2

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.

Niklas
  • 289
  • 1
    This is not true, unless you assume that $T$ is divergence-free. For example, in $\mathbb R^2$ with its Euclidean metric $g=dx^2+dy^2$, take $X=\partial/\partial x$ and $T = x ,dx\otimes dx$. Then $X$ is a Killing vector field, but $T(X,\cdot) = x,dx$, which is not divergence-free. – Jack Lee Jun 06 '17 at 16:47
  • Thank you! Do you know how to prove the statement under additional condition that $T$ is divergence-free? – Niklas Jun 06 '17 at 17:02

1 Answers1

2

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.

Jack Lee
  • 46,803