paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0011050 | We prove the following two facts: CASE: MATH for every MATH. CASE: MATH where MATH is a homomorphism similar to the case of MATH. First note that as MATH centralizes MATH, MATH for MATH. Recall (compare REF) that MATH, MATH are isomorphic. As MATH acts on MATH as holomorphic homothetic transformations (compare REF ), MATH leaves MATH invariant. If MATH, then MATH. As MATH is a characteristic vector field for MATH, MATH. This concludes that MATH. On the other hand, if we note MATH, then MATH where MATH is a positive constant number. As MATH from REF and MATH from REF, we obtain that MATH. Equivalently, MATH. This shows REF. Now, we prove REF. From REF, MATH . Since MATH by the definition, MATH acts through holomorphic isometries of MATH. |
math/0011050 | Let MATH be the group of all holomorphic isometries of MATH. As the NAME field MATH is Killing, it generates a MATH-parameter subgroup MATH of (holomorphic) isometries on MATH. Then MATH. Since MATH is compact, the closure MATH of MATH in MATH is a MATH-torus MATH. CASE: When the MATH-parameter group MATH is not a circle, we can find a sequence of circles MATH which approaches to MATH. Denote by MATH the vector field induced by MATH for each MATH. Then there exists MATH with MATH . To prove this, suppose that for all MATH there exists a sequence of points MATH such that MATH. As MATH is compact, there exists a point MATH such that MATH. If we note that MATH converges to MATH, then MATH, which is impossible. Now, let MATH (respectively MATH) be the lift of the circle MATH (respectively MATH ) to MATH. Recall the NAME form MATH on MATH where MATH (compare REF). Note that each MATH leaves invariant MATH and so it satisfies: MATH . As MATH is NAME, MATH is constant on MATH. Suppose that every MATH is an isometry, MATH . MATH for all MATH. Then, MATH being a contradiction, and so some MATH is a nontrivial homothety; MATH for a nonzeo constant MATH. Hence every MATH is a nontrivial homothetic transformation. By REF , the circle MATH is purely conformal. Then the results REF of I follow from Theorem A. We prove II. Let MATH be a lift of MATH to MATH which induces a vector field MATH. Noting that MATH is an isometry of MATH because MATH is parallel, we can write MATH where MATH. Since MATH is constant on MATH for each MATH, we may put MATH for some constant MATH. By the definition MATH, so is its lift MATH on MATH. As MATH, MATH . We may normalize MATH so that (compare REF): MATH . In particular, the group MATH is isomorphic to MATH. As in the proof of Theorem A, we have a contact form MATH on MATH. However, comparing REF with REF, this will be defined as MATH where MATH such that MATH. (Compare REF.) Noting also that MATH in this case, we have that (compare REF) MATH . On the other hand, we can define a MATH-form MATH on MATH (that is, a NAME contact structure, see CITE) by MATH for MATH with respect to MATH such that MATH. If we note MATH, then the distribution MATH is the same as that of MATH. As MATH is isomorphic, MATH is well defined. We compare these contact forms first. By MATH and REF, we compute MATH so that MATH . Moreover, MATH on MATH from REF. Thus there is a map MATH with MATH. Hence, REF implies: MATH . Finally, recall from REF that MATH because MATH in REF. Thus, from REF we derive: MATH . Using REF MATH on MATH. Therefore MATH has the parallel NAME form MATH such that MATH for MATH on MATH. This proves II. |
math/0011050 | Suppose MATH is not spherical. Then the set MATH is a non-empty, open subset of MATH. As MATH is a MATH invariant (see for example, CITE), the set MATH is preserved by any NAME of MATH. If a contact form MATH is replaced by MATH, then the norm of the NAME tensor with respect to the NAME form associated to MATH satisfies the equality MATH (compare CITE). Choose MATH on MATH. Then we obtain a pseudo-Hermitian structure MATH defined on MATH such that MATH . If MATH, then MATH. The above relation shows that MATH. By REF, MATH on MATH. Then we can check that for each element MATH of MATH, the transformation REF can be reduced to the following: MATH . Thus the bundle of such coframes MATH with MATH gives rise to a principal bundle, restricted to MATH: MATH for which we note that MATH . On the other hand, as MATH is of order MATH (as a subgroup of MATH), the manifold MATH has a MATH-structure. Hence, by REF, for any fixed MATH, the orbit map MATH is a proper embedding, and the orbit MATH is closed in MATH. As any automorphism of MATH produces, by differentiation, an automorphism of MATH, we may consider the subgroup MATH. Restricting to NAME of MATH and using REF, we derive that the orbit MATH is closed in MATH and that MATH is homeomorphic to MATH. Now recall that we have a closed subgroup MATH in MATH. Denote with MATH its lift to MATH (MATH contains the differentials of the flows associated to the MATH-action). Hence, MATH is closed in MATH. Now, if we can prove that this orbit is contained in a certain compact subset MATH, we see that MATH has to be compact, in contradiction with it being homeomorphic with MATH by the proper embedding. To construct MATH, set MATH so that MATH. From REF , MATH is a quotient (continuous) map and MATH is locally compact (as a manifold), MATH is compact, hence we may find a connected, compact subset MATH such that MATH and MATH. Hence MATH. In particular, as MATH, we obtain MATH. But MATH acts properly discontinuously, so only a finite number of translated MATH meet MATH. Since MATH is connected, we may write MATH and note that MATH is compact. Hence, the closure MATH is compact. But a priori, it might exit MATH. We prove that this is not the case and, in fact, MATH following CITE. Then the inverse image MATH of the bundle REF is the one desired. To this end, let MATH be the vector field on MATH induced by the considered MATH-action MATH. We first prove: MATH . Indeed, by absurd, if MATH on MATH, then MATH on MATH. For any MATH, as MATH an infinitesimal contact transformation, we have MATH, in contradiction with MATH being strictly pseudo-convex, its NAME form MATH is positive definite, in particular non-degenerate. If we note that MATH can be chosen arbitrary in MATH, we may suppose MATH for some suitable, fixed MATH. We consider the non-empty set MATH and show that: MATH . Here is a simple argument of general topology. Observe first that, by MATH with MATH, MATH tends to MATH when MATH approaches the boundary of MATH in MATH. Now let MATH and choose a sequence MATH which converges to MATH. We have MATH. But MATH as MATH, hence MATH. This means that MATH is not a boundary point of MATH, thus MATH. This implies MATH, yielding MATH and proving REF. Since MATH for MATH, the equality MATH implies MATH. Obviously, MATH. This ends the proof of REF . |
math/0011055 | By applying REF to manifold MATH (the second picture of MATH in REF ) we see that MATH is compact NAME. Also since MATH is slice in MATH if MATH extended to a diffeomorphism MATH, then MATH would be slice also. But this contradicts the inequality of REF (here MATH, MATH, and MATH) |
math/0011055 | By REF manifold MATH is compact NAME domain, by REF it embeds as a domain in a minimal NAME surface MATH with MATH. If MATH were diffeomorphic to MATH, the generator of MATH would be represented by a smooth embedded sphere (since MATH is a slice knot the generator of MATH represented by a smooth sphere) with MATH self intersection, violating REF . |
math/0011055 | MATH. If MATH then MATH is just the blackboard push-off of MATH with MATH right half twist, but this is a projection of a Legendrian link and MATH and MATH hence MATH . If MATH and MATH, then MATH is the blackboard push-off with MATH left handed half twist. Knot MATH has MATH cusps, and MATH. We can produce a Legendrian picture by placing all half twists near cusps as in REF . Thus MATH. If MATH, then MATH and again MATH is the blackboard push-off with MATH left half twist. By above trick of placing half left twist near cusps, we can get rid of MATH left twist, and we are left with MATH left twist each of which contributes MATH to MATH as in REF . Hence MATH . |
math/0011055 | The first positive NAME double is obtained by connecting MATH and MATH by a left handed cusp as in REF which contributes MATH to MATH, hence MATH . So by iteration we get MATH. But REF says that any slice knot MATH must have MATH. For the second part, observe that since MATH, by REF MATH is compact NAME. By REF we can imbed MATH into NAME surface MATH. Then by REF we can not have a smoothly embedded MATH-sphere MATH representing MATH, since MATH. |
