Killing vector field and symmetric tensor field. Can you tell me where I can find a proof of the following fact: Let $M$ be a Riemannian manifold and $T$ be a symmetric tensor field on $M$. Furthermore let $X$ be a Killing vector field on $M$. Then the covector field $T(X,.)$ is divergence-free.
 A: Unless $T$ is divergence-free, this is false. (See my comment below the original question.)
If we assume $\operatorname{div} T = 0$, I think this is easiest to prove using index notation. The fact that $X$ is a Killing field means $\operatorname{Sym}(\nabla X)^\flat = 0$, which in index notation means
$$
X_{i;j} + X_{j;i} = 0.
$$
On the other hand, $\operatorname{div} T = 0$ means
$$
T_{ij;}{}^j = 0.
$$
(I'm using the summation convention here, and an index after a semicolon is one resulting from applying the total covariant derivative.)
Now the divergence of $T(X,\cdot)$ is
\begin{align*}
(T_{ij}X^i)^{;j}
&= T_{ij;}{}^j X^i + T_{ij} X^{i;j}\tag{1}\\
&= 0 + \tfrac 1 2 \left( T_{ij} X^{i;j} + T_{ji} X^{j;i}\right)\tag{2}\\
&= \tfrac  1 2\left( T_{ij} X^{i;j} + T_{ij} X^{j;i}\right)\tag{3}\\
&= \tfrac  1 2 T_{ij}  \left( X^{i;j} + X^{j;i}\right)\\
&= 0.
\end{align*}
In (1), I used the fact that covariant derivatives satisfy a product rule with respect to tensor product and commute with contractions. In (2), I interchanged the dummy indices $i$ and $j$ in the second term, and in (3), I used the fact that $T$ is symmetric.
