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math/0102190 | The equivalence of REF follows at once from REF . To prove the equivalence of REF , we have only to recycle some standard arguments (see, for example CITE, CITE). First, suppose that MATH , and let MATH be the unique differential operator such that REF holds, or, equivalently, such that MATH . By REF , the first term here belongs to MATH for all MATH ; by REF , so does the second term. Since generically MATH contains no function of the form MATH , it follows that the expression on the left of REF is zero, that is, MATH . Conversely, suppose MATH . Differentiating with respect to MATH , we get MATH for all MATH and for all MATH . Fix any MATH such that the functions MATH and MATH are regular at MATH . By REF , the right hand side of REF belongs to MATH . By REF , the functions MATH on the left hand side of REF span MATH ; hence the operator MATH maps MATH into MATH . By REF we then have MATH . |
math/0102190 | The only thing left to see is that REF ensures that the differential operator MATH has rational coefficients. The argument is like that in the proof of REF above: if we set MATH then we have MATH . Because the NAME functions MATH are rational in both variables MATH and MATH, the operators MATH, hence also MATH, have rational coefficients; and because MATH is differential, MATH has polynomial coefficients. It follows that MATH has rational coefficients. |
math/0102190 | In the proof all operators are supposed to be in the variable MATH : we omit MATH from the notation. Applying REF with MATH , MATH , we get that MATH . Also, MATH if and only if MATH , so applying REF again with MATH , MATH , we get MATH . But in view of REF coincide; hence the Proposition. |
math/0102190 | According to the definitions, we have to show that MATH is isomorphic to MATH . In fact these two right ideals coincide, for we have (denoting the formal adjoint of an operator MATH by MATH) MATH . |
math/0102190 | According to CITE REF , MATH is equal to MATH times the coefficient of MATH in the series MATH . The lemma follows easily from this. |
math/0102190 | We may assume that MATH is diagonal and that MATH is a NAME matrix, that is, the off-diagonal entries of MATH are given by MATH (compare REF : the MATH are necessarily distinct). By REF , there is a unique MATH such that MATH has the same eigenvalues as MATH; then MATH is conjugate to a pair MATH with MATH another NAME matrix. Thus MATH and MATH differ only in their diagonal entries. From the non-vanishing of the NAME determinant MATH , we find that there is a unique polynomial MATH as in the Lemma such that MATH. The Lemma follows. |
math/0102190 | This follows from NAME 's lemma (see REF ). |
math/0102191 | If MATH is any compact subset of MATH, then MATH is a compact subset of MATH; furthermore, any compact subset of MATH is contained in one of the form MATH. We have MATH . |
math/0102191 | Because MATH is a NAME subgroup, there is a compact subset MATH of MATH, such that MATH. However, from REF , we know that MATH is finite. Therefore MATH is finite. |
math/0102191 | CASE: Let MATH be a compact subset of MATH, such that MATH and MATH. Choose a corresponding compact subset MATH of MATH, as in REF . Then MATH and, similarly, MATH. CASE: Let MATH be a compact subset of MATH, such that MATH and MATH. Then MATH and, similarly, MATH. |
math/0102191 | CASE: We have MATH, and, because MATH is a proper map, we have MATH as MATH in MATH. Because MATH, we know that MATH is homeomorphic to the half-line MATH (with the point MATH in MATH corresponding to the endpoint MATH of the half-line), so, by continuity, it must be the case that MATH. Then REF implies that MATH is a NAME subgroup, but we provide the following direct proof that avoids any appeal to REF . From the definition of MATH, we have MATH. Therefore MATH so MATH is a NAME subgroup (by taking MATH in REF ). |
math/0102191 | By assumption, we may choose the same fundamental domain MATH for the NAME groups of MATH and MATH. Let MATH and MATH be the NAME projections; then MATH is the restriction of MATH to MATH. For simplicity, assume, without loss of generality, that MATH (for example, assume MATH is connected). Then, for any compact subset MATH of MATH, we have MATH is finite (see REF). Thus, MATH acts properly discontinuously on MATH (see REF), as desired. |
math/0102191 | First, let us note that every connected subgroup of MATH is closed (see REF), so we do not need to show that MATH is closed. Second, let us note that REF is a consequence of REF. To see this, let MATH be a maximal compact subgroup of MATH that contains MATH. Then a standard argument shows that MATH is a maximal compact subgroup of MATH. (Because all maximal compact subgroups of MATH are conjugate, there is some MATH, such that MATH is a maximal compact subgroup of MATH that contains MATH. Since MATH, we know that MATH normalizes MATH, so we may assume MATH; thus, MATH normalizes MATH. Then MATH contains MATH. Because MATH is compact, this implies that MATH normalizes MATH. So MATH is a maximal compact subgroup of MATH.) Therefore MATH . Similarly, MATH. Since MATH, we conclude that MATH, as desired. Assume MATH is semisimple. We have an NAME decomposition MATH; let MATH and MATH. Assume MATH is a one-parameter subgroup. Let CASE: MATH be the Real NAME Decomposition of MATH (see REF); CASE: MATH; and CASE: MATH be the closure of MATH. (REF implies that MATH is conjugate to a subgroup of MATH.) Assume MATH is abelian. We may write MATH as a product of one-parameter subgroups: MATH . Let MATH be the Real NAME Decomposition of MATH (see REF). Note that MATH, MATH, and MATH commute, not only with each other, but also with every MATH, MATH, and MATH (see REF). Let MATH and let MATH. (REF implies that MATH is conjugate to a subgroup of MATH.) The general case. From the NAME decomposition CITE, we know that there is a connected, semisimple subgroup MATH of MATH and a connected, solvable, normal subgroup MATH of MATH, such that MATH (and MATH is finite). Let MATH, so MATH is a connected, normal subgroup of MATH, and MATH is conjugate to a subgroup of MATH (compare CITE). By modding out MATH, we (essentially) reduce to the direct product of REF . |
math/0102191 | Because MATH is abelian and REF-connected, the exponential map is a NAME group isomorphism from the additive group of the NAME algebra MATH onto MATH. CASE: Let MATH. Because the exponential map is a NAME group isomorphism (hence a diffeomorphism), and because MATH is a closed MATH-submanifold of MATH, we know that MATH is a closed MATH-submanifold of MATH. Of course, MATH is contained in MATH, which is also a MATH-submanifold of MATH. Because the dimensions are the same, we know that MATH is open in MATH. Also, because MATH is closed in MATH, we know that MATH is closed in MATH. Therefore MATH (because MATH is connected). Finally, we know that MATH is a diffeomorphism from its domain MATH onto its image MATH. CASE: From REF, we have MATH and, similarly, MATH. Also, because MATH is bijective, we have MATH. Therefore MATH is connected. CASE: Because MATH is not compact (for MATH), we know, from REF, that MATH is trivial; so MATH is finite. Since MATH has no elements of finite order, we conclude that MATH is trivial. |
math/0102191 | Any principal bundle with a section is trivial CITE. If either the fiber or the base is contractible, then there is no obstruction to constructing a section CITE, so MATH is trivial: MATH. (The diffeomorphism can be taken to be MATH-equivariant, with respect to the natural MATH-action on MATH, given by MATH.) Then the conclusions on homotopy equivalence follow from the fact that MATH is contractible (that is, homotopically trivial). |
math/0102191 | Because MATH is REF-connected, we have MATH, so, from REF, we know that the sequence MATH is exact. Thus, MATH, so the desired conclusion is immediate. |
math/0102191 | We may assume the group MATH is nonabelian (otherwise, the desired conclusion is given by REF). Then, because MATH is solvable, there is a nontrivial, connected, proper, closed, normal subgroup MATH of MATH. Since MATH is simply connected (see REF), and MATH, we may assume, by induction on MATH, that MATH is diffeomorphic to some MATH. Therefore CASE: MATH is diffeomorphic to MATH and CASE: MATH is homotopy equivalent to MATH (see REF). Because MATH is MATH-connected, REF implies that MATH is MATH-connected; hence, MATH is a REF-connected, solvable NAME group, so we may assume, by induction on MATH, that MATH is diffeomorphic to some MATH. Thus, REF implies that MATH is diffeomorphic to MATH, as desired. |
math/0102191 | CASE: We may assume MATH is nontrivial, so MATH. Thus, by induction on MATH, using REF, we may assume that MATH is a closed, simply connected subgroup of MATH. Then, since MATH is homeomorphic to MATH, we see that MATH so REF implies that MATH is connected. CASE: Because MATH is solvable, there is a connected, closed, proper, normal subgroup MATH of MATH, such that MATH is abelian. We know that MATH is REF-connected (see REF), so, by induction on MATH, we may assume that every connected subgroup of MATH is closed and simply connected. From REF, we know that MATH is connected, so we conclude that MATH is closed, and MATH . From REF (with MATH in the place of MATH, and MATH in the place of MATH), together with REF, we conclude that MATH; that is, MATH is simply connected. So REF implies MATH is diffeomorphic to some MATH. Because both MATH and MATH are closed, it is not difficult to see that MATH is closed. CASE: Because MATH is solvable, there is a connected, closed, proper, normal subgroup MATH of MATH, such that MATH is abelian. We know that MATH is REF-connected (see REF), so MATH has no nontrivial, compact subgroups (see REF); thus, we must have MATH. Therefore, MATH is a compact subgroup of MATH. Then, since MATH is REF-connected (see REF), we may conclude, by induction on MATH, that MATH is trivial. |
math/0102191 | Because MATH is linear, it is a subgroup of some MATH. Replacing MATH by a conjugate, we may assume that MATH is contained in the group MATH of upper triangular matrices with positive diagonal entries (compare REF). The matrix entries provide an obvious diffeomorphism from MATH onto MATH, so MATH is REF-connected. Thus, REF implies that MATH is simply connected. |
math/0102191 | Because the action is proper, we know that the stabilizer of each point of MATH is compact. However, MATH has no nontrivial compact subgroups (see REF). Thus, the action is free. Because the action is free, proper, and MATH, it is easy to see that the manifold MATH is a principal fiber bundle over the quotient MATH CITE. Furthermore, the fiber MATH of the bundle is contractible (see REF), so REF implies that MATH homotopy equivalent to MATH. Therefore, the spaces MATH and MATH have the same homology. |
math/0102191 | Because MATH is solvable, it has a nontrivial, connected, closed, abelian, normal subgroup MATH. Since MATH is abelian and MATH is connected (see REF), we know that MATH is a REF-connected abelian group (see REF), so it is isomorphic to some MATH (see REF). We know MATH is closed (see REF). Also, since MATH is nontrivial, we have MATH, so we may assume, by induction on MATH, that MATH is diffeomorphic to some MATH. Now MATH is a principal MATH-bundle over MATH. Because MATH, this bundle is trivial (see REF): MATH is MATH-equivariantly diffeomorphic to MATH. Then MATH as desired. |