math/0011056 | We distinguish two cases; CASE: Suppose that there is a vertex MATH of MATH such that more than MATH edges connect MATH to antiparallel vertices. This implies that there exist more than MATH-edges connecting parallel vertices in MATH. Consider the subgraph of MATH consisting all vertices and those all MATH-edges of MATH. An NAME characteristic count gives that it contains more than MATH disk faces whose boundaries are MATH-edge cycles of MATH. By CITE, each disk face contains a NAME cycle. By REF , these all NAME cycles are, say, MATH-Scharlemann cycles. Construct a graph MATH in MATH as follows. Choose a dual vertex in the interior of the face MATH bounded by a MATH-Scharlemann cycle, and let the vertices of MATH be the vertices of MATH together with these dual vertices. The edges of MATH are defined by joining each dual vertex to the vertices of the corresponding NAME cycle in the obvious way. Let MATH be the number of MATH-Scharlemann cycles. Then MATH. MATH has MATH vertices and at least MATH edges because each NAME cycle has order at least REF by REF . Note that if MATH contains an NAME cycle of length two, then MATH contains a projective plane CITE. Again an NAME characteristic count guarantees that MATH has a disk face MATH. But MATH determines a MATH-edge cycle bounding a disk face in MATH which, as long as MATH, contains a NAME cycle. This contradicts the definition of MATH. CASE: As the negation of REF , suppose that each vertex MATH of MATH has at least MATH labels where edges connecting parallel vertices are incident. Let MATH-be the subgraph of MATH consisting of all vertices and edges connecting parallel vertices of MATH. By the assumption, every vertex of MATH has valency at least MATH. Let MATH be the reduced graph of MATH. By REF , any vertex has valency at least MATH in MATH. So, we can choose an innermost component MATH of MATH, and a block MATH of MATH with at most one cut vertex of MATH. Let MATH be the subgraph of MATH corresponding to MATH. Note that there is a disk MATH in MATH such that MATH. A vertex of MATH is a boundary vertex if there is an arc connecting it to MATH whose interior is disjoint from MATH, and an interior vertex otherwise. Remark that MATH has an interior vertex by CITE. If MATH contains a cut vertex of MATH, then let MATH be a label such that the cut vertex has an edge attached with label MATH there. Otherwise, let MATH be any label. Then each vertex of MATH, except a cut vertex, has at least two edges attached with label MATH. Thus we can find a great MATH-cycle. This implies that MATH has a NAME cycle and so MATH is even by CITE. Let MATH and MATH be the numbers of vertices, edges and faces of its reduced graph MATH. Also say MATH and MATH the numbers of interior vertices, boundary vertices and a cut vertex of MATH. Hence MATH and MATH or MATH. Since each face of MATH is a disk with at least MATH sides, we have MATH. Combined with MATH, we get MATH. Suppose that every interior vertex of MATH has valency at least MATH and that every boundary vertex except a cut vertex has valency at least MATH. Since a cut vertex has valency at least MATH, we have MATH. These two inequalities give us that MATH, a contradiction. Hence there are two cases. For the first case, assume that some boundary vertex MATH of MATH has valency at most MATH. There are at least MATH edges in MATH which are incident to MATH and connect MATH to parallel vertices. By REF , MATH, that is, MATH and MATH. But in the case of MATH there are at least MATH mutually parallel edges in MATH, which implies that MATH contains two NAME with disjoint label pairs. Then MATH contains a NAME bottle as in the proof of CITE. (Note that MATH is irreducible CITE. Hence the edges of a MATH-cycle cannot lie in a disk on MATH.) This contradicts to CITE. For the second case, assume that some interior vertex of MATH has valency at most MATH. Again we have MATH. Then we can use the same argument as above. |
math/0011057 | If MATH, then MATH contains a projective plane, which is impossible since MATH is hyperbolic. If MATH, then MATH contains a NAME band, which is also impossible. |
math/0011057 | Assume MATH is compressible in MATH. Let MATH be a compressing disk for MATH. Note that MATH is orientation-preserving on MATH (and hence MATH). Hence MATH bounds a disk MATH on MATH. Since MATH meets MATH, we can create a new projective plane by replacing MATH with MATH, which meets MATH fewer than MATH. This contradicts the minimality of MATH. Therefore MATH is incompressible in MATH. Next, assume that MATH is boundary-compressible. Then MATH would be compressible, or the core of MATH can be isotoped into MATH as an orientation-reversing loop. But, the latter case implies that MATH is boundary-reducible. |
math/0011057 | Suppose that MATH is compressible in MATH. Let MATH be a disk in MATH such that MATH and MATH does not bound a disk on MATH. Note that MATH is orientation-preserving on MATH. If MATH is non-separating on MATH, then we get a non-separating MATH-sphere in MATH by compressing MATH along MATH. This is clearly a contradiction. If MATH bound a disk on MATH, then we replace the disk with MATH, and get a new NAME bottle in MATH with fewer intersections with MATH than MATH. This contradicts the choice of MATH. Thus MATH is essential and separating on MATH. Compressing MATH along MATH gives two disjoint projective planes in MATH. Since MATH is irreducible, this is also impossible. Thus we have shown that MATH is incompressible. Next, let MATH be a disk in MATH such that MATH, MATH, where MATH is an essential (that is, not boundary-parallel) arc in MATH and MATH. If MATH joins distinct components of MATH, then a compressing disk for MATH is obtained from two parallel copies of MATH and the disk obtained by removing a neighborhood of MATH from the annulus in MATH cobounded by those components of MATH meeting MATH. Hence MATH is contained in the same component MATH, say, of MATH. If MATH, then MATH bounds a disk MATH in MATH together with a subarc of MATH. Then MATH gives a compressing disk for MATH in MATH. Therefore MATH. Then we can move the core of MATH onto an orientation-reversing loop in MATH by using MATH. This implies that MATH contains a properly embedded NAME band, which contradicts the fact that MATH is hyperbolic. |
math/0011057 | This follows from the fact that MATH and MATH are boundary-incompressible. |
math/0011057 | This follows from the fact that MATH is orientable and MATH is a torus. |
math/0011057 | Let MATH be a negative loop based at a vertex MATH. Then MATH is a NAME band. Since MATH and MATH can contain at most one NAME band and at most two disjoint NAME bands respectively, the conclusions follow. |
math/0011057 | This is CITE. |
math/0011057 | These follow from REF . |
math/0011057 | Let MATH be a family of MATH mutually parallel edges in MATH labelled successively. By REF , all MATH's are negative, and then they make orientation-preserving cycles in MATH. Note that an orientation-preserving loop in a projective plane is contractible. Therefore, we can choose an innermost cycle among them. Then the construction in CITE implies that MATH is cabled, a contradiction. |
math/0011057 | We distinguish two cases. CASE: Suppose that there is a vertex MATH of MATH such that more than MATH negative edges are incident to MATH. Remark that such negative edges are not loops because MATH. This implies that there exist more than MATH positive MATH-edges in MATH by the parity rule. Thus MATH has a larger number of edges than that of vertices. An NAME characteristic calculation gives that MATH contains a disk face. CASE: As the negation of REF , suppose that each vertex MATH of MATH has at least MATH positive edge endpoints. Let MATH be an extremal component of MATH. That is, MATH is a component of MATH having a disk support MATH such that MATH. First, assume that MATH is a single vertex. Then it must be vertex MATH. Thus only negative edges are incident to vertex MATH in MATH. If there is no negative loop at vertex MATH, then MATH can be the only extremal component of MATH. But then, MATH contains a MATH-face MATH as in REF has no positive level edges. Then MATH contains a NAME cycle by CITE, which is impossible by REF . If a negative loop is incident there, we can choose another extremal component of MATH, which has more than one vertex. Thus we can assume that MATH is not a single vertex. Choose a block MATH of MATH with at most one cut vertex. Then MATH is clearly a generalized web. |
math/0011057 | There is a possibility that MATH is not a circle. That is, MATH may contain a double edge, and also more than two edges of MATH may be incident to a vertex on MATH. Since we will find a generalized NAME cycle within MATH, we can cut formally the graph MATH along double edges of MATH and at vertices to which more than two edges of MATH are incident so that MATH is deformed into a circle. (See also CITE.) Thus we may assume that MATH is a circle. If MATH is a bigon, then the conclusion is obvious. Therefore, we assume that MATH has at least three sides. Suppose that MATH has a diagonal edge MATH, which has two distinct labels MATH and MATH at its endpoints as in REF . Since MATH is a MATH-face, MATH and MATH. Without loss of generality, we may assume that the labels appear in counterclockwise order around the boundary of each vertex, and that MATH. Formally, construct a new MATH-face MATH as follows. The edge MATH divides MATH into two disks MATH and MATH. We can assume that MATH lies on the right side of MATH when MATH is oriented from the endpoint with label MATH. If three labels MATH appear in this order around the corners of MATH, then discard MATH, and insert additional edges to the left of MATH, and parallel to MATH, until we first reach label MATH at one or both ends of an additional edge. See REF . The new MATH-face MATH is the union of MATH and some additional bigons. If three label MATH appear in this order, then discard MATH, and insert additional edges to the right of MATH as above. Then MATH is the union of MATH and