise to a nontrivial fixed conjugacy class. We conclude that MATH is of exponential growth, so this initial case follows from REF . Now assume that the restriction of MATH to MATH is hyperbolic. We choose constants MATH and MATH according to REF for the triple MATH. As in the proof of REF , we will distinguish several cases, and in each case we will find a collection MATH of subpaths having the desired growth and accounting for a definite positive fraction of the length of the circuit in question. For the inductive step, we distinguish three main cases, depending on the stratum MATH. CASE: MATH is a zero stratum. Then MATH is the collection of contractible components of MATH (see REF ). This implies that any nontrivial circuit in MATH is contained in MATH, so there is nothing to show in this case. CASE: MATH is an exponentially growing stratum. We fix some length MATH. Let MATH be a circuit in MATH with nontrivial intersection with MATH. If MATH is a zero stratum, we let MATH, MATH and MATH. Otherwise, let MATH and MATH. If MATH is a zero stratum, then it is the collection of contractible components of MATH (see REF ). Consider a subpath MATH of MATH that is contained in MATH. If MATH is maximal, i. e., if MATH is not a proper subpath of another subpath of MATH that is contained in MATH, then the edges preceding and following MATH in the edge circuit MATH are contained in MATH. This implies that MATH no matter whether MATH is a zero stratum or not. Hence, MATH. As in the proof of REF , we will decompose MATH into subpaths whose growth we understand. We consider several cases. CASE: MATH. In this case, there will be segments of length at least MATH in MATH, and the inductive hypothesis and REF show that they have the desired growth. Hence, it suffices to show that those segments account for some definite fraction MATH of the length of MATH, where MATH does not depend on the choice of MATH. An elementary computation will verify this. Let MATH be the total length of all segments of length at least MATH in MATH, MATH the total length of the remaining segments in MATH, and let MATH. Then our assumption implies MATH. Moreover, if MATH denotes the number of segments in MATH, we have MATH and MATH. We want to find a lower bound for MATH. Using the inequalities derived so far, we conclude that MATH . Hence, we only need to find a lower bound for MATH. Let MATH be the length of the longest path whose endpoints are vertices and whose length is strictly less than MATH. Then MATH and MATH, and we conclude that MATH independently of MATH. CASE: MATH. In this case, significant growth will occur in MATH, and as in the proof of REF , we distinguish two subcases depending on whether forward or backward growth dominates. CASE: MATH. In analogy with REF , we only need to show that MATH-legal segments of MATH-length at least MATH account for a definite fraction of the length of MATH, which can be accomplished with a computation very similar to the one in REF . CASE: MATH. As in the proof of REF , we consider two subcases. CASE: MATH. We define MATH to be the set of subpaths left after removing from MATH the maximal MATH-legal subpaths of MATH-length greater than MATH, and we obtain MATH from MATH by removing subpaths with fewer than five MATH-illegal turns. REF (with MATH) and REF show that the paths in MATH have the desired growth under backward iteration. An argument very similar to the one in the proof of REF shows that the sum of the lengths of the paths in MATH accounts for a definite positive fraction of the length of MATH, so we are done in this case. CASE: MATH. Only finitely many circuits MATH fall into this category, and the same argument as in the proof of REF shows that they have the desired growth under forward iteration. CASE: MATH is a polynomially growing stratum. Recall (see REF) that MATH contains only one edge MATH, and that basic paths of height MATH are of the form MATH, MATH, or MATH, where MATH is a path in MATH with endpoints in MATH. We fix some MATH. Let MATH be a circuit in MATH with nontrivial intersection with MATH. Using REF , we obtain a splitting of MATH by subdividing MATH at the initial endpoints of all occurrences of MATH and at the terminal endpoints of all occurrences of MATH. The subpaths of MATH obtained in this way are either basic paths of height MATH or paths in MATH, and the endpoints of all subpaths are contained in MATH. We first show that all basic paths of height MATH have the desired growth under sufficiently high iterates of MATH. Let MATH be a basic path of height MATH. Since a basic path of the form MATH can be turned into a basic path of the form MATH by reversing its orientation, we only have to distinguish two cases. CASE: MATH with MATH. If MATH, the inductive hypothesis and REF prove the claim, so it suffices to consider the case MATH. The endpoints of MATH are equal, and we denote by MATH the circuit defined by MATH. In general, MATH may be shorter than MATH because initial and terminal edges of MATH may cancel. However, we have MATH and MATH. Moreover, the growth of MATH under iterates of MATH provides a lower bound for the growth of MATH under iterates of MATH, so the inductive hypothesis proves the claim in this case. CASE: MATH with MATH. As in the previous case, we may assume that MATH. We first show that MATH. Suppose otherwise. Then the endpoints of MATH cannot be equal because MATH is atoroidal. This implies that the endpoints of MATH are distinct and MATH starts and ends at the terminal endpoint of MATH. However, this is impossible as it implies that MATH is a basic path of the form MATH (see REF ). We conclude that MATH. As there are only finitely many paths of length less than MATH, we conclude that the circuits in this category have the desired growth under forward iteration. We have shown that basic paths of height MATH have the desired growth, as do paths in MATH if their length is at least MATH. This leaves us with those subpaths in the splitting of MATH that are contained in MATH and whose length is less than MATH, but we can safely disregard them because there are at least as many basic paths of height MATH as there are subpaths in MATH. This completes the proof. |
math/9906015 | To abreviate, we denote MATH. Then, MATH where MATH and MATH denote the partial derivatives with respect to MATH and MATH respectively. But, according to the above computation, MATH and MATH. Therefore, MATH where the last determinant has to be considered with respect to any oriented orthonormal frame of MATH. |
math/9906015 | Let MATH be an arbitrary extension of MATH. Then the map MATH given by MATH is a diffeomorphism. In particular, it is transverse to the submanifold MATH. By the Transversality Theorem, it follows that for almost any MATH, the map MATH given by MATH is also transverse to MATH. Since MATH has codimension REF, this implies that MATH is finite. To construct the required extension MATH, we piece together MATH near MATH and MATH on the interior as follows. Let MATH be such that MATH. Let MATH be a smooth function such that MATH if MATH and MATH if MATH and let MATH. We claim that there are MATH as above and MATH such that for any MATH with MATH, MATH . Suppose that the claim is not true. Then, if for each MATH we consider MATH, MATH and MATH, there are MATH, MATH and MATH with MATH such that MATH . Thus MATH so that MATH. By taking subsequences if necessary, we can suppose that MATH and MATH. Thus, we arrive to MATH, in contradiction with the hypothesis. Now, we can choose MATH, where MATH is any one of the points with MATH for which MATH is finite. |
math/9906015 | Let MATH be an extension of MATH such that MATH for any MATH, being MATH. Then we can extend MATH to MATH by putting MATH . As in REF , it follows that MATH defines a smooth MATH-form on MATH. Moreover, since MATH is closed on MATH, MATH and by NAME Theorem, MATH where MATH denotes a small ball centered at MATH in the interior of MATH and such that MATH if MATH. In particular, MATH being MATH the degree of the map MATH. |
