What are the group extensions of $\mathbb{Z}_2$ by $U(1)$?

Question

Given the group extension $1\rightarrow U(1)\overset{f}{\rightarrow}G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$, what are the possibilities for $G$?

I understand how to solve this for $1\rightarrow \mathbb{Z}_{2n}\overset{f}{\rightarrow}G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$, with $\mathbb{Z}_{2n}=\langle \bar{z}\mid\bar{z}^{2n}=1\rangle$ and $\mathbb{Z}_2=\langle \bar{a}\mid\bar{a}^2=1\rangle$. There exist $a,z$ such that $f(\bar{z})=z,g(a)=\bar{a}$. Then $\langle z\rangle$ is a normal subgroup of $G$ and $a,z$ generate $G$. There exist $0\leq i,j<2n$ such that $a^2=z^i$ and $za=az^j$. We have $z^{1+i}=za^2=az^ja=a^2z^{j^2}=z^{i+j^2}$ thus $j^2\equiv 1 \mod 2n$ and $j\equiv\pm 1\mod 2n$. This yields two cases, commutative $za=az$ and noncommutative $za=az^{–1}$.

In the commutative case, if $a^2=z^{2m+1}$, then consider the element $a^{-1}z^{m+1}$ and its square $(a^{-1}z^{m+1})^2=a^{-2}z^{2m+2}=z^{-2m-1+2m+2}=z$. Since $z$ has order $2n$, we see that $a^{-1}z^{m+1}$ has order $4n$ and $G\cong\mathbb{Z}_4$.

In the commutative case, if $a^2=z^{2m}$, then consider the element $a^{-1}z^m$ and its square $(a^{-1}z^m)^2=a^{-2}z^{2m}=z^{-2m+2m}=1$. The elements $a^{-1}z^m$ and $z$ generate $G\cong\mathbb{Z}_2\oplus\mathbb{Z}_{2n}$.

In the noncommutative case, repeatedly applying $za=az^{-1}$ gives $z^ia=az^{-i}$. Note that $a$ and $a^2=z^i$ commute, thus $az^i=az^{-i}$ and $2i\equiv 0 \mod 2n$. We have $i=0$ or $i=n$, which is to say, $a^2=1$ (the dihedral group) or $a^2=z^n$, $a^4=1$ (the dicyclic group).

I wish to understand what happens when we have $U(1)$ instead of $\mathbb{Z}_{2n}$

Answer 1

In any case, $G/U(1)\cong\Bbb Z_2.$

If the sequence is right but not left split, then $G\cong U(1)\rtimes_\varphi \Bbb Z_2.$

Since $\rm Aut(U(1))\cong\Bbb Z_2,$ we are just looking at embeddings $\varphi: \Bbb Z_2 \to \Bbb Z_2.$ So the identity. So there's only one non-trivial example.

If it's left split, we have $G\cong U(1)×\Bbb Z_2.$

Answer 2

I worked out the answer I was looking for. Thank you, @non-euclidean_geometry for your answer, which got me on the right track and encouraged me to look at the automorphisms of $U(1)$.

Motivation

My purpose is to understand Gregory Moore's Quantum Symmetries and K-Theory, notably his Section 2.6 Realizing the 10 classes using the CT groups. The 10 CT groups are the extensions of $H$ by $U(1)$, where $H$ is one of the five subgroups of $\mathbb{Z}_2\times\mathbb{Z}_2$. I need to understand them well (intuitively, concretely) so that I can understand further how their $\mathbb{Z}_2\times \mathbb{Z}_2$-graded representations (which impose `structure: even vs. odd, linear vs. antilinear $2\times 2$ matrices) manifest 8-fold and 2-fold Bott periodicity, and how that relates to $\mathbb{Z}_2$-graded representations (even vs. odd) of real and complex Clifford algebras. So first I want to understand the group extensions of $\mathbb{Z}_2$ by $U(1)$.

My personal interest is to see if I can model human experience (unconscious, conscious, consciousness) with Bott periodicity, as I discuss here. Quantum symmetries, notably CPT symmetries (charge conjugation, parity, time reversal) are metaphysical. For example, charge conjugation lets us consider, is a particle (of matter) the same as a hole (of anti-matter) or not? In other words, is there a difference between a particle (what is) and a hole (what is not)?

