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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 and MATH . (This implies that MATH and MATH must have the same sign.) From REF , we conclude that MATH for every MATH. Thus, there is a constant MATH, such that MATH for every MATH. Then MATH so MATH . Therefore MATH . |
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 here as REF . CASE: REF appears here as REF . CASE: REF is a special case of 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 here as REF . CASE: REF is a special case of REF (with MATH). CASE: REF is a special case of REF (with MATH). |
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 applies). Then, because MATH for every MATH, we see that MATH for every nonzero MATH. Thus, the projection of MATH to MATH is nontrivial. However, because MATH, we have MATH. We know that MATH (because MATH), so, because each of MATH and MATH differs from MATH by MATH, we conclude that MATH, so REF applies. Assume REF . Let MATH be the projection of MATH to MATH. Because MATH, we know that MATH projects nontrivially to MATH. However, because MATH, we also know that MATH. Therefore MATH. Then, because neither MATH nor MATH differs from MATH by a multiple of MATH, we conclude that MATH, so REF applies. Assume REF . We may assume MATH (otherwise REF applies). Assume MATH. Because MATH, the projection of MATH to MATH is nontrivial. However, because MATH, this projection intersects neither MATH nor MATH. Therefore MATH. Then, because neither MATH, MATH, nor MATH differs from MATH by a multiple of MATH, we conclude that MATH, so REF applies. Assume MATH. This means MATH for every MATH, and MATH. Because MATH, we know that MATH projects nontrivially to MATH. Because MATH, we know that MATH. Thus, it is easy to see that if MATH projects nontrivially to MATH or MATH then either REF or REF applies. Thus, we may assume MATH. If MATH, then MATH, so REF applies. Otherwise, REF applies. Assume REF . Assume the projection of MATH to MATH is trivial. Because MATH for every MATH, we know that MATH for every MATH. Thus, if the projection of MATH to MATH is nontrivial, then MATH, and we see that REF applies. If not, then MATH, so either REF or REF applies. Assume the projection of MATH to MATH is nontrivial. Let MATH be the projection of MATH to MATH. Because MATH for every MATH, we know that MATH. Then, because MATH, MATH, and MATH all differ by multiples of MATH, we conclude that MATH. Therefore REF applies. Assume REF . Let MATH be the projection of MATH to MATH. Because MATH, we see that MATH and MATH. Therefore MATH. Because no other roots differ by a multiple of MATH (and MATH), we conclude that MATH. Thus, REF applies. Assume REF . Assume MATH. Let MATH be the projection of MATH to MATH. From the assumption of this subcase, we know MATH. However, because MATH for every MATH, we know that MATH and MATH. Therefore MATH, so REF applies. Assume MATH. We may assume MATH (otherwise REF applies). Therefore, MATH projects nontrivially to MATH. However, because MATH, for every nonzero MATH, we know that MATH and MATH. Because MATH, MATH, and MATH all differ by multiples of MATH, we conclude that MATH, so REF applies. Assume REF . Assume MATH. Because MATH is the only root that differs from MATH by a multiple of MATH, we must have MATH. Thus, there is some MATH, such that MATH and MATH, but the projection of MATH to MATH is zero. This contradicts the fact that MATH. Assume MATH. Because MATH, MATH, and MATH all differ by multiples of MATH, we must have MATH. Then, because MATH for every nonzero MATH, we conclude that MATH. Let MATH be the projection of MATH to MATH. Because MATH and MATH differ by MATH, we know that MATH. Assume MATH. Because MATH, there is some MATH, such that MATH and MATH. For every such MATH, because MATH, we know that MATH. Thus, we see that MATH. We know that MATH, that MATH projects trivially to MATH, and that MATH is the only root that differs from MATH by a multiple of MATH, so we conclude that MATH for every MATH. We now see that REF applies. Assume MATH. This is similar to the preceding subsubcase (indeed, they are conjugate under the NAME reflection corresponding to the root MATH); we see that REF applies. Assume REF . By considering the projection of MATH to MATH, and noting that MATH for every MATH, we see that MATH. The only other pair of roots that differ by a multiple of MATH is MATH. Thus, we see that REF applies. Assume REF . By considering the projection of MATH to MATH, we see that MATH. Because MATH for every MATH, but MATH is the only root that differs from MATH by a multiple of MATH, we conclude that MATH projects trivially into every root space except MATH, MATH, and MATH. Thus REF applies. Assume