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In Proc. Amer. Math. Soc. 5 (1954), 753-768 (https://doi.org/10.1090/S0002-9939-1954-0087028-0), Taylor points out that there are three compact one-dimensional Lie groups with two components:

  1. The product of the circle group $\mathbb{T}$ with the alternating group $\mathbb{Z}_2$,
  2. The orthogonal group $O(2)$, and
  3. The multiplicative group of unit quaternions $a + b i + c j + d k$ where $(a^2 + b^2)(c^2 + d^2)=0$ which Taylor gives the label $G^3$ to distinguish it from the previous two.

What are the irreducible complex representations of this third group $G^3$? Is there a more well-known geometric characterization of this third group?

creillyucla
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    The third group seems to consist of the quaternions of the types $e^{it}$ and $e^{it} j$, $t\in[0,2\pi)$. It seems to be a quotient group of the semidirect product $\Bbb{T}\rtimes\langle j\rangle$, where $j$ acts on the circle group by complex conjugation. A quotient group is needed, because we need to identify $j^2$ with $-1\in\Bbb{T}$. Anyway, I think this point of view helps with the search for irreducible complex reps. – Jyrki Lahtonen Sep 03 '24 at 16:21
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    An ingredient is the quaternion product rule $jz=\overline{z}j$ for all $z\in\Bbb{C}$. – Jyrki Lahtonen Sep 03 '24 at 16:26
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    My first guess would be that there are several pairwise non-isomorphic 2-dimensional irreducible reps with the toral part acting diagonally as the appropriate power and its conjugate. Don't know if the action of $j$ allows further modifications? – Jyrki Lahtonen Sep 03 '24 at 16:35
  • @JyrkiLahtonen I follow most of your comments, but for example I do you understand your statement about the necessity of a quotient group and the need to identify $j^2$ with -1. Your guess about the group's irreps closely match that of $O(2) \cong C_{\infty v}$. The group $G^3$ is a double cover of $O(2)$. Is it possible they have the same irreps (i.e. identical character tables)? Also, does the fact that $G^3$ fails to split in sense given by Taylor imply that it is not a quotient group? – creillyucla Sep 03 '24 at 19:55
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    All I am trying to accomplish with that quotient group is that I started with the semidirect product $G:=\Bbb{T}\rtimes C_4$, the latter generated by $j$. Then the product of quaternions gives a surjective homomorphism from $G$ to $G^3$. I think the kernel of that mapping is the cyclic group of order two generated by $(-1,j^2)$. But I'm prepared to be very wrong about some details. – Jyrki Lahtonen Sep 03 '24 at 20:15
  • What does that double cover map look like, exactly? – Jyrki Lahtonen Sep 03 '24 at 20:16
  • In the parameterization $(x,\pm1)$ used by Taylor where $x \in [0,1)$ and $\pm1 \in \mathbb{Z}_2$, the double cover map is simply $(x,\pm1) \to (\phi(2x),\pm1)$ where $\phi: \mathbb{R} \to [0,1)$ is the natural mapping. – creillyucla Sep 03 '24 at 20:29
  • related: https://physics.stackexchange.com/questions/826619/projective-representations-and-reduction-of-half-integer-spin-representations-un – creillyucla Sep 04 '24 at 09:43

1 Answers1

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Your group "$G^3$" is equivalent to the pin group ${\rm Pin}(2)$. You can dig into that for maybe the closest to what you would call a "geometric" interpretation.

Edit: I am using the convention that $v^2=-\|v\|^2$ for vectors $v\in V\subset C\ell(V)$, where $V=\Bbb R^2$ has the usual positive-definite Euclidean norm. Here, ${\rm Pin}(2,0)$ is the subgroup of the Clifford algebra $C\ell(2,0)$ generated by unit vectors. If you use the negative-definite norm, you get ${\rm Pin}(0,2)$ as the corresponding subgroup of $C\ell(0,2)$, and here ${\rm Pin}(0,2)\cong{\rm O}(2)$. An isomorphism extends from sending $\exp(i e_1e_2)\mapsto \exp(i\theta)$ and $e_1\mapsto R$ where $R\in{\rm O}(2)\setminus{\rm SO}(2)$ is arbitrary.

Many sources use the opposite convention $v^2=+\|v\|^2$ to define $C\ell(V)$, which swaps the role of $C\ell(p,q)$ and $C\ell(q,p)$ associated to pseudo-Euclidean spaces $V=\Bbb R^{p,q}$, and in particular swaps the meaning of the pin groups ${\rm Pin}(p,q)$ and ${\rm Pin}(q,p)$. In any given source you may have to infer which convention they're using from context, unfortunately.

I believe the $(-)$ convention traces its lineage to quaternions since $C\ell(2,0)\cong\Bbb H$, but it turns out the $(+)$ convention may be more natural. First of all, it's more natural to view $\Bbb H$ as either of $C\ell_0(3,0)$ or $C\ell_0(0,3)$ than $C\ell(2,0)$, and second of all IIRC the interplay between the algebra and geometry is more intuitive on the $(+)$ way (hence the alternative development of Clifford algebras called "geometric algebra"). I just learned it the $(-)$ way first is all.

Define $e(\theta):=\exp(i\theta)$. The irreps of $\Bbb T$ are $V_n$ ($n\in\Bbb Z$) defined by the action $e(\theta)\cdot v=e(n\theta)v$. Notice, then, that if this $V_n$ is a $\Bbb T$-subirrep of a $G^3$ then $jV_n\cong V_{-n}$, so the irreps of $G^3$ are

$$ W_n \cong \begin{cases} V_0 & n=0 \\[4pt] V_n\oplus V_{-n} & n>0 \end{cases} $$

as $\Bbb T$-reps for $n\ge0$. For $n>0$ the element $j$ acts as $\bigl[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\bigr]$. For $n=0$ the element $j$ can act as either of $\pm1$, so there are two corresponding $G^3$-irreps (call them, say, $V_0^\pm$).

More generally, if $N\trianglelefteq G$ is a normal subgroup then $G$ permutes $N$-subreps of any $G$-rep.

coiso
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  • Excuse if I am mistaken, but doesn't your characterization omit the existence of two type-0 irreps, one symmetric and the other antisymmetry under $j$?

    Also, are you sure that G3 is isomorphic to Pin(2)? I ask because I had read that Pin(2) was isomorphic to O(2), and I believe that G3 and O(2) are not isomorphic as groups.

     – creillyucla Sep 18 '24 at 07:57
  • @creillyucla You're right about two $n=0$ irreps. Also, I mixed up $N$ and $G$ in my last sentence, now fixed. But which one of ${\rm Pin}(2,0)$ or ${\rm Pin}(0,2)$ is isomorphic to ${\rm O}(2)$ depends on sign conventions (see my edit). – coiso Sep 18 '24 at 17:13