paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010184 | We will choose MATH smaller than MATH and hence can pick a point MATH at distance MATH from MATH. Consider a minimizing segment MATH from a point MATH to MATH. We apply comparison to the geodesic triangle with sides MATH. This can be done although the side MATH is of course not minimizing. We obtain (for both angles be... |
math/0010184 | The NAME domain MATH is convex and therefore contains the convex hull of MATH in MATH. Since MATH is not hyperspherical, the convex hull fills out the whole component of MATH on the concave side of MATH. This is a standard ball or cusp, according to whether MATH is spherical or horospherical, and it embeds into MATH vi... |
math/0010184 | CASE: MATH is the closed convex hull of MATH in MATH and therefore belongs to MATH. This implies the first part of the assertion. CASE: Note that as soon as MATH contains a neighborhood of a point of MATH, then it contains a neighborhood of the entire leaf MATH and thus MATH. We are using here that MATH is connected. T... |
math/0010184 | A shortest segment from MATH to MATH corresponds to a point MATH. Let MATH denote the direction at MATH of the perpendicular to MATH. The NAME domain MATH contains the convex hull of MATH and MATH which in turn contains, locally at MATH, the cone over the disc of radius MATH around MATH in MATH. This shows the first as... |
math/0010184 | Using REF , we choose MATH sufficiently small so that MATH. According to REF any two shortest segments from MATH to MATH have angle MATH, and hence by REF there can be at most two of them. Regarding the second part, the assertion is clear for smooth points MATH. Suppose therefore that MATH is singular and that there ar... |
math/0010184 | Suppose that there exists a unit speed segment MATH of length MATH emanating in the perpendicular direction to the convex side of MATH. Moreover, consider another such segment MATH of length MATH which connects MATH to the nearest point of MATH. We then have MATH, and, due to the convexity of MATH, MATH. The segments M... |
math/0010184 | The first part of the assertion is proven as for REF . Note that MATH and MATH may be chosen monotonically decreasing as MATH decreases. Thus, to obtain the second part, we may decrease MATH, if necessary, below the value MATH from REF , and then decrease MATH and MATH accordingly. |
math/0010184 | CASE: The orientation reversing involution MATH on MATH cannot have fixed points because there are no singular edges with cone angle MATH. CASE: We have MATH. We apply the NAME Theorem to MATH and note that, for MATH sufficiently small, the contribution of its smooth part to the curvature integral is a small negative n... |
math/0010184 | We choose MATH sufficiently small so that MATH, compare REF . This rules out in REF the possibility of cut points near MATH with a unique minimizing segment to MATH. |
math/0010184 | We use the constant MATH from REF and accordingly the constants MATH and MATH from REF . Suppose that MATH and MATH. REF imply for MATH that MATH is a closed totally geodesic surface contained in MATH. Then MATH is a closed totally geodesic surface as well, and it follows that MATH. Since MATH is non-orientable, the NA... |
math/0010184 | Suppose that MATH separates and is not totally geodesic. We choose MATH smaller than the constant MATH in REF. By combining REF, there exists an embedded umbilic tube MATH with MATH as boundary component and the following properties: MATH is a ball if MATH is spherical, a cusp if MATH is horospherical, and a neck if MA... |
math/0010184 | The assertion follows by applying angle comparison to the triangle with sides MATH. This triangle has two minimizing sides, namely twice MATH, and the non-minimizing side MATH. We apply comparison at the two angles adjacent to MATH. |
math/0010184 | Suppose that one of the segments MATH is minimizing. We then can apply angle comparison to the geodesic triangle MATH which has MATH as two of its sides and as third side a minimizing segment MATH. REF implies that the comparison angle at MATH is MATH with MATH. By choosing MATH sufficiently small, we obtain that MATH.... |
math/0010184 | We may assume that MATH, say MATH. This is relevant only in the positive curvature case. CASE: We denote the cone angle at MATH by MATH. The NAME polyhedron MATH associated to MATH can be regarded as a convex polyhedron in the model space MATH with singular axis of cone angle MATH, compare REF. The minimizing segments ... |
