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math/0106063 | CASE: This is a special case of REF, but we provide a concrete, explicit proof (using the methods of REF). Define CASE: MATH by MATH, CASE: MATH, CASE: MATH; and CASE: MATH by MATH for MATH. Then CASE: MATH is a MATH-form of MATH (compare REF), CASE: MATH is a vector-space lattice in MATH (compare REF), CASE: MATH is a homomorphism, CASE: MATH is defined over MATH (with respect to the MATH-form MATH) (see REF below), and CASE: MATH (compare REF). Thus, REF (together with REF) implies that MATH is an arithmetic lattice in MATH. Now let us show that MATH . This can be verified directly, by finding an appropriate collection of MATH-polynomials, but let us, instead, show that MATH is dense in MATH. Define MATH as in REF, but allowing MATH to range over all of MATH, instead of only MATH. Then MATH (see REF), so MATH. Furthermore, MATH is dense in MATH . Similarly, there is a dense subgroup MATH of MATH, such that MATH (see REF). Since MATH, we know that MATH is dense in MATH, so MATH is dense in MATH. Therefore MATH is defined over MATH (see REF). CASE: By calculation, one may verify, directly from the definition of MATH, that the subgroup MATH is contained in MATH. Then, since every element of MATH is unipotent, it is obvious that MATH has nontrivial unipotent elements. So the NAME REF implies that MATH is not compact. CASE: Define a nondegenerate MATH-Hermitian form MATH on MATH by MATH. Then MATH is an isotropic vector for MATH. On the other hand, because MATH is nondegenerate, and MATH is MATH-dimensional, there are no REF-dimensional totally isotropic subspaces (see REF). Thus, the maximal totally isotropic subspaces are one-dimensional, so MATH (compare REF ). |
math/0106063 | Let MATH be a lattice in MATH, such that MATH is not compact. We know, from the NAME REF , that MATH is arithmetic. Since MATH is not compact, this implies there are CASE: a group MATH, defined over MATH, and CASE: an isogeny MATH, such that MATH is commensurable with MATH (see REF). From REF, we see that there are only two possibilities for MATH. Furthermore, because MATH is not compact, we must have MATH (see REF). Assume MATH, for some central division algebra MATH of degree MATH over MATH, with MATH. Because MATH is prime, there are only two possibilities for MATH and MATH. Assume MATH and MATH. Because MATH, we have MATH, so MATH. Thus, MATH. Therefore MATH. Assume MATH and MATH. We have MATH. Then MATH is cocompact (see REF ). This is a contradiction. Assume MATH, for MATH as in REF, with MATH, MATH, and MATH. If MATH, then MATH, so REF applies. Thus, we may assume that MATH and MATH. Since MATH, we have MATH, so MATH, where MATH is the (unique) NAME automorphism of MATH over MATH, and MATH is a MATH-Hermitian form on MATH. Since MATH is not cocompact, we know that MATH is isotropic (compare REF ), so there is a basis of MATH with MATH (see REF ). Therefore MATH is as in REF. |
math/0106063 | CASE: It is easy to see that CASE: MATH (see REF); CASE: MATH and MATH are subrings of MATH (even though MATH is not a ring homomorphism); CASE: MATH is a MATH-form of MATH; CASE: MATH is a vector-space lattice in MATH; CASE: if we define MATH by MATH, then MATH is defined over MATH (with respect to the MATH-form MATH (see REF)); and CASE: MATH. So MATH is an arithmetic lattice in MATH (see REF ). CASE: If MATH is not compact, then there is a nontrivial unipotent element MATH in MATH (see REF). Then MATH is an eigenvalue of MATH (indeed, it is the only eigenvalue of MATH), so there is some nonzero MATH with MATH. Hence MATH. Since MATH and MATH, we conclude that MATH has a zero divisor. Thus, it suffices to show that MATH is a division algebra. (That is, every nonzero element of MATH is invertible.) For convenience, define MATH by MATH (see REF). We know that MATH, for all MATH. It is easy to see that MATH. Note that if MATH, but MATH, then MATH is invertible. For example, if MATH, then MATH. Since MATH (assuming MATH), we have MATH, as desired. The other cases are similar. For any MATH, with MATH, we have MATH is invertible, so MATH is invertible. CASE: If MATH, for some MATH, then one can show that MATH (compare REF). From this, it is easy to see that MATH contains unipotent elements. This implies that MATH is not cocompact. |
math/0106063 | The conclusions follow from the observations in REF. |
math/0106063 | We have MATH. We may assume MATH. Let MATH . Then MATH is isomorphic to MATH. (Indeed, MATH is conjugate to MATH, simply by permuting the basis vectors.) Therefore, MATH. There is an obvious embedding of MATH in MATH: MATH so MATH . We have MATH. By making the argument of REF more precise, one shows that the existence of a large MATH-split torus implies the existence of a large collection of isotropic vectors, indeed, an entire totally isotropic subspace; that is, a subspace MATH, such that MATH for all MATH. More precisely, there is totally isotropic subspace whose dimension is MATH. [Probably should prove this!] Since MATH is the maximum dimension of a totally isotropic subspace of MATH, the desired conclusion follows. |
math/0106063 | We may assume there is some nonzero MATH, with MATH. (Otherwise, let MATH and MATH.) Because MATH is nondegenerate, there is some MATH with MATH. Multiplying MATH by a scalar, we may assume MATH. For MATH, we have MATH and MATH . Thus, we may assume MATH (by replacing MATH with MATH, where MATH). Now the matrix of MATH, with respect to the basis MATH, is the MATH matrix REF. By induction on MATH, we may assume there is an appropriate basis MATH for MATH. Let MATH. |
math/0106063 | Because MATH acts transitively on the space of unit tangent vectors of MATH (see REF), we see that MATH acts transitively on the space of all geodesic rays in MATH. Therefore, MATH acts transitively on MATH. We wish to show, for any MATH, that MATH is parabolic in MATH. Let MATH be a geodesic ray, and let MATH be the initial point of this ray. From REF, we know that there is a MATH-split torus MATH of MATH, such that MATH acts simply transitively on the flat MATH. Because MATH, we know that MATH is a one-parameter subgroup of MATH; write MATH. Then MATH. Let MATH so MATH is a parabolic subgroup of MATH. Note that, for MATH, we have MATH, for an appropriate choice of the matrix norm MATH. Thus, for any MATH, we have MATH (because MATH is constant - independent of MATH). Thus, we see that MATH is equivalent to MATH if and only if MATH, as desired. |