some additional bigons. Remark that there is no generalized NAME cycle among additional edges and MATH. Repeat the above process for every diagonal edge which is not level, then get a new MATH-face MATH and a graph MATH in MATH. All diagonal edges of MATH are level, and all boundary edges are MATH-edges, especially label MATH can appear on both ends of a boundary edge. Such boundary edges are called level MATH-edges to distinguish them from level MATH-edges. See REF . From now on, we assume that there is no NAME cycle in MATH. MATH contains a level MATH-edge. Suppose not. Then MATH contains a NAME cycle by CITE. Let MATH be the disk face of MATH, which is not a bigon. We will show that for any edge of MATH on MATH, it has labels only on MATH at both ends. There are consecutive vertices MATH on MATH such that MATH is not a base of a level MATH-edge for MATH, but MATH and MATH are base of level MATH-edges. Possibly, MATH. See REF . Let MATH be the MATH-edge connecting MATH with MATH for MATH. Let MATH (MATH, respectively,) be the label of the end of MATH at MATH (MATH, respectively,) and MATH the family of mutually parallel edges containing MATH. The number of edges in MATH is denoted by MATH. Also, let MATH be the edge of MATH lying on MATH. We define MATH and MATH similarly as above. CASE: If MATH is even, then MATH. Furthermore, if MATH, then either the MATH-edge MATH is level, or its label on the other end is MATH or MATH. CASE: If MATH is odd, then MATH. Furthermore, if MATH, then MATH is a level MATH-edge. These follow from REF and the definition of MATH. If MATH, or MATH for some MATH, then each MATH has labels on MATH. Suppose MATH. Then MATH, since MATH cannot have MATH edges, except the case MATH, by REF . Thus MATH and MATH, since there cannot be a level edge or a MATH-cycle. When MATH, we have the same conclusion. For, MATH or MATH, and if two then MATH by REF . The same argument runs until we got that MATH and MATH for all MATH. Then it is easy to see that MATH has labels on MATH at both ends (see REF ). In particular, when MATH, the corner on MATH at MATH contains label MATH. Similarly for the case where MATH. Suppose that MATH as in REF . Assume MATH. By REF , MATH, and so MATH. If MATH, then it contains either a level edge or a MATH-cycle. Thus MATH. Hence MATH and MATH. This implies that MATH. If MATH, then a similar argument shows that MATH and MATH, which implies MATH. When MATH, then MATH. Then we get the desired result by REF . Therefore, we assume MATH. Furthermore, we can assume that MATH and any MATH by REF . If MATH for all MATH, then each MATH has labels on MATH. Note that MATH and MATH. Consider MATH. Since MATH contains neither level edge nor MATH-cycle, MATH. This implies MATH. Then MATH. Thus we have MATH. Hence MATH or MATH for all MATH. The result immediately follows from this observation. Thus we can assume that MATH and MATH for some MATH. (Possibly, MATH.) Then MATH. See REF . Also, MATH by REF . Then MATH. Thus we have MATH and MATH for MATH, and MATH. CASE: MATH is even. Then MATH, since MATH. See REF . Thus MATH or MATH by REF . For, if MATH then MATH contains a MATH-cycle. Indeed, MATH, since MATH. Hence MATH, and MATH or MATH again. If MATH, then MATH as above. Thus we can conclude that MATH, and that MATH, and MATH, MATH for MATH. Also MATH or MATH, and hence MATH or MATH, and MATH. Since MATH and MATH, we see MATH and so MATH. Thus MATH for MATH, and MATH for MATH and MATH or MATH. CASE: MATH is odd. Then MATH, and MATH similarly for the case where MATH is even. See REF . Also, MATH by REF . Then we see that MATH and MATH for MATH. Thus MATH and MATH for MATH. Since MATH and MATH, we have MATH and MATH. Then MATH and therefore MATH. By REF , the only possibility is MATH. Thus we have shown that MATH gives a generalized NAME cycle. This completes the proof of REF . |
math/0011057 | This follows immediately from REF . |
math/0011057 | If MATH contains a NAME cycle, then we can get a new NAME bottle in MATH which meets MATH fewer than MATH CITE. REF implies REF . Here, we need the fact that the distance between two NAME fillings creating projective planes is at most one CITE. REF is a restatement of REF . For REF , if not, such a family contains a generalized MATH-cycle. |
math/0011057 | Assume for contradiction that MATH because of REF . Let MATH and MATH be the vertices of MATH, where MATH can be a base of negative loops. Since MATH and MATH have the same degree MATH, if MATH has a loop, then so does MATH. Then there would be a trivial loop. Thus we can see that MATH has no loops. Then MATH consists of at most two families of mutually parallel edges; one is a family of positive edges, and the other is that of negative edges. If MATH, then MATH contains more than MATH mutually parallel negative edges by REF . Then an easy NAME characteristic calculation shows that MATH contains a MATH-face and no level edges. As in REF of the proof of REF , this gives a contradiction. If MATH, then MATH has at most two positive edges. Otherwise, there would be two edges which are parallel in both MATH and MATH. But this implies that MATH is cabled by CITE. Thus MATH contains at least four negative edges. Similarly, MATH contains a MATH-face and no level edges, a contradiction. |
math/0011057 | Assume for contradiction that MATH contains a generalized web MATH, possibly with an exceptional vertex MATH among boundary vertices of MATH. Let MATH denote a disk support of MATH. REF guarantees the existence of a label MATH such that MATH contains no positive level MATH-edges. Consider MATH consisting of all vertices and MATH-edges of MATH. Since every boundary vertex of MATH, except MATH, has degree at least MATH, it has at least two edges attached with label MATH. We remark that MATH may be disconnected. Choose an innermost component MATH of MATH (in MATH), and let MATH be its block with at most one cut vertex of MATH. Let MATH and MATH be the numbers of vertices, edges, and disk faces of MATH, respectively. (We view MATH as the graph in a disk.) Also let MATH and MATH be the numbers of interior vertices, boundary vertices of MATH and a cut vertex of MATH in MATH, respectively. Hence MATH and MATH or MATH. Since MATH has neither a level MATH-edge nor a generalized MATH-cycle, each face of MATH is a disk with at least MATH sides. Thus we have MATH. Combined with MATH because it has only disk faces, we get MATH. On the other hand we have MATH because each boundary vertex of MATH, except a cut vertex of MATH, has at least two edges attached with label MATH. These two inequalities give us that MATH, a contradiction. |
math/0011057 | Any interior vertex of MATH has degree at least six, since MATH. Also, any boundary vertex of MATH, except an exceptional one, has degree at least four. Therefore MATH contains a pair of parallel edges by CITE. Then these edges are both level as above. |
math/0011057 | CASE: Suppose that three positive MATH-edges, say, are incident to MATH. Since all negative loops based at vertex MATH in MATH are mutually parallel, there are at least MATH negative loops at vertex MATH. This is impossible by REF . CASE: Let MATH and MATH be the vertices of MATH. By REF , MATH contains a pair of parallel (positive) edges MATH. Here, MATH is a MATH-edge for MATH, and MATH connects a vertex MATH to another MATH (possibly, MATH). Then MATH gives a negative loop based at MATH with the label pair MATH. Remark that all negative loops at MATH are mutually parallel. If MATH has another MATH-edge MATH connecting MATH to MATH, then MATH and MATH are parallel, and so MATH. This means that any MATH-edge at MATH goes to either vertex MATH or MATH. |
math/0011057 | Let MATH be these pair edges, where MATH is a MATH-edge. Assume for contradiction that MATH is the consecutive MATH-edge next to MATH. As in the proof of REF, we may assume that MATH connects MATH to MATH and MATH and MATH connect MATH to MATH for some MATH. Let MATH and MATH be the endpoints of these edges. Let MATH be a properly oriented arc from MATH to MATH along MATH, and MATH similarly. See REF . In MATH, MATH is a negative loop at vertex MATH, and its endpoint MATH has label MATH, and MATH has MATH. Also, MATH and MATH are contained in the family of mutually parallel negative loops at vertex MATH in MATH, and MATH and MATH appear at the same end of this family. Otherwise, the family contains more than MATH edges, contradicting REF . Let MATH be the boundary component of MATH with label MATH (MATH). We may assume that MATH is oriented counterclockwise as in REF . Then the subarc MATH, with the induced orientation from MATH, contains at most MATH edge endpoints. Consider the annulus part of MATH between MATH and MATH, containing MATH and MATH. Then we see that the two oriented subarcs MATH and MATH contain the same number of points in MATH. See REF . But, the subarc MATH contains at least MATH edge endpoints. The second conclusion immediately follows from the first one. |
math/0011057 | CASE: If a disk face is bounded by only mixed edges, then the boundary cycle is a NAME cycle, which is impossible by REF . If the boundary cycle contains an odd number of level edges, we cannot put the labels correctly. CASE: Suppose that there is such a disk MATH. Since its interior contains no vertex of MATH, there is an outermost disk face of MATH whose boundary edges consist of mixed edges and at most one diagonal edge (or exceptional one), contradicting REF . |
