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math/0106054 | As the rational representation of REF is a faithful representation of the division algebra MATH, it follows that MATH must divide MATH (see REF ). |
math/0106054 | Let MATH, MATH, MATH, MATH be as in REF . Now MATH is a subfield of MATH, and MATH, and MATH, as MATH is uniformizable. Since the centralizer of MATH contains MATH, we conclude that MATH is a subfield of MATH. Because each maximal commutative subfield of MATH has dimension MATH over MATH (see REF ), we conclude that MATH. By REF we know MATH, and so MATH . Hence MATH and so MATH; thus MATH. Now MATH is isomorphic to an order in MATH, and since MATH is the largest order MATH for which MATH leaves MATH invariant, we must have MATH. |
math/0106054 | The proof is based on two straightforward remarks: CASE: MATH is a NAME extension with MATH (compare REF ). CASE: As MATH consists of the extensions to MATH of the distinct elements of MATH, precisely MATH distinct elements of MATH restrict to each of the MATH embeddings of MATH into MATH. CASE: By the first part of REF , MATH, so MATH. By the second part of REF , elements of MATH in distinct MATH restrict to give distinct embeddings of MATH. According to REF , MATH contains precisely MATH elements restricting to the same embedding of of MATH. According to REF , these elements restrict to the same embedding of MATH. As MATH, these are all the embeddings of MATH restricting to the given embedding of MATH, that is, these elements constitute a subset of the form MATH. This proves REF . CASE: Since elements of a set MATH restrict to the same embedding of MATH into MATH, REF shows that MATH contains at least MATH such sets MATH, and MATH . Since, by REF , MATH, we find MATH. Thus we must have that MATH. Therefore MATH, and it follows that MATH, which is the first part of REF . Now we know that each MATH contains MATH elements, each of which restricts to the same embedding of MATH into MATH. By REF then, elements from distinct MATH must restrict to distinct embeddings of MATH into MATH. This is the second part of REF . |
math/0106054 | We first prove REF . Let MATH be a simple sub-MATH-module. Let MATH be the largest MATH-algebra for which MATH leaves MATH invariant. Let MATH be the fraction field of MATH; thus MATH is an order of MATH. Let MATH; MATH; and MATH. For some MATH, the direct sum MATH is a MATH-submodule of finite index. Now for all MATH, MATH is a simple sub-MATH-module isogenous to MATH. As a sub-MATH-module of a uniformizable abelian MATH-module, MATH is also abelian and uniformizable REF , and for MATH we have MATH. The action of MATH on MATH induced by MATH leaves both MATH and the period lattice MATH of MATH invariant. For each MATH, MATH, and MATH has finite index in the period lattice of MATH. By REF , we know both that MATH spans MATH over MATH and that MATH spans MATH over MATH. As MATH has finite index in MATH, we find that MATH. Moreover, MATH . Now as each MATH is simple, we can re-order the MATH so that for some MATH, we have an isogeny MATH . Thus MATH is isogenous to MATH, which proves REF . We will now show that MATH. First, we show that the centralizer of MATH in MATH is MATH. As a REF-dimensional MATH-vector space, MATH, for some MATH. For any MATH, there is a MATH so that MATH . If MATH commutes with all of MATH, then in fact, MATH. Because MATH spans MATH over MATH, it follows that MATH. For some non-zero MATH, MATH has MATH, and therefore MATH. In other words, MATH. Now let MATH be the center of MATH. From the diagonal mapping MATH, we see that the image of MATH is in the center of MATH and in particular in the centralizer of MATH. Therefore, MATH by the above paragraph. Thus there is a field MATH such that MATH. Let MATH and MATH. By the general theory of semi-simple algebras (see CITE, pp. REF), since REF MATH is a division algebra and REF MATH is a subfield of MATH which is its own centralizer, it follows that MATH . Thus we have MATH and so MATH . Since MATH by REF , it follows that MATH. Therefore MATH and MATH. On the other hand, MATH . Now MATH stabilizes MATH and is an order in MATH. Our choice that MATH be the largest MATH-subalgebra of MATH for which MATH leaves MATH invariant then implies that MATH and so MATH. Therefore, MATH and MATH . From REF we have MATH, and of course MATH. Hence MATH and MATH is isogenous to MATH. For REF we prove the contrapositive of both directions. If MATH is not simple, it contains a proper simple sub-MATH-module MATH. Picking up the considerations and notation of REF , we see that MATH. As the action of MATH on MATH is obtained through MATH, we find from REF that we can partition MATH such that for all MATH and for all MATH, MATH . Thus for each MATH, MATH, the embeddings MATH agree on MATH. Since MATH, the set MATH can be written as MATH, for MATH equal to any of the elements MATH, MATH. Thus MATH is the union of sets MATH, and the criteria of REF are satisfied. Suppose now that there is a field MATH satisfying the equivalent criteria in REF . By REF it suffices to consider the case that the period lattice of MATH is isomorphic to the maximal order MATH of MATH. Let MATH, MATH, and let MATH and MATH be the integral closures of MATH in MATH and MATH respectively. By hypothesis, we know that MATH is the union of MATH subsets of MATH, each consisting of all MATH embeddings of MATH with given restriction to MATH, given by, say, MATH, that is, MATH . We may index the elements of MATH so that MATH . We define MATH to be the image in MATH of MATH under the conjugate embedding MATH as in REF, taking MATH. By NAME 's REF , MATH is the period lattice of a MATH-module MATH of NAME MATH with complex multiplications by MATH. Let MATH be the exponential function of MATH. Let MATH be given as in REF so that MATH, and let MATH be a direct sum so that MATH is a free MATH-module of maximal rank MATH lying inside MATH. Let MATH which has finite index in MATH. Then for MATH, by definition we have MATH . Keeping the convention REF in mind, for MATH we let MATH be the matrix MATH and then set MATH . From REF we see that MATH where MATH run through MATH. Since the image of MATH spans MATH, we know that MATH is invertible. It follows that for MATH, MATH where MATH. Let MATH denote projection onto the MATH-th component of MATH so that MATH . Now define an analytic map MATH by MATH . The function MATH then REF is entire; REF vanishes exactly on MATH with simple zeros; REF as a power series in MATH satisfies MATH according to REF; and REF inherits a functional equation from MATH. These properties make MATH the exponential function of the MATH-module MATH where MATH is given by MATH . As MATH for arbitrary MATH, by REF this MATH-module is uniformizable, since MATH is. Moreover, it is abelian since MATH is. By construction MATH is isomorphic to MATH. Now MATH has finite index. By REF there is a uniformizable abelian MATH-module MATH isogenous to MATH, which has MATH as its period lattice. However, uniformizable abelian MATH-modules are determined by their period lattices by REF , and so MATH. Therefore, MATH is isogenous to MATH and so contains a sub-MATH-module isogenous to MATH. |
math/0106054 | Suppose MATH is isogenous to MATH and MATH is isogenous to MATH with MATH and MATH both simple. From REF a it follows that the group MATH of MATH-module morphisms is non-trivial if and only if MATH and MATH are isogenous. By REF we know that MATH and MATH are MATH-modules of NAME. Because for the concerns of this theorem we fix our sub-MATH-modules only up to isogeny, by REF we can assume that MATH and MATH have period lattices isomorphic to the full rings of integers in their respective NAME. Now suppose that subfields MATH exist as in the statement of the theorem. Let MATH and MATH. Because MATH, we can reorder the coordinates of either MATH or MATH so that MATH . Let MATH be the ring of integers in MATH, MATH. We can assume without loss of generality that for, say, MATH the period lattices MATH and MATH of MATH and MATH are MATH . By REF, for MATH, MATH. So it follows that MATH, and, as uniformizable abelian MATH-modules are determined by their period lattices, the original MATH and MATH are isogenous (by REF ). For the other direction, we start with a given isogeny MATH, where each MATH has CM by the full ring of integers MATH in MATH. By REF, there is a MATH and an isogeny MATH so that MATH. These isogenies induce a map of MATH-modules MATH via MATH . This map lifts to a MATH-linear map MATH, MATH which is also multiplicative, as the following calculation shows: MATH . By REF , since the MATH are simple, the maps MATH extend to isomorphisms MATH. Thus MATH . Note also that the natural map MATH extends naturally to MATH via MATH . For each MATH, we verify that the following diagram commutes: MATH . Indeed, writing MATH with MATH and MATH, we see that MATH as MATH . Because MATH, MATH . Therefore it follows from REF that, for every MATH, MATH and MATH are conjugate and hence have the same eigenvalues. By choosing a primitive element MATH over MATH, we see that, up to a fixed permutation of coordinates, MATH equals MATH for all MATH, and therefore MATH. |
math/0106054 | By MATH-linearity, we need only check that the ``competing" expressions given by applying the product formula inductively to different factorizations MATH are equal. This verification is straightforward. |
math/0106054 | It is clear that for MATH, MATH . Since MATH is torsion-free, multiplication by MATH is injective. |
math/0106054 | The direct sum decompositions are immediate. By definition, MATH. To calculate the dimension of MATH, we appeal to the following diagram, where here we assume that MATH is abelian of rank MATH: MATH . The labels indicate the MATH-dimensions of the quotients. By REF , MATH is the codimension of MATH in MATH. Thus MATH and MATH. |
math/0106054 | Write MATH, with the variables chosen so that MATH is upper triangular. If MATH then we can equate coefficients in REF with MATH to find the equality MATH . It shows that the MATH is uniquely determined and that the terms MATH as MATH, for any fixed value of MATH. Therefore, for any fixed values of MATH, the terms MATH tend to zero as MATH in the following equality: MATH . Consequently, we find that the terms MATH tend toward zero as MATH for any fixed value of MATH. Iterating this argument, we find that each series MATH is uniquely determined and entire. Thus MATH is uniquely determined and everywhere convergent. The MATH-linearity of MATH and the following recursive calculation show that MATH satisfies the required functional equation with respect to MATH: MATH . |
