TITLE: Riemann's formula for the metric in a normal neighborhood
QUESTION [38 upvotes]: I would love to understand the famous formula $g_{ij}(x) = \delta_{ij} + \frac{1}{3}R_{kijl}x^kx^l +O(\|x\|^3)$, which is valid in Riemannian normal coordinates and possibly more general situations.
I'm aware of 2 proofs: One using Jacobi fields [cf. e.g. S.Sternberg's "Curvature in Mathematics and Physics" from which the question title and formula is stolen :-) or cf. S.Lang's "Differential and Riemannian Manifolds"]. The other proof involves computing that $\partial_k\partial_lg_{ij}(x)$ shares some symmetries of curvature [cf. M.Spivak's "A Comprehensive Introduction to Differential Geometry, Vol. 2" where it is a several page "hairy computation" or cf. H.Weyl's 1923 edition of Riemann's Habilitationsvortrag (reprinted in a recent German book by Jürgen Jost) which I find uncomprehensible.]
Are you aware of any other proof? Are normal coordinates necessary?
While the Jacobi fields proof is short and elegant enough, it irks me that it requires "higher technology" not involved in the endproduct. Somehow the formula should be provable by pure calculus. Indeed, it is stated as an exercise in P.Petersen's "Riemannian Geometry": From the context I guess he thinks it should follow from the expression of $\partial_lg_{ij}$ as a sum of 2 Christoffel symbols and the simplified expression for curvature at $x=0$ where the Christoffel symbols vanish. Alas my attempts at this go in circles...
I find the situation quite amazing: Not many textbooks treat this fundamental and historic formula. (Estimating from the sample on my shelf it is $3/17.$ E.g. it seems it's not even in Levi-Civita's classic.)
Update/Scholium:
In classical language: The knackpoint seems to be a "differential Bianchi formula" for the Christoffel symbols at $0$. This follows from the geodesic equation. I see no other way yet.
A more modern approach minimizing (but not eliminating) the role of geodesics is in A.Gray's Tubes book. (Noted in comments. I'm waiting for www.amazon.de to deliver this treasure.)
$\bullet$ While geodesics are very geometric and normal coordinates are very practical, methinks the formula is a tad ungeometric. What I'm hoping/asking for is a coordinate-independent formula for the second derivative of $g$ in terms of a suitable "reference connection".

REPLY [6 votes]: I came to this post many years later, since I too was concerned about the absence of Riemann's formula in most texts, lengthy treatment in others, or reliance on more advanced techniques like Jacobi fields. I include here a direct concise proof which I think would be well suited for beginning students. We want to show that
$$
g_{ij}(x)=\delta_{ij}+\frac{1}{3} R_{k ij\ell}(o)x_kx_\ell+\mathcal{O}(|x|^3),
$$
where $x=(x_1,\dots, x_n)$ are normal coordinates centered at a point $o$ in a Riemannian manifold $M$.  By Taylor's theorem, we need to check that
\begin{eqnarray}
\tag{1}  g_{ij}(o)&=&\delta_{ij}\label{eq:1},\\ 
\tag{2} g_{ij,k}(o)&=&0\label{eq:2},\\
\tag{3} g_{ij,k\ell}(o)&=&\frac{1}{3}\big(R_{kij\ell }(o)+R_{\ell ij k}(o)\big)\label{eq:3},
\end{eqnarray}
where we use the notation $f_{,i}:=\partial_i f$, and $f_{,ij}:=\partial^2_{ij} f$.
\eqref{eq:1} follows immediately from the construction of normal coordinates, and \eqref{eq:2} is not difficult to establish either. \eqref{eq:3}, which is the heart of the matter, requires a bit more work.
In most sources, like Spivak, vol II or the  relatively recent  book by Jost, which gives a very comprehensive treatment of Riemann's lecture, \eqref{eq:3} is established via symmetry properties of $g_{ij,k\ell}$ which involve long computations; although 2011 Lecture Notes of John Douglas Moore gives a very nice and efficient proof of them. Instead I will demonstrate \eqref{eq:3} via a Bianchi type cyclic identity for derivatives of Christoffel symbols, which follows quickly from the geodesic equation (as the OP had mentioned in the "Update/Scholium" above). The only place I have seen this approach is in 2013 Lecture Notes of Christian Bär.
Proofs of (1) and (2)
All indices here range from $1$ to $n$, and any term which involves repeated indices stands for a sum over that index.
Let $\exp_o\colon T_o M\to M$ be the exponential map, and $U\subset M$ be a ball centered at $o$ such that $\exp_o\colon \exp_o^{-1}(U)\to U$ is a diffeomorphism. Let $e_i$ be an orthonormal basis for $T_o M$, i.e.,
$
g(e_i, e_j)=\delta_{ij}.
