Linearization of Gradient Flow

Question

As someone who has only "theoretical" knowledge in Riemannian geometry, I have a hard time trying to wrap my head around how to actually compute the so called "linearization" of a gradient flow on a manifold.

The setting is: We have a symplectic manifold $(M,\omega)$ with an almost complex structure $J:TM\rightarrow TM$ and an induced Riemannian metric $g(\cdot,\cdot) = \omega(\cdot,J\cdot)$. Now we are interested in solutions $u\in W^{1,2}(\mathbb{R}\times\mathbb{S}^1,M), (s,t)\mapsto u(s,t)$ of the following gradient flow: $$\frac{\partial u(s,t)}{\partial s} + J(u(s,t))\frac{\partial u(s,t)}{\partial t}= 0$$

In the papers I've come across, this flow is then linearized along a vector field $\zeta \in C^\infty(u^*TM)$ to yield the following linear differential operator: $$F(u)\zeta = \nabla_s \zeta + J(u)\nabla_t \zeta+\nabla_\zeta J(u)\frac{\partial u}{\partial t}, $$

the covariant derivatives $\nabla_s,\nabla_t,\nabla_\zeta$ being taken w.r.t. the metric $g$ above.

Now, my main question is: Why is $\nabla_\zeta \frac{\partial u(s,t)}{\partial t} = \nabla_t \zeta$?

At first, I had the idea to just see $\frac{\partial u(s,t)}{\partial t} \in T_{u(s,t)}M$ as a (linear) map $$f : W^{1,2}(\mathbb{R}\times\mathbb{S}^1,M) \rightarrow L^2(\mathbb{R}\times\mathbb{S}^1,TM), u(s,t)\mapsto \frac{\partial u(s,t)}{\partial t}$$ and then just differentiate it like every other map on a manifold: Fix $u\in W^{1,2}$ and for $\zeta \in C^\infty(u^*TM)$ set $\gamma:(-\epsilon,\epsilon)\rightarrow W^{1,2}$ with $\gamma(0)(s,t) = u(s,t)$, $\dot{\gamma}(0)(s,t) = \zeta(s,t) \in T_{u(s,t)}M$. Then: $$df (\zeta) = \left.\frac{d}{d\tau}\right|_{\tau=0} f(\gamma(\tau)) = \left.\frac{d}{d\tau}\right|_{\tau=0} \frac{\partial \gamma(\tau)(s,t)}{\partial t} = \frac{\partial}{\partial t}\left.\frac{d}{d\tau}\right|_{\tau=0} \gamma(\tau)(s,t) = \frac{\partial}{\partial t}\zeta(s,t)$$ However, I have no idea how (or if) this is connected to $\nabla_t\zeta$. I am grateful for any tipps regarding my original question or my try to solve it.

Answer 1

I'm answering this question two years later, because I stumbled upon it while googling a related question (I think I'm studying the same thing you were two years ago. You probably have the answer by now, but I'm including it for closure).

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You asked why $\nabla_\zeta \partial_t u = \nabla_t \zeta$.

First, recall that you are linearizing at a solution $u$ of the flow equation. And this linearization $F$ acts on vectors $\zeta \in \mathcal{C}^\infty(u^* TM)$. Such vectors can be written $\zeta = \partial_s u_s$, where $s \mapsto u_s$ is a path of cylinders, with $u_0 = u$. Hence, $\nabla_\zeta$ is the same thing as $\nabla_s$.

Now, recall from classic Riemannian geometry that: given a curve $\gamma = \gamma(t)$ and a vector field $V$ along $\gamma$, one can define its covariant derivative along $\gamma$, $D_t V$. And, if one can extend $V$ to some $\tilde{V}$ on a neighbourhood of the curve, we have $D_t V = \nabla_{\dot{\gamma}} \tilde{V} =: \nabla_t \tilde{V}$.

And we have the property that if $s \mapsto \gamma_s$ is a variation of $\gamma$ (and $\gamma$ is not too bad), then $D_s \partial_t \gamma_s = D_t \partial_s \gamma_s$ (see, for example, chapter 6 of Lee's Riemannian Manifolds).

In our case, our original "curve" is $u$, and its variation is some path $s \mapsto u_s$ (in the space of such cylinders) with $\zeta = \partial_s u_s$. Therefore, $\nabla_\zeta \partial_t u_s = \nabla_s \partial_t u_s = \nabla_t \partial_s u_s = \nabla_t \zeta$.