Riemann tensor with all indices lowered on a 2D manifold 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
 A: 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.
