$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Sp{Sp}$Let $\omega=\sum dx_i\wedge dy_i$ be the standard symplectic structure of $\mathbb{R}^{2n}=\mathbb{R}^{n}\times \mathbb{R}^n$.

We consider the following two
extensions of $\Sp(2n,\mathbb{R})$, the group of linear isomorphisms of $\mathbb{R}^{2n}$ preserving $\omega$:

1) Let $G$ be the group of all $A\in \GL(2n,\mathbb{R})$ which map all isotropic subspaces to isotropic subspaces. (A closed subgroup of the general linear group containing the symplectic group $\mathrm{Sp}(2n,\mathbb{R})$.)

2) Let $H$ be the group of all elements in $\GL(2n,\mathbb{R})$ which map all symplectic subspaces to symplectic subspaces.

A non-linear construction as above can be introduced on a symplectic manifold $M$: $G(M)$ is the group of all diffeomorphisms of $M$ whose derivatives map isotropic subspaces to isotropic subspace. And similarly $H(M)$ is the group of all diffeomorphisms whose linear parts map symplectic subspaces to symplectic subspaces.

Is there a terminology for Lie groups $G$, $H$, $\bar{H}$ and their Lie algebras and also the corresponding structures on symplectic manifolds? Is there any relation between $G$ and $\bar H$? What can be said about their first fundamental groups, as it is customary to compute the first fundamental groups of classical Lie groups?

Inspired by the concept of symplectic vector fields as Lie algebra of symplectomorphism group, what can be said about the Lie algebra of $G(M)$ and some dynamical interpretations on a symplectic manifold?

  • $\begingroup$ Does "symplectic subspace" mean "subspace on which the symplectic form is non-degenerate"? (I originally asked what $\bar H$ was, but deleted it when I realised you meant the closure.) $\endgroup$
    – LSpice
    Commented May 23, 2020 at 21:28
  • $\begingroup$ Yes The symplectic subspace are meant as you mentioned.$H$ is not closed subset of general linear group so we consider its closure $\bar H$ $\endgroup$ Commented May 23, 2020 at 21:32
  • 1
    $\begingroup$ Is there an easy example to show that $H$ isn't closed? $\endgroup$
    – LSpice
    Commented May 23, 2020 at 21:33
  • 1
    $\begingroup$ +1 for this comment Note that the limit of a sequence of symplectic subspaces is not necessarilly symplectic subspace (In grassmanian topoloy) but the limit of isotropic subspace is again an isotropic subspace. For example The 2 dimensional space $(x,y,\epsilon y, \epsilon x)$ is a symplectic subspace only when $\epsilon \neq 0$ $\endgroup$ Commented May 23, 2020 at 21:40
  • $\begingroup$ @LSpice I think that "diffeomorphisms whose linear part/ derivative" was correct and better than with plural. Googling "functions whose image" yields thrice more occurrences than "functions whose images". $\endgroup$
    – YCor
    Commented May 23, 2020 at 23:39

1 Answer 1


Short answer: $G = H$ is the group of conformal symplectic linear maps. What follows is a proof of this (which I've simplified slightly from what I originally wrote):

1) $G$ is the group of conformal symplectic linear maps

First, some notation. We write $\{e_i\} \cup \{f_j\}$ for the standard basis for $\mathbb{R}^{2n}$, i.e. with $\omega(e_i,e_j) = 0$, $\omega(f_i,f_j) = 0$, and $\omega(e_i,f_j) = \delta_{i,j}$.

Suppose $\phi \in G$. We begin by singling out the isotropic subspaces $A = \mathbb{R}^n \times \{0\}$ and $B = \{0\} \times \mathbb{R}^n$. The following lemma is one way you might think about the group $\mathrm{Sp}(2n,\mathbb{R})$ in the first place: as being defined by its action on an isotropic subspace and a choice of isotropic complement.

Lemma: Given $\phi$, there is a unique element $\psi \in \mathrm{Sp}(2n,\mathbb{R})$ such that the following three properties hold:

  • $A = (\psi \circ \phi)(A)$, with $(\psi \circ \phi){\big|}_{A} = \mathrm{id} \colon A \rightarrow A$
  • $B = (\psi \circ \phi)(B)$

Given this lemma, we have that $\psi \circ \phi = \begin{pmatrix}\mathrm{id} & 0 \\ 0 & T\end{pmatrix}$, and it suffices to find conditions on $T \in \mathrm{GL}(n,\mathbb{R})$ so that isotropic subspaces are preserved by $\psi \circ \phi$. It suffices to consider just two cases:

  • For $i \neq j$, the span of $e_i$ and $f_j$ is isotropic, so its image under $(\psi \circ \phi)$, the span of $e_i$ and $Tf_j$, must also be isotropic. This proves that $T$ is diagonal.
  • Also for $i \neq j$, we can use the subspace spanned by $e_i + f_j$ and $e_j + f_i$, which will give us the condition that the diagonal entries of $T$ are all equal. Hence $T = c \cdot \mathrm{id}$ for some constant $c \neq 0$.

Finally, we see that with $T = c \cdot \mathrm{id}$, $(\psi \circ \phi)^*\omega = c \cdot \omega$, which clearly preserves isotropic subspaces. So this property completely characterizes $G$.

In terms of nomenclature, $G$ might be called the group of conformally symplectic linear operators, and the nonlinear theory would fall under the umbrella of (locally) conformal symplectic geometry, which is slightly more general than what you ask (i.e. $\omega$ itself does not need to be symplectic, only (locally) conformally symplectic, which is the more natural setting). I know only a little about the theory, and am far from an expert on the history, so I'll say nothing more that there exists literature on the subject.

As for computations of classical invariants, we see that $G \cong \mathbb{R}^* \times \mathrm{Sp}(2n,\mathbb{R})$, so there's nothing interesting to say about computing its standard invariants that can't already be said for $\mathrm{Sp}(2n,\mathbb{R})$.

2) $G=H$

Suppose $\omega(v,w) \neq 0$ and $\omega(v,x) = 0$. Then $\omega(v,w+tx) \neq 0$ for all $t \in \mathbb{R}$, so $v$ and $w+tx$ always span a symplectic subspace. Hence, if $\phi \in H$, since $\phi$ preserves symplectic subspaces, we require $$0 \neq \omega(\phi(v),\phi(w+tx)) = \omega(\phi(v),\phi(w)) + t\omega(\phi(v),\phi(x))$$ for all $t \in \mathbb{R}$. Hence, we must have $\omega(\phi(v),\phi(x)) = 0$ whenever $\omega(v,x) = 0$. In particular, $\omega$ preserves isotropic subspaces, so $H \subseteq G$. Meanwhile, every conformal symplectic linear map certainly preserves symplectic subspaces, so $G \subseteq H$. Hence $G=H$.

Remark: Even though the limit of symplectic subspaces may not be symplectic, as in the comments, this is not enough to show $H$ is closed, since if a sequence of elements of $H$ realizes this collapsing of a symplectic subspace onto a non-symplectic one, it can (and must by what I've just written) limit to a singular matrix.

  • $\begingroup$ Thank you very much for your attention to my question abd your very interesting answer $\endgroup$ Commented May 26, 2020 at 9:54

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