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CT groups

I want to describe the equivalence of the category of {$(φ, χ)$}-representations of the {$CT$} group and the category of the graded representations of the corresponding Clifford algebra.

Definitions

A {$\mathbb{Z}_2$}-graded group is a pair {$(G,\phi)$} where {$G$} is a group and {$\phi :G\rightarrow\mathbb{Z}_2$} is a homomorphism.

• Consequently, a {$\phi$}-twisted extension of {$G$} is an extension of the form {$1\rightarrow U(1)\rightarrow G^{\textrm{tw}}\overset{\pi}{\rightarrow} G\rightarrow 1$} where {$G^\textrm{tw}$} is a group such that {$\tilde{g}$} is linear {$\tilde{g}z=z\tilde{g}$} if {$\phi(g)=1$} and {$\tilde{g}$} is anti-linear {$\tilde{g}z=\bar{z}\tilde{g}$} if {$\phi(g)=-1$}, where {$\tilde{g}\in G^{\textrm{tw}}$} is any lift of {$g\in G$} and {$z\in G^{\textrm{tw}}$} is any phase {$|z|=1$}.
• "the group of operators representing the {$G$}-symmetries of a quantum system form an extension, {$G^\textrm{tw}$}, of {$G$} by {$U(1)$}
• the pullback construction defines {$G^\textrm{tw}:=\{(S,g)|\pi(S)=\rho(g)\}\subset\textrm{Aut}_\mathbb{R}(\mathcal{H})\times G$} where {$\pi :\textrm{Aut}_\mathbb{R}(\mathcal{H})\rightarrow\textrm{Aut}_\textrm{qtm}(\mathbb{P}\mathcal{H})$} and {$\rho :G\rightarrow \textrm{Aut}_\textrm{qtm}(\mathbb{P}\mathcal{H})$}
  • {$\textrm{Aut}_\textrm{qtm}(\mathbb{P}\mathcal{H})$} is the group of quantum automorphisms of a system with Hilbert space {$\mathcal{H}$}.
    • A quantum automorphism is a pair of bijective maps with real linear {$s_1:\mathcal{O}\rightarrow\mathcal{O}$} mapping observables (self-adjoint operators) and {$s_2:\mathcal{S}\rightarrow\mathcal{S}$} mapping states such that probability measures are preserved {$P_{s_1(O),s_2(\rho)}=P_{O,\rho}$}.
    • The group can be identified with the group of (suitably continuous) maps {$s:\mathbb{P}\mathcal{H}\rightarrow\mathbb{P}\mathcal{H}$} such that the overlap function (also known as the transition probability function) is invariant under {$s$}, namely {$\textrm{Tr}P_{s(l_1)}P_{s(l_2)}=\textrm{Tr}P_{l_1}P_{l_2}$}
  • {$\textrm{Aut}_\mathbb{R}(\mathcal{H})$} is the group whose elements are unitary and anti-unitary transformations on the Hilbert space {$\mathcal{H}$}

A bigraded group {$G$} has a homomorphism {$G\rightarrow\mathbb{Z}_2\times\mathbb{Z}_2$}, which is to say, a pair of homomorphisms {$(\phi,\chi)$} from {$G$} to {$\mathbb{Z}_2$}.

• A {$(\phi,\chi)$}-representation of {$G$} is a complex {$\mathbb{Z}_2$}-graded vector space {$V=V_0\oplus V_1$} and a homomorphism {$\rho :G\rightarrow\textrm{End}(V_\mathbb{R})$} such that

{$\rho(g)=\left\{\begin{matrix} \mathbb{C}-linear & \phi(g)=+1\\ \mathbb{C}-anti-linear & \phi(g)=-1 \end{matrix}\right.$}

{$and$}

{$\rho(g)=\left\{\begin{matrix} \textrm{even} & \chi(g)=+1\\ \textrm{odd} & \chi(g)=-1 \end{matrix}\right.$}

Define the group {$M_{2,2}=\left< \bar{T},\bar{C}\; |\; \bar{T}^2=\bar{C}^2=\bar{T}\bar{C}\bar{T}\bar{C}=1\right>\cong\mathbb{Z}_2\times\mathbb{Z}_2$}

{$M_{2,2}$} has 5 subgroups: {$M_{2,2}, \left<\bar{T}\right>, \left<\bar{C}\right>, \left<\bar{T}\bar{C}\right>, \{\}$}

The {$CT$} groups are the {$\phi$}-twisted extensions of {$M_{2,2}$} and its subgroups. They are bigraded groups.