math/0106063 | Assume, for simplicity, that MATH. (See REF for a general proof, using the theory of real roots.) REF Assume MATH is the group of diagonal matrices. Let MATH . Then MATH . These generate MATH. CASE: If MATH, then an appropriate sequence is MATH . Thus, we may assume MATH. Furthermore, we may assume that MATH and MATH are maximal unipotent subgroups. Thus, replacing the pair MATH with a conjugate under MATH, we may assume MATH . We may now take the sequence MATH . CASE: This follows from REF This follows from REF. |
math/0106063 | CASE: This follows from REF see REF. CASE: Assume MATH is simple, and MATH. It is straightforward to verify the desired conclusion if MATH is classical. The general case follows from the theory of real roots. For example, CASE: MATH, MATH, and MATH contain MATH. CASE: MATH, MATH, and MATH contain MATH if MATH. CASE: MATH is not simple see REF. CASE: MATH is isogenous to MATH, which contains MATH. CASE: MATH contains MATH. CASE: MATH is isogenous to MATH see REF. |
math/0106063 | CASE: NAME a matrix is equivalent to finding a basis consisting of eigenvectors. CASE: One direction is obvious. The other follows from the fact that MATH is dense in MATH (see REF). CASE: We may assume MATH consists of diagonal matrices. Then the eigenvalues of any MATH are simply the matrix entries of MATH. If MATH, these matrix entries are rational. CASE: If MATH, then, for each eigenvalue MATH of MATH, the equation MATH is a linear equation with rational coefficients, so the solution space is spanned by rational vectors. Thus, each eigenspace of MATH is spanned by rational vectors. Therefore, MATH is diagonalizable over MATH. Since this is true for each MATH, and any set of commuting diagonalizable linear transformations can be simultaneously diagonalized, we conclude that MATH is diagonalizable over MATH. |
math/0106063 | If MATH is a torus in MATH, and MATH is defined over MATH, then MATH is a MATH-torus in MATH. Conversely, it is not difficult to see that any MATH-torus of MATH is contained in a torus of the form MATH. Thus, the desired conclusion follows from the fact, which will be proved in REF, that MATH (see REF). |
math/0106063 | Given isogenies MATH that are defined over MATH, let MATH . The projection maps MATH defined by MATH are polynomials. |
math/0106063 | Let us assume MATH and MATH. (The general case follows from this by Restriction of Scalars.) From the proof of REF, we see that, after replacing MATH with an isogenous group, we may assume that MATH is a polynomial with rational coefficients. Assume MATH and MATH. Define MATH by MATH. Being a polynomial, the function MATH is defined on all of MATH. Because the coefficients are rational, there is some nonzero MATH, such that MATH. Thus, letting MATH be the principal congruence subgroup of MATH of level MATH, we have MATH. Because MATH is a lattice in MATH (and MATH is discrete), we know that MATH is a lattice in MATH. Because MATH, we conclude that MATH is commensurable with MATH (see REF). |
math/0106063 | Suppose MATH is classical. Then, for the natural embeddings described in REF, it is easy to see that MATH (compare REF ); thus, it is easy to construct MATH. The method of REF yields a cocompact lattice MATH in MATH, MATH, or MATH (see REF). A similar method applies to the other classical groups, as is explained in REF. We omit the proof for exceptional groups. |
math/0106063 | From REF, we have MATH. Suppose MATH. The NAME REF implies that MATH is arithmetic. Since MATH, this implies that (up to isogeny) there is an embedding of MATH in some MATH, such that MATH is commensurable with MATH (see REF). (The point here is that, because MATH is not compact, REF shows there is no need for the compact groups MATH and MATH of REF.) From the description of arithmetic subgroups of MATH that appears in REF (and the fact that MATH is odd), we conclude that we may take MATH, and that there is a symmetric bilinear form MATH on MATH, such that MATH is defined over MATH, and CASE: MATH has signature MATH on MATH (because MATH), but CASE: no MATH-dimensional subspace of MATH is totally isotropic (because MATH). A theorem of number theory asserts: CASE: If MATH is any nondegenerate, symmetric bilinear form on MATH, such that CASE: MATH is defined over MATH, CASE: MATH, and CASE: MATH is isotropic over MATH (that is, MATH for some nonzero MATH), then MATH is also isotropic over MATH (that is, MATH for some nonzero MATH). (This is related to, but more difficult than, the fact that every integer is a sum of four squares.) From REF, we know that there is a nontrivial isotropic vector MATH. Then, because MATH is nondegenerate, there is a vector MATH, such that MATH and MATH. Let MATH. Because the restriction of MATH to MATH is nondegenerate, we have MATH. This direct sum is obviously orthogonal (with respect to MATH), and the restriction of MATH to MATH has signature MATH, so we conclude that the restriction of MATH to MATH has signature MATH. Thus, there is an isotropic vector in MATH. From REF, we conclude that there is an isotropic vector MATH in MATH. Then MATH is a MATH-dimensional totally isotropic subspace of MATH. This is a contradiction. |
math/0106063 | Let us assume that MATH, and that MATH. There is no difficulty in generalizing the proof to any arithmetic lattice, but, for a general lattice, REF is rather difficult, and we will not deal with this. CASE: We prove the contrapositive: suppose MATH. (We wish to show MATH is not compact.) By definition of MATH, we know that there is a nontrivial MATH-split torus MATH in MATH. Then MATH is infinite, so there is some MATH, and some (rational) eigenvalue MATH of MATH, such that MATH; for definiteness, let us assume MATH. Choose some nonzero MATH, such that MATH. Multiplying by an integer to clear the denominators, we may assume MATH. Now MATH . Thus, because MATH, we conclude that MATH is an accumulation point of the orbit MATH. Hence, the NAME Compactness REF asserts that the image of MATH in MATH is not compact. Since this is the image of the natural inclusion MATH, we conclude that MATH is not compact. CASE: Because MATH is assumed to be arithmetic, this follows from the NAME REF . CASE: We prove the contrapositive: suppose MATH is a nontrivial unipotent element of MATH. (We wish to show that MATH.) Since MATH, the NAME REF implies that there is a nontrivial polynomial homomorphism MATH with MATH. Then, letting MATH be a nontrivial MATH-split torus in MATH, we see that MATH is a nontrivial MATH-split torus in MATH, so MATH. CASE: Obvious. CASE: See REF. |
