paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003087 | According to REF we have to show that there is an automorphism MATH such that MATH. But for MATH we can chose the operator existing according to REF. |
math/0003087 | According to REF there is an automorphism MATH of MATH, such that MATH. Now the spectrum of an operator is invariant under automorphisms, and REF shows that also the NAME multiplicities are invariant under automorphisms, that is, MATH and MATH have the same eigenvalues and NAME multiplicities, and the unitary equivalen... |
math/0003087 | We construct the MATH inductively: Since the range of MATH is all of MATH, if MATH is finite, and MATH, if MATH is infinite (compare CITE) there is a projection in MATH, such that MATH. Suppose now that for MATH the MATH are pairwise orthogonal with MATH (MATH). Setting MATH the restricted algebra MATH is again a type ... |
math/0003087 | Since MATH and MATH we can arrange the eigenvalues MATH such that MATH and MATH and MATH. Now chose the MATH for MATH as in the proof of REF and set MATH for MATH, MATH for MATH, and MATH such that MATH . Then MATH commutes with MATH and have the stated properties. |
math/0003087 | Let MATH REF a family of orthogonal projections in MATH with MATH and MATH (such a family exists according to REF). Then we define MATH which is a non-singular positive selfadjoint operator affiliated with MATH, has eigenvalues MATH and MATH . Then MATH are the modular objects corresponding to MATH, where MATH is the c... |
math/0003087 | Since MATH is s.a., there exists a measure space MATH, a unitary MATH and a real valued measurable function MATH, such that for every MATH with MATH . Define now MATH . Then MATH is the wished conjugation. |
math/0003087 | CASE: Let MATH. Then MATH and MATH for every MATH. CASE: Let MATH. Then MATH . This shows that MATH, and the rest follows from standard calculation. |
math/0003087 | This follows immediately from REF. |
math/0003087 | CASE: Let MATH be the spectral measure of MATH. Then we decompose MATH in the direct sum of the following three orthogonal subspaces: MATH where MATH, MATH, MATH. Then MATH (MATH). From MATH we see MATH REF and, since MATH, also equality holds. CASE: In MATH we set MATH. Then MATH and MATH and, since MATH, MATH . CASE:... |
math/0003087 | Set MATH. Then an easy calculation shows that also MATH is a trace with MATH. Since the trace is unique we get MATH, that is, MATH for all MATH. |
math/0003087 | Since MATH is positive, we can examine the unitary group MATH. This equality gives MATH. Since also MATH and MATH commute (what is shown by an easy computation), MATH is also a unitary group in MATH, that is, MATH . Now the assertion follows from the uniqueness of the generator of a group. |
math/0003091 | The proof is essentially borrowed from CITE where a more general situation is considered. We specialize terminology to our case and give more details when seems appropriate. Since MATH splits over a virtually nilpotent subgroup, MATH acts without edge inversions on a simplicial tree MATH with no proper invariant subtre... |
math/0003091 | Assume first that the splitting of MATH given by REF is MATH. Think of MATH as glued from MATH, MATH, and MATH. Since the splitting is nontrivial, both MATH and MATH have infinite index in MATH so MATH, MATH are noncompact. Hence MATH, MATH are noncompact since they are glued from noncompact spaces along compact subset... |
math/0003091 | Being the fundamental group of a pinched negatively curved manifold, MATH is torsion free and any abelian subgroup of MATH is finitely generated CITE. Therefore, every element in MATH is a power of a primitive element (an element of a group is called primitive it is not a proper power). Fix an arbitrary MATH and find M... |
math/0003091 | Arguing by contradiction find MATH that converges to MATH while MATH does not approach MATH. Pass to a subsequence so that MATH converges to MATH. Write MATH and MATH and pick preimages MATH of MATH, and MATH of MATH. Since MATH is uniformly bounded, a result in CITE implies that MATH subconverges to MATH in the equiva... |
