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math/0007039 | We have MATH. Also, from REF, we have MATH for every MATH. Also, MATH whenever MATH. Assume MATH. Then there is some MATH such that MATH and MATH. We have MATH. Therefore MATH . Thus, MATH. Assume MATH. We may assume MATH for otherwise it is clear that MATH. (So we have MATH.) Thus, there is some MATH, such that MATH a... |
math/0007039 | The theorem is obtained by merging the statement of REF with the statement of REF , and eliminating some redundancy (see REF). Specifically: CASE: REF appears here as REF . CASE: REF appears here as REF . CASE: REF appears here as REF . CASE: REF appears here as REF . CASE: REF appears here as REF . CASE: REF appears h... |
math/0007039 | It is clear that each of the given subgroups is normalized by the indicated torus. We now show that the list is complete, and that no larger subtorus of MATH normalizes MATH. Assume MATH is nontrivial. We proceed in cases, determined by REF . Assume REF . We may assume MATH is neither MATH nor MATH (otherwise REF appli... |
math/0007039 | For MATH, we wish to approximately calculate MATH. We write MATH with MATH and MATH. Writing MATH, we always assume either that MATH or that MATH and MATH (perhaps replacing MATH with MATH - because MATH, this causes no harm). Because MATH normalizes MATH, we know that MATH is a subgroup that is listed in REF , and we ... |
math/0007039 | We use the notation of REF : MATH, MATH, MATH, and MATH. We need only consider the cases in REF for which MATH (now called MATH) is normalized by the kernel of some (reduced) positive root. Here is a list of them. CASE: MATH: REF , and REF . CASE: MATH: REF . CASE: MATH: REF , and REF . CASE: MATH: CASE: MATH: REF , an... |
math/0007039 | By passing to a subgroup of MATH, there is no harm in assuming MATH. We use the notation of the proof of CITE. For each MATH, clearly MATH, so MATH. NAME MATH is a line perpendicular to MATH, and MATH is logarithmically close to this line, it is clear that MATH contains all but a bounded subset of the region MATH. Ther... |
math/0007039 | We begin by showing that MATH (compare CITE). Let MATH be the projection of MATH to MATH. Because MATH, we just need to show that MATH. Because MATH does not intersect MATH (or MATH, either, for that matter), and MATH has codimension MATH in MATH, this is immediate. When MATH is even, there is a subgroup of dimension M... |
math/0007039 | Replacing MATH by a conjugate under MATH, we may assume MATH. Therefore MATH for every MATH. (Thus, in particular, we have MATH.) For the projection MATH, we have MATH. (There cannot exist a linearly independent MATH; otherwise, replacing MATH by some linear combination with MATH, we could assume MATH, which is impossi... |
math/0007039 | For MATH, here is the construction of REF subalgebras of MATH of this type. Let MATH and MATH. Choose MATH, MATH, and MATH, such that MATH . Now, choose MATH, such that MATH and MATH . Define MATH as in REF, and let MATH. Then MATH and MATH, but, from REF, we have MATH. Thus, we may let MATH be the subalgebra generated... |
math/0007043 | The first statement follows from the second. It is enough to treat the case MATH; the general case follows by trivial induction. There is a cartesian diagram MATH . By the fundamental theorem of symmetric functions the fibre product is MATH . So we get an étale trivialization of MATH. Any two trivializations are relate... |
math/0007043 | By REF MATH is a locally trivial fibre bundle over MATH with fibre MATH, thus MATH. There is a natural morphism MATH defined on MATH-valued points by sending MATH to MATH. MATH is obviously invariant under the action of MATH by permuting the factors in the MATH, and the induced morphism from the quotient to MATH induce... |
math/0007043 | By REF we get MATH . By REF this implies MATH where the MATH rund through the MATH with MATH. The result follows by REF . |
math/0007043 | Any MATH induces a map MATH as follows. Let MATH. For all MATH let MATH. For each MATH we have MATH, so MATH; furthermore MATH. We define MATH by MATH . Then MATH and MATH . Now fix MATH with the above properties. Let MATH. We claim that MATH . For MATH define MATH by letting MATH be the sum over all MATH with MATH. Fo... |
math/0007043 | If MATH is a variety with a natural map to MATH for some MATH, we will write MATH for the preimage of the locus of MATH with MATH. Let MATH with the reduced structure. In CITE it is shown that MATH has a cell decomposition. Her formula for the numbers of cells of different dimensions implies that MATH . We now determin... |