math/0106054 | Note that MATH has no linear terms and that the functional equation holds: MATH . |
math/0106054 | That MATH defines a MATH-module follows from the hypothesis that MATH does and that MATH where MATH is the MATH identity matrix. That MATH satisfies the appropriate functional equation MATH follows from the functional equations for MATH and for the MATH. Moreover, since the MATH have zero linear terms, the linear terms of MATH are precisely MATH, as required for the exponential function of a MATH-module. |
math/0106054 | This proposition follows from the following remark, whose verification is immediate. The first identity there shows that the isomorphism class of the extension is independent of the representatives chosen for the classes in MATH. The second shows that the isomorphism class is independent of the generators chosen for the span in MATH. |
math/0106054 | If MATH is inner, then choose MATH in the first part of the preceding lemma. Similarly, if there is a self-isogeny MATH of MATH such that MATH then comparing entries first in the lower right-hand corner gives that the lower right-hand entry MATH in MATH satisfies MATH, that is, MATH. Then considering entries along the bottom row shows that, up to a non-zero scalar multiple MATH, MATH where MATH is the bottom row of MATH. Now if MATH, then MATH. However, since the MATH-motive MATH is a free MATH-module, we would have MATH. Thus the bottom line of MATH would consist of zeros. In that case, MATH would have an infinite kernel and could not be an isogeny. Therefore MATH, and MATH is inner, as claimed. |
math/0106054 | CASE: We adopt the notation of REF for our strictly reduced quasi-periodic extension. The plan is to start with a non-trivial algebraic relation on a proper sub-MATH-module MATH of MATH and conclude that the image of MATH lies in a proper sub-MATH-module of MATH. Choose coordinates for MATH so that MATH is upper-triangular. Assume that we have a non-trivial relation holding on the coordinates of MATH, which is smallest with respect to the reverse lexicographical ordering on monomials in MATH. Since the underlying group is MATH-linear, this minimal relation has the following form for all MATH (see REF): MATH in which all the MATH, MATH. Now we make a series of observations based on the fact that MATH is a sub-MATH-module, that is .that, for all MATH, MATH as well. We show by contradiction that the variables of MATH are not involved in this minimal relation. Since the relation REF is minimal and the effect of MATH on the variables MATH is multiplication by MATH (plus a sum affecting the variables of MATH), we apply MATH to obtain another algebraic relation and, on comparing maximal monomials with respect to our ordering, we conclude that, if the variables MATH were involved in REF, then for each MATH for a fixed MATH. Thus, since MATH, each MATH is a monomial of degree MATH. However MATH is a perfect field, and we can write MATH with the MATH. Again comparing terms in the relations REF, but this time for the variables in MATH, we find that MATH where MATH and MATH. In particular, for each variable MATH, MATH where MATH and MATH. Assume for the moment that MATH. Let MATH, MATH. Notice now that the MATH lack linear terms. Since MATH is upper-triangular, MATH has no linear terms unless MATH. From REF we see that MATH . Because MATH, it follows that MATH, that is, MATH does not have a linear term. Similarly from REF we see that MATH . Since MATH, we see as above that MATH. Proceeding by induction we find that MATH for MATH, and so none of the MATH involve linear terms. But in that case, since the MATH are MATH-linear, the relation REF would not be minimal; we could extract the MATH-th root and have a equation of smaller degree for elements of MATH. Therefore we are reduced to the case MATH, and we find that REF has the form MATH . However this means that MATH and thus MATH are inner, whereas the given MATH are MATH-linearly independent modulo the inner biderivations. Consequently MATH. Since MATH, we see that the minimal relation REF actually involves only variables from MATH after all. Therefore the projection of MATH in MATH cannot be surjective. CASE: Assume now that MATH do not represent linearly independent classes in MATH. Then by an automorphism of MATH involving only a linear change of coordinates, we may assume that MATH is inner. Then, as we have seen above, we may conjugate by a matrix leaving the terms corresponding to MATH invariant to see that MATH has a direct factor corresponding to MATH. If there are further inner biderivations lying in MATH, we may proceed to find further direct factors of MATH. At any rate, in this case, the direct complement of these factors form a proper sub-MATH-module of MATH which projects onto MATH. Thus MATH is not a minimal extension of MATH in this case. |
math/0106054 | According to the identity of the first part of REF , the elements of MATH corresponding to MATH form a direct summand of MATH. Let MATH be a direct summand of MATH. We need to show that the space MATH of biderivations associated to elements of MATH is contained in MATH. The above short exact sequence becomes MATH or, since MATH, MATH . Choose a MATH-basis MATH for MATH and extend to a basis for MATH with the biderivations MATH. Augment MATH with MATH to a MATH-basis for MATH and finally extend with biderivations MATH to obtain a basis MATH which produces MATH. Now the part of MATH corresponding to MATH is a direct summand of MATH. NAME the original exact sequence by it, we may assume that MATH, that is, that MATH, and take as our objective to show that, in this case, MATH. According to the identity of the first part of REF , the quasi-periodic extension of MATH corresponding to MATH is also a direct summand of MATH. NAME by it, we may assume that also that MATH, that is, that MATH. But then MATH are representatives of linearly independent classes in MATH, and therefore by REF in this case, the original extension is minimal. Consequently, since MATH projects onto MATH, MATH, and MATH, as desired. |
math/0106054 | By isogeny MATH and MATH have the same dimension and rank; furthermore, MATH and MATH have the same rank because MATH . In light of the dimensions calculated in REF , it suffices to show that MATH and MATH are injective. Since MATH is an isogeny, there is an isogeny MATH and a non-zero element MATH so that MATH . Clearly MATH, and so we need only show that MATH and MATH are injective. Suppose that MATH is inner (or strictly inner). Then MATH for some MATH with MATH (or MATH) and for all MATH. Let MATH. Since MATH, we see that MATH . Since MATH has no MATH-torsion, we can cancel the common right factor to see that MATH, which is strictly inner if and only if MATH is so. |
math/0106054 | According to the preceding proposition, the MATH are representatives of a basis for MATH. Since MATH we see that MATH is an isogeny from MATH to MATH. |
math/0106054 | Let MATH arise from the MATH-biderivations MATH. Then according to REF , the MATH are linearly independent modulo MATH. As in the proof of REF , MATH . Moreover, MATH . Thus, according to REF , MATH . Consequently, the MATH span MATH, and we conclude via REF . |
math/0106054 | Let MATH be an isogeny and let MATH denote the period lattices for MATH. Then MATH is a MATH-sublattice of MATH of finite index. Thus the MATH-spans of the lattices are the same. We saw in the preceding result that such an isogeny MATH induces an isogeny MATH of the related minimal quasi-periodic extensions. The claim follows on appealing to REF. |
math/0106054 | CASE: We first demonstrate that MATH satisfies the functional equation of MATH, proceeding as in REF . Using MATH, we see that MATH where the last equality follows exactly as in REF, using the definitions of MATH and MATH. CASE: Next we show that the composite MATH is an entire function. For that we choose a system MATH of coordinates for MATH. Given MATH, set MATH, MATH. Thus from the definitions of the various twists involved, we see that MATH . Our goal is to express MATH as a MATH-linear combination of MATH with coefficients which decrease sufficiently rapidly with MATH. We make some preliminary observations. Let MATH. Recall REF and use the fact MATH which is verified recursively, to see that MATH . Next, we write MATH as MATH . Combining REF we then have MATH . Finally, since for each MATH, MATH, we have MATH, if we set MATH, we can write MATH where each MATH. Then we divide by MATH and substitute this expression in for MATH in the term MATH to obtain MATH where each MATH has entries which are the sum of terms of the form MATH or MATH, where the MATH are coefficients of MATH. Continuing in this way, we can progressively eliminate the first MATH degree terms to find an expression MATH where each MATH has coefficients which are sums of terms of the form MATH for which MATH, for MATH, each MATH is an entry of the constant matrix MATH, and MATH. Now we substitute the expressions from REF with MATH into the terms of REF involving MATH. According to REF, when we multiply MATH by the preceding REF and apply MATH, we obtain MATH . When MATH, the above differential multiplied by MATH is regular along MATH, since then the uncancelled MATH occurs in the numerator. Therefore the contribution towards the residue at MATH of such terms is nil, and we can concentrate on the terms with MATH. We obtain that MATH where MATH is a rational function in MATH and certain twists MATH on MATH without poles along MATH. Because MATH is constant on MATH, we conclude from REF that MATH . It follows that MATH and so, using the fact that we have already shown that this function satisfies the proper functional equation, by REF it will be shown equal to MATH once we verify that the right-hand side is entire. To this end we need to estimate MATH, which we do by estimating the terms appearing in REF. Since we have the bound MATH on the factors occurring in REF, where MATH is an upper bound for the absolute values of the entries of the matrices MATH, we see that MATH where MATH is an upper bound on the coefficients MATH in REF. Here we have used the fact that MATH in REF, whereas MATH, MATH. From this estimate it is clear that REF is entire, since MATH. One can also show that the power series in REF is entire by appealing to REF. |