math/9906015 | Let MATH. For any MATH, we have that MATH, where MATH. Thus, we can define MATH and MATH, so that MATH is a right-handed orthonormal frame of MATH. Now, we can consider REF-forms on MATH defined by MATH, for any MATH. Since MATH, by taking differentials we see that MATH. Moreover, we have that MATH . But from the fact that MATH, we deduce that MATH . In particular, MATH and it is not difficult to see that MATH defines a REF-form on MATH. This gives, by NAME Theorem, that MATH where MATH denotes the disk centered at MATH of radius MATH in MATH. To conclude the proof, we just have to show that for any MATH, MATH . On one hand, if we put MATH, it is easy to see that the left hand side is equal to MATH, in accordance with the sign of MATH. If we compute this, we get MATH and using the isometry between MATH and MATH, MATH . On the other hand, if we suppose that MATH for MATH, we have that MATH is equal to MATH depending on the sign of MATH . Now, MATH and thus, MATH . Finally, note that MATH which implies the desired result. |
math/9906015 | Suppose that this is not true. Then, for each MATH, there are MATH and pairs MATH such that MATH. By taking subsequences if necessary, we can suppose that MATH and MATH. If MATH, we arrive to MATH, in contradiction with REF . Otherwise, let MATH. If we denote by MATH a frame for MATH, we have for any MATH, MATH . Since MATH we have after substitution and division by MATH, MATH . This would imply that MATH in contradiction with REF . |
math/9906015 | Let MATH be an orthonormal oriented frame of MATH and consider the map MATH given by MATH . By the Transversality Theorem, we have that for a residual subset of curves MATH and vector bundles MATH with the corresponding MATH . NAME topologies, the curve MATH meets the hypersurface MATH transversely at a finite number of points: MATH with MATH and MATH. Since MATH, MATH and MATH depend continuously on MATH and MATH, we can suppose that MATH and MATH are generic in the above sense. In particular, for MATH small enough, the same can be said if we consider the intersection of MATH with MATH, where MATH . Then, REF gives that MATH and taking limit when MATH, MATH . On the other hand, note that MATH. By using the same argument as in the proof of REF , MATH. If we apply NAME Theorem, MATH where MATH denotes the disk centered at MATH of radius MATH in MATH. Again we refer to the proof of REF to claim that MATH . In particular, MATH . To conclude the proof, we just have to compute the integral on MATH. We parameterize MATH by considering the curves: MATH, MATH, MATH and MATH, for MATH. Then, we have that MATH . Note that MATH and MATH for any MATH. This gives that MATH . |
math/9906018 | CASE: Let MATH, MATH both be as in REF. So for each MATH, MATH for some MATH, MATH. Hence, if MATH, MATH and each is in MATH as defined by MATH (as MATH). As this holds for every MATH, all generators of MATH as defined by MATH are in MATH as defined by MATH. As the situation is symmetric we finish. CASE: Similar proof. The first phrase follows from REF , and check the second. |
math/9906018 | Remember that (by CITE) there is MATH, a generating sequence for MATH. Let for MATH, MATH exemplify MATH, MATH; by CITE, without loss of generality CASE: MATH. Without loss of generality for MATH if MATH, MATH if MATH. We define MATH by: MATH and for MATH: MATH . Clearly CASE: MATH. MATH . Let MATH, so MATH and hence for some MATH and MATH from MATH (hence MATH), we have MATH. So by the definition of the MATH's we have: CASE: if MATH, and MATH, then MATH. However, CASE: MATH (for our fixed MATH) belongs to MATH [as MATH implies MATH which is in MATH]. Together we get REF . For any MATH for some MATH, MATH. The family MATH of sets MATH for which this holds (that is, for each MATH there is MATH such that MATH satisfies: CASE: MATH for MATH, CASE: MATH is an ideal of subsets of MATH, CASE: if MATH (for MATH) are in MATH, MATH for each MATH then MATH is in MATH. We shall show their satisfaction below. This suffices for REF [as MATH; why? just prove that MATH by induction on MATH. For successor use MATH. For singular, let MATH be such that MATH is strictly increasing continuous with limit MATH, and MATH; by the induction hypothesis MATH, MATH are in the ideal, by REF we know that MATH is in the ideal and by the induction hypothesis MATH so by REF MATH is MATH; note MATH as MATH is singular. As MATH, we have covered all cases]. Now why REF holds? We shall use MATH from above freely. For REF : if MATH as MATH and MATH; if MATH), MATH then for some MATH, MATH, hence MATH; this shows MATH. For REF : (trivially MATH;) if MATH, MATH and MATH, choose, for MATH, MATH such that MATH. Now let MATH be defined by MATH, so by an assumption on MATH and MATH, for some MATH, MATH, now MATH is as required by MATH. For REF : let MATH; by assumption for each MATH for some MATH, MATH. Now MATH is MATH-directed, hence for some MATH, MATH. By MATH for some MATH, MATH and MATH necessarily MATH. Now for each MATH; if MATH trivially MATH, so assume MATH; now for some MATH, MATH; so MATH and MATH, hence MATH, hence by their definitions MATH. So MATH is as required, that is, we have proved REF . Now REF follows from REF [using REF for MATH we can get there MATH, so by REF for some club MATH of MATH, MATH . Together MATH exemplify MATH is MATH, as required]. |
math/9906018 | CASE: Trivially the second and third terms are equal (see REF ). Let MATH be defined as in the second term, so MATH. So by REF without loss of generality MATH, so MATH. Using REF's notation, MATH exemplify MATH. CASE: By REF . CASE: This follows by the proof of REF, but as I was asked, we repeat the proof of REF with the required changes. Without loss of generality MATH [why? if MATH then MATH, so the conclusion is trivial, if not let MATH, so MATH and MATH]. We let MATH. Remember that (by CITE) there is MATH, a generating sequence for MATH. Let for MATH, MATH exemplify MATH, MATH; by CITE without loss of generality CASE: MATH. Without loss of generality for MATH if MATH, MATH if MATH. For any MATH we define a function MATH by: MATH and for MATH: MATH . Let MATH vary on MATH. Clearly CASE: MATH. If MATH (both in MATH of course) then MATH. Let MATH, so MATH, hence for some MATH and MATH from MATH, (hence MATH) we have MATH. So by the definition of the MATH's we have: CASE: if MATH, and MATH, then MATH. However CASE: MATH (for our fixed MATH) belongs to MATH [as MATH implies MATH which is in MATH]. Together we get REF . For any MATH for some MATH we have MATH. The family MATH of sets MATH for which this holds, that is, for each MATH there is MATH such that MATH, satisfies: CASE: MATH for MATH, CASE: MATH is an ideal of subsets of MATH, CASE: if MATH (for MATH) are in MATH, MATH for each MATH then MATH is in MATH. We shall show their satisfaction below. Why REF + REF + REF suffice for REF? As MATH; why? just prove that CASE: MATH by induction on MATH. For successor use MATH. For singular, let MATH be such that MATH is strictly increasing continuous with limit MATH, and MATH; by the induction hypothesis MATH, MATH are in the ideal, by REF we know that MATH is in the ideal and, as said above, MATH so by REF MATH is MATH; note MATH as MATH is singular. As MATH implies MATH has no inaccessible accumulation point, we have covered all cases in the induction, so MATH holds. Now note that MATH, so from MATH we get MATH and by the definition of MATH we are done. Next, why REF hold? We shall use MATH from above freely. For REF : let MATH; if MATH as MATH and MATH; if MATH (MATH), then MATH let MATH hence MATH . Hence MATH; this shows MATH. For REF : (trivially MATH); if MATH, MATH and MATH, choose for MATH such that MATH. Now let MATH be defined by MATH, so MATH and MATH and MATH hence MATH. For REF : let MATH; by assumption for each MATH for some MATH, MATH. Now MATH is MATH-directed, hence for some MATH, MATH. Without loss of generality MATH. Now for each MATH; if MATH trivially MATH, so assume MATH. Now, for some MATH, MATH; so MATH and MATH, hence MATH, hence by their definitions MATH. So REF hold and hence MATH is as required, that is, we have proved REF . We finish by MATH is MATH-directed. Assume MATH for MATH, by REF for each MATH for some MATH we have MATH. But MATH is MATH-directed hence for some MATH we have MATH . By REF we have MATH hence by the previous sentence MATH, so MATH is a MATH-upper bound of MATH, as required. |
math/9906021 | Consider the spectral representation associated with a vector MATH and perform a straightforward computation: MATH . |
math/9906021 | The proof is a direct computation. For details, we refer to REF . |
