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Suppose I have a topological space $(G,\tau)$ and am interested in whether there exists a topological group $(G,*,\tau)$. In other words, can we assign a binary operation $*$ to $(G,\tau)$ which satisfies the group axioms, and under which the functions $x \mapsto x^{-1}$ and $(x,y) \mapsto x*y$ are continuous? If so, how many non equivalent ways can we do this? And what would a sensible form of equivalence be? For example we can give $\mathbb R$ the standard additive group structure, but we could equally inter it "backwards" in the group and define $x*y = \begin{cases} x \cdot y & \mbox{if } x,y \le 0\\ -(x \cdot y) & \mbox{if } x \le 0 \mbox{ and } y \ge 0 \\ -(x \cdot y) & \mbox{if } x \ge 0 \mbox{ and } y \le 0\\ x \cdot y, & \mbox{if } x, y \ge 0 \end{cases}$

Here negation is used in the context of $\mathbb R$. It does not represent the inverse with respect to $*$.

Clearly $\tau$ has to satisfy a lot of conditions for this to even stand a chance at being possible. In a topological group the map $x \mapsto a*x$ is a homeomorphism for each $a \in G$ and a result of this is the topology "looks the same" at any given point. If this is not true then there definitely is no $(G,*,\tau)$. But I'm sure this condition isn't enough, so are there any properties of the topology that can be reasonable computed which necessitate a compatible structure to exist?

There is a similar question, which I know has a lot of theory behind it: Can we determine whether a topological group $(G,*,\tau)$ admits a Lie group structure by only looking at $(G,*,\tau)$? There is yet another question I found on StackExchange which is asking the opposite of me: How many agreable topologies can we assign an abstract group which make the group operations continuous? I would guess that problem is easier than mine. But I am not interested in either of these at the moment.

Daron
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  • The extreme case: If you have a discrete topological space, any group structure you define on that set will make it into a topological group. I like this question though. +1 – J. Loreaux Sep 19 '13 at 23:06
  • Each topological group can be given the structure of a uniform space in a couple (conceptually identical) natural ways, so a non-uniform space cannot be a topological group. – dfeuer Sep 19 '13 at 23:46
  • Also, any space that has the fixed point property also can't be a topological group, since then for every $g \not = 1$ there would be some $h \in G$ such that $gh = h$. So that rules out a lot of spaces as well - for instance, any compact and convex subspace of Euclidean space. Also, here is a paper that shows (without too much machinery) that no sphere of even dimension can be a topological group: http://www.unizar.es/acz/05Publicaciones/Revistas/Revista62/p075.pdf – Elle Najt Nov 04 '13 at 06:53
  • (To summarize: The author uses degree theory to show that the only nontrivial group which can act freely on $S^{2n}$ is $Z_2$ - the degree of the action of g and its inverse must be 1 which means that each group element acts with degree plus or minus 1 (as degree is a homomorphism onto Z). The identity is 1, and the since the action of G is free every other element has degree -1. This means that the kernel is trivial and the map is onto, so G is Z_2. Crucial is that continuous maps of an even dimensional sphere to itself with no fixed points have degree -1 - I'm not sure why that is true.) – Elle Najt Nov 04 '13 at 07:09
  • I am sorry but what you defined does not seem to be a group: $20=0=30$? Is "$\cdot$" supposed to be "$+$"? If it is then still $-41 = 3 = 21$? – Nex Nov 03 '15 at 10:44
  • From algebraic topology: the fundamental group of a topological group is abelian. This rules out lots of spaces, for example all wedge sums of (non-simply connected) topological spaces, e.g. the $\infty$ shape. – freakish Jun 30 '17 at 13:02

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