paper stringlengths 9 16 | proof stringlengths 0 131k |
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gr-qc/9902044 | From the above remarks on the invertible character of the NAME transformation we may infer the identity MATH whence we obtain MATH . After the application of REF , this becomes MATH . Since the identity MATH obviously holds, REF reduces to the expected result, namely MATH . |
gr-qc/9902044 | We firstly write down the equations of motion with respect to the conjugate momenta MATH: MATH . Owing to REF , this reduces to MATH which is equivalent to REF . |
gr-qc/9902044 | Owing to REF , NAME 's equations for the momenta MATH, MATH reduce to MATH which are identical to the NAME - NAME REF . The conclusion is then readily inferred owing to REF , and REF . |
gr-qc/9902044 | We firstly examine the NAME formalism for the Lagrangian MATH. We introduce auxiliary degrees of freedom and their conjugate momenta through the standard recursion relations MATH . We then obtain the canonical Hamiltonian corresponding to MATH, namely MATH . Now we turn to the NAME formalism for the Lagrangian MATH. On... |
gr-qc/9902044 | Taking into account REF , and REF we obtain successively MATH where the last right-hand side may be written as a function MATH, which could be formally given by REF . |
gr-qc/9902044 | It is sufficient to prove that an arbitrary choice of the functions MATH leads to a Hamiltonian formulation that is canonically equivalent to the constrained NAME Hamiltonian formulation of Subsection REF. We consider the standard NAME variables MATH defined as previously by MATH . We introduce the generalised NAME con... |
gr-qc/9902044 | Firstly, we expand the MATH terms in the NAME action REF, making use of formul REF, viz. MATH . Solving the second-class constraints amounts to eliminating variables MATH and MATH according to the strong equations MATH . Hence we obtain successively MATH and MATH . Summing up the above results we have MATH where we hav... |
gr-qc/9902080 | Let MATH be an arbitrary point. From a theorem in CITE there exists a causal neighbourhood MATH of MATH. Now, consider the wave REF . This is a linear, diagonal second order hyperbolic system for MATH. Hence, from the theory for hyperbolic equations (see for example, CITE or CITE) we know that it has a solution MATH th... |
gr-qc/9902080 | It is simply a matter of going through the same calculations as in the previous section to arrive at a similar wave equation for MATH as REF . By the theory for hyperbolic equations, this equation will also have a solution locally. Proceed as in the proof of the previous theorem. |
gr-qc/9902080 | By rescaling, it suffices to assume that MATH is a unit timelike- or spacelike vector. We start by proving uniqueness; suppose that MATH and MATH. For the case when MATH is timelike we obtain, MATH where we have used that MATH. In the spacelike case, the same calculations give MATH . This proves the uniqueness part so ... |
gr-qc/9902080 | An outline of the existence part of the proof is as follows. We start by solving the second of REF for MATH so that MATH. Then we evolve this initial data using the second equation of REF in such a way that MATH. Next we calculate the normal derivative of MATH using the first equation of REF and use the so obtained NAM... |
math-ph/9902027 | As MATH the set of orbits cover MATH. Suppose MATH and MATH. We have MATH, MATH and MATH for some elements MATH. From this we find MATH and so MATH but by symmetry we also have MATH and so the two orbits coincide. Thus two orbits either are disjoint or coincide and we have a partition. |
math-ph/9902027 | We have MATH if and only if MATH if and only if MATH. |
math-ph/9902027 | Let MATH and define MATH by associating to MATH the coset MATH. One has MATH if and only if MATH that is, if and only if MATH, and this if and only if MATH. Thus the correspondence is well defined and injective. It is obviously surjective, and so bijective. One has MATH and so MATH defines an equivalence of actions. |
math-ph/9902027 | Suppose MATH establishes the equivalence. Let MATH. For MATH one has MATH and so MATH. Thus MATH and we conclude that MATH. On the other hand, given MATH one has MATH. As MATH is an isomorphism one has MATH meaning that MATH which with the previous inclusion implies that MATH. |