Group Extensions

Wikipedia: Metacyclic group A metacyclic group is an extension of a cyclic group by a cyclic group. A metacyclic group is metabelian and supersolvable.

• Metabelian group A metabelian group is a group whose commutator subgroup is abelian. The commutator subgroup (also, the derived subgroup) is the group generated by the commutators, the elements of the form {$[g,h]=g^{-1}h^{-1}gh$}.
• Supersolvable group A supersolvable group is a group that has an invariant normal series where all factors are cyclic groups.

Math Stack Exchange. Extensions of Finite Cyclic Groups.

• In the group extension {$1\rightarrow <i>\rightarrow Q_8 \rightarrow Q_8/<i>\rightarrow 1$}, the quaternion group {$Q_8$} is not a semi-direct product. Note here that {$Q_8/<i>$} has the cosets {$\{1,j,-1,-j\}\cong\mathbb{Z}_4$} and {$\{i,-k,-i,k\}\cong\mathbb{Z}_4i$}.
• https://math.stackexchange.com/questions/3460265/extension-of-a-group-g-by-a-commutative-group-f
• https://math.stackexchange.com/questions/1429542/loop-group-of-u1
• https://math.stackexchange.com/questions/3379017/general-formula-for-group-extensions-by-an-abelian-group
• https://math.stackexchange.com/questions/2601555/extensions-of-finite-groups-by-compact-lie-groups

Group Extension Example: Extending {$\mathbb{Z}_2$} by {$\mathbb{Z}_2$}

Consider the group extension {$1\rightarrow\mathbb{Z}_2\rightarrow X\rightarrow\mathbb{Z}_2\rightarrow 1$}.

This says that {$f:\mathbb{Z}_2\rightarrow X$} is injective, that {$g:X\rightarrow\mathbb{Z}_2$} is surjective, and that {$\textrm{Img}(f)=\textrm{Ker}(g)$}.

Write {$\mathbb{Z}_2=\{1,-1\}$}, and similarly, write {$1, -1$} as the elements of the subgroup {$\mathbb{Z}_2$} embedded in {$X$}. We know that {$g(1)=1, g(-1)=1$}.

Suppose {$x\in X,x\neq 1,-1$}. We know that either {$g(x)=1$} or {$g(x)=-1$} and so {$1=g(x)^2=g(x^2)$}. This means that {$x^2\in\textrm{Ker}(g)=\textrm{Im}(f)=\mathbb{Z}_2$}, which is embedded in {$X$}. We have two possibilities: {$x^2=1$} and {$x^2=-1$}.

We know that {$(-1)x\neq 1$} (otherwise {$x=-1$}) and {$(-1)x\neq -1$} (otherwise {$x=1$}). Likewise {$x(-1)\neq 1,-1$}. Thus we know that {$((-1)x)^2=(x(-1))^2=1$}. Consequently, {$1=(-1)x(-1)x=x(-1)x(-1)$} and so {$[x(-1)](-1)x(-1)x=(-1)x(-1)x[x(-1)]$}, thus {$x(-1)=(-1)x$}. This means that we can write {$-x=(-1)x=x(-1)$} and the group {$\{1,-1,x,-x\}$} is commutative.

We know that {$\mathbb{Z}_2=X/\mathbb{Z}_2$} and this means that {$|X|=4$}, thus {$X=\{1,-1,x,-x\}$}.

There are two possibilities for {$X$}.

• If {$x^2=–1$}, then {$x$} generates {$X=\{1,x,x^2=-1,x^3=-x\}=\mathbb{Z}_4$}.
• If {$x^2=1$}, then we can set {$X=\{1=(1,1),-1=(-1,1),x=(1,-1),-x=(-1,-1)\}$}, which is {$\mathbb{Z}_2\times\mathbb{Z}_2$}.

Group Extension Example: Extending {$\mathbb{Z}_2$} by {$\mathbb{Z}_4$}

• Math Stack Exchange. Group extension of {$\mathbb{Z_4}$} by {$\mathbb{Z_2}$}.

Group extension example: Extending {$\mathbb{Z}_n$} by {$\mathbb{Z}$}

Math Stack Exchange. Group extensions of cyclic groups.

