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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 $F_*$?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $F_*$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ You're right. Thanks Ted :) $\endgroup$ – Antonios-Alexandros Robotis Jun 20 at 17:23

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