Can the elements of the double tangent bundle $TTM$ be characterized as doubly parameterized curves?

Question

I've been thinking about the double tangent bundle $TTM$ after reading this answer, particularly the passage describing an element of $TTM$:

For the sake of visualization, I like to imagine a doubly parametrized curve $(s,t)↦γ(s,t)$ from $\mathbb{R}^2→M$, so that its image forms a sort of "2-D grid" on the manifold $M$ (for example, think of $M=\mathbb{R}^2$ and $γ(s,t)=(s,t)$; then the image of $γ$ is exactly the "coordinate grids" on the plane). I've been thinking about the structure of the double tangent bundle $TTM$.

If an element of $TM$ can be characterized as an equivalence class of curves $I\to M$ for some interval $I$, can an element of $TTM$ be characterzed as equivalence class of doubly parametrized curves $I^2\to M$?

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Why, after reading the post, I continue to have this doubt:

The post showed formally that an element of $TTM$ has the form $[t\mapsto [s\mapsto \gamma(s,t)]]$, while the argument for thinking of elements of $TTM$ as doubly parametrized curves was -it seems to me- a heuristic argument.

If we were dealing with curves, as opposed to equivalence classes of curves, then from the map $t\mapsto (s\mapsto \gamma(s,t))$ we could construct a map $(t,s)\mapsto \gamma(s,t)$, but it is not clear to me how to construct that map when thinking of equivalence classes.

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Edit: I'm writing this to better explain what I don't understand about the linked post.

Result 1: given a chart $(U,\alpha)$ in $M$, there is a chart $(TU,T\alpha)$ in $TM$ given by $$T\alpha: TU\to \alpha(U)\times E :[\gamma]\mapsto \bigg((\alpha\circ\gamma)(0),\frac{d}{dt}\bigg|_{t=0}(\alpha\circ\gamma)\bigg).$$ where $\gamma$ is defined as any member of $[\gamma]$. We know $T\alpha$ is well-defined.

So far so good.

Result 2: given a chart $(U,\alpha)$ in $M$, there is a chart $(T^2U,T^2\alpha)$ in $T^2M$ given by $$T^2\alpha : T^2U\to \alpha(U)\times E^3 : [\Gamma]\mapsto \bigg((\alpha\circ\gamma)(0,0), \frac{\partial (\alpha\circ\gamma)}{\partial s}\bigg|_{(0,0)}, \frac{\partial (\alpha\circ\gamma)}{\partial t}\bigg|_{(0,0)}, \frac{\partial^2 (\alpha\circ\gamma)}{\partial t\partial s}\bigg|_{(0,0)}\bigg)$$

where $\gamma$ is defined as follows:

• Take any $\Gamma\in[\Gamma]$, so that $\Gamma : I\mapsto TM : t\mapsto [\gamma_t]$.
• For any $t\in I$, take any $\gamma_t\in[\gamma_t]$, so that $\gamma_t : I\mapsto M : s\mapsto \gamma_t(s)$.
• Let $\gamma(s,t) = \gamma_t(s)$.

Here is what I struggle to understand: why is $T^2\alpha$ well-defined? i.e. why does $T^2\alpha([\Gamma])$ not depend on our choices for $\Gamma$ and $\gamma_t$?

Answer 1

Fix $p\in M$ and let $\mathcal{C}_{2,p}$ be the set of all smooth maps $\gamma$ from some open neighbourhood of $(0,0)\in\Bbb{R}^2$ into $M$ such that $\gamma(0,0)=p$. For each chart $(U,\alpha)$ of $M$ around the point $p$, we define a relation $\sim_{(U,\alpha)}$ by saying that $\gamma_1\sim_{(U,\alpha)}\gamma_2$ if \begin{align} \begin{cases} \frac{\partial(\alpha\circ\gamma_1)}{\partial s} \bigg|_{(0,0)} &= \frac{\partial(\alpha\circ\gamma_2)}{\partial s} \bigg|_{(0,0)},\\ \frac{\partial(\alpha\circ\gamma_1)}{\partial t} \bigg|_{(0,0)} &= \frac{\partial(\alpha\circ\gamma_2)}{\partial t} \bigg|_{(0,0)},\\ \frac{\partial^2(\alpha\circ\gamma_1)}{\partial t\partial s} \bigg|_{(0,0)} &= \frac{\partial^2(\alpha\circ\gamma_2)}{\partial t\partial s} \bigg|_{(0,0)}. \end{cases} \end{align} Clearly, $\sim_{(U,\alpha)}$ is an equivalence relation. In fact, if $(V,\beta)$ is any other chart around $p$, then the equivalence relations $\sim_{(U,\alpha)}$ and $\sim_{(V,\beta)}$ are the same, by the chain rule. Explicitly, it follows from the fact that for any such map $\gamma$, we have \begin{align} \begin{cases} \frac{\partial(\beta\circ\gamma)}{\partial s}\bigg|_{(0,0)}&=D(\beta\circ\alpha^{-1})_{\alpha(p)}\left[\frac{\partial(\alpha\circ\gamma)}{\partial s}\bigg|_{(0,0)}\right],\\ \frac{\partial(\beta\circ\gamma)}{\partial t}\bigg|_{(0,0)}&=D(\beta\circ\alpha^{-1})_{\alpha(p)}\left[\frac{\partial(\alpha\circ\gamma)}{\partial t}\bigg|_{(0,0)}\right],\\ \frac{\partial^2(\beta\circ\gamma)}{\partial t\partial s}\bigg|_{(0,0)}&= D^2(\beta\circ\alpha^{-1})_{\alpha(p)}\left[\frac{\partial(\alpha\circ\gamma)}{\partial t}\bigg|_{(0,0)}, \frac{\partial(\alpha\circ\gamma)}{\partial s}\bigg|_{(0,0)}\right] D(\beta\circ\alpha^{-1})_{\alpha(p)}\left[\frac{\partial^2(\alpha\circ\gamma)}{\partial t\partial s}\bigg|_{(0,0)}\right]. \end{cases} \end{align} Since all these equivalence relations are equal, we shall simply denote it as $\sim_{2,p}$. You could then define an element of the quotient $\mathcal{C}_{2,p}/\sim_{2,p}$ to be an element of the second tangent bundle (over the point $p$). You may notice that these definitions were made precisely to match up nicely with my formula $(\ddot{\smile})$ in the other answer.

Having said all this though, you should note that in reality, we don’t really care about what the ‘elements’ are but rather the various vector bundle structures and fiber bundle structures. For instance, $TTM$ has two vector bundle structures over $TM$, and it has a fiber bundle structure over $M$ (but generally not a vector bundle structure); and more generally speaking, if $(E,\pi,M)$ is a vector bundle, then $TE\to E$ is the usual tangent vector bundle (often called the primary vector bundle structure of $TE$), $(TE,T\pi,TM)$ also a vector bundle (called the secondary vector bundle structure of $TE$), and finally we also have a fiber bundle $TE\to M$, and all these projections fit together into a nice commutative diagram.