math-ph/9902027 | Consider the action map MATH taking MATH into MATH. Consider also on MATH the action MATH and let MATH be the space of orbits under this action endowed with the quotient topology. Now MATH so MATH is constant on the orbits and thus factors through MATH: where MATH is the canonical map to the quotient. Since MATH maps M... |
math-ph/9902027 | Apply the NAME Identity for covariant derivatives MATH to any section of MATH. Using REF and the definition of MATH, one arrives at the conclusion. |
math-ph/9902027 | Let MATH be vector fields in an open set such that at any point MATH they form a basis for MATH. Use this basis to locally trivialize MATH. One has MATH and MATH for some functions MATH. REF now means that MATH . Let now MATH and MATH then REF becomes MATH . Refer to REF as MATH and now add MATH to MATH and subtract MA... |
math-ph/9902027 | By the universal property of the tensor algebra, MATH extends to a unique algebra homomorphism MATH. Now MATH vanishes on the ideal I, and so descends to an extension MATH to MATH showing existence. Uniqueness follows immediately from the fact that MATH is a set of generators for MATH. |
math-ph/9902027 | The map MATH given by MATH is MATH-linear and, by REF , is anti-symmetric. Thus it descends to a map MATH. The direct sum of these maps is an algebra homomorphism MATH, which is obviously surjective as MATH generates MATH. An element MATH is in the kernel of MATH if and only if each MATH-homogeneous part is in the kern... |
math-ph/9902027 | Since finite tensor products of the MATH form a basis for MATH, the finite NAME products of the MATH generate MATH. By what was said above, any finite product of the MATH is a linear combination of terms of the form MATH with MATH. The image of MATH in MATH is MATH. As these images are linearly independent in MATH, the... |
math-ph/9902027 | Consider the map MATH given by MATH. Since the MATH are fermionic, one calculates that MATH. By REF this map extends to an algebra homomorphism MATH. Let MATH be a NAME basis for MATH and MATH a NAME basis for MATH. Assume MATH and MATH disjoint and consider a total order on MATH in which any element of MATH is less th... |
math-ph/9902027 | Consider the map MATH defined by MATH. One has in MATH that MATH which by REF algebras means that the map extends to an algebra homomorphism MATH. REF shows now that this homomorphism establishes a bijection between complex linear bases of the two algebras. |
math-ph/9902027 | Let MATH, then MATH which, using the fact that MATH is a homomorphism and that MATH, reduces to MATH which is equal to MATH. That the image of MATH under MATH is all of MATH can be deduced from REF as this formula defines an orthogonal reflection in a hyperplane, and by the NAME theorem any element of MATH is a product... |
math/9902018 | From the definition of finite type invariants it is sufficient to construct a hyperbolic three-manifold MATH satisfying MATH for a given MATH. By CITE, there exists a hyperbolic NAME link MATH with MATH components, that is, the exterior of MATH is hyperbolic and each proper sublink of MATH is trivial. Let MATH be the c... |
math/9902020 | It is straightforward to verify (by considering all possible patterns for the last four entries of MATH) that the involution MATH interchanging MATH and MATH increases the number of runs by REF in half of all permutations, and, being an involution, decreases the number of runs by REF in the other half of the permutatio... |
math/9902020 | Recall from proof of REF that involution MATH makes pairs of permutations, and each pair contains two elements whose numbers of runs differ by REF. Note that half of these pairs consist of two elements with MATH and the other half consist of two elements with MATH. As MATH is invariant to MATH, it suffices to consider ... |
math/9902020 | By induction on MATH. For MATH, the statement is true by REF . Now suppose we know the statement for MATH. To prove that MATH, we need to group all MATH-half-ascending permutations in pairs, so that the MATH values of the two elements of any given pairs differ by one, and show that the set of permutations consisting of... |
