Projective Extensions and the Little Group Method - Examples

Direct computation gives $I^2=J^2=-1_2$, $IJ=K=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$, $JI=-K$ and $K^2=-1_2$, so the eight matrices $\pm 1_2,\pm I,\pm J,\pm K$ are distinct and form a subgroup $M\le {\mathrm{GL}}_2(\mathbb{C})$. Listing them, $1_2,I,J,K$ have second columns $(0,1),(0,-i),(1,0),(i,0)$ up to sign, so no two of the eight coincide; the assignment $i\mapsto I$, $j\mapsto J$, $k\mapsto K$, $-1\mapsto -1_2$ preserves the relations of [F1], so $M\cong Q_8$ with centre $\{\pm 1_2\}$ and quotient $M\mathrm{/}\{\pm 1_2\}\cong Q$ via $I\mapsto x$, $J\mapsto y$.

Define $P:Q\to {\mathrm{GL}}_2(\mathbb{C})$ by $P(x^a{y}^b):=I^a{J}^b$ for $a,b\in \{0,1\}$; this is well defined because every element of $Q$ is uniquely $x^a{y}^b$. Step 1.1 further gives $IK=I(IJ)=I^{2}J=-J$, $KI=J$, $JK=I$ and $KJ=-I$. Hence, using $P(xy)=K$, the products on basis elements are $P(x{)}^2=I^2=-1_2$, $P(y{)}^2=J^2=-1_2$, $P(xy{)}^2=K^2=-1_2$, $P(x)P(y)=IJ=K=P(xy)$, $P(y)P(x)=JI=-K=-P(xy)$, $P(x)P(xy)=IK=-J=-P(y)$, $P(y)P(xy)=JK=I=P(x)$, $P(xy)P(x)=KI=J=P(y)$, $P(xy)P(y)=KJ=-I=-P(x)$, and $P(1)P(q)=P(q)=P(q)P(1)$. Comparing with $qr$ in $Q$ shows $P(q)P(r)=\alpha (q,r)P(qr)$ for all $q,r\in Q$, where $\alpha (1,q)=\alpha (q,1)=1$, $\alpha (x,x)=\alpha (y,y)=\alpha (xy,xy)=\alpha (y,x)=\alpha (xy,y)=\alpha (x,xy)=-1$ and $\alpha (x,y)=\alpha (y,xy)=\alpha (xy,x)=1$.

The function $\alpha$ is a normalized two-cocycle: comparing $(P(q)P(r))P(s)=\alpha (q,r)\alpha (qr,s)P(qrs)$ with $P(q)(P(r)P(s))=\alpha (r,s)\alpha (q,rs)P(qrs)$ using step 2.1 and cancelling the invertible matrix $P(qrs)$ gives $\alpha (q,r)\alpha (qr,s)=\alpha (r,s)\alpha (q,rs)$ for all $q,r,s\in Q$, and $\alpha (1,q)=\alpha (q,1)=1$ holds by definition, matching [F4].

The cocycle central extension $E_{\alpha }$ of [F2] contains the eight elements $Q\times \{\pm 1\}=\{(q,z):q\in Q,z=\pm 1\}$; the map $\phi (q,z):=s(q)z$ with $s(1)=1_2$, $s(x)=I$, $s(y)=J$, $s(xy)=K$ satisfies $\phi ((q,z)(r,w))=s(qr)\alpha (q,r)zw=s(q)s(r)zw=\phi (q,z)\phi (r,w)$, because $P(q)P(r)=\alpha (q,r)P(qr)$ by step 2.1; it is injective on the eight elements and its image is $M$, so $Q\times \{\pm 1\}\cong M\cong Q_8$. Also $P$ is faithful, since $P(q)=1_2$ forces $q=1$ by the distinctness of the eight matrices in step 1.1.

