I'm looking for an example of a non-Euclidean non-compact Riemannian manifold $(M,g)$ such that we could define a non-constant affine function $f:M\rightarrow \mathbb{R}$, namely its gradient vector field is a killing vector field.
7
-
$\begingroup$ Looks unlikely, as the minimum of $f$ is a sink of the gradient, if we ignore degeneracies. $\endgroup$– Fan ZhengCommented Jan 16, 2016 at 18:56
-
$\begingroup$ @FanZheng I changed my question a little bit. $\endgroup$– Morteza AzadCommented Jan 16, 2016 at 19:08
-
$\begingroup$ What does "affine function" mean in this context? $\endgroup$– Deane YangCommented Jan 16, 2016 at 19:18
-
1$\begingroup$ @MortezaAzad probably you mean noncompact? $\endgroup$– Fan ZhengCommented Jan 16, 2016 at 20:37
-
2$\begingroup$ This really is not a research-level question. You should ask it on MathStackExchange or work it out yourself, assuming that you have had the first couple of weeks of a course in differential geometry. $\endgroup$– Robert BryantCommented Jan 16, 2016 at 22:55
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
1
Consider the manifold $\mathbb{S}^2\times \mathbb{R}$ with the product metric. Define $f:\mathbb{S}^2\times \mathbb{R}\to \mathbb{R}$ as $f(z,t)=t.$ Then it is clear that $Hess(f)=0,$ that is, $\nabla f$ is a Killing vector field.
Note that if $M$ is compact then such a function doesn't exist. Indeed, you have $\Delta f\equiv 0.$ Thus,
$$0=\frac 12 \int_M \Delta f^2=\int_M f\Delta f+\int_M |\nabla f|^2=\int_M |\nabla f|^2,$$ from where one gets that $f$ must be constant.
-
$\begingroup$ A harmonic function on a compact manifold is constant. QED. $\endgroup$ Commented Jan 16, 2016 at 23:36