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math/0003087
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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.
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math/0003087
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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 equivalence follows from REF.
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math/0003087
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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 factor (compare CITE) with the dimension function MATH where MATH . With the same argument as above there is again a projection MATH, such that MATH. Then MATH and MATH (MATH).
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math/0003087
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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.
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math/0003087
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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 cyclic and separating vector corresponding to to non-singular operator MATH (MATH is the modular operator corresponding to MATH), and MATH has the same eigenvalues and multiplicities like MATH (see the proof of REF). This means that they have the same unitary invariants in the type MATH NAME factor MATH . According to REF there is a unitary MATH such that MATH and MATH commutes with MATH (Since MATH and MATH both are modular objects with modular conjugation MATH we have MATH and MATH). Now we are in exactly the same situation as in the proof of CITE and can show like there that there is also a unitary MATH such that MATH commutes with MATH, MATH, and MATH, whence MATH is a solution with the stated properties.
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math/0003087
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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.
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math/0003087
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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.
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math/0003087
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This follows immediately from REF.
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math/0003087
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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: In MATH we chose according to REF an ONB MATH. Setting MATH REF we see that MATH is an ONB in MATH. Define now the following conjugation MATH in MATH REF MATH . Then we can calculate with MATH: MATH . By linear continuation follows for MATH . CASE: Now we can set MATH. With this definition we deduce from REF MATH and setting MATH from REF MATH . Also we see from REF MATH and the proof is complete.
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math/0003087
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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.
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math/0003087
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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.
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math/0003091
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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 subtree, no global fixed point, and exactly one orbit of edges. We seek to construct a MATH-action on a (perhaps another) tree with the above properties such that every maximal parabolic subgroup of MATH fixes a vertex. Fix an edge MATH of MATH and denote its stabilizer in MATH by MATH. Let MATH be the maximal virtually nilpotent subgroup of MATH that contains MATH. First, note that any maximal parabolic subgroup MATH other than MATH fixes a vertex of MATH. (Indeed, every edge stabilizer of MATH is a conjugate of MATH, in particular, it lies in a conjugate of MATH, hence it must have trivial intersection with MATH. So, unless MATH fixes a vertex of MATH, we get that MATH splits over the trivial group which is impossible because MATH is noncyclic virtually nilpotent, and hence one-ended.) In particular, if MATH is not parabolic, then the MATH-action on MATH satisfies the desired properties. Now assume that MATH fixes a vertex MATH of MATH. Then the fixed-point-set of MATH contains the segment joining MATH and a vertex of MATH. In particular, MATH lies in MATH, the stabilizer of an edge MATH adjacent to MATH. Since MATH is conjugate to MATH, it is virtually nilpotent. Hence MATH lies in a maximal parabolic subgroup containing MATH which is MATH. Thus the splitting of MATH over MATH satisfies the desired properties. It remains to consider the case when MATH is parabolic and MATH does not fix a vertex of MATH. Let MATH be the unique MATH-invariant minimal subtree of MATH. Let MATH be an element of MATH such that MATH is an edge of MATH (such a MATH exists since there is only one orbit of edges). Then the stabilizer of MATH in MATH is MATH. Note that the group MATH (which is the stabilizer of MATH in MATH) is infinite because otherwise MATH splits over a finite group MATH and MATH cannot since noncyclic nilpotent groups are one ended. Since MATH, we conclude that MATH is infinite, therefore MATH and, hence, MATH because any maximal parabolic subgroup is equal to its normalizer. The above argument has several implications as follows. MATH must be an edge of MATH. (Indeed, using that there is only one orbit of edges we find MATH such that MATH is an edge of MATH. By above MATH must belong to MATH and since MATH stabilizes MATH, MATH is an edge of MATH.) MATH is equal to the setwise stabilizer of MATH. (Indeed, if MATH, than MATH takes the edge MATH of MATH to an edge of MATH. Hence, again MATH.) MATH . Any edge of MATH lies in a unique MATH-image of MATH. (Indeed, any edge of MATH is MATH-equivalent to MATH, so it is an edge of a tree that is a MATH-image of MATH. Uniqueness of this tree is deduced as follows. Suppose, arguing by contradiction, that two trees that are in MATH-image of MATH share an edge. Applying an element of MATH we can assume these two trees are MATH and, say, MATH for some MATH. This common edge is of the form MATH for some MATH, so by above MATH. Hence, MATH is the common edge of MATH and MATH. So MATH is MATH-image of an edge of MATH, or equivalently, MATH is an edge of MATH. By the same argument, MATH, so MATH which contradicts to MATH.) Let MATH be the set of MATH-images of MATH. Construct now a graph MATH with the vertex set MATH where we deem two vertices MATH adjacent iff MATH, MATH, and MATH. First, prove that MATH is a (simplicial) tree. Indeed, to show that MATH is connected it suffices to find an arc that joins two arbitrary vertices MATH. The shortest arc that joins MATH and MATH in MATH can be uniquely written in the form MATH where MATH, MATH and the arc MATH is contained in the tree MATH (any edge of MATH lies in a unique tree from MATH). Then the arc MATH joins MATH and MATH in MATH. Second, show that MATH has no circuits. Indeed, suppose MATH is a circuit in MATH with MATH. Let MATH be the arc in MATH that connects MATH to MATH. Then MATH is a circuit in MATH, a contradiction. The group MATH acts on MATH without edge inversions. Any maximal parabolic subgroup now fixes a vertex. In particular, MATH is the stabilizer of MATH. Since MATH is not virtually nilpotent, there exists a vertex MATH of MATH whose stabilizer MATH is not a subgroup of MATH. (If the stabilizers of two adjacent vertex groups of MATH were subgroups of MATH, then, since MATH has only one MATH-orbit of edges, the groups MATH and MATH would lie in a conjugate of MATH. Hence MATH would lie in the conjugate of MATH.) Now look at the edge MATH that joins vertices MATH and MATH. The stabilizer MATH of MATH is a subgroup of MATH, so it is not equal to MATH. If MATH, then MATH has to stabilize MATH as well, hence MATH fixes a point of MATH which is not the case. Thus, MATH is not equal to the stabilizers of MATH and MATH. Now if any vertex of MATH is MATH-equivalent to MATH, then MATH has one orbit of edges and MATH is the desired splitting. Otherwise, MATH has two orbits of vertices represented by MATH and MATH and hence MATH has two orbits of edges MATH, MATH. This defines a splitting MATH over MATH as needed.
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math/0003091
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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. Now look at the NAME sequence for homology with MATH-coefficients. MATH . The map MATH can be written as MATH where MATH is the covering onto MATH, and MATH is the homeomorphism onto MATH. Show that MATH is injective. This is clear if the cohomological dimension of MATH is MATH. Otherwise, by the exact sequence of the pair MATH it suffices to show that MATH which is true by CITE. (As stated, CITE only applies when the complement of MATH in MATH is connected. However, if MATH is nonconnected, it has two noncompact components each equal to MATH and the result again follows since by the relative NAME sequence MATH is a sum of two copies of MATH.) By exactness MATH and we get a contradiction with the fact that MATH is homotopy equivalent to the closed MATH-manifold MATH. Similarly, if MATH, then we get the following NAME sequence with MATH-coefficients CITE. MATH . The map MATH can be written as MATH where MATH is the homeomorphism onto MATH, and MATH is the covering of MATH onto MATH. It remains to show that MATH is injective which is clear if the cohomological dimension of MATH is MATH. Otherwise, look at the exact sequence of the pair MATH with MATH-coefficients. Again, by CITE MATH, hence the inclusion MATH induces an injection on MATH-homology. Therefore, MATH is nonzero as wanted an we get a contradiction as before.
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math/0003091
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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 MATH so that MATH for some MATH. By passing to the appropriate root we can assume MATH is primitive. There always exists a sequence MATH that converges to MATH. Then both MATH and MATH converge to MATH. Hence MATH for large MATH by CITE. Since MATH is primitive, MATH so MATH as wanted.
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math/0003091
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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 equivariant pointed NAME topology. Now since the sectional curvature is nonpositive, MATH and MATH where MATH is the infimal displacement of the point MATH by the isometries of MATH. It follows easily from the definition of equivariant NAME convergence that MATH converges to MATH so we get a contradiction.
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math/0003091
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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 is a smooth manifold with a MATH . Riemannian metric MATH of NAME curvature of MATH is within MATH. Let MATH be the NAME 's smoothing of MATH CITE. Since MATH is a complete Riemannian manifold of pinched negative curvature which is homotopy equivalent to MATH, CITE implies that MATH has finite volume so CITE implies that MATH is the interior of a compact manifold with boundary and the number of boundary components is the same as the number of ends of MATH (or, alternatively, the number of maximal parabolic subgroups of MATH). Since MATH converge to MATH in unpointed MATH-topology, MATH tends to MATH. Also MATH is NAME with NAME constant approaching MATH, hence the volumes of MATH are uniformly bounded CITE. Thus the volume of MATH is finite. Being a quotient of MATH, the manifold MATH also has finite volume so MATH has finite index in MATH. Hence MATH by REF.
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math/0003091
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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 diameters of the MATH-thick parts of MATH are MATH; thus, for all large MATH, the MATH-thick part of MATH contains a point that lies in MATH. Furthermore, by continuity of injectivity radius MATH contains a point of injectivity radius MATH for all large MATH. Since MATH, the MATH-thick part is connected, so for all large MATH, MATH contains a point of injectivity radius MATH. This point subconverges to a point in MATH of injectivity radius MATH which is a contradiction. Now a lower bound on the injectivity radius (and hence on the diameter of the MATH-thick part) is provided by the NAME lemma CITE. As for an upper bound assume there is a sequence of manifolds MATH in with MATH and MATH. Pass to a subsequence so that MATH converges to MATH. Since the MATH-thick part of MATH is compact, the injectivity radius of MATH is bounded above which contradicts the continuity of the injectivity radius.