REF . We may assume MATH (otherwise REF applies). Thus, there is some root MATH, such that the projection of MATH to MATH is nontrivial. However, because MATH for every nonzero MATH, we know that MATH. Thus, MATH. Because MATH and MATH for every nonzero MATH, we know that MATH and MATH. If MATH or MATH, we obtain REF . If MATH, we obtain REF . Assume REF . Because MATH and MATH, we must have MATH. Then, because MATH does not differ from MATH by a multiple of MATH, we conclude that MATH. Because MATH, but no root differs from MATH by a multiple of MATH, we conclude that MATH. Because MATH is one-dimensional, this implies MATH, so MATH. Since MATH and MATH, we conclude, from the inequality MATH, that MATH. This is a contradiction, because MATH does not differ from MATH by a multiple of MATH, and MATH (because, as shown above, MATH). |
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 have MATH. This leads to the various cases listed in the statement of the theorem. CASE: We have MATH for MATH and MATH for MATH, so MATH is a NAME subgroup. CASE: We have MATH and MATH, so MATH whenever MATH. Then, because MATH, we see that MATH whenever MATH. Therefore MATH. This completes the proof if MATH (that is, if MATH). If MATH, then MATH and MATH. We have MATH, MATH and MATH . Thus, MATH. We conclude that MATH. CASE: Replacing MATH by a conjugate under MATH, we may replace MATH with a similar subgroup MATH with MATH. Thus, MATH with MATH. Then CITE implies MATH is a NAME subgroup. CASE: The NAME reflection corresponding to the root MATH conjugates MATH to a subgroup of type REF . CASE: The NAME reflection corresponding to the root MATH conjugates MATH to a subgroup of type REF . REF implies MATH is a NAME subgroup. REF implies MATH is a NAME subgroup. CASE: We have MATH and MATH. We conclude that MATH. CASE: From the proof of REF , we know that MATH, that MATH whenever MATH, and that MATH whenever MATH and MATH. (In particular, MATH.) Furthermore, we have MATH. Therefore MATH . The desired conclusion follows. CASE: From REF , we know MATH. Then, because MATH and MATH, we have MATH . CASE: Assume REF . (The other case, REF , is conjugate to this one by the NAME reflection corresponding to the root MATH.) From REF , we have MATH. Then, because MATH, we have MATH . CASE: From the proof of REF , we know MATH. The proof is completed as in REF . CASE: Because MATH, it is easy to see that MATH and MATH . Then it is not difficult to see that MATH for every MATH. So MATH. CASE: We have MATH for MATH and MATH for MATH, so MATH is a NAME subgroup. CASE: MATH is conjugate (via an element of MATH) to MATH. From CITE, we have MATH for every MATH. Therefore MATH for every MATH. |
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 , and REF . Note that in each of the cases with MATH, there is a (reduced) positive root MATH, such that MATH. Assume MATH. Assume REF . From REF , we know that MATH is a NAME subgroup. Assume REF . There is some MATH, such that MATH. Then, because MATH normalizes MATH, we must have MATH. This is a contradiction. Assume REF . Let MATH. Because MATH is normalized by MATH, there is some nonzero MATH, such that MATH normalizes MATH; thus, MATH. Then, because MATH, but MATH for every nontrivial MATH, we conclude that MATH. However, MATH, and either MATH or MATH, so either MATH or MATH. This is a contradiction. Assume MATH. There is a positive root MATH, such that MATH. If MATH, then, from REF , we know that MATH is a NAME subgroup. Suppose MATH. Clearly MATH. Also, MATH where MATH if MATH (that is, if MATH for every MATH) and MATH if MATH. The smallest value of MATH relative to MATH is obtained by taking MATH, resulting in MATH. Then, since MATH for MATH, we conclude that MATH. Because MATH is normalized by the nontrivial subgroup MATH of MATH, we know that MATH. Therefore, we may now assume MATH. We show that MATH. For MATH, we have MATH. For MATH, we have MATH . All that remains is to show that MATH for every MATH. Because MATH for every MATH (see CITE) and MATH, we have MATH . Assume MATH. The NAME reflection corresponding to the root MATH conjugates each of REF to a subgroup with MATH. Thus, we may now assume MATH. If MATH, then the NAME reflection corresponding to the root MATH conjugates MATH to a subgroup with MATH. If MATH, then the NAME reflection corresponding to the root MATH does not change MATH, but conjugates MATH to a subgroup MATH with MATH. Then (as we already observed) the NAME reflection corresponding to the root MATH conjugates MATH to a subgroup with MATH. Assume MATH. Because MATH must be normalized by the nontrivial subgroup MATH of MATH, we see that MATH cannot be of type REF or REF . Assume REF . Because MATH must be normalized by the nontrivial