math/0010184 | Let MATH denote the endpoints of MATH and MATH its cone angle. We study the NAME polyhedron MATH which we regard as a convex polyhedron in MATH. CASE: The edges of MATH are almost vertical. By REF, near MATH the interior points on faces of MATH correspond to smooth cut points, and the points on boundary edges correspon... |
math/0010184 | Let MATH and MATH be the maximal minimizing subsegments of MATH emanating from MATH in the two antipodal singular directions. Suppose that both have length MATH. Due to our diameter assumption, there exists a segment MATH of length MATH starting in MATH. We regard MATH also as segments in the NAME polyhedron MATH and d... |
math/0010184 | Let us consider the cut locus MATH with respect to the entire boundary. Note that MATH. The discussion of REF applies since we can replace boundary components of MATH by nearby equidistant umbilic surfaces. We choose MATH arbitrarily and then MATH as the constant provided by REF . Combining REF and the description of t... |
math/0010184 | The lower bound MATH on cone angles yields an upper bound MATH for the diameter of the turnover MATH. We are done by REF if MATH. We suppose therefore that MATH. We certainly have MATH because MATH is MATH-thin and denote MATH. We consider the family of embedded umbilic surfaces MATH, MATH, equidistant to MATH with dis... |
math/0010184 | Let MATH be a compact core of MATH. The boundary MATH is the union of compact pairs of pants (which are a compact core of MATH) together with a collection MATH of tori and annuli, corresponding to the boundary of a neighborhood of edges in MATH. Either MATH is NAME fibred or MATH is an atoroidal pared manifold. We reca... |
math/0010184 | We assume that MATH contains a spherical MATH-suborbifold MATH which is essential. Since MATH is irreducible, MATH bounds a discal MATH-orbifold MATH, which contains MATH, for MATH sufficiently large. Let MATH denote the holonomy representation of MATH. We have that MATH fixes a point of MATH, because MATH is contained... |
math/0010184 | Suppose that the lemma is false. Then, by passing to a subsequence and changing the indices of the MATH, we can assume that the curves MATH represent a fixed class MATH independent of MATH. Let MATH and MATH denote the holonomy representation of MATH and MATH, respectively. Since the curves MATH are compressible in MAT... |
math/0010184 | Consider first a vertex MATH. We choose a small positive number MATH and denote by MATH the three singular points at distance MATH from MATH. We take the convex hull of MATH inside the closed ball MATH, and denote by MATH the interior of the convex hull. Its closure MATH can be obtained by doubling a hyperbolic simplex... |
math/0010184 | Suppose that MATH contains singular vertices and consider a diverging sequence MATH of distinct vertices in the same MATH-orbit. We fix a base point MATH and denote by MATH the direction of the segment MATH. We may assume without loss of generality that the MATH converge, MATH and that MATH. Applying triangle compariso... |
math/0010184 | The Euclidean cone structure MATH on MATH (with cone angles strictly less than the orbifold angles of MATH) lifts to a Euclidean cone structure MATH on the universal cover MATH. The Euclidean cone manifold MATH has cone angles MATH, for some constant MATH. In addition, the fundamental group MATH acts isometrically on M... |
math/0010184 | Let MATH be the orbifold obtained from MATH by removing the open edge MATH from the branching locus. (This change of the orbifold structure amounts to putting the label MATH on MATH.) Since all edges of MATH adjacent to an endpoint of MATH have label MATH, MATH is still an orbifold. The fundamental group of MATH is fin... |
math/0010184 | We first prove the irreducibility of MATH. Let MATH be a spherical REF-suborbifold. It bounds a discal REF-suborbifold in MATH, since MATH is irreducible. If it does not bound a discal REF-orbifold in MATH, then a neighborhood MATH is contained in the interior of a discal REF-suborbifold of MATH. This is impossible, si... |
math/0010184 | Assume that MATH does not collapse. After choosing base points MATH in the thick parts of MATH and passing to a subsequence, MATH converges geometrically to a pointed hyperbolic cone REF-manifold MATH with finite volume. MATH cannot be compact because MATH. We use the finite cover MATH of REF and denote by MATH the lif... |