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math/0003091
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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 between MATH and MATH take the MATH-thick part of MATH . NAME close to the MATH-thick part of MATH. Hence for all large MATH the MATH-thick part of MATH is contained in the MATH-image of the MATH-thick part of MATH. Now let MATH be so small that MATH-thin part of MATH consists only of cusps. Thus, the MATH-image of any MATH-tube of MATH lies in a MATH-cusp of MATH. Hence if MATH be a closed geodesic in a tube of MATH, then MATH represents a parabolic element. By algebraic reasons, MATH implies that MATH takes parabolics to parabolics. So MATH represents a parabolic element which is a contradiction.
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math/0003091
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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 a splitting which is impossible. So assume MATH is uniformly bounded. Let MATH the the NAME fundamental domain for the action of MATH on MATH. There exists a sequence MATH such that MATH. Then MATH subconverges in the pointwise convergence topology. Suppose first that MATH is precompact, so that passing to subsequence we can assume that MATH converges to MATH. Then MATH also subconverges in the pointwise convergence topology, in other words, passing to subsequence, we deduce that MATH converges in MATH for any MATH. Since MATH is a closed subgroup the limit lies in MATH. So MATH converges in MATH. Since the space MATH is discrete, MATH are all equal for large MATH. So MATH lie in the same conjugacy class for large MATH, a contradiction. If MATH is not precompact, then passing to subsequence we can assume MATH converges a parabolic fixed point. (The closure of MATH at infinity is just finitely many parabolic fixed points.) Choose a MATH-invariant set of mutually disjoint horoballs. Passing to subsequence, we assume that MATH lies in one horoball MATH for all MATH. Since MATH is not virtually nilpotent, for each MATH there is a generator MATH such that the horoballs MATH and MATH are disjoint. As MATH is finite we can pass to to subsequence so that MATH and MATH are disjoint for all MATH and some MATH. Since MATH is uniformly bounded, the distance between MATH and MATH is uniformly bounded. On the other hand, this distance has to converge to infinity because it is bounded below by the distance from to MATH to the horosphere MATH. This contradiction completes the proof.
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math/0003091
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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 an index MATH subgroup of MATH for each MATH. Same proof as before gives that MATH are all equal for large MATH. In particular, MATH and MATH are isometric for all large MATH. But there exists a MATH-sheeted cover MATH, hence MATH. Thus, MATH, a contradiction.
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math/0003091
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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 cusp under consideration. We denote by MATH, MATH, MATH the lifts of MATH, MATH, MATH to the universal cover which are in bounded distance from a horosphere about MATH. Let MATH be a geodesic asymptotic to MATH with MATH and assume that MATH intersects MATH and MATH in the points MATH and MATH, respectively. Note that MATH and MATH. Since MATH, one can find MATH such that MATH. Now MATH. Thus, by REF MATH . Similarly, since MATH, one can find MATH with MATH. Again, MATH so MATH . So MATH. Hence, by the triangle inequality MATH . Furthermore, if MATH, then MATH is ``closer" to infinity than MATH as desired.
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math/0003091
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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 bounded above and the result follows because the projection MATH is volume non-increasing. We fix a MATH, and an arbitrary MATH-cusp of MATH. Now we seek to obtain a uniform upper bound on MATH. Let MATH so that MATH is ``closer" to infinity than MATH. Let MATH be the distance between between MATH and MATH; note that MATH. By above, the volume enclosed between MATH and MATH is bounded above on MATH by a constant MATH depending only on MATH. The same is then true for the volume enclosed between MATH and MATH which is equal to MATH where MATH is the quotient of a horosphere at MATH-level, and MATH, MATH. By REF, pushing along radial vector field gives a MATH-Lipschitz map MATH. Thus, MATH (in this proof we always equip MATH with the Riemannian metric induced by the inclusion into MATH). Hence by the NAME 's theorem (which applies since the NAME function is a MATH-Riemannian submersion) we have MATH . Thus MATH is uniformly bounded above over MATH. Now pushing along radial vector field gives a MATH-Lipschitz diffeomorphism MATH so MATH. Hence MATH and we get a uniform upper bound for the volume of the MATH-cusp under consideration. Thus, we get a uniform upper bound on the volume of MATH.
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math/0003091
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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 MATH converges to MATH in MATH-topology uniformly on compact subsets. By CITE we can take MATH. As we shall prove below, for each small enough MATH and large enough MATH, there exists a diffeomorphism MATH that is equal to MATH when restricted to the MATH-thick part of MATH. Note that the continuity of injectivity radius implies that, given MATH, the maps MATH are defined on the MATH-neighborhood of the MATH-thick part of MATH for all MATH and some positive integer valued function MATH. Now the diffeomorphism MATH enjoys the desired properties. Thus, it remains to construct MATH. Use REF to find a MATH such that MATH-thin part of each MATH consists of cusps; we assume MATH. Fix a cusp of MATH and the corresponding cusps of MATH. Using REF, we can make so MATH is so small that MATH is closer to infinity than MATH, and MATH; MATH is closer to infinity than MATH, and and MATH. (Here subindex MATH indicates that the quotients of the horospheres lie in a cusp of MATH.) Let MATH be the radial vector field defined on the MATH-thin part of MATH, and let MATH be the MATH-neighborhood of the region between MATH and MATH. Using exponential convergence of geodesics, one can easily see that for any MATH there exists MATH, depending only on MATH and independent of MATH, such that for any MATH and any MATH lying in the same MATH-cusp of MATH with MATH, the angle at MATH formed by MATH and the tangent vector to the geodesic segment MATH is MATH. Fix MATH and fix the corresponding MATH. Assume MATH is large enough so that the embeddings MATH are defined on the MATH-neighborhood of MATH. By continuity of injectivity radius the domains MATH converge to some compact set in NAME topology and we can find a smooth domain MATH which is NAME close to the set and satisfies MATH for all large MATH. Now use MATH to pullback all metrics to MATH. We want to show that the region bounded by MATH and MATH is diffeomorphic to MATH. It suffices to produce a MATH nowhere vanishing vector field on MATH that is transverse to both MATH and MATH. We shall construct such a vector field as a controlled approximation of MATH, the MATH-pullback of the radial vector field MATH. Choose a harmonic atlas MATH on MATH as in CITE in which the MATH-pullbacks of MATH converge to MATH in MATH-topology. Fix a partition of unity associated with the atlas. Let MATH be a point in the same cusp and MATH. Now we construct a MATH vector field MATH on MATH by defining it on each chart neighborhood MATH and then gluing via the partition of unity. Look at the MATH-neighborhood of MATH equipped with the metric MATH. The preimage of MATH under MATH is the disjoint union of copies of MATH. Pick a copy closest to the origin, join each of its points to the origin by rays, and then project the rays to MATH via MATH. Now the tangent vectors at the endpoints of the obtained geodesic segments joining MATH with points of MATH form a vector field on MATH. Gluing these local data via the partition of unity gives MATH. Note that by construction MATH is a nowhere vanishing vector field such that the angle formed by MATH and the exterior normal MATH to MATH or MATH is within MATH. Now on each chart MATH is a solution of the geodesic equation. Since metrics converge in MATH topology, NAME symbols converge in at least MATH topology, so by standard ODE results CITE MATH converges in MATH-topology to some MATH vector field MATH. So the angle, measured in the metric MATH, formed by MATH and MATH is within MATH for all MATH large enough, and some MATH. Now a standard differential topology arguments implies that the region between MATH and MATH is diffeomorphic to MATH as needed. Finally, we are ready to define MATH. Let MATH be the compact manifold obtained from MATH by chopping off cusps along all surfaces MATH; we think of MATH as a bounded domain in MATH. Define MATH, and define MATH as the following composition. First, map MATH diffeomorphically to the interior of MATH by a map which is the identity on the MATH-neighborhood of MATH. Then map MATH to MATH by MATH. Next use the above argument to map MATH diffeomorphically onto MATH by a map which is the identity on the MATH-neighborhood of MATH. Last, map MATH diffeomorphically to MATH, again keeping MATH fixed.
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math/0003091
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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 manifold of NAME curvature MATH. By CITE MATH is isometric to the real hyperbolic space and we are done. (Alternatively, one can repeat the argument below appealing to CITE to deduce that MATH is a MATH metric.)
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math/0003091
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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-Kähler. Any quaternionic-Kähler is NAME so MATH is smooth NAME manifold and convergence is in MATH topology. Hence, sectional curvatures of MATH converge to sectional curvatures of MATH, so MATH is quarter-pinched, and we are done by CITE. Second, assume that each MATH is NAME. Adapting the argument in CITE to negative curvature, we deduce that the holomorphic sectional curvatures of MATH converge to MATH uniformly on MATH. Look at the "curvature MATH-tensor" MATH for NAME metric of holomorphic sectional curvature MATH defined in terms of MATH and almost complex structure MATH of MATH CITE. Since sectional curvature can be written in terms of holomorphic sectional curvature CITE and the curvature MATH-tensor can be written in terms of sectional curvature, the MATH-tensor MATH of MATH is getting close to MATH uniformly on MATH when MATH. Taking traces we conclude that NAME tensor of MATH is getting close to MATH (see CITE). Now MATH subconverges to a MATH-Riemannian metric on a finite volume manifold MATH hence MATH converges to zero. Then the proof of CITE implies that the limiting metric MATH is a weak solution of the NAME equation, hence MATH is a MATH . NAME metric. Also MATH has NAME curvature within MATH hence MATH and we are done by CITE.
