Gradient of solution to heat equation under evolving metric

Question

The following simple question came to me when I was studying the heat equation on a Riemannian manifold: Suppose $M$ is a closed Riemannian manifold and $g_t$ is a smooth family of Riemannian metrics on $M$. It's a well-known question the study the evolving heat equation on $M$: \begin{equation} \dfrac{\partial}{\partial t}f=\Delta_{g(t)} f \end{equation} Suppose that $V\subset TM$ is a fixed sub-bundle of the tangent space of $M$. Given that at time $t=0$, the gradient of $f_0$ belongs to $V$. Is it true that the gradient of $f$ remains in $V$ the whole time? I have a feeling that this is true because of the equalizing nature of the heat equation, but I don't know how to formally prove or disprove it. Does anyone have any thoughts or helpful references? Please apologize if this is a trivial matter.

Answer 1

Let me expand a bit on my comment to try to answer your question.

In general, there is no relationship between the metrics $g(t)$ and the sub-bundle $V$, so it is possible to find examples where the gradient is contained in $V$ at an initial time but is not contained within it for future times.

To get a positive answer, we need to assume that the sub-bundle $V$ interacts with the metric in a nice way. For instance, if $V$ is the tangent bundle of a totally geodesic foliation of $M$, then it might be possible to show that $\nabla f \subset V$ is preserved if you deform the metric by Ricci flow and the function by a heat equation. To be honest though, I don't know if this is the case or not. However, there is one situation where the answer is affirmative.

Consider the case where the function $f$ is invariant under the action of some group of isometries $G$. At first, this might seem unrelated to sub-bundles, but the idea is that when the group is infinite, this condition implies that the $\nabla f$ must be perpendicular to the infinitesimal transformations induced by the isometries (picture $O(1)$ spinning a round $\mathbb{S}^2$ around its axis and $f$ only a function of the polar angle). The subset of $TM$ perpendicular to the infinitesimal transformations is not necessarily a bundle (as can be seen with this example), but $f(t)$ will remain invariant under the group action under Ricci flow (so $\nabla f$ remains in a subset of $TM$).

To see this, we use the uniqueness of the Ricci flow. For compact manifolds, this was proven by Hamilton and for complete manifolds with bounded curvature, Ricci flow uniqueness is a theorem of Chen-Zhu [1]. These results imply that the isometries at the initial time are preserved in forward time, and when combined with the linearity of the heat equation, it follows that $f(t)$ is preserved under $G$ for positive time as well.

Since you are asking this question about Ricci flow, I'm guessing that you might be interested in the case where the heat flow runs in reverse time (since that's how the $\mathcal{W}$-functional is defined). In this case, the flow \begin{equation} \dfrac{\partial}{\partial t}f=\Delta_{g(t)} f \end{equation} no longer preserves mass so you need an extra term to correct for that. After fixing that issue, you can use the same argument as before to show that $f$ remains invariant under the group $G$, except now we use Kotschwar's result showing backwards uniqueness for the Ricci flow [2].

[1] Chen, Binglong; Zhu, Xiping, Uniqueness of the Ricci flow on complete noncompact manifolds, J. Differ. Geom. 74, No. 1, 119-154 (2006). ZBL1104.53032.

[2] Kotschwar, Brett L., Backwards uniqueness for the Ricci flow, Int. Math. Res. Not. 2010, No. 21, 4064-4097 (2010). ZBL1211.53086.