# The covariant derivative from the contravariant derivative

I know that the following is true: $$V^{\mu}_{~~~~~;\nu} = \frac{\partial V^{\mu}}{\partial x^{\nu}} +\Gamma^{\mu}_{~\sigma\nu} V^{\sigma}.$$

Also, by definition, we have that $V_{\rho} = g_{\rho\nu}V^{\nu}$.

I would like to show that $$V_{\rho;\nu} = \frac{\partial V_{\rho}}{\partial x^{\nu}} - \Gamma^{\sigma}_{~\rho\nu} V_{\sigma}$$

I have attempted to do this by taking $(g_{\rho\nu}V^{\nu})_{;\mu}$ and using the Leibniz rule and the fact that $g_{\rho\nu;\mu}=0$ but I am missing the minus sign!

$\def\m{\mu} \def\n{\nu} \def\r{\rho} \def\s{\sigma} \def\t{\tau} \def\G{\Gamma}$While the method in @DanielC's answer is often the one followed it is possible, straightforward, and instructive to approach this in the way you have outlined in your question statement. We have \begin{align*} V_{\r;\n} &= (g_{\r\m}V^\m)_{;\n} \\ &= \underbrace{g_{\r\m;\n}}_0 V^\m + g_{\r\m}V^\m_{;\n} \\ &= g_{\r\m}(V^\m_{,\n} + \G^\m_{\ \s\n}V^\s) \\ &= g_{\r\s}V^\s_{,\n} + \G_{\r\s\n}V^\s \\ &= (g_{\r\s}V^\s)_{,\n} - g_{\r\s,\n}V^\s + \G_{\r\s\n}V^\s \\ &= V_{\r,\n} - (g_{\r\s,\n} - \G_{\r\s\n})V^\s \\ &= \ldots \\ &= V_{\r,\n} - \G^\t_{\ \r\n}V_\t. \end{align*} In the omitted steps one will need to use the well-known identity $$g_{\r\s,\n} = \G_{\s\r\n} + \G_{\r\n\s},$$ where $\G_{\r\s\t} = g_{\m\r} \G^\m_{\ \s\t}$.
This is not the way to do it. You need to compute $$\nabla_\mu \left( V^\alpha V_\alpha\right)$$ using the Leibniz rule and the fact that there is a scalar inside the bracket.
• After using this method, I now have a '-' sign in front of $V_{\rho ,\nu}$. Could you expand your answer a bit please to help me see what I am missing? – user10000654 May 31 '18 at 6:33