# What are the fixed-point free subgroups of $\mathrm U(2)$?

Let $$\mathrm U(2)\subset\Bbb C^{2\times 2}$$ be the group of unitary $$(2\times 2)$$-matrices. I wonder the following:

Question: What are the maximal fixed-point free subgroups of $$\mathrm U(2)$$?

A group $$\Gamma\subset\mathrm U(2)$$ is fixed-point free if there are no non-trivial $$T\in\Gamma$$ and non-zero $$v\in\Bbb C^2$$ with $$Tv=v$$. For example, $$\mathrm U(2)$$ itself is not fixed-point free, as

$$T:=\begin{pmatrix} -1 & 0 \\ \phantom+0 & 1 \end{pmatrix}\in \mathrm U(2)\setminus\{\mathrm{Id}\}$$

has the fixed point $$(0,1)\in\Bbb C^2$$.

I know that $$\mathrm{SU}(2)\subset\mathrm U(2)$$ is fixed point free (it acts like the multiplicative group of unit quaternions on $$\Bbb H\cong\Bbb C^2$$). It is also maximal, since it acts regularly on the unit sphere $$\Bbb S^1(\Bbb C)\subset\Bbb C^2$$. There are also the conjugates of $$\mathrm{SU}(2)$$. But are there any others?

Update

Thanks to the enlightening answer by YCor, I understand that the problem in this form seems to be intractable. By that answer I also realized that what I am more interested in are those maximal fixed-point free subgroups that have "interesting" finite subgroups, that is, other than, say, $$\{\mathrm{Id}\}$$ and $$\{\pm\mathrm{Id}\}$$. Or also, closed subgroups (in the topological sense).

YCor's answer already established that besides $$\mathrm{SU}(2)$$ we have the maximal torus $$\mathrm U(1)\times\mathrm U(1)$$ as another maximal subgroup. Both are closed and have "interesting" finite subgroups.

• Just to complement the edit, the maximal torus $U(1)^2$ is doesn't act freely on $\mathbf{C}^2-\{0\}$, although it has large subgroups with this property. Also, it is not exactly maximal: its normalizer is maximal: this is an overgroup of index 2 of the maximal torus, consisting of all diagonal or anti-diagonal matrices.
– YCor
Dec 7, 2019 at 9:55
• Also since you're interested in groups in regard to their finite subgroups, it's useful first describing finite subgroups of $U(2)$ acting freely on $\mathbf{C}^2-\{0\}$ (i.e., with no non-identity element having eigenvalue 1). This should be doable...
– YCor
Dec 7, 2019 at 10:01

I can answer the question in the text "are there any others", the answer is yes, and I expect many, although I don't claim to answer the question in the title "What are the fixed-point free..." since it's not a full description.

Restated in more standard terms, the question is to understand the class $$\mathcal{L}$$ of subgroups of $$\mathrm{U}(2)$$ acting freely on $$\mathbf{C}^2$$ and especially describe the set $$\mathcal{L}_\max$$ maximal elements in $$\mathcal{L}$$.

The OP already observed that $$\mathrm{SU}(2)\in\mathcal{L}_\max$$.

If $$X$$ is the closed subset of $$\mathrm{U}(2)$$ consisting of elements with $$1$$ as eigenvalue, observe that $$\mathcal{L}$$ consists of subgroups $$G$$ such that $$G\cap X=\{\mathrm{id}\}$$. Clearly every element of $$\mathcal{L}$$ is contained in a maximal one.

There are obvious cyclic subgroups in $$\mathcal{L}$$, namely the subgroup generated by any diagonal matrix $$(u,v)$$ with $$|u|,|v|=1$$ and $$u,v$$ not roots of unity. Choosing $$uv\neq 1$$, and embedding into a maximal subgroup, we obtain elements of $$\mathcal{L}_\max$$ that are not included in $$\mathrm{SU}(2)$$.

I also claim that any generic (in the topological or measure-theoretic meaning) pair freely generates a subgroup in $$\mathcal{L}$$. Indeed, for every nontrivial word $$w(x,y)$$, the $$Y_{w}$$ of pairs $$(g,h)$$ such that $$w(g,h)\in X$$ is Zariski-closed and is not the whole set of pairs (by evaluation to a free pair in $$\mathrm{SU}(2)$$). Hence the union over $$w$$ of $$Y_w$$ has measure zero, and is an $$F_\sigma$$ subset. For this we see that there are dense subgroups in $$\mathcal{L}_\max$$, and pushing a little it can be checked that it has dense subgroups of cardinal $$2^{\aleph_0}$$.

I guess there is no reasonable way to list elements of $$\mathcal{L}_\max$$. For instance, I expect that the cardinal of $$\mathcal{L}_\max$$ is the largest possible, namely $$2^{2^{\aleph_0}}$$; still I can check that this is the cardinal of $$\mathcal{L}$$. (But at the moment I don't even have a proof that $$\mathcal{L}_\max$$ modulo conjugation has cardinal $$\ge 2^{\aleph_0}$$.)

• Thank you for that answer. If I am not mistaken, then the first maximal subgroup you constructed is $\mathrm U(1)\times \mathrm U(1)\subset\mathrm U(2)$ (the maximal torus is $\mathrm U(2)$). Reading you answer (and understanding the problem with the quetion in that general form), I think I might be actually interested in maximal fixed-point free subgroups that have non-trivial finite subgroups. Dec 1, 2019 at 15:34
• No, there is no "first maximal subgroup" I constructed: I did a first way to obtain such subgroups, but they are not unique, and some of these subgroups in $\mathcal{L}_\max$ thus obtained are dense. Certainly the construction does not yield $U(1)^2$ because the latter is not in the class $\mathcal{L}$.
– YCor
Dec 1, 2019 at 22:19
• BTW A group "with no non-trivial finite subgroup" is known as a "torsion-free" group.
– YCor
Dec 1, 2019 at 22:20
• I know, you have not done it explicitly, but every subgroup generated by the diagonal matrices $(u,v)$ is contained in a maximal torus (I believe). At least this is what I meant. The hint to the term "torsion free" was quite useful. Thank you. Dec 2, 2019 at 22:52