math/0106054 | The MATH-linear maps MATH on MATH are linearly independent over MATH. Indeed by NAME, for MATH sufficiently large MATH and MATH. Further the MATH are trivial on MATH. Thus we can choose MATH as representatives of a basis modulo MATH dual to the maps MATH. That the MATH are defined over MATH follows from the fact that the MATH-motive MATH is obtained by extending scalars on a MATH-motive which is initially defined over MATH (as in REF ). |
math/0106054 | Fix MATH and MATH. In the following, for precision we distinguish between the function MATH on MATH and the scalar MATH. As MATH from REF, it follows from REF that MATH . From REF it follows that MATH. Thus as MATH, the remark immediately following REF shows that the poles of MATH contained in MATH lie along the support of the divisor MATH . Moreover, we establish MATH . This equality follows from REF . The required estimates are obtained exactly as in REF , (taking MATH), and the sum on the left of REF is taken over all of the poles of MATH contained in MATH. Combining REF we find that MATH . As MATH, it follows that MATH is regular at MATH for MATH, MATH. REF bc combined with the calculation of REF shows that MATH . Because MATH and MATH are defined over MATH, the constant MATH and MATH. |
math/0106054 | MATH. |
math/0106054 | Since MATH is itself a MATH-module of NAME, it is isogenous to a power of a simple MATH-module of NAME. Thus there is a smallest field MATH which satisfies the criteria of REF . Let MATH be the subgroup of MATH corresponding to MATH. Certainly MATH. However, if MATH, then since MATH is the union of cosets of MATH, it must be the case for all MATH that MATH is monic if and only if MATH is monic. Thus MATH, MATH, and MATH is simple. |
math/0106054 | Certainly REF are equivalent according to the discussion in REF. Assuming that MATH and MATH are isogenous, then by REF the NAME MATH of MATH and MATH is a subfield of MATH and simultaneously satisfies the criteria of REF for both MATH and MATH. Furthermore, if we let MATH be the preferred embeddings for the NAME of MATH, then for some MATH we have MATH . Now let MATH be the least common multiple of MATH and MATH, and let MATH. We claim that MATH . Indeed, suppose MATH and MATH are monic with MATH. It follows that MATH and MATH. However, by REF we can choose MATH so that MATH. Then certainly MATH, and so by the hypotheses on MATH from REF , we must have MATH . Therefore MATH, implying that MATH. Using REF we find that MATH, completing REF implies REF . Now suppose that MATH for some MATH and MATH. For any MATH it follows that MATH. Thus we can specify without loss of generality that MATH for some MATH. Let MATH be the subset MATH . Certainly MATH is a subgroup of MATH by the bracket relations. We claim that MATH . Indeed, if MATH and MATH, then MATH . Furthermore, as in REF, MATH . We let MATH be the fixed field of MATH, and thus MATH is the NAME of both MATH and MATH. Because MATH and MATH are surjective, if follows from REF that MATH . Because MATH, it follows that MATH . Thus MATH, and by REF MATH is isogenous to MATH. |
math/0106054 | If the values of REF are linearly dependent over MATH, then, by NAME 's Theorem of the Sub-MATH-Module REF, the point corresponding to REF lies in the tangent space at the origin of a sub-MATH-module MATH of MATH. REF shows that MATH is a minimal extension of MATH. Therefore, by minimality, MATH is contained in the tangent space of a proper sub-MATH-module MATH of MATH. As MATH is simple, MATH is zero, contrary to our choice of a non-zero MATH. |
math/0106054 | By NAME 's Theorem of the Sub-MATH-Module REF, any linear dependence relation would give a proper sub-MATH-module MATH of MATH with MATH containing the point whose coordinates are given by tuples of the above values. Since strictly quasi-periodic extensions are minimal, we know by REF that MATH projects onto a proper sub-MATH-module MATH of MATH. In that case, we can apply NAME 's NAME result REF , to conclude that, for some fixed MATH, there are non-zero endomorphisms MATH such that the projection of MATH is contained in the sub-MATH-module of MATH given by MATH. In particular, MATH. This identity is contradicted by our hypothesis that MATH are linearly independent over MATH. |
math/0106054 | Note from REF that MATH is spanned by the coordinates of the periods from MATH. We use the CM structure to show that MATH is spanned by the coordinates of any non-zero period of MATH: Define MATH, in the notation of REF. By REF , we may assume that the period lattice of MATH is MATH for some MATH. We need to verify that, for each MATH among the MATH-biderivations MATH underlying the quasi-periodic extension MATH, MATH lies in the MATH-span of the coordinates of MATH and the MATH. Recall that MATH extends from MATH to MATH in such a way that MATH and thus in particular MATH. Now let MATH be a MATH-biderivation. Then the fact that MATH shows thus that MATH is itself the quasi-periodic function associated to the MATH-biderivation MATH. Therefore the values of this function at MATH will lie in the MATH-span of the coordinates of any non-zero period. Hence MATH is spanned by the coordinates of any non-zero period of MATH. Since the MATH-action on MATH is scalar, NAME 's Theorem of the Sub-MATH-Module REF, implies that any MATH-linear relation on the coordinates of a fixed period MATH actually will hold on MATH for some proper sub-MATH-module MATH of MATH. However MATH is a minimal extension of the MATH-module MATH. So the period onto which MATH projects in MATH would lie in a proper sub-MATH-module of MATH. Since MATH is simple, MATH projects onto MATH; however such a MATH, contrary to our hypothesis. Therefore the coordinates of MATH and the quasi-periods MATH form a MATH-basis for MATH, as claimed. |
math/0106054 | We know from the preceding result that MATH is a MATH-basis for the MATH-vector space MATH . Since each MATH is a minimal extension of MATH, then by REF, MATH is a minimal extension of MATH. We know by NAME 's Theorem of the Sub-MATH-Module REF that any MATH-linear relation on MATH gives rise to a proper sub-MATH-module of MATH, which, by minimality, projects onto a proper sub-MATH-module MATH of MATH. By the simplicity and non-isogeneity of the MATH, we know that the only proper sub-MATH-modules for MATH have trivial projections onto some factor, say MATH. But then the underlying period of MATH must be zero. That forces the whole period of MATH to vanish, contrary to our choice of MATH. |
math/0106054 | It is clear that the claim holds for every choice of representatives if it holds for any choice. If MATH, then MATH, since MATH. Therefore we may always choose representatives of MATH of the form MATH. REF says that all MATH-equivalence classes among these values have cardinality MATH when MATH. Thus there are exactly MATH classes among them. According to REF , and REF , these representatives span MATH, a MATH-space of dimension MATH. Therefore any choice rep-MATH of MATH-representatives among the MATH gives a MATH-basis for MATH. Thus one choice of MATH will be a disjoint union of rep-MATH, taken over non-isogenous MATH. Consequently, the claim of the theorem is equivalent to the statement that MATH and coordinates of non-zero periods from non-isogenous MATH are MATH-linearly independent. We proceed to prove this assertion. NAME 's Theorem of the Sub-MATH-Module REF, shows that, if there were a MATH-linear relation on the coordinates of such periods, then there would have to be a proper sub-MATH-module MATH of a finite product of the form MATH for which MATH would contain the point MATH, where each MATH is some non-zero period of MATH, MATH and MATH denotes the NAME module. Here the MATH occur exactly when some MATH is involved in the supposed MATH-linear relation. Moreover, since MATH, MATH and/or MATH appear exactly when MATH and/or MATH are involved in the relation. Compare REF. For ease of exposition, we simply assume that to be the case here. By REF, the above product is minimal. Therefore MATH projects onto a proper sub-MATH-module MATH of the corresponding product MATH for which MATH would contain the projection MATH of the point of MATH of MATH. Again the point MATH in MATH projects non-trivially onto the factors because, as we saw in REF , the non-zero periods of MATH are produced from - and project to - non-zero periods of MATH. For the conclusion of the proof we keep in mind the following three remarks: CASE: The underlying simple MATH, MATH have been taken to be non-isogenous. CASE: The NAME module MATH has period an algebraic multiple of MATH, and MATH does not have CM. Therefore it is a simple MATH-module which is not isogenous to any of the soliton MATH-modules MATH. CASE: The trivial MATH-module MATH is also simple and not isogenous to MATH nor any MATH. Now MATH . Since the factors of MATH are non-isogenous and simple, the only proper sub-MATH-modules have tangent spaces which project trivially to the tangent space of at least one factor. Thus MATH does so, as MATH is a proper sub-MATH-module. However this is contradicted by the facts that MATH and our choice of product MATH ensures that MATH has a non-zero entry in every factor of MATH. We conclude that there is no non-trivial MATH-linear relation on the set in question. |
