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I found the following in a textbook: On a 2-dimensional manifold the Riemann tensor with all indices lowered takes the form:

$R_{abcd} = R{g_{a}[_{c}g_{d}]_{b}}$

However I cannot see why this is true! I am sure it is correct but no amount of fiddling on my part gives the desired above result. Help would be much appreciated

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  • $\begingroup$ is $g$ the metric tensor? $\endgroup$
    – tired
    Nov 28, 2016 at 14:46
  • $\begingroup$ Yes, I believe so. $\endgroup$ Nov 28, 2016 at 14:56

1 Answer 1

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Since the Riemann tensor is anti-symmetric under exchange of the first two components, as well as under exchange of the last two components, we must have $a \neq b$ and $c \neq d$. Since $a,b,c,d$ can only take the values $1,2$ in two dimensions, this tells us that the only non-zero components of the Riemann tensor are $R_{1212}$ and permutations thereof.

If we work in Riemann normal coordinates at a given point, we obtain $$R_{abcd} = 2 R_{1212} g_{a[c} g_{d]b}$$ (just check all possible values of $a,b,c,d$ and use $g_{ab} = \delta_{ab}$ in normal coordinates). Taking the trace of both sides to compute the scalar curvature $R$ we obtain $$ R = R_{abcd} g^{ac} g^{bd} = 2 R_{1212} g_{a[c} g_{d]b} g^{ac} g^{bd} = 2R_{1212}$$ Combining this with the result above we have $$ R_{abcd} = R g_{a[c} g_{d]b} $$ and since this last equation is tensorial, it holds independent of the chosen coordinate system.

Disclaimer: the calculation may be off by some overall factors, which I did not bother to check.

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