math/0003091 | Let MATH. By REF and CITE we can assume by passing to a subsequence that MATH converges to MATH in pointwise convergence topology and MATH converges to MATH in equivariant pointed NAME topology. By CITE we know that the MATH-action on MATH is free. By CITE MATH converges to MATH in the pointed MATH topology. Now MATH i... |
math/0003091 | Arguing by contradiction, let MATH be a sequence of manifolds with the diameter of the MATH-thick part going to infinity for some MATH. Find points MATH such that MATH subconverges to MATH. By REF the MATH-thick part of MATH is compact hence it lies in the open ball MATH for some MATH. Pass to subsequence so that diame... |
math/0003091 | Arguing by contradiction find a sequence of manifolds MATH each containing a closed geodesic of length MATH. By REF and CITE MATH subconverges in the pointed NAME topology. Denote the limiting manifold by MATH. By REF the MATH-thick part of MATH contains the MATH-thick part of MATH. For large MATH . NAME approximations... |
math/0003091 | Let MATH lie in different conjugacy classes. This defines a sequence MATH of free isometric actions of MATH on the universal cover MATH of MATH. For a finite generating set MATH of MATH, let MATH and MATH. As in CITE, we can choose a sequence of points MATH such that MATH. If MATH, we get an action on a MATH-tree hence... |
math/0003091 | Arguing by contradiction assume there exists an injective homomorphism MATH which is not onto. By CITE the manifold MATH has finite volume hence MATH must be of finite index, say MATH, in MATH. Iterating MATH, we get a sequence of free isometric actions MATH of MATH on the universal cover MATH of MATH such that MATH is... |
math/0003091 | To prove REF find MATH with MATH. Let MATH be the points obtained by pushing MATH along radial geodesics. Then by the triangle inequality MATH where the latter inequality holds because NAME functions are MATH-Lipschitz. It remains to prove MATH. Let MATH be a point of the ideal boundary of MATH corresponding to the cus... |
math/0003091 | First, show that the volume of the MATH-thick part is uniformly bounded above on MATH for any MATH. Indeed, observe that the diameter of MATH is bounded above by REF. Hence, MATH is in the image of a ball in MATH of some uniformly bounded above radius. By NAME volume comparison the volume of the ball is uniformly bound... |
math/0003091 | Renumerate the sequence so that MATH now come with even indices while odd indices correspond to NAME 's smoothings of MATH. We still denote the obtained sequence by MATH. Since MATH converges to MATH in the pointed MATH-topology there are MATH-Lipschitz smooth embeddings MATH with MATH such that MATH-pushforward of MAT... |
math/0003091 | Arguing by contradiction find a sequence MATH of manifolds with fundamental group isomorphic to MATH and sectional curvatures within MATH. By the main theorem, we can assume that MATH converges in MATH topology to a MATH-Riemannian manifold MATH. Now the universal cover MATH of MATH is a complete MATH-Riemannian manifo... |
math/0003091 | Arguing by contradiction find a sequence MATH of finite volume NAME or quaternionic-Kähler manifolds with fundamental group isomorphic to MATH and sectional curvatures within MATH. By REF, we can assume that MATH converges in MATH topology to a MATH-Riemannian manifold MATH. First, assume that each MATH is quaternionic... |
math/0003092 | As an element of the ordinary MATH-theory of MATH, the index bundle may be represented as a difference of complex vector bundles. We need to represent MATH as the difference of two genuine MATH-bundles over MATH, and so adapt the standard argument to the context of MATH-equivariant MATH-theory (compare CITE) as follows... |
math/0003092 | Points of MATH fall into two categories depending on whether they are MATH-fixed or not. We consider the former case first, making use of the MATH-equivariance of the NAME map. Then we modify the argument to deal with the rest of the points in MATH. Suppose MATH is MATH-fixed. The NAME model for the solutions to MATH a... |
math/0003096 | Any MATH in the kernel of our homomorphism must preserve each light-line and so has each light-line as an eigenspace. This forces MATH to be a multiple of the identity matrix MATH and then MATH gives MATH. Thus the main issue is to see that our homomorphism is onto. However, by NAME 's Theorem, MATH is generated by ref... |
