Let $G$ be a diffeomorphism of $\mathbb{R}$. Then the map $\mathbb{R}\times\{0\}\rightarrow\mathbb{R}\times\{0\}$ given by $(x,0)\mapsto (G(x),0)$ may be extended to a symplectomorphism $(x,\xi)\mapsto (G(x),(G')^{-1}(x)\xi)$ of $T^*\mathbb{R}$. (Here, $(x,\xi)$ are symplectic coordinates in $T^*\mathbb{R}$.)

In other words, a map on the zero section of $T^*\mathbb{R}$, of the form given above (i.e., which is identifiable with a diffeomorphism of $\mathbb{R}_x$), may be lifted to a symplectomorphism of the cotangent bundle. (This could certainly be stated in higher dimensions, with $G'$ replaced by the Jacobian matrix of $G$.)

Now let $\mathcal{C} = \{\xi_1 = 0\}\subset T^*\mathbb{R}^2$. We take $(x_1,\xi_1;x_2,\xi_2)$ as coordinates on $T^*\mathbb{R}^2$. Suppose we start with the map $\Psi:\mathcal{C}\rightarrow\mathcal{C}$ given by $(x_1,0;x_2,\xi_2)\mapsto (\varphi(x_1,x_2,\xi_2),0;x_2,\xi_2)$. Here, $\varphi$ is differentiable in all three components and for each $(x_2,\xi_2)$, $x_1\mapsto\varphi(x_1,x_2,\xi_2)$ is invertible. $\Psi$ is a pre-symplectomorphism of the coisotropic submanifold $\mathcal{C}$, with respect to the form $d\xi_2\wedge dx_2$. Think of $\varphi$ as a diffeomorphism of $x_1$ depending smoothly on $(x_2,\xi_2)$: $\varphi(x_1,x_2,\xi_2) = G_{(x_2,\xi_2)}(x_1)$. My question is: Can $\Psi$ be extended to a symplectomorphism of $T^*\mathbb{R}^2$? That is: Does there exist a symplectomorphism $\Phi$ of $T^*\mathbb{R}^2$ such that $\Phi|_\mathcal{C} = \Psi$?

A first attempt would be $\Phi:(x_1,\xi_1;x_2,\xi_2)\mapsto (G_{(x_2,\xi_2)}(x_1),(G'_{(x_2,\xi_2)})^{-1}(x_1)\xi_1;x_2,\xi_2)$, where the prime indicates differentiation in the $x_1$ variable. But this isn't a symplectomorphism.

Any help would be greatly appreciated!


1 Answer 1


The answer is negative already at infinitesimal level. If your extension was possible, then the Jacobian matrix would be in general an upper triangular block matrix (with 2x2 blocks). And it is easy to check that such a matrix can never be symplectic, unless it is a diagonal block matrix: Indeed, if we denote $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, if $$\begin{pmatrix}A&B\\0&C\end{pmatrix}^T\begin{pmatrix}J&0\\0&J\end{pmatrix}\begin{pmatrix}A&B\\0&C\end{pmatrix}=\begin{pmatrix}J&0\\0&J\end{pmatrix}$$ then $A^TJA=J$, $C^TJC=J$ (i.e. $A$ and $C$ are symplectic) and $B^TJA=0$ which implies $B=0$.
So I think, your map can be extended only if $\phi$ only depends on $x_1$

  • $\begingroup$ It isn't so obvious to me that every upper triangular block matrix which is symplectic must be a diagonal block matrix. Could you please explain this? $\endgroup$
    – Rohan
    Mar 28, 2017 at 18:28
  • $\begingroup$ Sorry, it was a bit quick. I added some details to the answer. $\endgroup$
    – Vincent H
    Mar 28, 2017 at 19:46
  • $\begingroup$ Thanks. But isn't $\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$ a symplectic matrix? $\endgroup$
    – Rohan
    Mar 28, 2017 at 20:44
  • $\begingroup$ It is indeed symplectic but not of the form $\begin{pmatrix}A&B\\0&C\end{pmatrix}$ with $A$, $B$ and $C$ 2x2 matrices. $\endgroup$
    – Vincent H
    Mar 29, 2017 at 7:02
  • $\begingroup$ I don't see why this should be true only for 2x2 blocks (or blocks larger than 1x1). $\endgroup$
    – Rohan
    Mar 29, 2017 at 13:20

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