math/0007043 | In CITE the authors use virtual NAME polynomials MATH in order to show that there exists a universal power series MATH such that MATH . To do this they only use the basic property of virtual NAME polynomials that MATH for MATH a closed subvariety of MATH. So their proof shows that there is a universal power series MATH... |
math/0007043 | MATH odd implies MATH odd and MATH odd, therefore REF implies that MATH. By REF MATH depends only on the numerical equivalence class of MATH. Therefore we can assume that MATH of MATH. We write MATH with the negative of the intersection form. We note that MATH. By REF MATH is the coefficient of MATH in MATH . F or MATH... |
math/0007045 | We argue by induction on MATH. If MATH is empty, so is the statement of the proposition. Otherwise there are two cases. The lucky case: If MATH we can compute the MATH integral first CITE, and we need to show that MATH where we know that MATH is link relation equivalent to MATH via MATH-flavored link relations. Multipl... |
math/0007045 | By induction on MATH (and row expansion of the relevant determinants) one establishes the equality MATH . After that, the theorem follows from simple observations regarding the determinant MATH and the minors MATH of the matrix MATH. Namely, that MATH, that MATH is MATH, that MATH is triangular with ones on the diagona... |
math/0007045 | In CITE it is shown that MATH in MATH, and thus using REF , MATH . The MATH-invariance of MATH is used in asserting the equality between the contractions MATH and MATH. If MATH-invariance is not assumed, the contraction map MATH may not descend to a map MATH. |
math/0007045 | The first assertion follows immediately from REF . The second assertion follows from the first and from the equality MATH, which follows from the fact that the two-legged part of MATH is MATH. |
math/0007045 | Clearly, MATH. And so, using REF twice and the fact that MATH, we get MATH . |
math/0007045 | REF follows from the following computation: MATH . The operator MATH commutes with translation by MATH (the map MATH), because MATH has ``constant coefficients". Thus REF is a consequence of REF . Alternatively, it can be proven directly along the same lines. |
math/0007045 | Indeed, MATH . |
math/0007045 | REF follows immediately from the generalized form of the Wheeling Theorem, REF ', and the known multiplicative property of the NAME integral MATH. REF follows from REF using the following lemma, which is of independent interest. |
math/0007045 | MATH . |
math/0007045 | MATH . |
math/0007045 | The first assertion, REF is a simple assembly of REF . The second assertion follows by applying MATH to the definition of MATH REF , using Wheeling and substituting the result into REF . |
math/0007045 | REF follows immediately from REF (right below) and REF (on page REF), which assert that MATH and that MATH respectively. |
math/0007045 | This is REF. It also follows from the formalism of [-I - III] noting that CASE: Formal Gaussian integration behaves correctly under iteration CITE. CASE: The covariance matrix of MATH is the linking matrix of MATH and if MATH denotes the numbers of positive/negative eigenvalues of the covariance matrix of MATH, then MA... |
math/0007045 | The MATH-framed NAME link MATH is the MATH-marked double MATH of REF-framed unknot MATH, and hence MATH . It remains to undo the MATH framing by using REF on each component: MATH . |
math/0007045 | To get MATH from MATH, we need to apply MATH and MATH, and to ``open up" the MATH-component, which amounts to multiplication (using the product MATH) by MATH (recall that the NAME integral of an open unknot is trivial). The latter operation can more easily be performed first, by MATH-multiplying MATH by MATH, as in REF... |
math/0007045 | Both sides of the required equality are clearly made of local contributions, one per each shackling element or framing fraction MATH. Thus it is enough to prove the proposition at the locale where all the actors act. Ergo we may as well assume that the link MATH in question is a straight line MATH marked MATH, and then... |
math/0007045 | Recall that the MATH lens space MATH is obtained from MATH by surgery over the MATH-framed unknot MATH. Thus we can use REF to compute MATH, remembering also that by REF , MATH: MATH . |
math/0007045 | As noted in CITE, the lens spaces MATH and MATH are not homeomorphic but the NAME symbols MATH are equal (see REF ), and thus their LMO invariants are equal. |
