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Let $X$ be a non-empty set. It is known that we can give a group structure on $X$. Now let $X$ be a non-empty topological space. Then can we give a group structure on $X$ so that it becomes a topological group w.r.t. its original topology ?

Asaf Karagila
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    No, of course you need hypothesis. For instance, a topological group is T1 if and only if one of its points is closed. In which case, it is also a Tychonov space. Moreover, there is Birkhoff-Kakutani theorem: a T1 first countable topological space is metrizable (plus stuff). –  Mar 25 '18 at 07:12
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    Some deep stuff related to this is https://en.wikipedia.org/wiki/Hilbert%27s_fifth_problem It basically says that if your topological group is locally Euclidean, it is already a Lie group. – orangeskid Mar 25 '18 at 08:56
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    Something obvious: the translatoions should be homeomorphisms, so the group of homeomorphisms is transitive. And something deeper: t $\pi_1(X)$ is abelian for a topological group. – orangeskid Mar 25 '18 at 08:57

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No, you cannot do that for all spaces $X$.

If $X$ has the structure of a topological group, it implies a lot of extra facts about it, and those give necessary conditions that $X$ should fulfill.

Some examples of such properties:

  1. If $X$ is $T_0$ it must also be $T_{3\frac{1}{2}}$ (Tychonoff). (it's uniformisable)

  2. $X$ is homogenous: for every $x, y \in X$ there is a homeomorphism $h:X \to X$ such that $h(x) = y$.

  3. $X$ does not have the fixed point property (any non-unit multiplication shows this)

  4. If $X$ is compact it is dyadic and thus ccc.

  5. If $X$ is first countable and $T_0$ it is metrisable. (Birkhoff metrisation theorem).

So e.g. $X= [0,1]^n$ cannot be made into a topological group, because of both 2 and 3. The Sorgenfrey line fails 5. The infinite cofinite topology fails 1.

So many spaces cannot have a structure of a topological group.

@orangeskid mentioned an algebraic topology reason of possible failure: $\pi_1(X)$ is Abelian when $X$ is a topological group. This makes the wedge sum of circles $S^1 \vee S^1$ another example, I believe.

Henno Brandsma
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  • Thanks ... yes I realized the two obstacles : fixed point property and abelian fundamental group ... and thanks to you and all others for poonting out other obstacles. Do you know of any work in this direction of giving topological conditions on a space so that it can be made into a topological group ? –  Mar 25 '18 at 09:44
  • @misao I don’t think there is a general theorem for when $X$ can be made into a topological group. Maybe special cases are known like for manifolds and Lie groups. It’s probably very hard. – Henno Brandsma Mar 25 '18 at 09:47
  • I think $[0, 1]^\omega$ is a good counter-example as well. It's homogeneous (!) but it has fixed point property, so is not a topological group. – Jakobian Apr 26 '22 at 13:46