# Is there a nonabelian topological group operation on the reals?

Inspired by A binary operation, closed over the reals, that is associative, but not commutative. That question asks for a noncommutative semigroup operation on $\Bbb R$, for which right projection is a continuous solution. If you ask for inverses as well, that is, a nonabelian group operation, then you can use bijection tricks from any other nonabelian group of cardinality $\frak c$, for example the group $M_2(\Bbb R)$ of $2\times 2$ real matrices. But this will not usually give a continuous group operation on $\Bbb R$ because the bijection is usually exotic.

To "upper bound" the properties needed, according to @PseudoNeo, if we require that the group operation be not just continuous but $C_1$, then it is necessarily of the form $x\ast y=\phi^{−1}(\phi(x)+\phi(y))$ for some $C_1$ diffeomorphism $\phi$, which is manifestly abelian. (Anyone have a reference for this result?)

My question lies between these extremes:

Is there a nonabelian topological group operation on the reals? That is, a group operation $\ast$ such that $\ast:\Bbb R\times\Bbb R\to\Bbb R$ is continuous and so is ${}^{-1}:\Bbb R\to\Bbb R$.

I believe L.E.J. Brouwer showed (in his thesis "Over de Grondslagen der Wiskunde", in Dutch (1907), in a relatively unknown chapter not published elsewhere, AFAIK) that Hilbert's problem on "continuous groups" (i.e. topological groups) being Lie groups was true for the one-dimensional case, the real line. It is believed that Brouwer added this to add some "gravitas" to the thesis (which also contained some more controversial stuff on the foundations of maths).

This answer provides a argument in the same vein. It is pretty elementary, but leaves some details to check.

There's a wonderful link to an elementary disproof in the comments. Here's a completely absurd approach.

1) Gleason-Montgomery-Zippin: every topological group whose underlying space is a manifold is actually a Lie group (with respect to some smooth structure on the topological space). Herein lies the complete absurdity - this is an incredibly hard theorem with more than a decade of work going into it.

2) Invoke the Lie group / Lie algebra correspondence - simply connected Lie groups are in canonical bijection with Lie algebras of the same dimension. There is only one 1-dimensional Lie algebra, hence only one Lie group structure on $\Bbb R$.

Now that one has this, one has the ability to prove a variety of very strong theorems about group structures on compact manifolds (invoking the classification of compact Lie groups); one thing that's particularly nice is that, for instance, $T^n$ admits one and only one group structure, and I think of someone was more careful than I want to be right now they could probably get similar results for $SO(n)$, $U(n)$, etc.

Note that this is already false in dimension 2: $\Bbb R^2$ can be given the group structure of $\text{Aff}(1)$, the affine transformations of the line, which is not abelian.

• Am I correct in thinking that the net result of this is the statement that a topological group on $\Bbb R$ is the image of $+$ under some homeomorphism $\phi$? It's not clear to me whether the "smooth structure" implies that $\phi$ is smooth, or if it's just the transition maps that are smooth. – Mario Carneiro Jun 20 '16 at 16:48
• @MarioCarneiro $\phi$ needn't be smooth in the standard smooth structure (in which $f(x) = x$ is smooth), no. What you call the net result is correct. – user98602 Jun 20 '16 at 16:52
• Fun fact: Both $SO(4)$ and $U(n)$ for any $n\geq 2$ admit multiple smooth Lie group structures: $SO(4)$ is diffeomorphic to $SO(3)\times SU(2)$ and $U(n)$ is diffeo to $S^1\times SU(n)$! (I don't know for a fact, but would be willing to guess that these "exotic" Lie structures are the only exotic ones on these manifolds.) (Also, I believe that for $SU(n), Sp(n)$ and $SO(odd)$ the standard Lie structures are the only ones, but I'm not sure about $SO(even)$. Specifically, $Spin(4n)$ has center $Z = \mathbb{Z}_2\oplus \mathbb{Z}_2$ and quotients by different $\mathbb{Z}_2\subseteq Z$ .... – Jason DeVito Jun 20 '16 at 17:34
• @JasonDeVito: Oh, of course I knew these but they completely slipped my mind! I looked at their cohomology rings, said "eh, looks indecomposable enough", and moved on a little too quickly... – user98602 Jun 20 '16 at 17:36
• ....give rise to different Lie groups (at least sometimes), but I don't know about the underlying manifold type. – Jason DeVito Jun 20 '16 at 17:37