$\DeclareMathOperator{\vol}{vol}$I've been working through the computation of the first variation of volume presented in Jost's Riemannian Geometry and Geometric Analysis (page 196 in the sixth edition, section titled: Minimal Submanifolds), and I've been getting caught up in all the notation, and I've been having a lot of trouble exactly understanding how to interpret the partial derivatives in this context.
I'll start with the setup: let $M$ be a smooth submanifold of $N$ and let $F:M\times(-\epsilon,\epsilon)\to N$ be a local variation with compact support. For small enough $t$ we have that $\Phi_t(\cdot):=F(\cdot,t)$ is a diffeomorphism from $M\to M_t\subseteq N$. Now let $\{e_1,\dots,e_m\}$ be an orthonormal frame on $M$. Using this diffeomorphism we can write $$\vol(M_t)=\int_{M}\left\langle\Phi_{t*}e_1\wedge\cdots\wedge\Phi_{t*}e_m,\Phi_{t*}e_1\wedge\cdots\wedge\Phi_{t*}e_m\right\rangle^{\frac{1}{2}}\eta_{M},$$ where $\langle{\cdot,\cdot}\rangle$ is the induced inner product on $\bigwedge^m(TN)$, that is $\langle{v_1\wedge\cdots\wedge v_m,w_1\wedge\cdots\wedge w_m}\rangle=\det(\langle{v_i,w_j}\rangle)$, and $\eta_M$ denotes the Riemannian volume form on $M$.
Then we differentiate this with respect to $t$ to find that $$ \left.\frac{d}{dt}\vol(M_t)\right|_{t=0}=\left.\sum_{i=1}^{m}\int_{M}\frac{\left\langle\Phi_{t*}e_1\wedge\frac{\partial}{\partial t}\Phi_{t*}e_i\wedge\cdots\wedge\Phi_{t*}e_m,\Phi_{t*}e_1\wedge\cdots\wedge\Phi_{t*}e_m\right\rangle}{\|\Phi_{t*}e_1\wedge\cdots\wedge\Phi_{t*}e_m\|}\eta_M\right|_{t=0}$$
- My first question is about the notation $\frac{\partial}{\partial t}\Phi_{t*}e_i$. Should I interpret this as follows: Let $\gamma:(-\epsilon,\epsilon)\to TN$ be given by $\gamma(t)=\Phi_{t*}e_i\in T_{\Phi_{t}(p)}M$. Then does $\frac{\partial}{\partial t}\Phi_{t*}e_i$ simply mean $d\gamma\left(\frac{\partial}{\partial t}\right)$, where we naturally identify $T(T_qN)\cong T_qN$?
He goes on consider the vector field $X:=\left.\frac{\partial}{\partial t}\Phi_{t}\right|_{t=0}$. I'm assuming the interpretation of this vector field is the same as before.
My main confusion is with this next part:
To compute $\frac{\partial}{\partial t}\Phi_{t*}e_i$ at $t=0$ we consider a curve $c_i(s)$ in $M$ with $c_i(0)=p$ and $c_i'(0)=e_i$ and let $c_i(s,t):=\Phi_t(c_i(s))$. Then $$\left.\Phi_{t*}e_i=\frac{\partial}{\partial s}c_i(s,t)\right|_{s=0}.$$
- How do I justify this? It definitely lives in the right tangent space since $c_i(0,t)=\Phi_t(p)$, but why does this coincide with the pushforward $d\Phi_t(e_i)$. I'm probably missing something pretty fundamental.
Carrying on with the computations we have $$\left.\frac{\partial}{\partial t}\Phi_{t*}e_i\right|_{t=0}=\left.\frac{\partial}{\partial t}\frac{\partial}{\partial s}c_i(s,t)\right|_{s=t=0}=\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}c_i(s,t)\right|_{s=t=0}=\left.\nabla^N_{\frac{\partial}{\partial s}}X\right|_{s=0} =\nabla^N_{e_i}X, $$ where $\nabla^N$ is the Levi-Civita connection on $N$.
- My question is why do the partial derivatives commute in this case? Again, how should these mixed partials be understood, and what justifies this computation? Even intuitively, it doesn't make sense to me that they should.