What I am learning from reading Moore is that this boils down to the ways of integrating two groups: a circle $U(1)$ and a pair of switches $\mathbb{Z}_2\times\mathbb{Z}_2$. This brings to mind architect Christopher Alexander's patterns, whereby recurring activity evokes structure, and structure channels activity. The circle models recurring activity and the switches model structure. There are not many ways that they can fit together. Those are the contexts for patterns that relate activity and structure.

I note further that the short exact sequence can be interpreted as relating four levels of knowledge: $1\overset{whether}{\rightarrow}U(1)\overset{what}{\rightarrow}G\overset{how}{\rightarrow}\mathbb{Z}_2\times\mathbb{Z}_2\overset{why}{\rightarrow}1$. The short exact sequence carves up the mind in terms of knowledge: whether knows nothing $1$, what knows something $U(1)$, how knows anything $G/U(1)$, why knows everything $\mathbb{Z}_2\times\mathbb{Z}_2/G\cong 1$. $G$ is being carved up as what $U(1)$ and how $\mathbb{Z}_2\times\mathbb{Z}_2$. Mentally, the switch $how\rightarrow what$ precedes and controls the switch $why\rightarrow whether$. Another way to say this is that Consciousness (why) is the brake that keeps the Conscious (how) from imposing its model prematurely on the Unconscious (what).

Notation

Given the short exact sequence $1\rightarrow U(1)\overset{f}{\rightarrow} G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$.

The injectivity of $f$ means that $N=f(U(1))\cong U(1)$. $N$ is a normal subgroup of $G$. $G/N\cong\mathbb{Z}_2$.

Define $\mathbb{Z}_2=\{1,\bar{a}\}$. The surjectivity of $g$ implies there exists an $a\in G$ such that $g(a)=\bar{a}, g(a^2)=\bar{a}^2=1, a^2\in N$. Thus there exists a rotation $\theta\in N$ such that $a^2=\theta$. Let $H=\{1,a\}$

Note that $a\notin \ker g=N$. We have $G=N\sqcup aN$ and $G=NH$.

Automorphism $\phi_a$

The map $\phi_a(z)=aza^{-1}$ defined for all $z\in N\cong U(1)$ is an automorphism.

Here I need to assume that $f$ is continuous. Otherwise, as described here, there are uncountably many highly discontinuous homomorphisms, and I suppose that includes many automorphisms.

If we assume $f$ is continuous, then there are only two automorphisms of $U(1)$: $z\rightarrow z$ and $z\rightarrow z^{-1}$. This fact is proven here. Informally, we note that the continuous homomorphisms from a circle into a circle have to wind around the circle, forwards or backwards, an integer number of times. Of these homomorphisms, only two are invertible, namely, the identity homomorphism but also the inverse homomorphism.

This means that, for all $z\in N$, either $\phi_a(z)=aza^{-1}=a$ or $\phi_a(z)=aza^{-1}=a^{-1}$

One commutative case

Suppose for all $z\in N, aza^{-1}=a, az=za$. We know that $N$ and $H$ are commutative and that the elements of $N$ commute with those of $H$ and vice versa. Thus $G$ is a commutative group.

We know that $a^2=\theta\in N$. There exists $\theta^{\frac{1}{2}}\in N$ such that $(\theta^{\frac{1}{2}})^2=\theta$. Consider the element $b=a \theta^{-\frac{1}{2}}\notin N$. We have that $b^2=(a \theta^{-\frac{1}{2}})^2=a^2\theta^{-1}=1$. Define $B=\{1,b\}$.

$G=N\sqcup Na = N\sqcup Nb=NB$. For any two elements $g_1,g_2\in G$, there is a unique decomposition $g_1=z_1b_1, g_2=z_2b_2, g_1g_2=z_1z_2b_1b_2$ where $z_1,z_2\in Z, b_1,b_2\in B$. So $G\cong N\times B \cong U(1)\times\mathbb{Z}_2$.

Two noncommutative cases

For all $z\in N$, $aza^{-1}=z^{-1}$ and $az=z^{-1}a$. We know that $a^2=\theta\in N$. Thus $a\theta=\theta^{-1}a, a^3=a^{-1},a^4=1$. This means that $\theta^2=1$, so $\theta=1$ or $\theta=-1$, which is to say, $a^2=1$ or $a^2=-1$ (the rotation halfway round the circle).

