Transformation of Christoffel symbol

Question

I don't quite get the result that I should get.

We have $$\Gamma^\bar{\mu}_{\bar{\alpha}\bar{\beta}}=\frac{1}{2}g^{\bar{\mu}\bar{\nu}}\left(\partial_\bar{\alpha}g_{\bar{\beta}\bar{\nu}}+\partial_\bar{\beta}g_{\bar{\alpha}\bar{\nu}}-\partial_\bar{\nu}g_{\bar{\alpha}\bar{\beta}}\right).$$

Furthermore we find $$\partial_\bar{\alpha}g_{\bar{\beta}\bar{\nu}}=g_{\epsilon\tau}\left( x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}}+x^\epsilon_{, \bar{\beta}} x^\tau_{,\bar{\nu}\bar{\alpha}}\right)+x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\beta}}\partial_\lambda g_{\epsilon\tau}.$$

Here, I've written $x^\tau_{\bar{\nu}}=\partial_{\bar{\nu} }x^\tau.$

Substituting the partial derivatives of the metric in the Christoffel symbol I get two terms. One is $$\partial_\lambda g_{\epsilon\tau} \left(x^\epsilon_{, \bar{\beta}}x^\tau_{, \bar{\nu}} x^\lambda_{,\bar{\alpha}}+x^\epsilon_{, \bar{\alpha}}x^\tau_{, \bar{\nu}} x^\lambda_{,\bar{\beta}}-x^\epsilon_{, \bar{\alpha}}x^\tau_{, \bar{\beta}} x^\lambda_{,\bar{\nu}}\right). $$ By renaming the indices I obtain $$x^\epsilon_{, \bar{\beta}}x^\tau_{, \bar{\nu}} x^\lambda_{,\bar{\alpha}}\left(\partial_\lambda g_{\epsilon\tau}+\partial_\epsilon g_{\lambda\tau}-\partial_\tau g_{\lambda\epsilon}\right) .$$

Now, for the second term I get $2g_{\epsilon\tau}x^\epsilon_{,\bar{\beta}\bar{\alpha}}x^\tau_{,\bar{\nu}}$ after using the interchangeability if the second derivatives, the symmetry of the metric tensor and switching dummy variables in one term.

So, at the end I have $$\Gamma^\bar{\mu}_{\bar{\alpha}\bar{\beta}}=g^{\bar{\mu}\bar{\nu}} g_{\epsilon\tau}x^\epsilon_{,\bar{\beta}\bar{\alpha}}x^\tau_{,\bar{\nu}}+\frac{1}{2}g^{\bar{\mu}\bar{\nu}}x^\epsilon_{, \bar{\beta}}x^\tau_{, \bar{\nu}} x^\lambda_{,\bar{\alpha}}\left(\partial_\lambda g_{\epsilon\tau}+\partial_\epsilon g_{\lambda\tau}-\partial_\tau g_{\lambda\epsilon}\right).$$

However, this deviates what I read in most forums and books. In the first and second term the co- and contravariant metrics should share one dummy variable. Does anyone have a clue?

Answer 1

Starting with $$\Gamma^\bar{\mu}_{\bar{\alpha}\bar{\beta}}=\frac{1}{2}g^{\bar{\mu}\bar{\nu}}\left(\partial_\bar{\alpha}g_{\bar{\beta}\bar{\nu}}+\partial_\bar{\beta}g_{\bar{\alpha}\bar{\nu}}-\partial_\bar{\nu}g_{\bar{\alpha}\bar{\beta}}\right),$$

we have $$\partial_\bar{\alpha}g_{\bar{\beta}\bar{\nu}}=g_{\epsilon\tau}\left( x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}}+x^\epsilon_{, \bar{\beta}} x^\tau_{,\bar{\nu}\bar{\alpha}}\right)+x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\alpha}}\partial_\lambda g_{\epsilon\tau},$$

$$\partial_\bar{\beta}g_{\bar{\alpha}\bar{\nu}}=g_{\epsilon\tau}\left( x^\epsilon_{, \bar{\alpha} \bar{\beta}} x^\tau_{,\bar{\nu}}+x^\epsilon_{, \bar{\alpha}} x^\tau_{,\bar{\nu}\bar{\beta}}\right)+x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\beta}}\partial_\lambda g_{\epsilon\tau},$$

$$\partial_\bar{\nu}g_{\bar{\alpha}\bar{\beta}}=g_{\epsilon\tau}\left( x^\epsilon_{, \bar{\alpha} \bar{\nu}} x^\tau_{,\bar{\beta}}+x^\epsilon_{, \bar{\alpha}} x^\tau_{,\bar{\beta}\bar{\nu}}\right)+x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\beta}}x^\lambda_{,\bar{\nu}}\partial_\lambda g_{\epsilon\tau}.$$

Substitution gives us $$\Gamma^\bar{\mu}_{\bar{\alpha}\bar{\beta}} =\frac{1}{2}g^{\bar{\mu}\bar{\nu}}g_{\epsilon\tau}\left(x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}} +x^\epsilon_{, \bar{\beta}} x^\tau_{,\bar{\nu}\bar{\alpha}} +x^\epsilon_{, \bar{\alpha} \bar{\beta}} x^\tau_{,\bar{\nu}} +x^\epsilon_{, \bar{\alpha}} x^\tau_{,\bar{\nu}\bar{\beta}} -x^\epsilon_{, \bar{\alpha} \bar{\nu}} x^\tau_{,\bar{\beta}} -x^\epsilon_{, \bar{\alpha}} x^\tau_{,\bar{\beta}\bar{\nu}}\right)$$ $$+\frac{1}{2}g^{\bar{\mu}\bar{\nu}}\partial_\lambda g_{\epsilon\tau}\left(x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\alpha}} +x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\beta}} -x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\beta}}x^\lambda_{,\bar{\nu}}\right).$$