math/0011057 | By REF , MATH. If MATH, then REF give a contradiction. We have just shown that the case MATH is impossible. |
math/0011057 | Assume that MATH. The reduced graph MATH is a subgraph of the graphs shown in REF . If MATH is even, there are at most MATH positive edges by REF , and at most MATH mutually parallel negative edges by REF . Thus MATH, and then MATH and MATH. Similarly, if MATH is odd, we have that MATH and MATH. It follows that MATH contains at least two positive edges. First, assume that MATH is a subgraph of REF . This family of mutually parallel positive edges in MATH contains a generalized MATH-cycle, contradicting REF . Next, assume that MATH is a subgraph of REF . Since each family of mutually parallel negative edges contains at most MATH edges, we can assume that the labels are as in REF . Then the family of mutually parallel positive edges contain a MATH-cycle with label pair MATH. |
math/0011057 | REF is shown in CITE. CASE: If there are four such labels, there are two MATH-cycles with disjoint label pairs. Then MATH contains a NAME bottle as in the proof of CITE. (Recall that MATH is irreducible.) By REF , MATH must contain a NAME bottle MATH which meets the core MATH of the attached solid torus MATH in a single point, since MATH. Then the torus MATH is essential in MATH. Otherwise MATH would be either a lens space containing a NAME bottle or a prism manifold, both of which are not toroidal. This essential torus meets MATH in two points. This contradicts the minimality of MATH. REF follows from REF , since a family of MATH mutually parallel positive edges contains either an extended MATH-cycle or two MATH-cycles with disjoint label pairs. |
math/0011057 | This follows from the same argument as in the proof of REF . We use REF instead of REF . |
math/0011057 | Assume for contradiction that MATH contains a generalized web MATH, possibly with an exceptional vertex MATH of MATH. Let MATH denote a disk support of MATH. Let MATH be a label of MATH. Suppose that MATH has no MATH-cycle with a label MATH. Consider MATH, consisting of all vertices and MATH-edges of MATH as in the proof of REF . Choose an innermost component MATH of MATH (in MATH), and let MATH be its block with at most one cut vertex of MATH. Since MATH has neither MATH-cycle with label MATH nor extended MATH-cycle, each face of MATH is a disk with at least MATH sides. Then the same calculation as in the proof of REF gives a contradiction. Therefore MATH has MATH labels which are labels of NAME. Note that the existence of an NAME in MATH guarantees that MATH is even CITE and so MATH. This contradicts REF . |
math/0011057 | By REF , MATH. Then REF give a contradiction. |
math/0011057 | The reduced graph MATH is a subgraph of the graph shown in REF. Here, MATH denotes the number of edges in each of the families of mutually parallel edges. Then MATH. We have that MATH and MATH for MATH by REF . Thus, MATH. This inequality implies that MATH and all MATH's are non-zero. All non-loop edges of MATH are negative. Assume not. Let MATH and MATH be the vertices of MATH. If MATH is even, then MATH, which is absurd. If MATH is odd, then MATH. Thus MATH. But then, each of six families at MATH, say, consists of two edges, and we have a MATH-cycle in the family of loops at vertex MATH. Without loss of generality, we can assume that MATH. Let MATH. Since MATH, we see MATH. Then MATH. Hence MATH. Thus the loop family around MATH, say, contains a generalized MATH-cycle. |
math/0011057 | Assume that MATH. Since MATH, MATH can have only positive edges, and so MATH has only negative edges. Note that MATH has at most three families of mutually parallel edges by [REF ]. Let MATH, MATH and MATH be such three families, and MATH, MATH and MATH be the number of edges of each family. Without loss of generality, we can assume that MATH has at least one edge. First assume that MATH is even. If MATH, then MATH contains a generalized NAME by the symmetry of labels around the vertex. Thus MATH. Similarly we have MATH, MATH, and therefore MATH. Since MATH is even and MATH, we have MATH and MATH. Then MATH has two or three level edges with different labels, contradicting REF . Assume now that MATH is odd. Since MATH is integral, MATH must be even. Thus MATH by REF , and similarly for MATH. Then MATH. Hence we have MATH and MATH. This implies that MATH contains two NAME cycles of length three, which is a contradiction. |
math/0011059 | We give only the derivation of REF because the other follows by symmetry. Conditioning on MATH implies that: MATH, MATH, MATH and MATH. From REF we obtain that MATH and by simple integration by parts it emerges that: MATH from which follows REF. |
math/0011059 | By NAME expansion, one gets that MATH and MATH are solutions to REF and rewriting system REF in terms of the functions MATH and MATH it emerges that MATH . The conclusion arises by direct substitutions. In fact, by the first of system REF we have MATH . Furthermore MATH by using respectively the second and the first equation of system REF. |
math/0011059 | We start without assuming REF and by noting that MATH is a solution to REF. Moreover, the process MATH only if there occur at least one NAME event up to time MATH, thus MATH that gives one of the conditions to solve REF. We now search for solutions of the following type: MATH . Thus the function MATH satisfies the following partial differential equation MATH . For a generic function MATH a solution to REF not available. If the intensity function MATH satisfies the following ordinary differential equation MATH then REF becomes MATH and the solution to it can be written explicitly. With the initial condition MATH the solution to REF is REF and then MATH. Following NAME REF , by the change of variable MATH we transform REF into the following standard NAME 's equation MATH whose general integral is MATH . The function MATH is the modified NAME function of first kind and order k while MATH is the second type NAME function of order REF with the unpleasant property that MATH, so we put MATH. In terms of MATH the solution is of the following form MATH . From REF it follows that MATH and so, there is no constant value MATH that satisfies REF. To turn around this problem, we note that if MATH is a solution to REF so is its partial derivative with respect to MATH. Hence we search for solutions of the form MATH . Form the following two equalities (see NAME, REF) MATH and MATH REF implies that MATH . Observe that for MATH as in REF we have MATH thus the equality in the above equation is attained by taking MATH and MATH. The solution is finally MATH that is non negative and satisfies the conditions required. It also verifies REF and this is a boring calculus' exercise. |
math/0011060 | Recall that a collection of sets is a MATH-matroid if and only if it satisfies the symmetric exchange axiom, REF : MATH . Set, without loss of generality, MATH in REF above (there is no loss of generality since we are not concerned with signs or orderings). Set MATH, and MATH such that MATH is REF. Thus some other MATH must be MATH; let MATH. Now, from the defining equation for MATH, we have MATH and so we obtain MATH, which is more than we need. |
math/0011060 | The first six statements are immediately clear from the construction of MATH. From consideration of determinants, which can be more readily seen, the final part is correct up to sign. But the term MATH appears in some sense `early' in the construction of MATH from MATH and cannot then change sign, so this is the correct sign also. |
math/0011060 | Throughout, MATH and MATH. Define MATH if MATH and MATH otherwise. Thus MATH wherever MATH is defined, and MATH. Thus, the elements of the skew-symmetric matrix MATH satisfy: MATH . (for MATH). It is easy to see that the lemma holds for determinants rather than NAME of minors, so MATH. Each term of MATH, rewritten in terms of the MATH, corresponds to several terms of MATH; thus we need check only that one of these has the same sign in MATH as in MATH. Let MATH and write MATH. We divide the proof into the four cases MATH, MATH, MATH, MATH. We shall divide these each into sub-cases depending on whether MATH and MATH are odd or even. First we take MATH; we may assume MATH is even (as otherwise MATH). Now, take MATH which has positive sign in MATH; this contains the signed term MATH where MATH . Now we consider our sub-cases. If both MATH are even, then MATH has positive sign in MATH, and in fact all the MATH are positive, so the term has positive sign in MATH as well. If MATH is odd but MATH is even then MATH has negative sign in MATH, and all the MATH are positive except for MATH, so again MATH has the correct sign. Similarly, if MATH is odd but MATH is even then MATH has negative sign, and all the MATH are positive except for MATH. Finally, if both MATH are odd, then MATH has positive sign in MATH, and all the MATH are positive except for MATH and MATH. This disposes of the first case. For the second case, take MATH. Once again MATH is even in the non-trivial case. Now MATH has positive sign in MATH, and MATH yields the term MATH . Similarly to the first case, this is MATH when MATH are both even or both odd, and MATH when exactly one of MATH is even. However, MATH has positive sign in MATH exactly when MATH are both even or both odd. This disposes of the second case. Now take MATH. Here the non-trivial case has MATH odd. Suppose first that MATH is odd. Take MATH which has positive sign in MATH . Now take MATH . This contains the term MATH . Now, MATH has positive sign in MATH exactly when MATH is odd also. Since MATH, MATH is negative, and all the MATH are positive except for MATH, which appears exactly when MATH is odd. This disposes of the sub-cases where MATH is odd. The remaining cases, for MATH even and for MATH, are similar. |
