Let $\left( M, g \right)$ be a Riemannian manifold. Then we have the Riemann curvature tensor $R$. I know that somehow you can define the curvature operator:
$$ R_x: \Lambda^2 T_xM \rightarrow \Lambda^2 T_xM $$
But how do you define it to be a self-adjoint operator? How does it work on the basis elements?
Maybe you will recommend me books, where is it defined in detail?
Thank you very much!
I assume you're familiar with the following symmetries of the curvature tensor:
The first two symmetries tell you that $R$ can be considered as acting 2 bivectors rather than 4 vectors, since they imply that $R(X,Y,Z,W)$ depends only on $X\wedge Y$, $Z\wedge W.$ Thus we can view the curvature tensor as a bilinear form $$R : \Lambda^2 TM \otimes \Lambda^2 TM \to \mathbb R.$$ Since the metric $g$ on $TM$ induces a natural fibre metric on $\Lambda^2 TM$ via the formula $$\langle X \wedge Y, V \wedge W \rangle = g(X,V)g(Y,W) - g(X,W)g(V,Y),$$ we can turn this bilinear form into a linear operator in the usual way: $$\langle R(X\wedge Y), Z\wedge W\rangle = R(X \wedge Y, Z \wedge W).$$ Finally, the third symmetry above tells us that the bilinear form we started with was symmetric; so the operator corresponding to it via the inner product is self-adjoint (with respect to this inner product).