Let $\Bbb S^0(\Bbb H)\cong \mathrm{SU}(2)$ denote the multiplicateive group of unit quaternions. This group is non-abelian.

Of course, the subgroups generated by as $\def\<{\langle}\def\>{\rangle}\<1,i\>$, $\<1,j\>$ and $\<1,k\>$ are commutative and also isomorphic to $\Bbb S^0(\Bbb C)\cong\mathrm U(1)$ (the multiplicative group of unit complex numbers). And there are probably other such subgroups conjugate to the ones named above. And of course, any subgroup of such a conjugate is also abelian.

Question: Are there other abelian subgroups of $\Bbb S^0(\Bbb H)$?

  • $\begingroup$ Any cyclic subgroup is obviously abelian. $\endgroup$ – freakish Nov 13 at 9:00
  • $\begingroup$ @freakish I suppose that any cyclic subgroup is a subgroup of a conjugate of $\langle 1,i\rangle$. Or not? $\endgroup$ – M. Winter Nov 13 at 9:01
  • $\begingroup$ That would imply that entire $\mathbb{S}^0(\mathbb{H})$ is a union of conjugates of $\langle 1,i\rangle$. Hmm, I'm not sure, although I do doubt it. $\endgroup$ – freakish Nov 13 at 9:04
  • $\begingroup$ @freakish I would not be surprised. This sounds like the Hopf fibration that covers $\Bbb S^3$ by disjoint $\Bbb S^1$. $\endgroup$ – M. Winter Nov 13 at 9:07

Slup's answer tells us that all abelian subgroups are contained in maximal abelian subgroups which in turn are products of $n$ circles. The number $n$ is the rank of the group, hence also the number of nodes in the dynkin diagram which is $1$ for $SU(2)$ (or more generally $n = N-1$ for $SU(N)$). It remains to see why we have $n= 1$ in this case. I will type a very 'quaternionic' (as opposed to $SU(N)$-ic) answer to this. (The answer will also point out in which circle group the given abelian group lies and why.)

Let $u$ be a unit quaternion that is completely imaginary, i.e. $u \in \mathbb{R}i \oplus \mathbb{R}j \oplus \mathbb{R}k$. The subspace $\mathbb{C}_u := \mathbb{R} \oplus \mathbb{R}u$ is a subalgebra of $\mathbb{H}$ isomorphic to $\mathbb{C}$ via the isomorphism $u \mapsto i$. We even have that there is an exponential map $\exp: \mathbb{R}u \to \mathbb{C}_u$ that maps the imaginary axis in $\mathbb{C}_u$ to the unit circle in $\mathbb{C}_u$ in the standard way: $\exp$ is defined by the same Taylor series as it always is and convergence works just as always since $\mathbb{H}$ is, from a topological perspective, just $\mathbb{R}^4$.

Now this unit circle in $\mathbb{C}_u$, which I will denote $S^1_u$ is one of the maximal abelian subgroups that we were talking about. Concretely I claim:

Claim: let $A$ be any abelian subgroup of $\mathbb{S}(\mathbb{H})$ containing an element $\alpha 1 + \beta u$ for some $\alpha, \beta \in \mathbb{R}$ then $A \subset S^1_u$.

The proof is very simple: pick arbitrary $x$ not in $S^1_u$ and show that it does not commute with $\alpha 1 + \beta u$.

To simplify this: since we are dealing with unit quaternions $x \not\in S^1_u$ is the same as saying $x \not\in \mathbb{C}_u$. We know that $x$ is of the form $\gamma 1 + \delta u + \epsilon v$ with $\gamma, \delta, \epsilon$ in $\mathbb{R}$ and $v$ a unit quaternion that is both purely imaginary AND not in $S^1_u$. To show that $\alpha 1 + \beta u$ does not commute with $x$ it suffices to show that $u$ does not commute with $v$ since everyting inside $\mathbb{C}_u$ does commute with eachother.

Now we just need to remember that for purely imaginary quaternions $u$ and $v$ we have that $\Re(uv) = \Re(vu), \Im(uv) = - \Im(vu)$ and that $\Im(uv) = 0$ if and only if $u$ and $v$ are real scalar multiples of each other. (Here I use notation $\Re(a + bi + cj + dk) = a$ and $\Im(a + bi + cj + dk) = bi + cj + dk$ for the real and imaginary part of a quaternion respectively.)

These facts are so useful in any computation with quaternions that I expect you have seen them before.