math/0106061 | Let MATH be the simply-connected algebraic group with NAME algebra MATH. MATH descends to MATH, preserving MATH, the NAME subgroup with NAME algebra MATH. Thus MATH descends to the Flag Manifold MATH, and in particular preserves the big cell MATH. This induces an action of MATH on MATH - the NAME algebra of vector fields on MATH, denoted MATH. It follows from this that the embedding of MATH into MATH induced by the left action of MATH on MATH is equivariant. Let MATH be a basis of MATH such that MATH, and MATH. Let MATH be the set of simple roots, and set MATH. Then MATH is a basis for MATH. We coordinatize MATH by MATH, with MATH dual to MATH, and MATH by MATH, with MATH dual to MATH. Then MATH acts by sending MATH, MATH. We introduce a NAME algebra isomorphic to MATH with generators MATH and commutation relations MATH . This change of basis for MATH simply corresponds to the change of basis on MATH from a root basis to one on which MATH acts diagonally. Under the identification MATH, MATH corresponds to the NAME algebra of MATH. In this basis for MATH, MATH, MATH. Similarly, we introduce a NAME algebra isomorphic to MATH with generators MATH, and commutation relations MATH . Here, MATH. If MATH, let MATH be the polynomial vector field on MATH induced by MATH. We write MATH . We will use the notation MATH to denote the vector of MATH given by the following expression MATH where MATH means that we replace MATH by MATH in the polynomial MATH. It follows from the untwisted NAME construction CITE that MATH where MATH is some constant depending on the level MATH. It suffices to verify the lemma for these vectors, since MATH generate MATH. Because the embedding MATH is equivariant, it follows that MATH. Therefore the embeddings of MATH are already equivariant, and it remains to prove that MATH. To shorten notation, we denote the image of MATH under MATH, MATH, by MATH, and suppress the vacuum vector. Let MATH . Using the fact that MATH is a vertex algebra homomorphism, and that MATH, we have: MATH . However, we know from the NAME construction that MATH, so that MATH, which implies that MATH as desired. This proves that MATH is equivariant. The equivariance of MATH is shown similarly. |
math/0106061 | The homomorphism MATH is MATH - equivariant, so that MATH inherits a structure of MATH - twisted MATH - module. Let MATH. Consider the twisted fields MATH. Using the associativity property for twisted modules (see CITE) and the fact that MATH is a vertex algebra homomorphism, we get (the vacuum MATH is suppressed for brevity): MATH . This implies that the fields MATH have the same OPE as REF , and hence their NAME coefficients generate MATH. |
math/0106061 | The NAME subalgebra MATH of MATH is spanned by MATH. When the level MATH is understood, we write elements MATH in the form MATH, where MATH is the orthogonal projection of MATH to MATH, and MATH is dual to MATH. Unraveling the homomorphism MATH, we see that MATH acts on MATH as the operator MATH has a basis of monomials of the form MATH where MATH and MATH are non-negative integers. Now MATH . An easy computation shows that the above monomial lies in the MATH weight space. The formal character of MATH is therefore given by: MATH which is the character of MATH - the NAME module. |
math/0106061 | It is clear that MATH is injective. REF for MATH implies that we can solve for MATH in terms of MATH and MATH. In other words, MATH where in the summation, MATH in the root lattice of MATH, and MATH are polynomials. Now, MATH has a basis of monomials of the form MATH where MATH. Thus we see that MATH is surjective. |
math/0106061 | The degree of the operator MATH is MATH. It follows that repeated application of such operators to the vacuum can never result in a vector with degree MATH. |
math/0106061 | The formal character MATH of a category MATH module MATH can be uniquely expressed as a linear combination of characters of irreducible highest-weight modules MATH. Let MATH denote the coefficient with which MATH occurs in MATH. Suppose that MATH has a proper submodule. Then either MATH or MATH has a singular vector MATH. Suppose first that MATH, and that MATH has weight MATH. It follows that MATH. Now, MATH is a quotient of MATH, which by REF has the same character as the NAME module MATH. It follows that MATH. By REF , MATH. REF from CITE then implies that MATH contains a singular vector of weight MATH, and hence degree MATH, contradicting the following Lemma. The case where MATH is treated in the same way. |
math/0106061 | Suppose that MATH has weight MATH. By REF, there exists a sequence of positive roots MATH, and a sequence of positive integers MATH such that MATH for MATH and MATH. Now MATH, so MATH for some MATH. Let us choose the smallest such MATH. We have that MATH and since real and imaginary roots are orthogonal, MATH, contradicting the assumption that MATH is generic. |
math/0106063 | CASE: From the Chain Rule, and the fact that MATH, we have MATH . Also, because MATH, we know that MATH. We conclude that MATH; thus, the linear transformation MATH satisfies the polynomial equation MATH. Suppose MATH. (This will lead to a contradiction.) Since the polynomial MATH has no repeated roots, we know that MATH is diagonalizable. Furthermore, because MATH and MATH are the only roots of MATH, we know that MATH and MATH are the only possible eigenvalues of MATH. Thus, because MATH, we conclude that MATH is an eigenvalue; so we may choose some nonzero MATH, such that MATH. Let MATH be the geodesic with MATH and MATH. Then, because MATH is an isometry, we know that MATH is also a geodesic. We have MATH and MATH . Since a geodesic is uniquely determined by prescribing its initial position and its initial velocity, we conclude that MATH. Therefore, MATH, so MATH is a fixed point of MATH, for every MATH. This contradicts the fact that the fixed point MATH is isolated. CASE: Define MATH, so MATH is a geodesic. Because MATH is an isometry, we know that MATH is also a geodesic. We have MATH and, from REF, MATH . Since a geodesic is uniquely determined by prescribing its initial position and its initial velocity, we conclude that MATH, as desired. |
math/0106063 | To simplify the proof slightly, let us assume that MATH (see REF). Because MATH is compact, we know there is a MATH-invariant NAME metric on MATH (see REF). Then, because MATH is finite, and normalizes MATH, it is not difficult to see that we may assume this metric is also MATH-invariant (see REF). (This conclusion can also be reached by letting MATH and MATH, so MATH is a compact subgroup of MATH, such that MATH.) Thus, MATH is an involutive isometry of MATH. Suppose MATH is a fixed point of MATH, with MATH. Then MATH, so we may write MATH, for some MATH. Since MATH centralizes MATH (and MATH is an automorphism), we have MATH . On the other hand, we know MATH (because MATH is involutive), so we conclude that MATH. Since MATH, and MATH, we have MATH, so MATH. Since MATH, we conclude that MATH. (There is a neighborhood MATH of MATH in MATH, such that, for every MATH, we have MATH.) Therefore MATH, so MATH; thus, MATH. |
math/0106063 | For MATH, we have MATH. The stabilizer MATH of a point in MATH acts transitively on the unit tangent vectors at that point. So MATH acts transitively on the unit tangent vectors of MATH. |
math/0106063 | Suppose MATH is a connected, abelian, normal subgroup of MATH. For each function MATH, let MATH . That is, a nonzero vector MATH belongs to MATH if CASE: MATH is an eigenvector for every element of MATH, and CASE: the corresponding eigenvalue for each element of MATH is the number that is specified by the function MATH. Of course, MATH for every function MATH; let MATH. (This is called the set of weights of MATH on MATH.) Each element of MATH has an eigenvector (because MATH is algebraically closed), and the elements of MATH all commute with each other, so there is a common eigenvector for the elements of MATH. Therefore, MATH. From the usual argument that the eigenspaces of any linear transformation are linearly independent, one can show that the subspaces MATH are linearly independent. Hence, MATH is finite. For MATH and MATH, a straightforward calculation shows that MATH, where MATH. That is, MATH permutes the subspaces MATH. Because MATH is connected and MATH is finite, this implies MATH for each MATH; that is, MATH is a MATH-invariant subspace of MATH. Since MATH is irreducible, we conclude that MATH. Now, for any MATH, the conclusion of the preceding paragraph implies that MATH, for all MATH. Thus, MATH is a scalar matrix. Since MATH, this scalar is a MATH root of unity. So MATH is a subgroup of the group of MATH roots of unity, which is finite. Since MATH is connected, we conclude that MATH, as desired. |
math/0106063 | Let MATH be a connected, normal subgroup of MATH. Because MATH, it is not difficult to show that MATH is completely reducible: there is a direct sum decomposition MATH, such that MATH is irreducible, for each MATH (see REF). The proof of REF (everything except the final paragraph) shows that MATH consists of scalar multiples of the identity, for each MATH. Hence MATH. Since MATH is connected, but (by assumption) MATH is finite, we conclude that MATH is trivial. |
math/0106063 | Let MATH be the NAME closure of MATH. Then MATH is semisimple. (For example, if MATH is irreducible in MATH, then MATH is also irreducible, so REF implies that MATH is semisimple.) Because MATH is connected, we know that the normalizer MATH is NAME closed (see REF). Therefore MATH, so MATH is a normal subgroup of MATH. Hence (up to isogeny), we have MATH, for some closed, normal subgroup MATH of MATH (see REF). Then MATH is almost NAME closed (see REF ). |
math/0106063 | Because MATH is discrete, there is a nonempty, open subset MATH of MATH, such that MATH. Since MATH is second countable (or, if you prefer, since MATH is MATH-compact), there is a sequence MATH of elements of MATH, such that MATH. Let MATH . Then MATH is a NAME fundamental domain for MATH in MATH (see REF). |
math/0106063 | See REF for REF. The uniqueness of MATH follows from REF and the uniqueness of the NAME measure MATH. |
math/0106063 | CASE: From REF, we have MATH. Thus, MATH has finite volume if and only if MATH. CASE: Obvious. CASE: We have MATH, for every MATH, so, from REF, we see that MATH . |
math/0106063 | CASE: |
math/0106063 | Let MATH be a NAME fundamental domain for MATH in MATH, such that MATH. Define a MATH-finite NAME measure MATH on MATH by MATH for every NAME set MATH in MATH, such that MATH. The proof of REF shows that MATH (see REF). Since MATH, this implies that MATH . Since MATH has finite volume, we conclude that MATH, as desired. |
