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In 4D spacetime, we may choose a locally inertial frame at point P, that is we always have a transformation such that $g_{{\mu'}{\nu'}}(P) = \eta_{{\mu'}{\nu'}}$ and its first derivatives vanish. However, 20 second derivatives cannot be made to vanish. My question is, if we did this in an $n$-dimensional spacetime, how many second derivatives cannot be made to vanish? I know this counts the number of independent components of the Riemann curvature tensor. I know a similar question is here, and that they get $n^2 (n^2-1) / 12$, but I'm not sure where this comes from?

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The $n^2(n^2-1)/12$ comes from the symmetries of the Riemann tensor and the algebraic Bianchi identity.

$R_{abcd}$ is antisymmetric in $ab$ and in $cd$. This means that these pairs of indices can take

$$m=\binom{n}{2}=\frac{n(n-1)}{2}$$

different values. For example, for $n=4$, they take the values $01,02,03,12,13,23$; for other values like $11$, the component is zero.

$R_{abcd}$ is also symmetric when you swap $ab$ with $cd$. This means that, if we didn't take the algebraic Bianchi identity into account, there would be

$$\frac{m(m+1)}{2}=\frac{n^4-2n^3+3n^2-2n}{8}$$

independent components. For $n=4$, there are 21 of them:

$$0101, 0102, 0103, 0112, 0113, 0123$$ $$0202, 0203, 0212, 0213, 0223$$ $$0303, 0312, 0313, 0323$$ $$1212, 1213, 1223$$ $$1313, 1323$$ $$2323$$

Finally, the algebraic Bianchi identity,

$$R_{abcd}+R_{acdb}+R_{adbc}=0$$

provides

$$\binom{n}{4}=\frac{n^4-6n^3+11n^2-6n}{24}$$

relationships between these components. (In this identity, the indices must be all different; otherwise, it reduces to the previous symmetry and antisymmetry relations. So we're choosing 4 indices out of $n$.) Subtracting off this number reduces the number of independent components to

$$\frac{n^4-2n^3+3n^2-2n}{8}-\frac{n^4-6n^3+11n^2-6n}{24}=\frac{n^4-n^2}{12}=\frac{n^2(n^2-1)}{12}.$$

Summarizing this argument, the number of independent components is

$$\frac{1}{2}\binom{n}{2}\left[\binom{n}{2}+1\right]-\binom{n}{4}=\frac{n^2(n^2-1)}{12}.$$

Here is a table of the number of independent components of the Riemann tensor for various dimensions up to 26, the maximum that I think physicists care about:

$$\begin{array}{cc} 1 & 0 \\ 2 & 1 \\ 3 & 6 \\ 4 & 20 \\ 5 & 50 \\ 6 & 105 \\ 7 & 196 \\ 8 & 336 \\ 9 & 540 \\ 10 & 825 \\ 11 & 1210 \\ 12 & 1716 \\ 13 & 2366 \\ 14 & 3185 \\ 15 & 4200 \\ 16 & 5440 \\ 17 & 6936 \\ 18 & 8721 \\ 19 & 10830 \\ 20 & 13300 \\ 21 & 16170 \\ 22 & 19481 \\ 23 & 23276 \\ 24 & 27600 \\ 25 & 32500 \\ 26 & 38025 \\ \end{array}$$

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  • $\begingroup$ Fantastic solution, thanks a lot! $\endgroup$ – starseed_trooper Dec 17 '20 at 13:39
  • $\begingroup$ Great answer. The part I don't get: why can just deduce the number obtained using Bianchi identity from the previous result? Asked a question here about that topic: math.stackexchange.com/questions/4060075/… $\endgroup$ – zabop Mar 13 at 10:59

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