User simon salamon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:45:29Z http://mathoverflow.net/feeds/user/3975 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34088/a-geometric-interpretation-of-the-levi-civita-connection/34129#34129 Answer by Simon Salamon for A geometric interpretation of the Levi-Civita connection? Simon Salamon 2010-08-01T17:28:42Z 2010-08-01T17:28:42Z <p>To understand the existence and uniqueness of the LC connection, it is not possible to sidestep some algebra, namely the fact (with a 1-line proof) that a tensor $a_{ijk}$ symmetric in $i,j$ and skew in $j,k$ is necessarily zero. The geometrical interpretation is this: once one has the $O(n)$ subbundle $P$ of the frame bundle $F$ defined by the metric, there exists (at each point) a unique subspace transverse to the fibre that is <em>tangent</em> both to $P$ and to a coordinate-induced section <code>$\{\partial/\partial x_1,\ldots,\partial/\partial x_n\}$</code> of $F$.</p> http://mathoverflow.net/questions/13188/do-hyperkahler-manifolds-live-in-quaternionic-kahler-families/33846#33846 Answer by Simon Salamon for Do hyperKahler manifolds live in quaternionic-Kahler families? Simon Salamon 2010-07-29T20:33:05Z 2010-07-29T20:33:05Z <p>What you suggested makes sense. You propose to replace the $P^1$ fibre by the <em>twistor space</em> of an HK manifold M, so that the big total space would not only display separately the complex structures of M, but allow deformations of M to be parametrized by X. I think the real question is whether there exist sensible examples over a compact QK base like X$=S^4$ in which a consistent choice of complex structure on the varying HK manifolds is therefore not possible. I am not sure. The problem is that the construction looks a bit unwieldly, and experience dictates that it is more natural to look for bundles whose fibres are HK. In this sense, your idea is very close to a known (but in some sense simpler) construction that goes under the heading "Swann bundle" or "C map".</p> <p>Let me add two comments in support of your question. First, the concept of a manifold foliated by HK manifolds (like $T^4$ or K3) is very powerful. This is most familiar in work on special holonomy, but here's a more classical construction: the curvature tensor at each point of a Riemannian 4-manifold can be used to construct a singular Kummer surface and an associated K3 (the intersection of 3 quadrics in $P^5$), but the complex structure is fixed so not twistorial. Second, escaping from quaternions, one sees <em>twistor space fibres</em> in the following situation: each fibre of the twistor space $SO(2n+1)/U(n)$ parametrizing a.c.s.'s on the sphere $S^{2n}$ can be identified with the twistor space of $S^{2n-2}$!</p> http://mathoverflow.net/questions/33056/diffeomorphism-group-of-the-unit-sphere-of-complex-n-space/33564#33564 Answer by Simon Salamon for Diffeomorphism group of the unit sphere of complex n-space Simon Salamon 2010-07-27T19:32:55Z 2010-07-27T19:32:55Z <p>A slightly more elementary question, whose answer may be easier to grasp, is to determine the largest subgroup of the <em>isometry</em> group $O(2n)$ of $S^{2n-1}$ that acts holomorphically on $\mathbb{C}^n$. The answer is the unitary group $U(n)$, that is of course also a subgroup of the automorphism group $PU(n,1)$. The unitary group (and also $SU(n)$) is is one of a fairly restricted list of compact Lie groups that act effectively and transitively on a sphere. </p> http://mathoverflow.net/questions/33438/projection-of-the-co-derivative-co-derivative-of-the-projection/33557#33557 Answer by Simon Salamon for projection of the co-derivative == co-derivative of the projection ? Simon Salamon 2010-07-27T19:11:22Z 2010-07-27T19:17:44Z <p>Homework or not, it is not true that your covariant derivative in N is already parallel to the submanifold