## Archive for the 'Differential geometry' Category

### If you want to kill your productivity, move to a new house!

I know things have been extremely slow lately, but I was moving last week (and well, that means packing everything, moving things to the new house, cleaning the old house, unpacking things at the new one… well, you get the picture).

My goal for this week is presenting Aleksandrov’s solution to  the Minkowski problem (see an earlier post I did introducing this problem). So I am going to leave you a problem as a preview, it is a sort of discrete version of the Minkowski problem:

Let $n_1,...,n_k$  be a family of non-coplanar unit vectors in $\mathbb{R}^3$ and let $\alpha_1,...,\alpha_k$ be positive numbers such that

$\sum \limits_{i=1}^k \alpha_in_i=0$

Then, show that there exists a convex closed polyhedron with exactly $k$ faces with normal vectors given by $n_1,...,n_k$ and corresponding areas $\alpha_1,...,\alpha_k$. Plus, this polyhedron is unique up to translation.

This is fact is not surprising (since it is not hard to check that any polyhedron has this property), but the proof is far from trivial. As you may guess, the proof cannot be constructive, it will use a continuity argument to show that there must be at least one such polyhedron. I will present this in my next post.

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

### Quick post: A theorem and an open problem about Gauss curvature

While I finish my next post on the Dirichlet problem for the Monge-Ampere equation, I thought I could mention two neat things I have learned from reading Jerry Kazdan’s survey on Prescribing the curvature of a Riemannian manifold, a short book that I recommend strongly (its a bit out of date, but hey!, geometry has advanced quite a bit in the last couple of years!)

First, a cute solution to an easily-stated problem, as  observed by Wallach and Warner in 1970

Theorem:  Given a compact smooth 2-d manifold and $\Omega$ is a 2-form such that $\int_M \Omega = 2\pi \chi(M)$, then there exists a smooth Riemannian metric $g$ on $M$ such that $\Omega = KdA$, where $K$ is the Gauss curvature of $g$ and $A$ is the area form of $g$.

Proof: Pick your favorite metric $g_0$ on $M$, we will  prove that the metric we are looking for is in fact conformal to $g_0$, for any smooth function $u$, define $g_u = e^{2u}g_0$. Using the well known formulas for the Gauss curvature and area form of $g_u$, one arrives at the identity

$K_udA_u=K_odA_0-(\Delta_0 u) dA_0$

where the sub-indexes $0$ and $u$ refer to the object corresponding to the metric $g_0$ and $g_u$, in particular, $\Delta_0$ is the Laplace operator for $(M,g_0)$. Then one wants to pick $u$ such that

$K_odA_0-\Delta_0 u dA_0=\Omega$

or $(\Delta_0 u) dA_0 = K_0dA_0-\Omega$

but, using the metric $g_0$ we can write $\Omega = f(x)dA_0$ for some $f$ (or equivalently, using the Hodge star operator given by $g_0$) thus the $u$ we want is the solution of the equation

$\Delta_0 u = K_0-f (**)$

but since $\int_M \Omega =\int_MfdA_0= 2\pi\chi(M)=\int_MK_0dA_0$ we have that $\int_MK_0-fdA_0=0$ thus by standard elliptic theory (or Fredholm theory, etc) there exists a smooth function $u$ solving equation (**). Thus the metric $g_u$ is the one we were looking for and the theorem is proved.

Now, an easily stated problem whose solution is not likely to be as short, and has led to a lot of research in the last 2 decades. I  read about it for the first time in an interview with Louis Nirenberg, which can be found here.

Open problem:  Given a two dimensional Riemannian manifold, can we embed it isometrically (even just locally) in three dimensional Euclidean space?

Recall that Gauss curvature solely determines the local geometry of a 2d manifold, so its not surprising that this problem is equivalent to the following problem involving the  (…tataaaaa! )  Monge-Ampere equation:

Open problem restated: Given a $C^\infty$ function $K(x,y)$, find (even just locally) a function $u(x,y)$ which solves the equation

$\frac{u_{xx}u_{yy}-u_{xy}^2}{\left (1+u_x^2+u_y^2 \right)^2} = K(x,y)$

but mind you, this is not your grandpa’s Monge Ampere equation, for it is not an elliptic equation unless $K$ is strictly positive. For general $K$, it is an equation that varies  between hyperbolic or elliptic according to the sign of $K$, so you run into real problems in the set of points where $K$ vanishes .  If on the other hand $K$ is strictly negative, Kazdan says that the equation above is solved using tools from non-linear hyperbolic equations, of which sadly I know nothing. The case $K >0$ is then dealt with the techniques from the last two and the next post, which I should finish in a day or two.