paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0010271 | Applying REF , we get that MATH and MATH commute with each MATH. Since MATH are linearly independent, each MATH is a linear combination of the MATH's and so MATH and MATH commute with each MATH. |
math/0010271 | Applying REF (and using its notation) we get that if MATH, then MATH and so MATH and MATH. Next as MATH for MATH, we get MATH for MATH. On the other hand, since MATH we have MATH so that MATH and therefore MATH . Hence there must be a gap in the spectrum of MATH that contains the interval MATH. Next, if we select any M... |
math/0010271 | Suppose that MATH exposes MATH in MATH so that there is a number MATH such that MATH and MATH if MATH and MATH. Write MATH and suppose MATH is a sequence in MATH such that MATH. As MATH is compact, we may assume that MATH converges to a vector MATH. Since MATH, MATH exposes MATH and MATH, we must have that MATH. Hence,... |
math/0010271 | Write MATH for the closure of MATH and observe that since MATH is an isolated point in MATH, MATH. Further, if MATH denotes the closed convex hull of MATH, then MATH. Indeed, if we had MATH, then MATH would be an extreme point of MATH. But in this case we would get that MATH by NAME 's ``converse" to the NAME - NAME th... |
math/0010271 | CASE: Suppose MATH is an isolated extreme point in MATH. By REF , there is a unique projection MATH such that MATH. Next, applying REF and parts MATH and MATH of REF , we get linearly independent spectral pairs-MATH such that MATH . Since MATH, we have MATH . Since these spectral pairs are linearly independent, a strai... |
math/0010271 | If MATH is abelian and has finite dimension, then each projection in MATH is diagonal. Hence MATH contains a finite number of projections and since every extreme point in MATH is the image under MATH of a projection MATH, MATH is finite. For the converse, if MATH has a finite number of extreme points, then each extreme... |
math/0010271 | Suppose that the closure of the isolated points of MATH is not all of MATH. By CITE each open subset of MATH is a locally compact NAME space. Hence, there is a nonempty open set MATH in MATH with no isolated points. Since MATH is countable, MATH is countable and we may write MATH. Set MATH for each MATH. As MATH is NAM... |
math/0010271 | Write MATH, let MATH denote the map from MATH to MATH defined by the formula MATH and write MATH. It is clear that MATH. Next suppose MATH is countable and fix MATH and MATH. For each extreme point MATH in MATH write MATH so that MATH is a closed face of MATH. Now each MATH is nonempty and so contains extreme points. S... |
math/0010271 | We have that MATH for some spectral pair MATH . By REF . Now suppose that MATH has dimension greater then zero. In this case we must have MATH and so MATH. Hence, MATH is transverse to the isotrace slices of MATH and so MATH is true. The remaining assertions now follow easily from MATH. |
math/0010271 | Fix a closed proper line segment MATH in the boundary of MATH. As MATH is in the boundary of MATH, it is in the boundary of MATH and so is contained in a minimal face MATH in MATH. Since faces in MATH of dimension greater than zero are transverse to the isotrace slices by REF , MATH must be two dimensional. Hence MATH ... |
math/0010271 | As MATH is isolated among the extreme points of MATH, there are distinct two dimensional faces MATH and MATH of MATH such that MATH by REF . Next, since MATH and MATH have maximal dimension in MATH, they must be exposed and so we may write MATH . Further, since the MATH's intersect MATH in adjoining line segments, we m... |
math/0010271 | Write MATH and MATH. To show that MATH is abelian, it suffices to show that MATH is central for each spectral pair of the form MATH. Indeed in this case, by a standard argument in measure theory, we get that the spectral projections for MATH and MATH are central and so MATH and MATH commute. So now let MATH denote such... |
math/0010272 | If MATH, where MATH, then MATH . The fact that MATH is indeed the NAME involution follows from the details of the calculation in CITE. |
math/0010272 | Every cusp on MATH can be obtained from MATH by means of MATH and the NAME involution. It remains to recall now the value of MATH at MATH, which is computed by means of the NAME involution. |