math/0010184 | By assumption, MATH and hence also MATH are closed. The lemma follows from the fact that, for closed orbifolds, smallness is a property independent of the labels of the branching locus. |
math/0010184 | Assume for instance that MATH is abelian. This means that the image of MATH is contained in a diagonalizable subgroup MATH. Therefore MATH preserves the corresponding NAME fibration on MATH. It follows easily from this that MATH is a link and MATH is NAME fibred (compare CITE). |
math/0010184 | Let MATH be the subset of representations in MATH that are such lifts. By our previous discussion, MATH contains a neighborhood of the endpoint MATH of MATH. Openness of MATH. REF imply that perturbations of representations in MATH still take values in MATH. Moreover, perturbations of holonomy representations are induc... |
math/0010184 | Recall that MATH is a closed connected subset of the real algebraic curve MATH. Moreover, MATH is compact because it is contained in the subset of conjugacy classes of MATH-valued representations. As an algebraic curve, MATH is homeomorphic to a graph with finitely many vertices. It follows that MATH is a compact graph... |
math/0010184 | By folding these orbifolds along their symmetry axis, one obtains index-two ramified coverings over the orbifolds with vertex isotropies MATH. The angles of the quotient orbifolds satisfy the assumptions in the proposition. The quotients have the same universal cover, and the assertion follows from the previous proposi... |
math/0010184 | Let MATH, for MATH or MATH. The proof has three main steps. In REF we show that the dimension of the NAME tangent space of MATH at MATH equals the number of meridians. In REF we prove that MATH is a smooth point of MATH. In REF we check that the differential forms MATH are a basis for the cotangent space. CASE: Computa... |
math/0010184 | Assume that the assertion were false. Then there exist MATH, MATH and a sequence of cone manifolds MATH with diameter MATH, curvature in MATH and MATH-thick links such that MATH is MATH-thin, and there exist points MATH for which the conclusion of the lemma does not hold. The fact that MATH is MATH-thin and has MATH-th... |
math/0010184 | Assume the contrary. By hypothesis, MATH is closed and small. Hence every turnover in MATH bounds a discal suborbifold. Using the bilipschitz homeomorphism from MATH to the local model, it follows that the sum of cone angles of the turnover is close to MATH. This contradicts the MATH-thickness of links. |
math/0010184 | The main point is to prove that MATH is irreducible. For this we have to show that MATH is not contained in a discal suborbifold. Since the pairs MATH and MATH are homeomorphic, this amounts to showing that MATH is not contained in a discal suborbifold, and we can use the metric properties of MATH. If MATH contains a s... |
math/0010184 | We may assume that MATH, and moreover MATH. By the triangle inequality we have MATH so MATH . This shows REF follows because MATH. |
math/0010184 | This follows because MATH and MATH is contained in the ball of radius MATH around MATH. |
math/0010184 | Let MATH be an arbitrary point. By maximality, there exists a point MATH such that MATH. By REF we have MATH and MATH. Thus MATH. |
math/0010184 | For every ball MATH intersecting MATH it holds MATH. On the other hand, the points MATH are separated from each other, since MATH. Thus the number of such MATH is bounded above by: MATH . Here MATH denotes the volume of the ball of radius MATH in REF-space of constant curvature MATH, and the last inequality follows fro... |
math/0010184 | If MATH we choose MATH. Suppose that MATH. Then there exists MATH with MATH. The assertion is trivial if MATH and MATH are disjoint. We assume therefore also that MATH. Then by REF : MATH . Hence MATH. |
math/0010184 | Let MATH be an auxiliary REF-Lipschitz function which vanishes in a neighborhood of MATH and satisfies MATH. We put MATH on MATH and extend it trivially to MATH. Then MATH is MATH-Lipschitz. Let MATH. Then at most MATH functions MATH are non-zero in MATH, and all of them have NAME constant MATH. The claim follows since... |
math/0010184 | It suffices to find a constant MATH such that every MATH-dimensional simplex MATH contains a point MATH at distance MATH from both MATH and the image of MATH. To push MATH into the MATH-skeleton we compose it on MATH with the central projection from MATH. This will increase the NAME constant by a factor bounded in term... |