  • $\begingroup$ It is not true that for "purely imaginary quaternions $u$ and $v$" that their imaginary parts anti-commute in general. Rather, this is true iff $u$ and $v$ are perpendicular, when viewed as elements of $\mathbb{R}^3$. Consider, e.g. $u = i$ and $v = i + j$. $\endgroup$ – Jason DeVito Nov 29 at 14:36
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    $\begingroup$ In this example $uv = -1 + k$ so that $\Im(uv) = k$ and $vu = -1 - k$ so that $\Im(vu) = -k$. Thus $\Im(vu) = -\Im(uv)$ as I wrote, what is the problem? $\endgroup$ – Vincent Nov 30 at 10:31
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    $\begingroup$ You are of course right that we do not have $uv = -vu$ unless $\Re(uv) = 0$, the latter condition being equivalent to being orthogonal in $\mathbb{R}^3$. But I never claimed that $uv = -vu$ for purely imaginary quaternions in general! $\endgroup$ – Vincent Nov 30 at 10:32
  • $\begingroup$ Oh, I see that I misinterpreted what you wrote! Thanks for the clarification. I'd be happy to delete my original comment, since it no longer serves a purpose. $\endgroup$ – Jason DeVito Nov 30 at 12:51

The following answers the question only partially.

There are some general facts of Lie group theory that can be applied here. Let $G$ be a connected and compact Lie group. A maximal torus of $G$ is a maximal abelian, connected subgroup of $G$.

Theorem. Let $G$ be a connected and compact Lie group and $T$ be a maximal torus of $G$. Then $$G= \bigcup_{g\in G}gTg^{-1}$$

Now if $G$ is a connected and compact Lie group, then a maximal torus $T$ is isomorphic to $(S^1)^{\times n} = \underbrace{S^1\times ...\times S^1}_{n\,times}$. Note that there exists a topological generator of $(S^1)^{\times n}$ i.e. there exists $t\in (S^1)^{\times n}$ such that $\{t^n\,|\,n\in \mathbb{Z}\}$ is dense in $(S^1)^{\times n}$.

We prove now the following.

Fact. Let $G$ be a connected and compact Lie Group and $A$ be an abelian subgroup of $G$ contained in containing a maximal torus $T$. Then $A=T$.

Proof. Closure of an abelian subgroup in $G$ is an abelian subgroup, so we may assume that $A$ is closed. Then $A$ is compact and $T\subseteq A$ is the connected component of the identity. Hence $T$ is open in $A$. These imply that $A/T$ is a finite abelian group. Next $T$ is a divisible abelian group. Thus $A = T\times B$, where $B$ is some finite group isomorphic with $A/T$. Suppose that there exists nontrivial $b\in B$ of order $m>0$. Let $t$ be a topological generator of $T$. Since $T$ is divisible group, there exists $t'\in T$ such that $t'^m = t$. Then $a = (t',b)\in A$ is a topological generator of $\langle T,b\rangle$. By Theorem there exists a maximal torus $T'$ of $G$ which contains $a$. Then $\langle T,b\rangle\subseteq T'$. Hence $T\subsetneq T'$ and this is contradiction with maximality of $T$. Therefore, $B$ is a trivial group and $A = T$.

Note that $S^3 = \mathbb{S}^0(\mathbb{H})$ is a compact, connected Lie group. All maximal tori are conjugate and cover $S^3$ by virtue of Theorem (it can also be viewed as an application of the existence of Hopf fibration as stated in M.Winter's comment). By Fact there are no abelian subgroups strictly containing a maximal torus of $S^3$. On the other hand $S^1$ is a maximal torus of $S^3$. This is precisely one of the copies of $U(1)$ indicated in the question.

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    $\begingroup$ It is absolutely true that all abelian subgroups of $SU(2)$ are subgroups of circle groups, and in that sense the question is a full answer and not just partial. However something is wrong in the proof. Obviously the 'fact' as stated is wrong: the finite groups from the original question are abelian but not circles. $\endgroup$ – Vincent Nov 13 at 10:26
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    $\begingroup$ Thanks for the edit! Now I see what you mean $\endgroup$ – Vincent Nov 13 at 10:40
  • $\begingroup$ I am sorry. The statement of the fact was incorrect. $\endgroup$ – Slup Nov 13 at 10:40
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    $\begingroup$ There is one last minor issue: 'By Fact there are no abelian subgroups containing a maximal torus of $S^3$' in the end of the post must be 'By Fact there are no abelian subgroups strictly containing a maximal torus of $S^3$. $\endgroup$ – Vincent Nov 13 at 10:41
  • $\begingroup$ @Vincent Thank you for this fix. $\endgroup$ – Slup Nov 13 at 10:42

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