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math/0003092
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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. By standard arguments there exists a MATH-linear morphism MATH which is onto the cokernel of MATH. This morphism extends to a MATH-equivariant morphism MATH, where the product bundle MATH has the product action of MATH; the MATH-action on the space of quaternions MATH is via right quaternionic multiplication. Given any MATH and MATH set MATH and extend by linearity. Perturbing the family of NAME operators MATH by the morphism MATH produces a MATH-equivariant epimorphism MATH . Through this we have represented MATH in MATH-equivariant NAME as the difference of the kernel bundle of MATH and the product MATH-bundle MATH. Considered as a complex bundle, the kernel bundle MATH splits as a sum MATH, for dimensional reasons, where MATH is a trivial complex bundle, and MATH is a MATH-bundle with MATH and MATH. In fact, MATH splits in the category of MATH-bundles over MATH, in such a way that MATH is a trivial MATH-bundle. To construct this splitting note that any vector bundle over MATH with fibre dimension greater than REF (over MATH) admits a nowhere vanishing section MATH. On any MATH-bundle MATH such a section MATH gives rise to a trivial MATH-invariant sub-bundle MATH of complex rank REF, spanned by MATH and MATH. Moreover, MATH has a MATH-invariant complement in MATH; the latter can be taken to be perpendicular to MATH with respect to some (compatible) hermitian inner product on MATH. For the case at hand we choose the standard hermitian structure on MATH and the MATH-inner product on the fibres of MATH. Denote the resulting MATH-equivariant isomorphism by MATH. The bundle MATH admits a structure of a quaternionic line bundle; we use this to construct a MATH-equivariant morphism MATH, injective everywhere except at MATH chosen points on MATH. Let MATH be a MATH-invariant subset with MATH elements; such exists for any MATH since MATH-action on MATH has fixed points (for example MATH). We choose a section MATH of the bundle MATH which vanishes only at the points of MATH and intersects the zero section transversely. Then the sections MATH and MATH (defined from MATH as above) endow MATH with a structure of a quaternionic line bundle over the complement of MATH. Dividing MATH by the square of its (quaternionic) norm produces a nowhere vanishing section MATH of MATH over the complement of MATH and this section MATH induces the required bundle morphism MATH. Note that close to any MATH, the norms of linear maps MATH are bounded below by some positive constant times distance from MATH to MATH. The morphisms MATH and MATH together define a MATH-equivariant morphism MATH which is injective on all the fibres except over the points of MATH where the kernels can be identified with a copy of MATH. We think of the pair MATH as the family of NAME operators associated to the stabilized NAME equations (which are defined below). Note that by construction of MATH and MATH, the associated family of NAME operators has nontrivial kernels only at the points of MATH, thus proving the last statement of the proposition. To finish the construction of the stabilized equations, we need to globalize the perturbation terms MATH and MATH. Let MATH be the MATH-orthogonal projection (where we treat MATH as the origin of the above affine spaces), MATH the orthogonal projection to the complement, and MATH the orthogonal projection to the kernel of MATH. The morphism MATH defines a map MATH given by MATH . Similarly, MATH gives rise to MATH which is well defined up to gauge change by MATH; it is completely determined by the appropriate choice of MATH. We define the stabilized NAME equations via a map MATH which is the sum of the original NAME map and the stabilization term given by MATH where MATH depends smoothly on the MATH-norms of MATH and MATH and on the norm of MATH in such a way that it is equal to REF in a small neighborhood of MATH and equal to REF in a slightly bigger neighborhood; notice that MATH is invariant under the action of the gauge group MATH as well as under the action of MATH. It is clear from the nature of the perturbation terms that the moduli space of solutions to the stabilized NAME equation MATH still contains the torus of reducibles MATH. This proves the proposition.
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math/0003092
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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 around MATH is given by a MATH-equivariant map MATH, where MATH corresponds to the harmonic REF-forms, MATH to the self-dual harmonic REF-forms, and the quaternions represent the kernel and the cokernel of the perturbed NAME operator at MATH. Note that the leading term of MATH is a quadratic polynomial map which we will make non-degenerate by a perturbation. Denote by MATH, MATH the quadratic parts of the components of MATH. In principle these maps from MATH can contain three sorts of terms: quadratic in the first or the second variable, or bilinear. Which terms really appear is determined by the MATH-equivariance. Recall that MATH acts on MATH by right quaternionic multiplication and on the spaces of forms by multiplication by MATH, whereas MATH acts by complex multiplication on MATH and trivially on the spaces of forms. This forces MATH to be quadratic in the second (quaternionic) variable and MATH to be bilinear. Note that MATH-equivariance imposes extra restrictions on these terms; clearly MATH. The second component satisfies MATH if we think of MATH as a linear map MATH. We choose a non-degenerate MATH-invariant quadratic map MATH (with the associated linear map an isomorphism, compare CITE) to perturb MATH. For all but finitely many MATH, the map MATH is non-degenerate in the above sense. Admissible perturbations of MATH are of the form MATH, where MATH is a MATH-linear map which anti-commutes with the MATH-action. The space MATH of such maps is REF-dimensional over MATH and its non-zero elements are isomorphisms. We choose the map MATH, MATH to be an isomorphism. Then for almost all MATH the map MATH, where we interpret MATH as a linear map MATH, is an isomorphism. Notice that MATH is itself an isomorphism for MATH; this follows from the construction of the linear perturbation MATH. Moreover, the norms of these linear maps are bounded from below by MATH for some positive MATH. This means that we can choose MATH small enough so that for MATH the perturbation term is dominated by the original (quadratic) map. The benefits of this perturbation are twofold; firstly, the only solutions to the perturbed equations close to MATH are the reducible ones. Secondly, for a generic MATH, the preimage of MATH under the perturbed NAME map consists of exactly one circle of solutions, hence the point MATH contributes MATH to the NAME invariant. Consider now a point MATH with MATH not MATH-fixed. Such a point has its MATH-image in MATH; to make the perturbation term MATH-equivariant in this case, we construct a MATH-equivariant perturbation at MATH and use the MATH-action to define the perturbation at its MATH-image. Given only MATH-equivariance for MATH and MATH in this case, the structure of these quadratic maps is not so restricted. Using additional properties of the NAME map, we still conclude that MATH is quadratic in the second variable and MATH is bilinear. However, the space of MATH-invariant quadratic polynomials MATH is four dimensional and MATH can be any MATH-linear map, so there is no canonical choice of a good perturbation. To gain the same control over the solution space as for MATH-fixed points, we endow the kernel and the cokernel with a quaternionic structure. The perturbation terms can then be constructed as above, using right multiplication by the quaternion MATH in place of the MATH-action. For the perturbation term MATH, the associated linear map is surjective and for all but finitely many MATH, the map MATH is an epimorphism in the above sense. The perturbation of the second component gives rise to an injective map MATH; again, for all but finitely many MATH the map MATH is a monomorphism. The remark about domination of the pertubation term MATH by MATH holds as above, and so do the conclusions about the solution space. We fix a small, generic MATH and define the perturbation term as a sum of terms localized near the points of MATH. For a point MATH define the perturbing map by MATH where MATH is the MATH-orthogonal projection and MATH is a MATH-valued function depending smoothly on the norms of the arguments (using MATH), that has support inside a small neighborhood of MATH (the projection of which by MATH is contained in the domain of the NAME map) and is equal to REF on a smaller neighborhood. The moduli space of solutions to the perturbed equations MATH still contains the torus of reducibles MATH and it is clear from above that this torus is isolated.
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math/0003096
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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 reflections in hyperplanes and inversions in hyperspheres: indeed, any Euclidean motion is a composition of reflections while a dilation is a composition of two inversions in concentric spheres. On the other hand, we have seen in REF that all these reflections and inversions are induced by reflections in MATH for MATH a space-like line. Such reflections are certainly in MATH and we are done.
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math/0003096
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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 MATH, for MATH, so that MATH. Applying the NAME Lemma to REF, we get, for each MATH, MATH with MATH. Thus, REF becomes, for fixed MATH, MATH from which we conclude that MATH whenever MATH and MATH. If we can choose MATH all distinct we quickly conclude that all MATH so that MATH: a contradiction. We must therefore have MATH and MATH with MATH and MATH not both zero. We now have MATH and MATH . The converse is more direct: let MATH and choose MATH an orthonormal basis of MATH with respect to MATH and set MATH. Thus MATH. Now MATH gives MATH while MATH forces MATH so that MATH is parallel to either MATH or MATH. Finally, MATH forces the second possibility to hold so that there is MATH such that MATH . Then MATH since MATH and similarly for MATH.
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math/0003096
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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 must have centre on the hyperplane orthogonal to MATH through MATH and so is stable under reflection in this hyperplane (which interchanges MATH and MATH). Moreover, there is a unique MATH-sphere (or possibly MATH-plane) of this kind whose tangent space at MATH is MATH. The tangent space to this sphere at MATH is therefore MATH which is tangent to MATH at MATH if and only if MATH.
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math/0003096
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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 constant, we have MATH where we have used REF to compute MATH. Thus we have two equations MATH where MATH is the anti-commutator in MATH. The first of these is our desired REF with MATH . Conversely, if REF holds, define MATH by REF so that REF holds and define MATH by REF. At each point MATH, MATH so that the enveloping sphere at MATH is defined by a MATH-plane lying in the hyperplane MATH. Moreover, reflection in this hyperplane permutes the light-lines spanned by MATH and MATH so that MATH where MATH is the corresponding NAME transformation. We will therefore be done if we can show that MATH is constant which, since REF holds by construction, amounts to establishing REF. However, differentiating REF gives MATH . REF follows immediately and the proof is complete.
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math/0003096
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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.
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math/0003096
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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 MATH is similar.
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math/0003096
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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.
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math/0003096
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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 of common umbilics prevents this possibility occurring for all MATH. So choose MATH and principal vectors MATH, orthogonal with respect to MATH, such that the tangential component of MATH is a non-zero multiple of MATH. Since MATH is conformal and NAME, MATH are orthogonal for MATH and principal for MATH so that MATH whence MATH . We therefore conclude that there are are functions MATH such that MATH . Since MATH and MATH are conformally equivalent, we get MATH and there are only two possibilities: either MATH which quickly gives MATH so that MATH are a NAME pair, or, MATH which, by REF , forces MATH to be non-full.
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math/0003096
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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 have the same adjoint action. Thus we are free to choose a convenient frame to establish the theorem. With MATH, we take MATH so that, by REF , MATH . Thus MATH, MATH and the vanishing of MATH is precisely the condition that MATH is a NAME pair of isothermic surfaces. Moreover, if MATH is isothermic with NAME transform MATH and MATH, we have MATH so that MATH .
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math/0003096
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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.
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math/0003096
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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 task best left to a computer algebra engine (compare CITE).
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math/0003096
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We must check that MATH. However, from REF, we know that MATH whence MATH . Finally, MATH so that MATH since MATH.
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math/0003096
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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-surface also.
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math/0003096
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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.
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math/0003096
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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.
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math/0003096
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If MATH then MATH is holomorphic on MATH and so is constant. Moreover MATH whence MATH.