subgroup MATH of MATH, we see that MATH for every MATH, so MATH. Thus, again using the fact that MATH is normalized by MATH, we see that MATH, and the projection of MATH to MATH is one-dimensional. For every MATH, we see that MATH (because MATH). Thus, we conclude that MATH. Therefore MATH is conjugate under MATH to a subgroup of type REF (considered in NAME REF below). Assume MATH. If MATH, then REF implies that MATH is a NAME subgroup. Because MATH is normalized by the nontrivial subgroup MATH of MATH, we know that MATH. Assume MATH. We have MATH . We have MATH and, for MATH, we have MATH. The largest value of MATH relative to MATH is obtained by taking MATH (and MATH small), which yields MATH. Because MATH for MATH, we conclude that MATH. Assume MATH. We have MATH . We have MATH and MATH (note that MATH). The smallest value of MATH relative to MATH is obtained by taking MATH, which results in MATH. Because MATH for MATH, we conclude that MATH. Assume MATH. The NAME reflection corresponding to the root MATH conjugates REF to a subgroup MATH with MATH (of type REF with MATH). Thus, we may now assume MATH. If MATH, then the NAME reflection corresponding to the root MATH conjugates MATH to a subgroup with MATH. Now assume MATH. The NAME reflection corresponding to the root MATH does not change MATH, but conjugates MATH to a subgroup MATH with MATH. Then (as we already observed) the NAME reflection corresponding to the root MATH conjugates MATH to a subgroup with MATH. |
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. Therefore MATH contains all but a bounded subset of MATH, so MATH is a NAME subgroup. |
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 MATH. (For example, the MATH subgroup of MATH. More general examples are constructed in CITE.) Let us show that if MATH is odd, then MATH. (Our proof is topological; we do not know an algebraic proof.) Suppose that MATH (this will lead to a contradiction). Because MATH, we have MATH. Thus, there is a MATH-dimensional real subspace MATH of MATH and a real linear transformation MATH, such that MATH and MATH are linearly independent over MATH, for every nonzero MATH (compare CITE). Thus, if we define MATH by MATH; then MATH, MATH, and MATH are linearly independent over MATH, for every nonzero MATH. Thus (writing MATH): there is a MATH-dimensional real subspace MATH of MATH and real linear transformations MATH, such that MATH, MATH, and MATH are linearly independent over MATH, for every nonzero MATH. There is no harm in assuming MATH (under its natural embedding in MATH). Let MATH, where MATH, and define a continuous map MATH by MATH, so MATH is a vector bundle over MATH. Then MATH, where MATH is the tangent bundle of MATH, MATH is a trivial line bundle, and MATH is the canonical bundle of MATH. (To see this, note that the subbundle MATH is the total space of MATH CITE, the subbundle MATH has the obvious section MATH, and the subbundle MATH is isomorphic to MATH via the bundle map MATH.) Therefore, letting MATH be a generator of the cohomology ring MATH, we see that the total NAME class of MATH is MATH CITE, so MATH (because MATH is odd). Therefore, there do not exist three pointwise linearly independent sections of MATH CITE. Any linear transformation MATH induces a continuous function MATH, 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, because MATH, MATH, and MATH are linearly independent over MATH, for every MATH. This contradicts the conclusion of the preceding paragraph. |
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 impossible.) Thus, MATH. Because MATH for every MATH, MATH must be a totally isotropic subspace for the symplectic form MATH, so MATH. Therefore MATH. For MATH, here is an example that achieves this bound: MATH . For MATH, we claim that MATH only if either MATH or MATH. (In either case, it is clear from the definition of MATH that either MATH or MATH, respectively.) Suppose MATH, with MATH and MATH. There is some nonzero MATH, such that MATH. We must have MATH. (Otherwise, let MATH be minimal with MATH. Then MATH, contradicting the minimality of MATH.) Because MATH is pure imaginary, but MATH is real, we see that MATH is pure imaginary. On the other hand, MATH is real (and nonzero), and MATH is also real, so MATH is real. Because MATH, this is a contradiction. Now let MATH, and suppose MATH. (This will lead to a contradiction.) Because equality is attained in the proof above, we must have MATH and MATH. In particular, there exists MATH with MATH. For MATH, let MATH. Then MATH . Thus, this expression changes sign, so it must vanish for some MATH. This is a contradiction, because MATH for every MATH. |