Given a short exact sequence

{$1\rightarrow N\overset{f}{\rightarrow}G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$}

what are the possibilities for group $G$? In particular, I want to show that when $N=\mathbb{Z}_{2n}$, the only possibilities are the dihedral group and the dicyclic group, and when $N=U(1)$, the only possibilities are the pin groups $\textrm{Pin}_+(2)$ and $\textrm{Pin}_-(2)$.

Group Extension Example: Extending {$\mathbb{Z}_2$} by {$\mathbb{Z}_{2n}$}

Consider the group extension {$1\rightarrow\mathbb{Z}_{2n}\overset{f}{\rightarrow} G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$}

These are cyclic groups {$\mathbb{Z}_{2n}=<\bar{z}\mid\bar{z}^{2n}=1>$} and {$\mathbb{Z}_2=<\bar{a}\mid\bar{a}^2=1>$}.

The surjectivity of {$g$} implies there exists an {$a\in G$} such that {$g(a)=\bar{a}, g(a^2)=\bar{a}^2=1, a^2\in f(N)$}. Thus there exists {$0\leq i<2n$} such that {$a^2=z^i$}.

Note that if {$x\in G$}, then {$g(x)=\bar{a}^k$} for some integer {$k$}, thus {$g(x)=g(a^k)$} and {$g(xa^{-k})=1$}, {$xa^{-k}\in N$}, {$x\in Na^k$}. Since {$z$} generates {$N$}, consequently {$z,a$} generate {$G$}.

{$N$} is a normal subgroup of {$G$}, thus there exists a {$z'\in f(N)$} such that {$za=az'$}, where {$z'\neq 1$}, {$z'=z^j$} for some {$0<j<2n$}.

We have {$za=az^j$} and we can calculate {$za^2=az^ja=a^2z^{j^2}$} and, in general, {$za^k=a^kz^{j^k}$}.

When {$k=2$}, {$a^2=z^i$} commutes with {$z$} so {$z^{1+2i}=z^{2i+j^2}$}. Thus {$j^2\equiv 1 \mod 2n$}.

We can solve this writing {$j^2-1\equiv 0 \mod 2n, (j-1)(j+1)\equiv 0\mod 2n$}, thus {$j=\pm 1 \mod 2n$}.

Combinatorially, the number {$j=-1=2n-1 \mod 2n$}, if added {$2n-1$} times, shifts around the clock one tick smaller, starting from {$0$}, yielding {$0 - (2n-1) = 1 \mod 2n$}. Algebraically, {$(2n-1)(2n-1)=(4n^2-4n+1)=1\mod 2n$}.

We have that the generators {$a$}, {$z$} are such that {$z^{2n}=1$} and {$za=az$} or {$za=az^{-1}$}.

If {$za=az$}, then the two generators commute, so we have an abelian group, indeed, a finite abelian group of order {$4n$} with a cyclic subgroup of order {$2n$}. This group must be either {$\mathbb{Z}_{4n}$} or {$\mathbb{Z}_{2n}\times\mathbb{Z}_2$}.

If {$a^2=z^{2m+1}$}, then consider the element {$a^{-1}z^{m+1}$} and note that, since {$a$} and {$z$} commute, we have {$(a^{-1}z^{m+1})^2=a^{-2}z^{2m+2}=z^{-2m-1+2m+2}=z$}. Recall that {$z$} has order {$2n$}, so {$a^{-1}z^{m+1}$} has order {$4n$} and generates {$\mathbb{Z}_4$}.

If {$a^2=z^{2m}$}, then consider the element {$a^{-1}z^m$} and note that {$(a^{-1}z^m)^2=a^{-2}z^{2m}=z^{-2m+2m}=1$}. Note that {$a = z^m(a^{-1}z^m)^{-1}$} is generated by the elements {$z$} and {$a^{-1}z^m$}, which indeedy generate the group {$\mathbb{Z}_{2n}\times\mathbb{Z}_2$}.

Now consider the noncommutative case {$za=az^{-1}$}. Applying this repeatedly, we have {$z^ia=az^{-i}$}. But {$z^i=a^2$} so {$z^i$} commutes with {$a$} and we have {$az^i=az^{-i}$} and {$z^i=z^{-i}$} so {$z^{2i}=1$}. Thus {$2i\equiv 0 \mod 2n$}. This means that {$i=0$} or {$i=n$}, which is to say, {$a^2=1$} or {$a^2=z^n$}.