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math/0003096
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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 .
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math/0003096
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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 MATH is based if and only if MATH is.
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math/0003096
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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 vanishes automatically. This leaves the term involving MATH. However, when MATH, we have MATH and MATH so that MATH is skew-symmetric. Thus, since MATH is MATH-dimensional, we have MATH giving MATH. Thus MATH is holomorphic at zero if and only if MATH. The invertibility now follows by applying this result to MATH.
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math/0003096
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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 assertion follows at once from REF : MATH which is holomorphic at MATH if and only if MATH is holomorphic at MATH. However, MATH is holomorphic at MATH with value MATH there so REF applies.
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math/0003096
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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 at MATH. However, holomorphicity at MATH follows at once from REF and then we get holomorphicity at MATH from the twisting condition: MATH .
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math/0003096
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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, using MATH, we have MATH . Similarly, we have MATH .
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math/0003096
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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 is holomorphic at MATH and, from the right hand side, we see that is is holomorphic at MATH. Thus it is holomorphic on MATH and so constant. Evaluating at MATH now gives MATH that is MATH .
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math/0003096
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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 possibly at MATH. We therefore conclude that MATH has poles at MATH only, that is, MATH so that we have factorisations MATH . However, for MATH, MATH so factorisations of this kind are unique (recall REF !) and, comparing with REF, we get MATH .
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math/0003097
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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 that MATH whenever MATH: since MATH, and similarly for MATH, it follows that MATH if and only if MATH for MATH. In the second case, there is a MATH such that MATH whenever MATH: since MATH, and similarly for MATH, it follows that MATH if and only if MATH for MATH.
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math/0003097
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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 prove that the reverse inclusion holds for the degree-reverse lexicographic term order. Let MATH be homogeneous of degree MATH; then the monomials in MATH are ordered as follows: first the ones in MATH, if any, then the ones in MATH, and so on. Let MATH, then MATH for MATH sufficiently large, and MATH otherwise. In the same way, MATH for sufficiently large MATH, and MATH otherwise. So MATH.
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math/0003097
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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, hence MATH. If MATH has a multiplicative inverse MATH, then MATH the inclusion MATH is proved as follows. Suppose that MATH for all MATH, then MATH for all MATH, hence MATH, hence MATH.
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math/0003097
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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-hand side to be defined, it is necessary and sufficient that MATH . If this holds, then denoting by MATH the cardinality of the finite subset MATH, REF becomes MATH a well-know binomial identity. To prove REF, note that MATH, hence from REF we get that MATH .
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math/0003097
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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 .
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math/0003097
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MATH .
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math/0003097
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It follows from CITE that MATH for MATH, and that MATH for MATH. By CITE, MATH whenever MATH.
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math/0003097
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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.
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math/0003097
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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.
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math/0003097
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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 takes MATH to MATH, MATH to MATH, and so on. Then MATH, and MATH hence the result follows by applying REF .
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math/0003097
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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).
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math/0003097
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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 REF have total degree MATH. Hence, if MATH is lfg, so that each MATH is a polynomial, then only the various lcm's of elements in the support of MATH may contribute to MATH. The number of elements in the support of MATH is thus MATH. Conversely, if MATH is not lfg, suppose that MATH are polynomials, but that MATH is not. Using REF we see that MATH receives contribution from a finite number of terms stemming from lcm's of elements in the support of MATH, and from the non-polynomial MATH. Thus MATH is not a polynomial.
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math/0003097
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This follows from REF .
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math/0003097
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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.
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math/0003097
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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 integer, and MATH. Then MATH, and this is a basic clopen set of MATH.
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math/0003097
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Obvious.
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math/0003097
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This follows from the previous Lemma and from REF .
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math/0003097
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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.
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math/0003097
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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. The last assertion follows immediately from the fact that MATH is continuous.
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math/0003097
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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 .
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math/0003097
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The inclusions are obvious. To see that the strict ones are indeed strict, consider the following examples: MATH .
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math/0003097
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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 claim, we note that MATH, and assume by induction that we have shown that MATH exist. We note that the minimal generators which affect MATH are those of degree MATH, which each contribute with MATH, and also MATH-tuples MATH with MATH, MATH, and with MATH, which each contribute MATH. If we pick MATH elements of MATH, MATH elements of MATH, et cetera, then for the resulting lcm to be of total degree MATH it is necessary that MATH and that MATH for all MATH; thus only finitely many MATH are relevant. We thus have that MATH where MATH, MATH, MATH for MATH, MATH is the number of non-zero entries in MATH. The symbol MATH denotes a finite interval MATH of integers, where MATH is the maximal numbers of lcm's of MATH elements of degree REF, drawn from a set of cardinality MATH, MATH elements of degree REF, drawn from a set of cardinality MATH, and so on, which have total degree MATH. For instance, if MATH and MATH then MATH, if MATH and MATH then MATH. These finite intervals are added using interval arithmetic, so that MATH. We can the deduce that MATH for some integers MATH. Putting MATH we have the desired bound.
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math/0003097
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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.
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math/0003097
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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.
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math/0003097
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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 hence MATH.
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math/0003097
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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 course free to use more variables, if we so desire. The first part of the theorem is therefore demonstrated. To prove the second part, we proceeds as follows. We know by REF that MATH is a closed subset of MATH. We have that MATH is a closed map REF , hence MATH is closed in MATH. Now, REF shows that the set of polynomial MATH-graded NAME numerators is dense in the set of all MATH-graded NAME numerators; since this latter set is closed, the second part of the theorem follows.
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math/0003099
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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 structure is locally symmetric. In particular, the eigenvalues of MATH are all constant. Also, MATH implies that MATH. This, combined with the constancy of the eigenvalues of MATH implies that MATH is constant and equal to MATH for some real number MATH. This, in turn, implies that MATH. Consequently, MATH has at most two distinct eigenvalues. It follows that either MATH, in which case the structure has constant holomorphic sectional curvature MATH, or else that there is a symmetric frame reduction of MATH to a MATH-subbundle MATH on which MATH . Thus, the structure is a locally isomorphic to MATH where MATH.
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math/0003099
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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 blocks of the structure equation for MATH then shows that MATH must be zero, thus implying that the structure is locally symmetric. Now apply REF .
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math/0003099
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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 there are two real-analytic REF-forms MATH and MATH, taking values in MATH and MATH, respectively, and a real-analytic function MATH with the properties that MATH is a MATH-valued coframing on MATH, that REF are satisfied on MATH, and that there exists a MATH for which MATH. Since MATH, the equation MATH defines an integrable plane field of codimension MATH on MATH. After shrinking MATH to an open neighborhood of MATH if necessary, an application of the complex NAME theorem shows that there is a submersion MATH with MATH so that the leaves of this integrable plane field are the fibers of MATH and, moreover, that MATH for some function MATH that satisfies MATH. Since MATH, REF-form MATH is closed. Since MATH is MATH-semibasic and since, by definition, the fibers of MATH are connected, it follows that MATH is actually the pullback to MATH of a closed, positive MATH -form on the open set MATH, that is, a NAME structure on MATH. Let MATH be the unitary coframe bundle of this NAME structure. Define a mapping MATH as follows: If MATH, then MATH is surjective and, by construction, has the same kernel as MATH. Thus, there is a unique linear isomorphism MATH so that MATH. In fact, MATH is complex linear; using the standard identification MATH, one sees that MATH becomes MATH. The equation MATH implies that MATH is a unitary coframe for all MATH. Since MATH is a coframing, it follows that MATH is an open immersion of MATH into MATH. Shrinking MATH again if necessary, it can be assumed that MATH embeds MATH as an open subset of MATH. Thus, nothing is lost by identifying MATH with this open subset of MATH. The structure REF now become identified with the structure equations of the unitary coframe bundle MATH, implying that the underlying NAME structure on MATH is, in fact NAME, and that the structure function MATH takes on the value MATH at MATH, as desired. Further details are left to the reader. This completes the existence proof. Uniqueness in the real-analytic category now follows directly from REF . Now, while NAME states the uniqueness part of REF only in the real-analytic category, uniqueness can actually be proved using only ordinary differential equations (that is, the NAME theorem); the NAME or NAME Theorems are not needed. Thus, his uniqueness result is valid as long as the form MATH is sufficiently differentiable for MATH to exist as a differentiable bundle and for MATH, MATH, and MATH to be defined and differentiable. For this to be true, it certainly suffices for MATH to be MATH. Since NAME 's existence proof produces a real-analytic example, uniqueness then implies that any MATH . NAME structure is real-analytic.
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math/0003099
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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 contain a subgroup isomorphic to MATH that will act as unitary rotations on the subvector MATH. Acting by an element of the stabilizer of MATH, one can then reduce to the case where MATH is real and nonnegative while MATH. By definition, the resulting new MATH is an element of MATH. It is clear from the construction that this element is unique.
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math/0003099
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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 subset of MATH can be uniquely analytically continued to a symmetry vector field on all of MATH. Moreover if MATH is such a symmetry vector field, then, as discussed in REF, there is a unique vector field MATH on MATH satisfying MATH and MATH. Conversely, if MATH is a vector field on MATH satisfying MATH, then MATH where MATH is a symmetry vector field on MATH. In other words, the mapping MATH defines an embedding MATH that realizes MATH as the NAME algebra of vector fields on MATH whose flows preserve the coframing MATH. By the structure REF the flow of such a vector field must necessarily preserve the structure function MATH, which is a submersion onto its (connected) image. Applying NAME 's REF (see the Appendix), for any MATH the evaluation map MATH defined by MATH is a vector space isomorphism between MATH and the kernel of MATH. Let MATH denote this kernel. Then, by REF, the image MATH consists of the pairs MATH that satisfy MATH . By the first of these equations, MATH and MATH . Thus, the third equation is a consequence of the first and so can be ignored for the rest of this discussion. Let MATH. By MATH-equivariance, it is enough to show that the dimension of MATH is at least MATH at any point MATH where MATH and MATH are both real, so assume this for the rest of the argument. By the structure REF , the dimension of MATH is equal to the dimension of the space of solutions of the linear equations MATH for MATH and MATH. Consider the solutions of REF for which MATH and MATH are purely imaginary, that is, where MATH and MATH for some symmetric (real) matrix MATH and some MATH. Then the equations in REF reduce to MATH . The right hand side of the first equation of REF takes values in MATH and the right hand side of the second equation of REF takes values in MATH. Thus, this is MATH equations for the MATH components of MATH and MATH. Consequently, the space of solutions is at least of dimension MATH.