The factor set $\alpha$ is not a coboundary, so its class in $H^2(Q,{\mathbb{C}}^{\times })$ is nonzero: if $\alpha =\delta c$ for some $c:Q\to {\mathbb{C}}^{\times }$, then [A1] would give $\alpha (y,x)=c(y)c(x)c(yx{)}^{-1}=c(x)c(y)c(xy{)}^{-1}=\alpha (x,y)$, contradicting $\alpha (y,x)=-1\mathrm{\ne }1=\alpha (x,y)$ from step 2.1. Consequently the lifts of the two generators anticommute, $P(y)P(x)=-P(x)P(y)$, the representation $P$ of step 2.1 is a faithful projective representation of $C_2\times C_2$ with nontrivial factor set, and it corresponds by [F3] to the representation $D(q,z)=zP(q)$ of $E_{\alpha }$ whose restriction to $Q\times \{\pm 1\}\cong Q_8$ is the faithful two-dimensional representation of $Q_8$.

The example is complete: $Q_8\mathrm{/}\{\pm 1\}\cong C_2\times C_2$ by step 1.1, the faithful two-dimensional representation of $Q_8$ restricts on $Q\times \{\pm 1\}\cong Q_8$ to the projective representation $P$ of step 2.1 whose factor set has $\alpha (y,x)=-1\mathrm{\ne }1=\alpha (x,y)$ and is therefore not a coboundary by step 5.1, and the $\pm 1$-valued subgroup $Q\times \{\pm 1\}$ of the cocycle central extension $E_{\alpha }$ is isomorphic to $Q_8$ by step 4.1.

The centre of $Q_8$ is $Z(Q_8)=\{\pm 1\}$: the element $-1$ is central by [F1], while $i$, $j$ and $k$ are not central because $ij=k$ and $ji=-k\mathrm{\ne }k$, together with their cyclic analogues; since every element of $Q_8$ is one of $\pm 1,\pm i,\pm j,\pm k$ by [F1], these are all the central elements.

Every linear character $\lambda :Q_8\to {\mathbb{C}}^{\times }$ satisfies $\lambda (-1)=1$: from the relations of [F1], $ij{i}^{-1}{j}^{-1}=(ij)(ji{)}^{-1}=k\cdot (-k{)}^{-1}=k\cdot k=k^2=-1$, so $-1$ is a commutator, and multiplicativity gives $\lambda (ij{i}^{-1}{j}^{-1})=\lambda (i)\lambda (j)\lambda (i{)}^{-1}\lambda (j{)}^{-1}=1$.

The prescription $\theta (1)=1$, $\theta (-1)=-1$ defines a linear character of $Z(Q_8)$: products involving $1$ satisfy $\theta (1z)=\theta (z1)=\theta (z)=\theta (1)\theta (z)$, and the remaining product satisfies $\theta ((-1)(-1))=\theta (1)=1=(-1{)}^2=\theta (-1)\theta (-1)$. These are all four pairs in $Z(Q_8)\times Z(Q_8)$, so the map is multiplicative. It acts on the nonzero one-dimensional space $\mathbb{C}$, which has no nonzero proper subspace, hence is irreducible.

The character $\theta$ is $Q_8$-invariant: for $g\in Q_8$ and $z\in Z(Q_8)$ one has ${}^g\theta (z)=\theta (g^{-1}zg)=\theta (z)$ because $z$ is central, so ${}^g\theta =\theta$ for every $g$ and the inertia group of $\theta$ is all of $Q_8$ by [F2].

Therefore $\theta$ is invariant but does not extend to a linear character of $Q_8$: a linear extension would be a homomorphism $\lambda :Q_8\to {\mathbb{C}}^{\times }$ with $\lambda {\mathrm{\mid }}_{Z(Q_8)}=\theta$ by [F3] and [A1], in particular $\lambda (-1)=\theta (-1)=-1$, whereas step 1.2 forces $\lambda (-1)=1$ for every linear character. Hence no extension of $\theta$ to $Q_8$ exists, and the invariance established in step 3.1 is not sufficient for extendibility.

By [F4] the failure of extension recorded in step 4.1 is exactly the statement that the Clifford obstruction class of $\theta$ in $H^2(Q_8\mathrm{/}\{\pm 1\},{\mathbb{C}}^{\times })$ is nonzero; this is a genuine obstruction, since [F5] exhibits a nonzero class in that same cohomology group, and it shows that the invariance hypothesis alone cannot replace the split hypothesis of the little group method. The character $\theta$ is the central character of the faithful two-dimensional representation of $Q_8$, so the example is exactly the nonsplit counterpart of the extendible invariant types.