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 by MATH and MATH. (So MATH is a basis of MATH over MATH.) Note that, because MATH, we know that MATH and MATH must be linearly independent over MATH. Thus, these REF-dimensional examples do not exist when MATH. |
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 related by the action of the group MATH by reordering the factors in the MATH. This acts on the fibres of MATH by reordering the factors MATH in the MATH. Choosing an origin in MATH determines an origin in MATH, and the action becomes linear on on MATH, so MATH is an étale locally trivial vector bundle over MATH. Therefore, by NAME REF p. REF, it is locally trivial in the NAME topology. Thus we get MATH. |
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 induces a bijection on MATH-valued points. This implies MATH. |
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. For MATH with MATH let MATH with MATH where we write MATH. It is straightforward from the definitions that MATH and MATH are inverse to each other. |
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 determine MATH. First it is easy to see analoguously to the case of MATH in CITE that MATH is a locally trivial fibre bundle over MATH with fibre MATH. Therefore MATH . There is a natural morphism MATH given on MATH-valued points by MATH. MATH is obviously a bijection on MATH valued points. Thus we get MATH . Now we determine MATH. Let MATH . Then MATH. Furthermore we have a morphism MATH sending MATH to MATH, which is bijective on MATH-valued points. Thus MATH . |
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 in MATH, such that MATH . In the paper CITE they compare the wallcrossing on a rational ruled surface and its blowup in a point to determine MATH. We can translate their argument into our language, where it proves the theorem: Let MATH, MATH be ample on MATH with MATH and MATH odd. Write MATH and MATH. Write MATH. Then, as MATH is universal, we get, using REF , MATH . By definition it is obvious that MATH. The result follows by REF and the identity 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 is a locally trivial bundle over MATH with fibre MATH, thus MATH. Using REF , we only need to show that MATH . Let MATH be the lattice generated by MATH. MATH and MATH are a basis of MATH modulo MATH. Therefore MATH and MATH. By MATH and MATH, REF gives that MATH . Easy computations give MATH . The result now follows from the product formulas 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. Multiplication by MATH is well defined modulo link relations (exercise!), and thus MATH is also link relation equivalent to MATH via MATH-flavored link relations. By the definition of formal Gaussian integration the inner integral in REF is given by MATH . The map MATH kills all MATH-flavored link relations (because the MATH-marked strut MATH is MATH-invariant), and maps MATH-flavored link relations to MATH-flavored link relations. Therefore the inner integral in REF is link equivalent to MATH via MATH-flavored link relations. By the induction hypothesis we now find that the outer integral in REF vanishes. The ugly case: If MATH we consider MATH where MATH is an arbitrary scalar. Multiplication by MATH is well defined modulo link relations, and so the integrand here remains link equivalent to MATH. Thus by the lucky case, MATH vanishes for all MATH. On the other hand, the coefficient of every diagram that appears in MATH is a rational function in MATH which is non singular at MATH because MATH is regular. Thus it must be that MATH. |
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 diagonal, and that MATH is MATH. |
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 MATH. |
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. Thus MATH and hence MATH . It only remains to note that MATH. |
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, after choosing MATH and MATH as in REF (more precisely, as in REF ), the shackled MATH becomes the shackling element MATH (so MATH, the set of ``extra" labels, is MATH). We just need to compute MATH as in REF, starting from MATH. Set MATH and start crunching: MATH as required. |