math/0003096 | Suppose first that MATH. Choose an orthonormal basis MATH of MATH so that MATH and a basis MATH of MATH so that MATH . Write MATH for some MATH. Then REF reads MATH . The elements MATH REF are linearly dependent in MATH while MATH, for MATH, whence taking coefficients in REF gives: MATH . From REF, we see that each MAT... |
math/0003096 | REF means that MATH where MATH is reflection in the hyperplane orthogonal to MATH. Now fix MATH and restrict attention to a MATH-dimensional affine space containing MATH and MATH. Certainly, any MATH-sphere (or MATH-plane) tangent to MATH and MATH at MATH lie in this space. Now any MATH-sphere containing MATH and MATH ... |
math/0003096 | Suppose first that MATH where MATH is the NAME transformation induced by reflection in a non-degenerate hyperplane MATH. Let MATH be the frame of MATH given by MATH (compare REF ) and fix MATH with MATH. Up to a scaling, MATH is the reflection in MATH of MATH whence MATH so that MATH for some MATH. Since MATH is consta... |
math/0003096 | Choose a holomorphic coordinate MATH on MATH. We must show that MATH . As in REF , there is a function MATH with MATH so that MATH since both MATH and MATH vanish by the conformality of MATH. |
math/0003096 | For MATH a holomorphic coordinate on MATH, MATH . Now MATH is dual to MATH if and only if MATH is parallel to MATH or, equivalently, MATH for any normal MATH to MATH. Taking MATH in REF gives MATH and then REF asserts that MATH is parallel while REF asserts that MATH so that MATH is isoperimetric. The case of minimal M... |
math/0003096 | It suffices to check that the inversion MATH is isothermic. Now MATH and we put MATH. Then MATH . Thus, locally, MATH with MATH the NAME transform of MATH. |
math/0003096 | Let MATH be normal along MATH so that MATH is normal along MATH. The second fundamental form MATH of MATH along MATH is given by MATH and we know from REF that MATH whence MATH . It is not difficult to check that if MATH at some point then the same identity is also true for MATH at that point: MATH . Thus our exclusion... |
math/0003096 | The first thing to note is that the conditions on MATH are independent of the choice of frame: if MATH is another frame of MATH then MATH for MATH with MATH and MATH. Thus MATH for MATH, and setting MATH we readily compute that MATH . Thus MATH where the last equality follows since MATH and MATH are collinear and so ha... |
math/0003096 | Possibly after a NAME transformation, we may assume that MATH lie on a MATH so that their cross-ratio lies in MATH. Write MATH. Then MATH and, writing MATH, we see that MATH where MATH is the usual complex cross-ratio which is real if and only if the MATH are concircular. |
math/0003096 | The points MATH lie on a MATH-sphere or plane and so, after a NAME transformation, may be taken to lie on a copy of MATH where, as in the proof of REF , all NAME algebra cross-ratios reduce to complex cross-ratios. One now solves MATH to obtain, in turn, MATH and then checks that the remaining two equations hold: a tas... |
math/0003096 | We must check that MATH. However, from REF, we know that MATH whence MATH . Finally, MATH so that MATH since MATH. |
math/0003096 | Consider the configuration of REF surfaces obtained from REF starting with MATH and MATH. REF tells us that MATH and MATH while, from REF , we have MATH . Now, an obvious scaling symmetry of the cross-ratio gives MATH so that MATH . We conclude that MATH, that is, MATH so that, by REF , MATH is a generalised MATH-surfa... |
math/0003096 | Let MATH and contemplate the power series expansion of MATH: MATH with MATH. The twisting and reality REF force MATH whence MATH . Moreover, MATH so that MATH. Thus MATH has a simple pole at MATH if and only if all MATH for MATH which is the case precisely when MATH for some MATH. |
math/0003096 | We have MATH and differentiating with respect to MATH gives MATH that is, MATH. Now view MATH as a map MATH with NAME - NAME form MATH. Then, for MATH and MATH, we have MATH . The NAME - NAME equations for MATH give MATH . However, MATH since MATH for all MATH so we are left with MATH that is, MATH. |
