Why do we have that for a surface (dimension $2$) that
$$\text{Ric}(X, Y) = K \langle X, Y \rangle ,$$
where $K$ is the Gaussian curvature and $X, Y$ are vector fields?
[Math] The relationship between Ricci and Gaussian curvatures
curvaturedifferential-geometryriemannian-geometrysurfaces
Best Answer
For a surface $(M, g)$, there is only independent component of the curvature tensor $R$, namely, $$R_{1212} = - R_{1221} = -R_{2112} = R_{2121},$$ and this quantity (which depends on the choice of coordinates) is related to the Gaussian curvature $K$, by $$R_{1212} = -\det(g) K$$ (which is independent of coordinates). (Beware that this sign is a matter of convention, and not everyone uses the same one.)
This leads to a decomposition of $R$--- $$R_{abcd} = K(g_{ac} g_{bd} - g_{ad} g_{bc})$$ ---and taking an appropriate trace, say, with $g^{ac}$ gives the formula you mention: $$\text{Ric}_{bd} = K g_{bd}.$$