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 NAME remainder theorem. The overall sign of MATH and the MATH's (and therefore also of MATH) can be read from the coefficient of MATH in MATH (see REF ). It remains to check to what extent does MATH determine the MATH's. Thus we regard MATH as `known', and try to read out MATH and MATH. For the sake of simplicity we assume that MATH; that is, that MATH. This done, REF shows that MATH makes sense as an honest analytic function of MATH, and not merely as a formal power series. Assume for the moment that MATH. For large values of MATH it is easy to bound the integral in REF above and below by rational functions of MATH, and thus the growth rate of MATH as a function of MATH is determined by the exponential prefactor. Thus by observing the growth rate of MATH we can determine MATH, factor that term out, and therefore regard the integral in REF , MATH as known. As MATH varies, this is the integral of a fixed function against all possible NAME. Knowing all of these integrals we know what the function is. Thus we know all the quantities MATH and the value of MATH (recall that the MATH's are all distinct and greater than MATH, so no accidental cancellations can occur). This finishes the case of MATH. If MATH, then MATH is a sphere and therefore MATH. This leads to MATH, which grows like MATH. This growth rate is much slower than the minimal possible value of MATH for MATH, which is attained when MATH and MATH. Thus the case of MATH is easily separated from the case of MATH. |
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 or MATH. For otherwise both MATH and MATH, and therefore MATH, have bounded inverses, so that MATH. Let MATH. The set MATH is compact and hence there is an element, MATH, with maximal modulus. If MATH then MATH has modulus greater than MATH and hence does not belong to MATH. This shows that MATH. A similar argument for MATH shows that MATH, and so MATH. If MATH, then MATH and MATH have bounded inverses since their spectra are bounded away from MATH. Hence both MATH and MATH are surjective and injective and thus they have bounded inverses. |
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 operators all coincide. Let MATH denote the associated projection. Hence MATH. By the polar decomposition theorem there are partial isometries MATH such that MATH, MATH. Note that MATH. We have MATH, so that MATH commute with MATH. (This follows from MATH.) Now we have MATH, and since the range of MATH is dense in MATH, it follows that MATH, and MATH. Finally, MATH . We can extend MATH to a unitary map MATH. |
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 multiplicity one and MATH is not zero, we prove that the map MATH is a well-defined embedding. For MATH in MATH, we have MATH . This completes the proof of REF . |
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 the field MATH is different from MATH.] |
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 ``point-derivation" MATH given by MATH, and it is clear that MATH. |
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 trivial. It follows that MATH for all MATH, and this shows that the exponent of MATH is MATH that of MATH. The inequality MATH is immediate since MATH is surjective. |
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 a subalgebra of MATH, a product of NAME algebras MATH with MATH equal to the exponent of MATH; REF now follows since the MATH-exponent of the NAME subalgebra MATH can't exceed that of MATH. |
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-nilpotent vectors (the root vectors) and that MATH is generated by elements of order MATH (the images of root homomorphisms). By REF now follows from REF , and REF follows from REF . |
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 MATH for every nilpotent MATH. The assertion for unipotent elements follows in the same way. |
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 MATH with respect to these coordinate functions. Take for MATH the localization of the ring of integers of MATH at some prime lying over MATH, and fix some embedding of the residue field of MATH in MATH. If MATH denotes a prime element of MATH, we may multiply MATH be a suitable power of MATH and assume that all the coefficients of MATH are in MATH, and that not all are REF modulo MATH. Let MATH be the function obtained by reducing MATH modulo MATH. Then the distinguished open set determined by MATH is non-empty and so must meet the set of NAME elements in MATH. After possibly enlarging MATH and MATH, we may suppose that there is a NAME element MATH with MATH, and such that the coefficients of MATH (in the root-vector basis) lie in the residue field of MATH. It is then clear that any lift MATH of MATH to MATH has the desired property. |
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 acts non-trivially. Take MATH; we show that MATH for a suitable (and hence any) NAME element. First, use the lemma to find a finite extension MATH of MATH, a valuation ring MATH, and an element MATH such that MATH is NAME, and MATH is NAME. The discussion in REF shows that the lattice MATH is MATH graded as in REF with MATH. In the notation of REF , we have MATH and MATH. Note that MATH has degree REF. It follows from CITE (or more precisely, the proof of that Proposition) that there are elements MATH such that MATH is semisimple, MATH is a subalgebra isomorphic to MATH, and such that the grading of MATH determined by MATH is the same as that determined by the function MATH as in REF. Thus for each MATH we have MATH. Applying REF we see that MATH is surjective, from which it follows that MATH is surjective. REF shows that MATH. If we regard MATH and MATH as endomorphisms respectively of MATH and of MATH, then REF of that proposition yields MATH; thus we apply REF to conclude that MATH as desired. |
