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I wanna know why if $\alpha$ a 1-form on $X$, then the next map is a diffeomorphism.

$$T_{\alpha}:T^*X\to T^*X$$ $$\hspace{2.4cm}\beta \mapsto \beta + \alpha_{\pi(\beta)} $$

I know if $\alpha$ is a closed form then the map is a symplectic diffeomorphism, and if $\alpha$ isn't closed it isn't symplectic, but it is a diffeomorphism. I saw some info in this question Coordinate-free proof of non-degeneracy of symplectic form on cotangent bundle but I really don't catch it.

Gyadso
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  • Which part of the definition of diffeomorphism are you having trouble verifying? – Jason DeVito - on hiatus Dec 06 '20 at 23:12
  • I've been trying to check that the map has an inverse $T_{\alpha}^{-1} that is C^{\infty}, but I don't even know how to define it – Gyadso Dec 06 '20 at 23:52
  • (If you want me to be notified, you have to type @Jason somewhere in your comment). What if you just focus on one cotangent space? Can you define the inverse there? – Jason DeVito - on hiatus Dec 06 '20 at 23:58
  • Tangentially-related question for you: Do you understand why the graph of any smooth function $f\colon X\to Y$ (as a submanifold of $X\times Y$) is diffeomorphic to $X$? – Ted Shifrin Dec 07 '20 at 00:50
  • That's really more of a cotangentially related question ;). – Jason DeVito - on hiatus Dec 07 '20 at 01:19
  • smacks @Jason – Ted Shifrin Dec 07 '20 at 01:54
  • Guys, I understand the idea with the graph, I can use the projection to prove that. But with $T_{\alpha}$ I don't see how to use it. I've been trying to see how it works in each fiber, but really I don't know how to define the inverse. @JasonDeVito – Gyadso Dec 09 '20 at 14:39
  • @Luis: Say the fiber is $2$-dimensional, just to pick a nice concrete number. Then the cotangent space at a point is isomorphic to $\mathbb{R}^2$ with an isomorphism taking $\alpha_{\pi(p)}$ to some vector $v\in\mathbb{R}^2$, and then $T_\alpha$ is identified with the map $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $f(x)=x+v$. (If it helps, pick a concrete vector in $\mathbb{R}^2$). What does adding $v$ do geometrically? What's the inverse of that? – Jason DeVito - on hiatus Dec 09 '20 at 15:18

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