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math/0003099
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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, which is nonnegative. For MATH, write MATH in `block' form as MATH where MATH takes values in MATH-MATH complex matrices for MATH. Correspondingly, write MATH in `block' form as MATH where MATH takes values in MATH. Then the first equation of REF breaks into blocks as MATH . When MATH, this forces MATH, so that either MATH, in which case this places no restriction on MATH or else MATH, in which case MATH must be a purely imaginary multiple of MATH, say MATH for some MATH. In either case, MATH is purely imaginary. When MATH, the above equation can be written as MATH . Substituting this equation into the second equation of REF yields MATH which, by the definition of MATH and the purely imaginary nature of MATH, can be written in the form MATH . Now, if MATH, then this equation can be written in the form MATH which is MATH real equations for the MATH entries of MATH. In fact, the solutions of this equation can be written in the form MATH where MATH is any solution to MATH, an equation that defines the stabilizer subalgebra of MATH in MATH and so has a solution space of dimension MATH. If MATH, then the equation above simplifies to MATH. If MATH, then this implies that MATH while if MATH, the equation degenerates to an identity. In particular, it follows that REF impose no interrelations among the MATH, just the condition MATH if MATH, the condition MATH if MATH but MATH, and no condition on MATH if MATH. Moreover, once the MATH have been chosen subject to these conditions, the MATH for MATH are completely determined while the MATH are determined up to a choice of MATH if MATH or are freely specifiable if MATH. The desired dimension formulae follow immediately.
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math/0003099
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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 as REF-forms on MATH. The first step will be to prove the following identities for all MATH: MATH . Now, the first set of identities is just a calculation. The case MATH is obvious, so assume MATH. Using MATH and the structure equation, one has MATH . (The terms in MATH involving MATH cancel since MATH is constant on the fibers of MATH.) Taking the exterior derivative of this relation and dividing by MATH yields MATH . The second set is a little more complicated, but still just a calculation. The case MATH is trivial, and the case MATH follows from the fact that MATH, so MATH. Because MATH, the second identity for MATH is just the structure equation for MATH. Also, MATH verifying the formula when MATH. Thus, suppose from now on that MATH and compute (again ignoring terms involving MATH, which must cancel) MATH . Thus, the formulae REF are established. Now, I claim that the functions MATH satisfy the differential equations MATH . Granting REF for the moment, computation gives MATH . It remains to verify REF. This is a classical identity: Let MATH be free variables, let MATH be the MATH-th elementary function of the MATH, and let MATH be the MATH-th power function of the MATH, that is, MATH. For any constant MATH, one has MATH . Taking the logarithm and then computing the differential of both sides yields MATH . Expanding the right hand side out as a formal geometric power series in MATH and collecting like powers of MATH yields MATH . It follows that MATH . Thus, MATH which, after equating coefficients of like powers of MATH on each side, is REF.
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math/0003099
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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 that MATH, MATH by virtue of REF. Thus, MATH, so that MATH is the MATH-Hamiltonian vector field associated to MATH. Since the flow of MATH is both holomorphic and symplectic, MATH belongs to MATH, as claimed. Since the representative function MATH is constructed as a polynomial in MATH and MATH, which are invariant under the MATH-flow on MATH for any vector field MATH, it follows that MATH is invariant under the flow of any MATH, that is, MATH for any MATH. Thus, the MATH for MATH span a central subalgebra MATH. For any MATH, the nondegeneracy of MATH implies that MATH is the MATH-dual of MATH see REF. The map MATH has components given by MATH and MATH. By REF, each MATH can be written on each connected component of MATH as a weighted polynomial in MATH with constant coefficients. Thus, the kernel of MATH is the same the kernel of MATH where MATH, which establishes the stated MATH-complementarity. Now suppose that MATH is connected. For each MATH, each MATH, and each MATH, MATH . Since MATH is Hermitian symmetric, it follows that the dimension of MATH over MATH is the largest integer MATH so that the vectors MATH are linearly independent (over either MATH or MATH) and, moreover, that MATH. Let MATH be the (nonempty) open set consisting of those MATH for which MATH achieves the maximum value MATH. If MATH, then MATH for all MATH and, since MATH for all MATH, it follows that MATH. If MATH, then MATH vanishes identically, implying that MATH. If MATH, let MATH satisfy MATH. Then, on MATH, the vector field MATH is a linear combination of the independent vector fields MATH. Thus, there exist smooth real-valued functions MATH on MATH so that MATH . Consequently, MATH . However, the left hand side of this equation is a holomorphic vector field while the holomorphic vector fields MATH are linearly independent over MATH at each point of MATH. It follows that the functions MATH are real-valued holomorphic functions on MATH and hence must be constants. Since MATH is connected, the identity MATH must hold on all of MATH. In other words, the vector fields MATH are a basis of MATH, as desired. If MATH, then at any point MATH, the vectors MATH are linearly independent. If MATH, then MATH are linearly independent, implying (since MATH is diagonalizable) that MATH does not lie any sum of fewer that MATH distinct eigenspaces of MATH. By REF, this implies that the differential of the mapping MATH has kernel of dimension equal to MATH, so MATH, as desired.
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math/0003099
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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 should treat the cases MATH and MATH separately, but the argument for MATH differs from that for MATH by trivial notational changes, so I will not explicitly assume MATH but, rather, let the reader make the necessary modifications for the case MATH. By REF , the differentials MATH are linearly independent exactly where the vector fields MATH are linearly independent. In particular, the locus MATH where MATH vanishes is also where the holomorphic MATH-vector MATH vanishes. Thus, MATH is a closed, proper, complex analytic subvariety of MATH and so has real codimension at least MATH. Consequently, its complement MATH is a connected, open, dense subset of MATH. Since MATH for all MATH, a subbundle MATH can be defined over MATH by saying that MATH lies in MATH if and only if MATH. Then MATH is a smooth principal MATH-bundle over MATH. Pull the forms MATH and MATH and the functions MATH, MATH, and MATH back to MATH. By definition, for every MATH, the vectors MATH span MATH and, moreover, MATH. Thus, there exists a function MATH, a function MATH on MATH with values in the open set of symmetric MATH-MATH (real) matrices with MATH distinct eigenvalues, and a function MATH on MATH with values in Hermitian symmetric MATH-MATH matrices so that MATH . Write MATH in MATH-block form as MATH where, of course, MATH and MATH take values in skew-Hermitian matrices of dimensions MATH and MATH, respectively. The lower right-hand MATH-MATH block of the MATH equation in REF then becomes MATH . Consequently, the eigenvalues of MATH are constant on MATH. Let MATH be the characteristic polynomial of MATH, where the MATH are constants. Then, on MATH at least, MATH divides MATH. Using the Euclidean algorithm, write MATH where MATH and MATH are polynomials in MATH and where the degree of MATH is at most MATH. The coefficients of MATH and MATH are constant linear combinations of the coefficients in MATH and so are continuous. Since the coefficients of MATH vanish on MATH, which is dense in MATH, it follows that MATH vanishes identically on MATH. Thus, MATH divides MATH on all of MATH. Defining real-analytic functions MATH on MATH by MATH one sees that MATH on MATH, that is, that MATH. Of course the roots of MATH on MATH equal the eigenvalues of MATH on MATH and so are distinct and therefore real-analytic on MATH. Since the MATH are constant coefficient linear combinations of the MATH, it follows that there is a constant MATH so that MATH . Obviously, MATH is nonzero and MATH is nonvanishing on MATH. Since the roots of MATH are distinct on MATH, and the MATH are the elementary symmetric functions of these roots, it follows that these roots must be functionally independent on MATH, as claimed.
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math/0003099
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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. Thus, for example MATH but MATH. Also, MATH. The MATH-entry of the structure REF for MATH becomes (no sum over MATH) MATH . Meanwhile, the structure equation for MATH becomes MATH . Combining these equations yields MATH . Since MATH is nonzero on MATH, it follows that MATH as desired.
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math/0003099
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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 dense in MATH) have codimension MATH. Consequently, MATH, so that MATH. Moreover, the inequality MATH follows immediately from the definitions. Thus, by the very definition of MATH, it will suffice to show that, for each MATH, the polynomial MATH has a constant root MATH of multiplicity at least MATH. By REF , it suffices to to show this constancy in an open neighborhood of the point MATH for which there exists a MATH satisfying MATH. Thus, let MATH be a nonzero vector and let MATH be the constant speed geodesic satisfying MATH. Then MATH can be lifted uniquely to a curve MATH that satisfies MATH and MATH (that is, the coframe field MATH is parallel along MATH). Because MATH is a constant speed geodesic, MATH also satisfies MATH, where MATH is the parameter on MATH. Because the polynomial MATH is invariant under the action of the symmetry pseudo-groupoid, it suffices to consider only geodesics with initial velocities orthogonal to the subspace MATH that is the tangent to the orbit through MATH. Thus, I will assume that if MATH, then MATH is real and that if MATH, then MATH is orthogonal to MATH. For simplicity, set MATH, MATH, and MATH. Then these functions on MATH satisfy the initial conditions MATH and the system of ordinary differential equations MATH . Let MATH be any fixed element that satisfies MATH and MATH. Because of the latter equation, MATH preserves each of the eigenspaces of MATH, that is, the subspaces MATH. Consequently, MATH and MATH for all MATH. Conversely, if MATH preserves the eigenspaces of MATH and annihilates MATH and MATH for all MATH, then it satisfies satisfies MATH and MATH. Now, the above differential equations imply the differential equations MATH so that the quantities MATH satisfy a linear system of ordinary differential equations with vanishing initial condition at MATH. Consequently MATH and MATH vanish identically for all MATH, as does MATH (for trivial reasons). By the above characterization of those MATH that satisfy MATH, this implies that the subspace MATH that is perpendicular to MATH and MATH is necessarily an eigenspace of MATH that is perpendicular to MATH and MATH for all MATH. If MATH, then there is a well-defined eigenvalue MATH of MATH associated to MATH. In particular, MATH for all MATH. Of course, this implies that MATH, that is, that MATH for all MATH. There are now four cases to consider: If MATH is such that MATH and MATH, then MATH is either MATH or MATH, depending on whether MATH is zero or not. In either case, MATH, so MATH is a root of MATH of multiplicity at least MATH, as desired. If MATH is such that MATH and MATH, then MATH and MATH. In this case, of course, MATH, so MATH is trivially a root of of MATH of multiplicity at least MATH, as desired. If MATH is such that MATH but MATH, then MATH is either MATH or MATH, depending on whether MATH is zero or not. In either case, MATH, so MATH is a root of MATH of multiplicity at least MATH, as desired. Finally, if MATH and MATH, then MATH by the above condition on MATH. Thus MATH, so that MATH and, again, MATH is a root of MATH of multiplicity at least MATH, as desired.
