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On a compact Riemannian manifold $(M, g)$, it's easy to see that the gradient of a function (grad $f$ $= \nabla f$ in my notation) can't be a Killing vector unless it's the zero vector. In fact, for any constant $c$, if

$$\textstyle\frac{1}{2}\mathcal{L}_{\nabla f} g = cg$$

then tracing gives

$$\Delta f = cn$$

Integrating over $M$ implies $c=0$, and thus $f$ is constant by the maximum principle.

My question: It it possible on a compact Riemannian manifold for there to be two functions $f$ and $\alpha$ such that $\alpha \nabla f$ is a Killing vector? One can easily see that the equation here is

$$\alpha \nabla^2 f + \textstyle\frac{1}{2}(d\alpha\otimes df + df\otimes d\alpha)=0$$

After tracing we find

$$\alpha\Delta f + (d\alpha, df) = 0$$

It seems to me that the above maximum principle argument doesn't do much here since it's entirely possible that $\alpha$ could be zero at the critical points of $f$.

To reiterate, I'm interested whether in this situation $\alpha \nabla f$ must be zero or whether there is some example that shows that it need not be zero in general.

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    Hand-wavy answer: non-zero Killing vectors look like rotation fields near their zeroes, and $\alpha \nabla f$ can't look like this. – 2017-02-28
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    (near a critical point of $f$, that is.) – 2017-02-28
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    @AnthonyCarapetis Right; the Lie derivative $\mathcal{L}_{K}X$ of a vector field $X$ by a Killing field is $[K,X]=\nabla_{K}X-\nabla_{X}K$ which equals $-\nabla_{X}K$ at a zero of $K$, and $\nabla K$ is antisymmetric by the Killing condition. In other words, the derivative of the flow of $K$ at a zero is an antisymmetric endomorphism of the tangent space there i.e. an infinitesimal rotation. I'll think about whether we can use this to answer the original question. – 2017-02-28
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    @AnthonyCarapetis Your argument can be made rigorous. Run through the argument in my comment above at a zero of the vector but now plug in $K=\alpha \nabla f$. There are a couple cases depending on whether it's $\alpha$ or $\nabla f$ that's making $K$ zero, but regardless the only way the endomorphism is skew-symmetric is if it's the zero endomorphism. Now use the proof of the main theorem in Kobayashi's "Fixed Points of Isometries". You find that if the endomorphism is zero at a point, then there's an open set where $K$ is zero, but $K=0$ is a closed condition. – 2017-02-28
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    Of course the case where there are no zeros can be handled by the maximum principle by dividing by $\alpha$, so as long as the above reasoning is okay, we're done. If you care to write this up as an answer before I can muster the energy tomorrow, I'll accept it. – 2017-02-28

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Does not $\partial_\phi = \sin^2 \theta \, \mathrm{grad} \, \phi$ give an example of such a Killing vector on the sphere with line element $ds^2 = d\theta^2 + \sin^2 \theta \, d\phi^2$? It depends whether $\phi$ counts as a "function", I suppose.

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    If this counts then grad $\theta$ on the circle is a contradiction to the OP's initial assertion, so I'd say it doesn't. – 2017-02-28