math/0003096 | If MATH then MATH is holomorphic on MATH and so is constant. Moreover MATH whence MATH. |
math/0003096 | If REF holds then MATH . Conversely, if MATH then MATH so we can write MATH with MATH. Now put MATH and MATH. The uniqueness assertion is proved as in REF . |
math/0003096 | Set MATH and let MATH be its NAME - NAME form. By REF , we must show that MATH has a simple pole at MATH. However, in a punctured neighbourhood of MATH, we have MATH with MATH so that MATH . Since MATH is holomorphic at MATH, we immediately conclude that MATH has the same pole at MATH as MATH. Finally, MATH so that MAT... |
math/0003096 | Write MATH as a power series: MATH . Comparing coefficients of MATH, we see that MATH is holomorphic at zero if and only if CASE: MATH (this is the coefficient of MATH); CASE: MATH (these are the components of the coefficient of MATH). Now observe that MATH if and only if MATH and then, since MATH, MATH whence MATH van... |
math/0003096 | For the first assertion we must establish the reality condition for MATH and there are two cases. First, if MATH, we must show that MATH. However, in this case, MATH and MATH so this follows immediately. When MATH, we must show that MATH and, in this case, we have MATH while MATH . Thus MATH as required. The second ass... |
math/0003096 | Let MATH be such that MATH. The first part of REF assures us that MATH so all we need do is see that MATH given by REF defines an element of MATH. It is clear that MATH has the reality and twisting conditions as it is a product of maps with these conditions so the only issue is that of holomorphicity and invertibility ... |
math/0003096 | We have MATH and differentiating gives MATH whence REF. Further, MATH whence REF. For the second part, recall that under the isomorphism MATH, MATH acts on MATH by MATH. We must therefore show that, with MATH, we have MATH for MATH. For MATH, MATH anti-commutes with both MATH and MATH and so commutes with MATH. Again, ... |
math/0003096 | Since MATH, we have that MATH is holomorphic near MATH and so we may apply REF with MATH to conclude that MATH and, further, that MATH is holomorphic and invertible at MATH. Similarly MATH and MATH is holomorphic and invertible at MATH. Now contemplate MATH . Looking at the left hand side, we see that this expression i... |
math/0003096 | Multiply REF by MATH to get MATH and use REF to get MATH . Rearranging this and using REF yields MATH whence MATH . Temporarily denote by MATH the common value in REF. From the left hand side, we see that MATH is holomorphic except possibly at MATH while the right hand side tells us that MATH is holomorphic except poss... |
math/0003097 | CASE: It is clear the MATH. Since MATH are closed, MATH if and only if MATH. If MATH are closed ideals, and if MATH, MATH, then MATH, hence MATH. Thus the triangle inequality holds. CASE: Obvious. CASE: Let MATH, let MATH be monomial ideals, and suppose that either MATH or MATH. In the first case, there is a MATH such ... |
math/0003097 | Using the results of CITE, it is straight-forward to show that if MATH are lfg ideals such that MATH, then for any term-order MATH, MATH. Hence the first result follows. It is immediate that the identity REF implies continuity of REF. For all term orders, the LHS of REF is included in the Right-hand side, so we need to... |
math/0003097 | We put MATH . Then the MATH's are clopen ideals which form a fundamental system of neighbourhoods of zero. It follows CITE that for any subset MATH, the closure is given by MATH . Hence if MATH and MATH, then MATH where the crucial inclusion MATH is proved as follows. If MATH for all MATH, then MATH for all MATH, hence... |
math/0003097 | If we identify monomial ideals with their characteristic functions, and write MATH for intersections of ideals, and MATH for sum of ideals, then MATH and MATH correspond to component-wise minimum and maximum, and REF to the identity MATH where the sum is component-wise. The LHS of REF is always defined; for the Right-h... |
math/0003097 | We have that MATH, and that MATH. Hence the result follows from REF, once we have proved that that MATH for all MATH. But MATH, hence by NAME inversion MATH . |
math/0003097 | MATH . |