For each integer $k$ the formula ${\theta }_k(r^j):={\zeta }^{jk}$ is a well-defined homomorphism $A\to {\mathbb{C}}^{\times }$, because ${\zeta }^n=1$ makes it independent of the representative $j$ modulo $n$, and ${\theta }_k(r^{j+l})={\zeta }^{(j+l)k}={\theta }_k(r^j){\theta }_k(r^l)$. By [F3] every irreducible complex representation of the abelian group $A$ is one-dimensional, hence of this form, and by [F4] the $n$ functions ${\theta }_0,\dots ,{\theta }_{n-1}$ are distinct (they take the distinct values ${\zeta }^k$ at $r$); so $\widehat{A}=\{{\theta }_0,\dots ,{\theta }_{n-1}\}$ with ${\theta }_k={\theta }_{k^{\mathrm{\prime }}}$ exactly when $k\equiv k^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}n)$.

The generator $s$ acts on $\widehat{A}$ by ${}^s{\theta }_k={\theta }_{-k}$: for all $j$, ${}^s{\theta }_k(r^j)={\theta }_k(s^{-1}{r}^{j}s)={\theta }_k(r^{-j})={\zeta }^{-jk}={\theta }_{-k}(r^j)$, using $s=s^{-1}$ and [F5]. Hence the $H$-orbit of ${\theta }_k$ is $\{{\theta }_k,{\theta }_{-k}\}$, of size one exactly when $k\equiv -k(\mathrm{m}\mathrm{o}\mathrm{d}n)$, i.e. $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$, and of size two otherwise; correspondingly $H_{{\theta }_k}=H$ in the first case and $H_{{\theta }_k}=1$ in the second.

Fixed case: if $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ then $H_{{\theta }_k}=C_2$ and $I_{{\theta }_k}=A⋊C_2=G$ by step 2.1, so [F2] lists the representations over ${\theta }_k$ as ${\tilde{\theta }}_k\otimes \mathrm{Infl}\sigma$ with $\sigma \in \mathrm{Irr}(C_2)$; the group $C_2$ has exactly its two linear characters, of degree $1$, and $[H:H_{{\theta }_k}]=1$, so this orbit contributes two linear characters of $D_{2n}$.

Non-fixed case: if $2k\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ then $H_{{\theta }_k}=1$ and $I_{{\theta }_k}=A=⟨r⟩$ by step 2.1, so the only $\sigma$ is the trivial character of the trivial group and [F2] gives the single representation ${\mathrm{Ind}}_A^G{\theta }_k$, of degree $[H:1]\cdot 1=2$, induced from $C_n$. The two members of the orbit induce isomorphic representations, since ${\theta }_k$ and ${\theta }_{-k}$ are conjugate under $s$ and [F2] uses one orbit representative; so each orbit of size two contributes exactly one irreducible of degree two.

Counting: by [A1] exactly $f=\gcd (2,n)$ indices $k$ modulo $n$ satisfy $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$, so there are $2f$ linear characters by step 3.1 and the remaining $n-f$ characters of $\widehat{A}$ form $(n-f)\mathrm{/}2$ orbits of size two, contributing $(n-f)\mathrm{/}2$ irreducibles of degree two by step 3.2; the sum of squares of the degrees is $2f\cdot 1^2+\frac{n-f}{2}\cdot 2^2=2f+2(n-f)=2n=\mathrm{\mid }{D}_{2n}\mathrm{\mid }$ by [F1], and by [F2] this list is exactly $\mathrm{Irr}(D_{2n})$ with no repetitions.

The degenerate cases are included. For $n=1$ one has $f=1$ by [A1] and $D_2\cong C_2$ by [F1], and the list consists of the two linear characters of $C_2$, with no degree-two character. For $n=2$ one has $f=2$ and $D_4\cong C_2\times C_2$ by [F1], and the list consists of the four linear characters, with no degree-two character; both agree with the classification of step 4.1 since $(n-f)\mathrm{/}2=0$ in these cases.

