What do the fibers of the double tangent bundle look like?

Question

Consider the tangent bundle $\pi:TM\to M$ for some smooth manifold. As outlined in the Wikipedia page, we can then consider the double tangent bundle via the projection $\pi_*:TTM \to TM$, with $\pi_*$ the pushforward of the canonical projection $\pi$.

In the above page, they then proceed to mention that, given $$\xi =\xi^k \frac{\partial}{\partial x^k}\Bigg|_x\in T_x M, \qquad X =X^k \frac{\partial}{\partial x^k}\Bigg|_x \in T_x M,\tag A$$ and "applying the associated coordinate system" $\xi\mapsto (x^1,...,x^n,\xi^1,...,\xi^n)$ on $TM$, the fiber on $TTM$ at $X\in T_x M$ takes the form $$(\pi_*)^{-1}(X)=\left\{ X^k\frac{\partial}{\partial x^k}\Bigg|_\xi + Y^k\frac{\partial}{\partial \xi^k}\Bigg|_\xi : \,\, \xi\in T_x M,\,\, Y^1,...,Y^n\in\mathbb R \right\}.\tag B$$ I'm struggling to understand where this expression comes from.

I think I understand that $(x^1,...,x^n,\xi^1,...,\xi^n)$ is a (local) parametrisation for $TM$, and I can see that the fiber we are interested in is $T_X TM$, that is, the set of elements of $TTM$ above $X$, but I don't understand what the expression in the last equation represents.

If I were to write a fiber $T_x M$ of the tangent bundle, this would be the set of pairs $(x,v)$ with $v$ ranging across all (equivalence classes of) smooth curves $I\to M$ passing through $x$. In local coordinates, and focusing on the "curve component" of tangent vectors, I suppose we could write this as the set $$\pi^{-1}(x)=\left\{v^k \frac{\partial}{\partial x^k}\Bigg|_x : \,\, v^k\in\mathbb R\right\}.$$ By way of analogy, I'd guess $T_X TM$ to be the set of pairs $(X,V)$ with $V$ (equivalence classes of) curves $I\to TM$ passing through $X\in T_x M$. But even switching to local coordinates, I'm not sure how to go from this description to the expression in (B).

Answer 1

In the end, this is just a simple case of computing a tangent map and the fact that $\pi:TTM\to TM$ defines a second vector bundle structure on $TTM$ is not relevant for the problem at hand. (Partially things also get a bit complicated because you are using a notation that makes it complicated to distinguish between and object and its expression in local coordinates.)

Just start by thinking about how one constructs charts on the tangent bundle $TM$: You start with an open subset $U\subset M$ and a diffeomorphism from $U$ to an open subset $V\subset\mathbb R^n$, whose components are the local coordinates $x^i$. Denoting by $p:TM\to M$ the canonical projection, you get an induced diffeomorphism $p^{-1}(U)\to V\times\mathbb R^n$ that is used as a chart on $TM$. The first $n$ components of this are just the local coordinates $x^i$ and if you go through the construction, you see that indeed the tangent vector with coordinates $(x^1,\dots,x^n,\xi^1,\dots,\xi^n)$ exactly is $\sum_i\xi^i\tfrac{\partial}{\partial x^i}|_x$ where $x$ is the point with coordinates $x^i$. Now this shows that $(x^i,\xi^i)$ is a local coordinate system on $p^{-1}(U)\subset TM$, and hence given a point $\xi_x\in p^{-1}(U)$, any tangent vector at that point can be written as $\sum\alpha^i \tfrac{\partial}{\partial x^i}|_{\xi_x}+\sum_j\beta^j\tfrac{\partial}{\partial \xi^j}|_{\xi_x}$ for real numbers $\alpha^i$ and $\beta^j$.

At this point, it is best to forget about vector bundle structures and all that and just notice that $\pi:TTM\to TM$ is the tangent map (i.e. the derivative) of $p:TM\to M$. Since $p$ maps $p^{-1}(U)$ to $U$, this tangent map maps $Tp^{-1}(U)$ to $TU$ and hence can be expressed in the local coordinates we have constructed. But in these local coordinates, we simply get $p(x^1,\dots,x^n,\xi^1,\dots,\xi^n)=(x^1,\dots,x^n)$. This readily shows that the tangent map sends $\tfrac{\partial}{\partial x^i}$ to $\tfrac{\partial}{\partial x^i}$ and $\tfrac{\partial}{\partial \xi^j}$ to zero. Otherwise put, $$\pi\left(\sum_i\alpha^i \frac{\partial}{\partial x^i}|_{\xi_x}+\sum_j\beta^j\frac{\partial}{\partial \xi^j}|_{\xi_x}\right)=\sum_i\alpha^i \frac{\partial}{\partial x^i}|_x$$ and this equals $\xi_x$ if and only if $\alpha^i=\xi^i$ for all $i$, which is exactly what you want to see.