math/0010184 | We first show that MATH does not enter to far into the other sets MATH. There exists a constant MATH such that, if MATH is sufficiently large, then MATH for all MATH. By REF , MATH and we obtain: MATH . For the last estimate, we use that MATH and, since MATH, MATH. We now assume that MATH intersects MATH because otherw... |
math/0010184 | The inverse image under MATH of the open star of MATH is contained in MATH. Using REF repeatedly, we can homotope MATH to a map MATH which is locally NAME and satisfies MATH. More precisely, there is a universal constant MATH such that MATH is MATH-Lipschitz on MATH. It suffices to show that no MATH-simplex MATH is con... |
math/0010184 | Since the orbifold MATH is NAME, it is very good and there is a finite covering MATH by a manifold CITE. The boundary MATH is a union of tori. The JSJ splitting of MATH pulls back to the JSJ splitting of MATH. We have to show that no hyperbolic components occur in the JSJ splitting of MATH. We may assume that the bound... |
math/0010186 | We use induction on MATH. The result is certainly true for MATH. Now, using induction and matrix multiplication, MATH . |
math/0010186 | Using matrix multiplication and REF we obtain MATH . Therefore, denoting MATH, we obtain MATH . If MATH and MATH, MATH . Using REF in the previous recurrence, we obtain MATH which gives us REF. If the relations REF are satisfied, we obtain, for MATH, MATH where MATH. Furthermore, for MATH, MATH . Therefore, MATH which ... |
math/0010186 | We show first that the entries of MATH satisfy a relation of the form MATH and then will proceed to find these coefficients. From REF we observe that MATH, MATH and MATH. Now, the coefficients of MATH satisfy, for MATH, MATH where MATH. We evaluate, for MATH, MATH since MATH and MATH . Thus, MATH . Therefore, we obtain... |
math/0010186 | We showed that the entries of MATH are MATH, for any integer MATH. Thus, the entries on the principal diagonal of MATH are all REF. If MATH, then MATH. Assume that there is an integer MATH with MATH, such that MATH, for all MATH. Take MATH. Thus, MATH, a contradiction. Therefore, the integer MATH is the least integer M... |
math/0010186 | We prove by induction on MATH that the elements on the first row and first column of MATH are MATH . First we deal with the elements in the first row. The top equation is certainly true for MATH, if we define MATH. Now, MATH . Again by induction we prove the result for the elements in the first column. The case MATH is... |
math/0010186 | We observe that, since MATH, we have MATH, otherwise MATH, and by NAME 's Theorem the entry point MATH divides MATH. Therefore, MATH and MATH. Therefore, MATH must be REF. That is not possible, since MATH, which is not divisible by any prime. So, MATH and MATH. Thus MATH . By the previous NAME 's Theorem, MATH, therefo... |
math/0010186 | Let MATH. The entry point of the NAME sequence modulo MATH is MATH. Since MATH, REF shows the result in this case. Assume MATH. We know that in this case we must have MATH. Using the known formula (see for instance REF ) MATH taking MATH, and using NAME 's NAME Theorem, MATH, we obtain MATH since for the primes MATH, M... |
math/0010186 | If MATH, then the order of MATH is MATH. If MATH, then MATH. Therefore, MATH. Thus MATH. The bound is met for all primes MATH and all even integers MATH. |
math/0010186 | We have MATH which is MATH, unless MATH, in which case it is MATH. A similar analysis for MATH will produce its inverse. |
math/0010189 | Fix an orthogonal NAME basis MATH of MATH. Consider norm-convergent sums MATH for suitably selected sequences MATH. If the NAME basis of MATH is a normalized tight frame then the equality MATH is valid for every admissible choice of the coefficients MATH. In particular, one admissible selection is MATH and MATH for eac... |
math/0010189 | Suppose, MATH possesses an orthogonal standard NAME basis MATH. That means, there are two constants MATH such that the inequality MATH is fulfilled for every MATH. Obviously, MATH since MATH is supposed to be a NAME basis and, therefore, MATH by one of REF bases. Considering the lower estimate with the constant MATH sp... |
