### Prescribing the Ricci Curvature of Riemannian metrics

I would like to spend the next few posts talking about a problem I read about in chapter 5 of the book Einstein Manifolds by Arthur Besse (it turns out that the name Arthur Besse is made up as you can read in the preface of the book). The question that we want to address is the following

Given a compact smooth manifold without boundary $M$, when is it possible to find a Riemannian metric $g$ in $M$ satisfying $Ric(g)=r$ for a given Ricci candidate $r$?

It is of course too ambitious to try to answer this question in full generality but we can start by showing some examples of Ricci candidates for which this equation does not have a solution.

Trying to solve $Ric(g)=$ for $g$ amounts to solving a second order, quasilinear PDE on $g$, however, the main difficulty here is that the operator $g\mapsto Ric(g)$ is not elliptic.

A motivation for considering this problem comes from the question of existence of metrics with constant sectional curvature on $3$– manifolds (compact and without boundary). This of course has to do with the celebrated theorem of Richard Hamilton on the description of $3$– manifolds with positive Ricci curvature:

Theorem (Hamilton, 1982): Let $M$ be a connected, compact smooth $3$ dimensional manifold without boundary and suppose that $M$ admits a metric $g$ such that $Ric(g)$ is positive definite everywhere. Then $M$ also admits a metric with constant sectional curvature.

We will discuss some of the ideas involved in the proof of this theorem in future posts. A consequence of this result is that $M$ is diffeomorphic to the quotient of the $3$-sphere $S^{3}$ by a discrete group $G$.

Back to our original problem, recall that given a Riemannian metric $(M,g)$, the full curvature tensor is defined by

$Rm(X,Y,Z,W)=g(R(X,Y)Z,W)=\langle R(X,Y)Z,W\rangle$

Where

$R(X,Y)Z=\nabla_{X}\nabla_{Y}Z-\nabla_{Y}\nabla_{X}Z-\nabla_{[X,Y]}Z$

Here $\nabla$ is the Levi-Civita connection of $g$.

The Ricci tensor is then defined as

$Ric(X,Y)=\mathrm{tr}\left(Z\mapsto R(Z,X)Y\right)$.

Here are two basic properties of $Ric$:

1) $Ric$ is symmetric in $X$ and $Y$, i.e. $Ric(X,Y)=Ric(Y,X)$

2) In local coordinates $Ric$ looks as follows

$R_{ij}=\partial_{l}\Gamma_{ij}^{l}-\partial_{i}\Gamma_{lj}^{l}+\Gamma_{lm}^{l}\Gamma_{ij}^{m}-\Gamma_{im}^{l}\Gamma_{lj}^{m}$

We are using the summation convention (i.e we sum over repeated indices). The Christoffel symbols $\Gamma_{\alpha\beta}^{\gamma}$ are defined by

$\Gamma_{\alpha\beta}^{\gamma}=\frac{1}{2}g^{\gamma\nu}\left(\partial_{\alpha}g_{\beta\nu}+\partial_{\beta}g_{\alpha\nu}-\partial_{\nu}g_{\alpha\beta}\right)$

Where $\left(g^{\gamma\nu}\right)$ are entries of the matrix $g^{-1}$. This says that in local coordinates we can write schematically $Ric(g)=F(g,\partial g,\partial^{2}g)$ where $F$ is a $C^{\infty}$ function that depends linearly on the entries of $\partial^{2}g$. If we then want to solve $Ric(g)=r$ locally, property 2) tells us that we have to look at a system of the form

$F(g,\partial g,\partial^{2}g)=r$

Property 1) tells us that an admissible Ricci candidate $r$ has to be symmetric.