math/0106063 | From REF, we see that (up to isogeny and commensurability) there is a compact group MATH, such that we may embed MATH in some MATH, such that MATH is defined over MATH, and MATH. Let MATH be the (almost-)NAME closure of the subgroup of MATH generated by all of the unipotent elements of MATH. From the definition, it is clear that MATH is normalized by the NAME closure of MATH. Thus, the NAME Density REF implies that MATH is normalized by MATH. Because MATH has no unipotent elements (see REF), we know that MATH. Also, because MATH is not compact, we know that MATH contains nontrivial unipotent elements (see REF), so MATH is infinite. Therefore, under the simplifying assumption that MATH is simple, we conclude that MATH. Thus, MATH is the (almost-)NAME closure of a subset of MATH. Since MATH, this implies that MATH is defined over MATH (compare REF). Hence, MATH is a lattice in MATH, so MATH has finite index in MATH (see REF). Because MATH, we conclude that MATH (up to commensurability). |
math/0106063 | Assume MATH is arithmetic. It suffices to show that the maximal unipotent subgroups of MATH are precisely the subgroups of MATH of the form MATH, where MATH is the unipotent radical of a minimal parabolic MATH-subgroup of MATH. (For then the desired conclusion follows from REF.) Suppose MATH is a maximal unipotent subgroup of MATH. Then the NAME closure MATH of MATH is a unipotent MATH-subgroup of MATH. Hence, there is a parabolic MATH-subgroup MATH, such that MATH. Then MATH, so the maximality of MATH implies that MATH. The converse is similar (and uses REF). |
math/0106063 | The ideas of the preceding two examples generalize to other groups MATH. Let MATH be a minimal parabolic MATH-subgroup of MATH. We may write MATH (see REF, and note that MATH centralizes MATH). Since MATH is minimal, we know that MATH (see REF ), so there is a compact subset MATH of MATH, such that MATH (see REF). Similarly, there is a compact subset MATH of MATH, such that MATH (see REF); let MATH. If MATH, the subgroup MATH is isomorphic to MATH, and we define MATH to be a ray in MATH, as in REF above. In general, the group MATH is isomorphic to some MATH, where MATH. Then MATH is a polyhedral cone in MATH (with nonempty interior). (It is called the positive NAME chamber.) We define MATH to be the points of MATH that are within distance MATH of this NAME chamber (see REF). Define the NAME set MATH. Then, for an appropriate choice of MATH, one can show that some finite union MATH of translates of MATH is a weak fundamental domain for MATH. (REF implies that only finitely many translates are needed.) Because MATH centralizes MATH, the definition of MATH implies that there is a compact subset MATH of MATH, such that MATH (see REF). Thus, from a large distance, MATH simply looks like MATH. Therefore, from a large distance MATH looks like a finite union of translates of MATH. So the asymptotic cone at MATH is a MATH-simplex. |
math/0106063 | (The key observation is that the action of MATH on MATH is continuous, so the stabilizer of any function is closed.) |
math/0106063 | CASE: |
math/0106063 | Assume, for simplicity, that MATH . (A reader familiar with the theory of real roots and NAME chambers should have little difficulty in extending this proof to the general case.) Let MATH . Assume that MATH . (It is not difficult to eliminate this hypothesis; see REF.) By passing to a subsequence, we may assume MATH converges weakly, to some operator MATH; that is, MATH . Let MATH . For MATH, we have MATH so MATH. Therefore, letting MATH be the space of MATH-invariant vectors in MATH, we have MATH . We have MATH so the same argument, with MATH in the place of MATH and MATH in the place of MATH, shows that MATH . Because MATH is unitary, we know that MATH is normal (that is, commutes with its adjoint) for every MATH; thus, the limit MATH is also normal: we have MATH. Hence MATH so MATH. Thus, MATH . By passing to a subsequence of MATH, we may assume MATH (see REF). Then MATH, so MATH. Hence, for all MATH, we have MATH as desired. |
math/0106063 | Assume MATH is cyclic. Given a nonempty, compact, convex MATH-space MATH, choose some MATH. For MATH, let MATH . Since MATH is compact, MATH has an accumulation point MATH. It is not difficult to see that MATH is fixed by MATH (see REF). Since MATH generates MATH, this means that MATH is a fixed point for MATH. |
math/0106063 | Let us assume MATH is MATH-generated. (See REF for the general case.) Given a nonempty, compact, convex MATH-space MATH, REF implies that the set MATH of fixed points of MATH is nonempty. It is easy to see that MATH is compact and convex (see REF), and, because MATH is abelian, that MATH is invariant under MATH (see REF). Thus, MATH is a nonempty, compact, convex MATH-space. Therefore, REF implies that MATH has a fixed point MATH in MATH. Now MATH is fixed by MATH (because it belongs to MATH), and MATH is fixed by MATH (by definition). Hence, MATH is fixed by MATH. |
math/0106063 | Assume MATH is compact, and let MATH be a NAME measure on MATH. Given a nonempty, compact, convex MATH-space MATH, choose some MATH. Since MATH is a probability measure, we may let MATH . (In other words, MATH is the center of mass of the MATH-orbit of MATH.) The MATH-invariance of MATH implies that MATH is a fixed point for MATH (see REF). |
math/0106063 | CASE: |
math/0106063 | This proof employs a bit of machinery, so we postpone it to REF. (For discrete groups, the result follows easily from some other characterizations of amenability; see REF below.) |
math/0106063 | CASE: If MATH acts on MATH, and MATH is compact, then MATH is a nonempty, compact, convex MATH-space (see REF). So MATH has a fixed point in MATH; this fixed point is the desired MATH-invariant measure. CASE: Suppose MATH is a nonempty, compact, convex MATH-space. By replacing MATH with the convex closure of a single MATH-orbit, we may assume MATH is separable; then MATH is metrizable (see REF). Since MATH is amenable, this implies there is a MATH-invariant probability measure MATH on MATH. Since MATH is convex and compact, the center of mass MATH belongs to MATH (see REF). Since MATH is MATH-invariant (and the MATH-action is by linear maps), a simple calculation shows that MATH is MATH-invariant (see REF). |
math/0106063 | CASE: To avoid technical problems, let us assume MATH is discrete. The set of means on MATH is obviously nonempty, convex and invariant under left translation (see REF). Furthermore, it is a weak-MATH closed subset of the unit ball in MATH (see REF), so it is compact by the NAME REF . Therefore, the amenability of MATH implies that some mean is left-invariant (see REF). CASE: Suppose MATH acts continuously on a compact, metrizable space MATH. Fix some MATH. Then we have a continuous, MATH-equivariant linear map from MATH to MATH, defined by MATH . Therefore, any left-invariant mean on MATH induces a MATH-invariant mean on MATH (see REF). Since MATH is compact, the NAME Representation Theorem tells us that any continuous, positive linear functional on MATH is a measure; thus, this MATH-invariant mean is a MATH-invariant probability measure on MATH. |