math/0010189 | Since the sequence MATH is a standard normalized tight frame in MATH the frame operator is correctly defined and the equality MATH holds for any MATH. Moreover, the image of MATH is closed because MATH is closed by assumption. So, MATH is an isometric MATH-linear embedding of MATH into MATH with norm-closed image. To c... |
math/0010189 | Since MATH is a normalized tight frame we get MATH for any MATH by the reconstruction formula. The basis property forces MATH for any MATH and each fixed MATH. However, the right carrier projection of MATH equals the carrier projection of MATH for every MATH if calculated inside the bidual NAME algebra MATH. So MATH fo... |
math/0010189 | By supposition and REF we have the reconstruction formulae MATH for any MATH. Taking the MATH-valued inner product of MATH by MATH with respect to MATH and the MATH-valued inner product of MATH by MATH with respect to MATH simultaneously the right sides of REF become adjoint to one another elements of MATH. Since MATH ... |
math/0010189 | The set MATH is supposed to be standard frame for the NAME MATH-module MATH. NAME to the definition of module frames we have the inequality MATH valid for every MATH and two fixed numbers MATH. So the image of MATH inside MATH has to be closed since MATH is closed by assumption and the operator MATH is bounded from abo... |
math/0010189 | Since MATH is an orthogonal NAME basis of MATH with MATH has the property of a standard normalized tight frame. Writing down this property for the special setting MATH we obtain MATH . |
math/0010189 | Consider a normalized tight frame MATH of MATH. For this normalized tight frame the sum exists as a weak limit in MATH. Fixing a point MATH and applying the NAME space formula to the NAME space frame MATH we obtain MATH, CITE. Therefore, the sum is locally constant because the number obtained is precisely the dimension... |
math/0010189 | We have only to check a chain of equalities in MATH that is valid for our standard normalized tight frames. For an arbitrary fixed standard normalized tight frame MATH we have MATH in case one of the sums at either ends converges weakly. Since we can repeat our calculations for the other standard normalized tight frame... |
math/0010189 | By REF there is a standard isometric embedding of MATH into MATH induced by the frame transform MATH. In the context of that embedding MATH is an orthogonal summand of MATH, and the MATH-valued inner products on MATH and on MATH coincide. The corresponding projection MATH maps the standard orthonormal NAME basis MATH o... |
math/0010189 | Set MATH and MATH and define MATH. By assumption the MATH-valued inner products are preserved by MATH, and MATH extends to a unitary map between MATH and MATH by MATH-linearity. Fix MATH. Then the equality MATH is valid. So MATH and MATH for MATH. Applying MATH the equality MATH yields. Consequently, MATH splits into t... |
math/0010189 | If MATH is an invertible adjointable bounded MATH-linear operator on MATH and MATH is a standard normalized tight frame of MATH, then the sequence MATH fulfills the equality MATH for every MATH. Since MATH for every MATH (compare CITE) and since the sum in REF converges in norm, the sequence MATH is a standard frame of... |
math/0010189 | By REF there exists a standard normalized tight frame MATH for MATH and an adjointable invertible operator MATH on MATH such that MATH for any MATH. Moreover, there is another NAME MATH-module MATH and a standard normalized tight frame MATH for MATH such that the sequence MATH is an orthogonal NAME basis in MATH, see R... |
math/0010189 | Consider the operator MATH defined by MATH for MATH and for an orthonormal basis MATH of MATH. The operator MATH is a bounded MATH-linear, surjective and adjointable operator since MATH is supposed to be algebraically generated by MATH and the NAME MATH-module MATH is self-dual, compare CITE. By CITE the operator MATH ... |
math/0010189 | Let MATH and MATH be the frame bounds for the standard frame MATH. Then for MATH we obtain the inequality MATH by NAME 's theorem CITE and the frame property. The additional equality in the middle of this chain of two inequalities introduces a certain expression the comparison of which to both the ends of the chain of ... |
math/0010189 | Let MATH be any invertible operator onto some NAME MATH-module MATH with the property that the sequence MATH is a standard normalized tight frame of MATH. The existence of such an operator is guaranteed by REF setting MATH and MATH (compare the introductory considerations of the present section), or by REF . Set MATH a... |
