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I have read about this related question, Tautological 1-form on the cotangent bundle and have asked some professor and is still being confused.

So let $M=\mathbb{R^2}$, $N=\mathbb{R}$. $F:M\rightarrow N$ by $F(x_1,x_2)=x_1\in N$. Given $\alpha(y)=\xi(y)d y\in C^\infty(N; T^*N)$, we can pull it back as $$(F^*\alpha)(x_1,x_2)=\xi(x_1)d x_1,$$ which we can thinking of it as a field of arrow on the $\mathbb{R}^2$ plane pointing in the $x_1$ direction with its norm independent of $x_2$.

Now SUDDENLY, let's make $M\equiv T^*N= T^*\mathbb{R}=\mathbb{R}^2$ (therefore $F$ can also be called $\pi$), so $\xi$, as $\alpha(\partial_{x_1})$, is actually $x_2$. But to distinguish it from the coordinate, let's call it $\hat{x_2}$: $$F^*\alpha(x_1,x_2)=\xi(x_1)d x_1=\hat{x_2}(x_1)d x_1,$$ $\xi(x_1)$ or $\hat{x_2}(x_1)$ is still a function of $x_1$ (actually, a function of the 2-dimensional coordinate $(x_1,x_2)$, but no dependence on $x_2$). We define $$\theta=F^*\alpha\equiv \xi(x_1)d x_1$$ Can anybody tell me why the simplectic form on $T^*N=T^*\mathbb{R}=\mathbb{R}^2$ can be defined by the following way: $$\omega=d \theta=d\xi\wedge d x_1\equiv d x_2\wedge dx_1,$$ rather than $d \theta=\partial_{x_1}\xi(x_1)d x_1\wedge d x_1=0$? I haven been stuck on this problem for several days (not that that's the only thing I have been doing), really appreciate it if anyone can help me!

Ruairi
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  • The canonical one-form is not a pullback of a form from the base. Indeed, $\xi$ is the fiber coordinate, not a function of $x_1$. – Ted Shifrin Dec 03 '20 at 22:46
  • $\theta$ is the canonical $1$-form; it is not a pullback. Indeed, $\theta = \xi,dx_1$ and $\omega = d\theta = d\xi\wedge dx_1$. I have no idea where you got your pullback or $\xi = \partial_{x_1}(\alpha)$. – Ted Shifrin Dec 03 '20 at 22:59
  • Hi, for example in https://math.stackexchange.com/posts/252198/edit $\pi^* : T^_x M \to T^_{(x, \xi)} (T^* M)$ (same as the notes I am reading). So I guess I am confusing somethings. Could you give a clear description on this, am point out why I am wrong in my question? Thanks! – Ruairi Dec 03 '20 at 23:15
  • That's not even a final post, so I'm not going to look at it. This is talking about projection from the cotangent bundle of the cotangent bundle to the cotangent bundle. You are not. – Ted Shifrin Dec 03 '20 at 23:21
  • (Sorry grab the wrong link https://math.stackexchange.com/questions/252180/tautological-1-form-on-the-cotangent-bundle)Hi, I think I am talking exactly about that. My $M$ is actually $T^N$; $F$ is actually $\pi$ because it is projecting out the $x_2$, which is actually the "$\xi$" dimension. So in the language of that post, $\theta=\pi^\alpha$. – Ruairi Dec 03 '20 at 23:34

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