# Two Dimensional Riemann Tensor

1. Oct 18, 2009

### 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.

2. Oct 18, 2009

### 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 ?

3. Oct 18, 2009

### latentcorpse

i still dont understand the hint sorry.

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

4. Oct 19, 2009

### 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?

5. Oct 20, 2009

### 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...