math/0102191 | Since MATH, we know that the elements of MATH are simultaneously diagonalizable (over MATH), so their restrictions to the invariant subspace MATH are also simultaneously diagonalizable (compare CITE). Thus, MATH is a direct sum of weight spaces: MATH . For each weight MATH of MATH on MATH, we have MATH so MATH and MATH . The conclusion follows. |
math/0102191 | By REF , we may assume MATH. (So MATH and MATH (see REF).) From REF , we know that MATH and MATH have the same homology. Therefore MATH with equality if and only if MATH is compact CITE. Similarly, we have MATH . Combining these two statements, we conclude REF that MATH and, furthermore, REF that equality holds if and only if MATH is compact. |
math/0102191 | Suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) Because MATH acts properly discontinuously on MATH, the assumption on MATH implies that MATH also acts properly discontinuously on MATH (compare REF). So REF yields a contradiction. |
math/0102191 | It suffices to show that MATH is a NAME subgroup of MATH (see REF). We may assume, without loss of generality, that MATH (see REF); then MATH (see REF). REF implies that MATH is a NAME subgroup, so we may assume MATH; then MATH. Therefore MATH . Hence MATH, so, from REF , we see that, after replacing MATH by a conjugate subgroup, we may assume that MATH is normalized by MATH. Then, letting MATH and MATH in REF, we see that MATH. Since MATH, we have MATH, so this implies that MATH; therefore MATH contains MATH. Since MATH is a NAME subgroup (see REF), this implies MATH is a NAME subgroup, as desired. |
math/0102191 | Assume, without loss of generality, that MATH (see REF). Then MATH is torsion free, so MATH must act freely on MATH; therefore MATH is a compact manifold (rather than an orbifold). Because MATH is essentially the fundamental group of MATH (specifically, MATH), and the fundamental group of any compact manifold is finitely generated CITE, we know that MATH is finitely generated. From the NAME decomposition MATH, we see that MATH is homeomorphic to MATH, and REF asserts that MATH is homeomorphic to MATH, for some MATH. Obviously, we must have MATH, and we may assume MATH is not compact (otherwise, MATH is finite, so the desired conclusion is obvious), so REF implies that MATH. Thus, we conclude that MATH is connected at MATH. To complete the proof, we use a standard argument (compare CITE) to show that, because MATH is connected at MATH and MATH is compact, the group MATH has only one end. To begin, note that there is a compact subset MATH of MATH, such that MATH. Let MATH (compare CITE). Because MATH acts properly discontinuously on MATH, we know that MATH is finite; let MATH be a finite generating set for MATH, such that MATH. Suppose MATH, with MATH and MATH. (We wish to show there exist MATH and MATH, such that MATH; this establishes that MATH has only one end.) Because MATH is connected at MATH, there is a compact subset MATH of MATH, containing MATH, such that MATH is connected. Because MATH, we have MATH . Because MATH acts properly discontinuously on MATH, we know MATH and MATH are closed (and neither is contained in MATH), so connectivity implies that MATH: there exist MATH and MATH, such that MATH. Let MATH; then MATH, MATH, and MATH so MATH, as desired. |
math/0102191 | Suppose there is a compact subset MATH of MATH, such that each of MATH and MATH is infinite. (This will lead to a contradiction.) Let MATH be a (symmetric) finite generating set for MATH (see REF). We may assume MATH is so large that MATH for every MATH (see REF). We may also assume that MATH is convex and symmetric. Because MATH acts properly on MATH, there is a compact subset MATH of MATH, such that MATH (see REF). Furthermore, we may assume that MATH. Let CASE: MATH, CASE: MATH be the union of all the connected components of MATH that contain a point of MATH, and CASE: MATH. Then MATH is infinite (because MATH is infinite). Also, for any MATH and MATH, we have MATH, so MATH. Since MATH has only one end (see REF), this implies MATH is finite (see REF). Because MATH is infinite, we conclude that MATH. Because MATH separates MATH from MATH, and every connected component of MATH contains a point of MATH, we conclude that MATH. This contradicts the fact that MATH. |
math/0102191 | Suppose MATH acts properly discontinuously on neither MATH nor MATH. (This will lead to a contradiction.) From REF , we know there is a compact subset MATH of MATH, such that each of MATH and MATH is infinite. Then, since MATH, we may assume (by enlarging MATH) that each of MATH and MATH is infinite. This contradicts the conclusion of REF . |
math/0102191 | The desired conclusion is obtained by combining REF with REF. |
math/0102191 | We may assume MATH (see REF). From REF, we know that MATH is a NAME subgroup, so REF implies that MATH must be compact; thus, the trivial group MATH is a crystallographic group for MATH. However, since MATH (see REF), and MATH acts properly discontinuously on MATH, this contradicts REF. |
math/0102191 | Suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) By passing to finite-index subgroups, we may assume that MATH and MATH are connected. We will construct a volume form MATH on MATH that is exact: MATH. The integral of MATH over MATH is the volume of MATH, which is obviously not MATH, but NAME 's Theorem implies that the integral of any exact form over a closed manifold is MATH. This is a contradiction. Construction of a certain MATH-form MATH on MATH. Let MATH be the orthogonal complement to MATH, under the Killing form. Then MATH is a MATH-invariant subspace of MATH, such that MATH and MATH. Let MATH be the projection with kernel MATH, and let MATH be the corresponding left-invariant MATH-valued REF-form on MATH. The space MATH is a homogeneous principal MATH-bundle over MATH. It is well known CITE that MATH is the connection form of a MATH-invariant connection on this bundle, and that the curvature form MATH of this connection is given by MATH . Also, because MATH is abelian, the Structure Equation CITE implies MATH . By identifying the MATH-dimensional NAME algebra MATH with MATH, we may think of MATH and MATH as ordinary (that is, MATH-valued) differential forms. Because MATH and (hence) MATH are left-invariant, they determine well-defined forms MATH and MATH on MATH. (Note that MATH is a connection form on the principal MATH-bundle MATH over MATH, and the curvature form of this connection is MATH.) Because MATH is abelian, we have MATH for all MATH (compare CITE), so the horizontal form MATH determines a well-defined form MATH on the base space MATH. Construction, for a certain MATH, of a certain MATH-form MATH on MATH. Identifying MATH with MATH provides an ordering on the weights of MATH. Let CASE: MATH be the MATH weight space of MATH on MATH; CASE: MATH be the sum of the positive weight spaces of MATH on MATH; CASE: MATH be the sum of the negative weight spaces of MATH on MATH; CASE: MATH; and CASE: MATH be a nontrivial left-invariant MATH-form on MATH, such that MATH . Because MATH is left-invariant, and MATH, the form MATH determines a well-defined differential form MATH on MATH. (We remark that, because MATH, REF implies that the form MATH is unique up to a scalar multiple.) For a certain MATH, the wedge product MATH is a volume form on MATH. Let MATH. It suffices to show that the restriction of MATH to MATH is a symplectic form. It is obviously skew, so we need only show that it is nondegenerate. Thus, letting MATH we wish to show MATH. There is no harm in passing to the complexification MATH of MATH. Let MATH be a NAME subalgebra of MATH that contains MATH. Because MATH preserves the Killing form, centralizes MATH, and normalizes MATH, we know that MATH is MATH-invariant; thus, MATH is a sum of root spaces. Suppose there exists a nonzero element MATH of MATH, such that MATH belongs to some root space MATH. (This will lead to a contradiction.) There exists MATH, such that MATH for all MATH CITE. Because MATH, we have MATH. Then MATH, so MATH. We now know that MATH and MATH, so MATH. We therefore conclude, from the definition of MATH, that MATH. This is a contradiction. The form MATH is exact: we may write MATH. Let MATH be the connection form of a flat connection on MATH over MATH. (Since the principal bundle is trivial (see REF), it is obvious that there is a flat connection.) For any vector field MATH on MATH, let MATH be the lift of MATH to a vector field on MATH that is horizontal with respect to the flat connection MATH. Since MATH is abelian, there is a well-defined REF-form MATH on MATH given by MATH . Then MATH . For simplicity, assume that MATH . We have MATH. Let CASE: MATH be a basis of MATH, and CASE: MATH be the dual basis of MATH, with respect to the symplectic form MATH on MATH. Thus, MATH. Let MATH be a basis of MATH, write MATH and define MATH . MATH is independent of the choice of the basis MATH of MATH (with the understanding that MATH must be the dual basis of MATH). Let MATH for some MATH with MATH. Then MATH so MATH . Since MATH can be transformed into any other basis of MATH by a sequence of elementary operations as in REF, we conclude that MATH is independent of the choice of basis, as desired. We have MATH. NAME REF implies that MATH is centralized by MATH, so MATH is in the center of MATH. Let MATH be a NAME involution of MATH with MATH for MATH. NAME REF implies MATH. (From NAME REF, we see that MATH depends only on MATH and the chosen identification of MATH with MATH; MATH reverses the choice of identification.) Thus, MATH is a hyperbolic element of the center of MATH. By REF , this implies MATH, as desired. Completion of REF . Let MATH be the basis of MATH dual to MATH . We may assume MATH. Then, because MATH and MATH, we have MATH so MATH . Therefore MATH . From the choice of MATH, we have MATH, so MATH and MATH . Hence MATH . Since MATH, we have MATH, so the desired conclusion follows. MATH is exact. We have MATH . A contradiction. From REF , we know that MATH . On the other hand, REF implies that this integral is zero. This is a contradiction. |
math/0102191 | Suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) From REF, we know MATH. Let MATH and MATH be the two walls of MATH and, for MATH, let MATH. Because MATH is a ray (that is, a one-parameter semigroup), it is clear that MATH is a subgroup of MATH. From REF , we know that there is some MATH, such that MATH for every compact subset MATH of MATH. Since MATH, we have MATH, so this implies that MATH for every compact subset MATH of MATH. Also, because MATH, we have MATH, so MATH . Therefore MATH for every compact subset MATH of MATH. Hence, REF implies that MATH acts properly discontinuously on MATH. Then, because MATH (see REF), REF implies that MATH has a tessellation. This contradicts REF . |
