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.)


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".

  • 2
    $\begingroup$ It seems to me that it's first necessary to find a "simple" definition of the Riemann curvature tensor. I'm not sure how Levi-Civita defined it. $\endgroup$
    – Deane Yang
    Oct 27 '14 at 20:29
  • 2
    $\begingroup$ Martin, I find Christoffel symbols highly unenlightening, so that's not a simple definition to me. It seems to me that both normal co-ordinates (up to second order only) and the Riemann curvature tensor arise pretty naturally if you search for co-ordinates that simplify the 2nd order Taylor expansion of the metric as possible. But since the Hessian of the metric is a 4th order tensor, that's still a bit tricky. $\endgroup$
    – Deane Yang
    Oct 27 '14 at 22:53
  • 1
    $\begingroup$ Ryan, now I really counted my shelf and it is 3/17 :-) I guess what you're talking about amounts to the Jacobi fields proof. Slowly but surely methinks that's indeed the natural proof. Which of your books has it? $\endgroup$ Oct 28 '14 at 12:36
  • 2
    $\begingroup$ A. Gray's book Tubes contains in Section 9.1 (2nd Edition) a readable presentation of the Taylor expansion of a tensor in geodesic coordinates. It includes the case of the metric tensor as a special case (Corollary 9.8). $\endgroup$ Oct 28 '14 at 14:34
  • 1
    $\begingroup$ Also try this paper of Gray projecteuclid.org/download/pdf_1/euclid.mmj/1029001150 $\endgroup$ Oct 29 '14 at 19:28

Perhaps the simplest way to understand this formula is to think about how you would go about deriving it: Try to find the 'best' coordinates you can centered on a given point and see what doesn't change in such coordinates.

Suppose $g$ is a Riemannian metric on $M$ and $p\in M$ is fixed. Start by choosing a $p$-centered local coordinate system $x = (x^i)$ on $U\subset M$ and write $$ g = g_{ij}(x)\,\mathrm{d}x^i\mathrm{d}x^j. $$ Since $\bigl(g_{ij}(0)\bigr)$ is a positive definite matrix, you can make a linear change of coordinates in $x$ so that $g_{ij}(0) = \delta_{ij}$. Call such a $p$-centered coordinate system $0$-adapted to $g$ at $p$.

Now, ask what would be the effect of expressing $g$ in the coordinates $y=(y^i)$ that are related to the coordinate $x$ by $x^i = y^i + \tfrac12a^i_{jk} y^jy^k$ for some $a^i_{jk} = a^i_{kj}$. It is easy to see by Taylor series expansion that you can uniquely choose the $a^i_{jk}$ so that, when we write $$ g = \bar g_{ij}(y)\,\mathrm{d}y^i\mathrm{d}y^j, $$ we have, for all $i$, $j$, and $k$, $$ \frac{\partial\bar g_{ij}}{\partial y^k}(0) = 0. $$ (It's clear that this is the same number of equations as unknowns for the $a^i_{jk}$, one just has to check that the inhomogeneous system of equations has only the zero solution when the inhomogeneous part is set to zero.) Call such a system of $p$-centered coordinates $1$-adapted to $g$ at $p$. Thus, for a system of coordinates $y$ that is $1$-adapted to $g$ at $p$, one has $$ g = \left(\delta_{ij} + \tfrac12 \frac{\partial^2g_{ij}}{\partial y^k\partial y^l}(0)\, y^ky^l + R^3_{ij}(y)\right) \ \mathrm{d}y^i\mathrm{d}y^j, $$ where $R^3_{ij}(y)$ vanishes to order $3$ at $y=0$.

