What's the variation of the Christoffel symbols with respect to the metric?

Question

By the Leibniz rule, I expected it to be

$$\delta \Gamma^\sigma_{\mu\nu} = \frac 12 (\delta g)^{\sigma\lambda}(g_{\mu\lambda,\nu}+g_{\nu\lambda,\mu}-g_{\mu\nu,\lambda}) + \frac 12 g^{\sigma\lambda}(\partial_\nu (\delta g)_{\mu\lambda}+\partial_\mu (\delta g)_{\nu\lambda}-\partial_\lambda (\delta g)_{\mu\nu}) .\tag{1}$$

However, according to Sean Carroll,

$$\delta \Gamma^\sigma_{\mu\nu} = \frac 12 g^{\sigma\lambda}(\nabla_\nu (\delta g)_{\mu\lambda}+\nabla_\mu (\delta g)_{\nu\lambda}-\nabla_\lambda (\delta g)_{\mu\nu}).$$

In other words, the first term on the RHS of (1) is not there and partials on the second term are replaced by covariant derivatives. Why?

Answer 1

$ \Gamma^{a}_{bc} = \cfrac{1}{2}g^{ad}(\partial_{b}g_{dc} + \partial_{c}g_{bd} - \partial_{d}g_{bc}) \Rightarrow $
$ δ\Gamma^{a}_{bc} = \cfrac{1}{2}δg^{ad}(\partial_{b}g_{dc} + \partial_{c}g_{bd} - \partial_{d}g_{bc}) + \cfrac{1}{2}g^{ad}(\partial_{b}δg_{dc} + \partial_{c}δg_{bd} - \partial_{d}δg_{bc}) \Rightarrow$
$δ\Gamma^{a}_{bc} = -\cfrac{1}{2}g^{af}g^{de}(\delta g_{fe})(\partial_{b}g_{dc} + \partial_{c}g_{bd} - \partial_{d}g_{bc}) + \cfrac{1}{2}g^{ad}(\partial_{b}δg_{dc} + \partial_{c}δg_{bd} - \partial_{d}δg_{bc}) \Rightarrow $ $δ\Gamma^{a}_{bc} = -g^{af}(\delta g_{fe})\Gamma^{e}_{bc}+ \cfrac{1}{2}g^{ad}(\partial_{b}δg_{dc} + \partial_{c}δg_{bd} - \partial_{d}δg_{bc}) \Rightarrow $ $δ\Gamma^{a}_{bc} = -g^{ad}(\delta g_{de})\Gamma^{e}_{bc}+ \cfrac{1}{2}g^{ad}(\partial_{b}δg_{dc} + \partial_{c}δg_{bd} - \partial_{d}δg_{bc}) \Rightarrow $
$δ\Gamma^{a}_{bc} = \cfrac{1}{2}g^{ad}\big[\partial_{b}δg_{dc} + \partial_{c}δg_{bd} - \partial_{d}δg_{bc} - 2\delta g_{de}\Gamma^{e}_{bc}\big] \Rightarrow $
$δ\Gamma^{a}_{bc} =\cfrac{1}{2} g^{ad}\big[ \partial_{b}δg_{dc} -\Gamma^{e}_{bc}\delta g_{ed} -\Gamma^{e}_{bd}\delta g_{ec} $
$+\partial_{c}δg_{bd} -\Gamma^{e}_{cd}\delta g_{eb} - \Gamma^{e}_{cb}\delta g_{ed} - \partial_{d}δg_{bc} + \Gamma^{e}_{db}\delta g_{ec} + \Gamma^{e}_{dc}\delta g_{eb} \big] \Rightarrow$
$δ\Gamma^{a}_{bc} = \cfrac{1}{2}g^{ad}\big[\nabla_{b}δg_{dc}+\nabla_{c}δg_{bd}- \nabla_{d}δg_{bc}\big] $

Answer 2

• The difference between two connections is a tensor, so $\delta\Gamma$ is a tensor.

• Evaluate your variational formula in Riemannian normal coordinates at some arbitrary point $x_0$. Since the metric derivatives are zero at that point one gets $$ \delta\Gamma^\sigma_{\mu\nu}=\frac{1}{2}\eta^{\sigma\lambda}(\partial_\nu\delta g_{\mu\lambda}+\partial_\mu\delta g_{\nu\lambda}-\partial_\lambda\delta g_{\mu\nu}), $$ where all functions are evaluated at $x_0$

• In Riemannian normal coordinates, $\partial=\nabla$, so we can rewrite $$ \delta\Gamma^\sigma_{\mu\nu}=\frac{1}{2}g^{\sigma\lambda}(\nabla_\nu\delta g_{\mu\lambda}+\nabla_\mu\delta g_{\nu\lambda}-\nabla_\lambda\delta g_{\mu\nu}). $$

• This equation however is tensorial, so it must be valid at $x_0$ in other coordinates too, not just Riemannian normal coordinates.

• Since $x_0$ was arbitrary, this relation must then hold for any point.