The example is verified: the characters ${\theta }_k(r^j)=\exp (2\pi ijk\mathrm{/}n)$ of $C_n$ have $H$-orbits $\{k,-k\}$ by step 2.1; the fixed indices with $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ contribute two linear characters each by step 3.1; every other orbit contributes the single degree-two representation induced from $C_n$ by step 3.2; and by steps 4.1 and 5.1 these exhaust $\mathrm{Irr}(D_{2n})$ with the stated degree count, including $n=1$ and $n=2$.

The map $P$ with $P(1)=P(t)=1$ is a normalized projective representation with factor set $\alpha \equiv 1$: $P(1)={\mathrm{id}}_V$ and, since $t^2=1$ and $V$ is one-dimensional, $P(q)P(r)=1=\alpha (q,r)P(qr)$ for all four pairs $(q,r)\in Q\times Q$ with $\alpha (q,r)=1$.

Rephasing by $c$ gives $P_c(1)=c(1)P(1)=1$ and $P_c(t)=c(t)P(t)=i$, so $P_c$ is again normalized; and $P_c(t)P_c(t)=i\cdot i=-1$, while the defining relation for $P_c$ at the pair $(t,t)$ reads $P_c(t)P_c(t)={\alpha }_c(t,t)P_c(t^2)={\alpha }_c(t,t)P_c(1)={\alpha }_c(t,t)$, so ${\alpha }_c(t,t)=-1$; the pairs involving $1$ have ${\alpha }_c(1,q)={\alpha }_c(q,1)=1$, so ${\alpha }_c$ is the function with the single nontrivial value ${\alpha }_c(t,t)=-1$ and ${\alpha }_c\equiv ̸1$.

The rephasing formula of [F1] reproduces this value: ${\alpha }_c(t,t)=c(t)c(t)c(t^2{)}^{-1}\alpha (t,t)=i\cdot i\cdot c(1{)}^{-1}\cdot 1=-1$, using $c(1)=1$ and $\alpha \equiv 1$; so ${\alpha }_c=\delta c$ in the multiplicative coboundary notation, with $\delta c(t,t)=-1$ and $\delta c=1$ on the pairs involving the identity.

Rephasing is reversible and returns the genuine representation: with $c^{\mathrm{\prime }}(q):=c(q{)}^{-1}$, so that $c^{\mathrm{\prime }}(1)=1$ and $c^{\mathrm{\prime }}(t)=-i$, the rephased family $P_{c^{\mathrm{\prime }}}(q)=c^{\mathrm{\prime }}(q)P_c(q)$ has $P_{c^{\mathrm{\prime }}}(1)=1$ and $P_{c^{\mathrm{\prime }}}(t)=(-i)\cdot i=1=P(t)$; by [F1] its factor set is $\delta c^{\mathrm{\prime }}\cdot {\alpha }_c=1$, so it is the original multiplicative representation $P$ of step 1.1.

Both factor sets therefore have the same class in $H^2(C_2,{\mathbb{C}}^{\times })$, namely the zero class: $\alpha \equiv 1$ is the identity cocycle, and ${\alpha }_c=\delta c$ is a coboundary, so $[{\alpha }_c]=[\alpha ]=0$ by [F3]. To realize this as a Clifford obstruction, take $G=C_2$, $N=\{1\}$ and the unique irreducible representation $\rho$ of $N$ on $\mathbb{C}$. It is $G$-invariant, with inertia group $G$ and quotient $G\mathrm{/}N=C_2$. Both $P$ and $P_c$ restrict to $\rho$, and $P(ng)=\rho (n)P(g)$ and $P(gn)=P(g)\rho (n)$ (and the same identities for $P_c$) hold since $n=1$. Thus they are projective inertia operators for this specified type in [F3]. Their factor sets give its same vanishing obstruction, although ${\alpha }_c$ is not the constant cocycle.

The example is verified: the trivial one-dimensional projective representation of $C_2=\{1,t\}$ has $\alpha \equiv 1$; rephasing by $c(1)=1$, $c(t)=i$ produces $P_c(t)=i$ with ${\alpha }_c(t,t)=-1$, which is exactly $\delta c(t,t)=c(t)c(t)c(t^2{)}^{-1}$; and since $\delta c$ is a coboundary, the two factor sets lie in one cohomology class, the zero class, which step 3.2 confirms by rephasing back to $P$ with multiplier $-i$.