math/0102191 | Suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) To simplify the notation somewhat (and because this is the only case we need), let us assume MATH is a one-parameter subgroup of MATH. Because MATH and MATH are unimodular (recall that MATH is semisimple (see REF) and MATH is discrete), there is a MATH-invariant measure (in fact, a MATH-invariant volume form) on the homogeneous space MATH CITE. Thus, the natural representation MATH of MATH on MATH, defined by MATH is unitary. Because MATH is noncompact, and acts properly on MATH, we know that any compact subset of MATH has infinitely many pairwise-disjoint translates (all of the same measure), so we see that MATH . Therefore, MATH has no nonzero MATH-invariant vectors, so, because MATH is tempered, we know that there is some MATH, such that MATH for all MATH and all MATH-invariant MATH. Because MATH is compact, there is a compact subset MATH of MATH, such that MATH; let MATH be the image of MATH in MATH. From the choice of MATH, we know, for each MATH, that there is some MATH, such that MATH . Because MATH is a compact subset of MATH, there is a positive, continuous function MATH on MATH with compact support, such that MATH and, by averaging over MATH, we may assume that MATH is MATH-invariant. Fix some large MATH. Because MATH is compact, and MATH has infinite volume (see REF), there is some MATH-invariant continuous function MATH on MATH, such that MATH, MATH and MATH . We have MATH . However, because MATH, we know that MATH. This is a contradiction. |
math/0102191 | Assume MATH. By passing to a subsequence, we may assume MATH converges weakly, to some operator MATH; that is, MATH . Let MATH . For MATH, we have MATH so MATH. Therefore MATH. We have MATH so the same argument, with MATH in the place of MATH and MATH in the place of MATH, shows that MATH. Because MATH is unitary, we know that MATH is normal (that is, commutes with its adjoint) for every MATH; thus, the limit MATH is also normal: we have MATH. Therefore MATH so MATH. Thus, MATH . By passing to a subsequence of MATH, we may assume MATH (see REF). Then MATH, so MATH. Hence, for all MATH, we have MATH as desired. The general case. From the NAME Decomposition MATH, we may write MATH, with MATH and MATH. Because MATH is compact, we may assume, by passing to a subsequence, that MATH and MATH converge: say, MATH and MATH. Then MATH by REF . |
math/0102191 | By passing to a subsequence, we may assume MATH is contained in a single NAME chamber, which we may take to be MATH. Then, by passing to a subsequence yet again, we may assume, for every positive real root MATH, that either MATH or MATH is bounded. Let CASE: MATH be the set of positive real roots; CASE: MATH be the set of positive simple real roots; CASE: MATH; and CASE: MATH. There is a compact subset MATH of MATH, such that MATH, so, because MATH, we know that MATH is not trivial. For each real root MATH, let MATH be the corresponding root subspace of MATH. Then MATH . Now, for MATH, we have MATH if and only if MATH is in the linear span of MATH. Thus, we see that MATH is precisely the unipotent radical of the standard parabolic subalgebra MATH corresponding to the set MATH of simple roots CITE. Similarly, MATH is the unipotent radical of the opposite parabolic algebra MATH. Because MATH is simple, the unipotent radicals of opposite parabolics generate MATH CITE, so MATH, as desired. |
math/0102191 | Write MATH. From the Real NAME Decomposition REF, we may assume, after replacing MATH by a conjugate subgroup, that MATH, where MATH is a hyperbolic one-parameter subgroup, and MATH is a unipotent one-parameter subgroup, such that MATH and MATH commute with each other. We may assume MATH, so MATH is nontrivial. Since the growth of the hyperbolic one-parameter subgroup MATH is exponential, while that of the unipotent one-parameter subgroup MATH is polynomial, there is some MATH, such that MATH for large MATH. Since the function MATH is in MATH, it follows from REF that MATH is tempered, as desired. |
math/0102191 | We have MATH (see REF), so MATH is a connected, one-dimensional, unipotent subgroup. Hence, the NAME Lemma CITE implies that there exists a connected, closed subgroup MATH of MATH, such that MATH contains MATH, and MATH is locally isomorphic to MATH. Then MATH is a NAME subgroup of MATH (see REF), so there is a compact subset MATH of MATH, such that MATH (see REF). Also, we have MATH. Therefore, REF applies. |
math/0102191 | We may assume MATH (see REF), so MATH (see REF). CASE: If MATH, then REF applies. CASE: If MATH, then REF applies. CASE: If MATH and MATH, then REF implies that MATH is tempered, so REF applies. |
math/0102191 | Since MATH, the desired conclusion follows from REF ; since MATH is simple, it also follows from REF . However, we give a proof that requires only the special case described in REF , rather than the full strength of REF or REF. Suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) Let MATH and MATH be the two walls of MATH. From REF , we know that there exists MATH, such that MATH is finite, for every compact subset MATH of MATH. Because MATH, we have MATH. On the other hand, MATH interchanges MATH and MATH. Thus, the preceding paragraph implies that MATH is finite, for every compact subset MATH of MATH. For MATH as in REF, we have MATH, so the conclusion of the preceding paragraph implies that MATH acts properly discontinuously on MATH (see REF). Now REF implies MATH is compact; thus, MATH has a tessellation. This contradicts REF . |
math/0102191 | Since MATH and MATH, it is easy to construct a continuous, proper map MATH such that MATH, for all MATH (compare REF ). For example, choose two linearly independent elements MATH and MATH of MATH, and define MATH . If we identify MATH with its NAME algebra MATH, then MATH is a convex cone in MATH and the opposition involution MATH is the reflection in MATH across the ray MATH. Thus, for any MATH, the points MATH and MATH are on opposite sides of MATH, so any continuous curve in MATH from MATH to MATH must intersect MATH. In particular, for each MATH, the curve MATH from MATH to MATH must intersect MATH. Thus, we see, from an elementary continuity argument, that MATH contains MATH. Therefore, MATH is contained in MATH. |
math/0102191 | Suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) We may assume MATH (see REF). Let MATH be a (symmetric) finite generating set for MATH, and choose a compact, convex, symmetric subset MATH of MATH so large that MATH for every MATH (see REF). From REF , we know that MATH, so REF implies that MATH. Then, because MATH acts properly on MATH, we conclude that MATH is finite (see REF). Since MATH is a proper map, this implies that MATH is finite. Let MATH and MATH be the two components of MATH. Because MATH, we know that MATH. Then, because MATH interchanges MATH and MATH, we conclude that MATH. Therefore, MATH and MATH have the same cardinality, so they must both be infinite. So MATH . Because MATH has only one end (see REF), this implies there exist MATH such that MATH for some MATH. Then MATH, MATH, and MATH . Using the fact that MATH is symmetric and the fact that MATH contains the identity element MATH, we conclude that MATH therefore MATH intersects both MATH and MATH. Since MATH separates MATH from MATH, and MATH is connected, this implies that MATH intersects MATH; hence MATH. This contradicts the fact that MATH (see REF). |
math/0102191 | From REF, we see that MATH . Thus, from REF, we see that MATH for MATH (and, since MATH is diagonal, we have MATH for MATH). Therefore, the desired conclusions follow from the definitions of MATH and MATH. |
math/0102191 | Choose MATH, such that MATH. Because MATH for MATH, and MATH (since MATH is compact), we have MATH and MATH so REF holds. Similarly, we have MATH so REF holds. For MATH, we know, from REF, that MATH and MATH. Thus, letting MATH, and using REF, we see that MATH and MATH as desired. |
math/0102191 | CASE: Let MATH. From REF, we see that MATH for MATH, so, using REF, we have MATH . Therefore MATH, as desired. CASE: Because MATH is compact, we have MATH (compare proof of REF). Then the desired conclusions follow from REF. |
math/0102191 | Because MATH is compact, we have MATH and MATH for any MATH (compare proof of REF). Thus, the desired conclusion follows from REF . |
math/0102191 | Define MATH as in REF, and choose MATH, such that MATH. For any MATH, we see, using REF or REF, that MATH, so MATH . Note that MATH. Also, we have MATH and MATH, so it is obvious that MATH and MATH. Therefore MATH . Thus, from the definition of MATH, we conclude that MATH, as desired. |
math/0102191 | By hypothesis, there is a continuous, proper map MATH, such that MATH. Because MATH, we know that MATH is homeomorphic to some Euclidean space MATH (see REF). Suppose, for the moment, that MATH. (This will lead to a contradiction.) We know that MATH for MATH. Because MATH and MATH (see REF), we must also have MATH for MATH. There is no harm in assuming MATH; then MATH (because MATH), so we conclude that MATH for all MATH. This contradicts the fact that MATH for MATH. We may now assume MATH. Then, because MATH is homeomorphic to MATH, it is easy to extend MATH to a continuous and proper map MATH. From REF, we know that the curve MATH stays within a bounded distance from the wall MATH; say MATH for all MATH. We may assume MATH is large enough that MATH for all MATH. Then an elementary homotopy argument shows that MATH contains MATH so MATH. Because MATH, we conclude from REF that MATH is a NAME subgroup. |
math/0102191 | Let MATH or MATH. Then MATH is a maximal compact subgroup of MATH. From the NAME decomposition MATH and the definition of MATH, we conclude that MATH. In the notation of REF, we see (from REF ) that MATH, where MATH . Then, since MATH (see REF) and MATH for MATH, we conclude that MATH is a wall of MATH. Furthermore, we have MATH for MATH (see REF), so MATH for MATH (see REF). |
math/0102191 | For any (large) MATH, we see from continuity (more precisely, from the Intermediate Value Theorem) that there exists MATH, such that MATH . Then, by assumption and from REF, we have MATH so there is a compact subset MATH of MATH, such that MATH (see REF). Therefore MATH as desired. |
math/0102191 | From REF, we have MATH and MATH for MATH. Thus, letting MATH, and using REF, we have: MATH and MATH so MATH and MATH. The desired conclusions follow. |
math/0102191 | CASE: By passing to a subspace, we may assume MATH. Then, by modding out MATH, we may assume MATH is an isomorphism onto its image. Define MATH by MATH, and let MATH be any extension of MATH. For MATH, we have MATH so MATH. CASE: From REF, we have MATH and MATH, so MATH. |
math/0102191 | By continuity, the function MATH attains a non-zero minimum and a finite maximum on the unit sphere. Because MATH is homogeneous of degree zero, these values bound MATH on all of MATH. |
math/0102191 | Because MATH is a closed subgroup of MATH, we know that it acts properly on MATH (see REF). Thus, it suffices to show that MATH is compact. From REF , we see that there is no harm in assuming MATH, and that there is a closed, connected subgroup MATH of MATH, such that CASE: MATH is conjugate to a subgroup of MATH, CASE: MATH, and CASE: MATH, for some compact subset MATH of MATH. (Unfortunately, we cannot assume MATH: we may not be able to replace MATH with MATH, because there may not be a cocompact lattice in MATH. For example, there is not lattice in MATH, because any group with a lattice must be unimodular CITE.) It suffices to show that MATH is compact. (Because MATH is compact, and MATH is compact, this implies that MATH is compact, as desired.) We know that MATH acts properly on MATH (see REF), so MATH acts properly on MATH, with quotient MATH. Therefore, REF implies that MATH has the same homology as MATH; in particular, MATH . From the NAME decomposition MATH, and because MATH is homeomorphic to MATH (see REF), we know that MATH is homeomorphic to MATH. Since MATH is contractible, this implies that MATH is homotopy equivalent to MATH, so MATH and MATH have the same homology; in particular, MATH . Since MATH this implies that the top-dimensional homology of the manifold MATH is nontrivial. Therefore MATH is compact CITE, as desired. |