math/9906021 | The proof is standard and we provide it for the sake of completeness. See, for example, CITE for detailed exposition of similar results and further references. Throughout the proof, we assume that the function MATH is sufficiently smooth to justify integration by parts (local MATH is sufficient). Clearly this is the case under our assumptions on MATH (see, for example, CITE). To prove the bound REF with the constant independent of MATH let MATH where MATH are real-valued. For any MATH such that MATH when MATH when MATH we have MATH where MATH is the surface measure on MATH induced from MATH . The first term vanishes because MATH vanishes on MATH and MATH vanishes on MATH . Furthermore, by NAME 's formula MATH . The boundary term in this equality is also equal to zero. Substituting REF into REF, we find MATH . Therefore, MATH . A similar estimate holds for MATH . Combining these two estimates, we obtain the result of the lemma. |
math/9906021 | We have MATH since MATH and MATH satisfy the boundary conditions. Next note that MATH . We used the NAME inequality in the last step. |
math/9906021 | An interplay of the scales in space and in the spectral parameter plays an important role in the analysis. Let us assume that MATH . Take sufficiently large MATH such that MATH . By NAME 's formula we have MATH . In the above computation we used the definition of MATH and the fact that the function MATH satisfies MATH . Hence, we obtain MATH . Let us integrate REF from MATH to some larger value of MATH . Using REF , we see that MATH . According to REF, there exists a sequence MATH such that MATH . Let us set MATH and pick MATH and MATH in REF . We obtain MATH . Substituting REF into the last inequality, we find that there exists some constant MATH such that for MATH large enough, we have MATH . Now it remains to invoke REF and note that MATH for every MATH . Using the estimate REF and the relation between MATH and MATH we find MATH for sufficiently small MATH . The application of REF now completes the proof. |
math/9906021 | Recall that for every self-adjoint operator there is an associated spectral measure of maximal type, MATH such that for every MATH and any measurable set MATH implies MATH . A vector MATH is of the maximal type if for any measurable set MATH given that MATH . We will show that for any MATH of positive MATH-dimensional NAME measure, there exists a vector MATH with MATH . By the standard argument for the existence of vectors of maximal type (see for example, CITE), this would imply existence of the vector MATH as in the theorem. Pick some ball MATH such that MATH for energies MATH in a subset MATH of MATH of positive MATH measure (it is easy to see that such a ball exists, because of the MATH-additivity of MATH). We remark that for a wide class of operators MATH an arbitrary ball will do because of the unique continuation (solutions MATH cannot vanish identically on any ball), but there is no need to invoke these results. Pick a basis MATH in the NAME space MATH. Since MATH forms a basis, for every MATH there exists a MATH such that MATH . Consider the functions MATH on the set MATH . By REF , for every MATH there exists a MATH such that MATH . In particular, by MATH-additivity of MATH, there exists a MATH such that MATH for every MATH in a set MATH of positive MATH measure. By the results of NAME theory (see REF ), it follows that the measure MATH gives positive weight to the set MATH and hence to the set MATH . The case of the absolutely continuous spectrum corresponds to MATH in this case the application of NAME theory may be replaced by a well-known fact that a measure gives positive weight to a set of positive NAME measure when its derivative is positive a.e. in this set. |
math/9906021 | Suppose that MATH has positive MATH measure. By the remark after the proof of REF implies that MATH for every MATH and finitely supported MATH such that MATH . Proceeding as in the proof of REF , we can find a vector MATH such that MATH for any MATH in some set of positive MATH measure. This is not possible by NAME (see REF ) and therefore gives a contradiction. We remark that for MATH, this argument reduces to the well-known statement that a finite NAME measure MATH cannot have an infinite derivative on a set of positive NAME measure. |
math/9906021 | The argument repeats the proof of REF , except that we do not need REF . The analog of REF is proven directly by the observation that MATH and MATH . |
math/9906021 | Without loss of generality, assume MATH . We first establish the existence of a NAME set MATH, for which the following three properties are true: CASE: MATH. CASE: The restriction of the spectral measure MATH to MATH is MATH-H. CASE: There exists a constant MATH, such that for each MATH and MATH, the corresponding generalized eigenfunction MATH satisfies MATH . We shall establish REF - REF in two stages. First, note that the function MATH is a measurable function of MATH, which, by REF, is finite everywhere on MATH. Thus, since MATH, there is clearly a NAME subset MATH of positive MATH measure and a constant MATH, such that MATH for any MATH. That is, REF holds for MATH. Next, since the restriction of MATH to MATH is MATH-continuous, the above mentioned NAME theorem implies that there is a NAME subset MATH of positive MATH measure (so REF holds) such that the restriction of MATH to MATH is MATH-H (so REF holds). Let us now denote MATH, MATH, where MATH denotes the spectral projection over the corresponding set. Then MATH, and MATH, MATH are mutually orthogonal so MATH. Let MATH be the projector on the set of sites MATH with MATH. MATH is a function of the time parameter MATH to be chosen later. Given any vector MATH, we will routinely use the notation MATH. We have MATH by NAME theorem, and so MATH . For each MATH, we now define MATH such that we have MATH and thus MATH . Since MATH we obtain MATH which implies MATH proving REF. |
math/9906021 | By the results of the NAME theory, the spectral measure MATH is supported on the set of the energies MATH for which the decaying solution REF satisfies the boundary condition (namely, MATH coincides with the NAME boundary condition). Moreover, these decaying solutions, which we will denote by MATH, are exactly the generalized eigenfunctions in the sense of REF , if we normalize them by setting MATH. Fix MATH such that the results of the KLS theorem hold. REF implies that the generalized eigenfunctions MATH of the operator MATH satisfy MATH for every MATH given by REF. Pick an open energy interval MATH, such that MATH, and MATH. Let MATH, MATH. MATH is monotone on MATH. Assume, without loss, that MATH. The restriction of MATH to MATH is MATH-continuous, and by REF, MATH for any generalized eigenfunction MATH with MATH. Thus, by REF , for each MATH there is a constant MATH such that for all MATH . Since MATH, we can clearly choose such an interval MATH with MATH and MATH. Thus, REF follows. |
math/9906024 | Straightforward. |
math/9906024 | CASE: Immediate (by the MATH - completeness of MATH). CASE: Clause REF should be clear (remember REFMATH). For clause REF note that MATH, so in the game MATH the condition MATH is chosen by the anti-generic player. So if the conclusion fails, then for some MATH - name MATH for an ordinal we have MATH. Thus the anti-generic player can choose MATH so that MATH for some ordinal MATH, what guarantees it to win the play. CASE: Straightforward from REF. |