math/0106063 | We may assume MATH is reducible (otherwise, let MATH). Thus, there is some noncompact, connected, closed, normal subgroup MATH of MATH, such that MATH is not dense in MATH; let MATH be the closure of MATH, and let MATH. Because MATH, we know that MATH normalizes MATH, so MATH is a normal subgroup of MATH (see REF). Let MATH. By definition, MATH is open in MATH, so MATH is also open in MATH. On the other hand, we have MATH, and MATH is dense in MATH (by the definition of MATH), so we must have MATH. Therefore MATH is closed in MATH, so MATH is a lattice in MATH see REF. Because MATH is normal in MATH and MATH is semisimple (with trivial center), there is a normal subgroup MATH of MATH, such that MATH see REF. Let MATH be the projection of MATH to MATH. Now MATH normalizes MATH, and MATH centralizes MATH, so MATH must normalize MATH. Therefore REF implies that MATH is discrete (hence closed), so MATH is closed, so MATH is a lattice in MATH see REF. Because MATH is a lattice in MATH and MATH is a lattice in MATH, we know that MATH is a lattice in MATH. Thus, MATH has finite index in MATH see REF. By induction on MATH, we may write MATH and MATH, so that MATH is an irreducible lattice in MATH, for each MATH. |
math/0106063 | CASE: Define MATH by MATH. Then MATH is a covering map, so, for each MATH, there is a connected neighborhood MATH of MATH, such that the restriction of MATH to each component of MATH is a diffeomorphism onto MATH. Since MATH is an open cover of MATH, and MATH is compact, there is a connected neighborhood MATH of MATH in MATH, such that, for each MATH, there is some MATH, with MATH (see REF). Then MATH is injective, for each MATH. CASE: We prove the contrapositive. Let MATH be any nonempty, precompact, open subset of MATH. (We wish to show, for some MATH, that MATH is not injective.) If MATH is any compact subset of MATH, then, because MATH is not compact, we have MATH . Thus, by induction on MATH, we may choose a sequence MATH of elements of MATH, such that the open sets MATH are pairwise disjoint. Since MATH has finite volume, these sets cannot all have the same volume, so, for some MATH, the restriction MATH is not injective (see REF). |
math/0106063 | We have MATH . |
math/0106063 | Suppose MATH is a nontrivial, unipotent element of MATH. The NAME REF implies that there is a continuous homomorphism MATH with MATH . Let MATH . Then MATH . Thus, MATH is an accumulation point of MATH, so REF implies that MATH is not compact. |
math/0106063 | CASE: Since the image of MATH in MATH is precompact, there is a compact subset MATH of MATH, such that MATH (see REF). By enlarging MATH, we may assume that MATH. Then MATH is closed (since MATH, being discrete, is closed and MATH is compact), so MATH contains all of its accumulation points. In addition, since MATH is fixed by every element of MATH, we know that MATH. Therefore, MATH is not an accumulation point of MATH. CASE: To simplify the notation (while retaining the main ideas), let us assume MATH (see REF). Suppose MATH is a sequence of elements of MATH, such that MATH is not an accumulation point of MATH. It suffices to show that there is a sequence MATH of elements of MATH, such that MATH has a convergent subsequence. For each MATH, let CASE: MATH, such that MATH is minimal, CASE: MATH and MATH be the orthogonal projections, and CASE: MATH, such that MATH is minimal. By replacing MATH with MATH, for some MATH, we may assume MATH. Note that the minimality of MATH implies MATH so MATH . Let MATH be the convex hull of MATH and (thinking of MATH and MATH as column vectors) let MATH. From the minimality of MATH and MATH, we see that MATH (see REF), so MATH. Thus, perhaps after replacing MATH with MATH, we have MATH. Since MATH and MATH, we may assume, by replacing MATH with MATH, that MATH . Note that MATH . By combining this with REF, we see that MATH is a bounded sequence, so, by passing to a subsequence, we may assume MATH converges to some vector MATH. By assumption, we have MATH. Now, from REF, and the fact that MATH is bounded away from MATH, we see that MATH is bounded. Because MATH is also bounded, we conclude that MATH is bounded. Hence, by passing to a subsequence, we may assume MATH converges to some vector MATH. From REF, we know that MATH, so MATH. Since MATH and MATH, there is some MATH with MATH and MATH. We have MATH and, similarly, MATH, so MATH for all MATH. Thus, MATH, as desired. |
math/0106063 | Because MATH is a bundle over MATH, with fiber MATH, the conclusion is essentially a consequence of NAME 's Theorem. |
math/0106063 | REF implies that the natural inclusion map MATH is proper (see REF); hence, its image is closed. CASE: From REF, with MATH in the role of MATH, we know that MATH is a lattice in MATH. Then the desired conclusion follows from the observation that MATH is MATH-equivariantly diffeomorphic to MATH. |
math/0106063 | For illustration, let us prove REF in the special case where MATH is compact. Assume also that MATH has no compact factors (see REF); then MATH is generated by its unipotent elements (see REF), so it suffices to show that MATH is invariant under MATH, for every nontrivial unipotent element MATH of MATH. Because MATH is unipotent (see REF), we see that MATH is a polynomial function of MATH. (Write MATH, where MATH for some MATH. Then MATH.) Because MATH is compact and MATH, we know that MATH is compact, so MATH is bounded. A bounded polynomial must be constant, so we conclude that MATH for all MATH; in particular, MATH, as desired. In the general case, REF reduces to REF: MATH (or MATH) must be MATH-invariant, but MATH admits no nontrivial homomorphism to the abelian group MATH (or MATH) , so we conclude that MATH is MATH-invariant. REF is a special case of REF below (let MATH). |
math/0106063 | The NAME algebra MATH of MATH is a subspace of the NAME algebra MATH of MATH. Because MATH, we know that MATH normalizes MATH, so MATH is invariant under MATH. From REF, we conclude that MATH is invariant under MATH; thus, MATH is a normal subgroup of MATH. |
math/0106063 | For simplicity, let us assume that MATH is linear; more precisely, assume MATH, for some MATH. Let MATH be the vector space of all real MATH matrices, so MATH. For MATH and MATH, define MATH, so MATH is a continuous representation. If MATH, then MATH for every MATH, so REF implies that MATH. Therefore MATH. |
math/0106063 | The quotient MATH is finite, because it embeds in the finite group MATH, so MATH is a lattice in MATH (see REF ). Then, because MATH, REF implies MATH. |
math/0106063 | Because MATH is discrete, the identity component MATH of MATH must centralize MATH. So MATH is finite. On the other hand, MATH is connected. Therefore, MATH is trivial, so MATH is discrete. Hence MATH has finite index in MATH (see REF). |
math/0106063 | Let MATH . By definition, we have MATH (see REF). Since MATH has only finitely many connected components (see REF), this implies that MATH is a connected subgroup of MATH that contains a finite-index subgroup of MATH. Hence REF implies that MATH (see REF), so MATH, as desired. |
math/0106063 | Let MATH . It suffices to show MATH for every MATH. Suppose MATH. Then we may choose a subset MATH of MATH, such that MATH and MATH. Because the sets MATH all have the same measure, and MATH, the sets cannot all be disjoint: there exists MATH, such that MATH. By applying MATH, we may assume MATH. For MATH, we have MATH and MATH, so MATH . This contradicts the definition of MATH. |
math/0106063 | Because MATH is generated by its unipotent elements (see REF), it suffices to show that MATH is supported on the set of fixed points of MATH, for every unipotent element MATH of MATH. Let MATH be a unipotent element of MATH, and let MATH. Let MATH. Then MATH is nilpotent (because MATH is unipotent (see REF)), so there is some integer MATH, such that MATH, but MATH. We have MATH so MATH is a fixed point for MATH. Also, for each MATH, we have MATH (because, for MATH, we have MATH as MATH). Thus, MATH converges to a fixed point of MATH, as MATH. The NAME Recurrence REF implies, for MATH-almost every MATH, that there is a sequence MATH, such that MATH. On the other hand, we know, from the preceding paragraph, that MATH converges to a fixed point of MATH. Thus, MATH-almost every element of MATH is a fixed point of MATH. In other words, MATH is supported on the set of fixed points of MATH, as desired. |
math/0106063 | Let MATH be the maximal connected, compact, normal subgroup of MATH, and write MATH, for some closed, normal subgroup MATH of MATH. Then MATH has no compact factors, so we may apply REF to the restriction MATH. |
math/0106063 | Let MATH, and let MATH be a MATH-invariant probability measure on MATH. Note that MATH induces a continuous homomorphism MATH. Because MATH fixes MATH, MATH induces a MATH-equivariant map MATH with MATH. Then MATH is a MATH-invariant probability measure on MATH. Let us assume, for simplicity, that MATH has no compact factors (see REF). In this case, REF implies that MATH is contained in the set of fixed points of MATH. In particular, MATH is fixed by MATH, so MATH is invariant under MATH. |
math/0106063 | Because MATH is invariant under MATH, we conclude from REF that it is invariant under MATH. Therefore MATH is normal in MATH. |
math/0106063 | Let MATH. We know that MATH is finite (see REF ), so it suffices to show that MATH generates MATH. Here is the idea: think of MATH as a tiling of MATH. The elements of MATH can move MATH to any adjacent tile, and MATH is connected, so a composition of elements of MATH can move MATH to any tile. Thus MATH is transitive on the set of tiles. Since MATH also contains the entire stabilizer of the tile MATH, we conclude that MATH. Here is a more formal version of the proof. Suppose MATH. (This will lead to a contradiction.) Now MATH is open, for each MATH, and MATH. Since MATH is connected, this implies that there exists MATH, such that MATH (and MATH). From the definition of MATH, we must have MATH, for some MATH. Then MATH, so, by definition of MATH, we have MATH. Hence MATH. So MATH . This is a contradiction. |
math/0106063 | Because MATH is compact, there is a compact subset MATH of MATH, such that MATH (see REF). Let MATH be a precompact, open subset of MATH, such that MATH. Because MATH, we have MATH. Because MATH is precompact, and MATH acts properly discontinuously on MATH, we know that REF holds. Thus, MATH is a weak fundamental domain for MATH. |