math/0003097 | It follows from CITE that MATH for MATH, and that MATH for MATH. By CITE, MATH whenever MATH. |
math/0003097 | Suppose that MATH is a monoid ideal in MATH, then MATH is the characteristic function of MATH. This is an order ideal, that is, if MATH and MATH . , then MATH. It follows that the set of MATH's is the set of MATH such that CASE: MATH, CASE: If MATH and MATH then MATH. Since MATH, the result follows by NAME inversion. |
math/0003097 | From REF we have that MATH is the reduced NAME characteristic of some simplicial complex on MATH vertices. CITE showed that the absolute value of the reduced NAME characteristic of a simplicial complex on MATH vertices is MATH. |
math/0003097 | Let MATH be a permutation of MATH. Define MATH, and MATH. We let MATH act on monoid ideals in MATH in the obvious way. Then MATH and MATH are fix-points for the action of MATH on MATH, and MATH for all monoid ideals MATH. Hence MATH . Let MATH be the support of MATH, that is, MATH, and let MATH be a permutation which t... |
math/0003097 | For each total degree MATH, there are only finitely many multi-degrees in MATH of total degree MATH. Thus MATH and MATH are equivalent. The equivalence of MATH and MATH is parallel to REF and is proved in the same way (see CITE). |
math/0003097 | Write MATH . It is immediate that MATH is lfg if and only if MATH, which occurs precisely when every MATH is a polynomial. We note that if MATH has cardinality MATH, and the minimal and maximal total degree of elements in MATH is MATH and MATH, respectively, then MATH. Clearly, the terms of MATH contributing to MATH in... |
math/0003097 | This follows from REF . |
math/0003097 | Suppose that MATH, where MATH. Fix a total degree MATH. By the definition of the total degree filtration topology, there exists a MATH such that MATH for all MATH. Since for all MATH, MATH is a polynomial, this is true for MATH, as well. |
math/0003097 | We assume for simplicity that MATH. Suppose that MATH in MATH. Fix an integer MATH, and choose a MATH such that MATH for MATH. Thus for MATH we have that the MATH coefficient of MATH and MATH coincides. This shows that MATH. To show that this map is clopen, we pick a basic clopen subset MATH, where MATH is a positive i... |
math/0003097 | Obvious. |
math/0003097 | This follows from the previous Lemma and from REF . |
math/0003097 | By the previous Lemma, the set of characteristic functions of lfg monoid ideals is a closed subset of MATH. By REF , the mapping MATH is a closed mapping, hence MATH is a closed subset of MATH. From REF we have that MATH, hence it is closed in there. |
math/0003097 | By the previous lemma, MATH if and only if MATH. Since the endomorphism given by multiplication with a fixed element in a topological ring is continuous, MATH . If MATH, then fixing a total degree MATH, we get that there exists a MATH such that MATH for MATH. It then follows that MATH for MATH. The converse also holds.... |
math/0003097 | From REF we know that MATH . Then REF gives that MATH hence, using REF , we have that MATH . It is clear that MATH . As we remarked above, a MATH-homogeneous ideal have the same MATH-graded NAME series as its initial ideal, so MATH hence MATH . |
math/0003097 | The inclusions are obvious. To see that the strict ones are indeed strict, consider the following examples: MATH . |
math/0003097 | We claim that there are positive integers MATH such that if MATH is a finitely generated monomial ideal generated in degrees MATH satisfying REF, then MATH with MATH for MATH. Assuming the claim, it is clear that the total degree of MATH is MATH, since this is a bound of the lcm of all the generators. To establish the ... |
math/0003097 | Let MATH, where MATH is a lfg monomial ideal. Let MATH denote the ideal generated by everything in MATH of total degree MATH. Since the maximal degree of a lcm of the generators of degree MATH is MATH, it follows from REF that MATH. |
math/0003097 | Let MATH, and let MATH. Then MATH in MATH, with respect to the MATH-adic topology. Let MATH, and choose MATH such that for MATH, MATH. Then REF shows that there is a monoid ideal MATH in MATH with MATH as its MATH-graded NAME numerator. |
