In some coordinate system ,we can calculate the Christoffel symbols using the following procedure

Basis vectors $\Gamma^k_{ij}\vec{e_k}=\frac{\partial \vec{e_i}}{\partial x^j}$

multiply $\vec{e_l}$ both side $$ \frac{\partial \vec{e_i}}{\partial x^j}.\vec{e_l}=\Gamma^k_{ij}g_{kl} $$ interchanging $(i,l)$ and adding $$ \frac{\partial g_{il}}{\partial x^j}=\Gamma^k_{ij}g_{kl}+\Gamma^k_{lj}g_{ki} $$ Rearranging $$ \frac{\partial g_{il}}{\partial x^j}-\Gamma^k_{ij}g_{kl}-\Gamma^k_{lj}g_{ki}=0 $$ Rearranging the indices we can get the Christoffel symbols in terms of partial of the metric. Only thing I have to assume the torsion free condition. But in curved spacetime the change will happen in the first definition, $\Gamma^k_{ij}\vec{e_k}=\nabla_j \vec{e_i}$. Then will this procedure of deriving the Christoffel symbol work.


1 Answer 1


You seem to be mixing torsion-free and metric-compatible. In curved spacetimes you can still demand that the connection be metric-compatible and torsion-free, and get the unique Levi-Civita connection. You can also drop the torsion-free requirement if you like.

Let $(M,g)$ be a semi-Riemannian manifold and let $\nabla$ be an arbitrary connection on $TM$. We say that $\nabla$ is metric-compatible if $\nabla g=0$. On the other hand we define the torsion of $\nabla$ by $$T(X,Y)=\nabla_XY-\nabla_Y X-[X,Y]\tag{1}.$$

We say that $\nabla$ is torsion free if $T=0$. In local coordinates, $\nabla$ can be specificied by its action on the coordinate basis vector fields. In other words:

$$\nabla_{\mu}\partial_\nu=\Gamma^\sigma_{\phantom{\sigma}\mu\nu} \partial_\sigma\tag{2}.$$

The quantities $\Gamma_{\phantom\sigma\mu\nu}^\sigma$ are the connection coefficients and they specify $\nabla$. You can show that the components of the torsion tensor are $T^\sigma_{\phantom{\sigma}\mu\nu}=\Gamma^\sigma_{\phantom\sigma[\mu\nu]}$, i.e., they are the anti-symmetric part of the connection coefficients.

Now, if you demand that $\nabla$ be both metric-compatible and torsion-free, you can show that there is a single connection satisfying these conditions. Indeed, these conditions allow you to completely determine $\Gamma^\sigma_{\phantom\sigma\mu\nu}$

$$\Gamma^\sigma_{\phantom\sigma\mu\nu}=\dfrac{1}{2}g^{\sigma\rho}(\partial_\mu g_{\rho\nu}+\partial_\nu g_{\mu\rho}-\partial_\rho g_{\mu\nu})\tag{3}.$$

It is also straightforward to check that this quantity is symmetric in $\mu\leftrightarrow \nu$ showing that indeed $T=0$.

Now you can still demand a metric-compatible connection and drop the torsion free requirement. What you will find this time is that metric compatibility still fixes the symmetric part $\Gamma^\sigma_{\phantom\sigma(\mu\nu)}$ to be given by (3), but the anti-symmetric part $T^\sigma_{\phantom\sigma\mu\nu}=\Gamma^\sigma_{\phantom\sigma[\mu\nu]}$ now is non-zero and is not fixed by the metric-compatibility requirement. What happens now is that you don't have therefore a unique connection obeying the constraints you imposed.

  • $\begingroup$ I am not getting my answer. My question is can we generalize the same procedure for nonEuclidean geometry also? Take only metric compatibility $\endgroup$ Jan 22 at 16:56
  • $\begingroup$ Yes, I said that before equation (2). Apply $\nabla_\sigma$ to $g(\partial_\mu,\partial_\nu)$. On the one hand since $g(\partial_\mu,\partial_\nu)=g_{\mu\nu}$ is $\mathbb{R}$-valued, $\nabla_\sigma$ acts on it as $\partial_\sigma$. On the other hand using the product rule $$\nabla_\sigma(g(\partial_\mu,\partial_\nu))=(\nabla_\sigma g)(\partial_\mu,\partial_\nu)+g(\nabla_\sigma \partial_\mu,\partial_\nu)+g(\partial_\mu,\nabla_\sigma\partial_\nu).$$ Using metric compatibility the first term in the RHS vanishes. The other two are calculated using (2). $\endgroup$
    – Gold
    Jan 22 at 17:15
  • $\begingroup$ Combining the two results above we have the equation $$\partial_\sigma g_{\mu\nu}=\Gamma_{\phantom\rho\sigma\nu}^{\rho}g_{\rho\nu}+\Gamma_{\phantom\rho\mu\sigma}^\rho \partial_\rho g_{\rho\nu}.$$ One may then solve for the symmetric part $\Gamma^\sigma_{\phantom\sigma \mu\nu}$ by taking combinations of this equation after a few index exchanges. BTW, you may also find this whole procedure in any good GR textbook like Carroll's or Wald's. $\endgroup$
    – Gold
    Jan 22 at 17:17

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