M; indeed the normal component eliminated by the projection is the <em>second fundamental form</em> of M in N. Consider the simple example of the 2-sphere in $N=\mathbb{R}^3$; if X is a normal (=radial) vector field on the sphere then <code>$(\nabla^N_b a)\cdot X=-a\cdot(\nabla^N_b X)$</code> is proportional to the inner product $a\cdot b$. To prove the projection formula, it suffices to (i) observe that the projected connection respects the induced metric on M, and (ii) prove that it has zero torsion, which is related to the fact that the second fundamental form is <em>symmetric</em> in a,b.</p> http://mathoverflow.net/questions/15366/which-journals-publish-expository-work/15374#15374 Answer by Simon Salamon for Which journals publish expository work? Simon Salamon 2010-02-15T22:12:41Z 2010-07-25T23:36:58Z <p>You can certainly find such journals if you are not too career-minded. Although there are not many examples in recent issues, <a href="http://seminariomatematico.dm.unito.it/rendiconti/" rel="nofollow">Rendiconti Torino</a> has open access and a tradition of publishing occasional expository articles, though typically less than 30 pages. It will effectively referee such papers provided that they appear to have a novel approach (as is the case of yours) thta cannot be found elsewhere. Publishing in such a journal may be more satisfying than simply leaving a paper on a homepage or the arXiv (which you can still do anyway). [I am its director, but it's a non-profit concern and I feel that such publicity is in the community's interest.]</p> http://mathoverflow.net/questions/24490/chern-connection-vs-levi-civita-connection/33338#33338 Answer by Simon Salamon for chern connection vs levi-civita connection Simon Salamon 2010-07-25T23:09:00Z 2010-07-25T23:09:00Z <p>Further to a previous answer, you might like to refer to <em>arXiv:0911.5655</em> and references therein for some interesting results on Chern connections on non-K&auml;hler (and indeed non-integrable) almost-Hermitian manifolds.</p> http://mathoverflow.net/questions/4331/is-the-wedge-product-of-two-harmonic-forms-harmonic/33336#33336 Answer by Simon Salamon for Is the wedge product of two harmonic forms harmonic? Simon Salamon 2010-07-25T22:54:56Z 2010-07-25T22:54:56Z <p>Here's a counterexample from the theory of nilmanifolds, which <em>by their very nature</em> are not formal. Take a compact quotient $H^3/\Gamma$ of the Heisenberg group. It admits invariant 1-forms $e^1,e^2,e^3$ with $de^1=0=de^2$ and $de^3=e^1\wedge e^2$. Then $e^1,e^2$ are harmonic, but $e^1\wedge e^2$ is exact, so not harmonic. You can take a product with $S^1$ to get a complex (non-K&auml;hler) surface on which the same thing works, but not I am afraid Ricci-flat or Einstein. </p> http://mathoverflow.net/questions/28519/references-for-modern-proof-of-newlander-nirenberg-theorem/33334#33334 Answer by Simon Salamon for References for "modern" proof of Newlander-Nirenberg Theorem Simon Salamon 2010-07-25T22:02:40Z 2010-07-25T22:02:40Z <p>This is not quite an answer to your question, but you might consult the book by Donaldson and Kronheimer "The geometry of 4-manifolds". In chapter 2 they prove an integrability theorem for holomorphic vector bundles, the point being that this can be regarded as a simpler version of the Newlander-Nirenberg theorem, and (in my view) very suitable for your course. You might also want to mention the following simple example for instructional purposes: the nilpotent Lie group H^3 x R where H^3 is the Heisenberg group has an obvious left-invariant almost-complex structure whose Nijenhius tensor vanishes. Although not a <em>complex Lie group</em>, it is easy to find independent local complex coordinates z_1, z_2. I suspect that there are similar classes of almost-complex examples where the integration is