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Is it known for which topological groups $X$ there exists a positive integer $d$ and a surjective, continuous group homomorphism $([0,1]\text{ mod } 1)^d \to X$?

Certainly, any such $X$ must be compact and path connected. Also, it's set of characters must be countable, but I can't image that these three conditions are sufficient. Is there some known sufficient condition? Are there any further, more or less obvious necessary conditions which are known?

P.S. the set $X^{\wedge}$ of algebraic characters for such a group $X$ is countable because there exists an injection $X^{\wedge} \to (([0,1]\text{ mod } 1)^d)^{\wedge} : \gamma\mapsto\gamma\circ f$ where $f : ([0,1]\text{ mod } 1)^d \to X$ is the mystical continuous, surjective homomorphism.

Thomas Winckelman
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1 Answers1

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Well, if $f:T^d\to X$ is a surjective continuous homomorphism, then it induces a topological isomorphism $T^d/\ker f\to X$ where $\ker f$ is a closed subgroup of $T^d$. In particular, $\ker f$ is a Lie subgroup and so the quotient $X$ is a Lie group. Being a compact abelian Lie group, it must be a torus itself.

Or alternatively, using Pontryagin duality, the Pontryagin dual of $X$ must be a subgroup of $\mathbb{Z}^d$, the Pontryagin dual of $T^d$. Every subgroup of a free abelian group is free, so the Pontryagin dual of $X$ is isomorphic to $\mathbb{Z}^e$ for some $e\leq d$ so $X\cong T^e$.

(This is assuming you require topological groups to be Hausdorff. If you don't, there are a lot of other nasty possibilities--for instance, $X$ could be any divisible abelian group of cardinality at most $2^{\aleph_0}$, with the indiscrete topology.)

Eric Wofsey
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  • This is a powerful answer indeed. Although you've referenced one or two results I'm not familiar with, this is helpful too, since it aids me in directing my studies. Thank you. – Thomas Winckelman Mar 12 '21 at 03:18