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 that MATH hence MATH. Since MATH commutes with each element of MATH, we deduce that MATH hence MATH. This shows that MATH is connected, and that MATH so that MATH. Since MATH, we get MATH and REF follows. To see the second assertion of the corollary, one applies REF . |
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 desired. |
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 components of MATH are the weight spaces for the action of a MATH-subalgebra containing a regular nilpotent element, so that MATH is the highest weight. Since the regular nilpotent elements are dense in the nilpotent variety, REF follows; moreover, this argument shows that MATH when MATH is regular.] REF Since MATH is invertible in MATH for each MATH and since MATH by REF , it follows that MATH leaves MATH invariant. The result now holds by REF . |
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 whose residue field MATH has the property that MATH for a suitable MATH. Thus, we may suppose MATH. Since MATH, we may choose a lift MATH of MATH. Since each element of MATH is nilpotent, REF of the previous proposition now shows that MATH, hence also MATH which implies MATH. Since MATH is invertible in MATH, REF of the previous proposition shows that the exponential map determines a morphism of group schemes MATH over MATH. The condition MATH implies that MATH has degree MATH when regarded as a morphism of MATH-varieties MATH. Denoting by MATH the morphism obtained by base change, it follows that also MATH has degree MATH. The differential MATH satisfies MATH. It follows that MATH. Now, the exponential through MATH is the unique homomorphism MATH with degree MATH and whose differential at REF is MATH. Thus, MATH coincides with the truncated exponential, and the corollary is proved. |
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 choose a lift MATH of MATH. REF now yields a homomorphism of group schemes MATH given by the NAME exponential, where MATH is the nilpotence degree of MATH. We get then by base change a homomorphism MATH over MATH; note that by the formula defining MATH in REF, this base-changed homomorphism vanishes on the subgroup MATH, and coincides with the homomorphism MATH. It follows that MATH takes values in MATH (hence has rights to be called MATH), and is injective, as claimed. It is clear that the partition of MATH determined by the NAME block sizes of the unipotent element MATH on the natural module MATH is the same as the partition of MATH. In the cases CGREF, CGREF and CGREF with MATH, the unipotent classes of MATH and nilpotent classes of MATH are classified by these partitions (see CITE), so we get the claim on unipotent elements in these cases. In REF with MATH even, let MATH denote the full orthogonal group; thus MATH is the identity component of MATH, and has index REF. The unipotent elements of MATH all lie in MATH, and MATH. Again by CITE, the unipotent and nilpotent classes of MATH are classified by partition, so it is clear that each unipotent element of MATH lies in a suitable NAME subgroup. But any such subgroup, being connected, must lie in MATH. The rationality assertion is clear. |
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 algebra of MATH by CITE; the last isomorphism follows at once. |
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 sees that the subset MATH of all coalgebra homomorphisms is closed, and the subset MATH of all antipode preserving linear maps is closed. Then MATH is also closed. |
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 follows that MATH is an isomorphism, as claimed. |
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 statement for MATH). Thus, it suffices to show that MATH induces an isomorphism MATH. Since MATH and MATH are semisimple, MATH; since MATH and MATH are spanned over MATH by the roots, the map MATH on character groups induced by the homomorphism MATH is injective; since MATH is reduced, MATH has order prime to MATH. Thus, the lemma shows that MATH induces an isomorphism MATH, and the result follows. |
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. In REF , this follows from REF together with REF . In REF , this follows from CITE. In REF , we get the result again by REF together with REF . |