math/9902020 | It is straightforward to see that the map described is injective on the set of labeled lattice path, not necessarily satisfying REF . Assume that MATH and MATH are both descents of the permutation MATH. Let MATH, respective MATH, be the permutation when restricting to the MATH, respective MATH, first elements. Observe ... |
math/9902020 | To prove the theorem combinatorially, we construct a quasi-injection MATH . By quasi-injection we mean that there will be some elements of MATH for which MATH will not be defined, but the number of these elements will be less than that of elements in MATH which are not in the image of MATH. In particular, the restricti... |
math/9902020 | First suppose that MATH is even. For MATH, the statement is true. If MATH, then REF shows that MATH. The coefficients of MATH are just the binomial coefficients, which are certainly log-concave CITE, while the coefficients of MATH are the MATH, which are log-concave by REF and the remark thereafter. As the product of t... |
math/9902028 | REF shows how MATH is isotopic to MATH in MATH, and this completes the proof since MATH is unknotted. |
math/9902028 | Let MATH, the characteristic polynomial of MATH. First, for MATH, we give recursive formulas which reduce the calculation of MATH to that of MATH. According to REF , MATH is a rank MATH matrix of the form MATH where we have left out the dependence on MATH. Expand its determinant by the third column to obtain MATH . Exp... |
math/9902028 | The algebraic linking number of REF components of MATH is MATH CITE. This is easily calculated from REF : MATH and the lemma follows simply from this. |
math/9902028 | The fundamental group of the complement of MATH is normally generated by the boundary MATH of a normal disk to MATH. The fundamental group of MATH is an amalgamated free product, MATH and MATH is the product of a circle with the fiber bundle MATH where MATH is a REF-disk with MATH punctures. (Of course, MATH depends on... |
math/9902028 | Fix MATH and consider the double branched covers MATH of MATH. If MATH is smoothly isotopic to MATH, then REF follows. However, since MATH is symplectic, MATH CITE, and it follows from REF that the MATH are all distinct for different MATH. Thus REF can hold only if MATH. |
math/9902028 | In this case the double branched covers MATH cannot be diffeomorphic for different MATH. For, it follows from REF that MATH. Thus MATH is a diffeomorphism invariant of MATH, the sum of all its NAME invariants. The calculation of REF shows that these numbers are different for different MATH. |
math/9902032 | These expressions follow from the explicit form REF of the MATH-generators, and from REF . |
math/9902032 | This is a direct consequence of REF , and of the formul REF . |
math/9902032 | Let us denote by MATH the adjoint of MATH with respect to the pairing REF . Consider the symmetric operator MATH which exists whenever MATH. Notice that it has the same principal symbol as MATH. Now, the map MATH is obviously MATH-equivariant. REF just apply and yield MATH. |
math/9902032 | In view of the preceding corollary, we need only the commutation relations of the operators REF with the generators MATH of inversions given in REF in order to determine MATH. Straightforward calculation leads to MATH for MATH. Now, REF guarantees that the monomials MATH are independent; they are of degree MATH as diff... |
math/9902032 | The matrix realization of the MATH-generators REF is given by MATH where MATH is the canonical basis of MATH and MATH its dual basis associated with the metric MATH. A simple calculation yields the basis MATH of MATH dual to MATH with respect to the Killing metric REF . One gets MATH . Using the MATH-action, MATH, on M... |
math/9902032 | The explicit formula for MATH is obviously obtained by replacing MATH by MATH in REF . Applying then REF to that expression immediately leads to the sought result REF . |
math/9902032 | This follows immediately from REF of the NAME operator MATH. |
math/9902032 | This is straightforward from REF . |
