Geometrical meaning of the Ricci Tensor and its Symmetry - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:34:25Z http://mathoverflow.net/feeds/question/69374 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69374/geometrical-meaning-of-the-ricci-tensor-and-its-symmetry Geometrical meaning of the Ricci Tensor and its Symmetry Fly by Night 2011-07-03T00:21:00Z 2012-08-26T13:56:18Z <p>Let $M$ be a smooth, pseudo-Riemannian manifold with $\dim(M) \ge 2.$ Let $\nabla$ be <em>any affine connection</em> on $M$. No reason for it to be the Levi-Civita connection. All we assume is that it has zero torsion.</p> <p>Given two smooth vector fields $X,Y \in \mathfrak{X}(M),$ The curvature tensor, with respect to $\nabla$, is given by </p> <p>$R(X,Y) := \nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]},$</p> <p>where $R(X,Y) : \mathfrak{X}(M) \to \mathfrak{X}(M).$ The Ricci curvature tensor is given by the trace:</p> <p>$\mbox{Ric}(Y,Z) := \mbox{trace} \left[ X \mapsto R(X,Y)Z \right].$</p> <p>I've read that the Ricci curvature tensor measures the second order deviation between the volume of a $\nabla$-geodesic ball and a standard Euclidean geodesic ball. This explanation causes me problems. A geodesic ball, centre $x \in M$ and radius $r$ is given by following each $\nabla$-geodesic, that passes through $x$, a distance $r$ with respect to the pseudo-Riemannian metric on $M$. Besides $\nabla$, this depends only on $x \in M$ and $r \ge 0$.</p> <p>The volume element is expressed in terms of the <em>symmetric</em> bilinear form $h$ that is the pseudo-Riemannian metric. We have:</p> <p>$\mbox{Vol}_h(X_1,\ldots,X_n) := \sqrt{|\det\left( h_{i,j} \right)|}, \ \mbox{ where } \ h_{i,j} := h(X_i,X_j).$</p> <p>Again, recall that $\nabla$ <em>need not be the Levi-Civita connection</em> on the pseudo-Riemannian manifold $M$. In other words $\nabla h$ need not be identically zero.</p> <p>I know how to manipulate the tensor and connection notation. But my geometrical insight is lacking. I don't see how the symmetry, or non-symmetry, of $\mbox{Ric}$ should have any relation to the volume of a ball, which is determined with respect to a <em>symmetric</em> bilinear form.</p> <p>I would appreciate some information and some references as to how to improve my geometrical intuition. </p> http://mathoverflow.net/questions/69374/geometrical-meaning-of-the-ricci-tensor-and-its-symmetry/69383#69383 Answer by Giuseppe for Geometrical meaning of the Ricci Tensor and its Symmetry Giuseppe 2011-07-03T07:50:48Z 2011-07-03T07:50:48Z <p><strong>Disclaimer</strong>: This is not an answer but just a comment, but I needed some more space.</p> <p>Dear Fly by night, I have the same your problem with the reported statement, where do you have read it?</p> <p>Until now I have just read that in a $n$-dimensional Riemannian manifold $(M,g)$, for any $m\in M$, the value in $m$ of the scalar curvature $\mbox{Scal}_g$ gives the coefficient of the second order term in the ratio between the volume of the geodesic ball of radius $r$ centered at $m$, and the Euclidean volume of the ball of radius $r$ in $\mathbb{R}^n$.</p> <p>$$\frac{\mbox{vol}_g(B_m(r))}{\mbox{vol}_{\mbox{eucl}}(B(r))}=1-\frac{\mbox{Scal}_g(m)}{6(n+2)}r^2+o(r^2)$$</p> <p>One source is §3 in Chapter XV of Fundamentals of Differential Geometry by Serge Lang. Here he is only considering the Scalar Curvature of the Riemannian manifold $(M,g)$.</p> <p>So could you give me a reference? I'm curious!</p> http://mathoverflow.net/questions/69374/geometrical-meaning-of-the-ricci-tensor-and-its-symmetry/72081#72081 Answer by Robert Bryant for Geometrical meaning of the Ricci Tensor and its Symmetry Robert Bryant 2011-08-04T12:10:11Z 2011-08-13T21:58:46Z <p>NB: I'm combining my previous comments into an answer, because I believe that this is better than leaving them scattered.</p> <p>As another commenter has pointed out, the skew-symmetric part of the Ricci tensor is the obstruction to there being a $\nabla$-parallel volume form in the first place. To see this, consider the first Bianchi identity: $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$. Set $i=j$ and sum to get $R^i_{ikl}+R^i_{kli}+R^i_{lik}=0$, which becomes $R^i_{ikl}=R^i_{kil}-R^i_{lik}$. Now $\Omega = \frac12 R^i_{ikl}\ dx^k\wedge dx^l$ is the curvature of the connection induced by $\nabla$ on the top exterior power of the cotangent bundle, and $\frac12(R^i_{kil}{-} R^i_{lik})dx^k\wedge dx^l$ is the skew-symmetric part of the Ricci tensor. Thus, the vanishing of the skew-symmetric part of Ricci is equivalent to the flatness of this induced connection on the top exterior power.</p> <p>Assume now that the Ricci curvature is symmetric, so that there is a (local) $\nabla$-parallel volume form, say, $\Upsilon$. Then the Ricci curvature has the following interpretation: Let $\exp_p:T_pM\to M$ be the exponential map of $\nabla$ based at $p$. Then $$\exp^\ast_p(\Upsilon)=(1 - \tfrac13 R_{ij} x^ix^j + \cdots)\ dx^1\wedge dx^2\wedge\cdots\wedge dx^n,$$<br> where <code>$\exp^\ast_p\bigl(\mathrm{Ric}(\nabla)\bigr)_p = R_{ij}\, dx^idx^j$</code>. (Here, the $x^i$ are any linear coordinates on $T_pM$ centered at $0_p$ that are $\Upsilon$-unimodular at $0_p$.) Thus, Ric gives the deviation of the parallel volume form from the exponentially flat one. (This makes sense, even though you can't define 'geodesic balls' without a metric. You still compare the volume of open neighborhoods of $p$ with respect to the two 'natural' volume forms.)</p> http://mathoverflow.net/questions/69374/geometrical-meaning-of-the-ricci-tensor-and-its-symmetry/105532#105532 Answer by S.B for Geometrical meaning of the Ricci Tensor and its Symmetry S.B 2012-08-26T13:56:18Z 2012-08-26T13:56:18Z <p>you can study following article, it is help you to visualize the Riemannian curvatures include Riemann tensor, ricci tensor,... it is very useful:</p> <p><a href="http://www.yann-ollivier.org/rech/publs/visualcurvature.pdf" rel="nofollow">http://www.yann-ollivier.org/rech/publs/visualcurvature.pdf</a></p>