math/0010189 | By the definition of a canonical dual frame and by the results of REF above we have the equality MATH for every MATH. Applying the invertible positive operator MATH to this equality we obtain the identity MATH for MATH. Since the operator MATH is invertible on MATH we can replace MATH by MATH, and the sought equality t... |
math/0010189 | By REF there exists a standard NAME basis MATH of a NAME MATH-module MATH and an orthogonal projection MATH such that MATH for MATH. Since the sum MATH is norm-bounded we can define another adjointable operator MATH by the formula MATH for MATH. Then MATH and MATH for MATH. The following equality holds for any MATH: MA... |
math/0010189 | We begin with the proof of the first statement. The convergence of the sums in the inequality above follows from the properties of the frame transforms and of the frame operators. If the standard frames MATH and MATH are both dual frames of MATH then the equalities MATH are valid for every MATH. Subtracting one sum fro... |
math/0010189 | Suppose the frame MATH is standard with respect to both the inner products on MATH. For MATH we have two reconstruction formulae MATH and MATH. By the optimality principle we obtain the equality MATH that is satisfied for any MATH and MATH, see REF . Let MATH. Multiplying by MATH from the right and summing up over MATH... |
math/0010189 | Suppose MATH for MATH and an adjointable invertible operator MATH. Let us count the values of the adjoint operator MATH of MATH. We have MATH for every MATH by the dual frame property. Consequently, MATH equals the identity operator on MATH. |
math/0010189 | Considering the first pair of equivalent conditions the product of the two positive elements MATH and MATH of the C*-algebra MATH can only be positive if they commute. Consequently, MATH forces them to commute since MATH by construction, compare REF . Conversely, if MATH then by the equality MATH for MATH we obtain MAT... |
math/0010189 | Suppose, there exists an adjointable invertible bounded MATH-linear operator CITE MATH with MATH for every MATH. Continuing the operator MATH and its adjoint on the orthogonal complements of MATH and MATH, respectively, as the zero operator we obtain an adjointable bounded MATH-linear operator MATH on MATH that possess... |
math/0010189 | If we assume that the frame transforms MATH, MATH corresponding to the two initial standard normalized tight frames have the same range in MATH then the orthonormal projection MATH of MATH onto this range MATH maps the elements of the standard orthonormal basis MATH of MATH to both MATH, MATH, by the construction of a ... |
math/0010199 | Let MATH be another metric on MATH, then there exists a MATH-symmetric endomorphism MATH of MATH, such that MATH . We consider the family MATH with scalar curvature MATH and NAME curvature MATH. By a straightforward calculation in normal coordinates around a point MATH in MATH, one checks that the derivative of MATH is... |
math/0010199 | Let MATH denote the spinor bundle of MATH, which exists over all sufficiently small open subsets of MATH even if MATH is not spin. We equip MATH with the metric and connection induced by MATH. If the NAME characteristic MATH of MATH is non-zero, then MATH is even-dimensional, and the local spinor bundle splits as MATH.... |
math/0010199 | Since MATH is spin, we can construct the bundle MATH and the NAME operator MATH on MATH as in REF. By the NAME index theorem, the index of MATH is given by MATH because MATH equals the NAME class of MATH. Under the hypotheses of the theorem, we have MATH. Now, the theorem follows from REF . |
math/0010199 | Recall that for MATH, the real spinor representation MATH of MATH acts on a real vector space MATH of real dimension MATH CITE. The complex spinor representation arises as MATH on MATH. Complex conjugation induces a MATH-antilinear involution MATH on MATH, which commutes with MATH. If we restrict MATH and MATH to the N... |
math/0010199 | We start with the following basic case: Assume that MATH is a Riemannian locally symmetric space of compact type with MATH. Let MATH, , MATH be a local MATH-orthonormal base of MATH, and let MATH denote the real NAME volume element. We consider the real NAME operator MATH . Note that MATH for all MATH, so MATH acts on ... |