math/9906024 | Let MATH be a strictly increasing continuous sequence cofinal in MATH. By induction on MATH choose MATH such that CASE: MATH is above (in MATH) of all MATH for MATH, CASE: MATH, MATH, and MATH, CASE: if MATH then MATH and MATH. (The choice is clearly possible as MATH is MATH - generic.) Let MATH be an upper bound of MATH (remember clause MATH above); then also MATH. Now we are going to define a condition MATH. We let MATH, and for MATH, MATH, we let MATH be a MATH - name for the following object in MATH (for a generic filter MATH over MATH): CASE: If MATH is an upper bound of MATH in MATH, then MATH. CASE: If not REF , but MATH has an upper bound in MATH, then MATH is the MATH - first such upper bound. CASE: If neither REF nor REF , then MATH. It should be clear that MATH. Now, CASE: MATH. Why? By induction on MATH we show that MATH. Steps MATH and ``MATH limit" are clear, so suppose that we have proved MATH, MATH (and we are interested in the restrictions to MATH). Assume that MATH is a generic filter over MATH such that MATH. Since MATH, we also have MATH and MATH. Hence MATH is an upper bound of MATH. Therefore, MATH (see REF above) and we are done. The proof of the proposition will be finished once we show CASE: MATH. Why does this hold? By induction on MATH we show that MATH for all MATH. Steps MATH and ``MATH limit" are as usual clear, so suppose that we have proved MATH (for MATH), MATH (and we are interested in the restrictions to MATH). Assume that MATH is a generic filter over MATH such that MATH. Then also (by the inductive hypothesis) MATH and therefore MATH is an increasing sequence of conditions from the (MATH - complete) forcing MATH. Thus this sequence has an upper bound, and MATH is such an upper bound (see REF above), as required. |
math/9906024 | REF Straightforward (use the MATH - completeness of MATH). CASE: If MATH is MATH - generic, then our assertion is clear (remember clause REF ). So suppose that we are in the second case (so MATH). Let MATH be a strictly increasing continuous sequence cofinal in MATH. For MATH and MATH let MATH, MATH (clearly MATH). Since MATH is MATH - generic, we may inductively pick conditions MATH (for MATH) such that CASE: MATH, MATH, CASE: MATH, MATH, CASE: if MATH then MATH and MATH. Let MATH be stronger than all MATH's. Now apply REF. |
math/9906024 | By REF, the forcing notion MATH is MATH - complete, so we have to concentrate on showing clause REF for it. So suppose that MATH is large enough, MATH and MATH, MATH, MATH and MATH, and MATH satisfies MATH. Furthermore, suppose that MATH is a MATH - semi diamond and MATH is a MATH - candidate. We may assume that for each MATH CASE: if MATH is not a MATH - increasing sequence of members of MATH, then MATH for all MATH. [Just suitably modify MATH whenever the assumption of MATH holds - note that the modification does not change the notion of a candidate, the game from REF, etc.] Before we define a generic condition MATH for MATH over MATH, let us introduce notation used later and give two important facts. Let MATH and let MATH be generic over MATH. We define: CASE: MATH is such that if MATH is a function, MATH and MATH is a MATH - name, then MATH, otherwise it is MATH; CASE: MATH is defined by MATH provided MATH is a function, and MATH otherwise; CASE: MATH; CASE: MATH is MATH; CASE: MATH; CASE: MATH is MATH. Observe that MATH is such that MATH and MATH is such that MATH. Plainly, by MATH, Assume MATH. Then MATH is a MATH - semi diamond sequence for MATH and MATH is a MATH - candidate. Assume that MATH and MATH is MATH - generic for MATH over MATH. Let MATH be a generic filter over MATH, MATH. Then in MATH: CASE: MATH, CASE: MATH is a MATH - semi diamond for MATH, and CASE: MATH is a MATH - candidate. CASE: Will follow from REF . CASE: Assume that this fails. Then we can find a condition MATH, a MATH - name MATH for an increasing sequence of conditions from MATH, and MATH - names MATH for members of MATH such that MATH and MATH . Consider a play MATH of the game MATH in which the generic player uses its winning strategy and the anti-generic player plays as follows. In addition to keeping the rules of the game, it makes sure that at stage MATH: CASE: MATH (so MATH; remember the anti-generic player plays at REF), CASE: MATH decides the values of all MATH for MATH. Let MATH be such that MATH for sufficiently large MATH. Note that the sequence MATH is MATH - increasing. So, as MATH is normal and MATH and MATH is a semi diamond for MATH (by REF), we may find a limit ordinal MATH such that MATH. Then also MATH, and since the play is won by the generic player, we conclude MATH. But then taking sufficiently large MATH we have MATH a contradiction. CASE: Should be clear. Fix a bijection MATH. Also let MATH list all pairs MATH such that MATH, MATH and MATH is a MATH - name for an ordinal. Next, by induction, we choose a sequence MATH such that CASE: MATH for MATH, CASE: if MATH and MATH, then MATH and MATH, CASE: if MATH is a limit ordinal, then MATH is the closure of MATH, and if, additionally, MATH is such that MATH (and MATH, of course), then MATH and MATH is such that CASE: for every generic MATH over MATH such that MATH, and two successive members MATH of the set MATH such that MATH we have: if MATH has an upper bound in MATH, then MATH is such an upper bound, CASE: for each MATH, for some MATH and a MATH - name MATH we have: MATH, MATH, MATH and if MATH is generic over MATH and MATH, then MATH (It should be clear that there are no problems in the induction and it is possible to pick MATH as above.) From now on we will treat each MATH as a MATH - name for a member of MATH. Now we are going to define a MATH - generic condition MATH for MATH over MATH in the most natural way. Its domain is MATH and for each MATH . For every MATH, the generic player has a winning strategy in the game MATH. We will prove the claim by induction on MATH. For MATH this implies that MATH is well-defined (remember REF). Of course for MATH we finish the proof of the theorem. Suppose that MATH and we know that MATH is MATH - generic for MATH over MATH for all MATH. We are going to describe a winning strategy for the generic player in the game MATH. The inductive hypothesis is not used in the full strength in the definition of the strategy, but we need it in several places, for example, to know that MATH is well defined as well as that we have the MATH's below. Also note that it implies that MATH for all MATH, where MATH and MATH. Moreover, during the play, both players will always have legal moves. Why? By the inductive hypothesis we know that MATH is MATH - generic for all MATH. Therefore, if MATH is a successor or a limit ordinal of cofinality MATH, then we immediately get that MATH is MATH - generic (remember clause REF of the choice of the MATH's!), and thus REF applies. If MATH is a limit ordinal of cofinality MATH, then we may use REF. Let MATH be a MATH - name for the winning strategy of the generic player in MATH, and let MATH . Plainly, MATH is a club of MATH. Let the generic player play as follows. Aside it will construct sequences MATH and MATH so that CASE: MATH is a MATH - name for a member of MATH, MATH is a MATH - name for a member of MATH, MATH is a MATH - name for a member of MATH, and CASE: if MATH, MATH, and MATH, then after the MATH move (which is a move of the generic player) the terms MATH, and MATH are defined. So suppose that MATH and MATH is the result of the play so far. To clearly describe the answer of the generic player we will consider two (only slightly different) cases in the order in which they appear in the game. (Remember MATH is chosen by the anti-generic player and that all successor moves are done by the generic player.) REF MATH, MATH. First the generic player picks conditions MATH, MATH such that MATH, MATH and for each MATH we have MATH . [Why possible? By REF.] Now the generic player looks at MATH such that MATH. It picks MATH - names MATH so that MATH forces that MATH is a play according to MATH in which the moves of the anti-generic player are determined as follows. First, it keeps the convention that if MATH, then MATH is (a name for) the MATH - first legal answer to the play so far. Now, if MATH, then we have already the play up to MATH (it easily follows from the inductive construction that MATH indeed forces that it is a ``legal" play). The MATH move of the anti-generic player is stipulated as MATH, MATH, MATH, and next we continue up to MATH keeping our convention. If MATH, then the generic player lets MATH, MATH and then it ``plays" the game according to MATH up to MATH keeping our convention for all MATH. Next, the generic player picks a condition MATH and MATH - names MATH (for MATH, MATH, MATH) such that CASE: MATH, and for every MATH, and MATH CASE: for every MATH and MATH with MATH, the condition MATH decides the value of MATH, and MATH where MATH. Then it lets MATH (for MATH) be conditions such that MATH and for MATH . Finally, for MATH it plays MATH. CASE: MATH, MATH. The generic player proceeds as above, the difference is that now MATH ``belongs to" the generic player, and that it is a limit of