math/0106063 | This is similar to REF, but somewhat more involved. As before, let MATH. For each MATH, define a formal symbol MATH, and let MATH be the free group on MATH. Finally, let MATH, so MATH is a finite subset of MATH. We have a homomorphism MATH determined by MATH. From REF, we know that MATH is surjective, and it is clear that MATH. The main part of the proof is to show that MATH is the smallest normal subgroup of MATH that contains MATH. (Since MATH is finite, and MATH, this implies that MATH is finitely presented, as desired.) Let MATH be the smallest normal subgroup of MATH that contains MATH. (It is clear that MATH; we wish to show MATH.) CASE: Define an equivalence relation MATH on MATH by stipulating that MATH if and only if there exists MATH, such that MATH and MATH (see REF). CASE: Let MATH be the quotient space MATH. CASE: Define a map MATH by MATH. (Note that, because MATH, the map MATH is well defined.) Because MATH we see that MATH factors through to a well-defined map MATH. Let MATH be the image of MATH in MATH. Then it is obvious, from the definition of MATH, that MATH. In fact, it is not difficult to see that MATH (see REF). For each MATH, the image MATH of MATH in MATH is open (see REF), and, for MATH, one can show that MATH if MATH (compare REF). Thus, from the preceding paragraph, we see that MATH is a covering map over MATH. Since MATH is covered by translates of MATH, it follows that MATH is a covering map. Furthermore, the degree of this covering map is MATH. Because MATH is connected, it is not difficult to see that MATH is connected (see REF). Since MATH is simply connected, and MATH is a covering map, this implies that MATH is a homeomorphism. Hence, the degree of the cover is MATH, so MATH. This means MATH, as desired. |
math/0106063 | Let MATH be a maximal compact subgroup of MATH, so MATH is a REF-connected symmetric space, on which MATH acts properly discontinuously. Arguing as in the proof of REF, we see that MATH has a fundamental domain MATH that is an open subset of MATH, so REF implies that MATH is finitely presented. |
math/0106063 | It suffices to find a weak fundamental domain for MATH that is a connected, open subset of MATH. Assume, without loss of generality, that MATH is irreducible. In each of the following two cases, a weak fundamental domain MATH can be constructed as the union of finitely many translates of ``NAME Domains." (This will be discussed in REF.) CASE: MATH is ``arithmetic," as defined in REF; or CASE: MATH has a simple factor of real rank one (or, more generally, MATH). The (amazing!) NAME REF implies that these two cases are exhaustive, which completes the proof. |
math/0106063 | Because MATH is linear, we may assume MATH, for some MATH. Let us start with an illustrative special case. Assume MATH. For any positive integer MATH, the natural ring homomorphism MATH induces a group homomorphism MATH (see REF); let MATH be the kernel of this homomorphism. (This is called the principal congruence subgroup of MATH of level MATH.) It is a finite-index, normal subgroup of MATH (see REF). It suffices to show that MATH is torsion free, for some MATH. We show that MATH is torsion free whenever MATH. It is easy to see that MATH is divisible either by MATH or by some prime MATH. Thus, because MATH whenever MATH is a divisor of MATH, there is no harm in assuming that MATH is a prime power: say MATH. Furthermore, either MATH is odd, or MATH. Given MATH and MATH, we wish to show that MATH. We may write MATH where CASE: MATH, CASE: MATH, and CASE: MATH (that is, not every matrix entry of MATH is divisible by MATH). Also, we may assume MATH is prime (see REF). Thus, either MATH or MATH. Assume MATH. Noting that MATH and using the Binomial Theorem, we see that MATH as desired. Assume MATH. Using the Binomial Theorem, we have MATH . (Note that if MATH, then the congruence requires MATH (see REF).) Assume MATH. From REF, we know there is a torsion-free, finite-index subgroup MATH of MATH. Then MATH is a torsion-free subgroup of finite index in MATH. The general case. The proof is very similar to REF, with the addition of some commutative algebra (or algebraic number theory) to account for the more general setting. Because MATH is finitely generated (see REF), there exist MATH, such that every matrix entry of every element of MATH is contained in the ring MATH generated by MATH (see REF). Thus, letting MATH, we have MATH. Now let MATH be a maximal ideal in MATH. Then MATH is a field, so, because MATH is also known to be a finitely generated ring, it must be a finite field. Thus, the kernel of the natural homomorphism MATH has finite index in MATH. Basic facts of Algebraic Number Theory allow us to work with the prime ideal MATH in very much the same way as we used the prime number MATH in REF. |
math/0106063 | Let CASE: MATH be the subring of MATH generated by the matrix entries of the elements of MATH, and CASE: MATH be the quotient field of MATH. Because MATH is a finitely generated group (see REF), we know that MATH is a finitely generated ring (see REF), so MATH is a finitely generated extension of MATH. We may assume that MATH is a purely transcendental extension of MATH. Choose a subfield MATH of MATH, such that CASE: MATH is a purely transcendental extension of MATH, and CASE: MATH is an algebraic extension of MATH. Let MATH be the degree of MATH over MATH. Because MATH is finitely generated (and algebraic over MATH), we know that MATH. Thus, we may identify MATH with MATH, so there is an embedding MATH . Thus, by replacing MATH with MATH (and replacing MATH with MATH), we may assume that MATH is purely transcendental. If MATH is any element of finite order in MATH, then MATH, and MATH. There is a positive integer MATH with MATH, so every eigenvalue of MATH is a MATH root of unity. The trace of MATH is the sum of these eigenvalues, and any root of unity is an algebraic integer, so we conclude that the trace of MATH is an algebraic integer. Since MATH is the sum of the diagonal entries of MATH, we know MATH. Since MATH is algebraic, but MATH is a purely transcendental extension of MATH, this implies MATH. Since MATH is an algebraic integer, this implies MATH. Since MATH is the sum of MATH roots of unity, and every root of unity is on the unit circle, we see, from the triangle inequality, that MATH. There is a prime number MATH, such that MATH. From the NAME REF, we know that there is a nontrivial homomorphism MATH, where MATH is the algebraic closure of MATH in MATH. Replacing MATH with MATH, let us assume that MATH. Thus, for each MATH, there is some nonzero integer MATH, such that MATH is an algebraic integer. More precisely, because MATH is finitely generated, there is an integer MATH, such that, for each MATH, there is some positive integer MATH, such that MATH is an algebraic integer. It suffices to choose MATH so that it is not a divisor of MATH. There is a finite field MATH of characteristic MATH, and a nontrivial homomorphism MATH. Because MATH, there is a maximal ideal MATH of MATH, such that MATH. Then MATH is a field of characteristic MATH. Because it is a finitely generated ring, MATH must be a finite extension of the prime field MATH (see REF), so MATH is finite. Let MATH be the kernel of the induced homomorphism MATH. Then MATH is torsion free. Let MATH be an element of finite order in MATH. Then MATH so MATH. Thus, from REF and the fact that MATH, we see that MATH. Since the MATH eigenvalues of MATH are roots of unity, and MATH is the sum of these eigenvalues, we conclude that MATH is the only eigenvalue of MATH. Since MATH, we know that MATH is elliptic (hence, diagonalizable over MATH), so this implies MATH, as desired. |
math/0106063 | Assume MATH. Let CASE: MATH be some (large) natural number, CASE: MATH be the principal congruence subgroup of MATH of level MATH, CASE: MATH be a nontrivial MATH root of unity, for some MATH, CASE: MATH be an element of MATH, such that MATH is an eigenvalue of MATH, CASE: MATH, CASE: MATH be the characteristic polynomial of MATH, and CASE: MATH, so MATH is a nonzero eigenvalue of MATH. Since MATH, we know that MATH, so MATH, for some integral polynomial MATH. Since MATH, we conclude that MATH, for some MATH. Thus, MATH is divisible by MATH, in the ring of algebraic integers. The proof can be completed by noting that any particular nonzero algebraic integer is divisible by only finitely many natural numbers, and there are only finitely many roots of unity that satisfy a monic integral polynomial of degree MATH. See REF for a slightly different argument. |
math/0106063 | Consider a word of the form MATH, with each MATH and MATH nonzero. We wish to show MATH. From REF, we have MATH . Therefore MATH and so on: points bounce back and forth between MATH and MATH. (Thus, the colloquial name of the lemma.) In the end, we see that MATH. Assume, for definiteness, that MATH. Then, by applying REF in the last step, instead of REF, we obtain the more precise conclusion that MATH. Since MATH (recall that MATH is disjoint from MATH), we conclude that MATH, as desired. |
math/0106063 | Let CASE: MATH and MATH be linearly independent eigenvectors of MATH, with eigenvalues MATH and MATH, such that MATH, CASE: MATH and MATH be small neighborhoods of MATH and MATH in MATH, and CASE: MATH and MATH be small neighborhoods of MATH and MATH in MATH. By the same argument as in REF, we see that if MATH is sufficiently large, then CASE: MATH is MATH-contracting, and CASE: MATH is MATH-contracting (see REF). Thus, the NAME REF implies that MATH is free. |
math/0106063 | By passing to a subgroup of finite index, we may assume that MATH is torsion free (see REF). Hence, MATH has no elliptic elements. Not every element of MATH is unipotent (see REF), so we conclude that some nontrivial element MATH of MATH is hyperbolic. Let MATH and MATH be linearly independent eigenvectors of MATH. The NAME Density REF implies that there is some MATH, such that MATH (see REF). Let MATH, so that MATH is a hyperbolic element of MATH with eigenvectors MATH and MATH. From REF, we conclude that MATH is a nonabelian, free subgroup of MATH, for some MATH. |
