# Asking whether there is a compact Lie group containing affine symplectic group

The affine symplectic group is interesting and important in physics. However, the Lie group is noncompact. In order to have some good properties (Basically, we need some good behavior of Haar measure) that need in physics. So the question is that I hope to find another compact Lie group $$G$$ which affine symplectic group is a subgroup of $$G.$$ (Not just topologically, but the group operation needs to be preserved). The simple analog is that $$\mathbb{R}$$ is a noncompact Lie group but can fit into a 2-torus $$T^2.$$ One can see this: Can a compact Lie group have a non-compact Lie subgroup?

I have searched the literature, but I can not find many pieces of literature for affine symplectic group.

Any ideas and comments are very welcome.

• A small note: while there is a continuous injective group homomorphism $\mathbb{R}\to T^2$, it does not "really" embeds it as a topological group, in that the topology on $\mathbb{R}$ is not the one induced by the embedding. Indeed I don't think one can embed a non-compact group into a compact group as a topological subgroup. Nov 10, 2021 at 8:44
• It can't contain the symplectic group over a nonzero symplectic space. An easy reason is that in a compact Lie group, any connected abelian subgroup has the property that its centralizer has finite index in its normalizer.
– YCor
Nov 10, 2021 at 11:00
• @DenisNardin Isn't $\mathbb{Q}/\mathbb{Z}\to S^1$ an embedding of a non-compact group into a compact one? I do not see a locally compact example, though. Nov 10, 2021 at 12:04
• @AndreiSmolensky Oh yeah, sorry I was automatically making my groups locally compact (although note that usually $\mathbb{Q}/\mathbb{Z}$ is given the discrete topology, not the subspace one - exactly because we want it to be locally compact!) Nov 10, 2021 at 13:34

The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $$\mathrm{SL}(2,\mathbb{R})$$ as a Lie subgroup, and $$\mathrm{SL}(2,\mathbb{R})$$ cannot be a Lie subgroup of any compact Lie group.
To see this, note that any compact Lie group has a bïinvariant Riemannian metric, a property that is inherited by all of its Lie subgroups. However, $$\mathrm{SL}(2,\mathbb{R})$$ does not have a bïinvariant Riemannian metric because its adjoint representation on $${\frak{sl}}(2,\mathbb{R})\simeq\mathbb{R}^3$$ acts as $$\mathrm{SO}(1,2)\subset \mathrm{GL}(3,\mathbb{R})$$, which does not preserve any positive definite inner product on $$\mathbb{R}^3$$.