math/0106063 | CASE: Because MATH is amenable, there exists a left-invariant mean MATH on MATH (see REF). For a measurable subset MATH of MATH, let MATH, where MATH is the characteristic function of MATH. It is easy to verify that MATH has the desired properties (see REF). CASE: We define a mean MATH via an approximation by step functions: for MATH, let MATH . Since MATH is finitely additive, it is straightforward to verify that MATH is a mean on MATH (see REF). Since MATH is bi-invariant, we know that MATH is also bi-invariant. |
math/0106063 | Because of REF, we may replace MATH with MATH. CASE: By applying the construction of means in REF to almost-invariant vectors in MATH, we obtain almost-invariant means on MATH. A limit of almost-invariant means is invariant (see REF). CASE: Because the means constructed in REF are dense in the space of all means, we can approximate a left-invariant mean by a MATH function. Vectors close to an invariant vector are almost-invariant, so MATH has almost-invariant vectors. However, there are technical issues here; one problem is that the approximation is in the weak-MATH topology, but we are looking for vectors that are almost-invariant in the norm topology. See REF for a correct proof in the case of discrete groups (using the fact that a convex set has the same closure in both the strong (or norm) topology and the weak-MATH topology). |
math/0106063 | CASE: Normalized characteristic functions of NAME sets are almost invariant vectors in MATH (see REF). CASE: Let us assume MATH is discrete. Given MATH, and a finite subset MATH of MATH, we wish to find a finite subset MATH of MATH, such that MATH . Since MATH is amenable, we know MATH has almost-invariant vectors (see REF); thus, there exists MATH, such that CASE: MATH, CASE: MATH, and CASE: MATH, for every MATH. Note that if MATH were the normalized characteristic function of a set MATH, then this set MATH would be what we want; for the general case, we will approximate MATH by a sum of such characteristic functions. Approximating MATH by a step function, we may assume MATH takes only finitely many values. Hence, there exist: CASE: finite subsets MATH of MATH, and CASE: real numbers MATH, such that CASE: MATH and CASE: MATH, where MATH is the normalized characteristic function of MATH (see REF). For all MATH and MATH, and any MATH, we have MATH (see REF), so, for any MATH, we have MATH and MATH . Therefore MATH . Summing over MATH yields MATH . Summing over MATH, we conclude that MATH . Since MATH (and all terms are positive), this implies there is some MATH, such that MATH . Hence, MATH, for every MATH, so we may let MATH. |
math/0106063 | Let CASE: MATH be a closed subgroup of a discrete, amenable group MATH, CASE: MATH be a finite subset of MATH, and CASE: MATH. Since MATH is amenable, there is a corresponding NAME set MATH in MATH. It suffices to show there is some MATH, such that MATH is a NAME set in MATH. We have MATH and, letting MATH, we have MATH so there must be some MATH, such that MATH (and MATH). Then, letting MATH, we have MATH so MATH is a NAME set in MATH. |
math/0106063 | For convenience, we consider only the free group MATH on two generators MATH and MATH. Suppose MATH has a left-invariant finitely additive probability measure MATH. (This will lead to a contradiction.) We may write MATH, where MATH, MATH, MATH, and MATH consist of the reduced words whose first letter is MATH, MATH, MATH, or MATH, respectively. Assume, without loss of generality, that MATH and MATH. Then MATH . Then, by left-invariance, we have MATH . This contradicts the fact that MATH. |
math/0106063 | Let MATH. The action of MATH on MATH by linear-fractional transformations is transitive, and the stabilizer of the point MATH is the subgroup MATH so MATH is compact. However, the NAME Density REF implies there is no MATH-invariant probability measure on MATH. (See REF for a direct proof that there is no MATH-invariant probability measure.) So MATH is not amenable. |
math/0106063 | The NAME REF tells us that MATH contains a closed subgroup isogenous to MATH. Alternatively, recall that any lattice MATH in MATH must contain a nonabelian free subgroup (see REF), and, being discrete, this is a closed subgroup of MATH. |
math/0106063 | CASE: REF The structure theory of NAME groups tells us that there is a connected, closed, solvable, normal subgroup MATH of MATH, such that MATH is semisimple. (The subgroup MATH is called the radical of MATH.) Since quotients of amenable groups are amenable (see REF), we know that MATH is amenable. So MATH is compact (see REF). |
math/0106063 | We know MATH is contained in some NAME space MATH; let MATH be the space of all bounded measurable functions from MATH to MATH (where two functions are identified if they are equal a.e.) Since MATH is separable, this is a NAME space with respect to the semi-norms MATH for MATH (see REF). Since MATH is nonempty and convex, it is obvious that MATH is nonempty and convex. Also, since MATH is compact (hence bounded), the NAME Theorem implies that MATH is compact. Furthermore, the action of MATH by right-translation is continuous (see REF). It is not difficult to see that MATH is a nonempty, closed, convex, MATH-invariant subset (see REF). |
math/0106063 | Suppose MATH is a closed subgroup of an amenable group MATH. For simplicity, let us assume MATH is discrete (see REF). Given a nonempty, compact, convex MATH-space MATH, REF tells us that MATH is a nonempty, compact, convex MATH-space. Thus, the amenability of MATH implies that MATH has a fixed point MATH in MATH. The MATH-invariance implies MATH is constant. Thus, for any MATH, we have MATH so MATH is a fixed point for MATH. |
math/0106063 | The argument is similar to the proof of REF tells us that MATH is a nonempty, compact, convex MATH-space. By restriction, it is also a nonempty, compact, convex MATH-space, so MATH has a fixed point MATH (under the action by right-translation). Then MATH factors through to an (essentially) well-defined map MATH. Because MATH is MATH-equivariant, it is immediate that MATH is MATH-equivariant. |
math/0106063 | MATH is amenable (see REF). |
math/0106063 | REF . |
math/0106063 | For REF, see REF Let MATH be the collection of all finitely generated subgroups of MATH. We have a unitary representation of MATH on each MATH, given by MATH. The direct sum of these is a unitary representation on MATH . Any compact set MATH is finite, so we have MATH, for some MATH. Then MATH fixes the base point MATH in MATH, so, letting MATH be a nonzero function in MATH that is supported on MATH, we have MATH for all MATH. Thus, MATH has almost invariant vectors, so there must be a MATH-invariant vector in MATH. So some MATH has an invariant vector. Since MATH is transitive on MATH, an invariant function must be constant. So a (nonzero) constant function is in MATH, which means MATH is finite. Because MATH is finitely generated, this implies that MATH is finitely generated. |