math/0106063 | Assume MATH. Choose some nontrivial, hyperbolic element MATH of MATH, with eigenvalues MATH. We may assume, without loss of generality, that MATH. (If the eigenvalue MATH has multiplicity MATH, then we may pass to the MATH exterior power MATH, to obtain a representation in which the largest eigenvalue of MATH is simple.) Let us assume that the smallest eigenvalue MATH is also simple; that is, MATH. (One can show that this is a generic condition in MATH, so it can be achieved by replacing MATH with some other element of MATH.) Let MATH be an eigenvector corresponding to the eigenvalue MATH of MATH, and let MATH be an eigenvector for the eigenvalue MATH. Assume, to simplify the notation, that all of the eigenspaces of MATH are orthogonal to each other. Then, for any MATH, we have MATH in MATH, as MATH (see REF). Similarly, if MATH, then MATH. We may assume, by replacing MATH with a minimal MATH-invariant subspace, that MATH is irreducible in MATH. Then the NAME Density Theorem implies that there exists MATH, such that MATH. Then, for any small neighborhoods MATH, MATH, MATH, and MATH of MATH, MATH, MATH, and MATH, and any sufficiently large MATH, the NAME Lemma implies that MATH is free. |
math/0106063 | Let MATH. CASE: Let MATH be the MATH-span of MATH. Then MATH, being a vector space over MATH, is homeomorphic to MATH, for some MATH. On the other hand, we know that MATH, and that MATH is compact, so MATH is compact. Hence MATH, so MATH. CASE: There is a linear isomorphism MATH, such that MATH. Since MATH is compact, we conclude that MATH is compact. CASE: Let MATH be the standard basis of MATH. Because MATH contains a basis of MATH, there is a linear isomorphism MATH, such that MATH for MATH; then MATH. Because MATH is dense in MATH, and MATH is continuous, we know that MATH is dense in MATH. CASE: Because MATH is dense in MATH, we see that MATH is defined by a system of linear equations with rational coefficients. (Namely, for each vector MATH in some basis of MATH, we write the equation MATH.) Thus, from (MATH), we conclude that there is a basis MATH of MATH, such that each MATH. Then MATH is defined by the system of REF, , MATH. CASE: By assumption, MATH is the solution space of a system of linear equations whose coefficients belong to MATH. By elementary linear algebra (row reductions), we may find a basis for MATH that consists entirely of vectors in MATH. Multiplying by a scalar to clear the denominators, we may assume that the basis consists entirely of vectors in MATH. |
math/0106063 | It is easy to handle direct products, so the crucial case is when MATH is simple. This is easy if MATH is classical. Indeed, the groups in REF are defined over MATH (after identifying MATH and MATH with appropriate subgroups of MATH and MATH, in a natural way). For the general case, one notes that MATH is a finite-index subgroup of MATH (see REF), so it suffices to find a basis of MATH, for which the structure constants of the NAME algebra are rational. This is not a terribly difficult exercise for someone familiar with exceptional groups, but we omit the details. |
math/0106063 | CASE: If MATH is classical and isotypic, then REF will show how to construct a fairly explicit irreducible lattice in MATH, by applying REF. In the general case, this is a consequence of the NAME Theorem in NAME cohomology, and that is a subject we will not reach in this introductory text. CASE: Suppose MATH is an irreducible lattice in MATH. We may assume that MATH is not simple (otherwise, the desired conclusion is trivially true), so MATH is neither MATH nor MATH. Thus, from the NAME REF , we know that MATH is arithmetic. Then, since MATH is irreducible, the conclusion follows from the theory of arithmetic groups (see REF). Briefly, the only way to construct an irreducible arithmetic lattice is by applying REF. Thus, there is a simple subgroup MATH of some MATH, such that (modulo some finite groups) CASE: MATH is defined over some number field MATH, CASE: MATH is isomorphic to MATH, where MATH is the ring of integers of MATH, and CASE: MATH is isomorphic to MATH. Since any NAME conjugate MATH of MATH is of the same type as MATH (see REF), the conclusion follows. |
math/0106063 | CASE: This is the easy direction (see REF). CASE: We prove the contrapositive: suppose MATH is not compact. (We wish to show that MATH has a nontrivial unipotent element.) From REF (and the fact that MATH is a lattice in MATH (see REF)), we know that there exist nontrivial MATH and MATH, such that MATH. Because the characteristic polynomial of a matrix is a continuous function of the matrix entries of the matrix, we conclude that the characteristic polynomial of MATH is approximately MATH (the characteristic polynomial of MATH). On the other hand, similar matrices have the same characteristic polynomial, so this means that the characteristic polynomial of MATH is approximately MATH. Now all the coefficients of the characteristic polynomial of MATH are integers (because MATH is an integer matrix), so the only way this polynomial can be close to MATH is by being exactly equal to MATH. Thus, the characteristic polynomial of MATH is MATH, so MATH is unipotent. |
math/0106063 | Let MATH and MATH. (Our proof will not use the fact that MATH is a lattice in MATH.) The image of MATH in MATH is precompact. Let CASE: MATH be a sequence of elements of MATH and CASE: MATH be a sequence of elements of MATH. Suppose that MATH. (This will lead to a contradiction, so the desired conclusion follows from the NAME Compactness REF .) Replacing MATH by an integer multiple to clear the denominators, we may assume MATH. Then, since MATH for all nonzero MATH, we have MATH for all MATH. Therefore MATH . This is a contradiction. The image of MATH in MATH is closed. Suppose MATH, with MATH and MATH. (We wish to show MATH.) Let MATH be the standard basis of MATH (so each MATH). Replacing MATH by an integer multiple to clear the denominators, we may assume MATH. Then MATH . We also have MATH so we conclude that MATH for any sufficiently large MATH. Therefore MATH, so MATH. Completion of the proof. Define MATH by MATH. By combining REF, we see that the image of MATH is compact. Thus, it suffices to show that MATH is a homeomorphism onto its image. Given a sequence MATH in MATH, such that MATH converges, we wish to show that MATH converges. There is a sequence MATH in MATH, and some MATH, such that MATH. The proof of REF shows, for all large MATH, that MATH. Then MATH, so MATH. Therefore, MATH converges (to MATH), as desired. |
math/0106063 | Let MATH and MATH. Then MATH is a REF-dimensional vector space over MATH, and MATH is a vector-space lattice in MATH. Since MATH is both a MATH-basis of MATH and a MATH-basis of MATH, we see that MATH is a MATH-form of MATH. Therefore, MATH is a MATH-form of MATH, and MATH is a vector-space lattice in MATH. Now MATH is defined over MATH (see REF), so MATH is an arithmetic lattice in MATH. It is not difficult to see that MATH (see REF). Furthermore, because MATH is trivial, we see that the lattice MATH must be irreducible in MATH (see REF). |
math/0106063 | As above, let MATH be the conjugation on MATH. Let MATH. Let MATH, so MATH; thus, we may define MATH by MATH. (Note that MATH.) Then the above proof shows that MATH is an arithmetic lattice in MATH. (See REF for the technical point of verifying that MATH is defined over MATH.) Since MATH is compact, we see, by modding out MATH, that MATH is an arithmetic lattice in MATH. (This type of example is the reason for including the compact normal subgroup MATH in REF.) Let MATH be any nontrivial element of MATH. Since MATH, and compact groups have no nontrivial unipotent elements (see REF), we know that MATH is not unipotent. Thus, MATH has some eigenvalue MATH. Hence, MATH has the eigenvalue MATH, so MATH is not unipotent. Therefore, NAME 's REF implies that MATH is cocompact. Alternatively, this conclusion can easily be obtained directly from the NAME Compactness REF (see REF). |
math/0106063 | For convenience, let MATH. There are exactly REF distinct embeddings MATH, MATH, MATH, MATH of MATH in MATH (corresponding to REF roots of MATH); they are determined by: CASE: MATH (so MATH); CASE: MATH; CASE: MATH; CASE: MATH. Define MATH by MATH. Then, arguing much as before, we see that MATH is a MATH-form of MATH, MATH is defined over MATH, and MATH. |
math/0106063 | It is easier to work with the algebraically closed field MATH, instead of MATH, so, to avoid minor complications, let us assume that MATH is defined over MATH (as an algebraic group over MATH), and that MATH. This assumption results in a loss of generality, but similar ideas apply in general. Write MATH, where each MATH is simple. Let MATH. We remark that if MATH, then the desired conclusion is obvious: let MATH, and let MATH be the identity map. Let MATH be the NAME group of MATH over MATH. Because MATH is defined over MATH, we have MATH for every MATH. Hence, MATH must permute the simple factors MATH. We claim that MATH acts transitively on MATH. To see this, suppose, for example, that MATH, and that MATH is invariant under MATH. Then MATH is invariant under MATH, so MATH is defined over MATH. Similarly, MATH is also defined over MATH. Then MATH and MATH are lattices in MATH and MATH, respectively, so MATH is reducible. This is a contradiction. Let MATH be the stabilizer of MATH, and let MATH be the fixed field of MATH. Because MATH is transitive on a set of MATH elements, we know that MATH is a subgroup of index MATH in MATH, so NAME Theory tells us that MATH is an extension of MATH of degree MATH. Since MATH is the NAME group of MATH over MATH, and MATH for all MATH, we see that MATH is defined over MATH. Let MATH be coset representatives of MATH in MATH. Then MATH are the MATH places of MATH and, after renumbering, we have MATH. So with MATH, we have MATH . Let MATH be the identity map. For MATH, let MATH. Then MATH for all MATH, so MATH. In fact, it is not difficult to see that MATH, and then one can verify that MATH, so MATH is commensurable with MATH. |
math/0106063 | From REF, we see that MATH is isomorphic to the product of NAME conjugates of a simple group MATH. Since all of these NAME conjugates are of the same type (see REF), the desired conclusion follows. |
