# Christoffel symbols and metric compatibility

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.

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.

• I am not getting my answer. My question is can we generalize the same procedure for nonEuclidean geometry also? Take only metric compatibility Jan 22 at 16:56
• 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).
– Gold
Jan 22 at 17:15
• 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.
– Gold
Jan 22 at 17:17