## Archive for July, 2009

Don’t be scared, this has not been turned into a photo blog!  Today I received my copy of Singular Integrals I ordered on-line recently, which is funny given that I have been reading that book for a long time now (at least now I won’t have to borrow a copy from the library or from a friend).

In any case, this reminded me of something I heard once: basically, that throughout the early years of your career as a mathematician, there will be a  list of books (that will depend strongly on your research interests) that you must read completely and in full detail to the point where you are be able to reproduce their contents on command. So, wondering what that list should be for me, I piled up some books from my book shelf and took a picture.

…perhaps I should be reading some of those books instead of writing this blog post

### Solving the Monge-Ampere equation (continued… and finished).

I have been postponing this post for over a week due to lack of time, but finally here it is. This post ought to finish a series of past posts (here and here) where I have been describing the proof of existence of classical solutions to the Dirichlet problem for the Monge-Ampere equationvia the continuity method. Via the method of continuity  we reduced the question of existence of classical solutions to the problem of proving good a priori estimates for classical solutions, namely, we were trying to prove

Theorem (A priori estimate for the Monge-Ampere equation) Let $u$ be a smooth solution of

$det(D^2u)=\psi \mbox{ in } B_1$

$u= 0 \mbox{ on } \partial B_1$

There is a constant $C>0$ (depending only on $n$ and the $C^3$ norm of $\psi$) such that

$||u||_{C^{2,\alpha}}\leq C$

Last time we got almost there, using the maximum principle and the right barriers we proved the estimate

$||u||_{C^{1,1}(B_1)}\leq C$

Which is still not strong enough for our needs, so in order to finish the proof of the a priori estimate, we are going to use a powerful interior estimate for concave elliptic equations, proved independently by L.C. Evans and N. Krylov in the 80’s:
Theorem (Evans-Krylov) Let $u$ be a $C^{1,1}$ solution of the elliptic equation

$F(D^2u)=f(x)$ for $x \in B_1$

if $F$ is concave (or convex), then we have the following interior estimate

$||u||_{C^{2,\alpha_0}(B_{\frac{1}{2}})}\leq C||u||_{C^{1,1}(B_1)}$

where $C$ depends only on $F$, and $\alpha_0$ is a universal constant.

This a well known theorem, a couple of places where one can read it are the book of Gilbarg and Trudinger (last edition), or the book of Caffarelli and Cabre. More recently, Caffarelli and Silvestre have come up with a shorter proof, still based on the original ideas of Evans and Krylov, this proof is available in arxiv. Maybe I will talk about the proof in some other post, but for now I am just going to quote the result.

The Evans-Krylov theorem is an interior result, we need also control at the boundary, that is provided by a result of Krylov

Theorem (Krylov) Let $u$ be a solution to our equation, there is a universal $\beta$ and constant $C$ controlled by $|u|_\infty$ such that  for $x \in B_1$ and $y \in \partial B_1$ we have

$|D^2u(x)-D^2u(y)|\leq C|x-y|^\beta$

This actually a corollary of Krylov’s theorem, which is a more general and remarkable result about equations in non-divergence form with measurable coefficients, but again I want to focus on the Monge-Ampere equation, I will talk about Krylov’s theorem some other time, a good place to read about it is the last chapter of Kazdan’s book. With these two tools its a standard argument to show that for a constant $C$ controlled by the previous two, and for $\alpha=min(\alpha_0,\beta)$, we have

$|D^2u(x)-D^2u(y)|\leq C|x-y|^\alpha$

I won’t do it in detail, but the proof is not too hard: basically, if the two points are closer to each other than to the boundary, then the Evans-Krylov estimate (properly scaled) gives us the inequality above, otherwise, the two points are closer to the boundary than to each other, so by the estimate of Krylov we get the same inequality in this case, and thats it!

…and that finishes the proof!. I did not present the most general result to simplify the presentation(at least for the weaker a priori estimates, which is where I did most of the details), but one can work in a more general domain (as long as it is convex) and have arbitrary boundary conditions. A much more general result, which includes not only the Monge-Ampere equation but also the $k-$Hessian equation, was proven  in a  paper by Caffarelli, Nirenberg and Spruck.

### 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 $g\mapsto Ric(g)$.