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math/0003099
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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 interval MATH for MATH. Moreover, since MATH is positive for all MATH sufficiently negative, MATH has an odd number of real roots (counted with multiplicity) less than MATH. These considerations imply that MATH has at least MATH distinct real roots. If MATH has exactly MATH real roots, then the above parity conditions show that they must all have odd order. Since MATH has degree MATH, either REF or REF must hold. If MATH has exactly MATH real, distinct roots, then REF must hold. If MATH has exactly MATH real, distinct roots, then REF must hold.
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math/0003099
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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, and let MATH, so that there exists a real factorization of the form MATH with MATH for MATH. First, assume that MATH lies in MATH, the interior of MATH. Then MATH lies in MATH for MATH and, in particular, the MATH are all distinct and are not roots of MATH. It follows that the rational function MATH has a simple pole at MATH for MATH. Since MATH has degree MATH and is monic, there is a partial fractions expansion of the form MATH . Because of the way that the possible momentum cells were defined, the inequality MATH holds for MATH, so it follows that MATH for MATH. If MATH, define MATH so that MATH . By hypothesis, each root of MATH is a real root of MATH and so is not equal to any of the roots of MATH. Consider the element MATH defined by letting MATH be the diagonal matrix with entries MATH for MATH; letting MATH for MATH and MATH for MATH (if MATH); and letting MATH. The hypothesis that MATH have no MATH-term is then seen to be equivalent to the condition that MATH, while the condition that each MATH for MATH be a root of MATH is then equivalent to the condition that MATH . By REF , the NAME structure on a neighborhood MATH of MATH that has a unitary coframe MATH with MATH has MATH and MATH as its characteristic polynomials and satisfies MATH. This establishes existence for the interior points of MATH. It remains to treat the boundary cases, that is, cases in which one or more of the MATH are actually roots of MATH. Now, if MATH for any MATH, then MATH, so that MATH is a simple root of MATH. In such a case, necessarily, MATH (if MATH) since MATH and MATH (if MATH) since MATH. Consequently, each MATH is at most a double root of MATH and, if so, it must also be a root of MATH. It follows that the rational function MATH has a simple pole at MATH for MATH. Since MATH has degree MATH and is monic, there is a partial fractions expansion MATH where, in order to make the MATH unique, it is now necessary to add the condition that MATH if MATH. If MATH is such that MATH is not a root of MATH, then the inequality MATH holds so that MATH. If MATH is such that MATH is a simple root of both MATH and MATH, then MATH. If MATH is such that MATH, then MATH and MATH are both positive on MATH. Since MATH is a double root of MATH and a simple root of MATH, it follows that MATH which can only hold if MATH. In particular, MATH has been defined for MATH so that the above partial fractions expansion is valid. If MATH, again define MATH so that MATH . Again, each root of MATH is a real root of MATH but now it may also be a root of MATH. Define an element MATH by letting MATH be the diagonal matrix with entries MATH for MATH; letting MATH for MATH and MATH for MATH (if MATH); and letting MATH. It must now be verified that the element MATH does indeed have MATH and MATH as its characteristic polynomials. Now, the hypothesis that MATH have no MATH-term is again seen to be equivalent to the condition that MATH, and the condition that MATH for MATH or when MATH implies that MATH so that MATH as desired. It remains to verify that MATH is the reduced characteristic polynomial associated to MATH, that is, to compute the numbers MATH and MATH for each eigenvalue MATH of MATH according to the recipe of REF and show that MATH is a root of MATH of multiplicity exactly equal to MATH. This will be done by breaking it down into a number of cases. If MATH is not MATH for any MATH, then MATH is equivalent to MATH and this implies that MATH is an eigenvalue of MATH of some multiplicity MATH that satisfies MATH, so that MATH. Thus, MATH is a root of MATH and has multiplicity MATH, as desired. If MATH is not a root of MATH, then, by construction, it is a simple eigenvalue of MATH and also satisfies MATH, so MATH and MATH, so that MATH is not a factor of MATH, again, as desired. If MATH is a simple root of MATH and a simple root of MATH, then, by construction, MATH. The quantity MATH is calculated to be MATH . Thus, the recipe gives MATH, again as desired. If MATH is a simple root of MATH and a double root of MATH, then it can be assumed that MATH, so that MATH (and MATH). Since MATH, the recipe gives MATH, again, as desired. Finally, if MATH is a multiple root of MATH, then it cannot be a root of MATH at all, by the definition of the momentum cell MATH. Consequently, MATH by definition and calculation shows that MATH as well. Thus MATH, as desired. By REF , the NAME structure on a neighborhood MATH of MATH that has a unitary coframe MATH with MATH has MATH and MATH as its characteristic polynomials and satisfies MATH. This establishes existence for the boundary points of MATH.
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math/0003099
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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 with the same data MATH are analytically connected. Now, if MATH and MATH are connected NAME manifolds with the same data MATH and their reduced momentum images MATH and MATH have nontrivial intersection, then they contain points MATH and MATH so that MATH where MATH and MATH are the corresponding moduli maps. By REF , the germs of NAME structures around MATH and MATH are isomorphic. Since MATH and MATH are connected, the germ of the NAME structure around any MATH is analytically connected to the germ of the NAME structure around any MATH. Now, from REF , it follows that MATH lies in the interior of MATH and that MATH is a submersion onto its image, which is therefore open. The union of the open sets MATH as MATH ranges over the NAME structures that are analytically connected to any given MATH is a connected component of MATH. Since MATH is convex and hence connected, this union must be all of MATH. By REF , the union of all the sets MATH as MATH ranges over the NAME structures with data MATH is equal to the entire cell MATH. Since MATH is a nonempty subset of MATH for any such MATH, it follows that all of these are analytically connected, as desired.
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math/0003099
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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 while MATH for some MATH. Since MATH by hypothesis, there exists a MATH so that MATH, MATH, and MATH. Since the metric on MATH is complete, there will exist a unique curve MATH satisfying the initial condition MATH and the ordinary differential equations MATH . That is, MATH is the parallel transport of MATH along the curve MATH in MATH. The structure REF and the Chain Rule now imply that the two curves defined on MATH in MATH satisfy the same initial conditions and system of ordinary differential equations, so they are equal on MATH. Now, setting MATH yields MATH. Since MATH was arbitrary, MATH is a subset of MATH. To prove the second part, suppose first that MATH is simply-connected. Then any local isometry MATH defined on an open subset MATH extends uniquely to a global isometry of MATH. Thus MATH, the global holomorphic isometry group of MATH, acts transitively on the fibers of MATH. Now MATH and, by the first part of the proof, MATH is surjective. Since MATH is also a submersion whose fibers are MATH-orbits, it follows that MATH is a fibration. To treat the case where MATH is not simply-connected, pass to the universal cover MATH and note that MATH is invariant under the deck transformations MATH of the cover MATH (which form a discrete subgroup of MATH). Since MATH is a fibration, dividing by the (free) action of MATH yields that MATH is also a fibration. To prove the connectedness of the fibers of MATH it suffices to treat the case where MATH is simply-connected, so assume this. Again, MATH acts transitively on the fibers of MATH. Since MATH is the complement of a codimension REF complex subvariety in MATH, it is connected. The exact sequence of the fibration MATH and the contractibility of MATH then imply that the fibers of MATH are connected. By REF , the MATH-stabilizer of any point in MATH is a product of unitary groups and hence is connected. Since the MATH-orbit of any MATH has been shown to be connected, it follows that MATH must be connected as well. Consequently, all of its orbits in MATH are connected and these are the fibers of MATH.
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math/0003099
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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 and that MATH as in REF. Since MATH for MATH, the quadratic form MATH is positive definite on MATH. Since MATH has rational coefficients, is well-defined on MATH minus the union of the hyperplanes MATH, and is invariant under permutations of the coordinates MATH, it follows that there are unique rational functions MATH on MATH so that MATH . That is, setting MATH, the positive definite quadratic form MATH defined on MATH is of the form MATH. Since MATH is a diffeomorphism, MATH is positive definite on MATH. Moreover, the above formula on MATH can now be written as MATH showing that MATH is a Riemannian submersion when the target is given the Riemannian metric MATH.
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math/0003099
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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 real root of MATH. Again, let MATH be the bundle over MATH constructed in the course of the proof of REF . Because the structure group of MATH is MATH, it follows that REF-forms MATH for MATH are actually well-defined on MATH. Let MATH be the vector field on MATH that is MATH-dual to MATH. Then, by the relation MATH, it follows that MATH for MATH and that MATH . In particular, along an integral curve of MATH the functions MATH for MATH are constant while MATH is strictly increasing. Fix MATH and let MATH be the maximal forward integral curve of MATH with MATH. I claim that MATH cannot be finite. If it were, the fact that MATH is a unit speed vector field and that MATH is complete would imply that MATH approaches a limit MATH as MATH approaches MATH (after all, MATH). The limit point MATH could not lie in MATH since then MATH would not be maximal. By continuity, MATH must not lie in MATH. However, MATH for MATH while MATH. Since MATH this forces MATH to lie in MATH, a contradiction since MATH lies in MATH. Thus MATH, as claimed. In particular, the forward flow of MATH exists for all time on MATH. However, this leads to a contradiction: Along MATH, the element of arc is given by MATH . Let MATH. Since MATH has unit speed, the integral MATH must be infinite. However, this integral is bounded by MATH which converges, since MATH has degree MATH. This contradiction implies that MATH could not have been complete.