Edit (to address the issue on dimensions raised in your comment): This again mainly is an issue of notation. Since you have denoted points in $M$ by $x$ and local coordinates by $x^i$, I have tried to use similar notation for tangent vectors and this gets a bit misleading. What I have actually shown above is that the tangent vector $$ T_{Y_x}p\left(\sum_i\alpha^i\tfrac{\partial}{\partial x^i}|_{Y_x}+\sum\beta^j\tfrac{\partial}{\partial \xi^j}|_{Y_x}\right)=\sum_i\alpha^i\tfrac{\partial}{\partial x^i}|_x. $$ So for fixed $Y_x$, there is an $n$-dimensional affine subspace in $T_{Y_x}TM$ that gets mapped to $X_x\in TM$. But the actual pre-image of $X_x$ in $TTM$ is the union of all these affine subspaces for all the points $Y_x$, which form an $n$-dimensional vector space. So it looks like the product of $\mathbb R^n$ with an $n$-dimensional affine subspace of $\mathbb R^{2n}$ and hence has dimension $2n$ as expected. In the expression that you wrote in the question, the $n$-missing dimensions come from the free tangent vector $\xi\in T_xM$.

Even easier, if you extend what you know about local coordinates on $TM$ to $TTM$, you see that in the notation above, we get local coordinates $(x^i,\xi^j,\alpha^k,\beta^\ell)$ on an open subset of $TTM$ and in these coordinates, we get $\pi(x^i,\xi^j,\alpha^k,\beta^\ell)=(x^i,\alpha^k)$. Hence the pre-image of $(x^1,\dots, x^n,X^1,\dots,X^n)$ has $2n$ free parameters (the $\xi^j$ and the $\beta^\ell$).

Answer 2

Taking inspiration from the other answer, I'll try here to use a more rigorous/explicit notation to make sure I understand what's really going on.

Let $\pi:TM\to M$ be the canonical projection on the tangent bundle. Let $\phi:U\to\mathbb R^n$ be a chart, for some open $U\subset M$, and let $x\in U$. The double tangent bundle can be defined as the tangent bundle of $TM$ with canonical projection $\mathrm d\pi\equiv \pi_*:TTM \to TM$. Note that the local chart $(\phi,U)$ of $M$ induces the local chart $(\mathrm d\phi,TU)$ for $TM$. Finally, we also get a local chart for $TTM$ via the second differential: $(\mathrm d^2\phi,TTU)$.

Now, if we want to work in local coordinates, as done in the quote text, we have to work with the local representation of $\pi$ and $\mathrm d\pi$, that is, the maps $$ \tilde\pi\equiv \phi_* \pi\equiv \phi\circ\pi\circ \mathrm d\phi^{-1}, \qquad \tilde\pi:\mathbb R^{2n}\to\mathbb R^n, \\ \mathrm d\tilde\pi=\mathrm d\phi\circ\mathrm d\pi\circ(\mathrm d^2\phi)^{-1}, \qquad \mathrm d\tilde\pi:\mathbb R^{4n}\to\mathbb R^{2n}. $$ The relations between these maps can be summarised via the following commutative diagram:

We now observe the following:

1. Denoting with $\{\mathbf e_i^1\}_{i=1}^n,\{\mathbf e_j^2\}_{j=1}^n\subset\mathbb R^{2n}$ bases for (subspaces of) $\mathrm d\phi(TU)$, where $\mathbf e_i^1\equiv (\mathbf e_i,\mathbf 0_n)$ and $\mathbf e_i^2\equiv(\mathbf 0_n,\mathbf e_i)$, and with $\{\mathbf e_i\}_{i=1}^n$ a basis for $\mathbb R^n$, we have $$\tilde\pi(x,v)\equiv \tilde\pi(x^i\mathbf e_i^1 + v^j\mathbf e_j^2) = x^i\mathbf e_i\equiv x.$$