The dicyclic group is the noncommutative group of order {$4n$} generated by {$z,a$} where {$z^{2n}=1, a^2=z^n, za=az^{-1}$}.

The dihedral group is the noncommutative group of order {$4n$} generated by {$z,a$} where {$z^{2n}=1, a^2=1, za=az^{-1}$}.

The dicyclic group {$\textrm{Dic}_n$} is a subgroup of {$\textrm{Pin}_-(2)$} for which {$a^2=z^n$}, {$a^4=1$}.

Note that {$a\neq a^{-1}$} so {$az\neq za$}. {$\mathbb{Z}_{2n}$} is not in the center of {$X$} and this is not a central extension.

The dihedral group {$\textrm{Dih}_{2n}$} is a subgroup of {$\textrm{Pin}_+(2)$} for which {$a^2=1$}.

Note that {$a=a^{-1}$} so {$az=za$}. {$\mathbb{Z}_{2n}$} is in the center of {$X$}, which is to say, this is a central extension.

---

Group Extension Example: Extending {$\mathbb{Z}_2$} by {$U(1)$}

Terminology and notation

Consider the group extension {$1\rightarrow U(1)\overset{f}{\rightarrow} G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$}

This is a left split short exact sequence if there exists a morphism {$t:G\rightarrow U(1)$} such that {$tf$} is the identity on {$U(1)$}.

This is a right split short exact sequence if there exists a morphism {$u:\mathbb{Z}_2\rightarrow G$} such that {$gu$} is the identity on {$\mathbb{Z}_2$}.

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 {$g(a)=\bar{a}, g(a^2)=\bar{a}^2=1, a^2\in f(N)$}. Thus there exists {$\theta\in U(1)$} such that {$a^2=\theta$}.

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

Automorphism

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

But there are only two automorphisms of {$U(1)$}: {$z\rightarrow z$} and {$z\rightarrow z^{-1}$}. This fact is proven here: How to prove the group of automorphisms of {$S_1$} as a topological group is {$\mathbb{Z}_2$}? Informally, one way to prove it is to 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 we have two cases for the map {$\phi_a(z)=aza^{-1}$}. Either {$aza^{-1}=z,\forall z\in N$} or {$aza^{-1}=z^{-1},\forall z\in N$}.

Commutative case

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

{$G=N\sqcup Na = N\sqcup Nb=NH$}. For any two elements {$g_1,g_2\in G$}, there is a unique decomposition {$g_1=z_1h_1, g_2=z_2hz_2, g_1g_2=z_1z_2h_1h_2$} where {$z_1,z_2\in Z, h_1,h_2\in H$}. So {$G\cong N\times H \cong U(1)\times\mathbb{Z}_2$}.

Noncommutative case

For all {$z\in N$}, {$aza^{-1}=z^{-1}$} and {$az=z^{-1}a$}. We know that {$a^2=\theta\in N$}. Thus {$a\theta=\theta^{-1}a, a^3=a^{-1},a^4=1. This means that {$\atheta^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).

Visualizing

---

The sequence is right split

Define the map {$u:\mathbb{Z}_2\rightarrow G$} as follows. {$u(1)=1, u(\bar{a})=a$}. Then {$gu(1)=1, gu(\bar{a})=\bar{a}$}. This is the requisite morphism and so the sequence is right split.

This means that we have a semidirect product {$G=N\ltimes H$} where {$H=\{1,a\}$}. Recall that {$a\notin N$} and {$N\cap H=1$}.

Suppose the sequence is not left split

There does not exist a morphism {$t:G\rightarrow U(1)$} such that {$tf$} is the identity on {$U(1)$}.

The semi-direct product defines the group homomorphism {$\phi :\mathbb{Z}_2\rightarrow \textrm{Aut}(U(1))$} given by {$\phi: h\rightarrow \phi_h$} where {$\phi_h(n)=hnh^{-1}$}.

Since {$\textrm{Aut}(U(1))\cong U(1)$}, we are looking at embeddings {$φ:\mathbb{Z}_2\rightarrow U(1)$}.

Suppose the sequence is left split

If it's left split, since it is also right split, we have a direct product {$G\cong U(1)\times \mathbb{Z}_2=t(G)\times u(G)$}.