math/0011060 | Let MATH be of rank MATH, and suppose without loss of generality that the leftmost MATH columns of MATH form a basis of MATH. Since performing row operations on representations of classical oriented matroids does not alter the oriented matroid represented, we may assume that these MATH columns form an identity matrix in the first MATH rows, and that the rightmost MATH columns of the orthogonal complement form an identity matrix in the last MATH rows also. We swap these first MATH columns into the right-hand-side, and make the appropriate multiplications, obtaining a matrix MATH, where MATH . Now we see that MATH where MATH is the determinant of the appropriate MATH columns of MATH, and MATH if its argument has more or less than MATH elements. Now the result follows at once from REF . |
math/0011061 | Let MATH. We can assume without loss of generality MATH, MATH and MATH. Then there exists a neighborhood MATH of MATH and a real-analytic map MATH such that MATH. Extend MATH to a holomorphic function MATH, where MATH is some neighborhood of MATH in MATH. Define MATH by MATH, then MATH is a biholomorphism near MATH and MATH gives the complex coordinates with the required property. |
math/0011061 | It follows immediately from REF since Lagrangian submanifolds are totally real. |
math/0011061 | Certainly we can write MATH for some holomorphic MATH. Let MATH, then REF gives: MATH . From REF it follows that, along MATH, we have MATH, giving that MATH. REF implies that MATH . Therefore we obtain that MATH and MATH. The only holomorphic function satisfying this is precisely MATH. |
math/0011061 | We write the hermitian metric MATH that we are looking for as MATH, where MATH and MATH are real valued matrices, symmetric and antisymmetric respectively. In the basis MATH for MATH the corresponding NAME form can be written as a MATH matrix MATH . In order to prove the theorem we need to solve the following ``initial value" PDE problem: MATH . If we do the computations explicitly we see that REF form the following system of equations in the coefficients of MATH: MATH . Here the index MATH goes from REF to REF, while MATH are such that MATH. A solution is constructed in three steps: first we find one on MATH, then we extend it to MATH and finally to the whole MATH. Notice that for the first step we need only to look at REF , which do not involve derivatives with respect to MATH or MATH. For reasons that will become apparent later we do not assume that MATH is symmetric. Hence, we have four equations for twelve unknowns (nine from MATH and three from MATH). We choose arbitrarily all MATH's on MATH except MATH, with the only requirements that they satisfy the initial REF , they are real-analytic and they can be coefficients of a metric (for example, MATH). It is now easy to see that by differentiating REF by MATH and substituting into it equations from REF can be written in the form MATH where MATH and MATH are real-analytic coefficients, which depend on the way we arbitrarily extended the other MATH's. Notice that this is possible also because, with the given initial conditions, the coefficient of MATH in REF is different from zero near MATH. Now equations REF are four equations in the four unknowns MATH of the type whose solution is guaranteed to exist uniquely (at least locally) by the NAME theorem (as stated for example in CITE). The solution will also satisfy REF . In fact this is demonstrated by differentiating REF , MATH, MATH by MATH; REF , MATH, MATH by MATH and REF , MATH, MATH by MATH. From the results it follows that MATH on MATH. This shows that since REF holds on MATH it holds everywhere also on MATH. The second step is similar. We now extend this solution to MATH by looking at REF and the group REF . This time we have seven equations for twelve unknowns. We arbitrarily extend MATH and MATH as before. Then, for the symmetry of MATH, we also impose MATH and MATH. Differentiating REF by MATH, again we see that we can reduce the system to one which is solvable by the NAME theorem, where now the evolution variable is MATH and the initial domain is MATH. Notice that REF will still hold for this extended solution. To see this, first differentiate REF by MATH. Then substitute, into the result, REF , MATH differentiated by MATH and REF , MATH differentiated by MATH. Thus we obtain MATH which tells us that REF hold for all MATH since, by the first step, they hold for MATH. Again, the solution will satisfy also REF . This is shown by the same method as in the first step, except that we use REF instead of REF . The same procedure produces the third and last extension. We have ten equations for twelve unknowns. We impose MATH and MATH. Notice that, because of REF , MATH and REF , MATH, we cannot impose MATH. So let's treat them as separate unknowns, for the moment. As in the first and second step we find a solution to the system. Again, we must show that REF are still satisfied. To prove that REF holds we do exactly as in step two when we proved the same thing, except that we use REF , in place of REF respectively. We do the same to prove that REF holds, except that we use REF and we differentiate with respect to MATH instead of MATH. Notice now that from REF we also obtain REF . To prove that REF holds, we proceed as follows: differentiate REF by MATH, REF by MATH and REF by MATH. Then, by suitably combining the results, we obtain MATH which proves REF . The proof that also REF holds is just as in the previous steps. It remains to show that MATH. In fact it follows from the following: MATH where the first equality follows from subtracting REF , MATH and REF , MATH; the second from substituting REF , MATH and using the imposed symmetry of the other coefficients; the last one is just REF . The proof is now complete. |
math/0011061 | We apply REF to any standard complexification MATH of MATH. We view MATH as a triply periodic metric tensor in MATH, then we make sure that every choice involved in the three steps of the theorem is made to be triply periodic in the real part. Solutions will also be triply periodic in the real part, hence they define a NAME structure on MATH. REF also ensures that in this way we can describe locally all isometric special Lagrangian embeddings of MATH in some NAME manifold MATH. |
math/0011061 | In CITE NAME constructed examples of metrics on REF-torus which admit harmonic forms with MATH zeroes of index MATH and MATH of index MATH. Let MATH be one of these metrics and MATH the corresponding harmonic form with zeroes. As constructed by NAME, MATH is not real-analytic, but we can approximate it (in the MATH topology) with a real-analytic one MATH. The MATH-harmonic form MATH cohomologous to MATH will also approximate MATH and, by the stability of zeroes of non-zero index, MATH will have at least the same number of zeroes if the approximation is precise enough. To the pair MATH we can then apply REF to construct the NAME neighborhood MATH. This proves the first claim. CITE identified the moduli space of nearby special Lagrangian tori in MATH with a three dimensional submanifold MATH of MATH, the space of sections of the normal bundle. In fact, given MATH, the nearby special Lagrangian torus associated with MATH is just MATH. Via the identification MATH, MATH may also be interpreted as a submanifold of MATH. As NAME showed, its tangent space at the zero section is the vector space of harmonic REF-forms. Now let MATH be a curve in MATH, viewed in MATH, such that MATH and whose tangent vector at MATH is MATH, the harmonic form with zeroes. Then MATH in some MATH topology. Again, by the stability of zeroes of non-zero degree, this implies that, for sufficiently small MATH, MATH will have at least the same number of zeroes as MATH. Now if MATH is the section in MATH corresponding to MATH, the special Lagrangian submanifold MATH will obviously intersect MATH precisely at the zeros of MATH. This completes the proof. |
math/0011061 | We use the same notation as in REF . In particular let the initial metric MATH. In the following, MATH will stand short for MATH (so, for example, MATH will mean MATH, in real coordinates for MATH). Imposing the special Lagrangian condition on the horizontal slices corresponds to MATH for all t, where MATH is the value of the matrix MATH on MATH. A simple computation shows that the first one of these holds if and only if: MATH for all t. Now, since MATH is holomorphic, from this and from the NAME equations we deduce that: MATH which, by the definition of MATH, holds if and only if MATH for all MATH. This is only a condition on the initial data. Both conditions in REF are satisfied if and only if REF in the previous section become MATH . It is easy to see that the first equation of REF together with REF corresponds to the closure of MATH while the second one to the closure of MATH, so that MATH has to be harmonic with respect to to MATH. The first equation of REF gives also the independence of MATH on MATH. It is also easy to see that these conditions are sufficient to proceed to the construction of the NAME metric on MATH just by following the second step of REF . |