moves of the anti-generic player. Again, we look at MATH such that MATH. If MATH, then every condition in MATH stronger than all MATH (for MATH) forces that MATH is a legal play in which the generic player uses MATH. The generic player determines MATH, and MATH for MATH ``playing the game" as earlier (with the same convention that if MATH, then the MATH - th move of the anti-generic player is stipulated as the MATH - first legal move). If MATH, then (any condition stronger than all MATH for MATH forces that) MATH, MATH are increasing, and MATH and MATH is MATH - generic. So, by REF, the generic player may let MATH be the MATH - first such that for all MATH we have MATH, MATH. It also lets MATH. Then the generic player chooses MATH, and MATH for MATH ``playing the game" with the strategy MATH (and keeping the old convention for MATH). Next the generic player picks a condition MATH (stronger than all MATH for MATH), MATH - names MATH and sets MATH (for MATH) as in the previous case. Then it chooses conditions MATH and MATH such that MATH and MATH. [Why possible? If MATH is limit of cofinality MATH, use REF; otherwise we know that MATH is MATH - generic.] Next it defines conditions MATH (for MATH) so that MATH and for MATH . Finally, for MATH it plays MATH. Why does the strategy described above work? Suppose that MATH is a play of the game MATH in which the generic player used this strategy and let MATH and MATH be the sequences it constructed aside. (As we said earlier, the game surely lasted MATH steps and thus the sequences described above have length MATH.) Let us argue that REF MATH holds. Assume that a limit ordinal MATH (so in particular MATH) is such that CASE: MATH. We are going to show that MATH and for this we prove by induction on MATH that MATH. For MATH this is the desired conclusion. For MATH this is trivial, and for a limit MATH it follows from the definition of the order (and the inductive hypothesis). So assume that we have proved MATH, MATH, and let us consider the restrictions to MATH. If MATH then by the choice of REF , we know that MATH . Now look at the clause REF of the choice of the MATH at the beginning: what we have just stated (and MATH) implies that MATH thus, MATH, so we are done. Suppose now that MATH and let MATH. Look at what the generic player has written aside: MATH forces that MATH is a play according to MATH and MATH, so we are clearly done in this case too (remember the choice of MATH). |
math/9906024 | Assume MATH is as in REF, and MATH. Put MATH and MATH. CASE: If MATH, MATH, then MATH and MATH. CASE: If MATH, MATH, and MATH, then MATH is stronger than both MATH and MATH. CASE: If MATH is MATH - increasing and MATH, then MATH has a least upper bound MATH, and MATH is a least upper bound of MATH. CASE: By the definition of MATH, MATH . REF Should be clear. MATH is MATH - generic for semi - diamonds (see REF). Suppose that MATH is a normal filter on MATH, MATH. Let MATH be such that MATH, MATH be a MATH - semi diamond, and let MATH be a MATH - candidate. We have to show that the condition MATH is MATH - generic for MATH over MATH, and for this we have to show that the generic player has a winning strategy in the game MATH. Note that the set MATH is a club of MATH (so MATH). Now, the strategy that works for the generic player is the following one: At stage MATH of the play, when a sequence MATH has been already constructed, the generic player lets MATH and it asks: CASE: Is there a common upper bound to MATH ? If the answer to MATH is ``yes", then the generic player puts MATH; otherwise it lets MATH. Now it chooses MATH to be the MATH - first element of MATH stronger than all members of MATH and MATH. Why the strategy described above is the winning one? Let MATH be a play according to this strategy. Suppose that MATH is a limit ordinal such that MATH. So, MATH is stronger than all MATH (for MATH), and for cofinally many MATH we have MATH. Therefore, MATH and (by REF) MATH has an upper bound. Now look at the choice of MATH. The proposition follows immediately from REF. |
math/9906024 | CASE: Follows from REF . CASE: Assume MATH. Let MATH, MATH and MATH be as in REF , MATH. Let MATH (for MATH) and put MATH. Clearly MATH and MATH for each MATH (remember MATH). By the proof of REF, the condition MATH is MATH - generic for MATH over MATH . To show that MATH for MATH, it is enough to do this for MATH. So, pick any MATH, MATH and MATH as in REF , and let MATH. Let MATH be a MATH - name for the generic partial function from MATH to MATH, that is, MATH. We claim that CASE: MATH . Why? Let MATH. Take MATH such that MATH (remember MATH). If for some MATH we have MATH, then MATH . Otherwise, we let MATH and MATH, and we put MATH. Then clearly MATH is a condition stronger than MATH and MATH . It should be clear that MATH implies that there is no condition MATH which is MATH - generic for MATH for all MATH (remember REF). |
math/9906024 | Assume MATH is as in REF, MATH and MATH. We are going to show that the condition MATH is MATH - generic for semi-diamonds. So suppose that MATH and MATH are as in REF. For MATH, let MATH be such that MATH, MATH, MATH, MATH. Note that MATH. Let us describe the winning strategy of the generic player in the game MATH. For this we first fix a list MATH of MATH, and we let MATH is a limit of members of MATH. Suppose that we arrive to a stage MATH and MATH is the sequence played so far. The generic player first picks a condition MATH stronger than all MATH's played so far and, if possible, stronger than MATH. Then it plays a condition MATH above MATH such that CASE: if MATH, then MATH, and CASE: MATH, and MATH. The set MATH played a this MATH, where MATH is the first ordinal such that MATH and MATH for all MATH. Why is this a winning strategy? Let MATH be a play according to this strategy, and suppose that MATH is a limit ordinal such that MATH . Note that then CASE: if MATH then MATH is an unbounded subset of MATH, and CASE: MATH. We want to show that there there is a common upper bound to MATH (which, by the definition of our strategy, will finish the proof). First we choose a pre-condition MATH such that: CASE: MATH, CASE: if MATH, then we let MATH, if MATH, then MATH CASE: MATH, CASE: MATH. Why is the choice possible? As MATH ! Now we may extend MATH to a condition in MATH picking for each MATH large enough MATH and adding MATH to MATH (and extending MATH suitably). |
math/9906024 | First let us argue that MATH is MATH - lub - complete. So suppose that MATH are such that MATH, MATH. We claim that MATH is a condition in MATH. Clearly MATH is a tree, and MATH. By clause REF (for MATH's) we see that MATH, and in a similar way we justify that MATH satisfies other demands as well. Now suppose that MATH and MATH are as in REF, MATH. Choose inductively complete MATH - trees MATH and fronts MATH (of MATH) such that CASE: MATH, CASE: if MATH, then MATH and MATH, and CASE: if MATH then MATH, CASE: if MATH is limit, then MATH, CASE: if MATH is limit and MATH is an increasing sequence of conditions from MATH and MATH, and MATH, then for some MATH we have MATH . Now we let MATH. It should be clear that MATH. MATH is MATH - generic for MATH over MATH. We have to describe a winning strategy of the generic player in the game MATH. Let MATH be the club of limits of members of MATH. Let the generic player play as follows. Assume we have arrived to stage MATH of the play when MATH has been already constructed. If MATH then the generic player chooses MATH such that CASE: MATH, MATH, and MATH, MATH, CASE: MATH for some MATH, and lets MATH. If MATH then the generic player picks MATH satisfying REF + REF and such that CASE: if possible, then MATH and it takes MATH as earlier. Why is this a winning strategy? First, as MATH is MATH - lub - complete, the play really lasts MATH moves. Suppose that MATH is such that MATH . Let MATH. Note that (as MATH and by REF ) we have MATH and (by REF ) MATH is included in MATH. Therefore, by clause REF of the choice of the MATH, for some MATH we have MATH and MATH. But this immediately implies that it was possible to choose MATH stronger than MATH in REF (remember MATH). |
