the Ricci curvature in two dimension

Question

In two dimensions, we know that the Ricci curvature can be written in terms of the Gauss curvature K as $Ric(g) = Kg$.

Can anyone prove this?

Sorry if the question is too trivial :).

Answer 1

In two dimentions $R$ and $K$ being Riemann and Gaussian curvature tensors, the Theorema Egregium gives you for any basis $ (X,Y)$ of $ T_p(M) $, at p we have $$ K = \frac{R(X,Y,Y,X)}{|X|^2|Y|^2-g(X,Y)^2} $$ Now if $E_1, E_2 $ is an orthonormal basis for the tanjent space and $ R_{ijjkl} = R(E_i,E_j,E_k,E_l) $ then the previous formula gives you that $K = R_{1221} $ in terms of this basis. From antisymmetry, nonzero components of $R$ are $ R_{1221} = R_{2112} = -R_{1212} = -R_{2121} = K $. But in this basis you have $ Ric_{ij} = R_{1ij1} + R_{2ij2} $. Hence $ Ric_{11} = Ric_{22} = K $ and $ Ric_{12} = Ric_{21} = 0 $ Hence $$ Ric(E_1, E_2) = \sum_{i,j} Ric_{ij}E_{1i}E_{2j} = K(E_{11}E_{21} + E_{12}E_{22}) = Kg(E_1,E_2) $$ Thus $ Ric (X,Y) = Kg(X,Y) $ for all $ X,Y $