math/0010272 | Observe first that the relations among MATH imply that MATH satisfies the differential equations up to first order in MATH. More explicitly, by summing REF up for all triples MATH and using the fact that MATH, we obtain the necessary relation MATH . On the other hand, the differential equations lead to recursive relati... |
math/0010272 | Observe that this ring is MATH-graded by the subscripts of MATH. The quadratic relations REF imply that the dimension of each MATH-graded component of MATH is at most MATH. Indeed, a spanning set is constructed inductively by multiplying the previous spanning set by MATH and by adding any monomial of the correct MATH-w... |
math/0010272 | The differential equations for the MATH's imply the following differential equations for the MATH's: MATH . To derive the first set of differential equations, we use the formula for the derivatives of MATH and then collect terms with the common factor MATH (where MATH is the index of summation in REF ). We then split o... |
math/0010272 | By REF we can use MATH to map MATH to MATH. We observe that each MATH-graded component of degree MATH is mapped onto a space of dimension at least MATH, because polynomials of degree MATH in MATH and fixed MATH-grading can have an arbitrary singular part in their NAME expansions. The only exception is the zero graded c... |
math/0010272 | Let MATH be a solution to the system REF. Then the results of REF allow us to define a dimension one subscheme MATH of MATH cut out by the ideal MATH. Notice that points MATH with coordinates MATH, where MATH, lie on MATH. Moreover, MATH defines a map from the disjoint union of the neighborhoods of MATH non-singular po... |
math/0010272 | We need to show that the scheme MATH cut out by these quadratic relations is smooth. Let MATH be a closed point of MATH that is not a cusp. Let MATH be a MATH-point of MATH that reduces to MATH. NAME though MATH and MATH are defined only up to homothety (multiplication by MATH and MATH, respectively), we shall make a c... |
math/0010272 | This is immediate by using relations REF to eliminate all MATH's in REF. Details are left to the reader. |
math/0010273 | It suffices to prove the expansion for MATH, so we can assume MATH. This is a special case of a result proven in CITE using the NAME transform. NAME give a different proof using methods found in CITE. It also suffices to consider a single face, say MATH. NAME work in a product neighborhood MATH and ignore the cutoff MA... |
math/0010273 | Since MATH is unique up to MATH, independence follows from the invariance of MATH under MATH. To study the MATH dependence, for simplicity let us specialize to the case MATH, that is, a manifold with boundary MATH with boundary defining function MATH. It suffices to consider an index MATH where MATH, and show that MATH... |
math/0010273 | The leading terms in the boundary expansion of MATH are MATH where MATH. Observe that for any MATH where MATH and MATH. Provided that MATH we can set MATH so as to have MATH . The remaining terms MATH are obtained by an obvious induction, with the requirement that MATH ensuring that no zeroes occur in denominators. The... |
math/0010273 | The first property follows from a standard wave-front set argument. To prove the second, consider MATH and MATH. We can compute MATH by first pull MATH up to MATH through the right, then multiplying by the lift of MATH times the Riemannian density (in the right factor), then pushing forward to MATH through the left. Bu... |
math/0010273 | It is most convenient to argue in the model for MATH given by the upper half-space blown up at the origin. Sticking to our convention that MATH is the boundary defining function, we use coordinates MATH. The model Laplacian is MATH . With radial coordinate MATH, the defining functions for the faces of MATH are MATH and... |
math/0010273 | Let MATH so that MATH . Assuming that MATH, we can apply the argument from REF to get MATH such that MATH . Now If we define MATH then from REF the relation to MATH is MATH . By solving a linear system for the coefficients MATH in terms of the MATH we can rewrite REF as MATH . Suppose MATH satisfies MATH for MATH. This... |