math/0102191 | Let MATH and MATH. By assumption, we have MATH, for some MATH; let MATH. Because MATH and MATH are the two walls of MATH (see REF), we know that MATH acts properly on MATH (see REF); since MATH, this implies that MATH acts properly on MATH (see REF). Also, we have MATH (see REF), and there is a cocompact lattice in MATH (compare REF). Thus, the desired conclusion follows from REF . |
math/0102191 | CASE: Given MATH, write MATH, with MATH and MATH. We may assume that MATH (by replacing MATH with MATH if necessary (see REF)). It suffices to show MATH (for then MATH, so REF applies). Assume MATH is trivial. From REF, we see that MATH so MATH . From REF, we have MATH . From REF, we see that MATH for every nonzero MATH, so REF implies MATH . Also, because MATH (and MATH), we have MATH . Thus, MATH so MATH, as desired. The general case. From REF , we know MATH. Then, because MATH, we have MATH . Then, since MATH and MATH, we conclude that MATH, as desired. CASE: From REF, we see that MATH. Then, because MATH (see REF), REF implies that MATH has a tessellation. |
math/0102191 | Assume MATH. Because MATH for every MATH, it is obvious that REF holds. Assume MATH. If REF holds, then MATH so MATH is symplectic. The argument is reversible. |
math/0102191 | CASE: Given MATH, it suffices to show that MATH (see REF). Write MATH, with MATH and MATH. We may assume that MATH (by replacing MATH with MATH if necessary (see REF)). Let MATH be the real quadratic form MATH and let MATH be the MATH-subspace of MATH defined by MATH . For MATH, we have MATH. For MATH, we have MATH (because MATH and MATH is purely imaginary). Thus, the restriction of MATH to MATH is positive definite, so the desired conclusion follows from REF . We have MATH. From REF (with MATH), we have MATH and MATH . Then, from REF , we see that MATH, so MATH as desired. Completion of the proof. From REF , we have MATH . Also, from REF, we have MATH . Thus, it is easy to see that MATH so the desired conclusion follows from REF From REF, we see that MATH. Then, because MATH (see REF), REF implies that MATH has a tessellation. |
math/0102191 | CASE: We may assume MATH (by interchanging MATH and MATH if necessary). Assume MATH and MATH. For MATH, we know that MATH is not compact (see REF), so REF implies that there is a compact subset MATH of MATH, such that MATH. Then, letting MATH, we have MATH, so REF implies that MATH does not have a tessellation. This is a contradiction. Assume MATH and MATH. From REF , we know that there is a compact subset MATH of MATH, such that MATH. Therefore, REF (with MATH in the place of MATH) implies MATH. Then, because MATH (and MATH), we conclude that MATH is the graph of a homomorphism from MATH to MATH, as desired. Assume MATH. From REF , we know that MATH. Then, since MATH, we conclude that MATH is the graph of a homomorphism from MATH to MATH. Interchanging MATH and MATH yields the desired conclusion. CASE: We verify the hypotheses of REF , with MATH in the role of MATH. Let MATH be the image of MATH under the natural homomorphism MATH. Because MATH, we know that MATH is closed (see REF). It is well known (and follows easily from REF) that any closed subgroup acts properly on the ambient group, so this implies that MATH acts properly on MATH. From the definition of MATH, we have MATH, so we conclude that MATH acts properly on MATH; equivalently, MATH acts properly on MATH (compare REF). Because MATH, we have MATH. Also, we have MATH (see REF) and, from the definition of MATH, we have MATH. Therefore MATH . There is a cocompact lattice in MATH (compare REF). So REF implies that MATH has a tessellation. |
math/0102191 | Assume, for the Real NAME Decomposition MATH of each element MATH of MATH, that MATH and MATH belong to MATH. Let MATH be a maximal split torus of MATH. (Recall that a split torus is a subgroup consisting entirely of hyperbolic elements.) Then MATH is contained in some maximal split torus of MATH, that is, in some subgroup of MATH conjugate to MATH; replacing MATH by a conjugate, we may assume MATH. In other words, we now know that MATH is a maximal split torus of MATH. Given MATH, we have the Real NAME Decomposition MATH. By assumption, MATH; thus, MATH belongs to some maximal split torus MATH of MATH. A fundamental result of the theory of solvable algebraic groups implies that all maximal split tori of MATH are conjugate via an element of MATH CITE, so there is some MATH, such that MATH. Then MATH, being a subgroup of MATH, is a split torus. Thus, the maximality of MATH implies that MATH; let MATH. Then MATH . Since MATH is arbitrary, we conclude that MATH so MATH is compatible with MATH. The general case. Let MATH be the subgroup of MATH generated by the NAME Components of the elements of MATH. (Of course, since every element of MATH has a NAME Decomposition, we have MATH.) Then REF applies to MATH, so, replacing MATH by a conjugate, we may assume MATH, where MATH and MATH (see REF). Because MATH normalizes MATH (see REF), we know that MATH acts as the identity on MATH, for all MATH. Hence, REF implies that MATH acts as the identity on MATH, for all MATH; therefore MATH. Also, we have MATH. Thus, letting MATH, we have MATH . Because MATH and MATH is MATH-invariant, the adjoint action of MATH on MATH is completely reducible, so REF implies that there is a subspace MATH of MATH, such that MATH and MATH. Therefore, MATH, so MATH . Let MATH be the projection with kernel MATH, and let MATH. Then MATH so MATH. For any MATH, we know, from REF, that there exist MATH, MATH and MATH, such that MATH. Because MATH, we must have MATH and, because MATH, we have MATH. Therefore, MATH. We conclude that MATH, so MATH is compatible with MATH. |
math/0102191 | Let MATH, MATH, and MATH. Because MATH centralizes MATH, we have MATH . Also, because MATH, and MATH is normal, we have MATH, so MATH . Since MATH, we have MATH, so MATH and MATH . Thus, MATH as desired. |
math/0102191 | Because MATH is compatible with MATH, we have MATH, where MATH and MATH. We may assume that MATH, for otherwise REF holds. Therefore MATH. Because MATH is a sum of root spaces, this implies that there is a positive root MATH, such that MATH. Because MATH, we have MATH, so we must have MATH (otherwise we would have MATH, so MATH; hence REF holds). Therefore, MATH. Because MATH, we have MATH, so there is a nontrivial one-parameter subgroup MATH in MATH that is not contained in MATH. Because MATH centralizes MATH, we may write MATH where MATH is a one-parameter subgroup of MATH and MATH is a one-parameter subgroup of MATH. Furthermore, this decomposition is unique, because MATH. (In fact, MATH is the Real NAME Decomposition of MATH.) Define MATH by MATH for all MATH. CASE: For all MATH, we have MATH, which establishes one inclusion of REF. The other will follow if we show that MATH, so suppose MATH. Then REF implies that MATH, so it follows from REF (with MATH and MATH) that MATH, contradicting our assumption that MATH. CASE: Because MATH, we know that each of MATH and MATH normalizes MATH (see REF). Being in MATH, they also normalize MATH. Therefore, they normalize MATH. CASE: Suppose MATH. Because the intersection MATH is connected (see REF), and MATH, we must have MATH. Therefore MATH, so MATH. This contradicts our assumption that MATH. |
math/0102191 | CASE: Let MATH be the projection with kernel MATH, and let MATH be the NAME closure of MATH. From the structure theory of solvable algebraic groups CITE, we know that MATH is the semidirect product of a torus MATH and and unipotent subgroup MATH. Replacing MATH by a conjugate under MATH, we may assume that MATH. Since MATH we must have MATH, so MATH normalizes MATH (see REF). Then, since MATH, we conclude that MATH (see REF). CASE: From REF, we see that, by replacing MATH with a conjugate subgroup, we may assume MATH. Because MATH is a NAME subgroup (see REF), this implies MATH is a NAME subgroup. |
math/0102191 | Because MATH and MATH, there is some MATH, such that MATH. Let CASE: MATH be the element of MATH with MATH, CASE: MATH, and CASE: MATH. From REF, we see that CASE: MATH, CASE: MATH, and CASE: MATH, as desired. |
math/0102191 | CASE: Suppose there is a nonzero element MATH of MATH with MATH. Let MATH (see REF). We have MATH . Then, because MATH, it is easy to see that MATH. Also, we have MATH, so MATH, as desired. CASE: Suppose there is an element MATH of MATH, such that MATH, and MATH . Let MATH. Assume MATH. Because MATH, we must have MATH. Then, from REF, we know that MATH. So, from REF, we see that CASE: MATH, CASE: MATH whenever MATH, and CASE: MATH whenever MATH and MATH. This implies that MATH. Assume MATH. This is similar to REF . (In fact, this can be obtained as a corollary of REF by replacing MATH with its conjugate under the NAME reflection corresponding to the root MATH.) Assume MATH. Because MATH, REF implies there is some MATH, such that, letting MATH, we have MATH, MATH, and MATH. We show below that REF is satisfied with MATH in the place of MATH, so, from REF , we conclude that MATH. Thus, the desired conclusion follows from REF (with MATH). To complete the proof, we now show that REF is satisfied with MATH in the place of MATH. (This can be verified by direct calculation, but we give a more conceptual proof.) Because MATH, multiplication by MATH on the left performs a row operation on the first two rows of MATH; likewise, multiplication by MATH on the right performs a column operation on the last two columns of MATH. These operations do not change the determinant MATH: thus MATH . From REF and the definition of MATH, we see that MATH . Because MATH, we have MATH, so this simplifies to MATH . Thus, REF is equivalent to the condition that MATH. Then, since MATH we conclude that REF is also valid for MATH. |
math/0102191 | Assume MATH. Because MATH is a quadratic form of signature MATH on MATH, we know, from REF, that MATH. Thus, we may assume MATH, so there is some MATH, such that MATH. Assume there exists MATH, such that MATH and MATH. From REF, we see that MATH, with MATH and MATH. This contradicts REF. Assume there exists MATH, such that MATH. From REF, we see that MATH is an element of MATH, such that MATH, and MATH is a purely imaginary multiple of MATH. So REF applies (with MATH in the place of MATH). Assume MATH and MATH, for all MATH. From REF, we see that MATH, so the assumption of this subcase implies MATH. Thus, MATH so the desired inequality holds. If equality holds, then MATH and MATH. Thus, we may choose MATH, such that MATH, and MATH, such that MATH. From the assumption of this subcase, we know that MATH; thus, MATH. Therefore, REF applies, with MATH in the place of MATH. Assume MATH. Because MATH and MATH, the desired inequality holds unless MATH and MATH. Thus, we may assume there is some nonzero MATH and some MATH, such that MATH. Assume MATH. We may assume MATH (by replacing MATH with MATH, if necessary). Let MATH. From REF, we see that MATH, but MATH . Therefore MATH, so REF implies that MATH. This is a contradiction. Assume MATH. Let MATH be the element of MATH with MATH, and let MATH. Then MATH (see REF). Thus, by replacing MATH with the conjugate MATH (see REF), we may assume MATH. For any large real number MATH, let MATH be the element of MATH that satisfies MATH. Then, from REF, we see that MATH . Clearly, we have MATH. A calculation shows that MATH, and certain other MATH minors also have cancellation. With this in mind, it is not difficult to verify that MATH (see CITE for details). This is a contradiction. |