math/0011061 | Let MATH be a family of metrics on the standard REF-torus MATH of the following type: MATH with the only condition that MATH. REF and the comments that follow show that this family can be realized as a one parameter family of special Lagrangian submanifolds of some NAME manifold. We now show that in general the function MATH is not constant along this family. Choose as basis MATH for MATH the standard one. A computation shows that the forms MATH are harmonic and they satisfy REF for every MATH. Now, since the volume form is just MATH and the functions given depend only on MATH and MATH, we have the following: MATH where we also used the fact that MATH. Now, using also the condition on the determinant of MATH, this implies that MATH which in general, for arbitrary MATH and MATH depending also on MATH, is not constant in MATH. |
math/0011062 | The tangent space to the symplectic leaf through MATH is the image of MATH, which by definition is the inverse image under MATH of the tangent space to the conjugacy class through MATH. |
math/0011062 | Choose MATH and let MATH be the MATH component of MATH. Observe that the one-form MATH on MATH is left-invariant and takes the value MATH at MATH. Thus by REF. This says precisely that MATH is a Hamiltonian for the vector field MATH generated by MATH. |
math/0011062 | Immediate upon differentiating the map MATH. |
math/0011062 | Immediate from the definitions. |
math/0011062 | Observe MATH and MATH are the bisecting directions of the two components of the intersection MATH of the supersectors. (Recall from REF , for each MATH, MATH is asymptotic to MATH at MATH when continued within MATH.) Thus MATH is asymptotic to MATH within the component of MATH containing MATH. The exponentials dominate so we must have MATH unless MATH as MATH along MATH. This says, equivalently, that MATH. The argument for MATH is the same once the change in choice of MATH is accounted for. |
math/0011062 | For injectivity, suppose two connections MATH have the same NAME matrices and exponent of formal monodromy MATH. Let MATH be the associated formal isomorphisms (with the same normal form MATH) and let MATH for MATH and MATH. MATH is a holomorphic solution of the connection MATH asymptotic to MATH at MATH in MATH. Since the NAME matrices are equal, MATH on both components of the intersection MATH and so they fit together to define an isomorphism MATH between MATH and MATH on the punctured disc. By NAME 's removable singularity theorem it follows that MATH extends across MATH (and has NAME expansion MATH) and so is the desired element of MATH. NAME follows from a result of NAME (see REF), together with the (straightforward to prove) fact that any meromorphic connection germ is gauge equivalent to the germ of a meromorphic connection on the unit disc. |
math/0011062 | CASE: When continued in a positive sense, MATH becomes MATH on MATH, which will become MATH on continuing around, back to MATH. CASE: Observe that changing MATH to MATH corresponds to changing MATH to MATH and so, by REF , the canonical solution MATH changes to MATH, whence the result is clear. |
math/0011062 | By REF the local monodromy of MATH around zero is conjugate to MATH. However MATH has only one other pole in MATH: a first order pole at MATH (logarithmic/regular singularity). The connection MATH has residue MATH at infinity and this implies it has local monodromy conjugate to MATH (see for example, CITE). Clearly a simple positive loop around MATH is a simple negative loop around MATH. |
math/0011065 | For MATH . |
math/0011065 | The proof is similar to the proof of REF and is omitted. |
math/0011067 | The NAME group of MATH is generated by the MATH, MATH, such that in odd characteristic (respectively, even) MATH (respectively, MATH), and, for MATH, MATH. The element MATH is fixed by no element of MATH. Thus, MATH. |
math/0011068 | MATH. Suppose MATH are the vertices of a parallelogram. Then MATH implies MATH, so that MATH is not dps. Now suppose MATH, and MATH is interior to the line segment MATH. If MATH is the midpoint of the segment, then MATH, so MATH is not dps. Otherwise, we may assume that MATH is closer to MATH than to MATH. Then MATH will also be a lattice point on the line segment MATH, and MATH is the midpoint of MATH; again, MATH is not dps. MATH. For MATH, let MATH. Suppose MATH. Then MATH for MATH, and the line segment MATH contains the lattice points MATH. Thus, if REF holds and MATH, MATH, we have MATH. Suppose MATH. Then MATH for nonzero integers MATH, and MATH. Hence MATH, so MATH. Now the parallelogram condition in REF implies that MATH. MATH. If REF holds for MATH, and MATH with MATH, then MATH, and so MATH. This proves REF . |
math/0011068 | If MATH, then by the Pigeonhole Principle, there exist MATH so that MATH and MATH are component-wise congruent modulo REF. This means that MATH is also a lattice point, and it follows from REF that MATH is not a dps polytope. |
math/0011068 | For MATH, let MATH; for MATH, consider REF . Suppose now that MATH is a maximal dps polytope in MATH, MATH. Write MATH and define the (finite) set of differences MATH . Let MATH be a unimodular integer matrix such that if MATH, then MATH. (We shall construct such a MATH below.) We define the polytope MATH in MATH as follows. Let MATH and let MATH. If MATH, then MATH, hence MATH equals REF or REF. Thus, MATH lies either on the face determined by MATH, in which case MATH, or on the face determined by MATH, in which case MATH. It follows that MATH, so MATH. Now consider MATH; this consists of three disjoint sets of points: MATH where MATH. Since both MATH and MATH are dps, the sums in the first and the third set are distinct. For the second set, we suppose that MATH or equivalently, MATH . If MATH, then MATH, so MATH, which is the only possible way for REF to hold in a dps polytope. Otherwise, MATH, so MATH, a contradiction to the choice of MATH. Thus, MATH is a maximal dps polytope in MATH. We now construct a matrix MATH with the desired properties. First, let MATH and let MATH be the MATH matrix given below: MATH . (In words, the only non-zero entries in MATH are the diagonal, the superdiagonal, and the first entry in the second row.) It is easy to see that MATH is unimodular. We show now that for every MATH, at least one entry of MATH has absolute value greater than MATH. This implies that MATH, and will complete the proof. Write MATH and suppose that MATH is the smallest index such that MATH. (Such an index exists because MATH.) If MATH, then MATH, and hence MATH . If MATH, then MATH, so MATH and MATH. Finally, we remark that the same proof applies in the case MATH, if we take as our matrix the MATH submatrix at the upper left of MATH. |
math/0011072 | Let MATH, so MATH is MATH-avoiding if and only if MATH is MATH-avoiding. On the other hand MATH is an invertible function. Hence the theorem holds. |
math/0011072 | Immediately by REF where MATH is the set of all signed permutations with set symbols MATH and set signs MATH. So clearly MATH, also MATH by removing the sign MATH. Hence the theorem holds. |
math/0011072 | Let us define a function MATH by MATH . So MATH if and only if MATH avoids MATH, which is equivalent to MATH avoids MATH for all MATH. Hence MATH is a bijection, which means that the theorem holds. |
math/0011072 | By REF MATH. Let MATH, MATH, and let us consider the possible values of MATH: CASE: Let MATH, MATH, and MATH; so MATH if and only if MATH is MATH -avoiding, hence in this case there are MATH signed permutations. CASE: Let MATH; since MATH is MATH-avoiding, the symbols MATH appeared with sign MATH or MATH. Also the symbols MATH are MATH-avoiding, and can be replaced anywhere at positions MATH, hence there are MATH signed permutations, which means there are MATH signed permutations. So by the above two cases we obtain a recurrence relation satisfied by MATH for MATH, and MATH. Let MATH. By multiplying the recurrence by MATH, and summing up over all MATH, we obtain MATH where MATH is the generating function of MATH. Besides MATH, hence the theorem holds. |
math/0011072 | Immediately by the proof of REF , for MATH . Besides, MATH, and MATH, hence the corollary holds. |
math/0011072 | The theorem holds for MATH by REF . Now let MATH, so by REF therefore, MATH . Hence by induction the theorem holds. |
math/0011072 | To choose a path of MATH steps with MATH points in bad or good lines we have to: CASE: choose MATH places in the path. There are MATH possibilities. CASE: choose MATH points from bad or good lines. There are MATH possibilities. CASE: choose MATH points on free lines, where MATH. There are MATH possibilities. Hence, by REF the theorem holds . |
math/0011072 | By section MATH (symmetric operations) we have to prove the following two cases: CASE: Let MATH; for MATH, MATH; CASE: Let MATH; for MATH, MATH. The first, and the second cases are obtained immediately by REF , and REF , respectively. |
