I am reading some lecture notes on Riemmanian geometry where it is stated if we take $\partial_s g_{ij} = h_{ij}$ to be the variation of a Riemannian metric and choose normal coordinates for $g$, the definition of the Christoffel symbols implies the following:

$$ \partial_{s} \Gamma^{\ell}_{ik} = \frac{1}{2}g^{\ell m} (\nabla_{i}h_{km} + \nabla_{k}h_{im} - \nabla_{m}h_{ik}).$$

Could someone clarify why this is? I have that $\partial_s g^{ij} = -h^{ij}$ but don't follow this part of the proof. Where is the fact that we are using normal coordinates coming into play?

  • 2
    $\begingroup$ In $\Gamma_{ik}^l$ there are terms like $g_{km, i}=\partial_i g_{km}$, these are $0$ at the point $p$ where you do the calculation. Therefore, when taking $s$ derivative, you want the derivative fall on first derivative terms like $g_{km, i}$ - otherwise if you leave them alone (and do $s$-derivative on things like $g^{lm}$) you get zero terms at $p$. Compute $(g_{km, i})'s=h_{km, i}$, and since we use normal coor at $p$ we see this is just $\nabla_i h_{km}$ at $p$. $\endgroup$
    – Yuval
    Apr 8, 2019 at 4:47

1 Answer 1


One thing at a time: the goal is to show that if $\partial_sg_{ij}|_{s=0} = h_{ij}$, then $$\partial_s\Gamma_{ik}^\ell|_{s=0} = \frac{1}{2}g^{\ell m}(\nabla_ih_{km}+\nabla_kh_{im} - \nabla_mh_{ik})\tag{1}.$$As $(1)$ is tensorial, it suffices to check it in normal coordinates centered at an arbitrary point $p$ in your manifold. For normal coordinates centered at $p$, $g_{ij}(p) = \delta_{ij}$ and $\Gamma_{ik}^\ell(p) = 0$, so that covariant derivatives at $p$ reduce to partial derivatives at $p$ (but in general nowhere else). In summary, once such normal coordinates are in place, establishing $(1)$ is equivalent to establishing $$\partial_s\Gamma_{ik}^\ell|_{s=0}(p) = \frac{1}{2}(\nabla_ih_{k\ell}(p) + \nabla_k h_{i\ell}(p) - \nabla_\ell h_{ik}(p)) \tag{2}$$

Recall that the derivative of the inversion map $\iota\colon {\rm GL}_n(\Bbb R) \to {\rm GL}_n(\Bbb R)$, $\iota(A) = A^{-1}$, is given by ${\rm d}\iota_A(H) = -A^{-1}HA^{-1}$. This is a noncommutative version of $(1/x)' = -x'/x^2$. It means that if $\partial_sg_{ij}|_{s=0} = h_{ij}$, then $$\partial_sg^{ij}|_{s=0} = -g^{ik}(\partial_sg_{k\ell}|_{s=0})g^{\ell j} = -g^{ik}h_{k\ell}g^{\ell j} = -h^{ij} \tag{3}$$The first equality in $(3)$ really amounts to the $(i,j)$-th entry of a product of three matrices.

The product rule applied to $$\Gamma_{ik}^\ell = \frac{1}{2}g^{\ell m}(\partial_i g_{km} + \partial_kg_{im} -\partial_mg_{ik})\tag{4}$$yields that $$\begin{split} \partial_s\Gamma_{ik}^\ell = \frac{1}{2}(\partial_sg^{\ell m})&(\partial_i g_{km} + \partial_kg_{im} -\partial_mg_{ik}) \\ &+ \frac{1}{2}g^{\ell m}(\partial_i \partial_sg_{km} + \partial_k\partial_sg_{im} -\partial_m\partial_sg_{ik}),\end{split}\tag{5}$$as $\partial_s$ commutes with coordinate partial derivatives $\partial_i$, etc.

Make $s=0$ in $(5)$ to obtain $$\begin{split}\partial_s\Gamma_{ik}^\ell|_{s=0} = -\frac{1}{2}h^{\ell m}&(\partial_i g_{km} + \partial_kg_{im} -\partial_mg_{ik}) \\ &+ \frac{1}{2}g^{\ell m}(\partial_i h_{km} + \partial_kh_{im} -\partial_mh_{ik}),\end{split}\tag{6}$$Now, since $$\partial_ig_{km}(p) + \partial_kg_{im}(p)-\partial_mg_{ik}(p) = g_{m\ell}(p)\Gamma_{ik}^\ell(p) = \delta_{m\ell} \cdot 0 = 0 \tag{7}$$ and $$\nabla_ih_{km}(p) = \partial_ih_{km}(p)\tag{8}$$and similarly for other indices, evaluating $(6)$ at $p$ and substituting $(7)$ and $(8)$ gives us that $$\partial_s\Gamma_{ik}^\ell|_{s=0}(p) = \frac{1}{2}\delta^{\ell m}(\nabla_ih_{km}(p) + \nabla_jh_{im}(p) - \nabla_mh_{ik}(p)),\tag{9}$$which immediately simplifies to $(2)$.

Note: a more elegant (and arguably shorter) proof consists of simply applying $\partial_s|_{s=0}$ to both sides of the Koszul formula $2g(\nabla_XY,Z) = \cdots$ and using non-degeneracy of $g$ to solve for $2g(\dot{\nabla}_XY,Z)$, thereby determining the tensor $\dot{\nabla}$.


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