math/0106063 | ??? |
math/0106063 | Suppose a representation MATH of MATH has almost-invariant vectors. Then MATH, so MATH (see REF). Because MATH has NAME 's property, we conclude that MATH. This implies MATH (see REF), as desired. |
math/0106063 | ??? |
math/0106063 | CASE: |
math/0106063 | CASE: Since MATH is MATH-invariant, we have a representation of MATH on MATH; let us say MATH. Thus, the isomorphism MATH yields a representation MATH of MATH on MATH. It is not difficult to verify that MATH extends MATH (see REF). CASE: For MATH and MATH, let MATH . It is easy to verify that MATH is right MATH-equivariant (see REF), so we may think of MATH as a section of MATH (see REF). Let MATH . Now the map MATH is linear and MATH-equivariant (see REF), so MATH is a MATH-invariant subspace of MATH. Since MATH it is obvious that the evaluation map is bijective. |
math/0106063 | CASE: |
math/0106063 | To illustrate the idea of the proof, let us assume MATH is the span of a MATH-invariant section. Since MATH is noncompact, the NAME Ergodicity REF tells us that MATH has a dense orbit on MATH (see REF). (In fact, almost every orbit is dense.) This implies that any continuous MATH-invariant section of MATH is determined by its value at a single point (see REF), so the space of MATH-invariant sections is finite-dimensional (see REF). Since this space contains MATH (see REF), the desired conclusion is immediate. The general case is similar, but needs to take into account the action of MATH on a basis of MATH (because this action is no longer assumed to be trivial). See REF. |
math/0106063 | Assume there exists a nonzero MATH-invariant section MATH of MATH. Let MATH thus, MATH is a MATH-dimensional subspace of MATH that is MATH-invariant. Now, for MATH, let MATH . Since MATH, it is clear that MATH is MATH-invariant. Thus, it will suffice to show (by induction on MATH) that each MATH is finite dimensional. Since MATH, it is clear that MATH is MATH-invariant. Thus, since MATH centralizes MATH, REF implies that MATH is finite dimensional. Now, since MATH, we know that MATH is MATH-invariant. Then, since MATH centralizes MATH, REF implies that MATH is finite dimensional. |
math/0106063 | CASE: |
math/0106063 | Assume, to simplify the notation, that all of the eigenspaces of MATH are orthogonal to each other. Then, for any MATH, we have MATH, as MATH (see REF). Since the finite-index subgroups of MATH act irreducibly, there is some MATH, such that MATH (see REF). Thus, MATH as desired. |
math/0106063 | CITE. |
math/0106063 | CITE. |
math/0106063 | Let CASE: MATH be the set of all point masses in MATH, and CASE: MATH be a MATH-stationary probability measure on MATH that is in the class of NAME measure (see REF). We wish to show MATH, for a.e. MATH. In other words, we wish to show that MATH is supported on MATH. Note that: CASE: MATH is a closed, MATH-invariant subset of MATH, and CASE: because MATH is MATH-equivariant, we know that MATH is a MATH-stationary probability measure on MATH. Roughly speaking, the idea of the proof is that almost every trajectory of the random walk on MATH converges to a point in MATH (see REF). On the other hand, being stationary, MATH is invariant under the random walk. Thus, we conclude that MATH is supported on MATH, as desired. We now make this rigorous. Fix any MATH. Then MATH is a stationary probability measure on MATH. By mean proximality REF, we know, for a.e. MATH, that MATH . Then, by using the definition of MATH, we see that MATH . Thus, MATH . Since MATH is MATH-equivariant, this means MATH . For any MATH, since MATH is stationary, the map MATH is measure preserving; hence MATH . Since the LHS does not depend on MATH, but tends to MATH as MATH, we conclude that it is MATH. Since MATH is arbitrary, we conclude that MATH for a.e. MATH, as desired. |
math/0106063 | Assume, for simplicity, that MATH. Let MATH be a normal subgroup of MATH, and assume MATH is infinite. We wish to show MATH is finite. Since MATH has NAME 's Property MATH (see REF), it suffices to show that MATH is amenable (see REF). Suppose MATH acts by homeomorphisms on a compact, metrizable space MATH. In order to show that MATH is amenable, it suffices to find an invariant probability measure on MATH (see REF ). In other words, we wish to show that MATH has a fixed point in MATH. CASE: Because MATH is amenable, there is an (essentially) MATH-equivariant measurable map MATH (see REF). CASE: From REF, we know there is a closed subgroup MATH of MATH, such that the action of MATH on MATH is measurably isomorphic (a.e.) to the natural action of MATH on MATH. Since MATH acts trivially on MATH, we know that it acts trivially on MATH. Thus, the kernel of the MATH-action on MATH is infinite (see REF). However, MATH is simple (modulo its finite center), so this implies that the action of MATH on MATH is trivial (see REF). Since MATH, then the action of MATH on MATH is trivial. In other words, every point in MATH is fixed by MATH. Since REF , we conclude that almost every point in MATH is fixed by MATH; therefore, MATH has a fixed point in MATH, as desired. |
math/0106063 | We will assume the reader has some familiarity with manifolds of negative curvature. Assume, for simplicity, that: CASE: MATH is torsion free, so it is the fundamental group of the locally symmetric space MATH (where MATH is a maximal compact subgroup of MATH). CASE: MATH is compact. CASE: The locally symmetric metric on MATH has been normalized to have sectional curvature MATH. CASE: The injectivity radius of MATH is MATH. CASE: There are closed geodesics MATH and MATH in MATH, such that CASE: MATH, and CASE: MATH. Now MATH and MATH represent (conjugacy classes of) nontrivial elements MATH and MATH of the fundamental group MATH of MATH. Let MATH be the smallest normal subgroup of MATH that contains MATH. It suffices to show that MATH is nontrivial in MATH, for every MATH (see REF). Construct a CW complex MATH by gluing the boundary of a MATH-disk MATH to MATH along the curve MATH. Thus, the fundamental group of MATH is MATH. We wish to show that MATH is not null-homotopic in MATH. Suppose there is a continuous map MATH, such that the restriction of MATH to the boundary of MATH is MATH. Let MATH so MATH is a surface of genus MATH with some number MATH of boundary curves. We may assume MATH is minimal (that is, the area of MATH under the pull-back metric is minimal). Then MATH is a surface of curvature MATH whose boundary curves are geodesics. Note that MATH maps CASE: one boundary geodesic onto MATH, and CASE: the other MATH boundary geodesics onto multiples of MATH. Thus, we have MATH . This is a contradiction . |