math/0007045 | To be an integral homology sphere, a NAME fibered space must be of the form MATH discussed above. Furthermore, it must have MATH. For this to happen, MATH must equal MATH, the MATH's must be pairwise relatively prime, and then the MATH's and MATH are uniquely determined up to an overall sign by the MATH's, using the NA... |
math/0007049 | Observe that MATH since MATH. Suppose MATH and MATH. Then MATH is invertible, and MATH. The operator on the left hand side has MATH in its spectrum and so MATH. This is a contradiction. Similarly, MATH and MATH cannot occur. Hence, either MATH or MATH. It follows that MATH. If MATH then MATH is in the spectrum of MATH ... |
math/0007049 | We have MATH. Since a similarity transformation leaves the spectrum unchanged, we immediately obtain MATH. Now suppose MATH. A similar argument as in REF shows that MATH and MATH, and so MATH. Finally, if MATH is unitary then MATH, so MATH. |
math/0007049 | We have MATH. On taking the trace, this yields MATH, hence MATH. The second statement follows similarly. |
math/0007049 | CASE: Observe that MATH, and so MATH. Hence MATH commutes with MATH and MATH, and similarly for MATH. Thus we get MATH . REF : REF is equivalent to MATH, where we have MATH. Thus MATH is normal. It follows that MATH . Let MATH denote the associated projection. We then have that the closures of the ranges of these four ... |
math/0007052 | For MATH and MATH in MATH, we have MATH . On the other hand, we have MATH . Since these equations hold for all MATH in MATH, we have proved the proposition. |
math/0007052 | To prove the lemma, we use the operator MATH given in REF. For MATH in MATH, we have MATH . The operator MATH is the constant MATH on MATH, so that MATH. Hence, we conclude that MATH . |
math/0007052 | We substitute REF into MATH and use the relation REF. |
math/0007052 | For MATH in MATH, we have MATH . |
math/0007052 | Let MATH be an irreducible component of MATH. We consider the following embedding from MATH to MATH, which dose not always preserve their inner products: MATH . The map MATH is independent of the orthonormal basis of MATH which we chose, and commutes with the action of MATH . Since the irreducible component of MATH has... |
math/0007052 | For any MATH in MATH, we can find a local orthonormal frame MATH such that MATH for any MATH and MATH, so that MATH. Then we have MATH . |
math/0007052 | We remark that MATH and, for any MATH, MATH . We have known the following identities: MATH . Then MATH . This inequality leads us to the proposition. |
math/0007052 | For MATH in MATH, we have MATH . |
math/0007055 | Note that REF in conservation form become, respectively, MATH where MATH . Call MATH and, respectively, MATH the fluxes in the two systems REF. Then, the estimate MATH follows from straightforward computations and completes the proof. |
math/0007056 | Note that for MATH or MATH we have MATH for all MATH. To prove REF , let MATH and write MATH where MATH for all MATH. Then MATH by REF. We have MATH by assumption, so by induction MATH. The proof of REF is essentially the same. |
math/0007056 | For MATH, let MATH . We may define a map MATH by assigning, for each MATH-algebra MATH and each MATH, the value MATH . Since MATH is invertible in MATH, it follows from CITE that MATH is an isomorphism of MATH-group schemes. [Note that the assertion is valid over any field MATH provided only that the characteristic of ... |
math/0007056 | For each MATH-algebra MATH and each MATH one uses induction on MATH and the definition of the MATH to verify that MATH . It follows that MATH whence REF . For REF , let MATH denote the coordinate functions on MATH with MATH; thus MATH is a polynomial ring in the MATH. The tangent space to MATH at MATH contains the ``po... |
math/0007056 | Let MATH be the image of MATH; thus MATH is a MATH-Lie subalgebra of MATH. According to REF , MATH contains MATH. It follows that MATH. On the other hand, we have MATH . Thus MATH, and the proposition follows. |
math/0007056 | REF follows immediately from the description of MATH given by REF . For REF , recall CITE that the ring of (infinite) NAME vectors MATH is a strict MATH-ring (see CITE II. REF for the definition) and that MATH for all MATH. REF now follows at once. |
