Commuting covariant Derivatives of a variation

Question

I am trying to understand the proof of the First and Second Variation of Arclength formulas for Riemannian Manifolds. I want some verifaction that the following covariant derivaties commute. I find it intuitive but I want to also have a formal proof.

Some notation: Let $\gamma(t,s):\overbrace{[a,b]}^{t}\times \overbrace{(-\epsilon,\epsilon)}^{s}\rightarrow M$ be a variation of $\gamma_0(t)$ with $|\dot{\gamma_{0}|}=\lambda \ \forall t \in [a,b] $ and $V=\gamma_{*}\left(\frac{\partial}{\partial s}\right)$ tha variational vector field.

Then I would like to prove that $\frac{D}{ds}(\dot{\gamma_t})=\frac{D}{dt}V$ or in other words that $\frac{D}{ds},\frac{D}{dt}$ commute. Note that $\frac{D}{ds}$ is the covariant derivative along the map $\gamma$ therefore we will use the property for connections along maps that $\nabla^{\gamma}_X(Z\circ \gamma(p))=\nabla_{\gamma_*(X)}Z|_p$.

Indeed $\frac{D}{ds}\left( \dot{\gamma_t}\right)=\frac{D}{ds}[\gamma_*(\frac{d}{dt})\circ(\gamma(s,t))]=D_{\gamma_*\left(\frac{d}{dt}\right)}\gamma_*(\frac{d}{ds})=D_{\gamma_*\left(\frac{d}{ds}\right)}\gamma_*(\frac{d}{dt})$.

The last equality follows since $D_{\gamma_*\left(\frac{d}{dt}\right)}\gamma_*(\frac{d}{ds})-D_{\gamma_*\left(\frac{d}{ds}\right)}\gamma_*(\frac{d}{dt})=\left[\gamma_*(\frac{d}{dt}),\gamma_*(\frac{d}{ds}) \right]=\gamma_*([\frac{d}{ds},\frac{d}{dt}])=0$ since $\frac{d}{ds},\frac{d}{dt}$ are coordinate vector fields.

Answer 1

Recall : $$(1) \ Xf = \frac{d}{dt}\bigg|_{t=0}\ (f\circ c)(t)$$ where $c$ is a curve with $ c'(0)=X$.

$$ (2) \ [X,Y] f = X(Yf)-Y(Xf) $$

Hence if $T(s,t)$ is a variation, then $X=T_t,\ Y=T_s$ are coordinate fields. Here $p=T(s,t)\in M$ and $ (Xf)(p=T(s,t)) $ is a function on $M$.

$c(\varepsilon) = T(s+ \varepsilon,t)$ is a curve s.t. $c'(0)=T_s= Y(p)$

Hence $$ Y(Xf) = \frac{d}{d\varepsilon}\ (Xf)(q=T(s+\varepsilon,t)) $$

$d(\epsilon)=T(s+\varepsilon,t+\epsilon )$ is a curve with $d'(0)=X(q)$ so that $$ Y(Xf) = \frac{d}{d\varepsilon } \frac{d}{d\epsilon } f(T(s+\varepsilon,t+\epsilon )) $$

Hence $F=f\circ T$ so that $X(Yf)=F_{ts}$ and $Y(Xf)=F_{st}$. Since we already have $F_{st}=F_{ts}$, the proof is followed.

Answer 2

HK Lee's answer is 100% correct. I am going to answer my question just to have an answer in the notation that I am more familiar with i.e. with connections along maps.

Say we have $f:(N,h)\rightarrow (M,g)$ and we endow $M$ with a connection $\nabla$. Then there is a unique connection along $f$ denoted $\nabla^f$, mainly a map $T(N)\times f^*(TM)\rightarrow f^*(TM)$ satisfying the standard axioms for a connection. What is more, if $\nabla$ has no torsion then neither does $\nabla^f$. In this notation $\frac{D}{dt}=\nabla^{\gamma}_{\partial_t}$ and $\frac{D}{ds}=\nabla^{\gamma}_{\partial_s}$.

Then $$T^{\gamma}(\partial_t,\partial_s)=\frac{D}{dt}\gamma_*(\partial_s)-\frac{D}{ds}\gamma_*(\partial_t)-\gamma_*[\partial_t,\partial_s]$$

But since we are using the Levi-Civita connection on $M$ the tensor $T^{\gamma}$ vanishes, as well as the term $[\partial_t,\partial_s]$. Therefore the result follows immediately.