# If $K$ is a $(1,1)$-tensor field such that $\nabla K = 0$ then $\nabla_XKY = K\nabla_XY, \forall X,Y.$

Let $M$ be a Riemannian manifold with Levi-Civita connection $\nabla$ and $K$ a $(1,1)$ tensor field on $M$. I am trying to prove that if $K$ is parallel then $\nabla_X K(Y) = K(\nabla_XY)$ for every $X,Y$ vector fields on $M$.

I tried to use the definition of covariant derivative for tensor fields but it deals with the covariant derivative of $1$-forms, I mean, the formula is:

$$(\nabla_XK)(Y,\omega) = X(K(Y,\omega)) - K(\nabla_XY,\omega) - K(Y, \nabla_X \omega).$$

How can I conclude the claim?

Thanks

• The formula you wrote is not for $(1,1)$ tensors but for $(1,2)$ tensors (under a certain identification). – levap Jan 22 '17 at 23:37

## 1 Answer

The formula for the covariant derivative of a $(1,1)$ tensor $K$ is

$$(\nabla_X K)(Y) = \nabla_X(K(Y)) - K(\nabla_X Y).$$

If $K$ is parallel then $\nabla_X K = 0$ and so $\nabla_X(K(Y)) = K(\nabla_X Y)$ for all vector fields $Y$ on $M$.

• Easy, thank you! I probably made some confusion! Thanks. – L.F. Cavenaghi Jan 22 '17 at 23:42