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math/0003099
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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 that MATH where MATH. By REF , the momentum cell MATH must be bounded, which implies that REF obtains, namely MATH for MATH. For MATH, let MATH be the MATH-th face of this MATH-simplex, that is, the intersection of MATH with the hyperplane MATH (where the functions MATH are as defined in REF). Let MATH be the preimage of MATH. Evidently, each MATH is a closed, analytic subset of MATH and the union of the MATH is the complex subvariety MATH. Thus, MATH is a (non-empty) complex subvariety of MATH. For MATH, define functions MATH on MATH and then define vector fields MATH by MATH. By REF, the MATH satisfy MATH . Moreover, any MATH of these vector fields are linearly independent on MATH. Note that since MATH reaches its minimum of MATH along MATH, the vector field MATH vanishes along MATH. Since MATH is complete, the flows of the vector fields MATH are complete. I am going to show that the flow of each vector field MATH is periodic of period MATH by examining the rotation MATH along the fixed hypersurface MATH. Now, REF can be written as MATH where I have replaced MATH by MATH and am regarding MATH as a parameter, taken to be sufficiently small so that the series converges in a neighborhood of any given compact domain in MATH. Using REF, this can be written in the form MATH and the series can then be summed, yielding the equation MATH . Replacing MATH by MATH, this becomes MATH . The final expression is valid for all MATH, while the middle expression is valid away from the locus MATH in MATH. Since MATH, and since MATH has constant coefficients, REF implies MATH away from the locus MATH. Define a vector field MATH on MATH by MATH. This vector field depends polynomially on MATH and lies in MATH for all MATH. In fact, comparison with REF, the definition of MATH, shows that MATH, so it follows that MATH . By REF, the vector field MATH has representative function MATH given by MATH . The expression on the left is polynomial in MATH, so the expression on the right must be also. Since the flow of MATH is a holomorphic isometry, it follows that MATH where MATH takes values in MATH. In fact, by REF, MATH and the matrix on the right is visibly skew-Hermitian. When MATH, REF simplifies to the form in which it will be the most useful: MATH . Now fix MATH in the range MATH and let MATH be the vertex that lies on the intersection of the faces MATH for MATH, that is, MATH is the vertex that lies opposite the face MATH. Applying REF , choose MATH to satisfy MATH and then let MATH satisfy MATH. Then MATH since the differential of MATH vanishes at MATH. In particular, MATH is a zero of MATH for all MATH. Now, MATH is a root of MATH for all MATH since MATH lies on each MATH with MATH. Thus, the set MATH consists of the MATH where MATH. Consequently, since REF now simplifies to MATH it follows that MATH. In particular, REF becomes MATH . Now, any eigenvalue of MATH is a root of MATH and so, by REF , must be of the form MATH for some MATH. Let MATH denote the eigenspace of MATH belonging to the eigenvalue MATH. Then the above formula implies that MATH annihilates MATH and MATH and that, for MATH with MATH, MATH . Since the right hand side of REF is a polynomial in MATH, it now makes sense to substitute MATH for any MATH. When MATH, this gives MATH for MATH, while, if MATH, this gives MATH . In other words, for MATH in the range MATH, MATH where MATH is the orthogonal projection onto this eigenspace. Thus, the flow of MATH is periodic of period MATH and so, by REF, the flow of MATH is periodic of period MATH, as claimed. Now, further information can be got by evaluating MATH at MATH itself. Indeed, in the above formula, if MATH lies in MATH with MATH, then putting MATH gives MATH . In other words, using the projection notation already introduced, MATH . Since the flow of MATH has period MATH, each of the ratios MATH must be an integer for MATH. Since MATH when MATH, these ratios cannot be MATH. Thus, as the inverse of each such ratio is another such ratio, these integers must all be MATH. However, this is equivalent to MATH, which is impossible, since MATH is greater than either MATH or MATH. This contradiction establishes the proposition.
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math/0003099
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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-connected cover (which is complete) must be isometric to MATH, as claimed.
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math/0003099
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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 the definition of the bundle MATH and the foliation MATH, the forms MATH vanish when pulled back to MATH, so, by the above equations, so do the forms MATH. Consequently, the complex MATH-manifold MATH is totally geodesic in MATH, as claimed. Denoting pullback to MATH by a superscript MATH, the formulae MATH hold, where MATH takes values in MATH and MATH takes values in MATH. The notations MATH, MATH, and MATH are as previously established. The NAME form MATH induced on MATH by pullback from MATH then satisfies MATH. The pullbacks of the structure equations to MATH then imply MATH, so that MATH is the connection matrix of the torsion-free NAME connection of the induced NAME structure. The pullbacks further imply MATH where MATH and MATH. Since MATH, this defines MATH as in the statement of the proposition. Thus, by definition, the induced metric on MATH is NAME and has momentum polynomial MATH as claimed. Moreover, the pullback of the MATH equation implies MATH so that MATH is the MATH-valued function defined by the structure equations for MATH. Since the entries of MATH are all nonzero and the eigenvalues of MATH are distinct, the rank of the momentum mapping for MATH is MATH, implying that MATH. Finally, the pullback of the identity for MATH becomes MATH so that, setting MATH, the structure function for the metric on MATH takes the form MATH. The formula for MATH then becomes MATH where the last line uses the definition of MATH, the identity MATH, and the identity MATH . These formulae establish the proposition.
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math/0003099
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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 MATH whose image contains MATH and for which the induced metric MATH on MATH is complete. This completion MATH is unique up to diffeomorphism. Here is a sketch of the proof: Fix MATH and consider, for every MATH, the constant speed geodesic MATH defined by MATH. Let MATH be the parallel translation of MATH along MATH from MATH to MATH, and let MATH be the set of all such MATH. Since MATH is totally geodesic, when MATH is sufficiently small MATH is equal to MATH. It follows, by the real-analyticity of MATH, that MATH embeds MATH into MATH as a totally geodesic submanifold of MATH as long as MATH is less than the injectivity radius at MATH. From this, it is not hard to prove that MATH is an embedded submanifold of MATH. Moreover, the basepoint projection MATH defined by MATH is a totally geodesic immersion. Since each of the geodesics MATH lifts to a complete geodesic MATH in MATH for the induced metric MATH, all of the MATH-geodesics in MATH passing through MATH are complete. Thus, MATH is complete. The completeness of the metric then ensures that MATH contains MATH, as desired. Now apply this result to the leaf MATH and consider MATH. Since the induced metric on MATH is real-analytic and is NAME on an open set, it is NAME everywhere. Moreover, by REF , it has cohomogeneity MATH, equal to its complex dimension. Let MATH denote its momentum mapping. By REF and real-analyticity, MATH, since this holds on MATH. Since the rank of MATH is MATH, the proof of REF coupled with the remarks of REF show that MATH accounts for all of the infinitesimal symmetries of MATH, that is, that the full symmetry group of MATH is generated by the canonical torus action, even if one were to pass to its simply-connected cover. In particular, by REF , the fibers of MATH are connected and are the orbits of the canonical torus action. Now, MATH is a submersion outside some closed complex submanifold MATH. Since MATH is a submersion only when it is restricted to MATH, it follows that that MATH must lie in MATH. Since MATH is connected, and since it contains the MATH-leaf MATH, it follows from analyticity that MATH must be equal to MATH. Thus, MATH consists of the tangent planes to MATH. It follows that MATH is one-to-one on MATH. If MATH were not one-to-one on MATH, this would violate the connectedness of the fibers of MATH, so MATH is one-to-one everywhere. In other words, MATH is a submanifold of MATH, as claimed. Obviously, MATH is the closure of MATH in MATH. Now, turning to the geometry of the leaves of the foliation MATH, note that these leaves are defined by the equations MATH (since MATH and MATH are real on MATH). To prove that these leaves are totally geodesic, it would suffice to show that the imaginary part of MATH vanishes when one restricts to such a leaf. Thus, write MATH and MATH, where MATH, MATH, MATH, and MATH take values in MATH, MATH, MATH, and the space of real symmetric matrices, respectively. Since MATH and MATH are real-valued, the imaginary part of the equation for MATH becomes MATH . It then follows by linear algebra that on the open set MATH where MATH has no two eigenvalues that sum to zero, the components of MATH are linear combinations of the components of MATH. By the structure equation for MATH, the eigenvalues of MATH are independent on each leaf of MATH since MATH. Thus, the open set MATH intersects each MATH-leaf in a dense open set. Consequently, the components of MATH vanish on each MATH-leaf, implying that each MATH-leaf is totally geodesic, as desired. Now, let MATH be a MATH-leaf and let MATH be its geodesic completion. By construction, MATH meets each isometry orbit in MATH orthogonally. Thus, by REF , the map MATH is an isometry when MATH is given the induced metric. It remains to show that MATH is surjective. The completeness and real-analyticity of the induced metric MATH on MATH coupled with the analysis of the resolution of the singular cell metric in REF -MATH done at the end of REF, shows that MATH must be an isometric quotient of MATH for some MATH. The completeness of this mapping and the surjectivity of this resolution imply the desired surjectivity.
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math/0003099
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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 basis for the holomorphic vector fields on MATH, so there are unique holomorphic REF-forms MATH on MATH that satisfy MATH for MATH. (The introduction of the factor of MATH simplifies formulae to appear below.) Because the vector fields MATH are NAME, REF-forms MATH are closed. Write MATH where MATH and MATH are real MATH-forms. Then the defining equation above is equivalent to MATH . Since the MATH are a basis for the holomorphic REF-forms on MATH, the metric on MATH can be written in the form MATH where MATH and where the pullback of MATH to MATH is MATH . Now, MATH or, equivalently, MATH . Since MATH for MATH is tangent to the fibers of MATH, the coefficient of MATH in the above equation must vanish, that is, MATH. Thus, MATH . Define MATH so that MATH. Note, in particular, that MATH. The metric on MATH can now be written in the form MATH . Since MATH is totally geodesic in MATH, it follows from REF that MATH . Thus, MATH and so MATH . Now lift everything to the universal cover of MATH. Since the MATH are closed, there exist functions MATH on this universal cover so that MATH. Then MATH where MATH is the inverse matrix to MATH and MATH . These are the desired formulae.