math/9906030 | Associated to the polynomial MATH over MATH is its NAME polygon, the lower boundary of the convex hull of the points MATH for MATH. The integers MATH such that MATH are vertices of the NAME polygon are called the breakpoints, and the ratios MATH, where MATH are adjacent breakpoints, are called the slopes of the NAME polygon (they exist because MATH is divisible). We will eventually see that the valuations of the roots of MATH are the slopes of the constituent segments of the NAME polygon, and keeping this in mind will clarify the motivation of the following argument. Let MATH be the smallest ordinal with cardinality greater than that of MATH. We show that for any monic polynomial MATH over MATH, there exists a map MATH and a map MATH such that MATH has the following properties. (In the definition of MATH, we formally take MATH.) CASE: If MATH, then MATH, with equality only if both are equal to MATH. CASE: If MATH, then MATH is a root of MATH. CASE: If MATH, then the polynomial MATH has largest NAME slope strictly greater than MATH. Since there is no injective map from MATH to MATH, there exists some MATH such that MATH, and the corresponding MATH will be a root of MATH, proving the theorem. In other words, we will carry out a ``transfinite NAME 's algorithm" and show that it converges to a root of MATH. We prove the claim by transfinite induction. The induction step is self-evident for limit ordinals, so we may need only worry about non-limit ordinals as well as the base case. Both of these are subsumed in the following fact: if MATH, there exist MATH and MATH such that the largest slope of the NAME polygon of MATH is strictly greater than the largest slope of the NAME polygon of MATH. To prove the latter claim, write MATH with MATH. Let MATH be the largest breakpoint (we do not regard MATH itself as a breakpoint) and MATH the slope of the segment of the NAME polygon between MATH and MATH. Since MATH is algebraically closed, the polynomial MATH factors completely over MATH. Let MATH be any root of this polynomial, let MATH be the multiplicity of the root MATH, and put MATH. If we put MATH, then MATH . From this description we can read off the NAME slopes of MATH. First note that for any breakpoint MATH, the sum defining MATH consists of MATH plus various terms of valuation strictly larger than that of MATH. For MATH, we have MATH, and the coefficient of MATH in MATH is equal to the coefficient of MATH in MATH. In particular, we have MATH and MATH for MATH. In short, the NAME slopes of MATH less than MATH are the same as those of MATH, the slope MATH occurs with multiplicity MATH, and the remaining slopes are greater than MATH. This completes the proof of the claim and hence of the induction step. |
math/9906030 | Let MATH be elements which reduce to a basis over MATH of the MATH-vector space solutions of MATH modulo MATH. For each MATH, we wish to exhibit MATH which reduce to a basis over MATH of the set of solutions of MATH in MATH. We construct these by an inductive argument: given the MATH, note that MATH . Since MATH has algebraically closed residue field, there exists MATH such that MATH; we set MATH. The facts that the MATH are independent and that they span the solution space of MATH are straightforward. |
math/9906030 | By the previous lemma, the sequences MATH and MATH satisfying the given relations have the form MATH where the MATH and MATH depend only on the MATH and MATH. Clearly MATH and MATH are also of this form, with the set of MATH being the union of the sets of MATH and MATH in the former case, and the set of products MATH in the latter case. Thus it suffices to show that given a set of MATH, the set of sequences of the form MATH satisfies a twist-recurrence relation. Clearly we may alter the set as needed to ensure that the MATH are linearly independent over MATH, and that the MATH-module they span is saturated (that is, if MATH lies in the span, so does MATH), or in other words, the images of the MATH in MATH are linearly independent over MATH. If MATH is the resulting set, then the equation MATH is satisfied when MATH, and expanding this determinant along the bottom row gives us an equation of the form MATH; all that remains to be checked is that MATH and MATH are not divisible by MATH. Since MATH, we need only check that MATH is not divisible by MATH. In fact, since MATH are linearly independent mod MATH, MATH . |
math/9906030 | The key observation here is that a sequence MATH of elements of MATH is twist-recurrent if and only if the sequence MATH of NAME lifts of MATH to MATH is twist-recurrent. Of course one implication is obvious. For the other, note that in MATH, MATH for any MATH and any MATH such that MATH. Thus if we write MATH, we may select MATH such that MATH, and then MATH is a lift of MATH. Now MATH satisfies a twist-recurrence relation not depending on the MATH. Given the observation, the rest of the proof is straightforward. On one hand, if the sequences of components of the NAME vectors are twist-recurrent, so are the sequences of their NAME lifts, and the given sequence is simply a linear combination of these. On the other hand, if the sequence of NAME vectors is twist-recurrent, then the sequence of unit components is as well, as is its corresponding sequence of NAME lifts. Subtracting this sequence off and dividing by MATH gives a sequence of NAME vectors over MATH which is twist-recurrent, and the claim follows by induction. |
math/9906030 | Let MATH be the distinct slopes of the NAME polygon and MATH their multiplicities. Then we can factor MATH as MATH, where MATH is the monic polynomial of degree MATH whose roots are the roots of MATH which have valuation MATH (counting multiplicities). Similarly we can factor MATH as MATH. Now observe that MATH and so MATH . Moreover, MATH and MATH also obey the conditions of the lemma. Thus by descending induction, we have MATH for MATH. In other words, the proof of the statement of the lemma reduces to the case in which MATH and MATH have only one slope, which we may assume is REF. In this case, MATH and MATH for all MATH. Let MATH be a root of MATH, and choose MATH such that MATH. Let MATH so that MATH and MATH. Then clearly MATH. By the assumption that MATH is a root of MATH, we have MATH. That means that the NAME polygon of MATH has at least one slope greater than or equal to MATH, which is to say MATH is congruent to a root of MATH modulo MATH, as desired. The converse implication follows by an analogous argument. |
math/9906030 | We first establish that MATH is algebraically closed; that is, for any MATH, the polynomial MATH has a root in MATH. Since this assertion only depends on the values of MATH, we may rechoose the MATH to ensure that MATH where MATH denotes reduction modulo MATH. Then the polynomials MATH over MATH and MATH over MATH clearly satisfy the conditions of REF with MATH. Thus for some root MATH of MATH of highest slope MATH, there exists a root MATH of MATH in MATH, also of slope MATH, with MATH. Since the coefficients of MATH lie in the algebraically closed domain MATH, MATH does as well. Choose a lift MATH of MATH to MATH, making sure that MATH is divisible by MATH. We may reapply the above steps to MATH to get another ``approximate root", and so on. Each step increases the largest slope of the residual polynomial by at least MATH, so the approximate roots converge in MATH to a root of MATH. Moreover, since we made MATH divisible by MATH, the approximate roots also converge MATH-adically in MATH, which proves the claim. In particular, we now have that MATH contains MATH, and it remains to establish that for any MATH, MATH. Without loss of generality, we may assume MATH. We will prove the claim by constructing a sequence MATH of elements of MATH and monic polynomials MATH, for MATH, such that for MATH: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH is a root of MATH of slope at least MATH; CASE: there exists a root MATH of MATH such that MATH; CASE: MATH and MATH. Note that REF can be arranged for MATH by lifting a polynomial over MATH having MATH as a root (making sure that REF is also satisfied). The fact that REF holds is a consequence of REF . Observe that MATH and MATH. That is, MATH is congruent to an element of MATH modulo MATH. Repeating this process with MATH in place of MATH exhibits succesive elements of MATH congruent to MATH modulo MATH for all MATH, which implies that MATH. Thus MATH, as desired. |
