6
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Dec 7, 2019 at 9:55
  • $\begingroup$ 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... $\endgroup$
    – YCor
    Dec 7, 2019 at 10:01

1 Answer 1

5
$\begingroup$

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 $\ge 2^{\aleph_0}$.)

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – M. Rumpy
    Dec 1, 2019 at 15:34
  • $\begingroup$ 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 $\mathcal{L}$. $\endgroup$
    – YCor
    Dec 1, 2019 at 22:19
  • 3
    $\begingroup$ BTW A group "with no non-trivial finite subgroup" is known as a "torsion-free" group. $\endgroup$
    – YCor
    Dec 1, 2019 at 22:20
  • $\begingroup$ 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. $\endgroup$
    – M. Rumpy
    Dec 2, 2019 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.