math/0003097 | From REF we get that all polynomial MATH-graded NAME numerators can be obtained from ideals generated in finitely many variables. To prove the second assertion, we note that if MATH is lfg, MATH, and MATH is the ideal generated by everything in MATH of degree MATH, then MATH, and since MATH is finitely generated, MATH ... |
math/0003097 | It follows from a well-know classification by NAME (see CITE) that the set of (generating functions of) NAME functions of homogeneous quotients of polynomial rings with finitely many indeterminates is MATH . The function MATH can be realised as the NAME function of a quotient of MATH with a monomial ideal; we are of co... |
math/0003099 | Since the pullback of the scalar curvature to MATH is MATH, the hypothesis of constant scalar curvature is equivalent to MATH. Now, by the structure equations MATH so MATH implies that MATH vanishes identically. However, if MATH vanishes identically, then MATH, so that the curvature tensor is parallel. Thus, the struct... |
math/0003099 | Assume that the NAME structure is locally a nontrivial product. Then for some MATH, there is a MATH-subbundle MATH on which MATH is blocked in the form MATH where MATH takes values in MATH and MATH takes values in MATH. This forces MATH to be blocked in the corresponding form MATH . The vanishing of the off-diagonal bl... |
math/0003099 | Since the exterior derivatives of REF are identities, NAME 's conditions (that is, his generalization of the NAME conditions) are satisfied for these equations as structure equations of a coframing. Thus, by REF (see the Appendix), for any MATH, there exists a real-analytic manifold MATH of dimension MATH on which ther... |
math/0003099 | Consider any MATH. Act by an element MATH so as to reduce to the case where MATH is diagonal and its (real) eigenvalues are arranged in decreasing order down the diagonal. If there are integers MATH so that MATH, suppose that MATH is a maximal unbroken string with this property. Then the stabilizer of MATH in MATH will... |
math/0003099 | Let MATH satisfy the assumptions of the theorem, let MATH be the unitary coframe bundle, with canonical forms MATH and MATH, and let MATH be the structure function. Because MATH is simply-connected and the NAME structure MATH is real-analytic, any symmetry vector field of the structure defined on a connected open subse... |
math/0003099 | Because all the integers involved are invariant under the action of MATH, it suffices to prove this formula in the case that MATH lies in MATH. Maintaining the notation introduced above, this means that MATH where MATH takes values in MATH for MATH and has all of its entries equal to zero except possibly the top one, w... |
math/0003099 | Define REF-forms MATH and MATH . (The indexing is determined by `scaling weight' considerations.) The MATH are visibly MATH-semibasic, but they are also invariant under the MATH-action on MATH. Thus, they are the MATH-pullbacks of well-defined REF-forms on MATH. Consequently, they will, by abuse of language, be treated... |
math/0003099 | A computation like that done in the proof of REF shows that, for MATH, MATH . Using this identity, a calculation analogous to REF yields MATH . Thus, according to REF, each MATH is the representative function of a vector field MATH on MATH whose local flow is holomorphic. Letting MATH be any vector field on MATH so tha... |
math/0003099 | If MATH, then, in particular, MATH is constant, so MATH has constant scalar curvature. By REF , MATH is locally homogeneous, so all of the eigenvalues of MATH are constant. (By REF , there are at most two distinct eigenvalues.) In this case, MATH can be taken to be empty. Suppose from now on that MATH. Technically, I s... |
math/0003099 | Let MATH have roots MATH, counting multiplicity, and set MATH . Let MATH consist of the coframes MATH for which MATH. This MATH is a bundle over MATH with structure group MATH, where MATH is the group of unitary matrices commuting with MATH. All calculations will now take place on MATH. Adopt the index range convention... |
math/0003099 | The formula for MATH follows directly from REF, so the formula for MATH will follow from the equivalent statement MATH and this is what will be proved. By REF, the generic orbit of the symmetry pseudo-groupoid has codimension equal to MATH. By REF , the orbits of the symmetry pseudo-groupoid in MATH (which is open and ... |
