Let $F: N \to M$ be a smooth map of smooth manifolds (with dimensions). Let $F_{*,p}: T_pN \to T_{F(p)}M$ be the differential at $p \in N$. Let $F_*: TN \to TM$ be the map between tangent bundles given by $F_*(X_p)=F_{*,p}(X_p)$. This says $F_*$ is a smooth embedding if $F$ is a smooth embedding.

What are some sufficient conditions to say $F_*$ is smooth besides $F$ being a smooth embedding?

  • I'm not really interested in deducing $F_*$ to be a smooth embedding or topological embedding. I'm just hoping for smooth for 1.4 here.

  • Some guide questions:

    1. If $F$ were smooth but not a smooth embedding, then is $F_*$ no longer necessarily smooth?

    2. What if $F$ were smooth and injective?

    3. What if $F$ were a smooth non-injective local diffeo?

    4. What if $F$ were a smooth non-injective immersion but not local diffeo?

    5. What if $F$ were a smooth injective immersion but not a topological embedding (My understanding is smooth embedding = smooth injective immersion + topological embedding)?

    6. I think each $F_{*,p}$ is smooth as a map of manifolds, besides linear as a map of vector spaces. What does this mean for $F_*$?

  • 2
    $\begingroup$ Any derivative of a $C^{\infty}$ function is $C^{\infty}$ pretty much by definition. Using this you can deduce that locally the differential is smooth, and hence globally. $\endgroup$ – leibnewtz Jun 20 at 13:41
  • $\begingroup$ @leibnewtz Thanks! Your second sentence refers to guide question (6)? $\endgroup$ – Selene Auckland Jun 20 at 13:48
  • $\begingroup$ @leibnewtz Wait so to clarify, even if $F$ were not an embedding, $F_*$ is still smooth? $\endgroup$ – Selene Auckland Jun 20 at 13:49
  • 1
    $\begingroup$ Yup. The map $F_*$ is always smooth as long as $F$ is $\endgroup$ – leibnewtz Jun 20 at 14:02
  • $\begingroup$ @leibnewtz Thanks! Also, good intuition in your first sentence in your first comment. $\endgroup$ – Selene Auckland Jun 20 at 14:06

It is still smooth. If $F\in \mathscr{C}^\infty(M,N)$, then fixing $p\in M$ and local coordinates $(x^1,\ldots, x^n)$ centred at $p$ on a trivializing neighborhood $U\subseteq M$ for $TM\to $M, and fixing analogous coordinates on $(y^1,\ldots, y^m)$ centred at $F(p)$ on a neighborhood $V\subseteq N$ containing $F(U)$ trivializing $TN\to N$, we can write down $F_*: TM\to TN$ in local coordinates as a map $TU\to TV$.

In local coordinates, $F$ is given by an $m-$tuple of smooth functions, $y^i=F_i(x^1,\ldots, x^n)$ for $1\le i \le m$. And given the local trivialization condition we can view $TU\cong U\times \mathbb{R}^n$ and $TV\cong V\times \mathbb{R}^m$. Then $F_*:U\times \mathbb{R}^n\to V\times \mathbb{R}^m$ is $F\times L$ where $$L|_{\{x\}\times \mathbb{R}^n}=L_x:\{x\}\times\mathbb{R}^n\to \{F(x)\}\times\mathbb{R}^m$$ is a linear transformation and the transformations $L_x$ vary smoothly according to the choice of $x\in U$. So, denoting the variable in $U$ by $x$ and the variable in $\mathbb{R}^n$ by $y$, $F_*$ can be viewed as a map $F_*(x,y)=(F(x),L_x(y))$. All the components are smooth, and hence so is $F_*$.

  • $\begingroup$ I was expecting just some link or reference to a textbook since I didn't really show effort into understanding how $TM$ and $TN$ are smooth manifolds in the first place. Thanks! $\endgroup$ – Selene Auckland Jun 20 at 13:53
  • 1
    $\begingroup$ No problem. Reviewing that construction will surely be a useful exercise. $\endgroup$ – Antonios-Alexandros Robotis Jun 20 at 13:54
  • $\begingroup$ Of course. Someday, I'll return to it. $\endgroup$ – Selene Auckland Jun 20 at 13:55
  • 1
    $\begingroup$ Careful — the linear transformation varies smoothly in the $x$ variables; your notation suggests that it is a fixed linear map. $\endgroup$ – Ted Shifrin Jun 20 at 17:14
  • 1
    $\begingroup$ You're right. Thanks Ted :) $\endgroup$ – Antonios-Alexandros Robotis Jun 20 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.