# Do Killing vectors of this particular type exist?

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.

• Hand-wavy answer: non-zero Killing vectors look like rotation fields near their zeroes, and $\alpha \nabla f$ can't look like this. Commented Feb 28, 2017 at 1:29
• (near a critical point of $f$, that is.) Commented Feb 28, 2017 at 1:40
• @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. Commented Feb 28, 2017 at 1:52
• @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. Commented Feb 28, 2017 at 2:28
• 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. Commented Feb 28, 2017 at 2:30

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.
• 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. Commented Feb 28, 2017 at 0:14