As is commonly known, the connection 1-forms of a Riemannian manifold are skew-symmetric: $\omega^i_j=-\omega^j_i$. Until now, I have not actually thought to hard on this, but I think I've hit a snag.

Let's consider the metric $e^{-(x^2+y^2)}(dx^2+dy^2)$ on $\mathbb{R}^2$. Then, by computation, the Christoffel symbols of the Levi-Civita connection are $\Gamma^1_{11}=\Gamma^2_{12}=-\Gamma^1_{22}=x$ and $-\Gamma^2_{11}=\Gamma^1_{12}=\Gamma^2_{22}=y$. Organizing these into a matrix 1-form gives us $$\begin{bmatrix}xdx+ydy & ydx-xdy \\ xdy-ydx & xdx+ydy\end{bmatrix},$$ which is not skew symmetric on the diagonal (i.e. nonzero on the diagonal).

What have I done wrong here? Have I simply misinterpreted something? I've double checked the calculations, so I'm fairly certain it is not a computation problem.

To be clear, I know how to prove that they are skew-symmetric using the fact that their inner products will be constant. I simply want to know what I did wrong in this instance.

  • 7
    $\begingroup$ The connection forms are only skew-symmetric when you compute them with respect to an orthonormal frame. The coordinate frame $\{\partial/\partial x, \partial/\partial y\}$ is not orthonormal for your metric. $\endgroup$ – Jack Lee Nov 15 '15 at 21:16
  • $\begingroup$ Jack Lee is correct. Also, you might want to triple check your calculations. I think you need to multiply everything by $-1$. $\endgroup$ – Robin Goodfellow Nov 15 '15 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.