math/0010273 | That MATH follows immediately from MATH, so we concentrate on MATH. The asymptotic properties of MATH depend only on the behavior of MATH near MATH and/or MATH. So we can specialize to a particular coordinate neighborhood with coordinates MATH and assume that all functions are compactly supported within this neighborho... |
math/0010273 | Integrating by parts gives MATH . In taking the limit MATH, the boundary term disappears by REF , yielding MATH . Using MATH, we rewrite this as MATH . Thus we can estimate MATH . We take care of the MATH on the right by an estimate of the form MATH and the result follows. |
math/0010273 | Let us single out the operator appearing in the radiation condition by defining MATH . We first note that the only issue is to control MATH at the boundary. Because for MATH, MATH for any MATH, and by REF , we have MATH . So we will work in a product neighborhood of MATH in which the metric has the form REF . Since MAT... |
math/0010273 | NAME argued in CITE that the existence of a local parametrix in MATH as in REF shows that there can be no solution of MATH lying in MATH. So the point here is to use the radiation condition to argue that MATH would have to be MATH. Let MATH be a cutoff in MATH. Choose MATH with MATH for MATH, MATH for MATH, and MATH fo... |
math/0010273 | Let MATH. Suppose that MATH. Then we can define a new sequence MATH so that MATH and MATH . By REF , MATH is a bounded sequence in MATH. Since the inclusion MATH is compact, any subsequence of MATH has a subsequence converging in MATH. The limit MATH of and such subsequence satisfies MATH and by REF it also satisfies M... |
math/0010273 | Let MATH with locally defined right parametrix MATH as in REF . For notational convenience we'll consider only the global case, since the introduction of cutoff functions for the local case is a simple matter. Assume MATH is as stated. The idea is to apply the transpose of the equation MATH to MATH. To justify this req... |
math/0010273 | A different choice of boundary defining function could always be written MATH, where MATH and is strictly positive. Thus we are concerned with the difference MATH for MATH. For simplicity we can assume MATH, so that we want to show that MATH . We use an induction similar to the proof of REF . Defining the commutators M... |
math/0010273 | It is easy to verify that vector fields in MATH lift to vector fields in MATH by considering local coordinates. For example, in the interior of the crossover face the coordinates MATH are valid. The lifts of the vector fields MATH and MATH to MATH are MATH . At the boundary of the regular and crossover faces, we use co... |
math/0010273 | Extending MATH smoothly into the interior in some arbitrary way, we have MATH because MATH is the indicial root. As in the proof of REF , let us use coordinates MATH, MATH. By the lifts of vector fields computed there we see that MATH where MATH. This shows that the asymptotic expansion of MATH at the regular face may ... |
math/0010273 | The first part of the construction is the symbolic inversion along the lifted diagonal in MATH. This does not involve MATH and so can be done just as in CITE. We obtain MATH such that MATH . The kernel of MATH, lifted to MATH, lies in MATH. The next step in the construction is to try to solve away this error term at th... |
math/0010273 | For convenience let us use subscripts to denote lifts by the MATH's, for example, MATH. Multiplying REF on each side by MATH gives MATH . If MATH is a smooth half-density on MATH, then MATH where MATH is a smooth half-density on MATH. Let MATH be the blow-down. By computing the codimensions of the submanifolds to be bl... |
math/0010273 | With MATH, we will compute MATH from REF . MATH and MATH are most succinctly described with a chart of index sets akin to REF . Fortunately we can consider the regular and crossover faces separately. At the regular faces we have: MATH and at the crossover faces: MATH . Under MATH the faces which map to the interior are... |
math/0010273 | Let us start with the second formula in REF . Suppose that MATH . As in the proof of REF , we examine the behavior of the lifts to MATH. At the regular faces we have: MATH . In order to apply REF we consider MATH, which is the push-forward of a b-density on MATH with index set MATH at MATH, MATH, and MATH, and MATH oth... |
