# Example of a degenerate metric which doesn't have the Levi-Civita connection

The proof of existence of the Levi-Civita connection for pseudo-Riemannian manifolds uses heavily the fact that the metric is non-degenerate - so that $\nabla_XY$ is characterized by all the values $\langle \nabla_XY,Z\rangle$ via the Koszul formula.

I tried to come up with an example of a manifold with a degenerate metric which doesn't have the Levi-Civita connection, something like the flat metric $$\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$on $\Bbb R^2$, but I'm not sure how to come up with a contradiction (there should be an easier way than brute forcing my way through the Koszul formula). Is this the right way? Thanks.

• What do you mean by "a degenerate metric which doesn't have the Levi-Civita connection"? The term "the Levi-Civita connection" only has meaning once you've proved that such a connection exists and is unique, which requires a pseudo-Riemannian manifold. If you're asking whether a degenerate metric admits a connection that is symmetric and compatible with the metric, the answer is yes -- in fact, there are many of them. The thing that goes wrong in the case of a degenerate metric is the uniqueness. – Jack Lee May 24 '16 at 0:16
• Ah. I thought something would already go wrong with existence. This helps a lot, thanks! It would be nice to see some examples, but I don't recall seeing any Riemannian geometry book discussing this kind of "pathology" (granted, I'm familiar with only a few books in the subject). Do you know any specific reference for this? Thanks again :-) – Ivo Terek May 24 '16 at 0:37
• No, I don't know of any references. But I'll post an example below. – Jack Lee May 24 '16 at 1:18

If $g$ is a degenerate metric, then typically there will be lots of connections that are symmetric and compatible with $g$. For example, let $g$ be the constant-coefficient degenerate metric on $\mathbb R^2$ that you mentioned. Then a straightforward computation shows that the standard Euclidean connection $\nabla$ is compatible with $g$. Since it's symmetric (which has nothing to do with the metric), it would qualify as a "Levi-Civita connection" for $g$.
Now define a new connection $\widetilde\nabla$ by $$\widetilde\nabla_X Y = \nabla_X Y + \langle X,Y\rangle \frac{\partial}{\partial y}.$$ Then $\widetilde\nabla$ is still symmetric and compatible with $g$ (because the vector field $\partial/\partial y$ is orthogonal to everything with respect to $g$).