One encounters obstructions for solving this system right away. One of the main difficulties has to do with the fact that the Ricci tensor satisfies the differential Bianchi identity

$\delta Ric(g)=-\frac{1}{2}dR$

Where $\delta$ is the divergence operator respect to $g$ and $R$ is the scalar curvature of $g$ (the trace of the Ricci tensor). This says that if we define a 1-form $Bian(r,g)$ by $Bian(r,g)=\delta r+\frac{1}{2}d\mathrm{tr}(r)$, then in order for $g$ to satisfy $Ric(g)=r$ we must have

$Bian(r,g)=0$

To write $Bian(r,g)$ in coordinates, we start with

$Bian(r,g)_{k}=g^{ij}\left(\nabla_{i}r_{jk}-\frac{1}{2}\nabla_{k}r_{ij}\right)$

From

$\nabla_{a}r_{bc}=\partial_{a}r_{bc}-\Gamma_{ab}^{l}r_{lc}-\Gamma_{ac}^{l}r_{bl}$

(which is just the definition of $\nabla r$ in coordinates) and from the expression in local coordinates of the symbols $\Gamma$, we easily see that

$Bian(r,g)_{k}=g^{ij}\left[\partial_{i}r_{jk}-\frac{1}{2}\partial_{k}r_{ij}-g^{ls}r_{ks}\left(\partial_{i}g_{js}-\frac{1}{2}\partial_{s}g_{ij}\right)\right]$

As discussed in Besse’s book, Dennis DeTurck came up with examples of symmetric tensors that cannot satisfy the Bianchi identity respect to any metric. One of his examples is the following

Consider in $\mathbb{R}^{n}$ a symmetric tensor of the form

$r=x_{1}dx_{1}\otimes dx_{1}+\displaystyle{\sum_{1

The existence of a metric $g$ such that $Ric(g)=r$ implies as we saw before that $Bian(r,g)=0$. In particular, from our expression for $Bian$ we must have

$0= Bian(r,g)_{1}=g^{ij}\left[\partial_{i}r_{j1}-\frac{1}{2}\partial_{1}r_{ij}-g^{ls}r_{1s}\left(\partial_{i}g_{js}-\frac{1}{2}\partial_{s}g_{ij}\right)\right]=$

$=\frac{1}{2}g^{11}+x_{1}g^{l1}g^{ij}\left(\partial_{i}g_{js}-\frac{1}{2}\partial_{s}g_{ij}\right)$

This implies that on the hyperplane $x_{1}=0$, the metric $g$ must satisfy $g^{11}=0$ which is impossible for a Riemannian metric. It follows that for any point $p$ in the hyperplane $x_{1}=0$ the equation $Ric(g)=r$ has no solution near $p$. Notice that at these points the tensor $r$ is singular.

In the next post we will interpret the existence of examples like the one we have just discussed as a consequence of the non-ellipticity of the system $g\mapsto Ric(g)$.

#### 4 Responses to “Prescribing the Ricci Curvature of Riemannian metrics”

1. July 10, 2009 at 3:30 pm

Thanks for the nice post Antonio, I have a question, what do you mean by the tensor $r$ being singular?

2. 2 aache July 10, 2009 at 10:58 pm

What I mean by $r$ being singular at a point $p$, is that $r(p)$ as a map from $T_{p}M$ to $T^{*}_{p}M$ is not invertible. In this particular example for any point $p$ in $x_{1}=0$ we have $r(p)_{1i}=0$ for all $i=1,\ldots, n$, so the matrix representing $r(p)$ is singular. Thanks for pointing that out Nestor, I realized I should been more careful about it.

3. 3 Ngô Quốc Anh November 26, 2009 at 9:14 am

Hi, thanks for the post. Could you explain a little bit what differences between Ricci Tensor and Ricci Curvature?

4. 4 aache November 27, 2009 at 9:35 pm

Hi, actually the Ricci Tensor and the Ricci Curvature of a Riemannian metric are the same thing. You can get very valuable information about the full curvature tensor of a Riemannian metric from the Ricci tensor, so I guess that is why people sometimes call it Ricci Curvature. For instance, in dimension 3 the Riemann curvature tensor of a metric is completely determined by its Ricci tensor, that tells you that in dimension 3 a metric has constant sectional curvature if and only if it has constant Ricci Curvature, i.e., if and only if Ric(g)=kg for some constant k. In that case the sectional curvature of g is k/4. In higher dimensions this is not true, but the Ricci Tensor of a metric still has very important geometric information. You can find more information about the Ricci Tensor in dimension 4 here http://www.math.sunysb.edu/~claude/luminy.pdf