math/0010202 | By REF , MATH where MATH and MATH are as described above. Let MATH, MATH be a family of matrix units generating MATH; thus MATH . Consider the algebra MATH . It is clear that MATH; thus it is sufficient to prove that MATH is isomorphic to MATH for some MATH. Note that MATH is generated by the family MATH hence MATH . C... |
math/0010202 | Let MATH be given by MATH (MATH copies of MATH). Since MATH, the canonical conditional expectation MATH is known to be faithful (see for example, REF). By CITE, MATH . Moreover, by CITE, MATH . Hence MATH . Applying REF , we get that MATH for some MATH. It follows that MATH for some real NAME space MATH and one-paramet... |
math/0010202 | Clearly, only the statement about the centralizer needs to be proved, since the core is isomorphic to the centralizer, tensor MATH, as soon as the centralizer is a factor (see CITE). Also, we can restrict ourselves to dealing with a particular choice of the state MATH, satisfying the hypothesis of this theorem; indeed,... |
math/0010202 | The fact that MATH is of type MATH was proved in CITE; this fact also follows from the explicit description of MATH given in REF. For the convenience of the reader we give an expanded proof of the isomorphism MATH (see CITE). Let MATH be the dimension of MATH as a vector space. Choose a standard orthonormal system of v... |
math/0010202 | It is known that the fundamental unitary MATH lies in MATH (acting on MATH), and hence the mapping MATH gives rise to an isomorphism from MATH onto MATH (see for example, CITE). Moreover, the crossed product structure and the NAME state MATH on MATH give rise to a normal faithful conditional expectation MATH given by M... |
math/0010202 | Since MATH is a factor REF and MATH is isomorphic to MATH , MATH (see REF ). Similarly, since MATH and MATH is a factor, MATH. |
math/0010202 | The case MATH was obtained by NAME in CITE. Let MATH, and choose MATH so that MATH. Let MATH be the fundamental representation of MATH. By REF , for each MATH, we obtain an irreducible inclusion of type MATH factors MATH of index MATH. |
math/0010202 | The proof is a straightforward modification of CITE. Note that MATH (MATH), so we use the following identities (instead of the trace property of MATH used in REF ): CASE: MATH for analytic MATH and MATH; REF for analytic MATH . |
math/0010204 | REF , the first part of REF , and REF go back to CITE; they are also stated in CITE. The second part of REF is a direct consequence of the first part (observed in CITE). CASE: See CITE. CASE: This is REF (compare REF). CASE: As observed in CITE, this is immediate from REF . |
math/0010204 | REF . See CITE. CASE: For MATH, we have MATH. CASE: For MATH we have MATH if and only if MATH, if and only if MATH, if and only if MATH. Suppose now that there is no largest member of MATH contained in MATH. Then, by REF , there are MATH and MATH with MATH and MATH for which MATH and MATH such that no member of MATH co... |
math/0010204 | We must show that, if MATH and MATH are not adjacent, then MATH and, if they are adjacent, then MATH. We evaluate the expressions on each MATH and show they are equal. We begin with the case in which MATH or MATH. To be specific, let MATH. Suppose first that MATH and MATH are not adjacent. Then MATH and MATH. Now MATH,... |
math/0010204 | The MATH should satisfy the relations REF for MATH. Substituting MATH for MATH, we find relations for the coefficients of MATH with MATH. The constant part involves only the MATH. It follows from REF that these equations are satisfied. The coefficients of MATH lead to MATH . We focus on the consequences of these equati... |
math/0010204 | CASE: The equations are necessary as they appeared under REF . CASE: Use induction on MATH. If MATH, the equation coincides with REF . If MATH, then either REF or REF applies. REF are obvious. REF follows from REF by use of REF . Observe that, if MATH and MATH for some MATH, then MATH. |
math/0010204 | We need to show that the matrices MATH satisfy the relations in REF . It is clear from REF that they do for MATH of height MATH or MATH. We use induction on MATH, the height of MATH, and assume MATH. We first check REF . If this applies, the value MATH was determined in REF, and we are really checking the value did not... |
math/0010204 | Combine REF . |
math/0010204 | The proof is similar to that of REF . |
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