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math/0003099
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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 the momentum cell to choose a particularly good basis for MATH, one for which each of the vector fields of the basis has a periodic flow of period MATH. I will then attempt to find global holomorphic coordinates on the leaf that will carry these vector fields into the vector fields that generate the standard MATH-torus action defined above. Using these calculations as a guide and then comparing with the discussion at the end of REF of the `resolution' of boundary singularities of the cell metric in REF -MATH, I will finally arrive at a candidate for the metric in these good coordinates and finish by showing how completeness and real-analyticity give the conclusions of the theorem. Thus, suppose that MATH is simply connected and has a complete NAME metric of cohomogeneity MATH. As has already been remarked, the momentum cell must fall into REF -MATH, so that MATH . For notational simplicity, I will use the index range MATH and the abbreviation MATH REF for MATH in what follows. Use REF as the definition of the linear function MATH for MATH, as before, and set MATH as was done in REF during the analysis of the metric MATH on this momentum cell MATH, which is defined by the inequalities MATH and MATH. Note that this momentum cell contains only one vertex, namely the point MATH where all of the MATH vanish. Also as before, let MATH be the face defined by MATH for MATH. As was done in the proof of REF , set MATH and consider the MATH vector fields MATH defined by MATH. These vector fields are a basis of MATH and are linearly independent on MATH. Fix MATH, let MATH be the leaf of the foliation MATH passing through MATH, and let MATH be the leaf of the foliation MATH passing through MATH. On MATH, the map MATH is a local isometry when MATH is given the metric MATH. Recall the discussion and notation at the end of REF about the metric MATH on the ellipsoidal domain MATH. The map MATH is surjective and is isometric and submersive away from the hyperplanes MATH. Letting MATH be the point with coordinates MATH, it follows that there is a real-analytic map MATH from a neighborhood of MATH to a neighborhood of MATH satisfying MATH and MATH. This map is an isometry when MATH is endowed with the metric MATH. Since MATH is simply-connected and the metric MATH is both real-analytic and complete, it follows that MATH can be extended uniquely as a global isometry MATH and that it satisfies MATH. Since the rank of the differential of MATH at any MATH is equal to the number of nonzero entries MATH, it follows that the rank of the differential of MATH at any MATH is equal to MATH minus the number of faces MATH on which MATH lies. Since the rank of the differential of MATH at MATH is equal to the dimension of the span of MATH, it follows from this discussion that for any MATH, the nonzero elements in the list MATH are linearly independent. This observation will be useful below. Each MATH vanishes on MATH, which, since the flow of MATH is isometric, is a totally geodesic submanifold of MATH and, moreover, is a complex submanifold of MATH as well (since MATH is the real part of a holomorphic vector field on MATH). It also follows from the discussion in the previous paragraph that MATH is nonzero off of MATH. Let MATH. Then MATH is a totally geodesic complex hypersurface in MATH. One of the goals of this argument is to show that there are holomorphic coordinates MATH on MATH for which MATH and to find the expression for the induced NAME metric on MATH in these coordinates. (The choice of the coefficient MATH is dictated by the fact that the flow of the vector field MATH has period MATH. The proof of this periodicity follows the same lines as the corresponding proof in REF situation analyzed in REF . Since it only differs in details from that proof, the argument will be left to the reader.) Accordingly, let MATH be the holomorphic REF-forms on MATH that satisfy MATH . These forms extend meromorphically to MATH, with simple poles along the hypersurfaces MATH. Since the vector fields MATH . NAME, it follows that each MATH is closed. Note that, if the coordinates MATH are to exist as claimed, it will have to be true that MATH. Writing MATH, the above equations are equivalent to MATH . Again, if the coordinates MATH exist as claimed, it will follow that MATH. Since the MATH are a basis for the holomorphic REF-forms on MATH, the metric on MATH can be written in the form MATH where MATH and where the pullback of MATH to MATH is MATH . Now, the identity MATH implies MATH or, equivalently, MATH . Since MATH is tangent to the fibers of MATH, and since MATH is constant on those fibers, it follows that the coefficient of MATH in the above equation must vanish, that is, MATH. Thus, MATH . Define MATH so that MATH. Note, in particular, that MATH. The metric on MATH can now be written in the form MATH . Since MATH is totally geodesic in MATH, it follows from REF that MATH where MATH for MATH is defined on MATH so that the formula MATH holds. Thus, MATH. Using the definitions of MATH and MATH, it follows from the formula for MATH that MATH so that, in particular, MATH . Meanwhile, if the coordinates MATH exist as claimed, this will imply that MATH that is, there will exist constants MATH so that MATH . Since MATH would only be determined up to a multiplicative constant anyway by the above normalizations, one might as well take MATH, which will normalize the MATH up to a phase. These calculations suggest the following construction of a candidate for the leaf metric: Consider the system of REF relating the MATH variables MATH to the variables MATH. These formulae define a real-analytic mapping MATH from the open halfspace MATH defined by MATH in MATH-space into MATH-space. I claim that the mapping MATH is a diffeomorphism from MATH onto its image MATH and that this open image contains the primary orthant MATH, that is, the closed domain defined by MATH. Consequently, MATH has an inverse MATH, that is, the above equations can be solved real-analytically in the form MATH . Moreover, this inverse MATH takes MATH diffeomorphically onto the partially open simplex MATH defined by the inequalities MATH and MATH. To prove this claim, it is helpful to introduce the intermediate quantities MATH . These equations can be inverted in the form MATH thus showing that they define a diffeomorphism from MATH to the half-space MATH defined by MATH. Then the claim above amounts to showing that the mapping defined by MATH is invertible on the domain MATH and that its image has the desired properties. Consider the function MATH on MATH defined by MATH . Now, MATH is positive on MATH, so that MATH is a strictly increasing function on MATH for every MATH. Note that MATH for MATH and that MATH for MATH (since each of the MATH is positive). It then follows by the intermediate value theorem and the implicit function theorem that the equation MATH can be solved uniquely and real-analytically for MATH on an open set MATH containing the domain MATH. Thus, let MATH satisfy MATH and set MATH . Then, by construction, MATH so that MATH. Moreover, when MATH lies in MATH, MATH so the image point lies in MATH. The inversion of the original system is therefore MATH as was desired. Now define a metric on MATH as follows: First, define functions on MATH by MATH where MATH . Note that MATH is strictly positive on MATH. Moreover, since MATH where MATH is strictly positive on MATH, the formula for MATH defines a smooth function on MATH for all MATH and MATH. Moreover, the inequalities satisfied by the MATH show that the Hermitian matrix MATH is positive definite for all MATH. Let MATH denote the components of the inverse matrix and define MATH . This is an Hermitian metric on MATH. It is visibly invariant under the torus action generated by the real parts of the holomorphic vector fields MATH . Setting MATH yields MATH. Tracing through the construction above, one sees that, away from the complex hyperplanes MATH, the metric can be written in the form MATH where the inverse matrix MATH has the form MATH with MATH . Thus, the map from MATH to MATH defined by MATH is a Riemannian submersion from the complement of the hyperplanes MATH onto MATH. It not difficult now to trace through the construction and see that the restriction of the metric MATH to MATH is isometric to the metric MATH on MATH and is hence complete. It now follows without difficulty that MATH is complete on MATH. Note that this completeness is a consequence of the completeness of the metric MATH on MATH and so, by the discussion in REF, is valid for any MATH all of whose entries are non-negative, and not just for those whose entries are positive and strictly increasing. Moreover, looking back at the formula for the metric on MATH and comparing terms, one sees that MATH is locally and (therefore, by completeness) globally holomorphically isometric to MATH with its induced metric and that, under this isomorphism, the NAME form corresponding to MATH is simply MATH . This provides the desired `explicit' formula for the metric on the leaf MATH. (Bear in mind, though, that the function MATH, which is the crucial ingredient in the recipe for the metric, was found by abstractly solving an implicit equation.) As the reader can verify, the formula given above simplifies (after some trivial changes in notation) to the formula for MATH given before the statement of REF . The argument to this point shows that the metric MATH defined before the statement of REF is NAME for any MATH all of whose entries are positive and distinct. However, the property of being NAME is preserved in the limit as any of the entries of MATH vanish or become equal. (The curvature tensor is evidently analytic in MATH.) Consequently, the metric MATH is NAME and complete for any MATH with non-negative entries. Finally, returning to the notation used before the statement of REF , suppose MATH with MATH . Suppose first that these inequalities are strict and set MATH and then MATH for MATH, so that MATH and MATH. From the construction in the first part of the proof, it follows that the metric MATH satisfies MATH and has cohomogeneity MATH. Since the metric is complete, by REF , the momentum cell must fall into REF -MATH. Moreover, since any strictly decreasing sequence MATH satisfying MATH can be written in the above form for a unique MATH with MATH, it follows that such parameters account for all of the MATH-dimensional reduced momentum cells in REF -MATH. Thus, by REF , this formula gives all of the complete, simply-connected cohomogeneity MATH . NAME metrics of dimension MATH. Note that the origin is the unique fixed point of all of the vector fields in MATH, and it follows from REF that MATH . From this, it follows from REF that the NAME algebra of the symmetry group is MATH. Since the group of symmetries is necessarily connected, it follows that the flows in MATH generate the entire symmetry group. Now consider what happens as the MATH vary. The metric MATH depends analytically on MATH, so the formulae for MATH and MATH must remain true for all values of MATH. The vector fields in MATH all vanish at MATH, so it follows that MATH must vanish at MATH. Now, applying REF , one sees that, as MATH varies through MATH satisfying MATH, the values of the moduli pass through all of the values that can give rise to momentum cells in REF -MATH, with the one exception of MATH, since, in this case, there is no such cell. Consequently, as MATH varies in the primary orthant, the MATH account for all of the possible analytically connected equivalence classes that can contain a complete metric. Since these metrics are all complete, it follows from REF , and REF , that these contain all of the inhomogeneous complete NAME metrics on simply connected manifolds.
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