math/0003099 | Substituting MATH into REF yields MATH . Since MATH and MATH on MATH, it follows that MATH for MATH holds on MATH. Equivalently, for every MATH, the polynomial MATH has an even number of real roots (counted with multiplicity) greater than MATH and an odd number of real roots (counted with multiplicity) in each open int... |
math/0003099 | This will be a matter of checking cases. First, some generalities. Given polynomials MATH and MATH satisfying the hypotheses of the proposition and a MATH lying in a possible momentum cell MATH for MATH, define MATH. By hypothesis, all the roots of MATH are real and are roots of MATH as well. Define MATH . Let MATH, an... |
math/0003099 | It has been established that MATH and MATH and the momentum cell MATH are invariants of the analytically connected equivalence class. Moreover, by REF , every point of MATH lies in the image of the reduced momentum mapping of some NAME structure. To prove REF , it will thus suffice to show that any two NAME structures ... |
math/0003099 | Fix MATH. By REF , to prove that MATH is surjective it suffices to show that if MATH is any NAME structure containing a MATH with MATH, then MATH is a subset of MATH. Consider any MATH and choose a smooth path MATH with MATH and MATH. Choose a MATH and let MATH be the parallel transport of MATH along MATH. Thus MATH wh... |
math/0003099 | Recall the bundle MATH that was introduced in the proof of REF and the notation introduced there. On MATH, the matrix MATH is diagonal and MATH for MATH. The structure equation for MATH then becomes MATH . By REF , MATH so REF can be written in the form MATH . In other words, MATH . Now suppose that MATH lies in MATH a... |
math/0003099 | Suppose that MATH is connected and complete, with characteristic polynomials MATH and MATH but that its reduced momentum mapping takes values in an unbounded momentum cell MATH. Let MATH be the corresponding spectral product. The unboundedness of MATH implies that MATH is either MATH or MATH where MATH is the largest r... |
math/0003099 | In view of REF and the remark above, what has to be shown is that REF cannot occur for a complete NAME manifold when MATH. This will involve an interesting examination of the fixed points of the flow of the canonical torus action. Thus, suppose, to the contrary, that MATH is a complete NAME structure with MATH and and ... |
math/0003099 | A compact NAME manifold is necessarily complete and its reduced momentum image is necessarily compact. REF , and the fact that only REF has a compact momentum cell imply that MATH is impossible. When MATH, the momentum mapping is constant and the metric is therefore locally homogeneous, so that, by REF , its simply-con... |
math/0003099 | Return to the structure equations on MATH that were introduced in the proof of REF . Since MATH for MATH while MATH for MATH, it follows that MATH is nonvanishing on MATH and hence on MATH. REF can thus be written as MATH . Let MATH be a MATH-leaf, and let MATH be the bundle MATH, which is a MATH-bundle over MATH. By t... |
math/0003099 | Before beginning the proof, it will be useful to establish the following fact. If MATH is any real-analytic, complete metric on a manifold MATH and MATH is a connected, totally geodesic submanifold of some dimension MATH, then MATH can be `completed': There exists a MATH-manifold MATH and a totally geodesic immersion M... |
math/0003099 | First, it will be useful to take a different basis for MATH. Recall that the functions MATH are constant linear combinations of the functions MATH and vice versa. By REF , MATH for MATH, so the vector fields MATH defined by MATH for MATH are also a basis of MATH. Let MATH be a MATH-leaf. The vector fields MATH are a ba... |
math/0003099 | The structure of the proof will be as follows: I will first assume that I have a complete NAME metric that is not locally homogeneous and consider the induced metric on a completed MATH-leaf. Knowing by earlier discussions that the only possibility for this is in REF -MATH, I will use knowledge of the form of MATH and ... |
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