Older notes comparing with {$\mathbb{Z}_{2n}$}

Thus there exists a {$z'\in f(N)$} such that {$za=az'$}, where {$z'\neq 1$}, {$z'\in U(1)$}. We see that {$z^na=a(z')^n$} for all {$n\in\mathbb{Z}$}. In general, every element in {$U(1)$} can be written as {$z^{\lambda}$} where {$\lambda\in\mathbb{R}$}. This map {$z\rightarrow z'$} implies the continuous homomorphism {$z^r a\rightarrow az'^r$} from {$U(1)$} to {$U(1)$}. Topologically speaking, the homology group {$H_1(U(1))=\mathbb{Z}$} is given by the winding number.

Now likewise, there exists a {$z''\in f(N)$} such that {$z'a=az''$}. Consequently, {$z^2=za^2=az'a=a^2z''=zz''$} and {$z=z''$}. This implies that our map is an automorphism. The only inverses in {$\mathbb{Z}$} are {$\pm 1$}. Thus the only possible maps here are {$z\rightarrow z$} and {$z\rightarrow z^{-1}$}, which means that either we have commutativity {$za=az$} or we have {$za=az^{-1}$}.

Analyzing with 2x2 matrices

We can represent {$U(1)$} in terms of {$2\times 2$} matrices of real numbers. Suppose that we can represent all of {$G$} likewise and with {$U(1)$} thus embedded.

We represent {$z\in U(1)$} with the matrix {$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}$}.

Let {$a$} be represented as {$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$}.

Suppose {$az=za$}.

Then

{$$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$$}

{$a_{11}\cos\theta + a_{12}\sin\theta = a_{11}\cos\theta - a_{21}\sin\theta$} implies {$a_{12}=-a_{21}$} or {$\sin\theta = 0$}

{$-a_{11}\sin\theta + a_{12}\cos\theta = a_{12}\cos\theta + a_{11}\sin\theta$} implies {$a_{11}=0$} or {$\sin\theta = 0$}.

{$a_{21}\cos\theta + a_{22}\sin\theta = a_{11}\sin\theta + a_{21}\cos\theta$} implies {$a_{22}=a_{11}$} or {$\sin\theta = 0$}.

{$-a_{21}\sin\theta -a_{22}\cos\theta = a_{12}\sin\theta -a_{22}\cos\theta$} implies {$a_{12}=-a_{21}$} or {$\sin\theta = 0$}.

Note furthermore that {$a^2=z$} and {$\det z=1$}. This means that {$\det a=\pm 1$}.

Thus, if {$\sin\theta\neq 0$}, then {$a$} is represented by {$\pm\begin{pmatrix}0 & -1 \\ 1 & 0 \\ \end{pmatrix}$}, but then {$z$} is represented by {$\begin{pmatrix}-1 & 0 \\ 0 & -1 \\ \end{pmatrix}$} with a contradiction.

Thus {$\sin\theta = 0$} and {$z$} is represented by {$\pm\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}.

Suppose {$az^{-1}=za$}.

{$$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{pmatrix}=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$$}

{$a_{11}\cos\theta - a_{12}\sin\theta = a_{11}\cos\theta - a_{21}\sin\theta$} implies {$a_{12}=a_{21}$} or {$\sin\theta = 0$}

{$a_{11}\sin\theta + a_{12}\cos\theta = a_{12}\cos\theta - a_{22}\sin\theta$} implies {$a_{11}=-a_{22}$} or {$\sin\theta = 0$}.

{$a_{21}\cos\theta - a_{22}\sin\theta = a_{11}\sin\theta + a_{21}\cos\theta$} implies {$a_{11}=-a_{22}$} or {$\sin\theta = 0$}.

{$a_{21}\sin\theta +a_{22}\cos\theta = a_{12}\sin\theta +a_{22}\cos\theta$} implies {$a_{12}=a_{21}$} or {$\sin\theta = 0$}.

Thus {$a$} is represented by {$\begin{pmatrix}a_{11} & a_{12} \\ a_{12} & -a_{11} \\ \end{pmatrix}$}

Again, {$a^2=z$} and {$\det z = 1$} implies {$\det a=\pm 1$}. Thus {$-a_{11}^2-a_{12}^2=-1$} so {$a$} is represented by a matrix of the form {$\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}$}. Then {$a^2=z$} means that

{$\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}=\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{pmatrix}$}

{$\cos^2\psi + \sin^2\psi=\cos\theta$} implies {$\cos\theta = 1$}

{$\cos\psi\sin\psi - \cos\psi\sin\psi = -\sin\theta$} implies {$\sin\theta=0$}

Thus {$z$} is represented by {$\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}, which is to say, {$z=1$}.

