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From Lee's introduction to smooth manifolds:

Proposition $3.18$. The tangent bundle of a smooth $n$-manifold has a natural smooth structure that makes it into a $2n$-dimensional smooth manifold.

Since I wanted to know how the atlas is defined, I looked into the proof. It says that given a chart $$\varphi\colon U\subset M\to \varphi(U)\subset \mathbb R^n$$ we can define a function $$\displaystyle{\widetilde{\varphi}\colon TU}\to \varphi(U)\times\mathbb R^n$$by$$\widetilde{\varphi}(v^i\partial_i|_p)=(x^1(p),\ldots,x^n(p),v^1,\ldots,v^n)$$ and the set $\{\widetilde{\phi}:\phi\in A\}$ is the "natural" atlas. But $\widetilde{\varphi}$ is just the differential of $\phi$, i.e. $\widetilde{\varphi}=\mathrm{d}\phi$, isn't it? In other words, the natural charts on $TM$ are precisely the differentials of the charts on $M$, aren't they?

It is suspiciuous that after more than a year of dealing with differential geometry no one ever said that to me explicitly, so I wanted to make sure that I don't oversee something.

Filippo
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  • Depends how strict you want to be. If $\phi:U\to \phi[U]$, then strictly speaking $d\phi:TU\to T(\phi[U])$. As sets, $T(\phi[U])$ is not equal to $\phi[U]\times \Bbb{R}^n$, but since $\phi[U]$ is an open subset of $\Bbb{R}^n$, its tangent bundle has a very simple bijection $T\phi[U]\cong \phi[U]\times\Bbb{R}^n$. So, the natural mapping $\widetilde{\phi}$ is the composition of $d\phi$ with the above bijection. – peek-a-boo Feb 20 '22 at 18:15
  • @peek-a-boo Strictly speaking, the fibre at a point $p$ of the tangent bundle $TU, U \subseteq \Bbb R^n$ is the collection of linear derivatives at $p$. It could as well be $U \times \Bbb R^n$ where $p \times v$ is the directional derivative in the direction of $v$. So I don’t think you need to have a composition. – Ivin Babu Feb 21 '22 at 05:43
  • @IvinBabu Did you mean to write $(p,v)\in U\times\mathbb R^n$ instead of $p\times v$? – Filippo Feb 21 '22 at 07:04
  • @Filippo Yes. I guess both notations are in literature – Ivin Babu Feb 21 '22 at 07:30
  • @IvinBabu Okay, but I wouldn't call $(p,v)$ a directional derivative. – Filippo Feb 21 '22 at 08:16
  • @IvinBabu This might be of interest to you. It's an explanation of the relation between tangent vectors and directional derivatives. – Filippo Feb 22 '22 at 21:33
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    @Filippo Thanks! – Ivin Babu Feb 28 '22 at 15:45
  • @peek-a-boo Given some open $V\subset\mathbb R^n$, you seem to distinguish between $V\times\mathbb R^n$ and $TV$. What is your preferred definition of $TV$? – Filippo Mar 01 '22 at 09:02
  • Really, if you take any manifold $M$ and use any of the standard definitions of $TM$ and apply it to $V$ you'll clearly see that set theoretically $TV\neq V\times \Bbb{R}^n$. Yes, there's a very simple bijection, but they're not equal. For example, I like the definition of $TM$ as the set of all equivalences classes (under an appropriate equivalence relation) of smooth curves $\gamma:I\to M$ where $I\subset \Bbb{R}$ is an open interval containing the origin. – peek-a-boo Mar 01 '22 at 16:58
  • So, if you apply this definition to $V$, you see that $TV$ is the set of all equivalence classes of smooth curves $\gamma:I\to V$ (or equivalently curves $\gamma:I\to\Bbb{R}^n$ with $\gamma(0)\in V$). So, an element of $TV$ is an equivalence class $[\gamma]$. How can this ever be equal to an element of $V\times\Bbb{R}^n$? Clearly it's not. Having said all of this, the mapping $TV\to V\times\Bbb{R}^n$, $[\gamma]\mapsto (\gamma(0),\gamma'(0))$ is a bijection (an intuitive one since it tells us tangent vectors are determined by the base point and a direction). – peek-a-boo Mar 01 '22 at 17:04
  • @peek-a-boo ❝For example...❞ - As you just admitted, there are many common ways to define the set associated to the tangent bundle of a manifold. So the tangent bundle is defined up to a bundle isomorphism anyway. – Filippo Mar 01 '22 at 18:32
  • And if $M$ is an open subset of $\mathbb R^n$, then setting $TM=M\times\mathbb R^n$ and letting $$\begin{align}\mathrm{d}\phi\colon TM&\to TU\(x,v)&\mapsto(\phi(x),D\phi(x,v))\end{align}$$ be the usual differential for any chart $\phi\colon M\to U\subset\mathbb R^n$ (i.e. $D\phi(x,v)$ is the directional derivative) works perfectly fine. So even if we are very "strict" I don't see a problem. – Filippo Mar 01 '22 at 18:48
  • Yes, there are many ways of defining the set of the tangent bundle of a manifold. But what you're doing is using the same symbol $TV$ to denote two different things set theoretically, and I think in your mind it's such a simple identification that you're not distinguishing it (it's fine to not distinguish it, as long as you know why). – peek-a-boo Mar 01 '22 at 19:23
  • Let us continue this discussion in chat. – peek-a-boo Mar 01 '22 at 19:24

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