math/0106063 | Extend MATH to an automorphism MATH of MATH. Then MATH, so it is clear that MATH is isomorphic to MATH. Unfortunately, however, the automorphism MATH is not continuous (not even measurable) unless it happens to be the usual complex conjugation, so we have only an isomorphism of abstract groups, not an isomorphism of NAME groups. Thus, although this observation is suggestive, it is not a proof. It is easier to give a rigorous proof at the NAME algebra level: we will show that MATH is isomorphic to MATH. Let MATH be a basis of MATH, and let MATH be the structure constants with respect to this basis, that is, MATH . Because MATH is isogenous to a group that is defined over MATH (see REF), there is a basis MATH of MATH with rational structure constants. Write MATH with each MATH, and define MATH . Then MATH is a basis of MATH whose structure constants are MATH. These are obviously the structure constants of MATH, so MATH is isomorphic to MATH, as desired. |
math/0106063 | CASE: Define a MATH-linear bijection MATH by MATH . It is straightforward to check that MATH preserves multiplication, so MATH is a ring isomorphism. For MATH, we have MATH . Therefore, MATH. CASE: For MATH, define MATH by MATH. Then MATH is MATH-linear. For MATH, we have MATH if and only if MATH. So MATH is an arithmetic lattice in MATH. CASE: We prove the contrapositive. Suppose MATH is not cocompact. Then it has a nontrivial unipotent element MATH (see REF). So MATH is an eigenvalue of MATH; that is, there is some nonzero MATH, such that MATH. By definition of MATH, this means MATH. Hence MATH. Since MATH and MATH, this implies MATH has zero divisors, so it is not a division algebra. CASE: This is a concrete restatement of the fact that MATH is a division algebra if and only if MATH, for all nonzero MATH (see REF). CASE: We prove the contrapositive. Suppose MATH is not a division algebra. Then MATH (see REF). So MATH is not cocompact. (It has nontrivial unipotent elements.) |
math/0106063 | This can be proved directly, but let us derive it as a corollary of REF. For each lattice MATH in MATH, constructed in REF, we wish to show, for some arithmetic lattice MATH in MATH, constructed in REF or REF, that there is an isogeny MATH, such that MATH is commensurable with MATH. CASE: First, let us show that every lattice of type REF appears in REF. Given positive integers MATH and MATH, such that MATH is the only rational solution of the equation MATH, let MATH . One can show that MATH is the only rational solution of the equation MATH (see REF), so MATH is a cocompact, arithmetic lattice in MATH (see REF). As a subspace of MATH, the NAME algebra MATH of MATH is MATH (see REF). For MATH and MATH, we have MATH, so MATH is a quadratic form on MATH that is invariant under MATH. For MATH, we have MATH . After the change of variables MATH and MATH, this becomes MATH, which is a scalar multiple of the quadratic form in REF. Thus, after identifying MATH with MATH by an appropriate choice of basis, the lattice constructed in REF (for the given values of MATH and MATH) is commensurable with MATH. CASE: Similarly, every lattice of type REF appears in REF (see REF). |
math/0106063 | Let us consider the isogenous group MATH, instead of MATH. There are CASE: a symmetric, bilinear form MATH on MATH, and CASE: an isogeny MATH, such that MATH is commensurable with MATH. From REF of the proof of REF, we have a totally real number field MATH, such that MATH is compact for all MATH. On the other hand, since MATH is not cocompact, we know that each MATH is noncompact (see REF). Thus, MATH is the only place of MATH, so MATH. We may assume MATH. Because MATH is not cocompact, we know that MATH is isotropic over MATH (see REF). Therefore, the desired conclusion is obtained by a change of basis, and replacing MATH with a scalar multiple (see REF ). |
math/0106063 | CASE: See REF. CASE: Let MATH. Then MATH . From the NAME Density REF , we know that MATH is NAME dense in MATH, so we conclude that MATH. Also, we have MATH. Since MATH is irreducible on MATH, we know that the quadratic form on MATH that is invariant under MATH is unique up to a scalar multiple (see REF). Hence, there is some MATH with MATH. Since both MATH and MATH are defined over MATH, we must have MATH. |
math/0106063 | CASE: For MATH we have MATH so the desired conclusion follows from REF. CASE: This follows from REF below, because MATH is not a square in MATH. |
math/0106063 | The discriminant of a quadratic form MATH on MATH is defined to be MATH, for any basis MATH of MATH. This is not uniquely determined by MATH, but it is well-defined up to multiplication by a nonzero square in MATH, because MATH for any MATH. Note that MATH. Thus, if MATH is odd, then MATH has the same discriminant as MATH (up to multiplication by a square). |
math/0106063 | The main issue is to show that each point of MATH has a neighborhood MATH in MATH, such that MATH is isometric to an open subset of MATH. This is not difficult (see REF). We have MATH. If MATH is compact, then it is obviously complete. More generally, since MATH and MATH are complete, and their union is all of MATH, it seems rather obvious that every NAME sequence in MATH has a convergent subsequence. Hence, it seems to be more-or-less obvious that MATH is complete. Unfortunately, if MATH is not compact, then there is a technical difficulty arising from the possibility that, theoretically, the Riemannian isometry MATH may not be an isometry with respect to the topological metrics that MATH and MATH inherit as submanifolds of MATH and MATH, respectively. See REF for an indication of how to deal with this problem. |
math/0106063 | It is clear, from the definition of MATH, that we need only show MATH is a (closed, embedded) submanifold of MATH. Let MATH. (Then MATH is a subgroup of index at most two in MATH.) The natural map MATH is proper (compare REF), so MATH, being the image of MATH, is closed. Because MATH is obviously an immersion (and is a proper map), all that remains is to show that MATH is injective. This follows from the assumption on MATH (see REF). |
math/0106063 | From REF, we know that MATH is commensurable to MATH. By interchanging MATH and MATH, we see that MATH is also commensurable to MATH. By transitivity, MATH is commensurable to MATH. |
math/0106063 | We construct only MATH. (See REF for the construction of MATH, which is similar.) Define quadratic forms MATH and MATH on MATH by MATH . Let CASE: MATH; CASE: MATH, where MATH; CASE: MATH; CASE: MATH be the image of MATH in MATH; and CASE: MATH. Then MATH and MATH are noncocompact (arithmetic) lattices in MATH (see REF). By passing to finite-index subgroups, we may assume MATH and MATH are torsion free (see REF). Thus, MATH and MATH are hyperbolic MATH-manifolds of finite volume (see REF). Because MATH is a lattice in MATH, and MATH is normalized by the involution MATH of REF, we know that MATH is a totally geodesic hypersurface in MATH that has finite volume (see REF). Let us assume that MATH and MATH are connected. (See REF for a way around this issue, or note that this hypothesis can be achieved by passing to finite covers of MATH and MATH.) We know that MATH (because both subgroups are commensurable to MATH). By taking a little bit of care in the choice of MATH and MATH, we may arrange that MATH (see REF). Then MATH so there is an isometry MATH. If MATH is odd, then MATH is not commensurable to MATH (see REF), so REF implies that MATH is not arithmetic; thus, the corresponding lattice MATH is not arithmetic (see REF). When MATH is even, an additional argument is needed; see REF. |
math/0106063 | CASE: Let MATH. CASE: Write MATH, for some torsion-free lattice MATH in MATH. CASE: Let MATH be the resulting covering map. CASE: Let MATH. Because MATH has totally geodesic boundary, we know that MATH is a union of disjoint hyperplanes. (That is, each component of MATH is of the form MATH, for some MATH.) CASE: Let MATH be the closure of some connected component of MATH that contains a point of MATH. CASE: Let MATH (see REF), so MATH. By definition, MATH is an intersection of half-spaces, so it is (hyperbolically) convex; hence, it is simply connected. Therefore, MATH is the universal cover of MATH, and MATH can be identified with the fundamental group of MATH. Since MATH, we may define MATH as above, but with MATH in the place of MATH. From the uniqueness of the universal cover of MATH, we know that there is an isometry MATH, and an isomorphism MATH, such that MATH, for all MATH and MATH. Since MATH extends to an isometry of MATH, we may assume (after replacing MATH with MATH) that MATH and MATH. Hence MATH. It suffices to show (after replacing MATH by a conjugate subgroup) that the NAME closure of MATH contains MATH, for then REF implies MATH is commensurable with MATH. We may assume that the NAME closure of MATH contains MATH. We may assume MATH is one of the connected components of MATH. Since MATH has finite volume, this means that MATH . Let MATH be the NAME closure of MATH. From REF and the NAME Density REF , we know that MATH contains MATH. Then, since MATH is a maximal connected subgroup of MATH (see REF), we may assume that MATH. (Otherwise, the claim holds.) Because MATH has finite index in MATH (see REF), this implies that MATH contains a finite-index subgroup of MATH. In fact, MATH . This will lead to a contradiction. Assume MATH is connected. We may assume MATH. Then, by passing to a finite-index subgroup, we may assume that MATH (see REF). Define MATH by MATH . Then CASE: MATH centralizes MATH; and CASE: MATH. Since MATH has finite volume, we know that MATH also has finite volume. Therefore MATH has finite volume, so MATH is a lattice in MATH. But this contradicts the NAME Density REF (since MATH). Assume MATH is not connected. Let MATH and MATH be two distinct connected components of MATH. Replacing MATH by a finite-index subgroup, let us assume that each of MATH and MATH is invariant under MATH (see REF). To simplify the argument, let us assume that MATH is compact, rather than merely that it has finite volume. (See REF for the general case.) Thus, MATH is compact, so there is a compact subset MATH of MATH, such that MATH. Let MATH . Because MATH acts by isometries, we have MATH. Because MATH is negatively curved, there is a unique point MATH in MATH, such that MATH. The uniqueness implies that MATH is fixed by every element of MATH. Since MATH acts freely on MATH (recall that it is a group of deck transformations), we conclude that MATH is trivial. This contradicts the fact that MATH is compact. (Note that MATH is not compact.) |
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