math/0010273 | Let MATH as an element of MATH. From REF we immediately see that MATH . Now consider REF . By construction MATH, which is smooth up to the front face, and since MATH lifts to MATH we conclude that MATH . This means that MATH must annihilate all of the terms in the asymptotic expansion of MATH at the front regular face,... |
math/0010273 | First note that MATH lifts under MATH to MATH, as may be easily checked in local coordinates. Observe that MATH . Because of the infinite order of vanishing at the two front faces, we deduce that MATH . Let MATH be radial vector fields at the left/right regular and left/right crossover boundary faces of MATH, respectiv... |
math/0010273 | Let MATH and MATH. As in the proof of REF , we use a chart to describe the boundary behavior of MATH. For the regular faces we have: MATH and at the crossover faces: MATH . The behavior at the middle faces, together with REF , shows that the push-forward is well defined. And the form of MATH is deduced from REF . The s... |
math/0010273 | We will prove only the second of the three formulas. The first follows by similar argument because of the cancelation of the interior singularity as in REF . The third formula follows by an analogous but even simpler argument on MATH. Let MATH, where MATH is a section of MATH. As noted in the remarks following REF , th... |
math/0010273 | Since the index family MATH implies MATH decay at the left face, the composition is represented by a convergent integral. The necessary estimates can be obtained by differentiating under the integral using dominated convergence. |
math/0010273 | Using REF we solve for MATH and asymptotic to MATH near MATH, such that MATH. Then let MATH, which is in MATH by REF . Thus MATH has the stated properties. |
math/0010278 | All claims follow by an easy induction. Using the skein relation REF we express a link as the sum or difference of two links which are either of shorter word length or are ``less linked". |
math/0010278 | Let MATH, MATH be a braid with exponent sum MATH satisfying our conditions. The polynomial MATH can be written as MATH for some polynomials MATH of degree less than MATH in MATH. Now the coefficients of MATH in MATH is a (framing independent) NAME invariant, which depends on MATH and MATH. Since, by our condition, all ... |
math/0010278 | That the given dimensions form an upper bound, follows from REF . It remains to construct the following: Let MATH be the degree. We will show that there are braids in MATH such that the dimension of the coefficients of MATH is MATH. For a braid MATH the polynomial MATH satisfies: MATH . Furthermore, we have for arbitra... |
math/0010278 | By REF we know that the polynomials MATH and MATH are polynomials in MATH of degree less than or equal to MATH. Let MATH (MATH, respectively) be the exponent sum of the braid MATH (MATH, respectively) on MATH (MATH, respectively) strands. We know that the NAME polynomial MATH - with our reparametrization - is given by ... |
math/0010282 | To show REF it suffices to show that MATH sends the skein relation MATH to MATH. We have MATH because MATH and MATH. The last statement follows from the fact that if MATH is an oriented framed link and MATH its diagram such that the framing of MATH is the flat framing of MATH then MATH where the sum is taken over all c... |
math/0010282 | Part MATH follows from the fact that relations REF follow from the NAME relations, REF. If the trivial link MATH is linearly independent in MATH as is the case for MATH or more generally for MATH (that is, the product of a surface and the interval), and as is conjectured for any MATH-dimensional manifold, then we have ... |
math/0010282 | Any tangle of REF is symmetric with respect to MATH, MATH, and MATH axes. |
math/0010282 | The fact that elements of MATH and MATH are MATH-algebraic follows from the definition. For a MATH-bridge tangle, we start from the MATH-tangle which is equal to MATH for MATH or MATH, and then we add crossings one by one (compare Figure MATH, the figure is drawn to stress the fact that we deal with a REF-bridge tangle... |
math/0010282 | It is sufficient to show that MATH, MATH, and MATH using the fourth skein relations. MATH. MATH. MATH. |
