$R^{\nabla}(X,Y)Z = K\big(\langle Y,Z \rangle X - \langle X, Z \rangle Y\big)$ for some $K \in C^{\infty}(M)$

Question

Let $(M, \langle , \rangle )$ be a $2$-dimensional Riemannian manifold, $\nabla$ it's Levi-Civita connection. Show that there is a function $K \in C^{\infty}(M)$ such that:

$R^{\nabla}(X,Y)Z = K\big(\langle Y,Z \rangle X - \langle X, Z \rangle Y\big)$ for all $X,Y,Z \in \Gamma(TM)$

I don't really have a strategy, thus far I've only used the definition and tried to use that $\nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?

Answer 1

Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $\frac{n^2(n^2-1)}{12}$ indepedent non-zero components.

Hint II: Do you know what sectional curvature is?

Answer 2

Hint Fix $p \in M$. By definition, $R_p(X, Y) {\,\cdot\,} = -R_p(Y, X){\,\cdot\,}$ (for all $X, Y \in T_p M$), so we can view $R_p$ as an element of $\bigwedge^2 T_p^*M \otimes T_p M \otimes T^*_p M$. On the other hand, since $\nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){\,\cdot\,}$ is $g$-skew. So, we can view $R_p$ as an element of $\bigwedge^2 T_p^*M \otimes \operatorname{End}_{\operatorname{skew}}(T_p^* M)$. But since $\dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.