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Suppose that $\Phi_t$ is a the global flow associated with a vector field $X$ on a Riemannian manifold $M$ and that $Y$ is any other vector field. Suppose furthermore that $X$ is a Killing vector field. Is there any way to write $$ \operatorname{div} [(\Phi_t)_* Y] $$ that is simpler than just writing it out in coordinates?

Thank you.

EDIT: What about $$ \nabla_{\Phi_*Y}(\Phi_* Z)? $$ (where $Z$ is a vector field)

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Since $\Phi_t$ is an isometry, I think $\operatorname{div} [(\Phi_t)_* Y]$ is the same as $(\Phi_t)_* (\operatorname{div}Y)$. – user53153 Jan 15 '13 at 23:41
this is what I would expect, but I don't know how to prove it... – William Jan 15 '13 at 23:57
Oops, I should not push functions forward. Hopefully, the answer below is right. – user53153 Jan 16 '13 at 0:28
up vote 3 down vote accepted

Writing everything in coordinates should not be too bad. But I'll try without, writing $f$ instead of $\Phi_t$. Let $\mu$ be the volume form. Recall the coordinate-free formula for divergence $d(i_Y\mu)=(\operatorname{div} Y )\mu$. For any diffeomorphism $f$ we have $i_{f_*Y}((f^{-1})^*\mu) = (f^{-1})^* (i_Y\mu)$. Since $f$ is an isometry, $(f^{-1})^*\mu=\mu$. Thus, $i_{f_*Y}\mu = (f^{-1})^* (i_Y\mu)$. Applying $d$ to both sides, we get $$ (\operatorname{div} f_*Y) \mu = d(i_{f_*Y}\mu ) = d((f^{-1})^* (i_Y\mu)) = (f^{-1})^* (d (i_Y\mu)) = (f^{-1})^*((\operatorname{div} Y )\mu) $$ hence $\operatorname{div} f_*Y =(\operatorname{div} Y ) \circ f^{-1}$.

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thank you very much! – William Jan 16 '13 at 0:39
@Guillermo Actually, I messed up the direction of maps at first; the edited version seems to make more sense... – user53153 Jan 16 '13 at 0:52
This was a special case of a more general question, which is actually more related to the title: what happens if instead of $div$ I write the covariant derivative with respect to some vector field? – William Jan 16 '13 at 1:15
@Guillermo I'm more used to the "connection" language, so I can reply in this way: the Levi-Civita connection is intrinsic to a Riemannian manifold. This means that if $f:(M,g_M)\to (N,g_N)$ is an isometry, then the pullback of the Levi-Civita connection on $N$ is the Levi-Civita connection on $M$. Unwinding this, you should get $\nabla_{f_*Y}(f_*Z) = (f^{-1})_* (\nabla_Y Z)$ – user53153 Jan 16 '13 at 17:49

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