math/0010282 | Let MATH be the number of crossings of MATH. If MATH, then there exists an element MATH of MATH such that MATH is strongly equivalent to MATH. If MATH, then MATH is an alternating expression of MATH or MATH. Then, MATH is dot equivalent to MATH from REF . There is no irreducible element of MATH with more than four cros... |
math/0010282 | We prove the theorem by an induction on the number of crossings of the element. If MATH, then clearly the statement holds. Assume that the statement holds in the case MATH fewer than MATH. Consider the case MATH. If the element is reducible, then it is generated by elements with fewer crossings than MATH, which can be ... |
math/0010282 | If both elements of the pair are MATH-braids, then the statement follows from REF . The other cases are easy to be checked. For simplicity, we ignore trivial components in the consideration. |
math/0010282 | We need to show only for MATH and MATH. From Figure MATH, we have MATH . |
math/0010284 | We can follow the same argument as we did for REF . Let MATH be the number of isogeny classes of MATH-dimensional abelian varieties over MATH such that MATH modulo MATH. Then our bounds on MATH and MATH give us MATH . On taking the limit, the theorem follows. |
math/0010288 | We choose an open cover MATH for MATH of coordinate balls of radius MATH and assume this is such that on each ball we can choose coordinates MATH with MATH represented by MATH. Further we assume that in fact the balls MATH also cover MATH. In each MATH we look at the disks MATH. We may assume that as MATH varies these ... |
math/0010288 | The kernel of any degenerate MATH must be at least MATH-dimensional and so have non-trivial intersection with MATH, which is clearly a contradiction. MATH . |
math/0010288 | For any MATH, MATH and MATH satisfy the conditions of the above lemma on MATH with MATH as MATH is orientation preserving and approximately preserves the NAME vector field. MATH . Now, since MATH and MATH agree near MATH, we can apply NAME 's method to find an isotopy of MATH, fixed near MATH, which generates a symplec... |
math/0010290 | We prove only the invariance of MATH. It follows from REF that we have MATH . |
math/0010290 | We can easily show that MATH . So we have proved the lemma. |
math/0010290 | We can show that MATH . The last equation is realized by using the NAME operators. In fact, we can show that MATH . Here, the NAME operator MATH on the irreducible representation space MATH is defined by MATH and acts as the constant MATH on MATH. Thus we have proved the proposition. |
math/0010291 | This is an immediate consequence of the corresponding property for the law of the pinned sites, given in REF below. |
math/0010291 | REF . is very simple: In the square-well REF is equivalent to MATH which follows from NAME 's inequality. The MATH-pinning case is similar. REF is easy, too: For any local increasing function MATH (of the random set MATH) with support inside MATH, one has MATH . Translation invariance is a simple consequence of this. I... |
math/0010291 | We recall that the variance of the Gaussian field can be written MATH where MATH is the law of the random walk in MATH, with transition probabilities MATH, starting at MATH. Inserting this in REF, we get MATH . (Remember that MATH.) Taking the expectation with respect to MATH inside, we get MATH . The corollary then fo... |
math/0010291 | We suppose first, for simplicity, that MATH is connected. The changes for the general case are the same as those described in CITE, and we'll indicate their effects on our bounds at the end of the proof. Let MATH, and define MATH as the union of MATH and all its nearest neighboring cells in MATH; let MATH be the larges... |
math/0010291 | CASE: If MATH, FKG property of MATH implies MATH for any set MATH, see REF. Therefore MATH which proves the claim since the probabilities are bounded by MATH. CASE: Using the expansion REF, we can write MATH . We therefore have to bound the conditional probability. This can be done as follows: MATH where we used REF to... |
math/0010291 | This follows from a standard subadditivity argument, since MATH where MATH, and the inequality follows from the FKG property. The constant MATH, which depends only on MATH, takes care of the possible discrepancy between MATH and MATH. |
