Ellipticity of Ricci tensor, does it depend on coordinates?

Question

Well, I am afraid this is a silly question because I know the answer must be 'yes, it does'. But I don't see why. I put the problem in context.

The ricci tensor can be regarded as a differential operator acting of metrics (it is an expression that depends on second and first derivatives of the metric). You give the ricci a metric on your manifold $M$ an the ricci gives you a symmetric two tensor. It is non linear. $$ Ric: \{\text{metrics on } M\} \to \{\text{symmetric two-tensors on }M\} $$ So one says that the ricci tensor is elliptic at the metric $g$ if its differential at $g$ denoted by $L:=d_g Ric$ is elliptic. Note that since {metrics on $M$} are an open set of the vector space of symmetric two-tensors, we have $$ L: \{\text{symmetric two tensors}\} \to \{\text{symmetric two tensors}\} $$ and $L$ is a linear differential operator of order two.

Recall that $L$ is by definition elliptic if for every $\xi \in T^*_x M$ its principal symbol $$ p(\xi,x): \{\text{symmetric bilinear forms on } x \} \to \{\text{symmetric bilinear forms on } x \} $$ is an isomorphism for $\xi \ne 0$. In coordinates the principal symbol is obtained just by taking the part that involves two derivatives, and in that part changing every $\nabla_i$ you find in for $\xi_i$, the $i-th$ coordinates of $\xi$. It turns out that $p(\xi,x)$ does not depend or coordinates regarded as a endomorphism between vector spaces.

Indeed, let $z$ be a symmetric bilinear form on $T_x M$ and let $u$ be a symmetric two tensor on $M$ such that $u(x)=z$. Let $\phi$ be a smooth function on $M$ with $\phi(x)=0$ and $d_x \phi =\xi$. Then $p(x,\xi)z = L(\phi ^2 u)|_x$. This invariant definition is on Besse's Einstein Manifolds.

So the principal symbol regarded as a endomorphism between vector spaces does not depend on coordinates, so being elliptic should not depend on coordinates either. But the books, for example Besse's Einstein Manifolds, says that the ricci tensor is elliptic in harmonic coordinates, and it is not elliptic in any coordinates.

I must say that is never explicitly said that the ricci operator is not elliptic in any coordinates, but I deduce that because the elliptic regularity results are not true in any coordinates.

The question is ¿how is it possible that being elliptic depens on coordinates?

Thank you for any help.

Answer 1

As you observed, the ellipticity of the Ricci operator (or any partial differential operator) does not depend on a choice of coordinates. The Ricci operator is never elliptic.

If you assume that the coordinates $(x^i)$ are harmonic for the metric $g$, and you write the coordinate expression for $\text{Ric}(g)$ in these coordinates, some terms cancel because of the harmonicity condition, and you end up with an expression of the form $\text{Ric}(g) = P(g)$, where $P$ is an elliptic nonlinear second-order differential operator. This is what people mean when they say "the Ricci tensor is elliptic in harmonic coordinates." However, the expression $P(g)$ does not give the Ricci tensor for every metric $g$; only for those metrics for which the given coordinates happen to be harmonic.

The most important use for harmonic coordinates is in proving regularity theorems. For example, if you can show that the Ricci tensor is smooth in harmonic coordinates, then in those coordinates the metric satisfies the elliptic equation $P(g) = \text{Ric}$, and it follows from elliptic regularity that the metric itself is smooth in those coordinates. A similar argument can be used to show that every Einstein metric is real-analytic in suitable coordinates.

It is possible to use variations on this idea to prove existence of Einstein metrics, or metrics with prescribed Ricci tensor, in some cases. Basically, the idea is to write down a pair of equations, one of which says $P(g)=\text{something}$, and the other of which says that the given coordinates are harmonic coordinates for $g$. You can think of this as an overdetermined elliptic system for $g$, or as a coupled system of equations for both $g$ and the coordinates.

Actually, for most existence results, it's better to use a more global condition in place of harmonic coordinates, so that you're not limited to working in one coordinate chart. There are various such conditions (typically called "gauge-breaking conditions") that have been used. The classic one is called the DeTurck trick. Robin Graham and I explained the relationship between the DeTurck trick and harmonic coordinates in the introduction to this article.