Finally, consider what such a metric would look like in the coordinates $z = (z^i)$ that are defined by $y^i = z^i + \tfrac16 b^i_{jkl} z^jz^kz^l$ for some constants $b^i_{jkl} = b^i_{kjl} = b^i_{jlk}$. Now, there are $n^2(n{+}1)(n{+}2)/6$ unknowns $b^i_{jkl}$, but there are $n^2(n{+}1)^2/4$ quantities in the second-order Taylor expansion of $g = {\bar g}_{ij}(z)\mathrm{d}z^i\mathrm{d}z^j$, i.e., $$ g = \left(\delta_{ij} + \tfrac12 \frac{\partial^2{\bar g}_{ij}}{\partial z^k\partial z^l}(0)\, z^kz^l + {\bar R}^3_{ij}(z)\right) \ \mathrm{d}z^i\mathrm{d}z^j. $$ Thus, the equations $\frac{\partial^2{\bar g}_{ij}}{\partial z^k\partial z^l}(0)=0$, as linear equations for the $b^i_{jkl}$, are overdetermined by $$ n^2(n{+}1)^2/4 - n^2(n{+}1)(n{+}2)/6 = n^2(n^2{-}1)/12 $$ equations.

It is not hard to see that the corresponding homogeneous equations in the $b^i_{jkl}$ have only the solution $b^i_{jkl}=0$. In fact, the $b^i_{jkl}$ are uniquely determined by requiring that, when we compute the Taylor expansion about $z=0$ we get $$ g = \left(\delta_{ij} + \tfrac12 h_{ij,kl}\, z^kz^l + R^3_{ij}(z)\right) \ \mathrm{d}z^i\mathrm{d}z^j, $$ with $h_{ij,kl}+h_{ik,lj}+h_{il,jk}=0$ (which is $n^2(n{+}1)(n{+}2)/6$ independent equations on the $b^i_{jkl}$). Say that a system of coordinates $z = (z^i)$ centered at $p$ for which $g$ has its Taylor expansion at $p$ of the above form is $2$-adapted to $g$ at $p$. Two such coordinate systems at $p$ are related in the form $z^i = a^i_j\,\bar z^j + O(|{\bar z}|^4)$, where $a = (a^i_j)$ is an orthogonal matrix.

Thus, the $2$-adapted condition forces the $h_{ij,kl}$ to lie in a vector space of dimension $n^2(n^2{-}1)/12$, as explained above.

It's now a matter of linear algebra to show, as Riemann did, that these conditions imply that the $h_{ij,kl}$ can be written uniquely in the form $$ h_{ij,kl} = \tfrac13(R_{kijl}+R_{lijk}) $$ where $R_{ijkl}=-R_{jikl}=-R_{ijlk}$ and $R_{ijkl}+R_{iklj}+R_{iljk}=0$.

  • $\begingroup$ But how to see that we get the "modern" curvature tensor? $\endgroup$ Oct 28 '14 at 17:52
  • 5
    $\begingroup$ I'm not sure what you mean by 'modern' curvature; I thought you wanted Riemann's curvature. I didn't describe it more explicitly because I thought that I had given enough hints that you could work it out. Note that Riemann himself did not use Christoffel symbols, which didn't exist in 1854. I think that his choice of representation for the curvature (and a choice was necessary) was determined by two things: He wanted it to be the Gauss curvature on surfaces and he wanted it to be equivariant with respect to linear changes of variables. $\endgroup$ Oct 28 '14 at 18:44
  • $\begingroup$ If $g$ is a flat metric, then in suitable coordinates $g_{ij} = \delta_{ij}$. It follows from your answer that we can arrange $g_{ij} = \delta_{ij}$ up to order three. Can a similar analysis show that all the higher order terms can be made to disappear too? $\endgroup$ Feb 18 '19 at 15:14
  • $\begingroup$ @MichaelAlbanese: Under what hypotheses? If you assume that $g$ is flat, then, sure, we can show all of the higher order terms vanish in suitable coordinates. If you only assume that the Riemann curvature vanishes at a point, then you can make the terms all vanish to order 3 at that point, but maybe not to order $4$. Maybe I am not understanding your question. $\endgroup$ Feb 18 '19 at 16:06
  • $\begingroup$ @RobertBryant: I suppose it's not clear to me that if $g$ is flat, the process you have outlined in your answer will continue to work. For example, it's not obvious to me that we will be able to remove the third order terms. I know we can construct coordinates with $g_{ij} = \delta_{ij}$ more directly, I was just interested in the approach you were using. $\endgroup$ Feb 22 '19 at 21:47