math/0010291 | REF is proved in CITE. REF follows from REF by a standard argument, see CITE. The proof there is for the nearest neighbor random walk only, but it can be easily adapted to cover the more general case considered here. REF follows from REF and PREF REF follows from a standard local limit theorem: MATH for some positive M... |
math/0010291 | We use the standard approximation of MATH by tilting the measure and applying a local central limit theorem with error estimate. For MATH in a neighborhood of MATH we consider the tilted measure MATH where MATH . Clearly MATH and MATH . Therefore, the mapping MATH is an analytic diffeomorphism of a neighborhood of MATH... |
math/0010291 | Approximate MATH in REF by an appropriate quadratic function. |
math/0010291 | REF is well known. For the convenience of the reader we give a crude proof, sufficient for our purpose. MATH for some MATH by choosing MATH appropriately, for large enough MATH . CASE: Let MATH . Then, by standard large deviation estimates, MATH for MATH large enough, when MATH is chosen appropriately for example, MATH... |
math/0010291 | The proposition is an easy consequence of the one-dimensional result. Indeed, write the random walk in (dependent) components MATH where MATH are one-dimension random walks, possessing an exponential moment. The first time MATH when MATH leaves MATH is also the first time where one of the MATH leaves the interval MATH ... |
math/0010291 | Remark first that under the conditions of the lemma, MATH for the large enough MATH . This easily follows from irreducibility and aperiodicity. Therefore, MATH is well-defined. We first derive a simple exact expression for this expectation: MATH . This readily follows from a standard ``last exit - first entrance" decom... |
math/0010291 | We can take MATH . If MATH then MATH and there is nothing to prove. We write MATH for the law of the random walk MATH conditioned on MATH . (For simplicity, we neglect trivial parity problems.) Let MATH be the number of points visited by MATH on the torus (assuming MATH for simplicity to be even). Then it suffices to p... |
math/0010291 | We use the same notations as in the proof of REF : MATH denotes the law of the random walk of length MATH (on the torus), conditioned to start in MATH and to end in MATH . If MATH and MATH then MATH . MATH for MATH . Let MATH which for the region of summation is in MATH, and MATH . Then MATH . Therefore, we get MATH . |
math/0010291 | We rescale the random walk by defining MATH . This random walk depends on MATH through MATH . It takes values in MATH which we regard as a (discrete) subset of the continuous torus MATH with lattice spacing MATH . Remember the setting MATH. The transition probabilities of the MATH-chain are given by MATH . Here MATH is... |
math/0010300 | By a result of NAME, see CITE, MATH is a symplectic manifold, and the fibers are symplectic submanifolds. As has been observed before CITE, the assumption of relative minimality here implies that MATH is minimal and not ruled, because any pseudo-holomorphic sphere in MATH would have to be contained in a fiber because M... |
math/0010300 | We consider a NAME fibration over the MATH-disk MATH with precisely MATH singular fibers, such that with respect to a basepoint on the boundary of MATH the vanishing cycles of all the singular fibers can be identified with MATH. Then the monodromy of the fibration around the boundary of MATH is MATH. If this can be exp... |
math/0010300 | If MATH, then we have a surjective homomorphism MATH given by collapsing a boundary component to a point. We also have surjective forgetful homomorphisms MATH so it is enough to prove the claim in the case MATH. But this case is immediate from REF . |
math/0010300 | From the presentation of MATH due to CITE, it follows that the NAME of MATH is a finite cyclic group of order MATH if MATH is odd, and of order MATH if MATH is even. Let MATH be a nontrivial separating simple closed curve on the surface of genus MATH which is invariant under the hyperelliptic involution. Then MATH, and... |
math/0010300 | For MATH this follows from the result of Mess CITE that MATH is a free group (on infinitely many generators). CITE proved that for MATH the commutator subgroup of the NAME group is: MATH where MATH is the subgroup generated by all squares of elements of MATH, and MATH is the subgroup generated by the NAME twists along ... |