math/9902032 | The highest degree component of the eigenvalue REF is just the eigenvalue equation for MATH. Hence, a solution of REF is necessarily of the form MATH and MATH according to the decomposition REF . The remainder of REF reads now MATH . Since MATH, by REF , the coefficients MATH do not vanish. In view of the nilpotency of... |
math/9902032 | Any polynomial MATH is of the form MATH where MATH is harmonic, see REF . One has MATH since MATH and MATH in the NAME algebra MATH (see REF). At last, MATH is harmonic because MATH and one furthermore easily checks that MATH. Hence, MATH. The proof for MATH and MATH is analogous and will be omitted. |
math/9902032 | One proceeds just as in the proof of REF . The fact that one restricts MATH in the decomposition REF to those values for which MATH again insures that the coefficients in REF do not vanish. |
math/9902032 | The first inclusion easily follows from REF . As to the second one, we then proceed as in the proof of REF , using REF . |
math/9902032 | It has been shown in REF (see also REF) that such a map MATH is necessarily a differential operator with constant coefficients. We can thus apply REF : MATH is a polynomial in the operators REF . We conclude by using REF . |
math/9902032 | In REF , both factors in the first term, MATH and MATH, are non-negative in view of REF . Also REF yields MATH so that the second term is bounded from below by MATH, which is non-negative. We have just shown that MATH. Now, if MATH, one has MATH because of REF . The result follows since MATH in REF . |
math/9902032 | CASE: An eigenvector of the NAME operator MATH, with principal symbol MATH, is of the form MATH . (See REF .) The eigenvalue REF therefore leads to MATH, where MATH is the eigenvalue of the NAME operator MATH given by REF , and to MATH . In order to solve this equation for MATH and MATH, one needs to introduce the proj... |
math/9902032 | One can consider the inverse map MATH, which is given by MATH. Using the well-known composition formula for differential operators MATH one obtains MATH . Recall that the NAME boundary of a MATH-cochain MATH is given by MATH and observe that the preceding expression is therefore MATH . One sees that MATH satisfies REF ... |
math/9902042 | See REF , p. REF. |
math/9902042 | The usual proof of NAME 's theorem establishes a polynomial bound for the number of rational points of bounded exponential height. Hence, the height zeta function of MATH converges for MATH. (There is no need to invoke NAME 's theorem CITE which gives the precise asymptotics.) Therefore, there are real numbers MATH suc... |
math/9902042 | This follows from the invariance of the height under the action of maximal compact subgroups, see REF . |
math/9902042 | If MATH, the line bundle MATH is ample iff all MATH and MATH. Moreover, the ample line bundles MATH provide a basis of MATH. Hence, REF implies the existence of a real MATH such that the series converges absolutely when MATH, uniformly for all MATH such that MATH, MATH. For any MATH, the function MATH is another height... |
math/9902042 | Using a partition of unity on MATH for the MATH-topology, we may assume that MATH is a relatively compact open subset MATH, with local coordinates MATH and that the divisor MATH is given by the equation MATH for some integer MATH. The integral is then MATH for some functions MATH giving the metrics in our local trivial... |
math/9902042 | Choose local analytic coordinates on MATH such that MATH is defined by MATH. Then, MATH may be embedded in MATH with coordinates MATH and is given there by the equations MATH for MATH. We consider the chart MATH. Then, local coordinates on MATH are MATH and MATH is given by MATH if MATH. On MATH, the derivation MATH ha... |
math/9902042 | REF preceeding lemmas imply that for any multiindex MATH, the derivative MATH is bounded on MATH. Moreover, MATH tends to MATH at infinity. We thus may integrate by REF-times. |
math/9902042 | We decompose the domain of integration MATH into subdomains MATH for MATH and MATH . Then MATH . One has MATH and for MATH, MATH . For all MATH with MATH, the geometric series converges absolutely and we obtain MATH . Simplifying, we obtain REF. |
math/9902042 | As above, we have MATH . We first integrate over the set MATH of MATH such that MATH. MATH . If MATH, as MATH doesn't divide all the MATH, this is the integral of a non-trivial character on a compact group, hence MATH. For MATH, we get MATH. Therefore, as MATH and MATH for MATH, MATH . This implies the lemma. |