Still another approach to the Riemann normal coordinates expansion formula can be found in http://arxiv.org/abs/gr-qc/9712092 (A Closed Formula for the Riemann Normal Coordinate Expansion, by U. Mueller, C. Schubert and A. van de Ven). This approach indicates that the Riemann normal coordinates are the gravity analogue of the Fock-Schwinger gauge in gauge theory. Fock-Schwinger gauge (centered at the origin) is defined by the condition $$x^\mu A_\mu(x)=0. \tag{1}$$ In the local neighbourhood of the origin, the condition (1) can be solved in terms of the following integral representation $$A_\mu(x)=x^\nu\int\limits_0^1F_{\nu\mu}(s x)s ds, \tag{2}$$ which connects the gauge potential $A_\mu$ and the field strength tensor $F_{\mu\nu}$. As a result, the Taylor expansion coefficients of $A_\mu$ at the origin is expressed through the covariant derivatives of $F_{\mu\nu}$.

In analogy, Riemann normal coordinates centered at the origin can be defined by the conditions $$g_{\mu\nu}(0)=\delta_{\mu\nu},\;\;\;x^\mu g_{\mu\nu}(x)=x^\mu g_{\mu\nu}(0). \tag{3}$$ (the second condition is equivalent to the following condition on the Chrisoffel symbol $x^\mu x^\nu \Gamma_{\mu\nu}^\lambda(x)=0$, which determines the coordinate system locally up to a rigid rotation).

Clearly, (3) is the analog of (1). While the analog of (2), proved in the Mueller, Schubert and van de Ven paper, is \begin{eqnarray} && g(x)=\sum\limits_{k=0}^\infty\int\limits_0^1ds_1\,(1-s_1)\int\limits_0^1ds_2\,(1-s_2)\cdots \int\limits_0^1ds_k\,(1-s_k) \\ && \times \sum\limits_{l=0}^ks_1s_2^3\cdots s_l^{2l-1}s_{l+1}^{2k-2l-1} s_{l+2}^{2k-2l-3}\cdots s_k \\ && \times {\cal R}(s_1 s_2 \cdots s_l x,x){\cal R}(s_2 s_3 \cdots s_l x,x)\cdots {\cal R}(s_l x,x) \\ && \times {\cal R}(s_{l+1} x,x){\cal R}(s_{l+1}s_{l+2} x,x)\cdots {\cal R}(s_{l+1}s_{l+2}\cdots s_k x,x), \tag{4} \end{eqnarray} where $${\cal R}^\mu_{\;\nu}(x,y)=R^\mu_{\;\alpha\beta \nu}(x)y^\alpha y^\beta.$$

The generalization of (4) to the Fermi normal coordinates in tubular geometry is considered in http://arxiv.org/abs/1203.1151 (All order covariant tubular expansion, by P. Mukhopadhyay).

  • $\begingroup$ Thanks for this! I was hoping for something from physicists: They're often better at calculus :-) The Ansatz in Mueller et al seems to be from Atiyah/Bott/Patodi (but I neither have this nor the Amsterdamski et al paper at hand). And they didn't know about Gray's full formula in Gray projecteuclid.org/download/pdf_1/euclid.mmj/1029001150 $\endgroup$ Feb 3 '15 at 13:35
  • $\begingroup$ Meanwhile I'm much fascinated by the full Taylor expansion. Maybe it can be simplified? Maybe it produces some bombastic curvature identities? Or plug such in to simplify? ................ Above all: Is the volume conjecture still open? I.e.: If a Riem. mf. M has the same small ball volumes as Euclidean space, then M is flat. (A.Gray, Tubes 2nd ed. p.198) $\endgroup$ Feb 3 '15 at 13:41
  • $\begingroup$ The 2nd condition in (3) is the "extrinsic Gauss lemma" I mentioned (H.Weyl+Spivak's ansatz and the repere mobile proofs). Methinks it complicates things (worst in Spivak's "hairy calculation") $\endgroup$ Feb 3 '15 at 13:48
  • $\begingroup$ About the volume conjecture: in journals.cambridge.org/article_S0004972700031452 (Semi-symmetric ball-homogeneous spaces and a volume conjecture, by G. Calvaruso and L. Vanhecke) it is proved for semi-symmetric Riemannian manifolds. A nice paper about the volume conjecture by Gray himself is (in case you not already know it) link.springer.com/article/10.1007%2FBF02395060 (Riemannian geometry as determined by the volumes of small geodesic balls, by A. Gray and L. Vanhecke). I don't know whether the volume conjecture is still open in general. $\endgroup$ Feb 4 '15 at 5:06