math/0010303 | Since MATH is a two-generated simple group, MATH is an epimorphic image of a free group of rank REF. So it is enough to prove that MATH and MATH. The first equality follows since there is no nontrivial homomorphism of MATH into MATH. On the other hand, there is a canonical embedding MATH, so MATH. |
math/0010304 | Let MATH. We first show that MATH. Choose any MATH. By requirement REF, clearly MATH. So, MATH for some MATH. By REF , we can find two continuously decreasing sequences MATH and MATH in MATH such that MATH, MATH, and whenever MATH and MATH, then MATH and MATH. Standard back-and-forth arguments involving closed unbounde... |
math/0010304 | Clearly, MATH is doubly homogeneous and dense in MATH. We identify MATH. By REF , each MATH and each MATH REF is invariant under MATH. By homogeneity we get MATH for each MATH. Hence each MATH REF is an orbit of MATH in MATH. By REF , any outer automorphism MATH of MATH determines an automorphism MATH of MATH which per... |
math/0010304 | Choose MATH-sequences MATH and MATH such that MATH and MATH for each MATH and MATH, MATH, MATH, MATH. For each MATH, there is an isomorphism MATH permuting the colours as prescribed by MATH. Patching the MATH's together, we obtain the required isomorphism MATH. |
math/0010304 | CASE: Observe that MATH in MATH and MATH in MATH. Now continue by induction through MATH and construction of MATH, MATH. CASE: Put MATH and MATH REF to obtain the result. |
math/0010304 | Choose MATH and MATH, MATH with MATH and MATH. By assumption, there are nice embeddings MATH and MATH with MATH, MATH, MATH, MATH. First, from MATH we obtain two continuously decreasing sequences MATH and MATH such that: CASE: MATH and MATH; CASE: MATH for each MATH; CASE: MATH for each MATH. By REF , for each MATH the... |
math/0010304 | We will first consider the case MATH (although the case MATH will be quite similar). We first construct MATH which is MATH-homogeneous for MATH-intervals. Let MATH be the coloured linear ordering defined above. Let MATH be the set of all gaps of MATH. For each gap MATH of MATH, choose a copy MATH of the good MATH-set g... |
math/0010305 | CASE: Should be clear REF So we are given the sequence MATH. We choose the increasing sequence MATH of natural numbers by letting MATH, MATH be the first MATH such that CASE: for every MATH and MATH we have MATH and MATH, and MATH], and MATH mentions only MATH with MATH. Note that MATH is well defined as the sequence M... |
math/0010305 | So let MATH be a basis of MATH and as MATH is a separable metric space there is a sequence MATH of (pairwise distinct) members of MATH with MATH. Let MATH, so MATH converges to MATH and MATH. Assume MATH is as in the conclusion of REF , and we shall eventually get a contradiction. Let MATH be a subgroup of MATH generat... |
math/0010305 | Follows by REF . |
math/0010306 | Let MATH be a model of MATH with the strict order property. So, there are MATH-definable partial order MATH on MATH-tuples in MATH for some MATH and a sequence MATH of MATH-tuples in MATH such that MATH for MATH. By NAME 's Theorem, we can assume that MATH is a MATH-indiscernible sequence in MATH. Also, we can assume t... |
math/0010309 | Let MATH. Then MATH, where MATH and MATH denotes NAME multiplication by the one-form MATH. An easy computation establishes that MATH and that MATH. Then MATH, where MATH denotes the projective MATH-action. |
math/0010309 | We need to show that MATH defined in REF is independent of MATH. By REF , it suffices to show that MATH is a norm continuous family of projections in the MATH-algebra MATH. By NAME 's principle, one has the following identity whenever MATH, MATH . It follows that there is a constant MATH independent of MATH such that M... |
math/0010309 | By hypothesis, the twisted NAME map is an isomorphism. Therefore to compute the range of the trace map on MATH, it suffices to compute the range of the trace map on elements of the form MATH . Here MATH is an even parity MATH-cycle over MATH. By the twisted analogue of the MATH index theorem of CITE and CITE for ellipt... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.