math/0007056 | The lemma is clear if either MATH or MATH has infinite exponent, so assume otherwise. Suppose that MATH is a surjection with finite kernel. Let MATH be the exponent of MATH. Since MATH is Abelian, the map MATH defines a group homomorphism MATH; since MATH is connected and MATH is finite, this homomorphism must be trivi... |
math/0007056 | The first assertion is CITE. The second assertion follows immediately from the lemma. For the last assertion, it is proved in CITE that the group MATH is a subgroup of a product of NAME groups. A careful look at the proof in CITE shows that the exponent of MATH and this product may be chosen to coincide. Thus, MATH is ... |
math/0007056 | This follows from CITE. |
math/0007056 | We first prove REF . By REF , we are reduced to the case where MATH is indecomposable. Since MATH is good, CITE shows that MATH for MATH. Essentially the same arguments show that MATH. Since every root MATH satisfies MATH; REF now follow at once. Note that we have showed for all MATH that MATH has a basis of MATH-nilpo... |
math/0007056 | Let MATH be a regular nilpotent element in MATH; thus MATH is a representative for the dense MATH-orbit on MATH [see the discussion of NAME 's dense orbit theorem below in REF]. Since MATH, the proposition shows that MATH. Since the regular nilpotent elements form a single dense orbit in the nilpotent variety, we get M... |
math/0007056 | It suffices to show that MATH is surjective. Since MATH and MATH, that follows immediately from the assumption on kernel dimensions. |
math/0007056 | The NAME orbit on MATH meets MATH in an open set, so we may find a regular function MATH on MATH such that MATH implies MATH is a NAME element. Using the lattice MATH in MATH, we obtain coordinate functions [dual to the root-vector basis] on MATH; let MATH by a finite extension of MATH containing the coefficients of MA... |
math/0007056 | In view of REF , we may suppose that MATH satisfies the hypothesis of REF . With MATH as in the statement of the theorem, REF shows that the MATH-nilpotence degree of any element of MATH is MATH; to prove that equality holds, it suffices to exhibit a representation MATH of MATH as a MATH-Lie algebra in which some MATH ... |
math/0007056 | Put MATH and MATH . Then MATH is a linear subspace of MATH, and MATH. This shows that MATH is a connected subgroup of MATH. Denoting by MATH the NAME algebra of MATH, we have MATH since any MATH in MATH centralizes MATH by definition. Similarly, we have MATH. On the other hand, if MATH, then MATH since MATH; this shows... |
math/0007056 | In view of REF , we may suppose that MATH satisfies the hypothesis of REF and of REF . Let MATH be as in REF . Then that corollary together with REF imply that MATH whence equality holds. |
math/0007056 | Let MATH be a faithful MATH-representation of MATH. Since the image of MATH is a subalgebra of MATH, one gets MATH by CITE; thus we get a morphism MATH defined over MATH and satisfying MATH. A second application of the result of CITE shows that MATH for any rational representation MATH. Unicity of MATH is clear. |
math/0007056 | MATH is determined by its comorphism MATH; the proposition will follow if we show that MATH. Note that MATH and MATH (the latter by definition). The comorphism MATH is then MATH. The hypothesis implies that MATH; since MATH is free as a MATH-module, the natural map MATH is injective, and it then follows that MATH as de... |
math/0007056 | Note first that the coordinate algebras MATH and MATH are free as MATH modules. REF follows immediately from the proposition combined with REF . For REF combined with REF shows that MATH determines on base change a morphism MATH; the result then follows from the proposition. |
math/0007056 | This is proved in CITE. |
math/0007056 | CASE: We may suppose that MATH. The co-character MATH induces a grading on MATH by MATH; evidently MATH acts as a sum of homogeneous terms of positive and even degree. Since the NAME group of MATH permutes the weights of MATH, it follows that MATH, and REF is then immediate. [One can alternately argue that the graded c... |
math/0007056 | In view of the results of REF, we may suppose that MATH is an algebraic closure of the finite field MATH. The image of MATH in MATH is the MATH-Lie algebra of the unipotent radical of a NAME subgroup. Since MATH is an algebraic closure of MATH, there is a finite extension MATH of MATH and a valuation ring MATH in MATH ... |
math/0007056 | We have for each MATH: MATH whence the result. |