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 group schemes MATH (note that we must use here REF ), and hence (by ``restriction") a homomorphism of group schemes MATH. It is now easy to see that the assignment MATH is functorial in MATH, hence defines a morphism of schemes MATH. We get then also a morphism of varieties MATH; MATH-equivariance follows from REF . To prove injectivity, we essentially copy the proof of CITE. Suppose that MATH. Differentiating gives then MATH. Multiplying each homomorphism with MATH, one sees that MATH are equal. But then MATH and MATH coincide in MATH, and the injectivity of MATH follows by induction (note that MATH is an isomorphism of varieties when MATH; see CITE). |
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 claim this is the vanishing ideal of MATH. It suffices to see that if MATH is a set of pairwise commuting nilpotent elements in MATH, then the Abelian NAME algebra MATH which they span is contained in a NAME subalgebra; that is a consequence of the lemma which follows. |
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 apply the results of CITE. The result is well known if MATH. So now suppose that MATH, and let MATH. By the Theorem proved in CITE, there is a proper parabolic subgroup MATH of MATH with NAME decomposition MATH such that MATH and MATH [in general, MATH need not be a NAME element in MATH]. Thus, we have MATH. Since MATH, the image of MATH in MATH has dimension strictly less than that of MATH. We obtain by induction on MATH some MATH such that the image of MATH in MATH is conjugate via MATH to a subalgebra of a NAME subalgebra of MATH. Since MATH leaves MATH invariant, MATH is contained in a NAME subalgebra of MATH (which is in turn a NAME subalgebra of MATH). Since all NAME subalgebras of MATH are conjugate, we obtain the lemma. |
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-presentation MATH, the operations are as follows: if the move was to replace the generator MATH by MATH, one replaces all instances of MATH by MATH in MATH and the images of MATH, modifying them by MATH. If the move was addition of a generator MATH to MATH and MATH, one extends all MATH by MATH. If the move was deletion of MATH from MATH and MATH, one deletes all instances of MATH in the images of all MATH, and adds MATH to MATH. |
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-convex. |
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 is finitely MATH-presented, but not finitely presented. |
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 generated by MATH. A presentation for the HNN extension of MATH by MATH is MATH . |
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 MATH, so we may define MATH, a set of relators in MATH. Since MATH is normal in MATH, we also have MATH for each MATH. Consider now the presentation 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. If MATH is a split extension, then MATH for all MATH, and similarly all MATH (with MATH and MATH) are relations in MATH. If MATH is finitely presented, we may suppose MATH and again all relations given in REF are satisfied. We have shown that in the cases considered MATH is a quotient of REF . Let now MATH be a word in MATH equal to MATH in MATH. The relations MATH allow MATH to be written as MATH; then projecting to MATH gives MATH by applying relations in MATH. The same relations in REF will reduce MATH to a word in MATH; the corresponding element of MATH can be reduced to MATH using relations in MATH, and these same relations exist in MATH, so MATH and REF is a presentation of MATH. |
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 denote its coset representative. By the NAME method, MATH is generated by the finite set MATH, and a presentation of MATH is given by MATH where MATH is a rewriting of MATH as a word over MATH. Now each MATH induces naturally a monoid homomorphism MATH over MATH, and since MATH, a finite MATH-presentation for MATH is given by MATH . For the second statement of the proposition, let MATH be a finite MATH-presentation of MATH and let MATH be a finite generating set for MATH. Then MATH is a finite MATH-presentation of 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 mapping each MATH to itself; and similarly for each MATH. This MATH-presentation is in general not finite, but this can be remedied by introducing new generators MATH in bijection with MATH and new homomorphisms MATH in bijection with MATH: MATH where MATH is defined by MATH and MATH and MATH. Indeed the new generators MATH do not enlarge MATH, since MATH is a relation; also, MATH is a relation, for all MATH. |
