Confusion about pushforwards of Lie Brackets

Question

This is a very simple example that's bugging me, there's a basic gap in my understanding.

Let $\gamma : [0,1]^2 \rightarrow M$ be a curve, show that $[\frac{\partial \gamma}{\partial t}, \frac{\partial \gamma}{\partial s}] = 0$.

I can immediately solve it using the fact that $[\gamma_*(\frac{\partial}{\partial t}), \gamma_*(\frac{\partial}{\partial s})] = \gamma_*[\frac{\partial }{\partial s},\frac{\partial }{\partial s}]$, and the coordinate vector fields vanish in the bracket, but if I try and do things the long way round I start getting quite muddled.

If I try to do things in coordinates, I end up having to calculate stuff like $X(\frac{\partial \gamma^i}{\partial s})$ for $X = \left.\frac{\partial \gamma^j}{\partial t}\frac{\partial}{\partial x^j}\right|_{\gamma(t)}$, but $\frac{\partial \gamma^i}{\partial s}$ is a function of $t$ and $s$, so how can it make sense to differentiate it with the $x^i$ coordinates on $M$?

I suspect this has something to do with pullback bundles, indeed, if you have access to a connection then I "feel compelled" to want to write something like $X(\frac{\partial \gamma^i}{\partial s}) = \nabla_X(\frac{\partial \gamma^i}{\partial s}) = \gamma^*\nabla_\frac{\partial }{\partial t}(\frac{\partial \gamma^i}{\partial s}) = \frac{\partial^2 \gamma^i}{\partial t\partial s}$, but this is wrong (I'm not getting any $\frac{\partial \gamma^j}{\partial t}$ terms, and I shouldn't have to invoke a connection anyway).

Any explanation appreciated.

Following Andreas Cap's answer, I have wish to understand when this works. If $\gamma(s,t) = (x(s,t), y(s,t))$ is a diffeomorphism of $2$ dimensional manifolds, then it makes sense to pushforward the vector fields $\frac{\partial}{\partial s}, \frac{\partial}{\partial t}$. I can say

$\gamma_*(\frac{\partial}{\partial s}) = \left.\frac{\partial x}{\partial s}(s,t)\frac{\partial}{\partial x}\right|_{\gamma(s,t)} + \left.\frac{\partial y}{\partial s}(s,t)\frac{\partial}{\partial y}\right|_{\gamma(s,t)}$

$\gamma_*(\frac{\partial}{\partial t}) = \left.\frac{\partial x}{\partial t}(s,t)\frac{\partial}{\partial x}\right|_{\gamma(s,t)} + \left.\frac{\partial y}{\partial t}(s,t)\frac{\partial}{\partial y}\right|_{\gamma(s,t)}$

If I want to try and take the Lie bracket then I have $$\left[\frac{\partial x}{\partial s}(s,t)\frac{\partial}{\partial x}\right. + \left.\frac{\partial y}{\partial s}(s,t)\frac{\partial}{\partial y}\right., \left.\frac{\partial x}{\partial t}(s,t)\frac{\partial}{\partial x}\right. + \left.\frac{\partial y}{\partial t}(s,t)\frac{\partial}{\partial y}\right]$$

This is the same problem, $\frac{\partial x}{\partial t}(s,t)$ are functions of $s,t$ but we must differentiate them with $\frac{\partial}{\partial x}$ in this bracket.

Edit: I was being foolish, you are supposed to use $\gamma$ to write the reconcile the coordinates using the fact its a diffeomorphism, so the calculation indeed works when the pushforwards are well defined.

Answer 1

The problem is that there is no well defined operation of push forward of vector fields along smooth maps. Indeed, the "vector fields" that you are trying to use are certainly only defined on the image of $\gamma$. Even there, they may not be well defined if $\gamma$ is not injective. So it is not obvious how the initial question makes sense, the question being what the Lie bracket should be.

The expressions $\frac{\partial\gamma}{\partial t}$ and $\frac{\partial\gamma}{\partial s}$ certainly make sense as "vector fields along $\gamma$, i.e. as maps $\xi:[0,1]^2\to TM$ such that $p\circ\xi=\gamma$. (These can be viewed as sections of the pullback bundle $\gamma^*TM$ over $[0,1]^2$ but I don't think that there is a Lie bracket in this seeting.

A way to make sense of the original question is the setting of $\gamma$-related vector fields. This means that you suppose that there are vector fields $\xi$ and $\eta$ on $M$ such that $\xi\circ \gamma=T\gamma\circ \frac{\partial}{\partial t}$, and similarly for $\eta$ and $\frac{\partial}{\partial s}$. (This just mean that $\xi$ and $\eta$ are vector fields whose restriction to the image of $\gamma$ gives the objects you were trying to construct. Such fields will not exist in general, they certainly do if e.g. $\gamma$ is an embedding.) Then there is a general result saying that the brackets of $\gamma$-related vector fields are $\gamma$-related, which then shows that $[\xi,\eta]$ vanishes on the image of $\gamma$.

Answer 2

The right setting is indeed the pullback of the tangent bundle of $M$. Here's a brief summary.

If you have a map $\Phi: N \rightarrow M$, you can pull back the tangent bundle $T = T_*M$ to a bundle $T^\Phi$ over $N$. Given a connection $\nabla$ on $T_*M$, you can pull that back to a connection $\nabla^\Phi$ on $T^\Phi$. Observe that if $X$ is a tangent vector field on $N$, then $\Phi_*X$ is a section of $T^\Phi$. Therefore, given vector fields $X$ and $Y$ on $N$, $\nabla_X\Phi_*Y$ is a section of $T^\Phi$.

Now suppose the connection $\nabla$ is torsion-free. Then there is the following formula for the pushforward of the Lie bracket: $$ \Phi_*[X,Y] = \nabla^\Phi_X\Phi_*Y - \nabla^\Phi_Y\Phi_*X $$

This is the appropriate abstract setting for the standard derivation of the Jacobi equation for Jacobi fields on a Riemannian manifold.

I've written up the details here: https://www.math.nyu.edu/~yangd/papers/PullbackConnection.pdf