math/0011072 | By REF it is easy to obtain MATH, and REF immediately yields MATH, and MATH. Now let us prove MATH and MATH. CASE: Let MATH, MATH, and let us consider the possible values of MATH: CASE: Let MATH, MATH; MATH if and only if MATH. Hence in this case there are MATH signed permutations. CASE: Let MATH; since MATH is MATH-avoiding, the symbols MATH are not signed by MATH, and can be replaced anywhere at positions MATH. Hence, in this case there are MATH signed permutations. Therefore by the above three cases MATH staisfies the following relation: MATH . Besides MATH, and MATH, hence MATH holds. CASE: Let MATH, MATH such that MATH, and let us consider the possible values of MATH: CASE: Let MATH; MATH if and only if MATH. Hence in this case there are MATH signed permutations. CASE: Let MATH; MATH if and only if MATH is a signed permutation with symbols MATH and signs MATH. Hence, in this case there are MATH signed permutations. Therefore by the above three cases MATH staisfies the following relation: MATH . Besides MATH, and MATH, hence MATH holds. |
math/0011072 | Immediately by the proof of REF , and by REF of Main Theorem, we claim these cases. Cases MATH, MATH: Let MATH, MATH, and let MATH. Also let us define MATH to be the set of all MATH such that MATH is signed by either MATH or MATH. Now we define a function MATH by reversing all the MATH where MATH. Hence by definitions, MATH is a bijection, which means that MATH. On the other hand, immediately by Main Theorem part MATH we get MATH which means by case MATH that the theorem holds. |
math/0011072 | CASE: Let MATH, MATH, MATH, and let us consider the possible values of MATH: CASE: Let MATH, MATH; so MATH if and only if MATH. Hence in this case there are MATH signed permutations. CASE: Let MATH; since MATH is MATH-avoiding, the symbols MATH are not signed by MATH, and the symbols MATH are not signed by MATH. Hence there are MATH signed permutations, which means by REF of Main Theorem that there are MATH signed permutations. Therefore MATH staisfies the following relation: MATH . Besides MATH, and MATH, hence case MATH holds. CASE: Let MATH, MATH, and MATH such that MATH. Let us consider the possible values of MATH: CASE: Let MATH; MATH if and only if MATH. Hence in this case there are MATH signed permutations. CASE: Let MATH; since MATH is MATH-avoiding, all the symbols in MATH are not signed by MATH, and the symbols in MATH are not signed by MATH. Hence there are MATH signed permutations, which means by REF of Main Theorem that there are MATH signed permutations. Therefore MATH staisfies the following relation: MATH . Besides MATH, and MATH, hence case MATH holds. CASE: Similarly to the case MATH for MATH such that MATH, we consider the possible values of MATH, and get the same result. |
math/0011072 | Let MATH, MATH, MATH, and let us consider the possible values of MATH: CASE: If MATH where MATH, then MATH if and only if MATH. Hence in this case there are MATH signed permutations. CASE: If MATH then, since MATH avoids MATH, the symbols MATH are not signed by MATH, and the symbols MATH are not signed by MATH. Hence there are MATH signed permutations, which means by REF of Main Theorem that there are MATH signed permutations. Therefore MATH staisfies the following relation: MATH . Besides MATH, and MATH, hence the theorem holds. |
math/0011072 | By REF , and REF we get MATH, MATH. The rest follows from definitions, and by the first elements of MATH where MATH set of two signed patterns from REF . |
math/0011076 | This follows from REF. |
math/0011076 | By assumption, MATH for all MATH, MATH and MATH for all MATH. The conditions MATH for a pre-Hilbert module are obviously satisfied. Since MATH the norm that comes from the MATH- valued inner product equals the operator norm. Hence MATH is a NAME module. We have MATH. It is a general feature of NAME modules that any MATH may be written as MATH for some MATH CITE. Hence MATH. REF follows. Since MATH, these three sets are equal. |
math/0011076 | We have MATH and MATH for all MATH, MATH. REF shows that MATH is isometric, so that MATH is closed. Thus MATH is a concrete NAME MATH- module and MATH is an isomorphism with respect to the NAME MATH- module structure of REF . MATH is essential because MATH is generated by elementary tensors MATH with MATH and MATH. Suppose that MATH is a concrete NAME MATH- module. The map MATH is isometric (hence well-defined) by REF and equivariant. If MATH is essential, then the range of MATH is dense, so that MATH is unitary. We compute MATH for all MATH, MATH. That is, MATH. |
math/0011076 | It is clear that MATH is a MATH- subalgebra of MATH. Let MATH be the closed linear span of MATH. By construction, MATH is closed and MATH. REF implies MATH and hence MATH. If MATH, then MATH. Hence MATH is a closed ideal in MATH. Conversely, if MATH, then MATH by REF. Consequently, MATH iff MATH and MATH. We define a MATH- homomorphism MATH by MATH for MATH, MATH. If MATH is essential, then MATH is injective because if MATH, then MATH vanishes on the dense subspace MATH, so that MATH. In general, at least the restriction of MATH to MATH is injective because MATH implies MATH and hence MATH. We have MATH for all MATH. Hence MATH and MATH equals the map MATH. REF implies that MATH is essential iff MATH is essential. Therefore, if MATH is essential, then MATH extends uniquely to a strictly continuous, unital, injective MATH- homomorphism MATH from MATH to MATH. The range of MATH is contained in MATH because MATH is an ideal in MATH. Since MATH, the map MATH is an isomorphism onto MATH. |
math/0011076 | By REF is equivalent to MATH. REF asserts that all unitaries MATH are contained in MATH. Since any element of MATH may be written as a sum of four unitaries, the first two conditions are equivalent. In the proof of REF , we observed that MATH satisfies MATH and MATH iff MATH. Hence REF is equivalent to MATH as well. |
math/0011076 | By construction, MATH is a closed linear subspace and MATH. The assumption MATH implies MATH. Suppose that MATH is dense in MATH. Since MATH is dense in MATH, the subset MATH is dense in MATH. Therefore, MATH is dense in MATH. |
math/0011076 | It is evident that MATH is relatively continuous. Let MATH and MATH. Since MATH is equivariant, REF yield MATH . This implies MATH. Thus MATH is relatively continuous. The above computation shows MATH. Since MATH is dense in MATH and MATH is a NAME MATH- module, MATH is dense in MATH. It follows that MATH and MATH are dense subsets of MATH. Therefore, MATH. |
math/0011076 | Let MATH be a concrete NAME MATH- module. It is evident that MATH is complete and relatively continuous. REF asserts MATH. Conversely, let MATH be complete and relatively continuous. Define MATH. Then MATH. We claim that MATH. Let MATH, we want to show that MATH. Let MATH be an approximate identity as in REF . Since MATH is the closure of MATH, there is a sequence MATH with MATH in operator norm. REF implies that MATH for all MATH. Hence MATH because MATH is complete. Since MATH is an approximate identity for MATH, we have MATH for all elements MATH of a NAME MATH- module. In particular, MATH converges towards MATH. Together with REF, this means that MATH in the norm MATH. Hence MATH. This proves that MATH. If MATH is essential, then MATH is dense in MATH. Hence MATH is dense in MATH. Conversely, if MATH is dense, then MATH is essential by REF . The last assertion of the theorem follows from REF . |
math/0011076 | The bijection between complete, relatively continuous subspaces and concrete NAME MATH- modules in REF preserves inclusions. Hence MATH is the smallest complete subspace containing MATH. |
math/0011076 | By REF , we have MATH for a concrete NAME module MATH over MATH. Let MATH. By NAME 's Factorization Theorem, the map MATH extends to a linear operator MATH that is continuous with respect to the strict topology on MATH and the norm topology on MATH. We have MATH whenever there is MATH with MATH and MATH. Using REF, we conclude that MATH for all MATH, MATH. Furthermore, we have norm estimates REF. Since the map MATH is strictly continuous, we have MATH for MATH. Therefore, the action of MATH on MATH is continuous. Similarly, if MATH is an approximate identity of MATH, then MATH in the norm MATH. Hence NAME 's Factorization Theorem yields MATH. |
math/0011076 | Fix MATH. If MATH, then MATH is well-defined and MATH . Hence MATH extends to a bounded operator MATH. The problem is to show that MATH is adjointable. This means that the set MATH equals MATH. Since MATH is a closed subspace, it suffices to prove that MATH is dense. If MATH, MATH, then MATH and hence MATH. Elements of this form exhaust MATH by REF. Since MATH is dense in MATH, REF yields that MATH is dense in MATH as desired. |
math/0011076 | Since MATH is dense in MATH, we conclude that MATH is dense in MATH and that MATH is dense in MATH. Moreover, REF yields that the space MATH defined in REF equals the space of MATH- continuous operators. Hence the assertions of the first paragraph follow from REF if we take the homomorphism MATH defined there. Since MATH, the closed linear span MATH of MATH is an ideal in MATH. We may view MATH as an imprimitivity bimodule for MATH and MATH. That is, MATH and MATH are NAME equivalent. The last assertion follows immediately from REF and the definition of MATH. |