math/9902042 | In the integral, we replace MATH by REF, MATH by MATH and use the computation in REF . For MATH, we obtain MATH . If MATH is a positive integer, we have an inequality MATH which can be deduced for example, from the Prime Number Theorem. This gives us REF . |
math/9902042 | By a unitary change of variables, we may assume MATH. Thus, MATH . For any MATH, the MATH-th derivative of MATH is of the form MATH with MATH a polynomial of degree MATH whose coefficients are polynomials in MATH. Thus we can integrate by parts and get for any MATH an expression MATH which imply the lemma. |
math/9902042 | Choose a real number MATH. From the lemmas above, it follows that there exists MATH such that for any compact MATH, any MATH, and any real MATH, the product of the local NAME transforms at the character MATH converges to a holomorphic function of MATH which satisfies the inequality MATH . Hence, the sum over all non-tr... |
math/9902042 | We prove the three lemmas simultaneously. By a unitary change of variables, we may assume that MATH. Then one has MATH and MATH hence their volumes as in REF . If MATH doesn't divide MATH, we may change variables and even assume that MATH. Then, MATH and the last integral has already been calculated when we studied the... |
math/9902043 | There are MATH binary strings of length less than MATH. A fortiori there are at most MATH elements of MATH that can be computed by binary programs of length less than MATH, given MATH. This implies that at least MATH elements of MATH cannot be computed by binary programs of length less than MATH, given MATH. Substituti... |
math/9902043 | Place MATH pebbles at positions chosen from the total of MATH grid points - there are MATH choices. Choose two pebbles, MATH and MATH, from among the MATH pebbles - there are MATH choices. Choose a new pebble MATH on the straight line determined by MATH. The number of grid points on this line between MATH (or MATH) and... |
math/9902043 | Place MATH pebbles at positions chosen from the total of MATH grid points - there are MATH choices. Choose one pebble MATH from among the MATH pebbles - there are MATH choices. Choose a new pebble MATH on the (horizontal) grid line determined by MATH - there are MATH choices. There is a fixed algorithm that, on input M... |
math/9902043 | Place MATH pebbles at positions chosen from the total of MATH grid points - there are MATH choices. Choose two pebbles, MATH and MATH, from among the MATH pebbles - there are MATH choices. Place a new pebble MATH at one of the remaining grid points. Without loss of generality, let the triangle MATH have MATH as the lon... |
math/9902043 | Choose MATH pebbles at positions chosen from the total of MATH grid points such that REF is satisfied. Divide the unit square by a horizontal grid line into an upper and a lower half, each of which contains MATH pebbles - there are no grid lines containing two pebbles by REF . We write forbidding line for a line determ... |
math/9902049 | CASE: We have MATH, and, because MATH is a proper map, we have MATH as MATH in MATH. Because MATH, we know that MATH is homeomorphic to the half-line MATH (with the point MATH in MATH corresponding to the endpoint MATH of the half-line), so, by continuity, it must be the case that MATH. Therefore MATH, so MATH is a NAM... |
math/9902049 | Let MATH be the identity component of the NAME closure of MATH, and write MATH, where MATH is a subgroup of MATH and MATH is conjugate, via an element of MATH, to a subgroup of MATH. Replacing MATH by a conjugate, we may assume that MATH. Let MATH. Then MATH (compare CITE and CITE), so we have MATH. Because MATH and MA... |
math/9902049 | Because MATH is compatible with MATH, we have MATH, where MATH and MATH. We may assume that MATH, for otherwise we have MATH. Therefore MATH. Because MATH is a sum of root spaces, this implies that there is a positive root MATH, such that MATH. Because MATH, we must have MATH, for otherwise we would have MATH, so MATH.... |