math/9906030 | It suffices to note that the latter set is a ring (by REF ), and has residue ring MATH (by CITE). |
math/9906030 | The most difficult part of the proof is showing that MATH is actually a ring! If MATH then MATH for certain universal polynomials MATH. Suppose that MATH and MATH are twist-recurrent. To check MATH, we must verify the condition given above for each MATH. However, for given MATH, we may replace the infinite sum over MATH by a sum with MATH without changing the values of MATH for MATH. This expresses MATH as a polynomial in the MATH and MATH, and any fixed polynomial in sequences satisfying given twist-recurrence relations satisfies one as well (by REF ). Thus MATH is twist-recurrent. If MATH, then MATH for certain universal polynomials MATH. The argument that MATH is twist-recurrent given that MATH and MATH are is similar to that for multiplication, except one must additionally note that as over a sequence of the form REF , the number of terms in the outer sum is uniformly bounded. See the analogous section of CITE for a detailed description of why this is the case. Given that MATH is a ring, which evidently is MATH-adically complete, let MATH be the image of MATH in MATH. To show MATH, we need only show MATH as as subsets of MATH. But this follows from the description of MATH given in CITE, together with REF . |
math/9906037 | REF holds because the extension group MATH is a finite set. REF imply that the map MATH has finite fibers; REF follows from this fact. REF is trivial. REF is REF. REF are consequences of REF (see also CITE, p. REF). To prove REF, it suffices to follow the proof of REF. Finally, to prove REF, one can adapt the proof of REF to the present framework. |
math/9906037 | REF enumi CASE: It is clear that MATH is abelian and satisfies REF , III. REF, and III. REF , and REF imply that MATH satisfies REF - REF . It follows from the definitions that MATH and MATH are closed under extensions in MATH, so that MATH and MATH also satisfy REF - REF . Since subobjects and quotients of torsion sheaves are torsion sheaves, MATH is abelian. Finally, the simple fact that finitely generated torsion modules over principal ideal domains have NAME series implies that MATH satisfies REF . CASE: Let MATH be a coherent sheaf. One can define in an obvious way the torsion subsheaf MATH of MATH, and the quotient sheaf MATH is locally free. By NAME 's vanishing theorem REF , the extension group MATH vanishes. Thus the short exact sequence MATH splits and one gets the decomposition MATH. Conversely, given a decomposition MATH as in the statement of REF , one has MATH and MATH. CASE: By REF , an indecomposable coherent sheaf is either a torsion sheaf or a locally free sheaf. The classification of torsion modules over the principal ideal domains MATH and MATH leads to the fact that the indecomposables torsion coherent sheaves are the sheaves MATH, where MATH and MATH is a closed point of MATH. On the other hand, REF asserts that any locally free coherent sheaf is isomorphic to a direct sum MATH for some sequence MATH of uniquely determined integers. Thus the sheaves MATH are the indecomposable locally free coherent sheaves. |
math/9906037 | REF enumi CASE: It is well known (for details, see REF ) that there are short exact sequences of the form MATH whenever MATH. Setting MATH, we get MATH when MATH, and an easy induction shows that MATH for all integers MATH. A similar argument shows the validity of REF for MATH. Now take a closed point MATH and an integer MATH. By definition, the sheaf MATH is the cokernel of an homomorphism in MATH, so MATH . The above discussion shows that MATH and MATH generate the group MATH. Since MATH and MATH form a basis of MATH, REF holds. CASE: Using the NAME formula REF and standard results of sheaf cohomology (REF , and III. REF), we obtain MATH for all MATH. This proves REF when MATH and MATH are locally free sheaves of rank REF. Since the classes of MATH and MATH generate MATH, the general case follows by the biadditivity of both sides of REF . |
math/9906037 | We first prove that REF , and REF are necessary. Suppose that the sequence is exact and non-split. If one of the homomorphism MATH or MATH were the zero arrow, then the other one would be an isomorphism since the cokernel of MATH is indecomposable, and the sequence would split. Similarly, neither MATH nor MATH can vanish. The four maps MATH, MATH, MATH, and MATH are thus non-zero, and none of them is an isomorphism. In view of REF, it follows that MATH. For degree reasons we also have MATH. Therefore REF holds. Let us turn to REF . In the unique factorization domain MATH, one may consider a g.c.d. MATH of the polynomials MATH and MATH. Since every irreducible factor of a homogeneous polynomial is itself homogeneous (by uniqueness of the factorization), the polynomial MATH is homogeneous and define a homomorphism MATH. We get a factorization MATH . Since MATH is surjective, so must be MATH. The non-zero morphism MATH being injective by REF, it is an isomorphism, which implies that MATH. Thus MATH and MATH are coprime, and REF holds. Finally, the equality MATH implies that MATH. REF and NAME 's lemma then imply the existence of a non-zero homogeneous polynomial MATH such that MATH and MATH. Since MATH is a constant polynomial, which proves REF . In order to prove the converse statement, we now assume that REF , and REF are fulfilled. Over the affine subset MATH, our sequence of sheaves reads MATH where the maps MATH, MATH are the multiplications by MATH, MATH respectively. REF implies that the polynomials MATH and MATH are coprime, which ensures by NAME 's lemma that the MATH-linear map MATH is surjective. An analogous simple reasoning based on NAME 's lemma shows that REF imply that MATH. Thus our sequence of sheaves is exact over the open subset MATH. A similar argument can be used over MATH, and we conclude that our sequence of sheaves is exact. |
math/9906037 | REF is a consequence of the structure theorem for finitely generated torsion modules over a principal ideal domain. This assertion implies that the class of sheaves with support in some fixed finite set MATH of closed points of MATH is closed under the operations of taking subobjects, quotients, and extensions. REF follows therefore from REF. Since every torsion sheaf can be uniquely written as a direct sum of subsheaves belonging to subcategories MATH, REF follows from REF . Now let MATH be a locally free sheaf and MATH be a torsion sheaf. The vanishing of MATH is a direct consequence of the definitions in REF, while that of MATH follows from NAME 's vanishing theorem. Thus REF holds. Finally, let MATH, MATH be a partition, and MATH be a closed point of MATH. The sheaf MATH has rank MATH and degree MATH, and the extension group MATH vanishes. Using REF, we therefore get MATH . |
math/9906037 | The polynomial MATH represents a morphism MATH. By unique factorization, and up to a non-zero scalar, we can write MATH, for some positive integers MATH and some distinct irreducible homogeneous polynomials MATH. For each MATH, let MATH be the closed point of MATH corresponding to MATH. The homogeneous polynomial MATH defines an element of MATH whose cokernel is MATH, whence a canonical morphism MATH. A direct application of the NAME remainder theorem implies that the sequence MATH is exact over the affine subsets MATH and MATH, hence is exact. Taking the tensor product with the locally free sheaf MATH, we get an exact sequence MATH . Therefore, the cokernel of MATH is isomorphic to MATH. If the support of MATH is included in MATH, then MATH and MATH, up to a non-zero scalar, which entails the lemma. |
math/9906037 | We can write MATH as the direct sum of its torsion subsheaf MATH and a locally free subsheaf MATH of rank MATH. Write the maps MATH and MATH as MATH and MATH in the decomposition MATH. The morphism MATH cannot be injective, so MATH cannot be zero, so MATH is injective REF , and it follows that MATH is injective. Thus MATH must be isomorphic to a subobject of MATH. Under the isomorphism of categories described in REF , the sheaf MATH corresponds to the elementary MATH-module MATH, hence to a vector space of dimension MATH over the residue field MATH. This shows that MATH is isomorphic to MATH for some MATH. In the same way, and using REF , we see that the image of MATH is either MATH or isomorphic to MATH. Now if the sequence is not split, then MATH is not an isomorphism, which rules out the case MATH. The surjectivity of MATH then requires that MATH, that MATH, and that MATH. The equality MATH then splits into the two equalities MATH and MATH. Since MATH is injective, we get MATH and thus MATH. Finally we compute MATH which shows that MATH. |