Whereas {$a$} can be represented by any matrix of the form below, which is a rotation times a reflection.

{$$\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}=\begin{pmatrix}\cos\psi & -\sin\psi \\ \sin\psi & \cos\psi \\ \end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$$}

• Math Stack Exchange. How to prove the group of automorphisms of {$S_1$} as a topological group is {$\mathbb{Z}_2$}?
• Math Stack Exchange. Extension of {$\mathbb{Z}_2$} by {$SO(n)$}

{$CT$} groups

We want to calculate the {$CT$} groups {$X$} which are group extensions, as below, where {$U$} is a subgroup of {$M_{2,2}=<\bar{C},\bar{T}\;|\;\bar{C}^2=\bar{T}^2=(\bar{C}\bar{T})^2=1>\cong\mathbb{Z}_2\times\mathbb{Z}_2$}.

{$1\rightarrow U(1)\rightarrow X\rightarrow U \rightarrow 1$}

When {$U=\{1\}$}, the {$CT$} group is {$U(1)$}

Consider the trivial subgroup {$U=\{1\}$} of {$M_{2,2}$}. What is the group extension?

{$1\rightarrow U(1)\overset{f}{\rightarrow}X\overset{g}{\rightarrow} 1 \rightarrow 1$}

{$f$} is injective, thus {$U(1)\cong\textrm{Img}(f)$} but also {$\textrm{Img}(f)=\textrm{Ker}(g)=X$}, thus {$X\cong U(1)$} is the {$CT$} group.

When {$U\cong\mathbb{Z}_2$}, the {$CT$} group is {$\textrm{Pin}_+$} or {$\textrm{Pin}_-$}

{$(\phi,\chi)$}-representations of {$CT$}-groups

Identity: linear, even: {$\rho(I)=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

Time reversal, antilinear, even: {$\rho(\bar{T})=\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$}

Charge conjugation, antilinear, odd: {$\rho(\bar{C})=\begin{pmatrix}0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}

Parity, linear, odd: {$\rho(\bar{CT})=\begin{pmatrix}0 & -1 \\ 1 & 0 \\ \end{pmatrix}$}

Perspectives

Time reversal fixes absolute perspective and flips relative perspective.

{$\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}=\begin{pmatrix}x \\ -y \end{pmatrix}$}

Antilinear swaps absolute and relative perspectives.

• First perspective: antilinear, odd, squares to +1: {$\rho(\bar{C})=\begin{pmatrix}0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}
• Second perspective: antilinear, odd, squares to -1: {$\rho(i\bar{C})=\begin{pmatrix}0 & i \\ i & 0 \\ \end{pmatrix}$}
• Third perspective: linear, odd, squares to -1: {$\rho(\bar{C}\bar{T})=\begin{pmatrix}0 & 1 \\ -1 & 0 \\ \end{pmatrix}$}
• Fourth perspective: linear, odd, squares to +1: {$\rho(i\bar{C}\bar{T})=\begin{pmatrix}0 & i \\ -i & 0 \\ \end{pmatrix}$}
• Fifth perspective?: antilinear, even, squares to +1: {$\rho(\bar{T})=\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$}
• Sixth perspective?: antilinear, even, squares to -1: {$\rho(i\bar{T})=\begin{pmatrix}i & 0 \\ 0 & -i \\ \end{pmatrix}$}
• Seventh perspective?: linear, even, squares to -1: {$\rho(iI)=\begin{pmatrix}i & 0 \\ 0 & i \\ \end{pmatrix}$}
• Eighth perspective?: linear, even, squares to +1 {$\rho(I)=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

{$\rho(I)=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

Note that {$i$} is a choice from {$\pm i$} and similarly {$\rho(\bar{T})$} is a choice from {$\pm\rho(\bar{T})$}.

The main (and only) perspective {$\bar{C}$} is the swap of absolute and relative spaces. Then it is a matter of the context for the perspective, which grows more sophisticated.

Thoughts

CT groups are relating structure (two on-and-off switches) and recurring activity (the circle). Bott periodicity cycles through all the ways of relating structure and recurring activity, thus all the contexts for patterns.

Group extensions express a short exact sequence, a division of everything, notably the foursome, how perspectives carve up space and fit together.

Thus we are matching graded representations of Clifford algebras and bigraded representations of CT groups. We also have nongraded Clifford representations. Are these the three minds?