Let $T^{*}X$ be the cotangent bundle of $X$ and $\omega$ the tautological $2$-form. If $\sigma$ is a closed $1$-form defined on $X$, by defining $$\omega_{\sigma}=\omega+\pi^{*}\sigma,$$ I have proved that $(T^{*}X,\omega_{\sigma})$ is again a symplectic manifold. I want to know if $(T^{*}X,\omega)$ and $(T^{*}X,\omega_{\sigma})$ are symplectomorphic if $\sigma$ is exact.

I have also shown that if $\theta$ is a $1$-form defined on $X$, $\theta(X)=\{(x,\theta_{x})\mid x\in X\}$ is a Lagrangian submanifold of $(T^{*}X,\omega_{\sigma})$ if and only if $\sigma=d\theta$.

Then, I do believe that $(T^{*}X,\omega)$ and $(T^{*}X,\omega_{\sigma})$ are not symplectomorphic. If $\sigma$ is exact, $\sigma=d\theta$, then $\theta(X)$ is a Lagrangian submanifold in $(T^{*}X,\omega_{\sigma})$. Then, if $\varphi$ were a symplectomorphism, $\varphi(\theta(X))=\{(\varphi(x),\varphi(\theta_{x}))\mid x\in X\}$ would be a Lagrangian submanifold of $(T^{*}X,\omega)$. I am trying to prove that this in not, in fact, a Lagrangian submanifold, but I do not know how to do it.

Can anyone help me, please? Thanks in advance.

$\mathbf{EDIT}$: I think that by taking $f\colon (T^{*}X,\omega_{\sigma})\longrightarrow (T^{*}X,\omega)$ to be $f((x,\xi))=(x,\xi+\theta_{x}))$, $f$ is a symplectomorphism. I should prove that $f^{*}\alpha=\alpha+\pi^{*}\theta$ and I would be done. But does this hold? If I check it in the basis, considering $\alpha=\sum_{i=1}^{n}\xi_{i} dx_{i}$, $$f^{*}\alpha\big(\frac{\partial}{\partial x_{i}}\big)=\xi_{i},$$ but why is $$(\alpha+\pi^{*}\theta)\big( \frac{\partial}{\partial x_{i}}\big)=\xi_{i}?$$


1 Answer 1


You are right: given a smooth $1$-form $\theta$ on $X$, the diffeomorphism $f : T^*X \to T^*X : (x, \xi) \mapsto (x, \xi + \theta(x))$ you considered satisfies $f^{\ast}\alpha = \alpha + \pi^{*}\theta$ where $\alpha := \sum_{k=1}^n \xi_k dx^k$ is the tautological 1-form on $T^{\ast}X$.

This follows from the fact that for the map $f$ above we have $f^*\alpha = \sum_{k=1}^n (\xi_k + \theta_k)dx^k$, which in turn is rather clearly equal to $\alpha + \pi^*\theta$.

The expression above means that the 1-form $f^*\alpha$ at a point $(x, \xi) \in T^*X$ is $\sum_{k=1}^n (\xi_k + \theta_k(x))dx^k$. To see this, fix a point $(x, \xi) \in T^*X$ and a vector $v \in T_{(x, \xi)}T^*X$. Then we compute

$$ \begin{align} \notag (f^*\alpha)_{(x, \xi)}(v) &= \alpha_{f(x, \xi)}(f_{*}v) = \alpha_{(x, \xi + \theta(x))}(f_{*}v) \\ \notag &= \sum_{k=1}^n (\xi_k + \theta(x)_k) [dx^k(f_*v)] = \sum_{k=1}^n (\xi_k + \theta(x)_k) [(d(f^*x^k))(v)] \\ \notag &= \sum_{k=1}^n (\xi_k + \theta(x)_k) [(dx^k)(v)] = \left\lbrack \sum_{k=1}^n (\xi_k + \theta(x)_k) dx^k\right\rbrack(v). \end{align}$$

The moral of this calculation is a very simple way to compute pullbacks of forms: to evaluate $f^*\alpha$ at a point $(x, \xi)$ knowing a coordinate expression of $\alpha$, simply replace each occurrence of the $j$-th coordinate in $\alpha$ by the $j$-th coordinate of the point $f(x,\xi)$.

Here is another more invariant way to get the result (at first sight it might however appear somewhat formal, but it remains suggestive). Notice that any $1$-form $\theta$ on $X$ is a section $s_{\theta} : X \to T^*X$, namely $s_{\theta}(x) = \theta(x)$. The map $f(x, \xi) = (x, \xi + \theta(x))$, which is here expressed in coordinates, could be written invariantly as $f(\Xi) = \Xi + (s_{\theta} \circ \pi)\Xi$, where $\Xi \in T^*X$ is what we usually suggestively but somewhat abusely write $(x, \xi)$. The '$+$' which appears here is the addition in the fibers, which is invariantly well-defined because $T^*X$ is a vector bundle. It is thus tempting to write simply $f = Id + s_{\theta}\circ \pi$. It 'follows' (at least formally, but this could be explained rigorously) that $$f^*\alpha = (Id + s_{\theta}\pi)^*\alpha \; "=" \; Id^*\alpha + \pi^* s_{\theta}^* \alpha = \alpha + \pi^* s_{\theta}^* \alpha .$$

Now, the tautological 1-form $\alpha$ is uniquely and invariantly characterized by the property that $s_{\theta}^* \alpha = \theta$ for any 1-form $\theta$ on $X$ (both sides are $1$-forms on $X$). Consequently, $f^*\alpha = \alpha + \pi^*\theta$ (as $1$-forms on $T^*X$).


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