Why Does the Two-Dimensional Riemann Tensor Simplify to R g_{a[c}g_{d]b}?

[latentcorpse]

show that in two dimensions, the Riemann tensor takes the form $R_{abcd}=R g_{a[c}g_{d]b}$.

i've expanded the RHS to get

$R g_{a[c}g_{d]b}=\frac{R}{2!} [g_{ac} g_{db} - g_{ad} g_{cb}]=\frac{1}{2} R_e{}^e [g_{ac} g_{db} - g_{ad} g_{cb}]$
but i can't seem to simplify it down.

this is problem 4a in Wald's General Relativity p54.

 

[dextercioby]

There's a hint in the book. I'll rephrase it like that: how many non vanishing components does the Riemann tensor have in 2 dimensions ?

 

[latentcorpse]

i still don't understand the hint sorry.

if its in two dimensions then each index can be either 1 or 2 is that correct?
does that help?

 

[gabbagabbahey]

Use the result of problem 3(b) to calculate the number of independent components of the Riemann tensor in 2D...What does that tell you about the dimensionality of the vector space of tensors having the symmetries of the Riemann tensor?

 

[latentcorpse]

ok. so the number of independent components is 1.

im going to guess that this means the dimensionality of the space is 1 but I'm not at all sure why...