math/9902049 | Let MATH be the projection with kernel MATH. Since MATH we must have MATH, so MATH. Therefore, letting MATH be the NAME closure of MATH, we may assume that MATH contains MATH (see REF), by replacing MATH with a conjugate subgroup. So MATH is normalized by MATH (see REF). Therefore, since MATH, we conclude that MATH (se... |
math/9902049 | Let MATH be a maximal connected semisimple subgroup of MATH, and let MATH be a maximal compact torus of the NAME closure of MATH, the solvable radical of MATH. Replacing MATH by a conjugate torus, we may assume that MATH centralizes MATH. Let MATH be the NAME decomposition of MATH. From NAME 's nilshadow construction (... |
math/9902049 | Because MATH, it is easy to see that MATH is a MATH-definable subset of MATH. Furthermore, the map MATH is obviously MATH-definable. Therefore, MATH is a MATH-definable subset of MATH, and MATH CITE. |
math/9902049 | We may assume that MATH (see REF), so MATH. We prove the contrapositive: suppose that MATH. Let MATH. Then MATH is MATH-definable and MATH (see REF). Let MATH be the sphere at infinity, and let MATH be the set of accumulation points of MATH in MATH. For any nonempty MATH-definable set MATH, we have MATH, where MATH is ... |
math/9902049 | CASE: See REF . CASE: This follows from REF , and REF . CASE: We begin by showing that, instead of only sequences MATH and MATH, there are two continuous curves MATH and MATH, MATH, in MATH, such that MATH as MATH, and, for each MATH, we have MATH, for MATH. Fix MATH. For each MATH, and each MATH let MATH . The existen... |
math/9902049 | Let MATH. Replacing MATH by a conjugate, we may assume that MATH, where MATH is a one-parameter subgroup of MATH and MATH is a one-parameter subgroup of MATH, such that MATH centralizes MATH (compare REF). The subgroup MATH is contained in a subgroup MATH of MATH that is locally isomorphic to MATH CITE. Replacing MATH ... |
math/9902049 | Let MATH, so MATH is semisimple of real rank one. (The only real roots of MATH are MATH and, possibly, MATH.) Let MATH, so MATH is a maximal split torus of MATH, and note that the NAME algebra of MATH is MATH. The choice of the ray MATH determines a corresponding NAME chamber MATH in MATH. We may assume that MATH is a ... |
math/9902049 | We prove only the nontrivial direction of each conclusion. CASE: Suppose that MATH is a NAME subgroup of MATH. CASE: The proof of the claim in REF , with MATH in the role of MATH, shows that MATH contains MATH. CASE: From REF , we know that there is a compact set MATH, such that, for each MATH, there is some MATH, such... |
math/9902049 | Let MATH be the identity component of MATH. Because every element of the NAME group of MATH has a representative in MATH CITE, we see that MATH and MATH have the same positive NAME chamber MATH, and the NAME projection MATH is the restriction of the NAME projection MATH. Thus, the desired conclusion is immediate from R... |
math/9902049 | CASE: We may assume that MATH (see REF). If we identify the NAME algebra MATH of MATH with the connected component of MATH containing MATH, then MATH is a convex cone in MATH and the opposition involution MATH is the reflection in MATH across the ray MATH. If MATH and MATH are the two walls of the NAME chamber MATH, th... |
math/9902049 | If MATH, then MATH is a NAME subgroup (because MATH is a NAME subgroup) and we have either MATH or MATH. Thus, we may henceforth assume that MATH. Then, from the proof of REF , we see that MATH is not contained in the NAME closure MATH of MATH. Therefore MATH, so there are MATH with MATH and MATH. (If MATH, then MATH.)... |
math/9902049 | Let MATH be a minimal NAME subgroup of MATH. Assume that MATH and that MATH. Consider the possibilities given by REF . If MATH is of type REF, then MATH is listed in REF . If MATH is of type REF, then we may assume that MATH, by replacing MATH with a conjugate via an element of MATH; thus, MATH is the subgroup of MATH ... |
math/9902049 | It is easy to see from REF that the subgroup of MATH corresponding to each of the subalgebras listed in REF is a NAME subgroup. Each is minimal, because each is REF-dimensional (see REF). REF shows that no two of the listed subalgebras are conjugate. Thus, given a minimal NAME subgroup of MATH that is contained in MATH... |