math/0102191 | By passing to a subgroup, we may assume MATH. Let MATH be the projection of MATH to MATH; then MATH. Assume MATH for every MATH. Assume MATH is even. From REF, we know that MATH does not intersect MATH (or MATH, either, for that matter), so MATH . Therefore MATH as desired. (If equality holds, then we have Conclusion REF.) Assume MATH is odd. We have MATH. Suppose not: then MATH . (This will lead to a contradiction.) Let MATH, so MATH is a MATH-subspace of MATH. For each MATH, there is some MATH, such that MATH; define MATH. By the assumption of this case, we know MATH so MATH is uniquely determined by MATH; thus, MATH is a well-defined MATH-linear map. Also, again from the assumption of this case, we know that MATH . Because MATH, we have MATH . If MATH (that is, if MATH), this implies MATH, so MATH is defined on all of MATH. Because MATH is odd, this implies that MATH has a real eigenvalue, which contradicts REF. We may now assume MATH. Let CASE: MATH, where MATH, CASE: MATH be the projective space of the real vector space MATH, and CASE: MATH, for MATH, so MATH is a vector bundle over MATH. Define MATH by MATH. Any MATH-linear transformation MATH is a continuous function, such that MATH for all MATH; that is, a section of MATH. Thus, MATH, MATH, and MATH each define a section of MATH. Furthermore, these three sections are pointwise linearly independent over MATH, because REF implies that MATH, MATH, and MATH are linearly independent over MATH, for every nonzero MATH. On the other hand, the theory of characteristic classes CITE implies that MATH does not have three pointwise MATH-linearly independent sections (see CITE for details). This is a contradiction. Completion of the proof of REF . From REF , we see that the desired inequality holds. We may now assume MATH and MATH. Since MATH, we must have MATH, so MATH. Therefore MATH, so MATH as desired. Assume there is some MATH, such that MATH. Assume MATH. Since MATH, we must have MATH. Then MATH for every MATH (otherwise REF yields a contradiction); this implies MATH and MATH . Because MATH, we know that MATH; so MATH . Since MATH for every MATH, but MATH (see REF), we must have MATH for every MATH. Therefore MATH . Let MATH be the natural projection. Note that MATH . (If MATH, with MATH, then there is some MATH, such that MATH. We also have MATH, so, from REF, we see that MATH. Thus MATH. So MATH is REF-dimensional.) Because MATH for every MATH, and MATH is a NAME algebra, we see, from REF, that MATH must be a totally isotropic subspace for the symplectic form MATH, so MATH . Therefore MATH . This completes the proof if MATH: CASE: If MATH is even, then, because MATH, we have MATH. CASE: If MATH is odd, then MATH, so MATH. Now let MATH, and suppose MATH. (This will lead to a contradiction.) Because equality is attained in the proof above, we must have MATH, so there exists MATH with MATH. For MATH, let MATH. Then MATH . Thus, this expression changes sign, so it must vanish for some MATH. On the other hand, since MATH, we have MATH for every MATH. Thus REF yields a contradiction. Assume MATH. This is similar to REF . (In fact, this can be obtained as a corollary of REF by replacing MATH with its conjugate under the NAME reflection corresponding to the root MATH.) Assume MATH. Because MATH, REF implies there is some MATH, such that, letting MATH, we have MATH, MATH, and MATH. There is no harm in replacing MATH with MATH (see REF). Then REF applies (with MATH in the place of MATH). |
math/0102191 | Let MATH . From REF, and REF , we have MATH . This implies the desired inequality, unless MATH, MATH, and we have equality in both REF . This is impossible, because equality in REF requires MATH, but REF implies MATH. In the remainder of the proof, we assume that equality holds in REF, and that MATH is even. REF implies REF. Assume MATH. Because equality holds, REF implies REF. Then REF implies REF (because MATH). Since MATH (see REF) and equality holds in REF, we have MATH and MATH so REF hold. Let MATH. REF implies that MATH normalizes MATH, so, from REF , we see that MATH. On the other hand, MATH, so, from equality in REF, we conclude that MATH. Therefore MATH. Suppose MATH is any continuous group homomorphism, such that MATH normalizes MATH. From REF, we see that MATH, so MATH must be trivial. This implies that REF cannot apply here, so REF yields REF. Assume MATH. REF implies that MATH for every MATH. Suppose REF is false. Then there is some MATH, such that MATH. Also, because MATH, we may fix some MATH. Then, letting MATH, we see, from REF, that MATH and MATH, so MATH. This contradicts the conclusion of the preceding paragraph. Conclusion REF follows from REF. Conclusion REF can be established by arguing as in the last two paragraphs of REF establish REF. |
math/0102191 | CASE: Suppose there is an element MATH of MATH, such that MATH. Let MATH. Then, from REF, we see that MATH. Furthermore, MATH . Because MATH, we have MATH, so MATH; therefore MATH, so MATH so REF implies that MATH, as desired. CASE: Suppose there is an element MATH of MATH, such that MATH; in other words, we have MATH. Let MATH (see REF). Then MATH and MATH so MATH so REF implies that MATH, as desired. CASE: Suppose there is an element MATH of MATH, such that MATH, MATH, MATH, and MATH. Let MATH. From REF, we see that MATH (note that, because MATH, we have MATH). Then MATH and MATH . So MATH. Thus, REF implies that MATH, as desired. |
math/0102191 | Because MATH, we may assume that there exist nonzero MATH, such that MATH and MATH (otherwise, the projection to one of the factors of MATH is injective when restricted to MATH, so REF holds). Then MATH, so, by assumption, we have MATH. Because MATH and MATH are nonzero, this implies MATH. For all MATH, we have MATH. We may assume MATH (otherwise the desired conclusion is obvious). Then, since MATH, we conclude that MATH. Similarly, because MATH we must have MATH. Therefore MATH . Since MATH, this implies MATH, so MATH, as desired. We have MATH. Given MATH, REF asserts that MATH. By symmetry (interchanging the two factors of MATH), we must also have MATH. So MATH, as desired. Completion of the proof. From REF , we have MATH . If MATH, then REF holds; otherwise, REF holds. |
math/0102191 | Note that MATH (see REF) and MATH . We have MATH. Suppose not. Let MATH be the projection of MATH to MATH. We have MATH and, for every MATH with MATH, we have MATH (see REF), so REF implies that MATH. Therefore MATH. Then, because MATH, we know that MATH and MATH; thus, there exist MATH, such that CASE: MATH, MATH; and CASE: MATH, MATH. Therefore MATH is a nonzero element of MATH (see REF), so MATH. This contradicts REF. We have MATH. Suppose not: then, because MATH, there is some MATH, and, because MATH, there is some nonzero MATH, such that MATH. We must have MATH (otherwise REF yields a contradiction); thus MATH. We must have MATH (otherwise REF yields a contradiction). Thus, we see that MATH is nonconstant as a function of MATH, so REF yields a contradiction. The desired inequality. We have MATH as desired. In the remainder of the proof, we assume that MATH. We must have equality throughout the preceding paragraphs. We have MATH. Suppose not: then there is some MATH, such that MATH. Let MATH and MATH. Then, from REF, we see that MATH and MATH, and that MATH. From REF, we have MATH and MATH, so MATH . Therefore MATH, so MATH (see REF), so REF yields a contradiction. We have MATH. From REF , together with the fact that MATH we conclude that MATH. Therefore, MATH so MATH . Because MATH, we must have MATH. We have MATH. Let MATH be the projection of MATH to MATH. Then there exists MATH, such that MATH, and, in the notation of REF , we have MATH . Because MATH normalizes MATH (see REF), we know, from REF , that MATH. Since MATH, we know that MATH projects nontrivially (in fact, surjectively) to MATH. On the other hand, we know that MATH (otherwise REF yields a contradiction). Therefore MATH, so there must be a positive root MATH, such that MATH. Then MATH; since MATH, we must have MATH. Because MATH, we have MATH . Then, since MATH, we must have MATH . By inspection, we see that this implies MATH, so we conclude that MATH, as desired. We have MATH, MATH, and MATH. Since MATH, it suffices to show MATH. Thus, let us suppose MATH. (This will lead to a contradiction.) We have MATH (and recall that MATH), so there is some MATH, such that MATH and MATH. Because MATH, we have MATH so there is some MATH, such that MATH and MATH. Then MATH, with MATH (see REF), so REF yields a contradiction. We have MATH. Suppose not: because MATH, we conclude that there is some nonzero MATH, such that MATH normalizes MATH (see REF). If MATH, then, because MATH, there is some MATH, such that MATH and MATH. Then MATH, with MATH (see REF), so REF yields a contradiction. If MATH, then, since MATH, we must have MATH. There is some MATH with MATH. Then MATH, with MATH (see REF), so REF yields a contradiction. Completion of the proof. CASE: From REF , We know that MATH, and, from REF , that MATH. CASE: Since MATH, it suffices to show MATH: given MATH, we wish to show MATH. Because MATH, we know that MATH. Thus, all that remains is to show that MATH. If not, then choosing MATH with MATH, we see that MATH (see REF). So REF yields a contradiction. CASE: From REF, we know that MATH for every MATH. |
math/0102191 | Assume Conjecture REF is true, and suppose MATH is a crystallographic group for MATH. (This will lead to a contradiction.) Let MATH (see REF). From REF, we have MATH and MATH, where MATH. We may assume that MATH (see REF), and that MATH is compatible with MATH (see REF). Because MATH is not a NAME subgroup (see REF), the contrapositive of REF implies, for some MATH, that there does not exist a continuous curve MATH in MATH, such that MATH. Therefore, either REF (if MATH) or REF (if MATH) implies that MATH. We consider two cases. Assume that MATH acts properly discontinuously on MATH. REF (combined with the fact that MATH) implies that MATH has a tessellation. This contradicts either REF (if MATH) or Conjecture REF (if MATH). Assume that MATH does not act properly discontinuously on MATH. From REF , we know that MATH and MATH are the two walls of MATH, so REF (combined with the assumption of this case) implies that MATH acts properly discontinuously on MATH. Therefore, since Conjecture REF asserts that MATH does not have a tessellation, the contrapositive of REF (with MATH in the role of MATH) implies that MATH. Hence, the contrapositive of REF implies there is a continuous curve MATH in MATH, such that MATH. Thus, there is a compact subset MATH of MATH, such that MATH (see REF). Since MATH acts properly discontinuously on MATH, this implies that MATH acts properly discontinuously on MATH (see REF). This contradicts the assumption of this case. |
