A Riemannian metric $g$, in normal coordinates, has a Taylor expansion $$g_{ij}(x) \ = \ \delta_{ij} \ - \ \frac{1}{3}R_{iklj}x^kx^l \ - \ \frac{1}{3}R_{iklj;m}x^kx^lx^m \ + \ \frac{1}{180}\left(8R_{ilmk}R_{jpqk}-9R_{ilmj;pq}\right)x^kx^lx^px^q $$ $$\ + \ \frac{1}{90}\left(4R_{iklr;m}R_{jpqr}-R_{iklj;mpq}\right)x^kx^lx^mx^px^q\ + \ \cdots $$ This is proven here, and the proof was asked for earlier. Each term in the series is a product of $R$ and its covariant dervatives. $$\text{}$$ This series is hideous, and I would like to know if there is a cleaner or more symmetric way of expressing the coefficients. If not, can we at least say something about them which makes them seem more natural?

For instance, if $f$ is a function, its Taylor series $$f(x) \ = \ f(0) \ + \ f_i(0)x^i \ + \ \frac{1}{2!}f_{ij}(0)x^ix^j \ + \ \cdots$$ is pretty easy to remember, and its symmetries are that the coefficients are symmetric in the indices $i,j,...$. Hopefully $f$ being matrix-valued doesn't mean its Taylor series is suddenly a structureless mess, but there are even more symmetries, though they might be harder to see.

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    $\begingroup$ If you want to express the Taylor series of $g$ in terms of partial derivatives of $g$, it's just as neat and simple as the scalar case. The problem arises when you try to express it in terms of the curvature tensor and its covariant derivatives. The complexity doesn't really have anything to do with the Taylor series; it's just an expression of the fact that the curvature tensor and its coordinate derivatives are given by ugly formulas in terms of $g$ and its partial derivatives. $\endgroup$
    – Jack Lee
    Commented Aug 27, 2018 at 23:09
  • $\begingroup$ @JackLee Thanks. I should have thought more carefully before asking the question. So as far as you know, there's no pattern in the functions of $R$ and its derivatives? $\endgroup$
    – Pulcinella
    Commented Aug 28, 2018 at 8:40
  • $\begingroup$ I think Jack Lee is saying that they are actually pretty clean in terms of $R$ (as above) or $g$ (naively)... what's horrid is trying to express derivatives of $R$ in terms $g$.... But yeah, I hope there is another answer to your question. :) $\endgroup$ Commented Sep 26, 2018 at 20:26


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