math/9902049 | For each subalgebra MATH, we look at the restriction of the exponential map to MATH. This is a polynomial function from MATH into MATH. For convenience, we use MATH, MATH, or MATH as coordinates on MATH, and use the matrix entries as coordinates on MATH. The subalgebra of type REF with MATH and the subalgebra of type R... |
math/9902049 | CASE: Assume that MATH in MATH. We may assume that MATH (see REF). Let MATH be the NAME projection of MATH, with MATH, and write MATH, with MATH (because MATH). Let MATH so MATH and MATH. Then MATH, so there is a nonzero vector MATH; we may assume that MATH. Because MATH, we may write MATH, so MATH . Because MATH, we m... |
math/9902049 | CASE: Assume that neither REF nor REF holds. We have MATH, for otherwise there is an element of MATH for which MATH, MATH, and MATH are all zero, so REF holds. Assume that MATH for every element of MATH. Because MATH, we know that MATH is determined by MATH and MATH, so there exist MATH and MATH, such that, for every e... |
math/9902049 | This follows from the fact that there is an element of the NAME group of MATH that sends each element of MATH to its inverse, so each element of MATH is conjugate to its inverse under MATH. |
math/9902049 | For future reference we prove the stronger fact that for every real number MATH and every MATH, we have MATH, where MATH . Let MATH. We have MATH and MATH . Because MATH and MATH are linearly independent for all nontrivial MATH, we have MATH so we conclude that MATH. |
math/9902049 | CASE: Suppose that MATH. Then MATH is a positive real number. Set MATH . Then a matrix calculation, using the facts that MATH and MATH, verifies that MATH and MATH for every element of MATH, so MATH. CASE: Because MATH is reductive and MATH is the positive NAME chamber of a maximal split torus of MATH, we have MATH. Th... |
math/9902049 | Let MATH, where MATH is some nonzero element of MATH. For any large real number MATH, we see that MATH, but MATH for MATH. Thus, MATH. Furthermore, we have MATH, and we have MATH whenever MATH and MATH, so we conclude that MATH, as desired. For future reference, we note the stronger fact that for every real number MATH... |
math/9902049 | We may assume that MATH is not contained in any root space, for otherwise REF applies. We may also assume that MATH, and that MATH is not a NAME subgroup. The proof proceeds in cases, suggested by REF . Assume that MATH and that MATH, for every nonzero element of MATH. We show that MATH, for every MATH, so MATH is not ... |
math/9902049 | Let MATH. There is a positive root MATH with MATH, and a one-parameter subgroup MATH of MATH, such that MATH, where MATH is the NAME algebra of MATH (see REF). Note that MATH and MATH are contained in the NAME closure of MATH. It is not difficult to see that every nonabelian subalgebra of MATH contains MATH. Therefore,... |
math/9902050 | NAME assumed that MATH is reductive, but the same proof works with only minor changes. From REF , we see that we may assume that MATH. Then, from the NAME decomposition MATH, it is immediate that the homogeneous space MATH is homeomorphic to the cartesian product MATH. Because MATH is a simply connected, solvable NAME ... |
math/9902050 | Suppose that MATH does have a compact NAME form. Assume for the moment that MATH. Then MATH is a NAME subgroup of MATH (see REF), so we see from REF that MATH must be compact. But the dimension of every connected, co-compact subgroup of MATH is at least MATH CITE, and MATH. This contradicts the fact that MATH. Thus, we... |
math/9902050 | CASE: Because every element of the NAME group of MATH has a representative in MATH CITE, we see that MATH and MATH have the same positive NAME chamber MATH, and the NAME projection MATH is the restriction of the NAME projection MATH. Thus, the desired conclusion is immediate from REF This is an easy consequence of the ... |
math/9902050 | From REF , we have MATH. From REF, we have MATH. Thus, REF implies that MATH acts properly on MATH. We have MATH, MATH, and MATH, so MATH. Therefore, conclusions REF follow from REF. To show conclusion REF, suppose that MATH for some MATH. Because all maximal split tori in MATH are conjugate, we may assume that MATH no... |
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