math/0007056 | Using the results of REF, we may suppose that MATH is an algebraic closure of the finite field MATH. As in the proof of REF , we may find a number field MATH with valuation ring MATH and residue field MATH for which MATH lies in MATH (where MATH is the MATH-span of suitable NAME basis elements, as before). Thus we may ... |
math/0007056 | The essential point is proved in CITE for MATH; the generalization to MATH is immediate. One observes as in CITE that MATH where MATH is a polynomial with coefficients in MATH in the MATH with MATH. It follows that MATH is an isomorphism over MATH (see CITE). The equivariance assertion is clear. |
math/0007056 | REF is immediate. For REF , one must note that the condition MATH implies that MATH and MATH commute when MATH is regarded as a group by the NAME series. |
math/0007056 | Since all our group schemes are affine, the homomorphisms between them may be identified with comorphisms on coordinate algebras. The first two isomorphisms follow from this and the fact that for any homomorphism MATH, the comorphism MATH vanishes on MATH. Since MATH is infinitesimal, MATH identifies with the dual NAME... |
math/0007056 | Since the MATH are finite dimensional, we regard MATH as a subset of the affine space MATH of all MATH-linear maps. For each MATH, the map MATH given by MATH is clearly a morphism of varieties, and the set MATH of all algebra homomorphisms is the intersection of all MATH, hence is a closed subvariety. One similarly see... |
math/0007056 | The variety structure is evident from the previous remarks. The above identification is compatible with the action of MATH on itself by inner automorphisms, and on MATH by the adjoint representation; this action yields the structure of MATH-variety. |
math/0007056 | Condition MATH yields MATH-bases MATH of MATH and MATH of MATH and integers MATH for which MATH and MATH. These bases of MATH and MATH determine bases for the respective distribution algebras, and we have MATH . One may check that MATH where MATH is a MATH-linear combinations of basis elements of lower degree. It follo... |
math/0007056 | According to CITE, multiplication is an isomorphism MATH where MATH and MATH are the unipotent radicals of opposite NAME subgroups and MATH is a maximal torus. Moreover, (see CITE) MATH induces maps on these tensor factors; it is clear from the description of MATH that it induces an isomorphism MATH (with a similar sta... |
math/0007056 | This follows from the observations made in CITE. |
math/0007056 | In all three cases, the condition on MATH guarantees that it does not divide the order of the fundamental group (this is well-known, and may be checked by looking at the tables in CITE). So it suffices by REF to show that there is a semisimple group MATH isogenous to MATH, and an exponential-type representation MATH of... |
math/0007056 | In this case, we must first work with schemes in order to know that the map we define is a morphism. Recall from REF that there is an isomorphism of group schemes MATH. Thus for each MATH-algebra MATH, each MATH determines a homomorphism MATH. If MATH, one emulates the construction in CITE to obtain a homomorphism of g... |
math/0007056 | We mimic the argument in CITE. Let MATH (where the exponent MATH denotes the MATH-th NAME twist); MATH acts on MATH by MATH. We denote also by MATH the action of MATH on the algebra MATH of regular functions on MATH. There is a homomorphism MATH obtained by mapping MATH to the section MATH. The kernel is MATH and we cl... |
math/0007056 | There is a central isogeny MATH where MATH is a direct product of a torus and quasisimple groups satisfying the hypothesis of REF . This isogeny induces a bijection (not in general an isomorphism of varieties) on the nilpotent sets in the respective NAME algebras. Thus, we may replace MATH with MATH, so that we may app... |
math/0007062 | The standard proof that being finitely presented involves NAME transformations, and extends to MATH-presentations. One changes MATH into MATH by a finite number of ``NAME moves", which either replace a generator by a product or quotient of generators, or add or delete a generator MATH along with the relator MATH. For a... |
math/0007062 | Let MATH be a finite ascending MATH-presentation of MATH. Consider the group MATH . It is finitely presented, and the map MATH defined by sending MATH to MATH is a well-defined injective homomorphism, since the MATH induce injective homomorphisms of MATH. |
math/0007062 | It follows from the hypotheses that MATH is also MATH. Therefore MATH is not finitely presented, so cannot be hyperbolic. On the contrary, MATH is finitely presented and its presentation is MATH, so it is hyperbolic. Finally a quasi-convex subgroup of a hyperbolic group would be hyperbolic, so MATH cannot be quasi-conv... |