math/0106063 | CASE: |
math/0106063 | We wish to show MATH; that is, the (forward) MATH-orbit of MATH is dense in MATH, for a.e. MATH. We will show that MATH and leave the remainder of the proof to the reader (see REF). Given a nonempty open subset MATH of MATH, let MATH . Clearly, MATH. Since MATH (because the measure on MATH is MATH-invariant), this implies MATH is MATH-invariant (a.e.). Since the NAME Ergodicity REF implies that MATH is ergodic on MATH, we conclude that MATH (a.e.). This means that, for a.e. MATH, the forward MATH-orbit of MATH intersects MATH. Since MATH is an arbitrary open subset, and MATH is second countable, we conclude that the forward MATH-orbit of a.e. MATH is dense. |
math/0106063 | Since MATH is nontrivial, it contains some nonconstant MATH. Now MATH cannot be essentially constant both on almost every vertical line and on almost every horizontal line (see REF), so we may assume there is a non-null set of vertical lines on which it is not constant. This means that MATH . REF tell us we may choose MATH in this set, with the additional properties that CASE: MATH, and CASE: MATH is dense in MATH. Let MATH, so MATH . Now, for any MATH, there exist MATH and MATH, such that MATH . Then we have MATH so the MATH-invariance of MATH implies MATH . Since MATH is closed, we conclude that MATH (see REF); since MATH is an arbitrary element of MATH, this means MATH. Also, from the choice of MATH, we know that MATH is not essentially constant. |
math/0106063 | Let MATH be the largest MATH-invariant subalgebra of MATH, and suppose MATH. (This will lead to a contradiction.) REF tells us that the only MATH-invariant subalgebras of MATH are CASE: MATH, CASE: MATH, CASE: MATH, and CASE: MATH, so MATH must be one of these MATH subalgebras. We know MATH (otherwise MATH). Also, we know MATH is nontrivial (otherwise MATH), so REF tells us that MATH. Thus, we may assume, by symmetry, that MATH . Since MATH, there is some MATH, such that MATH is not essentially constant on vertical lines. Applying the proof of REF yields MATH, such that CASE: MATH, so MATH, and CASE: MATH is not essentially constant on vertical lines. This contradicts REF. |
math/0106063 | We should embed MATH as a subgroup of MATH, find the corresponding set MATH of defining polynomials, and determine the complex solutions. However, it is more convenient to sidestep some of these calculations by using restriction of scalars, the method described in REF. Define MATH by MATH. Then MATH and MATH are linearly independent (over MATH), so they form a basis of MATH. Thus, MATH is the MATH-span of a basis, so it is a MATH-form of MATH. Therefore, letting MATH, we see that MATH is a real form of MATH. Let MATH . Then we have an isomorphism MATH defined by MATH, so MATH . |
math/0106063 | Define a MATH-linear map MATH by MATH . It is straightforward to verify that MATH is an injective ring homomorphism. Furthermore, MATH is a MATH-basis of MATH. Therefore, the map MATH defined by MATH is a ring isomorphism (see REF). |
math/0106063 | From REF, we have MATH (see REF). |
math/0106063 | Let CASE: MATH, CASE: MATH, and CASE: MATH be the usual complex conjugations MATH, MATH, and MATH. We have MATH so, in order to calculate MATH, we should determine the map MATH on MATH that corresponds to MATH when we identify MATH with MATH under the map MATH. First, let us determine MATH. That is, we wish to identify MATH with MATH, and extend MATH to a MATH-linear map on MATH. However, as usual, we use the MATH-form MATH, in place of MATH. It is obvious that if we define MATH by MATH, then MATH is MATH-linear, and the following diagram commutes: MATH . Thus, it is fairly clear that MATH. Hence MATH as desired. |
math/0106063 | Let CASE: MATH be the embedding described in the proof of REF; CASE: MATH be the usual conjugation on MATH, CASE: MATH, and CASE: MATH be defined by MATH. Then MATH is MATH-linear, and the following diagram commutes: MATH . Thus, because MATH we see that MATH . Similarly, letting MATH the same calculations show that MATH . |
math/0106063 | Let CASE: MATH, CASE: MATH be the NAME automorphism of MATH over MATH, CASE: MATH, CASE: MATH, for MATH, CASE: MATH, CASE: MATH, and CASE: MATH defined by MATH. We know (from Restriction of Scalars) that MATH is an irreducible, arithmetic lattice in MATH (see REF). Since MATH is compact, we may mod it out, to conclude that MATH is an arithmetic lattice in MATH. Also, since MATH is compact, we know that MATH is cocompact (see REF). |
math/0106063 | Let CASE: MATH, CASE: MATH, CASE: MATH be the usual complex conjugation (that is, the NAME automorphism of MATH over MATH), CASE: MATH be the NAME automorphism of MATH over MATH, so MATH, CASE: MATH, CASE: MATH, for MATH, CASE: MATH, and CASE: MATH. Since MATH is compact, we see that MATH is a cocompact, arithmetic lattice in MATH. |
math/0106063 | Let CASE: MATH, CASE: MATH, CASE: MATH be the NAME automorphism of MATH over MATH, CASE: MATH, CASE: MATH, for MATH, CASE: MATH, and CASE: MATH. Let MATH be the NAME automorphism of MATH over MATH, and extend MATH to an embedding of MATH in MATH. Note that MATH is an imaginary extension of MATH (because the square root of MATH is imaginary), but MATH is real. This implies that the NAME automorphism of MATH over MATH is the usual complex conjugation, so MATH for MATH. Therefore MATH is compact, so MATH is a cocompact, arithmetic lattice in MATH. Since MATH, we have MATH (see REF). On the other hand, because MATH, the proof of REF shows that MATH (see REF). Hence MATH, so MATH is a cocompact lattice in MATH. |
math/0106063 | Every irreducible, arithmetic lattice can be constructed by restriction of scalars (see REF), so this is the assertion of REF. Alternatively, the proof of REF provides an argument that does not rely on restriction of scalars. |
math/0106063 | Choose a compact simple NAME group MATH, such that MATH is isotypic. Then MATH has an irreducible, arithmetic lattice MATH. (This follows from REF if MATH is classical, but we will not prove the general case.) The compact factor MATH implies that MATH is cocompact (see REF). By modding out MATH, we obtain a cocompact, irreducible, arithmetic lattice in MATH. |
math/0106063 | We may assume MATH is simple. (If MATH and MATH are cocompact, arithmetic lattices in MATH and MATH, then MATH is a cocompact, arithmetic lattice in MATH.) Then MATH is isotypic, so REF applies. |
math/0106063 | Let CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, with MATH CASE: MATH, and CASE: MATH. Then Restriction of Scalars implies that MATH is an irreducible, arithmetic lattice in MATH (see REF). Since MATH we have MATH. |
