Tangent map vs. differential on manifolds

Question

Let $E, E'$ be normed vector spaces, $A \subseteq E$ an open set and $f: A \to E'$ a differentiable map. For each point $x \in A$, the differential of $f$ is a linear transformation $Df|_x: E \to E'$. This means that we can consider the map \begin{align} Df: A &\to \mathcal L(E,E') \\ x &\mapsto Df|_x: E \to E' \end{align} Since $\mathcal L(E,E')$, we can differentiate $Df$ to obtain the second differential $D^2f|_x$ and so on.

Now, when we have manifolds $M$ and $M'$ (modeled on $E$ and $E'$, respectively), and a differentiable map $f: M \to M'$, the differential of $f$ at a point $p \in M$ is a linear transformation $Df|_p: TM|_p \to TM'|_{f(p)}$. Here $TM|_p$ denotes the tangent space to $M$ at $p$. The differential $Df|_p$ is defined locally using charts and we can show that this definition does not depend on the chosen charts. Now it is known that the second differential cannot be defined in this way so that it does not depend on the chosen charts. Instead, we can consider the map between the tangent bundles \begin{align} Tf: TM &\to TM' \\ (p,v) &\mapsto (f(p), Df|_p(v)) \end{align}

Here $TM$ denotes the tangent bundle, and $(p,v)$ is a point of the bundle, been $p \in M$ and $v \in TM|_p$. Since the bundles are manifolds (modeled on $E^2$ and $E'^2$, respectively), we can differentiate $Tf$. (My using of a different notation for $Tf$ and $Df|_p$ will be explained ahead.)

As far as I can tell, this is the approach usually taken to try and obtain a second order differential of a map $f$ between manifolds. But following the idea used on linear spaces, one could try to define a map \begin{align} Df: M &\to X \\ p &\mapsto Df|_p: TM|_p \to TM'|_{f(p)} \end{align} where $X$ would be (possibly) some a disjoint union of linear spaces $\mathcal L(TM|_p, TM'|_{f(p)})$.

Question 1 Is this approaching in defining $DF$ viable? Has it been worked out? If so, is it equivalent in some way to the above definition of $Tf$?

Question 2 Is it possible to endow the disjoint union $X$ with a differentiable structure so that we could differentiate $Df$? I would think we could identify $\mathcal L(TM|_p, TM'|_{f(p)})$ with the tensor product $T^*M|_p \otimes TM'|_{f(p)}$, but I don't know if we can then glue them in some canonical way, some kind of tensor product of bundles...

Answer 1

Your question can be answered using the language of vector bundles. Recall that a manifold $E$ equipped with a map $\pi : E \to M$ is a vector bundle over $M$ if:

1. Each fiber $E_p := \pi^{-1}(p)$ has the structure of a vector space.
2. Every point $p \in M$ admits a neighborhood $U$ and a homeomorphism $\varphi : U \times \mathbb{R}^k \to \pi^{-1}(U)$ (called a local trivialization of $E$) such that
   1. $\pi \circ \varphi$ is the projection map $U \times \mathbb{R}^k \to U$,
   2. $\pi$ restricts to a linear isomorphism between fibers $\{q\} \times \mathbb{R}^k \to E_q$.

An example of a vector bundle is the tangent bundle $TM$ with the projection map $\pi : TM \to M$.

I claim that the space $X$ in your question can be given the structure of a vector bundle over $M$. This perhaps isn't surprising if we think about what $X$ really is. For each $p$, we have a vector space $\mathcal{L}(T_pM, T_pM')$. The statement that $X$ is a vector bundle is just that we can give $X$ some differential structure that is compatible with the linear structure of the vector spaces $\mathcal{L}(T_pM, T_pM')$, as you referenced in Question 2.

It is surely possible to show that $X$ is a vector bundle by constructing the local trivializations by hand, but I think it is more enlightening if we take a more abstract approach.

We will need the following general principle, which is intentionally stated in a bit of a vague way. I won't give a proof (usually, one can just prove it in the specific cases that are needed).

General principle. Any "natural" construction in linear algebra gives rise to a corresponding construction for vector bundles over some fixed manifold $M$.

Here are some examples of this principle. Given vector bundles $E, F$ over a manifold $M$, we can form:

• the dual bundle $E^*$, where the fiber over $p$ is the dual space of $E_p$,
• the bundle $\mathrm{Hom}(E, F)$ (which would probably be written $\mathcal{L}(E, F)$ in your notation), where the fiber over $p$ is the vector space $\mathrm{Hom}(E_p, F_p)$ of linear maps from $E_p$ to $F_p$,
• the tensor product bundle $E \otimes F$, where the fiber over $p$ is the tensor product $E_p \otimes F_p$,
• etc.

Exercise. What would the local trivializations of the bundles $E^*$, $\mathrm{Hom}(E, F)$, and $E \otimes F$ be (in terms of the local trivializations of $E$ and $F$)?

Now, it is probably tempting to say that $X = \mathrm{Hom}(TM, TM')$. However, this does not make sense, since $TM$ and $TM'$ are vector bundles over different spaces! (What manifold would $\mathrm{Hom}(TM, TM')$ be a vector bundle over?) That said, if we look closely at what $X$ is, we see that we really want $X$ to be a vector bundle over the manifold $M$ (since we have a vector space for each point $p \in M$).

This discussion leads naturally to the concept of pullback bundles. Given a vector bundle $E$ over $M$ and a map $g : N \to M$ between manifolds, we can define the pullback bundle $g^*E$ to be the vector bundle over $N$ with fibers $(g^*E)_q = E_{g(q)}$ (and local trivializations that I won't write down). In other words, $g_*E$ is the vector bundle whose fibers correspond to the fibers of $E$, where this correspondence is specified by the map $g$.

Example. The tangent map $Tf$ associated to a map $f : M \to M'$ gives rise to a homomorphism of vector bundles $Tf : TM \to f^*TM'$. This just means that $Tf$ is a smooth map between vector bundles which restricts to a linear map $T_p M \to (f^*TM')_p = T_{f(p)}M'$ between fibers. Note that the $Tf$ here is different from the map $TM \to TM'$ you reference in your question (the codomains are different!).

This resolves our issue from earlier—we now have vector bundles $TM$ and $f^*TM'$ that live over the same space. Hence, we can define $$ X := \mathrm{Hom}(TM, f^*TM'), $$ which is a bundle over the manifold $M$. This is exactly what we want: the fiber of $X$ over a point $p$ is the vector space $\mathrm{Hom}(T_pM, T_{f(P)}M)$.