math/0007062 | The first claim is obvious: finite MATH-presentations with MATH are precisely finite presentations. There are only countably many finite MATH-presentations, but uncountably many finitely-generated groups, so ``most" groups are not finitely MATH-presented. Finally, REF shows that the ``lamplighter group" described there... |
math/0007062 | Let MATH be a finite MATH-presentation of MATH, and let MATH be a finite MATH-presentation of MATH. A finite MATH-presentation of MATH is MATH where it is understood that each MATH is extended to a homomorphism MATH by mapping each MATH to itself; and similarly for each MATH. Let now MATH be the subgroup of MATH genera... |
math/0007062 | Let MATH be a finite MATH-presentation of MATH; let MATH be a finite MATH-presentation of MATH; let MATH be an extension of MATH by MATH, given as MATH. Let MATH be a section of MATH to MATH; in case the extension splits, we suppose that MATH is a group homomorphism. Each relator MATH lifts through MATH to an element M... |
math/0007062 | MATH - see Subsection REF. |
math/0007062 | Let MATH be a finite MATH-presentation of MATH, and let MATH be a right transversal of the finite-index subgroup MATH of MATH. In view of REF , we may suppose MATH is normal in MATH, since any finite-index subgroup is a finite extension of its core, which is normal of finite index. We then have MATH. For MATH, let MATH... |
math/0007062 | If MATH is finite, then MATH is finitely MATH-presented by REF . Let us assume then that MATH is abelian. Let MATH be a finite MATH-presentation of MATH, and let MATH be a finite MATH-presentation of MATH. A MATH-presentation of MATH is MATH where it is understood that each MATH is extended to a homomorphism MATH by ma... |
math/0007062 | It suffices to prove the claim for a relatively free group, since the quotient of a finitely MATH-presented group by a finitely normally generated normal subgroup remains finitely MATH-presented. Let us then suppose MATH relatively free, and generated by MATH, and write MATH. For MATH and MATH, define the endomorphism ... |
math/0007062 | By CITE, every group in the variety MATH is the quotient of the free group in that variety (defined by the identity MATH) by a finite number of relations. |
math/0007062 | Write MATH, and MATH. Consider the group MATH. It is abelian and generated by MATH. The maps MATH are such that MATH is finitely generated (by MATH), and we may filter MATH along MATH. For each MATH, write MATH where MATH splits back into MATH and MATH does not. Then MATH lies inside MATH, so MATH is of the required fo... |
math/0007062 | Consider first MATH with its natural projection MATH, and set MATH. Since MATH has finite index in MATH, it is finitely generated, say by the set MATH. Our purpose is to find a quotient of MATH in which MATH is generated by MATH. For each MATH, let MATH be an expression of MATH over MATH. It then suffices to consider M... |
math/0007062 | Let MATH be regular branch on its subgroup MATH, and fix generating sets MATH for MATH and MATH for MATH. It loses no generality to assume MATH, since one may always replace MATH by MATH. Let MATH be the group given by REF for MATH. Let MATH be the natural lift of MATH to MATH; and let MATH be the natural lift of MATH ... |
math/0007062 | Since MATH is contracting, there is a constant MATH such that MATH whenever MATH. This implies, using the triangular inequality, that there are constants MATH and MATH such that MATH whenever MATH. Now levels can be ``collapsed" in a branch group: for any MATH we may consider the (same) action of MATH on MATH, with map... |
math/0007062 | We concentrate on the exact sequence MATH, for some finite group MATH. By REF , MATH. Taking MATH-invariants of the right-hand side collapses all MATH copies of MATH together, but we are left with the equation MATH, where MATH, a quotient of MATH, is a finite group. Then MATH is obtained from MATH by extension and quot... |
math/0007062 | Clearly MATH is perfect, being generated by two perfect groups, and finitely generated, being generated by two finite groups. Note now that MATH. The map MATH is given by MATH where in this last expression the MATH is at position MATH and the MATH is at position MATH. The conditions on MATH imply that it contains an el... |
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