math/0102191 | CASE: See REF (and REF), or REF. CASE: Let MATH, so MATH. By combining REF , and REF , we see that MATH . Also, we may assume MATH is compatible with MATH (see REF). Because MATH is not a NAME subgroup (see REF), REF implies that one of the following two cases applies. Assume there does not exist a continuous curve MATH in MATH, such that MATH. Since MATH, REF implies that MATH, and that MATH is of the form MATH (with MATH), where CASE: MATH; CASE: MATH for every MATH; CASE: MATH, for every MATH; CASE: MATH for every nonzero MATH; CASE: MATH; and CASE: MATH. We may assume that MATH. Because MATH (see REF), it suffices to show that MATH for all MATH. This is trivially true if MATH, as MATH in this case. Thus, we assume MATH. For any MATH with MATH, we know, from REF, that MATH; therefore, REF implies there exist MATH-linear maps MATH and MATH, such that MATH for all MATH. More concretely, we may say that there exist MATH and MATH, such that MATH for all MATH. Let MATH be the element of MATH with MATH, and let MATH be the conjugate of MATH by MATH. Then MATH satisfies the conditions imposed on MATH (note that MATH, like MATH, is compatible with MATH (see REF)), so there exist MATH and MATH, such that MATH for all MATH. Given MATH with MATH, let MATH. Because MATH, we have MATH, so, from REF , we see that MATH and MATH . Therefore MATH . Since MATH is arbitrary, this implies MATH. Thus, by replacing MATH with MATH, we may assume that MATH. This means that MATH for all MATH. From REF (and because MATH, so MATH), we know that MATH, so there is some nonzero MATH, such that MATH. (So MATH.) Then MATH, so we see, from REF, that MATH for every nonzero MATH. Now, since MATH there is some nonzero MATH, such that MATH. We have MATH . Because MATH is pure imaginary, we know that MATH, so this implies that MATH. Thus, replacing MATH by a conjugate under a diagonal matrix, we may assume MATH, as desired. Setting MATH, we have MATH. Since MATH (see REF), we have CASE: MATH, CASE: MATH, and CASE: MATH. Thus, in the notation of REF , we have MATH and MATH, so MATH, as desired. (Note that this is a direct sum of vector spaces, not of NAME algebras: we have MATH.) Completion of the proof of REF . For any MATH with MATH, we have MATH so MATH (see REF); therefore, REF implies there is a MATH-linear map MATH, such that MATH for all MATH. Then, because MATH (see REF), we must have MATH . Combining this with REF and the conclusions of REF , we see that MATH. Therefore MATH, so Conclusion REF holds. From REF, we see that Conclusion REF holds. Letting MATH, for any MATH, we see, from REF, that MATH and MATH . From REF , we know that MATH, so this implies that Conclusion REF holds. Assume there does not exist a continuous curve MATH in MATH, such that MATH. Since MATH, REF implies that MATH, and that MATH is of the form MATH, where CASE: MATH, CASE: MATH, and CASE: MATH for every MATH. Let MATH (so MATH). Let MATH be the sesquilinear form (or bilinear form, if MATH) on MATH defined by MATH . Let MATH . From REF, we see that the restriction of MATH to MATH is a (positive-definite) inner product. Let MATH be the MATH-orthogonal complement to MATH. As a form over MATH, MATH has signature MATH. Thus, as a form over MATH, MATH has signature MATH. Since MATH we conclude that MATH is a MATH-dimensional MATH-subspace on which MATH is negative-definite. Choose some nonzero MATH. Multiplying by a real scalar to normalize, we may assume MATH. Because MATH is transitive on the vectors of norm MATH, there is some MATH, such that MATH. Thus, letting MATH and MATH, we have MATH, so, by replacing MATH with the conjugate MATH, we may assume MATH. Then MATH . Assume MATH. By comparing REF (with MATH), we conclude that MATH . By comparing dimensions, we see that equality must hold; this establishes Conclusion REF. Assume MATH. Choose some nonzero MATH, such that MATH is MATH-orthogonal to MATH. Multiplying by a real scalar to normalize, we may assume MATH. By replacing MATH with MATH if necessary, we may assume MATH. Because MATH is MATH-orthogonal to MATH, we have MATH (see REF). Let MATH and MATH. Then MATH so MATH . Also, MATH so MATH. Thus, MATH and MATH are of opposite signs so, because MATH, we have MATH . Therefore, we may choose MATH, such that MATH . Let MATH . Then MATH and MATH . Hence, there is some MATH, such that MATH and MATH. Thus, replacing MATH with the conjugate MATH (compare REF), we may assume MATH. Therefore MATH . By combining this with REF and comparing with REF (with MATH), we conclude that MATH. By comparing dimensions, we see that equality must hold; this establishes Conclusion REF (because MATH). |
math/0102191 | CASE: This is REF . REF provides us with three possibilities. CASE: In each case, we have MATH (see REF, and REF). CASE: In each case, there is some MATH, such that MATH for MATH (see REF, and REF). Then REF implies either that MATH (if MATH) or that MATH (if MATH). |
math/0102191 | Suppose MATH, for some MATH. Because all maximal split tori in MATH are conjugate, we may assume that MATH normalizes MATH. Since all roots of MATH on both MATH and MATH are positive, MATH cannot invert MATH, so we conclude that MATH centralizes MATH; that is, MATH. In the notation of REF of the proof of REF , define MATH (compare REF). Then MATH, so we may assume MATH (because MATH, being a subgroup of MATH, obviously normalizes MATH). Write MATH. Then, because MATH, we must have MATH; hence MATH. For any basis MATH of MATH with MATH and MATH, we have MATH . Similarly for any MATH-orthonormal basis MATH of MATH. Because MATH, this implies MATH. Because MATH, we conclude that MATH, as desired. |
math/0102193 | Suppose we pick a site MATH, flip a coin, and adjust the height accordingly. Then the expected value of the new height at MATH is just MATH. Assume that we pick each site (other than MATH and MATH) with probability MATH, where MATH is the number of positions that can be picked. Then with primes denoting the updated variables, MATH when MATH, so that MATH where we have used MATH in REF , MATH in REF , and the trigonometric identity MATH in REF is justified if MATH, since by REF for each MATH, and MATH; when MATH it becomes an equality. To upper bound the right-hand side we use MATH. MATH where to get the last line we used MATH when MATH and MATH. Here MATH, so these conditions are satisfied whenever MATH. The lower bound is somewhat easier: use the bound MATH. |
math/0102193 | To obtain the upper bound, we consider a pair of coupled paths MATH and MATH such that MATH is the topmost path, MATH is the bottommost path. The sequences MATH and MATH are generated by the NAME chain using ``the same random moves", so that MATH and MATH are obtained from MATH and MATH respectively by sorting or unsorting (same random decision made in both cases) at the same random location MATH. One can check by induction that MATH. Let MATH; MATH if and only if MATH. We compute MATH; when it is small compared to the minimum possible positive value of MATH, it will follow that with high probability MATH. By choosing MATH to be slightly smaller than MATH, we make this minimum positive value not too small, and thereby get a somewhat improved upper bound. From REF and induction, we get MATH . But MATH. Thus after MATH steps MATH. We have MATH, and MATH. The optimal choice of MATH is MATH, but all that matters is that MATH as MATH while MATH. Substituting, we find that MATH steps are enough to ensure that the probability of coalescence is at least MATH. |
math/0102193 | Perturb the stationary distribution by an eigenvector associated with the second largest eigenvalue MATH, and run the NAME chain starting from this distribution. After MATH steps the variation distance from stationarity is MATH. If we run the NAME chain starting from the top and bottom states, after MATH steps the states are different with probability at most MATH. The coupling time bound on the variation distance gives MATH. |
math/0102193 | By REF , our function MATH satisfies the contraction property required by REF when MATH, and MATH . The constraint on MATH is satisfied when MATH. To get a bound MATH, observe that any path MATH can have at most MATH local extrema, so MATH. But MATH, so MATH. The maximal path maximizes MATH, giving MATH. Substituting into REF and simplifying, MATH, so the numerator becomes MATH for bounded values of MATH, giving our lower bound of MATH. |
math/0102193 | Let MATH. By induction MATH . By our assumptions on MATH, in equilibrium MATH. With MATH denoting MATH, we have MATH for each MATH. To get the last line we used our constraints on MATH and MATH: MATH, so when MATH is odd, MATH. When MATH is even, we need MATH as well, which is satisfied when MATH or MATH. From NAME 's inequality, MATH . As MATH, if MATH, then the probability that MATH deviates below MATH is at most MATH, and the probability that MATH in stationarity deviates above this threshold is at most MATH, so the variation distance between the distribution at time MATH and stationarity must be at least MATH. If we take the initial state to be the one maximizing MATH, then MATH when MATH . |
math/0102193 | The lower bound comes from considering the MATH-th threshold function. By NAME REF, after MATH steps the variation distance of just this one threshold function from stationarity is MATH. The variation distance from stationarity of the permutation itself is at least as large. The upper bound follows from REF when we take MATH. As MATH, after MATH steps the probability of any one given threshold function differing for the upper and lower permutations is MATH. The probability that the upper and lower permutations differ is at most the expected number of threshold functions for which they differ, which is at most MATH. Taking MATH but MATH, after MATH steps coalescence occurs with probability MATH. |
math/0102193 | We apply REF here to lower bound the mixing time of the lozenge tiling NAME chain proposed by NAME, NAME, and NAME, when the region is a regular hexagon with side lengths MATH. Our potential function has the required contraction property, with MATH. MATH. Since we are using nonlocal moves, MATH can be as large as MATH. Suppose that when the NAME chain picks a site on a path and tries to push it, there are MATH paths in the way. With probability MATH, and otherwise MATH. Conditioning on this site being selected, MATH, and in general MATH. Applying REF , we obtain a mixing time lower bound of MATH . |
math/0102193 | After MATH steps the probability that the two permutations are still different is at most the expected number of items that are in different positions in the two permutations, which is at most MATH. Setting this equal to MATH gives the desired bound. |
math/0102193 | What we have analyzed in REF is a pairwise coupling of a NAME chain on permutations, that is, an update rule that updates pairs of permutations. This pairwise update rule does not extend to a grand coupling, that is, there is no update rule defined on all permutations such that pairs of permutations evolve according to the above pairwise coupling. But let us look at the evolution of a threshold function of the two permutations. Normally both permutations are either sorted or reverse-sorted at a location, which corresponds to a push-down or push-up move in the threshold functions. The exceptional case where one permutation is sorted while the other is reverse-sorted occurs when in one permutation items MATH and MATH are in adjacent locations MATH and MATH, while the other permutation has items MATH and MATH at these locations, and either MATH or MATH. Any given threshold function will map MATH to either MATH or MATH, and then one of the two permutations will have either an up-slope or down-slope at locations MATH and MATH, and for that permutation an observer would be unable to tell whether a sort or reverse-sort operation was performed. Thus from the standpoint of an observer, it appears as if the two threshold functions were evolving according to the monotone grand coupling considered earlier, and our bound on the pairwise coupling time for permutations translates to a bound on the grand coupling time for lattice paths. Next we can convert the bound on grand coupling time for lattice paths into an upper bound on the grand coupling time for permutations for the straightforward sort/reverse-sort coupling. After MATH steps the pairwise permutation coupling (and hence the lattice path grand coupling) has coalesced except with probability MATH, so the permutation grand coupling has coalesced except with probability MATH. |
