# A converse of Cartan's automatic continuity theorem

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.
• 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).
– YCor
Mar 9, 2019 at 12:30
• 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).
– YCor
Mar 9, 2019 at 13:45

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: $$G=L\ltimes V$$.
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.