These are the pin groups $\textrm{Pin}_+(2)$ (when $\theta=1$) and $\textrm{Pin}_-(2)$ (when $\theta=-1$).

The relations $az=z^{-1}a$ and $a^2=\pm 1$ allow us to determine how the elements of the cosets $N$ and $Na$ satisfy the equations $NN=N, NaN=Na, aNN=Na,NaNa=N$. Given $z_1,z_2\in N$, we have $z_1z_2\in N, z_1az_2=z_1z_2^{-1}a\in Na, az_1z_2=z_1^{-1}z_2^{-1}a\in Na, z_1az_2a=z_1z_2^{-1}a^2\in N$.

Visualizing

I visualize this by imagining a round pancake in the plane which I can rotate $z$ or flip $a$. What happens if I flip it, then rotate it a bit, then flip it back?

• Rotating and flipping could be independent: $za=az$. (But that's not physical!)
• Rotating and flipping could match our experience in the physical world: $aza=z^{-1}$. Flipping across the $y$-axis, rotating by angle $z$, then flipping again across the $y$-axis, yields a rotation by $z^{-1}$ Flipping twice $a^2=1$ gets us back to where.
• Flipping twice $(a^2=-1)$ could rotate the pancake by $\pi$. Flipping across the $y$-axis, rotating by angle $z$, then flipping again across the $y$-axis, yields a rotation by $z^{-1}$ plus a rotation by $\pi$. Thus rotating half way around (rotating by $\pi$) would get us back where we started (but with the opposite orientation) much like with the spin of an electron.

Pin groups in terms of Clifford algebras

I find it instructive to understand the definition of the Pin groups in terms of the Clifford algebra $Cl(\mathbb{R}^2,Q)$ where $Q$ is the quadratic form such that either $e_1^2=e_2^2=1$ (yielding $\textrm{Pin}_+(2)$) or $e_1^2=e_2^2=-1$ (yielding $\textrm{Pin}_-(2)$) where $e_1, e_2$ are the basis elements of $\mathbb{R}^2$ and the generators of the Clifford algebra.

Consider the vectors $v=\lambda_1e_1+\lambda_2e_2\in R^2$ such that $Q(v)=\pm 1$. $Q(v)=\lambda_1^2e_1^2+\lambda_2^2e_2^2$. Thus if $e_1^2=e_2^2=1$ then we must have $\lambda_1^2+\lambda_2^2=1$ and $v=\cos\theta e_1+\sin\theta e_2$ for some $\theta\in [0,2\pi)$. If $e_1^2=e_2^2=-1$ then we must have $-\lambda_1^2-\lambda_2^2=-1$ and likewise $v=\cos\theta e_1+\sin\theta e_2$ for some $\theta\in [0,2\pi)$.

The elements of the pin group $\textrm{Pin}_+(2)$ are the products of such vectors $v_1,v_2,\cdots v_k$ for $k\geq 0$ when $e_1^2=e_2^2=1$. Similarly, when $e_1^2=e_2^2=-1$, the products define the pin group $\textrm{Pin}_-(2)$. These groups are subgroups of the Clifford algebra and they are likewise $\mathbb{Z}_2$ graded, having odd elements of the form $a_1 e_1 + a_2 e_2$ and even elements of the form $a_0 + a_{12}e_1e_2$ where $a_0,a_1,a_2,a_{12}\in\mathbb{R}$. Indeed, more specifically, the odd elements have the form $\cos\theta \; e_1+\sin\theta\; e_2$ and the even elements have the form $\cos\theta + \sin\theta\; e_1e_2$ as we can see by calculating as below (and noting that $e_1e_2=-e_2e_1$):