math/0102193 | Since we are interested in the probability that the random walk has not hit the diagonal, and the regions below and above the diagonal behave symmetrically, let us consider the state transition matrix MATH for the random walk above the diagonal (MATH), where the random walker reflects off the boundaries of the grid, and dies when it hits the diagonal. The matrix MATH resembles a stochastic matrix, except that for those rows corresponding to states next to the diagonal, the row-sum will be less than one. For the reader's convenience we proceed to diagonalize the matrix MATH; other triangular regions with different boundary conditions have been similarly diagonalized in for example, CITE. For MATH, let the function MATH be defined by MATH . For convenience let the values of the grid coordinates MATH and MATH range from MATH to MATH. Since MATH is an eigenvector of the nearest-neighbor random walk on MATH with transition probabilities MATH, and its eigenvalue is MATH . Since furthermore MATH, MATH, MATH, MATH, and MATH, it follows that MATH is also an eigenvector of MATH with eigenvalue MATH. Next we show that any two of these MATH eigenvectors are orthogonal, and that each is not identically zero: MATH . Since MATH and MATH the second term is zero. The first term is also zero unless both MATH and MATH, giving us orthogonality. If MATH, then we find that this inner product is MATH, so the eigenvectors are nontrivial. Hence, we have an orthogonal eigenbasis of the matrix MATH. Suppose that the random walker starts at MATH. Let MATH be the function which is MATH at the starting location and zero elsewhere. Let MATH denote the function which is MATH whenever MATH. We have MATH . We need a bound on MATH to bound this summation, and to this end consider the line passing through MATH and MATH. When MATH, the line is at least as high as MATH when MATH. If MATH, the line's slope increases towards MATH, so it continues to be above MATH when MATH, and by concavity of MATH for MATH, the line is also above MATH when MATH. Taking MATH (assume MATH - the lemma is trivial if MATH) and MATH, we have MATH . |
math/0102193 | As an upper lattice path MATH and lower lattice path MATH evolve together via the push down / push up coupling, let us look at the difference path MATH. If MATH goes up and MATH goes down, which we will denote MATH , then the difference path MATH goes up, which we denote with U. If MATH goes down and MATH goes up (MATH ), then MATH goes down REF . In the remaining two cases (MATH and MATH ) the difference path remains flat REF . We may view the difference path as a string of U, F, and D particles, and it is easy to check that the evolution of the difference path is a NAME process: If the particles at the updated site are UU=MATH , then they remain MATH =UU. If a UD=MATH is updated, the result is either MATH =FF or MATH =FF. If a UF is updated, the underlying paths might be MATH and then change to MATH =FU or MATH =UF, or the underlying paths might be MATH and then change to MATH =UF or MATH =FU. Likewise, if a FF is updated, there are four possibilities for the underlying paths, and in each case the updated configuration is FF. The other cases (DD, DU, DF, FU, and FD) are similar and related to the above cases by symmetry. We may summarize the update rules for the string of U's, D's, and F's as follows: pick a random adjacent pair, and with probability MATH exchange them; when a D and U are exchanged past each other, they both turn into F's. If we start with the top path and bottom path, then in the difference path every U will be to the left of every D. We will do a number of comparisons between a random permutation MATH and the difference lattice path MATH. For the MATH-th comparison (MATH), look at the locations of cards MATH, and in particular their relative order. Let MATH denote the first of these cards encountered in a left-to-right scan, and in general MATH, MATH, denotes the MATH-th such card encountered. Label the first MATH U's of the difference path with the numbers MATH. Similarly let MATH, MATH, denote the MATH-th card from the cards MATH to be encountered in a right-to-left scan of the permutation, and label the MATH-th to last D of the difference path with MATH. We leave the remaining particles of the difference path unlabeled. When we evolve the difference path via random exchanges, we will let labeled particles be exchanged past each other, but a labeled U or D may not be exchanged past an unlabeled U or D. This rule for the labels does not affect the evolution of the unlabeled difference path, but it is important for our understanding of it. Initially for each MATH, the position of card MATH in the difference path is weakly to the left of card MATH in the permutation, while the position of card MATH in the difference path is weakly to the right of card MATH in the permutation. We will pick the same random adjacent pair in the permutation as in the labeled difference path, and make the same decision as to whether or not to exchange the adjacent items. Consider the first time that the above invariant fails to hold, say that card MATH in the labeled difference path moves to the right of card MATH in the permutation. On the previous step, card MATH was in the same location in the difference path and the permutation. The exchange could not have been to the right of the card MATH, because exchanges in the permutation always succeed, nor could the exchange be to the left of card MATH, as any particle to the left of card MATH is either an F or a labeled U, and such exchanges succeed. Thus the invariant is maintained. Consider the locations of two cards MATH and MATH in a random permuation MATH, or two labels in the difference path MATH. Let the weighted gap between them be defined by MATH, where the sum is taken over positions MATH between the two cards, and is negative if card MATH occurs before card MATH. Within a random permutation MATH we have MATH . The area under the difference path is the sum of the locations of the D particles minus the sum of the locations of the U particles. The potential function (the weighted area) is MATH . Since we started with a random permutation and the dynamics are reversible, then even conditional upon all the past moves, the permutation is still uniformly random. In particular MATH is independent of the maximum height MATH of the difference path, so that MATH . Note that this gives another proof that coalescence is likely after MATH steps. Notice that the difference path never changes by more than one at a time, and only if a U or D particle moves. There are MATH U and D particles, each particle can move in one of two directions, and a given proposed exchange occurs with probability MATH. Thus MATH . MATH . Thus if MATH, w.h.p. MATH, so that the time until coalescence is w.h.p. at least MATH. |
math/0102193 | The first relation is just REF , which is where we use the multiplicity-one assumption. To compute the variance of MATH in stationarity, we let MATH denote the total height in column MATH (MATH), and write MATH and use a formula CITE for the covariances of the MATH's MATH to find (with the help of Maple) that MATH . To evaluate MATH we write MATH which with the help of Maple simplifies to MATH . |
math/0102193 | For random adjacent transpositions on permutations, for any given card MATH one can define an eigenvector with eigenvalue MATH based on the location MATH of that card: MATH . There is one linear dependency amongst these eigenvectors (MATH), and it appears that there are no other eigenvectors with eigenvalue MATH. Note that MATH . By symmetry considerations MATH when MATH. Since MATH we have that MATH when MATH. As before, to determine MATH we compute the coefficients of the MATH's in the eigenbasis decomposition of MATH. Since there is a linear relation amongst the MATH's there will be a one-parameter family of valid sets of coefficients - we just need one such valid set of coefficients. We could be methodical and use the NAME procedure to extract MATH orthogonal vectors from the MATH's and then use these to get a valid set of coefficients, but the guess-and-verify method is less messy. Consider the function MATH . Since the dot products with the MATH's are the same (up to the constant factor MATH), by linear algebra we conclude that MATH has the desired coefficients. Next we evaluate this eigenfunction at MATH and multiply by MATH to obtain MATH: MATH . |
math/0102199 | Note that if MATH and MATH are disjoint vertex sets then MATH . The factor MATH in the above expression appears since common boundary edges of MATH and MATH are not boundary edges of their union. Then MATH. The hypothesis can be used to bound the second term. The last term equals twice the total weight on edges with one endpoint in MATH and the other endpoint in MATH; this does not increase if we require the latter endpoint to be in a subset of MATH. Therefore MATH . |
math/0102199 | It suffices to prove this for two MATH-isolated cores MATH. Let MATH. Two applications of the lemma imply MATH, and one inequality is strict unless MATH. |
math/0102199 | Any MATH-island in MATH has boundary volume at least MATH and positive MATH-isolation, thus volume greater than MATH. This gives MATH. Similarly, each MATH-island has volume greater than MATH, so no MATH-island is a subset of an island in MATH. But since MATH, this implies that MATH. |
math/0102199 | For MATH, let MATH denote MATH, and let MATH denote the union of islands intersecting MATH. Then MATH . The bound on the first term of the second inequality holds since MATH is a union of islands. By REF we have MATH . Therefore, using that MATH and MATH are disjoint, we get MATH . By the anchored expansion property there are only finitely many such sets MATH containing MATH. The lemma follows. |