elementary. </p> http://mathoverflow.net/questions/12601/a-topological-consequence-of-riemann-roch-in-the-almost-complex-case/33330#33330 Answer by Simon Salamon for A topological consequence of Riemann-Roch in the almost complex case Simon Salamon 2010-07-25T21:37:49Z 2010-07-25T21:37:49Z <p>I may be repeating what has been said, but I think the point is this. The index theory always works in the "almost" case because one can set up a <em>2-step</em> elliptic complex with operator D + D^* where D is d-bar. Moreover I believe that, in real dimension 6 or more, there are <em>no</em> known obstructions to the existence of an <em>integrable</em> complex structure beyond those for an almost complex structure. The case of 4 real dimensions is special because we have Kodaira's classification of complex surfaces. (PS: I don't think it was really necessary to have so many math formulae above!) </p> http://mathoverflow.net/questions/33315/lists-of-computed-cohomologies/33325#33325 Answer by Simon Salamon for lists of computed cohomologies? Simon Salamon 2010-07-25T21:20:49Z 2010-07-25T21:20:49Z <p>Consolidating the previous comments, one can start by understanding the topology of a compact simple group. Here is an example: the Lie group Sp(3) of dimension 21. It has rank 3 and the same Betti numbers as the product S^3 x S^7 x S^11 of spheres of dimension 2i+1; the numbers i (here 1,3,5) are the <em>exponents</em> of Sp(3). This theory dates back to Hopf, and was nicely explained by Bott using Morse theory. The (trivial) Lie algebra cohomology ring is generated by the associated forms in dimension 3,7,11. All this theory can be extended to compact irreducible symmetric spaces G/H, which (if not Hermitian) will also have b_1=b_2=0.</p> http://mathoverflow.net/questions/41934/spin-structures-on-the-grassmannians/41937#41937 Comment by Simon Salamon Simon Salamon 2010-12-04T20:14:15Z 2010-12-04T20:14:15Z One can compute the first Chern class of the complex Grassmannian of $k$-planes in $C^N$ as follows. For the purpose of computing its Chern character, the holomorphic tangent bundle $T$ is a product of the tautological rank $k$ bundle $V$ and (formally) $N−V$. This leads to the formula $c_1(T) = (N−2k)v$ where $v=c_1(V)$ is a generator of $H^2$. So it looks like the Grassmannian is spin iff $N$ is even. http://mathoverflow.net/questions/46752/is-the-cotangent-bundle-to-a-kahler-manifold-hyperkahler/46755#46755 Comment by Simon Salamon Simon Salamon 2010-12-04T19:27:15Z 2010-12-04T19:27:15Z This answer is pretty complete, but it is worth reading the paper of Calabi in Ann. Ec. Norm. Sup. 12 (1979) for an explicit construction of the HK metric on the cotangent bundle of complex projective space. The precise form of the metric is not obvious, and his approach (subsequently generalized to other HSS's) was to find the Kaehler potential. As in applications of Yau's theorem in the compact case, the HK metric is indeed compatible with the underlying holomorphic symplectic structure. http://mathoverflow.net/questions/34088/a-geometric-interpretation-of-the-levi-civita-connection/34129#34129 Comment by Simon Salamon Simon Salamon 2010-08-02T09:11:27Z 2010-08-02T09:11:27Z Yes, but any such section $s$ that passes through $p\in P$ is unique to first order. If we set $a_{ijk} = \Gamma_{ij}^r g_{rk}$ then $s$ is tangent to $P$ at $p$ iff $a_{ijk}+a_{ikj}=0$, which forces the Christoffel symbols to vanish at the point in question. http://mathoverflow.net/questions/4331/is-the-wedge-product-of-two-harmonic-forms-harmonic/33336#33336 Comment by Simon Salamon Simon Salamon 2010-07-25T23:24:50Z 2010-07-25T23:24:50Z The differential graded algebra of invariant forms on a nilmanifold M is a <i>minimal model</i> for M in Sullivan's sense because the nilpotency makes d decomposable. Unless d kills everything invariant (like on a torus), M will not therefore be formal.