math/0106063 | There is no harm in adding several copies of the compact factor MATH, so we may assume that MATH is a power of MATH. Also, we may assume that MATH . Let CASE: MATH be square-free positive integers that are pairwise relatively prime; CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, with MATH, CASE: MATH with MATH, for MATH, CASE: MATH, and CASE: MATH. From the choice of MATH, we have MATH if and only if MATH. Hence, MATH. |
math/0106063 | This follows from REF, which states that MATH is a MATH-form of MATH. |
math/0106063 | Let MATH. It suffices to find an irreducible polynomial MATH of degree MATH, such that MATH has exactly MATH real roots. (Then we may let MATH, where MATH is any root of MATH.) Choose a monic polynomial MATH, such that CASE: MATH has degree MATH, CASE: MATH has exactly MATH real roots, and CASE: all of the real roots of MATH are simple. (For example, choose distinct integers MATH, and let MATH.) Fix a prime MATH. Replacing MATH with MATH, for an appropriate integer MATH, we may assume CASE: MATH, and CASE: MATH (see REF). Let MATH. From REF, we know that MATH, so the NAME REF implies that MATH is irreducible. From REF, we see that MATH, like MATH, has exactly MATH real roots (see REF ). |
math/0106063 | CASE: Let MATH be a totally real algebraic number field of degree MATH over MATH; that is, MATH has exactly MATH real places and no imaginary places (see REF). CASE: For MATH and MATH, define MATH . CASE: Because MATH is dense in MATH (see REF), we may choose, for each MATH, some MATH, such that the sign (positive or negative) of MATH is MATH for MATH. CASE: Then the signature of MATH is MATH, for MATH. |
math/0106063 | One case is fairly obvious: if MATH and MATH, then MATH . The other cases are completely analogous to the calculation that MATH (see REF). The key point is to verify that MATH is the MATH-linear map on MATH that corresponds to the antiinvolution MATH. |
math/0106063 | CASE: Let MATH be an algebraic number field with exactly MATH real places MATH and exactly MATH imaginary places (see REF). CASE: Because MATH is dense in MATH (see REF), there is some nonzero MATH, such that CASE: MATH, for each imaginary place MATH, and CASE: MATH if and only if MATH. CASE: Let CASE: MATH, CASE: MATH, CASE: MATH be the set of MATH diagonal matrices with entries in MATH, CASE: MATH, and CASE: MATH be the set of MATH diagonal matrices with entries in MATH. For MATH, we have MATH and MATH, so MATH hence, MATH induces an embedding MATH with MATH. Because MATH is dense in MATH, we see that MATH so we may choose MATH, such that the signature of MATH, thought of as a MATH real symmetric matrix, is MATH, for each MATH. CASE: From REF, we see that MATH. |
math/0106063 | Suppose MATH is an irreducible lattice in MATH, such that MATH is not compact. This will lead to a contradiction. The NAME REF implies that MATH is arithmetic, so REF implies that MATH can be obtained by restriction of scalars: there exist CASE: an algebraic number field MATH, CASE: a connected, simple subgroup MATH of MATH, for some MATH, such that CASE: MATH is isogenous to MATH, and CASE: MATH is defined over MATH, and CASE: a continuous surjection MATH with compact kernel, such that MATH is commensurable with MATH. From the classification of arithmetic lattices (see REF on p. REF), we see that (after replacing MATH by an isogenous group) the group MATH must be described in REF: there is a quaternion algebra MATH over MATH (such that MATH is a central simple division algebra), and a Hermitian form MATH on MATH, such that MATH. Now, because MATH is not compact, we have MATH (see REF), so there must be a nontrivial isotropic vector MATH in MATH (see REF on p. REF). We know, from REF, that there is a basis of MATH, for which the matrix of MATH is MATH for some MATH (see REF). Thus, we may assume MATH . There is some MATH, such that MATH is isogenous to MATH. Then MATH is isomorphic to MATH (see REF ). Since MATH is the image of MATH under the isomorphism MATH, we see that MATH . Thus, MATH is isomorphic to either MATH or MATH; it is not isogenous to MATH. This is a contradiction. |
math/0106063 | Let MATH be a basis of MATH over MATH, with MATH. Let MATH be the structure constants of MATH with respect to this basis. That is, for MATH, we have MATH . There is some nonzero MATH, such that MATH, for all MATH. Let MATH be the MATH-span of MATH. |
math/0106063 | Define a ring homomorphism MATH by MATH. Because MATH is a division algebra, it is obvious that MATH is an irreducible module for MATH, so MATH is irreducible for MATH. Thus, the NAME Density REF implies that MATH is isomorphic to MATH, where MATH (see REF). Since MATH is simple (see REF), we know that MATH is faithful, so MATH. |
math/0106063 | REF lists only one possibility for MATH. The reference is to REF, with MATH and MATH. |
math/0106063 | Because MATH has only one simple factor, we have MATH. Because MATH, we see, from REF, that MATH does not split over MATH; thus, MATH and MATH are negative (see REF ). For each MATH, we know that MATH is compact, so we see, from REF, that MATH and both MATH and MATH are negative. By a change of basis, we may assume MATH is diagonal (see REF); thus, there exist MATH, such that MATH is of the form REF. Since MATH is Hermitian, we must have MATH, so MATH. Since MATH, we know that MATH of the MATH's are of one sign, and the other MATH of the MATH's are of the opposite sign. Thus, by permuting, and perhaps replacing MATH with MATH, we may assume that MATH are positive, and MATH are negative. Since MATH is compact, for MATH, we see that MATH are all of the same sign. |
math/0106063 | CASE: Let MATH. Because symplectic forms are, by definition, nondegenerate, there is some MATH, such that MATH. By induction on MATH, there is a good basis MATH of MATH. Let MATH. CASE: If MATH is a basis as in REF, let MATH . CASE: The desired basis is a permutation of the basis for REF. Alternatively, use the same proof as REF, but with MATH. CASE: This follows from REF. |
math/0106063 | CASE: To diagonalize MATH, choose MATH to be an orthogonal basis. CASE: We prove the more precise result by induction on MATH. We may assume MATH is isotropic (otherwise, there is nothing to prove), so there is some nonzero MATH with MATH. Because MATH is nondegenerate, there is some MATH, such that MATH. Note that the expression MATH is linear in MATH, so it has a zero. Thus, by replacing MATH with MATH, for some MATH, we may assume that MATH. Let MATH. Then MATH, and, by induction on MATH, there is a good basis MATH of MATH. Let MATH. CASE: The diagonal entries in REF are a permutation of the diagonal entries in REF. Thus, the desired basis is a permutation of the basis for REF If MATH, and MATH, then MATH. CASE: The desired basis is a permutation of the basis for REF The desired basis is a permutation of the basis for REF. |