My question has been answered in comments by Liviu Nicolaescu:

The (almost) ultimate proof (for my taste) is via A. Gray's formula(e) for (symmetric) higher covariant derivative(s) of normal coordinate vector fields. Any exposition of normal coordinates lacking this formula is severely lacking. (I'd prefer symmetrized c.d. of differentials of normal coordinates...). It is valid for any symmetric connection and gives Riemann's formula by application to some parallel bilinear form (e.g. the Riemann metric). (Riemann's really original Habilitationsvortrag formula needs a little additional algebra, cf. Dedekind/Weber [below] or Spivak.)

The second best approach is using Jacobi fields (cf. e.g. Le Spectre (LNM 194) for higher order terms without detail. (For hardcore syntacticists: Strook, Intro to An. of Paths on a Riem. Mf.)).

Third best: Repere mobile, but this seems Riemannian. (Cf. e.g. Heat Kernels and Dirac Operators, or Atiyah/Bott/Patodi appendix, or Cartan's Geometrie des Espaces de Riemann.)

The historically first proof is due to Dedekind/Weber (Anmerkung in "Bernhard Riemann's Gesammelte Mathematische Werke und Wissenschaftlicher Nachlaß", 1876/1892) via proving Riemann's Commentatio formula (which is also in Spivak) for curvature. (I could streamline this quite amazingly by doing the Levi-Civita in cotangent space.)

H.Weyl (1919/1923) has the historically second proof in his German edition of Riemann's Habilitationsvortrag, based on the extrinsic Gauss lemma (i.e. e.g. Besse, Einstein Mfs, Thrm. 1.45), which is what Spivak has worked out, and possibly has inspired the repere mobile proof (explicit in Heat Kernels [above]) - I don't (and am unwilling to) comprehend Cartan's proof.)

The nicest classical Riemannian proof is in L.P.Eisenhart, Riemannian Geometry, 1925/1949

I still don't have an opinion on R.Bryant's answer. It is possibly exactly what Riemann did for his inaugural lecture 1854. But then, Riemann had at least in later work (Commentatio, 1861) introduced Christoffel symbols (written $p_{\iota,\iota^\prime,\iota^{\prime\prime}}$) and other formulas for curvature.

  • $\begingroup$ Robert's answer is in my view the closest in spirit to the name "normal co-ordinates". You try to normalize the co-ordinates at one point to make the metric and its derivatives look as simple as possible. It also has the advantage that you use only very basic calculus (partial derivatives and the chain rule) and implicitly some tensor algebra. $\endgroup$
    – Deane Yang
    Jan 30 '15 at 2:39
  • $\begingroup$ On the other hand, viewing normal co-ordinates as a map from the tangent space to the manifold using radial geodesics is a geometrically natural thing to do. Then Jacobi fields provide an elegant way to figure out what the metric looks like not just at the origin but also away from it. This combined with Sturm-Liouville theory applied to the Jacobi equation is a powerful tool for studying not just local but also global behavior of the Riemannian manifold. $\endgroup$
    – Deane Yang
    Jan 30 '15 at 2:43
  • $\begingroup$ Yeah. I feel a bit ashamed about calling the formula "a tad ungeometric". But then, it is indeed not about Riemannian geometry, but general symmetric connections. $\endgroup$ Feb 3 '15 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.