For $\textrm{Pin}_+(2)$:

odd $\times$ odd: $(\cos\theta_1\; e_1+\sin\theta_1\; e_2)(\cos\theta_2\; e_1+\sin\theta_2\; e_2)=\cos (\theta_2-\theta_1) + \sin(\theta_2-\theta_1)\;e_1e_2$

even $\times$ odd = odd: $(\cos\theta_3 + \sin\theta_3\;e_1e_2)(\cos\theta_4\;e_1 + \sin\theta_4\;e_2)=\cos(\theta_3-\theta_4)e_1+\sin(\theta_3-\theta_4)e_2$

odd $\times$ even = odd: $(\cos\theta_4\;e_1 + \sin\theta_4\;e_2)(\cos\theta_3 +\sin\theta_3\;e_1e_2) =\cos(\theta_3+\theta_4)e_1+\sin(\theta_3+\theta_4)e_2$

even $\times$ even: $(\cos\theta_1 + \sin\theta_1\;e_1e_2)(\cos\theta_2 + \sin\theta_2\;e_1e_2)=\cos(\theta_1+\theta_2)+\sin(\theta_1+\theta_2)e_1e_2$

For $\textrm{Pin}_-(2)$:

odd $\times$ odd: $(\cos\theta_1\; e_1+\sin\theta_1\; e_2)(\cos\theta_2\; e_1+\sin\theta_2\; e_2)=\cos (\pi - (\theta_2-\theta_1)) + \sin(\pi -(\theta_2-\theta_1))\;e_1e_2$

even $\times$ odd = odd: $(\cos\theta_3 + \sin\theta_3\;e_1e_2)(\cos\theta_4\;e_1 + \sin\theta_4\;e_2)=\cos(\theta_3-\theta_4)e_1+\sin(\theta_3-\theta_4)e_2$

odd $\times$ even = odd: $(\cos\theta_4\;e_1 + \sin\theta_4\;e_2)(\cos\theta_3 +\sin\theta_3\;e_1e_2) =\cos(\theta_4-\theta_3)e_1+\sin(\theta_4-\theta_3)e_2$

even $\times$ even: $(\cos\theta_1 + \sin\theta_1\;e_1e_2)(\cos\theta_2 + \sin\theta_2\;e_1e_2)=\cos(\theta_1+\theta_2)+\sin(\theta_1+\theta_2)e_1e_2$

We see that in both cases the even elements form a subgroup, which is the circle group, and in fact, the spin group $\textrm{Spin}(2)$.

We see that the odd elements likewise map to the circle but they don't form a group because they don't include the identity $1$ and they aren't closed.

Thus each pin group is a double cover of the special orthogonal group $SO(2)\cong U(1)$. (When $n>2$, then $\textrm{Pin}_+(n)$ is a double cover of $O(n)$, and likewise for $\textrm{Pin}_-(n)$, but here we are dealing with $n=2$).

We see that when $e_1^2=1$, the generator $e_1=e_1\cos 0 + e_2\sin 0$ functions as the "flip" $a$, sending odd elements to even elements and vice versa. For then $e_1 (e_1\cos\theta + e_2\sin\theta) = \cos\theta + e_1e_2\sin\theta)$ and $e_1(\cos\theta + e_1e_2\sin\theta)=e_1\cos\theta + e_2\sin\theta$. And when $e_2^2=-1$, we can multiply on the right by $e_2=e_1\cos\frac{\pi}{2}+e_2\sin\frac{\pi}{2}$, yielding $(e_1\cos\theta + e_2\sin\theta)e_2 = -\sin\theta + e_1e_2\cos\theta) = e_1\cos(\theta +\frac{\pi}{2})+e_1e_2\sin(\theta +\frac{\pi}{2})$ and $(\cos\theta + e_1e_2\sin\theta)e_2=-e_1\sin\theta + e_2\cos\theta=e_1\cos(\theta+\frac{\pi}{2})+e_2\sin(\theta+\frac{\pi}{2})$.

Indeed, we have two cases: $a^2=e_1^2=1$ and $a^2=e_1^2=-1$.

Conclusion

This is the intuition that I have been seeking. Thank you, @non-euclidean_geometry, for setting me on the right path.

Now I better understand the Pin groups and see concretely how the Clifford algebra is relevant here.

I am also gleaning that the case $a^2=1$ models a direct relationship between the unconscious $a$ and the world $1$. Whereas the case $a^2=-1$ models the situation where the conscious $a$ and $-a$ is interposing between the unconscious $-1$ and the world $1$.

I wonder why Gregory Moore doesn't seem to consider the commutative case but I think I will figure that out as I learn more.