Let $G$ be a compact real Lie group. We say that $G$ has property $(*)$ if every abstract automorphism of $G$ is continuous. A theorem of Cartan says that if $G$ has perfect Lie algebra, it has property $(*)$. It the converse true?

I think that $G$ having property $(*)$ is equivalent to its identity component having property $(*)$ (because multiplication is continuous, and any connected component is some fixed element times the identity component). Therefore, we may assume that $G$ is connected. Therefore, $G$ is the product of its commutator subgroup $[G, G]$ and its center $Z(G)$ (and the intersection $I=[G, G]\cap Z(G)$ is finite). If $G$'s Lie algebra is not perfect, then the commutator subgroup has positive codimension so its center has positive dimension. Now a discontinuous automorphism of the center can be extended to a discontinuous automorphism of the whole group iff it fixes $I$.

So the question reduces to:

  • is my reasoning in the second paragraph correct?
  • does every compact abelian Lie group of positive dimension have a discontinuous automorphism fixing a given finite set? The center does not have to be connected.
  • $\begingroup$ As regards your last question, yes: in a compact abelian Lie group, any countable subset is contained in a countable direct summand (for the abstract group structure). $\endgroup$
    – YCor
    Mar 9, 2019 at 12:30
  • $\begingroup$ Your reduction to the connected case is not correct. The connected case is easy (find a discontinuous automorphism of $Z(G)^\circ$ that is identity on $Z(G)^\circ\cap [G,G]$. In general, one needs to extend such an automorphism and this requires some argument and I don't think you're giving one (you need to construct an automorphism). $\endgroup$
    – YCor
    Mar 9, 2019 at 13:45

1 Answer 1


Proposition : let $G$ be a compact Lie group (with unit component $G^\circ$) whose Lie algebra is not perfect; let $G_\delta$ be the underlying discrete group. Then there is a semidirect decomposition $$G_\delta=L\ltimes V,$$ where $[G^\circ,G^\circ]\subset L$, the subgroup $L$ is dense in $G$, and $V\subset Z(G^\circ)$ is an abelian normal subgroup, isomorphic to a vector space over $\mathbf{Q}$, with $Z(G^\circ)/V$ countable.

Remark: if $G$ is connected, the above is a direct product decomposition: $G=L\ltimes V$.

A positive answer to your question follows from the proposition:

Corollary: $G$ has discontinuous automorphisms. More precisely, $|\mathrm{Aut}(G^\delta)|=2^{2^{\aleph_0}}$ (while $|\mathrm{Aut}(G)|\le 2^{\aleph_0}$ as topological group, with equality iff $G^\circ$ is non-abelian).

Proof of the corollary: we have $\dim_{\mathbf{Q}}(V)=2^{\aleph_0}$; view $V$ as $L$-module: this action factors through the finite group $G/G^\circ$. Hence $V$ is a direct sum of $2^{\aleph_0}$ irreducible submodules, which are finite-dimensional, and there are finitely many isomorphism types. Hence (after choosing a decomposition and some isomorphism type occurring with multiplicity $2^{\aleph_0}$), any permutation of this set yields an automorphism of the $L$-module $V$. Since any automorphism of the $L$-module $V$ extends to an automorphism of $G^\delta=L\ltimes V$ by putting identity on $L$, we deduce $\mathrm{Sym}(2^{\aleph_0})$ embeds as a subgroup of $\mathrm{Aut}(G^\delta)$. $\Box$

Proof of the proposition: start with $H=G/[G^\circ,G^\circ]$; this is a compact Lie group with $H^\circ$ abelian and nontrivial. There exists a countable subgroup $D$ of $H^\circ$ such that $DH^\circ=H$ (just lift elements of $H/H^\circ$ and take the subgroup it generates). Then $D\cap H^\circ$ is an $H$-submodule of $H^\circ$; note that the $H$-module structure factors through $H/H^\circ$ and hence is action of a finite group. Let $D'$ be a dense, countable divisible submodule of $H^\circ$ (take $D'$ to contain the torsion subgroup, and then use that in the quotient which is a rational vector space, every countable submodule is contained in a countable direct summand). Then $D'$ is normal in $H$, and define $E=DD'$, and let $L$ be the inverse image of $E$ in $G$; it is dense. Then $LZ(G^\circ)=G$, and since by construction $L\cap Z(G^\circ)$ is divisible and is a (countable) $G$-submodule of $Z(G^\circ)$ containing the torsion subgroup. Let $V$ be a direct summand of $L\cap Z(G^\circ)$ in $Z(G^\circ)$, as $G$-module. Then $G=L\ltimes V$ with the required properties.


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