2. It follows that, denoting with $\{\mathbf e_i^3\}_{i=1}^n,\{\mathbf e_i^4\}_{i=1}^n$ linearly independent vectors to span the rest of $\mathbb R^{4n}$, we have $$ \mathrm d\tilde\pi((x,v),(\xi,\eta)) \equiv \mathrm d\tilde\pi( x^i\mathbf e_i^1 + v^j\mathbf e_j^2 + \xi^k\mathbf e_k^3 + \eta^\ell\mathbf e_\ell^4 ) \\= \tilde\pi(x^i\mathbf e_i^1 + v^j\mathbf e_j^2) + (\xi^k \partial_{\mathbf e_k^1}+\eta^\ell \partial_{\mathbf e_\ell^2})\tilde\pi(x^i\mathbf e_i^1 + v^j\mathbf e_j^2), $$ where we used the following notation for the differential of a generic smooth map $f:\mathbb R^n\to\mathbb R^m$: $$\mathrm df(\mathbf x,\mathbf v)\equiv \mathrm df(x^i\mathbf e_i + v^j \mathbf f_j) = (f(\mathbf x), v^j \partial_j f(\mathbf x)) \equiv f^i(\mathbf x)\mathbf e_i' + v^j \partial_j f^k(\mathbf x) \mathbf f_k',$$ if $\mathbf e_i',\mathbf f_i'$ are bases for the output space of $f$.

3. Observe that $$\partial_{\mathbf e_i^1}\tilde\pi(x^i\mathbf e_i^1 + v^j\mathbf e_j^2) = \mathbf e_i, \qquad \partial_{\mathbf e_i^2}\tilde\pi(x^i\mathbf e_i^1 + v^j\mathbf e_j^2) =0, $$ and thus $$ \mathrm d\tilde\pi((x,v),(\xi,\eta)) \equiv \mathrm d\tilde\pi( x^i\mathbf e_i^1 + v^j\mathbf e_j^2 + \xi^k\mathbf e_k^3 + \eta^\ell\mathbf e_\ell^4 ) = x^i\mathbf e_i^1 + \xi^j \mathbf e_j^2 \equiv (x,\xi). \tag X$$ To match this with some geometric intuition, remember that $x^i$ parametrise a point in $U\subseteq M$, $v^i$ parametrise a tangent vector to this point, $\xi^i$ parametrise another tangent vector to the point, corresponding to the "horizontal component" of the path along $U$ we are considering, and finally $\eta^i$ parametrise tangent vectors/changes to the former tangent vectors. In particular, $\mathbf e_i^1$ and $\mathbf e_i^3$ correspond to "horizontal" directions, whereas $\mathbf e_i^2$ and $\mathbf e_i^4$ correspond to "vertical" ones.

4. Having laid all of this groundwork, we get back to the question at hand: what do the fibers look like (in local coordinates)? In our notation, the answer is pretty straightforward: $$\mathrm d\tilde\pi^{-1}(x^i\mathbf e_i^1 + \xi^j \mathbf e_j^2) =\left\{ x^i\mathbf e_i^1 + v^j\mathbf e_j^2 + \xi^k\mathbf e_k^3 + \eta^\ell\mathbf e_\ell^4 : \,\, v^i,\eta^i\in\mathbb R \right\} \\\text{or, equivalently}\\ \mathrm d\tilde\pi^{-1}(x,\xi) = \{((x,v),(\xi,\eta)) : \,\, v,\eta\in\mathbb R^n\} \tag Y. $$

Equations (X) and (Y) are the main achievements of the above discussion. With this in mind, let me observe a few additional things to relate this expression with the notation given in the Wikipedia page, taking ideas from @peek-a-boo from the discussion in the comments:

1. In "partial derivatives notation", we would write $\mathbf e_i^2=\frac{\partial}{\partial x^i}\Big|_x$ and $\mathbf e_i^3=\frac{\partial}{\partial x}\Big|_{x,v}$, and $\mathbf e_i^4=\frac{\partial}{\partial v^i}\Big|_{x,v}$. In this notation, the fibers we are looking look like $$\mathrm d\tilde\pi^{-1}(x,\xi) = \left\{ v^i \frac{\partial}{\partial x^i}\Bigg|_{x} + \eta^i \frac{\partial}{\partial v^i}\Bigg|_{x,v} + \xi^i \frac{\partial}{\partial x^i}\Bigg|_{x} : \,\, v^i,\eta^i\in\mathbb R \right\},$$ where it is however worth stressing that there are two copies of $\frac{\partial}{\partial x^i}$, which refer to orthogonal subspaces, although in this notation they look the same because they both represent displacements of the variable $x$. However, in this context we are effectively considering two independent ways to displace $x$, and thus we want to keep these separate in this expression.

2. The above can equivalently be written a bit less explicitly by just prescribing $v$ to vary in the fiber $T_x M$, thus writing $$\mathrm d\tilde\pi^{-1}(x,\xi) = \left\{ \eta^i \frac{\partial}{\partial v^i}\Bigg|_{x,v} + \xi^i \frac{\partial}{\partial x^i}\Bigg|_{x} : \,\, v\in T_x M, \eta^i\in\mathbb R \right\}.$$ This expression also has the advantage of not featuring two "equal but different" copies of $\frac{\partial}{\partial x^i}$, and this is the notation used in the Wikipedia page.