math/0106063 | REF are immediate from REF Choose MATH, such that MATH contains an isotropic vector MATH, and let MATH. Then MATH, so the desired conclusion follows from the proof of REF. |
math/0106063 | We proceed by induction on the codimension of MATH in MATH. We may assume MATH (for otherwise MATH is an extension of MATH); let MATH. Suppose MATH. Choose MATH, and define MATH by MATH, for MATH and MATH. Then MATH is a linear bijection, MATH, for every MATH, and MATH, for every MATH. Since MATH, we know, from the induction hypothesis, that MATH extends to an isometry of MATH. This isometry also extends MATH. Suppose MATH. We must have either MATH or MATH. By interchanging MATH with MATH (and replacing MATH with MATH) if necessary, we may assume MATH. Since MATH, and we may assume MATH (for otherwise there is nothing to prove), we have MATH . Therefore MATH so MATH. Hence MATH. By comparing dimensions, we conclude that MATH . This implies that MATH so MATH where, for convenience, we define MATH . Now fix some MATH. There exists MATH, such that MATH so MATH . For any scalar MATH, we have MATH so, for an appropriate choice of MATH, we have MATH . We may extend MATH to MATH by defining MATH . |
math/0106063 | Let MATH be a codimension-one subspace of MATH, By induction on MATH, we may assume that the restriction MATH extends to an isometry MATH of MATH. Then, by composing MATH with the inverse of MATH, we may assume that MATH, for every MATH. Hence, REF applies. |
math/0106063 | Define MATH by MATH. Then MATH is an automorphism of MATH, so MATH is a (left or right) NAME measure on MATH. By uniqueness, we conclude that there exists MATH, such that MATH. It is easy to see that MATH is a continuous homomorphism. |
math/0106063 | CASE: NAME measure is finite on compact sets (see REF ). CASE: We prove the contrapositive. Let MATH be a compact subset of nonzero measure. Because MATH is continuous, and the continuous image of a compact set is compact, we know MATH is compact. Since MATH is not compact, then there exists MATH; thus MATH is disjoint from MATH. Continuing, we construct, by induction on MATH, a sequence MATH of elements of MATH, such that MATH is a collection of pairwise disjoint sets. They all have the same measure (since MATH is MATH-invariant), so we conclude that MATH . |
math/0106063 | MATH is continuous. It suffices to show that MATH is continuous at MATH (see REF). Thus, given a neighborhood MATH of MATH in MATH, we wish to find a neighborhood MATH of MATH in MATH, such that MATH. Choose a neighborhood MATH of MATH in MATH, such that CASE: MATH is symmetric (that is, MATH), and CASE: MATH. Let MATH, and MATH. Because MATH, it is clear that MATH, so we need only show that MATH contains a neighborhood of MATH in MATH. To this end, note that MATH is the union of countably many translates MATH of MATH, so MATH is the union of countably many translates of MATH; therefore, MATH is not a null set. Hence MATH contains an open subset of MATH (see REF). So MATH contains a neighborhood of MATH. MATH is MATH. Assume MATH. For simplicity, let us assume MATH. There are neighborhoods MATH of MATH in MATH and MATH of MATH in MATH, such that the exponential map, MATH is a diffeomorphism from MATH onto MATH. Furthermore, by shrinking MATH, we may assume that MATH is star-shaped; that is, if MATH and MATH, then MATH. Choose a neighborhood MATH of MATH in MATH, such that CASE: MATH, for all MATH, and CASE: MATH, and let MATH . From REF, we know there is some MATH, such that MATH . By rescaling, we may assume, without loss of generality, that MATH . Let MATH be the unique element of MATH with MATH, and define MATH by MATH . Then CASE: MATH is a MATH homomorphism, and CASE: MATH. Let MATH be the unique element of MATH with MATH. Then MATH . Since MATH and MATH each belong to MATH, we conclude that MATH, so MATH. Therefore MATH . Proceeding by induction on MATH, we see that MATH . Because MATH and MATH are homomorphisms, this implies that MATH for all MATH and all MATH. Since MATH is dense in MATH and MATH is continuous (see REF), we conclude that MATH is MATH, as desired. The general case. For simplicity, let us assume that MATH is a closed subgroup of MATH. Let MATH be a basis of the NAME algebra MATH of MATH, and, for each MATH, define a homomorphism MATH by MATH . CASE: From REF, we know that each MATH is MATH. Hence, the map MATH, defined by MATH is MATH. CASE: The Inverse Function Theorem implies that the map MATH, defined by MATH is a local diffeomorphism at MATH; there is a neighborhood MATH of MATH in MATH and a neighborhood MATH of MATH in MATH, such that MATH is a diffeomorphism from MATH onto MATH. By construction, we have MATH, so we conclude that MATH is MATH on MATH. Hence, MATH is MATH on all of MATH (see REF). |
math/0106063 | This can be proved algebraically, by combining NAME Theory with the elementary fact that every real polynomial of odd degree has a real zero (see REF), but we use a bit of complex analysis. Suppose MATH has no root. Then MATH is holomorphic on MATH. Furthermore, because MATH as MATH, it is easy to see that MATH is bounded on MATH. Hence, NAME 's Theorem asserts that MATH is constant. This contradicts the fact that MATH is not constant. |
math/0106063 | The map MATH defined by MATH is a surjective ring homomorphism whose kernel is MATH. |
math/0106063 | Given MATH, with MATH, let MATH. Conversely, given a subgroup MATH of MATH, let MATH be the fixed field of MATH. |
math/0106063 | CASE: MATH is a root of the monic polynomial MATH. CASE: Suppose MATH, where MATH with each MATH. Writing MATH (in lowest terms) with MATH, we have MATH . Since MATH is relatively prime to MATH (recall that MATH is in lowest terms), we conclude that MATH, so MATH. |
math/0106063 | Suppose MATH, where MATH, with each MATH, and MATH. Let CASE: MATH, CASE: MATH, and CASE: MATH. Then MATH is a monic, integral polynomial, and MATH . |
math/0106063 | Suppose MATH is an idea of MATH. (We wish to show that MATH is finitely generated.) For MATH, let MATH . Then MATH is an ideal of MATH, and we have MATH, so there is some MATH, such that MATH, for all MATH. For each MATH, let MATH be a finite set of polynomials of degree MATH, such that MATH generates MATH. Then MATH generates MATH. For any MATH, there exists MATH, such that MATH, and MATH. Then MATH, so we may assume, by induction, that MATH. |
math/0106063 | Note that MATH has only one proper ideal, namely MATH, so it is obviously NAME. Now use REF to induct on MATH. |
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