$
Then the normal coordinates $x_i\colon U\to \mathbf{R}$ (with respect to $e_i$) are given by
$$
\exp_o(x_i(p) e_i)=p.
$$
The mapping $x\colon U\to\mathbf{R}^n$, given by $x:=(x_1,\dots, x_n)$ identifies $U$ with a ball centered at the origin in $\mathbf{R}^n$, which  we again denote by $U$. Let $E_i(x):=\partial_i|_x$ be the coordinate vector fields on $U$. Then $g_{ij}\colon U\to\mathbf{R}$ are given by
$$
g_{ij}(x):=g(E_i(x),E_j(x)).
$$
Since $E_i(o)=e_i$ we immediately obtain \eqref{eq:1}. To see \eqref{eq:2} note that, by the definition of normal coordinates $x_i$, the geodesics in $U$ passing through $o$ and another point $x$ of $U$ are given by $\gamma(t):=tx$. Since $\gamma$ is a geodesic,
$$
\gamma_k''(t)+\Gamma_{ij}^k(\gamma(t))\gamma_i'(t)\gamma_j'(t)=0,
$$
which yields
\begin{equation}\label{eq:4}
\tag{4}\Gamma_{ij}^k(t x)x_i x_j=0.
\end{equation}
Setting  $t=0$, observing that $x_i$, $x_j$ may assume any values,  and recalling that $\Gamma_{ij}^k=\Gamma_{ji}^k$, we obtain
\begin{equation*}\label{eq:Gamma0}
\Gamma_{ij}^k(o)=0.
\end{equation*}
Since
$
\Gamma_{ij}^k=\frac{1}{2} g^{k\ell}(g_{\ell i,j}+g_{\ell j,i}-g_{ij,\ell}),
$
and $g^{ij}(o)=\delta_{ij}$, we have
\begin{equation*}\label{eq:8}
0=\Gamma_{ij}^k(o)=\frac{1}{2} \big(g_{k i,j}(o)+g_{k j,i}(o)-g_{ij,k}(o)\big).
\end{equation*}
Adding the above equation to itself, after a cyclic permutation of indices yields \eqref{eq:2}.
Proof of (3)
Since
$
\nabla_{E_i}E_j=\Gamma_{ij}^kE_k,
$
$$
g_{ij,k}=g(\nabla_{E_k}E_i, E_j)+g(E_i, \nabla_{E_k}E_j)=\Gamma_{k i}^\ell g_{\ell j}+\Gamma_{k j}^\ell g_{i\ell}.
$$
Differentiating again, and using \eqref{eq:2}, yields
\begin{equation}\label{eq:5}
\tag{5}g_{ij,k\ell}(o)=\Gamma_{k i,\ell}^j(o)+\Gamma_{k j,\ell}^i(o).
\end{equation}
Next we differentiate \eqref{eq:4} at $t=0$ to obtain
$$
\Gamma_{ij,\ell}^k(o)x_i x_jx_\ell=0,
$$
a homogeneous polynomial of degree $3$ which vanishes identically. The coefficient of each term $x_i x_jx_\ell$ is the sum of all $6$ permutations of lower indices of $\Gamma_{ij,\ell}^k(o)$. Since $\Gamma_{ij,\ell}^k=\Gamma_{ji,\ell}^k$, we obtain
\begin{equation*}\label{eq:cyclic}
\Gamma_{ij,\ell}^k(o)+\Gamma_{j\ell,i}^k(o)+\Gamma_{\ell i,j}^k(o)=0.
\end{equation*}
Now note that, since
$
R_{ijk}^\ell=\Gamma_{ik,j}^\ell-\Gamma_{jk,i}^\ell+\Gamma_{ik}^p\Gamma_{pj}^\ell-\Gamma_{jk}^p\Gamma_{ip}^\ell,
$
\eqref{eq:2} yields that
$$
R_{ijk\ell}(o)=\Gamma_{ik,j}^\ell(o)-\Gamma_{jk,i}^\ell(o).
$$
Here we have also used the fact that $R_{ijk}^\ell(o)=R_{ijk\ell}(o)$ due to \eqref{eq:1}. The last two displayed equations yield
\begin{eqnarray*}
R_{ik\ell j}(o)+R_{i\ell kj}(o)
= \Gamma_{i\ell,k}^j(o)+\Gamma_{ik,\ell}^j(o)-2\Gamma_{\ell k,i}^j(o)
=-3\Gamma_{\ell k,i}^j(o).
\end{eqnarray*}
The last equality together with \eqref{eq:5} and symmetries of $R$ now yields
\begin{eqnarray*}
g_{ij,k\ell}(o)
&=&-\frac{1}{3}\big(R_{\ell ik j}(o)+\require{cancel}\cancel{R_{\ell kij}(o)}+R_{\ell jk i}(o)+\require{cancel}\cancel{R_{\ell kji}(o)\big)}\\
&=&\frac{1}{3}\big(R_{\ell i jk}(o)+R_{kij\ell}(o)\big),
\end{eqnarray*}
as desired.