I am wondering if it is possible to have a nontrivial homomorphism from a connected topological group, such as a connected Lie group, to a discrete group. Here, we do not impose any restriction on the homomorphism, such as continuity. It seems that the answer is no. However, I am not sure how to prove that. Perhaps, there might be some counterexample. Thanks in advance!


Certainly! For instance, let $G$ be any topological group, and let $H$ be $G$ with the discrete topology. Then the identity map is a homomorphism from $G$ to $H$. Or more generally, take your favorite homomorphism from $G$ to any other group, and then put the discrete topology on the target group.

The point is that discreteness is a property of topological groups, not a property of groups. If a topological group is discrete, this tells you absolutely nothing about its group structure (and therefore tells you absolutely nothing about what topological groups can have discontinuous homomorphisms to it).

One thing you can say, though, is that there are no nontrivial homomorphisms from connected Lie groups to finite groups. To prove this, note that if $G$ is a Lie group, then there is a neighborhood $U$ of the identity such that every element of $U$ has an $n$th root for all $n$ (this follows from the fact that any element sufficiently close to the identity has the form $\exp(X)$ for some $X$ in the Lie algebra of $G$, and then $\exp(X/n)$ is an $n$th root). If $f:G\to H$ is a homomorphism and $H$ is finite, it follows that $f$ is trivial on $U$, since if $n=|H|$ the only $n$th power in $H$ is the identity. If $G$ is connected, it follows that $f$ is trivial, since any neighborhood of the identity generates the whole group.


Let $G$ be a connected Lie group; let think of it as a group which turns out to possess a connected Lie group structure. As Eric pointed out, there is always $G$ with the discrete topology.

However one has (1) every countable quotient of $G$ is solvable. (2) the only finitely generated quotient of $G$ is $\{1\}$.

Actually (1) can be improved (see (1') below), but at least there are infinite countable abelian quotients (e.g., $\mathbf{R}$ admits $\mathbf{Q}$ as a group quotient).

Proof of (1) First if $G$ is simple (in the Lie sense), its only nontrivial normal subgroups are contained in the countable center, and hence its only countable quotient is $\{1\}$. Hence if $G$ is semisimple, its only countable quotient is $\{1\}$ as well. In general, one has $G=RS$ with $S$ semisimple, $R$ solvable. In any countable quotient, $S$ has a trivial image, and hence it's a quotient of $R$, thus solvable (and is a quotient of $R$).

Proof of (2): using (1) this would be solvable, and any nontrivial finitely generated solvable group has a nontrivial finite quotient, and finite quotients were discarded (Eric gave the argument).

(1') actually every countable quotient of $G$ is nilpotent. Sketch of the argument: we can suppose that $G$ has a minimal dimension for the property of having a non-nilpotent finite quotient. By the proof of (1), $G$ is solvable; we can suppose that it's simply connected. Being non-nilpotent (and with a little work), it contains a Lie subgroup of the form $H=V\rtimes\mathbf{R}$, with $V$ of dimension $\le 2$ and $\mathbf{R}$ acting irreducibly and non-trivially, or of the form $W\rtimes\mathbf{R}$ with $W$ a 3-dimensional Heisenberg group and the action on $W/[W,W]$ being by rotations. Let $G/N$ be a countable quotient. Then (excluding the last case) $N\cap V$ is a subgroup of countable index, $\mathbf{R}$-invariant and it's easy to deduce that $N\cap V=V$. Hence the normal subgroup generated by $V$ in $G$ is contained in $N$, and this contains a normal closed Lie subgroup, contradicting the minimality. In the case of $W\rtimes\mathbf{R}$, one deduces by the same argument that the projection of $W\cap N$ in $W/[W,W]$ is onto, and hence that $W\cap N$ contains $[W,W]$, and then conclude in the same way.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.