math/0011076 | Let MATH, MATH. Since MATH is essential, MATH. Hence MATH. The same simple computation that yields REF shows that MATH where MATH is defined by REF. As a result, MATH and MATH . By definition, MATH is the closed linear span of MATH. The discussion of MATH above shows that the closed linear span of MATH equals MATH. Hence MATH equals the closed linear span of MATH. REF yields that the map MATH is an isometry of NAME modules over MATH. Hence MATH is relatively continuous and satisfies REF. |
math/0011076 | The isomorphism MATH maps MATH onto the linear span of MATH. |
math/0011076 | The assertion follows from REF . |
math/0011076 | Let MATH. Then MATH is the completion of MATH. Since MATH is compatible with completions, MATH is the completion of MATH. The linear span of MATH is a dense subspace of MATH by REF . Hence the completion of MATH equals MATH. This proves MATH. Conversely, let MATH and MATH. Then MATH is the completion of MATH. Hence MATH is the closed linear span of MATH. Let MATH be the closed ideal generated by MATH. NAME 's Factorization Theorem and REF show that MATH . Let MATH be the closed ideal generated by MATH. If MATH, then we may view MATH as a NAME module over MATH. Hence MATH. Conversely, if MATH, then MATH and hence MATH. Therefore, MATH . If MATH, then MATH for all MATH by REF. Therefore, MATH is annihilated by the canonical map MATH. If MATH is exact, this implies that MATH. Hence MATH and thus MATH. Suppose that MATH is not exact and that MATH is an invariant ideal for which MATH is strictly smaller than the kernel MATH of the map MATH. View MATH as a NAME module over MATH and let MATH be the associated continuously square-integrable NAME module according to REF . Then MATH. This implies MATH for all MATH by REF. Hence MATH. However, MATH is not contained in MATH. Hence MATH. |
math/0011076 | The proof is a straightforward exercise in topology. We may assume without loss of generality that MATH and MATH are quasi-compact, not just relatively quasi-compact. For MATH, MATH, there are open neighborhoods MATH, MATH of MATH and MATH such that MATH for all MATH outside a compact subset of MATH, because the action on MATH is proper. By quasi-compactness, for fixed MATH finitely many of the open sets MATH cover MATH. Let MATH be their union and let MATH be the intersection of the corresponding MATH. Then MATH is an open covering of MATH. Finitely many of these sets suffice to cover MATH. Let MATH be their union and let MATH be the intersection of the corresponding open neighborhoods MATH of MATH. These sets have the desired properties. |
math/0011076 | Let MATH be open and relatively compact and let MATH. By NAME 's Factorization Theorem, MATH is a closed linear subspace of MATH. The subset MATH is open and relatively compact by REF . We may write elements of MATH in the form MATH, MATH with MATH, MATH. REF yields MATH . Hence MATH. It follows that MATH. Since MATH is dense in MATH, MATH is dense in MATH. REF yields MATH, so that MATH is a dense, relatively continuous subspace and MATH is square-integrable. Since MATH, it follows that MATH for all MATH with some MATH. Hence the norms MATH and MATH are equivalent on MATH. Let MATH be a dense, complete, relatively continuous subspace. We claim that MATH contains MATH, so that MATH. REF implies that MATH. Since MATH is dense in MATH, MATH is dense in MATH with respect to the norm MATH and hence also with respect to the norm MATH because these two norms are equivalent on MATH. Since MATH is complete, it follows that MATH. Hence MATH. If MATH, then MATH and MATH by REF . Assume conversely that MATH and MATH. We claim that MATH. Since MATH is a dense MATH- ideal, there is an approximate identity MATH for MATH with MATH for all MATH. Thus MATH for all MATH. Let MATH be the completion of MATH. REF applied to MATH yields that MATH in the norm MATH. Hence MATH as asserted. Therefore, any relatively continuous subset MATH is contained in MATH. |
math/0011076 | Let MATH be a NAME MATH- module. By REF , there is a unique dense, complete, relatively continuous subset MATH. Hence there is no difference between continuously square-integrable NAME MATH- modules and NAME MATH- modules. REF shows that isomorphism classes of NAME modules over MATH and NAME MATH- modules correspond to each other bijectively. Since MATH is unique, we have MATH for all MATH. Hence MATH by REF . |
math/0011082 | Let MATH. Using the exact sequence MATH and the fact that MATH, we see that it suffices to find a canonical isomorphism MATH . To analyze the quotient, we begin with the exact sequence MATH . Using MATH, we get an exact sequence MATH where MATH is the inclusion. Applying MATH then gives an exact sequence MATH . As for the term MATH, the fact that MATH via the natural inclusion shows that MATH and it is easy to check that this isomorphism is compatible with the inclusion of MATH in MATH given above. Thus we have identified MATH with MATH. |
math/0011082 | Note that MATH. Applying MATH to the exact sequence MATH then gives the result. |
math/0011082 | If MATH, then MATH is normal along MATH, so that MATH there, and MATH is locally free of rank one in a neighborhood of MATH. Now suppose that MATH. If MATH is normal at MATH, then again MATH is locally free of rank one in a neighborhood of MATH. Suppose that MATH is not normal at MATH. We first find locally a rank two subbundle of MATH corresponding to the two branches of MATH. There is a small analytic NAME neighborhood MATH of MATH in MATH such that MATH, where MATH, MATH is connected, and MATH. By hypothesis, if MATH is sufficiently small, we can write MATH, where MATH and MATH are smooth and meet transversally along some component MATH of MATH. Since MATH is finite and flat and the local degree of MATH at MATH is MATH, MATH is an isomorphism of MATH onto MATH. Corresponding to MATH, we have MATH, where the rank of MATH is MATH. The restriction of MATH to MATH defines line bundles MATH on MATH and hence on MATH. Here MATH is the preimage in MATH of the set of points MATH where MATH. Since MATH has a nonzero section for each MATH, and has exactly one if MATH, it is easy to see that there is a nonzero map MATH. There is an effective divisor MATH on MATH such that the map MATH extends to a map MATH which does not vanish in codimension one. The quotient is thus of the form MATH, where MATH is a codimension two subscheme of MATH (or empty). Clearly MATH restricts to the trivial line bundle on every fiber MATH and hence is pulled back from a line bundle on MATH. After shrinking MATH, we may assume that this line bundle is trivial. There is a homomorphism from MATH to MATH which is an isomorphism away from MATH. Thus as before MATH and as before we may assume that the line bundle MATH is trivial. By semistability, the scheme MATH is the preimage of a subset MATH of MATH, which is either empty or of codimension MATH. Thus in the complement of MATH, there is an exact sequence MATH . For MATH, MATH is a direct sum of two line bundles of degree zero. It follows that the extension MATH is split for every MATH. For each MATH, let MATH be a NAME neighborhood of MATH. Now MATH . By the above discussion about MATH, if MATH is the corresponding global section of MATH, then the image of MATH is zero in the fiber of MATH over MATH for every MATH. To see that MATH is actually zero, it suffices to prove that MATH is supported along MATH and that its length is one there. Each line bundle MATH defines a morphism MATH from MATH to MATH, and the morphism MATH is given by MATH. Let MATH, so that MATH. Then MATH is the pullback to MATH via MATH of MATH. The assumption that MATH and MATH meet transversally is the assumption that MATH is smooth at the origin and hence that MATH as reduced divisors. By flat base change, MATH . It is a standard result that MATH is supported at MATH and has length one there. It follows that MATH is supported along MATH and has length one there. Hence MATH. Thus MATH. It follows that, at least over the complement of the set MATH of codimension at least two, MATH, which is a direct summand of MATH, is MATH. Hence, in the complement of the codimension two subset MATH of MATH, MATH is a line bundle MATH over the normalization MATH of MATH. Since MATH is normal, MATH extends uniquely to a rank one reflexive sheaf MATH on MATH. We have constructed MATH and an isomorphism from MATH to MATH over MATH, where now MATH is some global subscheme of MATH of codimension at least two. Since the codimension of MATH in MATH is at least two, if MATH is the inclusion, then it follows from REF that MATH. A similar result holds for MATH, since MATH is reflexive. Thus the isomorphism from MATH to MATH over MATH extends to give an isomorphism from MATH to MATH. The final statement follows since every rank one reflexive sheaf on a smooth scheme is a line bundle. |
math/0011082 | To prove REF , let MATH be the torsion module corresponding to MATH, that is, MATH. Then, by the equivalence of categories given by the NAME correspondence. MATH . To see REF , the group MATH classifies the set of isomorphism classes of extensions MATH, and via the NAME correspondence this set is the same as the set of isomorphism classes of extensions MATH, which is classified by MATH. But MATH via the resolution MATH. |
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