Symmetric Ricci Tensor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:39:26Z http://mathoverflow.net/feeds/question/68960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68960/symmetric-ricci-tensor Symmetric Ricci Tensor Fly by Night 2011-06-27T19:47:58Z 2011-06-27T21:34:21Z <p>Let $M$ be a pseudo-Riemannian manifold. Assume that $M$ comes with a zero-torsion affine connexion $\nabla.$ There is no need for $\nabla$ to be the Levi-Civita connexion. Recall that the curvature tensor of $\nabla$ is given by $R(X,Y)Z = \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z,$ while the Ricci tensor is given by the trace of the curvature tensor: $\mbox{Ric}(Y,Z) = \mbox{trace}\left[ X \mapsto R(X,Y)Z \right].$</p> <p>The connexion $\nabla$ is called locally equi-affine if around each point $x \in M$ there is a parallel volume form, i.e. a non-vanishing, skew-symmetric, $n$-form such that $\nabla\omega = 0.$ </p> <p>In Nomizu &amp; Sasaki's "Affine Differential Geometry", in Prop 3.1 on page 14, they say that a zero-torsion connexion has symmetric Ricci tensor if and only if it is locally equi-affine.</p> <p>In fact, there is a differential one-form $\tau$ such that $\nabla_X\omega = \tau(X)\omega$ for any vector field $X \in \mathfrak{X}(M).$ Thus, we see that $\nabla$ has symmetric Ricci tensor if and only if $\tau \equiv 0.$</p> <p>However, in Section 4 of page 237, they say that $\nabla$ has symmetric Ricci tensor if and only if the exterior derivative $\mbox{d}\tau \equiv 0.$</p> <p>Clearly only one of them is correct, but the proof to Prop 3.1. is so sparing with details that I can't follow it. It's one of those proofs that if you can't prove the statement thenyou can't understand the proof.</p> <p>So, is it $\tau \equiv 0$ or $\mbox{d}\tau \equiv 0$? </p> http://mathoverflow.net/questions/68960/symmetric-ricci-tensor/68972#68972 Answer by Willie Wong for Symmetric Ricci Tensor Willie Wong 2011-06-27T21:18:17Z 2011-06-27T21:34:21Z <p>They are both correct. <code>:-)</code></p> <p>I gave a somewhat detailed write-up last year on my <a href="http://williewong.wordpress.com/2009/12/09/parallel-volume-forms/#more-231" rel="nofollow">blog</a>, but the gist of the argument is that if $\tau = 0$, then $d\tau = 0$. On the flip side, if $d\tau = 0$, locally you can lift $\tau = du$ for some function $u$, and by replacing $\omega$ with $e^{-u}\omega$ you construct the desired equi-affine volume form. </p> <p>In other words, the second statement applies to arbitrary volume forms, while the first statement that $\tau$ vanishes applies to a <em>special</em> volume form, the one that gives the equi-affine condition.</p> <hr> <p>Just to make clear that there's no contradiction. We first note that for <em>any</em> volume form (non-vanishing differential form of top degree), the expression $\nabla_X \omega = \tau(X)\omega$ holds for some one-form $\tau$ <em>depending on</em> $\omega$. Since we are interested in <em>local</em> statements, assume now our manifold is diffeomorphic to the unit ball.</p> <p>The first statement says that "$\exists \omega$ a volume form such that $\tau = 0$."</p> <p>The second statement say that "$\forall \omega$ a volume form, the corresponding $d\tau = 0$."</p> <p>The two statements are equivalent using what I sketched in the second paragraph above. </p>