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 of MATH by MATH, and MATH for all other MATH. Then the following is a finite MATH-presentation of MATH: MATH . Indeed write MATH. Then MATH where the MATH are arbitrary words over MATH, and MATH is given by MATH. |
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 form. The NAME multiplier is obtained from MATH by extending by MATH (which has finite rank), and restricting to MATH, both operations preserving the claimed form of MATH. |
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 MATH . |
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 to MATH. Let MATH be a generating set of MATH (so MATH is generated by MATH), and let MATH be the natural lift of MATH ; it maps MATH to MATH, where the wide tilde is applied to all MATH factors of MATH. Note that MATH satisfies the contracting condition for the same constant MATH as MATH. Since MATH is regular branch, there is an embedding MATH, from which for each generator MATH of MATH we may choose a word MATH such that MATH with the MATH in position MATH. Now write MATH for some MATH. These elements' images in MATH are trivial, since MATH is a lift of MATH. Furthermore, since MATH is contracting, one may replace MATH by its iterates under all MATH, where MATH is the projection on the MATH-th factor, and still obtain a finite set of relations. Let MATH be the quotient of MATH by this sets' normal closure. Then MATH is finitely presented and surjects onto MATH (since MATH in MATH). Let MATH and MATH be the images of MATH and MATH in MATH, and note that MATH lifts again to MATH on MATH, because the new relators MATH map to other new relators. The data are summed up in the following diagram, which should be viewed as a ``chair with MATH and MATH coming forward": MATH . Since MATH contains MATH, it has finite index in MATH. Since MATH is finitely presented, MATH too is finitely presented. Similarly, MATH is finitely presented, and we may express MATH as the normal closure MATH in MATH of those relators in MATH that are not relators in MATH. Clearly MATH may be chosen to be finite. We now use the assumption that MATH is contracting, with constant MATH. Let MATH be the set of words over MATH of length at most MATH that represent the identity in MATH. Set MATH, which clearly is finite. We now consider MATH as a set distinct from MATH, and not as a subset of MATH. We extend each MATH to a monoid homomorphism MATH by defining it arbitrarily on MATH. Assume MATH, and let MATH be a representation of MATH as a word in MATH. We claim that the following is a MATH-presentation of MATH: MATH . For this purpose, consider the following subgroups MATH of MATH: first MATH, and by induction MATH . We computed MATH. Since MATH acts transitively on the MATH-th level of the tree, a set of normal generators for MATH is given by MATH, where MATH is any choice of MATH for MATH. We also note that MATH. We will have proven the claim if we show MATH. Let then MATH represent the identity in MATH. Applying to it MATH times the map MATH, we obtain MATH words that are all of length at most MATH, that is, that belong to MATH. Then since MATH, we get MATH, and REF is a presentation of MATH. As a bonus, the presentation REF expresses MATH as the subgroup of MATH generated by 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 given by MATH-fold composition of the original map MATH. The resulting group action is still branch. However, the result of this process is that the constant MATH above can be replaced by any power of itself, say MATH, at the cost of enlarging the branching number of the tree. The generating set then now be replaced by a ball of sufficiently large radius, so that the constant MATH becomes MATH. We have reached a ``canonical situation", where the maps MATH and MATH satisfy MATH for all MATH. Assume now by contradiction that MATH is finitely presented, say MATH with MATH the canonical map, and assume that the greatest length among the relators is minimal. All MATH being trivial in MATH, satisfy a fortiori MATH, so MATH is well defined. By the NAME process, a presentation of MATH is MATH . By our assumptions that MATH and MATH is minimal, we must have MATH for all relations, so MATH is free. However a free group may not contain commuting subgroups with trivial intersection, like MATH and MATH. This is our required contradiction. |
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 quotient by finite-rank abelian groups, and the claimed result follows. |
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 element MATH moving MATH but not MATH. The computation MATH shows that MATH contains MATH, since MATH is perfect; then MATH contains too MATH so MATH contains MATH, and since MATH is MATH-transitive it contains MATH. (Explicitly, we have MATH.) |
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