math/9906037 | Let MATH be the set of pairs MATH consisting of non-zero homogeneous polynomials of degree MATH and MATH respectively. The cardinality of MATH is MATH. One can also count the number of elements in MATH by factoring out a g.c.d. MATH of MATH and MATH. For a fixed degree MATH, there are MATH possibilities for MATH up to a non-zero scalar, and we thus get the relation MATH . The lemma then follows by induction on MATH. |
math/9906037 | REF enumi CASE: Since the extension group MATH vanishes by REF , any short exact sequence of the form MATH necessarily splits, and the product MATH in MATH is a scalar multiple MATH. It remains to compute the corresponding NAME number. Since MATH by REF, this NAME number is equal to the number of vector subspaces of dimension MATH in a vector space of dimension MATH over MATH, namely to MATH. CASE: By REF , the extension groups MATH vanish. Thus any short exact sequence of the form MATH splits, and the product MATH in MATH is a scalar multiple of MATH. Let us put MATH in the above short exact sequence, and write MATH according to this decomposition. Then MATH, because all spaces MATH vanish by REF. It follows that MATH is an automorphism. The number of suitable embeddings MATH is therefore equal to MATH, and the NAME number we are looking for is equal to MATH. CASE: By REF , any short exact sequence of the form MATH either splits, in which case MATH, or there exists MATH such that MATH. In the first case, we write MATH and MATH, where MATH, MATH, and MATH. Since MATH by REF, the existence of a left inverse of MATH requires that MATH be an automorphism. Similarly, MATH is an automorphism. The map MATH may be arbitrarily chosen and then the map MATH should be equal to MATH. Thus the set of suitable pairs MATH is in one-to-one correspondence with MATH, and the desired NAME number is MATH. This yields the term MATH in the NAME product. In the second case, the number of epimorphisms MATH such that MATH is MATH by REF . Since MATH, the NAME number MATH is MATH, whence the term MATH in the NAME product. CASE: and REF They follow from the vanishing of both MATH and MATH (seeLemma REF). CASE: This follows from REF , and REF with the same reasoning as for REF. |
math/9906037 | The isomorphism between the category MATH and the category of MATH-modules of finite length gives rise to an isomorphism between their NAME algebras. Thus MATH is isomorphic to the NAME algebra studied in REF and III of CITE. REF therefore follows from Paragraphs III REF , III. REF, III. REF , and III REF of CITE. It is well-known that MATH is the MATH-algebra of polynomials either in the complete symmetric functions MATH or in the elementary symmetric functions MATH, for MATH (see REF and I REF of CITE). This fact implies the first assertion in REF . The second one follows from REF of CITE and its proof. |
math/9906037 | Following Paragraphs I REF and I REF, we define generating series in MATH by MATH . By Formulae I REF and III REF in CITE, we have in MATH . Taking the inverse images by MATH, one obtains the relations in REF . As for REF , it follows from the equality MATH (use REF in CITE and REF ). |
math/9906037 | REF follows from REF. Let MATH be the subalgebra of MATH generated by the subalgebras MATH with MATH. Then MATH is the algebra of polynomials in the indeterminates MATH with coefficients in MATH. It is easy to see that MATH belongs to MATH, which proves the algebraic independence of the elements MATH. The algebraic independence of the other two families can then be deduced from REF . This completes the proof of REF . |
math/9906037 | REF can be proved by a tedious case by case examination, using Relations REF - REF . To prove REF , one first compute MATH using REF, and then MATH using REF. |
math/9906037 | REF comes from the fact that MATH is a subcategory of MATH closed under extensions REF and from the definition of the product MATH in MATH. REF follows from REF. Finally, REF imply REF . |
math/9906037 | We will use the generating series MATH and MATH. We set MATH and compute, using Relations REF : MATH . In view of REF, we therefore have MATH after expansion of the rational functions MATH in powers of MATH. Now in the formal power series ring MATH, one has MATH where the product in the left hand side runs over all closed points of MATH. The previous equality follows from the calculation of the zeta function of MATH (see REF for a proof). Therefore, MATH which is equivalent to our assertion. |
math/9906037 | REF implies that the multiplication MATH induces an isomorphism of MATH-modules from MATH to MATH. REF implies that MATH is a subalgebra of the NAME algebra MATH, obviously equal to MATH. Since the MATH-modules MATH and MATH are free (see REF), the multiplication MATH in the NAME algebra MATH induces an isomorphism of MATH-modules from MATH to MATH. REF shows that MATH is a polynomial algebra on each of the three set of indeterminates: MATH, MATH, or MATH. Thus the families MATH, MATH, and MATH are three bases of the MATH-vector space MATH. Both assertions of our proposition follow now from REF. |
math/9906037 | It is asserted in REF that the map MATH induced by the multiplication of MATH is surjective. The defining relations of MATH imply easily that the map MATH induced by the multiplication of MATH is also surjective. The algebra MATH is generated by the pairwise commuting elements MATH for MATH, which shows that the monomials MATH, for MATH, span the MATH-vector space MATH. Similarly, the family of elements MATH span the MATH-vector space MATH. Finally, an easy induction shows that the products MATH, performed in the ascending order of MATH and for MATH, span the MATH-vector space MATH. Consequently, the elements MATH where MATH, MATH, and MATH, span the MATH-vector space MATH. Now observe that the definition of MATH implies the existence of an automorphism MATH of the MATH-algebra MATH such that MATH for all MATH with MATH. (Using REF, one can easily see that MATH is the automorphism of MATH that lifts the translation along the fundamental weight to the braid group of the extended affine NAME group of MATH.) On the other hand, observe, as a consequence of the NAME theorem for MATH REF , that the set of elements MATH is linearly independent over MATH. Using this and the automorphism MATH, one proves the linear independence over MATH of the family of elements MATH. This family is therefore a basis of MATH, which entails simultaneously all the assertions of the lemma. |
math/9906037 | By definition, MATH is the subalgebra of MATH generated by the elements MATH and MATH, for MATH and MATH. REF follows therefore from the definition of the elements MATH REF and from the fact that the scalars MATH do not vanish in the field MATH for any MATH. REF states that for all integers MATH and MATH, one has MATH . Applying a well-chosen power of the automorphism MATH defined in the proof of REF , one immediately sees that REF holds more generally for each MATH. Together with REF, this shows that the multiplication map induces a linear isomorphism MATH. By Formulae REF, we have the following relations in MATH: MATH or, equivalently, for each MATH: MATH . Together with REF , this implies that MATH is a polynomial algebra on each of the three set of indeterminates: MATH, MATH, or MATH. Thus the families MATH, MATH, and MATH are three bases of the MATH-vector space MATH. REF follows from this and from REF. |
math/9906041 | It suffices to examine the effect of the formula on monomials MATH and for MATH . The first two terms produce MATH if MATH else MATH . In the sum, the (typical) term for MATH is MATH . A simple calculation shows this is the image under MATH of the corresponding term in REF . |
math/9906041 | REF follow immediately from REF . It was shown in REF that MATH where MATH is the conjugate under MATH of multiplication by MATH acting on MATH. Together with REF this proves REF . |
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