Using the interchangeability of the second derivatives, the first term becomes: $$\frac{1}{2}g^{\bar{\mu}\bar{\nu}}g_{\epsilon\tau}\left(2x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}}+x^\epsilon_{, \bar{\beta}} x^\tau_{,\bar{\nu}\bar{\alpha}}-x^\epsilon_{, \bar{\alpha} \bar{\nu}} x^\tau_{,\bar{\beta}}\right)$$ $$=g^{\bar{\mu}\bar{\nu}}g_{\epsilon\tau}x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}}+\frac{1}{2}g^{\bar{\mu}\bar{\nu}}g_{\epsilon\tau}x^\epsilon_{, \bar{\beta}} x^\tau_{,\bar{\nu}\bar{\alpha}}-\frac{1}{2}g^{\bar{\mu}\bar{\nu}}g_{\epsilon\tau}x^\epsilon_{, \bar{\alpha} \bar{\nu}} x^\tau_{,\bar{\beta}}.$$

Switching the indices $\epsilon,\tau$ in the last term cancels both last terms out because of the metric's symmetry. We are then left with: $$g^{\bar{\mu}\bar{\nu}}g_{\epsilon\tau}x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}} =g^{\mu\nu}x^\bar{\mu}_{,\mu}x^\bar{\nu}_{,\nu}g_{\epsilon\tau}x^\epsilon_{, \bar{\beta} \bar{\alpha}} x^\tau_{,\bar{\nu}} =g^{\mu\nu}\delta^\tau_\nu x^\bar{\mu}_{,\mu}g_{\epsilon\tau}x^\epsilon_{, \bar{\beta} \bar{\alpha}} =g^{\mu\tau}g_{\epsilon\tau}x^\bar{\mu}_{,\mu}x^\epsilon_{, \bar{\beta} \bar{\alpha}} =\delta^{\mu}_{\epsilon}x^\bar{\mu}_{,\mu}x^\epsilon_{, \bar{\beta} \bar{\alpha}} =x^\bar{\mu}_{,\mu}x^\mu_{, \bar{\beta} \bar{\alpha}}.$$

Now we have obtained the following: $$\Gamma^\bar{\mu}_{\bar{\alpha}\bar{\beta}} =x^\bar{\mu}_{,\mu}x^\mu_{, \bar{\beta} \bar{\alpha}} +\frac{1}{2}g^{\bar{\mu}\bar{\nu}}\partial_\lambda g_{\epsilon\tau}\left(x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\alpha}} +x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\beta}} -x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\beta}}x^\lambda_{,\bar{\nu}}\right).$$

Switching the indices $\epsilon\leftrightarrow \lambda$ in the second term of the remaining bracket and $\epsilon\rightarrow\lambda\rightarrow\tau\rightarrow\epsilon$ in the third term, we find: $$\frac{1}{2}g^{\bar{\mu}\bar{\nu}}\partial_\lambda g_{\epsilon\tau}\left(x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\alpha}} +x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\beta}} -x^\epsilon_{,\bar{\alpha}}x^\tau_{,\bar{\beta}}x^\lambda_{,\bar{\nu}}\right)$$ $$=\frac{1}{2}g^{\bar{\mu}\bar{\nu}}x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\alpha}} \left(\partial_\lambda g_{\epsilon\tau} +\partial_\epsilon g_{\lambda\tau} -\partial_\tau g_{\epsilon\lambda}\right).$$

As before, we now introduce the barless indices: $$g^{\mu\nu}x^\bar{\mu}_{,\mu}x^\bar{\nu}_{,\nu}x^\epsilon_{,\bar{\beta}}x^\tau_{,\bar{\nu}}x^\lambda_{,\bar{\alpha}} =\frac{1}{2}g^{\mu\nu}\delta^\tau_\nu x^\bar{\mu}_{,\mu} x^\epsilon_{,\bar{\beta}}x^\lambda_{,\bar{\alpha}} =x^\bar{\mu}_{,\mu} x^\epsilon_{,\bar{\beta}}x^\lambda_{,\bar{\alpha}} \cdot\frac{1}{2}g^{\mu\tau}.$$

At last, we are left with: $$x^\bar{\mu}_{,\mu} x^\epsilon_{,\bar{\beta}}x^\lambda_{,\bar{\alpha}} \cdot\frac{1}{2}g^{\mu\tau}\left(\partial_\lambda g_{\epsilon\tau} +\partial_\epsilon g_{\lambda\tau} -\partial_\tau g_{\epsilon\lambda}\right) =x^\bar{\mu}_{,\mu} x^\epsilon_{,\bar{\beta}}x^\lambda_{,\bar{\alpha}} \cdot\Gamma^\mu_{\lambda\epsilon}.$$

Now we have obtained the following: $$\Gamma^\bar{\mu}_{\bar{\alpha}\bar{\beta}} =x^\bar{\mu}_{,\mu}x^\mu_{, \bar{\alpha}\bar{\beta}} +x^\bar{\mu}_{,\mu}x^\lambda_{,\bar{\alpha}}x^\epsilon_{,\bar{\beta}} \cdot\Gamma^\mu_{\lambda\epsilon}.$$

This is the sought result.