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Let $M$ be a smooth, pseudo-Riemannian manifold with $\dim(M) \ge 2.$ Let $\nabla$ be any affine connection on $M$. No reason for it to be the Levi-Civita connection. All we assume is that it has zero torsion.

Given two smooth vector fields $X,Y \in \mathfrak{X}(M),$ The curvature tensor, with respect to $\nabla$, is given by

$R(X,Y) := \nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]},$

where $R(X,Y) : \mathfrak{X}(M) \to \mathfrak{X}(M).$ The Ricci curvature tensor is given by the trace:

$ \mbox{Ric}(Y,Z) := \mbox{trace} \left[ X \mapsto R(X,Y)Z \right].$

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 $r \ge 0$.

The volume element is expressed in terms of the symmetric bilinear form $h$ that is the pseudo-Riemannian metric. We have:

$\mbox{Vol}_h(X_1,\ldots,X_n) := \sqrt{|\det\left( h_{i,j} \right)|}, \ \mbox{ where } \ h_{i,j} := h(X_i,X_j).$

Again, recall that $\nabla$ need not be the Levi-Civita connection on the pseudo-Riemannian manifold $M$. In other words $\nabla h$ need not be identically zero.

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 symmetric bilinear form.

I would appreciate some information and some references as to how to improve my geometrical intuition.

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    $\begingroup$ What relationship are you assuming between the pseudo-Riemannian metric and the connection? You appear to saying there is none. Then why is the metric needed at all? $\endgroup$
    – Deane Yang
    Commented Jul 3, 2011 at 11:11
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    $\begingroup$ I agree with Deane Yang that there has to be some relation between the connection and the metric in order to have any relation between the exponential balls of a connection and the metric. However, perhaps you mean to measure the volume of balls using a volume form which is parallel for the connection. Most connections have no such volume form, and even when a volume form is parallel for a connection, so is any constant multiple, so the question wouldn't make sense then either. $\endgroup$
    – Ben McKay
    Commented Jul 3, 2011 at 14:49
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    $\begingroup$ Indeed, you can view $Ricci(v,v)$, up to a constant factor, as the average sectional curvature of planes containing a unit vector $v$. And, yes, a trace is an average. If you have a symmetric $n$-by-$n$ matrix $A$, then the trace of $A$ is equal to $n$ times the average of $v\cdot Av$ over all unit vectors $v$. $\endgroup$
    – Deane Yang
    Commented Jul 3, 2011 at 18:38
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    $\begingroup$ That's what I thought: the trace sums the eigenvalues, and that should give an average. Although I think that's why the normalised Ricci tensor is defined (divide the ordinary one through by the dimension of M). I can understand the geometry behind $\mbox{Ric}(X,X)$. But then what does $\mbox{Ric}(X,Y)$ mean, and why should it differ from $\mbox{Ric}(Y,X)$ for $X≠Y$? I've been using all of these tensors to prove some nice little theorems. But one of my colleagues asked me: "But what does that mean, geometrically?" And I had to wave a white flag with a red face! $\endgroup$
    – Fly by Night
    Commented Jul 3, 2011 at 21:19
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    $\begingroup$ I disagree about having to rewrite anything. The original statement of my question was a very good illustration of my ack of understanding. Several patient and kind volunteers have identified problems with my original post, which identifies problems with my (lack of) understanding. This comment thread, and the one below, have identified and addressed problems in my understanding, and have suggested ways of combating that misunderstanding. I would have said that that was the aim of a question: to identify and to tackle one's misunderstanding. $\endgroup$
    – Fly by Night
    Commented Jul 5, 2011 at 2:30

3 Answers 3

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NB: I'm combining my previous comments into an answer, because I believe that this is better than leaving them scattered.

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.

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 $$ \exp^\ast_p(\Upsilon)=(1 - \tfrac13 R_{ij} x^ix^j + \cdots)\ dx^1\wedge dx^2\wedge\cdots\wedge dx^n, $$
where $\exp^\ast_p\bigl(\mathrm{Ric}(\nabla)\bigr)_p = R_{ij}\, dx^idx^j$. (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.)

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Disclaimer: This is not an answer but just a comment, but I needed some more space.

Dear Fly by night, I have the same your problem with the reported statement, where do you have read it?

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 $\mathbb{R}^n$.

$$\frac{\mbox{vol}_g(B_m(r))}{\mbox{vol}_{\mbox{eucl}}(B(r))}=1-\frac{\mbox{Scal}_g(m)}{6(n+2)}r^2+o(r^2)$$

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 $(M,g)$.

So could you give me a reference? I'm curious!

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  • $\begingroup$ For the Levi-Civita connection of a pseudo-Riemannian metric, you can get the above formula by just expanding the the metric in normal coordinates and compute. $\endgroup$
    – Willie Wong
    Commented Jul 3, 2011 at 13:27
  • $\begingroup$ Thanks for your comment Giuseppe. Please see my comment above. My source does (secretly) assume Levi-Civita. So I guess we should forget the geodesic ball example. However, I'd till like to have some geometrical interpretation of the trace: does it "average things out"? What, geometrically, does the Ricci Tensor mean? What, geometrically, does its (non-)symmetry mean? $\endgroup$
    – Fly by Night
    Commented Jul 3, 2011 at 18:36
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    $\begingroup$ You already have the answer about the non-symmetric part. The first Bianchi identity shows that the skew-symmetric part of the Ricci tensor is a 2-form $\Omega$ that is equal (up to a universal constant) to the trace of the full curvature tensor. This $\Omega$ is the curvature of the induced connection on the top exterior power of the (co)tangent bundle, and hence this vanishes if and only if there is a $\nabla$-parallel volume form. Alternatively, the integral of this 2-form over a compact oriented surface $S$ in $M$ is the holonomy of the connection around $\partial S$. $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2011 at 2:08
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    $\begingroup$ In indices: First Bianchi is $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$. 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 induced connection on the top exterior power, and $\frac12(R^i_{kil}-R^i_{lik})dx^k\wedge dx^l$ is the skew-symmetric part of the Ricci tensor. I guess the universal constant is $1$. $\endgroup$
    – Robert Bryant
    Commented Jul 5, 2011 at 12:13
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    $\begingroup$ One more comment: If you assume 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 $\exp^\ast_p(\Upsilon)$ can be written as $(1 - \frac13 R_{ij} x^ix^j + \cdots)$ times a constant coefficient volume form on the vector space $T_pM$, where $\exp^\ast_p\bigl($Ric$(\nabla)\bigr)_p = R_{ij}\, dx^idx^j$. Thus, Ric gives the deviation of the parallel volume form from the exponentially flat one. $\endgroup$
    – Robert Bryant
    Commented Jul 6, 2011 at 17:16
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you can study following article, it is help you to visualize the Riemannian curvatures include Riemann tensor, ricci tensor,... it is very useful:

http://www.yann-ollivier.org/rech/publs/visualcurvature.pdf

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