math/0102199 | We start by constructing regions MATH, which are islands or unions of islands, together with bridge structures MATH connecting these islands if they are disjoint. First, for each MATH, label each MATH-island with volume in MATH as a level MATH region, and for these regions MATH, set MATH. Define the waters of a level MATH region MATH as MATH. Then, for MATH (in this order), consider a maximal matching of pairs of level MATH regions whose waters intersect, and label the union MATH of each matched pair MATH of regions a level MATH region. Set MATH to be the union of MATH, MATH and a shortest (minimal length) bridge connecting MATH and MATH. For every level MATH region MATH and vertex MATH in MATH or in its waters, consider the bridge structure MATH given by the union of MATH and a shortest bridge connecting MATH and MATH. We have MATH . In the second expression the first term is an upper bound on the length of a shortest bridge connecting MATH and MATH plus MATH. The first factor in the sum is an upper bound on the number of pairs of level MATH regions contained in MATH; the second factor is an upper bound of the length of a shortest bridge connecting such a pair. Since then MATH, it follows by REF that each vertex is contained in only finitely many regions or their waters. Therefore, every island is contained in a maximal region, that is a region which is not contained in any other regions. Call the union of a maximal region and its waters a country of level of the maximal region. Call the region itself the land of the country. This construction clearly satisfies the properties claimed in the proposition. |
math/0102199 | For a small MATH, let MATH. We have MATH . For sufficiently large MATH, this is bounded above by MATH, which is summable, so MATH eventually a.s. |
math/0102199 | The quantities MATH, MATH, MATH, MATH do not change if they are considered (for the walk) in the graph MATH instead of the graph MATH, so we will do this. Denote MATH, the NAME and NAME kernels of the walks on MATH, and MATH, respectively. Recall that MATH, so we have MATH and so from REF we get MATH and these inequalities also hold with MATH replaced by MATH everywhere. For the walk on MATH started at MATH, the probability of moving into MATH from the outside in one step is given by the function MATH which equals MATH outside MATH, and REF in MATH. Thus the chance of moving into MATH from the outside after MATH steps in MATH is given by MATH, and therefore MATH . In NAME kernel notation, this can be written as an inner product MATH . The norms are all MATH, and the last inequality follows from the NAME inequality and the norm bounds. Since MATH, the last norm is bounded above by MATH, which equals MATH. The first claim of the lemma now follows from REF . For the expected value, write MATH . Since MATH, the norm bound on the last formula and REF give the second claim of the lemma. Finally, denote MATH the random walk on MATH. Then MATH and summing twice on or under the diagonal and extending the range of MATH gives the upper bound MATH . By the NAME property this equals MATH . The third claim of the lemma follows if we write this as an inner product and use norm bounds, as before. Omitting the estimates for the first MATH steps gives the proof for general MATH. REF implies that we can use MATH. |
math/0102199 | For a positive MATH, let MATH, and let MATH be the event that the land of a country of level MATH is visited by time MATH. It suffices to prove that only finitely many of these events happen, which will follow if MATH for every large MATH. Consider MATH so large that the starting point of the walk is not contained in any level MATH country, and MATH. Let MATH denote the first hitting time of a country MATH. Let MATH denote the event that the land of the country MATH is visited by time MATH, and let MATH denote the event that the land of the country MATH is ever visited. The event MATH implies MATH and MATH, and therefore MATH . Summing over level MATH countries we get MATH . For fixed MATH, the events in the inner summand of REF are disjoint, so the second factor is bounded above by MATH. If MATH is a level MATH country with land MATH, then MATH . Therefore by the NAME Property and REF , MATH is not more than MATH . Then by REF , MATH and it suffices to prove that MATH . We apply MATH to both sides and use its monotonicity to transform the above to MATH . This certainly holds for all large MATH if MATH is large. |
math/0102199 | Between times MATH and MATH the walker is on an island with diameter bounded above by the volume MATH of the largest land visited by time MATH. Thus we have MATH . Dividing by MATH, and using the lemma together with REF proves the corollary. |
math/0102199 | Let MATH be a graph with the anchored expansion property and MATH-bounded geometry, and consider the construction of countries from REF . Using the notation of the previous section, we can decompose the inverse lim inf speed MATH as MATH . MATH in the above expression is defined in REF , and the last equality follows from REF . Let MATH denote the time spent in the islands up to time MATH. By REF we have MATH . The first factor in the last expression is the inverse of the lim inf speed in the graph MATH, for which we have the bound REF . Thus in order to show that MATH is greater than a constant a.s. it suffices to find constants MATH so that MATH . The equality holds since MATH is non-decreasing. For each MATH, if MATH is contained in a level MATH country MATH, then set MATH, otherwise set MATH. Set MATH, and for MATH define MATH . Also, for MATH, define the time spent in the land between stopping times: MATH . We will use the rough bound MATH . Since each vertex is contained in at most finitely many countries, we have MATH for all but finitely many MATH. So for REF it suffices to find summable MATH such that MATH . Now fix MATH, and suppose that MATH is on the inner vertex boundary of a level MATH country with land MATH. If MATH, then this means that MATH is in the ocean and MATH. REF with MATH gives MATH . If MATH, then MATH is contained in the land MATH, and this bound still holds (although it is very rough) by REF applied to a general starting point. By the NAME property, this implies that for all MATH, we have MATH, where MATH denotes the standard MATH-field at the stopping time MATH, that is the MATH-field generated by information available up to time MATH. Define MATH . Since MATH is a martingale, we can write MATH . Therefore, if MATH, then NAME 's inequality gives MATH . Thus if we set, for example, MATH, then it is clear from looking at the expression of MATH that the conditions of REF are satisfied. We thus have proved that the speed is greater than a constant depending on MATH and MATH only. It remains to give a bound on the constant in terms of MATH and MATH. Since MATH can be large when MATH is small, in order to get a reasonable bound, we need to deal with countries at or below some minimal level MATH separately. The value of MATH will be determined later, for now just assume that MATH. Then by REF the land of countries of level up to MATH is contained in MATH. Let MATH denote the time the random walk spends in MATH by time MATH, and let MATH denote the time the walk spends in the land of countries of level greater than MATH by time MATH. We then have MATH. Note that the sequence MATH is a subsequence of MATH, where MATH is the time of the MATH-th visit to MATH. By REF , the lim sup of this sequence, and thus the lim sup of the first one, is at most MATH. The bound REF on MATH, REF , and the bound REF on MATH from the first part of the proof imply MATH . We now want to choose a MATH so that the last sum is small, say each term MATH is at most MATH. From REF , since MATH, we have MATH with MATH. A simple computation shows that there is a constant MATH so that the right hand side of REF is at most MATH if MATH, where MATH is chosen so that MATH . Using this choice of MATH, from REF we conclude that MATH . |
math/0102199 | Fix a vertex MATH, and let MATH for MATH to be determined later. Let MATH, let MATH be the union of islands with volume at least MATH, and define the territory of such an island MATH as the set of vertices MATH with MATH . REF implies that there are only finitely many MATH for which MATH is contained in the territory of an island of MATH. Consider large MATH for which REF this does not happen and REF the right hand side of REF is at least MATH. REF and the definition of MATH ensures that the inner vertex boundary of the territory of an island is a subset of the ocean, MATH. REF implies that MATH, and, as shown in REF, MATH. By REF , islands of MATH with volume less than MATH do not intersect MATH, so MATH, and we have MATH. Let MATH denote the union of islands of MATH which are at distance at most MATH from MATH, and let MATH denote the transition kernel of the walk on MATH. Note that MATH is the sum of the probabilities of paths of length MATH starting at MATH and ending at MATH. Each such path stays in MATH or visits MATH. The total probability of the first kind of paths is at most MATH (regarded as MATH if MATH), so for all MATH we have MATH . The first term on the right satisfies MATH . The first inequality follows from NAME, the second from REF and the fact that MATH has NAME constant at least MATH. Suppose that there is a union MATH of MATH islands in MATH so that the territory of the first intersects the territory of all the others. Then there is bridge structure MATH interconnecting MATH with MATH . In the second expression, second term in parentheses is an upper bound on the number of vertices in a bridge connecting two islands with intersecting territories, and the first MATH is an upper bound for the number of vertices needed for the connection to MATH. REF then implies that there are finitely many MATH such that MATH exists. Thus for all large MATH, it is possible to MATH-color islands in MATH so that islands of the same color have disjoint territories. For such MATH, the probability of hitting some island is bounded by MATH times the maximal probability of hitting some island of a given color. Suppose that the walk starts at a vertex MATH on the inner vertex boundary of the territory of an island MATH. By construction, this means that MATH . Also note that another application of REF shows that for some MATH and all large MATH, MATH cannot contain islands with volume larger than MATH. For such n, by REF the probability of hitting MATH is bounded by MATH . In the first MATH steps the walker has at most MATH occasions to be at the inner vertex boundary of some island of a given color. Thus by the NAME property we get the bound MATH . There exists a MATH so that the exponents of REF are at most MATH with MATH. The statement of the theorem follows. |
math/0102200 | There is a unique solution MATH supported on MATH for REF . It can be obtained inductively; set MATH, then MATH . The solution is nonnegative (equivalently, MATH) if and only if MATH for all MATH. |
math/0102200 | MATH is a closed, bounded, convex subset of a finite dimensional vector space, so it equals the closed convex hull of its extreme points (for example, for a bit of overkill, by the NAME Theorem). Thus it suffices to prove that all extreme points of MATH are supported on MATH or on simple paths. Indeed, let MATH be extreme point. Consider the directed graph on MATH where MATH is an edge iff MATH (equivalently, if MATH and MATH). Consider the set MATH of vertices which are connected to MATH by a path directed towards MATH in this graph. First suppose that MATH. Summing the node law REF over elements of MATH yields MATH . The definition of MATH implies that for MATH we have MATH; summation yields MATH . Adding REF , the trivial inequality MATH, and REF yields MATH and therefore MATH. The second case is when MATH, so there is a directed path MATH satisfying the assumptions of REF and that MATH if MATH follows MATH in MATH. Thus there exists a MATH supported on MATH, and for a small MATH, the function MATH is nonnegative hence an element of MATH. As MATH is an extreme point, MATH, and the proof is complete. |
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