General literature does not seem to offer a characterisation of topological groups among all topological spaces. Of course, being completely regular (uniform) is necessary, but separation properties, or indeed any sort of "niceness" like pseudo-metrisability are not sufficient, since topological groups, for instance, cannot have fixed point property.
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6$\begingroup$ By "uniform" I think you mean that the space can be given a uniform structure, which is necessary as you say. Metrizability or any kind of separation axiom is not necessary; the indiscrete topology on any group gives a topological group. $\endgroup$– Tom GoodwillieCommented Feb 23, 2013 at 12:13
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2$\begingroup$ Much easier is: which spaces are homeomorphic to a subset of a topological group. $\endgroup$– Gerald EdgarCommented Feb 23, 2013 at 14:39
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1$\begingroup$ By the separation axiom complete regularity, I mean it in the weaker sense of Kelly and Willard - without the T0 axiom. It is this property which is equivalent to uniformisability (in the class of all topological spaces). This, I hope, every topological group (which is not assumed T0, I emphasise) must satisfy, as does the indiscrete topology. But then I regret having written metrisability and not pseudo-metrisability in my question. $\endgroup$– N UnnikrishnanCommented Feb 23, 2013 at 18:30
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2$\begingroup$ @N Unnikrishnan: As for subsets, every $T_{3\frac12}$ space is homeomorphic to a subspace of $\mathbb R^\kappa$ (or $(S^1)^\kappa$ if you prefer a compact group) for some $\kappa$, no homogeneity is needed. I believe that likewise every completely regular space can be embedded into an appropriate abelian topological group whose Kolmogorov quotient is, say, $\mathbb R^\kappa$. $\endgroup$– Emil JeřábekCommented Feb 23, 2013 at 18:42
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4$\begingroup$ @Unnikrishnan: I think you're using the "accept" checkmark all wrong. You should reserve it till you have an answer you're really happy with, or until a long time goes by and it seems you're not getting any new answers. Don't just shift it to whatever the newest answer is $\endgroup$– David WhiteCommented Feb 24, 2013 at 4:19
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5 Answers
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There is a homological criterion that is often helpful, to rule out the possibility for a topological space to admit a continuous group structure (even H-space structure):
The rational cohomology ring of a connected topological group (or H-space) $G$ is a connected graded-commutative Hopf algebra over $\mathbb{Q}$ and if $H^i(G;\mathbb{Q})$ is finite dimensional for all $i \ge 0$, then, by a theorem of Borel, $H^\ast(G;\mathbb{Q})$ is the tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators.
For example $H^\ast(\mathbb{C}P^n;\mathbb{Q})=\mathbb{Q}[X]/(X^n),\; \deg x=2$ isn't of this form. Hence $\mathbb{C}P^n$ is no topological group.
For the theorem (and variations thereof) and further examples see Hatcher: Algebraic Topology, Section 3.C.
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10$\begingroup$ Along the same lines, the fundamental group must be abelian. Note that these are restrictions on the (weak) homotopy type of the space. $\endgroup$ Commented Feb 23, 2013 at 12:17
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2$\begingroup$ Is this condition sufficient? Are there "fake groups" or "fake H-spaces" with the right structure on cohomology but no actual group structure? $\endgroup$– mkreiselCommented Feb 23, 2013 at 14:51
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5$\begingroup$ @mkreisel: $S^n$ for $n\neq 1,3,7$ odd are counter-examples. For, $H^\ast(S^n)=\mathbb{Q}[X]/(X^n),$ $\deg x=n$ is exterior, but only $S^1,S^3,S^7$ admit a H-space structure (Hatcher, 3.C). $\endgroup$– RalphCommented Feb 23, 2013 at 15:48
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Every topological group is homogenous - this rules out spaces like $[0,1]$, $[0,\omega_1]$ or $\beta \mathbb{N}$.
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$\begingroup$ Among necessary conditions, I should have added homogeneity as well, as Cleft pointed out. But, is every topological group strongly homogeneous, i.e., can any two points be mapped onto each other by a self-homeomorphism? Or is it ever so? $\endgroup$ Commented Feb 23, 2013 at 18:15
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This is really a long comment on the answer by Cleft accepted by the OP.
Note that there are homogeneous spaces with the fixed point property (e.g. the Hilbert cube) and there are first countable non-metrizable homogeneous spaces (e.g. Alexandroff's double arrow space). Under additional axioms, there are also homogeneous spaces (even compact ones) which are hereditarily separable but not hereditarily Lindelof and the other way around: some that are hereditarily Lindelof but not hereditarily separable. And the list goes on.
So homogeneity is far from characterizing topological groups (even in the class of compact spaces). On the other hand, a (kind of vague) question due to Kunen (I think) is: can one say something interesting about compact right topological groups that cannot be said about compact homogeneous spaces? (besides things like "do not have the fixed point property" or "admits a group operation which is continuous in one variable").
answered Feb 23, 2013 at 18:30
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1$\begingroup$ I did not in the least mean that cleft's answer gave a characterisation which we sought, when I accepted it. But the existence of homogeneous spaces with fixed point property was a really great reminder, in the least. But pardon me, how are first countability, hereditary separability and hereditary Lindeloffness related to being topological groups? Also, I shall be thankful if you tell me where to look for Kunen's question. $\endgroup$ Commented Feb 23, 2013 at 18:52
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$\begingroup$ A first countable topological group is metrizable. A compact group is hereditarily Lindelof iff it is hereditarily separable iff it is metrizable. As for the question, I remember reading about it in some paper by Jan van Mill, but can't remember which; anyway there wasn't anything else to read about it, it was just a comment towards the end of the paper (if I remember correctly). $\endgroup$ Commented Feb 23, 2013 at 23:08
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@N Unnikrishnan: Let $G$ be a topological group, and $a\ b\in G$. Consider $h:G\rightarrow G$ defined by: $$\forall_{x\in G}\quad h(x) := a\cdot x^{-1}\cdot b$$
Then $h$ is a homeomorphism such that $h(a)=b$ and $h(b)=a$. This shows that (in your terminology above) every topological group is strongly homogeneous.
In general, the above homeomorphism $h_{a\ b} := h$ is not an involution. Indeed, in general, it is not its own inverse (with respect to composition)--actually, $h_{b\ a}$ is the inverse of $h_{a\ b}$ (it is an involution in the Abelian case though since then $h_{a\ b}=h_{b\ a}$).
answered Feb 24, 2013 at 7:46
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$\begingroup$ Dear Prof Holsztynski W, when I was student I studied your articles about shape theory (a categorical study) . By the way I remeber that for compact (T2 I seem) connected group the (weak) shape funcor is a equivalence. ams.org/journals/tran/1974-194-00/S0002-9947-1974-0345064-8 $\endgroup$ Commented Feb 24, 2013 at 11:00
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$\begingroup$ Hi Buschi, thank you for your kind words. I am rusty, but I think I can sketch a complete argument to support your statement about the equivalence. It would then work for all compact groups (not just for connected). K.Borsuk and I only started along this line. However, the general categorical approach gives a good foundation for this question. It was later James Keesling who seriously studied compact groups in the context of shape theory. (Also, another q. on MO--about Cech cohomology--is strongly (and unknowingly?) related to shape theory; I'll try to find it, to post an "answer"). $\endgroup$ Commented Feb 24, 2013 at 20:05
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A necessary condition for a Hausdorff compact space to admit the structure of a topological group is the Suslin condition (I hope I am using proper terminology)
every family of pair-wise disjoint open sets is countable.
This is so because Hausdorff compact topological groups admit Haar measure.
answered Apr 29, 2013 at 16:41