# Can any symplectomorphism be called a canonical transformation?

I just want to make sure I am thinking clearly about canonical coordinates and transformations in Hamiltonian mechanics.

Suppose we have a Hamiltonian system $(M, \omega, H)$ — where $M$ is the phase space ($\dim(M)=2n$, although $M$ is not necessarily a cotangent bundle globally), $\omega$ is the symplectic structure (non-degenerate, closed 2-form), and $H$ is a function on $M$ serving as the Hamiltonian. Now suppose we have two overlapping coordinate charts $\phi\colon U \rightarrow V \subset \mathbb{R}^{2n}$ and $\psi: U \rightarrow W \subset \mathbb{R}^{2n}$, for some open $U \subset M$. The coordinate transformation $\psi\circ\phi^{-1}\colon V \rightarrow W$ is a symplectomorphism, since it is just the identity map expressed in different coordinates: more precisely, $\psi\circ\phi^{-1}\colon (V, \phi_{*}\omega) \rightarrow (W, \psi_{*}\omega)$ trivially satisfies $\phi_{*}\omega = (\psi\circ\phi^{-1})^{*}\psi_{*}\omega$. But we would not call such a coordinate transformation canonical unless $\phi$ and $\psi$ were both canonical coordinates (or Darboux charts) to begin with, right? For example, one defining criterion for canonical transformations often given in physics texts is that the Jacobian matrix of the transformation be a symplectic matrix.. the so-called symplectic condition. Here, $(\psi\circ\phi^{-1})_{*}$ is a symplectic matrix only if both $\phi$ and $\psi$ are canonical coordinates. So can we conclude that not all symplectomorphisms are canonical transformations?

## 1 Answer

Symplectomorphisms is one$^1$ possible definition of canonical transformations (CT), used by e.g. V.I. Arnold, cf. e.g. this Phys.SE post.

Now to be more precise: symplectomorphisms on a $2n$-dimensional symplectic manifold $(M,\omega)$ come in different versions:

• with or without explicit time dependence. (In this Phys.SE answer we discuss for simplicity only the case without explicit time dependence.)

• locally vs. globally defined.

• active vs. passive picture.

I) In mathematics, an active global symplectomorphism is a map $F:M\to M$ such that $F^{\ast}\omega=\omega$. Such map $F$ is manifestly independent of coordinate systems.

However, because of the Darboux theorem, we may (and will for simplicity) choose an atlas $$M~=~\bigcup_a U_{(a)}$$ of Darboux neighborhoods $$M~\supseteq~ U_{(a)}~\stackrel{\phi_{(a)}}{\longrightarrow} ~V_{(a)}~\subseteq~ \mathbb{R}^{2n}$$ from now on.

In Darboux coordinates $$\phi_{(a)}^I:~U_{(a)}~\to~\mathbb{R} ,\qquad I~\in~ \{1,\ldots, 2n\} ,$$ the Poisson bracket matrix $$\{\phi_{(a)}^I,\phi_{(a)}^J\}~=~J^{IJ}, \qquad I,J~\in~ \{1,\ldots, 2n\}.$$ is constant.

Let $$f_{(ba)}~=~\phi_{(b)}\circ F\circ \phi^{-1}_{(b)}$$ denote the cooresponding map from (a subset of) $V_{(a)}\subseteq \mathbb{R}^{2n}$ to (a subset of) $V_{(b)}\subseteq \mathbb{R}^{2n}$.

A symplectomorphism $F$ then satisfies

$$\sum_{K,L=1}^{2n} \frac{\partial f^I_{(ba)}(z_{(a)})}{\partial z_{(a)}^K} J^{KL} \frac{\partial f^J_{(ba)}(z_{(a)})}{\partial z_{(a)}^L}~=~J^{IJ}, \qquad I,J~\in~ \{1,\ldots, 2n\}.$$

II) In contrast in physics, a symplectomorphism is often formulated as a passive coordinate transformation $f$ from (a subset of) $\mathbb{R}^{2n}$ to (a subset of) $\mathbb{R}^{2n}$, such that

$$\sum_{K,L=1}^{2n} \frac{\partial f^I(z)}{\partial z^K} J^{KL} \frac{\partial f^J(z)}{\partial z^L}~=~J^{IJ}, \qquad I,J~\in~ \{1,\ldots, 2n\}.$$

This latter notion is coordinate dependent.

III) It seems that OP's examples are conflating the two cases I & II.

--

$^1$ Be aware that several non-equivalent definitions of CTs appear in the literature, cf. e.g. this Phys.SE post.

• Ok then, going with Arnold's definition, it follows that any (passive) coordinate transformation on a symplectic manifold is a canonical transformation? – luis Dec 12 '17 at 23:55
• $\uparrow$ No . – Qmechanic Dec 13 '17 at 0:06
• But any such coordinate transformation, being the identity on a symplectic manifold, is a symplectomorphism. Since symplectomorphism=canonical transformation it follows... sorry obviously I'm missing something. – luis Dec 13 '17 at 0:51
• I updated the answer. – Qmechanic Dec 13 '17 at 18:22