Projective Extensions and the Little Group Method

Relation [F1] at $q=1$ reads $P(1)P(r)=\alpha (1,r)P(r)$, that is $P(r)=\alpha (1,r)P(r)$ by [A1]; multiplying by the inverse of $P(r)$ gives $\alpha (1,r)=1$. Symmetrically, [F1] at $r=1$ gives $P(q)P(1)=\alpha (q,1)P(q)$, so $\alpha (q,1)=1$.

Multiplying the relation of [F1] on the left by $P(q)$ and applying it twice, $P(q)(P(r)P(s))=\alpha (r,s)P(q)P(rs)=\alpha (r,s)\alpha (q,rs)P(qrs)$. Applying it in the other order, $(P(q)P(r))P(s)=\alpha (q,r)P(qr)P(s)=\alpha (q,r)\alpha (qr,s)P(qrs)$.

By associativity in $\mathrm{GL}(V)$, the two expressions of step 1.2 are equal; since $P(qrs)$ is invertible, cancelling it gives $\alpha (q,r)\alpha (qr,s)=\alpha (r,s)\alpha (q,rs)$.

Reading the additive data of [F2] in multiplicative notation for the abelian group ${\mathbb{C}}^{\times }$ with trivial $Q$-action — sums become products, negatives become inverses and $0$ becomes $1$ — the cocycle equation $g\cdot f(h,k)-f(gh,k)+f(g,hk)-f(g,h)=0$ becomes $\alpha (h,k)\alpha (g,hk)=\alpha (gh,k)\alpha (g,h)$, which after renaming $(g,h,k)$ as $(q,r,s)$ is the equation of step 2.1, and $f(1,g)=f(g,1)=0$ becomes $\alpha (1,q)=\alpha (q,1)=1$. Thus $\alpha$ is a normalized two-cocycle on $Q$ with values in ${\mathbb{C}}^{\times }$ in the multiplicative form of the published convention.

$P_c(1)=c(1)P(1)={\mathrm{id}}_V$, so $P_c$ is normalized.

Using [F1] and [A1], $P_c(q)P_c(r)=c(q)c(r)P(q)P(r)=c(q)c(r)\alpha (q,r)P(qr)=c(q)c(r)c(qr{)}^{-1}\alpha (q,r)P_c(qr)$.

Step 1.2 exhibits ${\alpha }_c(q,r)=c(q)c(r)c(qr{)}^{-1}\alpha (q,r)$ as the factor set of $P_c$; it takes values in ${\mathbb{C}}^{\times }$, and it is normalized because ${\alpha }_c(1,q)=c(1)c(q)c(q{)}^{-1}\alpha (1,q)=1$ and ${\alpha }_c(q,1)=c(q)c(1)c(q{)}^{-1}\alpha (q,1)=1$ by [F2].

Read in multiplicative notation for the trivial action, the published coboundary of [F3] is $\delta c(g,h)=c(g)c(h)c(gh{)}^{-1}$; step 2.1 therefore says ${\alpha }_c=(\delta c)\cdot \alpha$, the factor set of $P$ multiplied pointwise by the coboundary of $c$.

By [F4], multiplying a normalized two-cocycle by a coboundary does not change its class in $H^2(Q,{\mathbb{C}}^{\times })$, so $[{\alpha }_c]=[\alpha ]$: rephasing preserves the cohomology class. Conversely, if ${\alpha }^{\mathrm{\prime }}=(\delta c)\cdot \alpha$ for a normalized $c$, then steps 1.2 and 2.1 show that ${\alpha }^{\mathrm{\prime }}$ is exactly the factor set of the rephased family $P_c$; so every factor set cohomologous to $\alpha$ is realized by a normalized rephasing of $P$.

On basis elements, $(u_q{u}_r)u_s=\alpha (q,r)\alpha (qr,s)u_{qrs}$ and $u_q(u_r{u}_s)=\alpha (r,s)\alpha (q,rs)u_{qrs}$ by [F1], and these scalars are equal; since the product is bilinear and the $u_q$ span $A$, the product is associative.

On basis elements $u_1{u}_q=\alpha (1,q)u_q=u_q$ and $u_q{u}_1=\alpha (q,1)u_q=u_q$ by [F1]; by bilinearity $u_{1}a=a=a{u}_1$ for every $a\in A$, so $u_1$ is a two-sided identity.

Conversely, let $P$ be a normalized projective representation of $Q$ on a nonzero finite-dimensional space $V$ with factor set $\alpha$. Define $a\cdot v$ for $a={\sum }_q{\lambda }_q{u}_q$ by $a\cdot v:={\sum }_q{\lambda }_{q}P(q)v$. This is well defined and $\mathbb{C}$-bilinear by [A1], and it satisfies $u_1\cdot v=P(1)v=v$ and $(u_q{u}_r)\cdot v=\alpha (q,r)u_{qr}\cdot v=\alpha (q,r)P(qr)v=P(q)P(r)v=u_q\cdot (u_r\cdot v)$ by [F2]; both sides extend by linearity to arbitrary $a\in A$, so [F5] makes $V$ a left $A$-module.

Suppose $M$ has finite dimension over $\mathbb{C}$ and let $W\le M$ be an $A$-submodule. On $M$ define $P(q)m:=u_{q}m$, using the scalar action $\lambda m=(\lambda u_1)m$. These operators are $\mathbb{C}$-linear since $u_q(\lambda u_1)=(\lambda u_1)u_q$. The module law gives $P(1)={\mathrm{id}}_M$ and $P(q)P(r)=\alpha (q,r)P(qr)$; [F1] and the cocycle identity at $(q,q^{-1},q)$ give $P(q)P(q^{-1})=P(q^{-1})P(q)=\alpha (q,q^{-1}){\mathrm{id}}_M$, so $P(q{)}^{-1}=\alpha (q,q^{-1}{)}^{-1}P(q^{-1})$. This holds also for $M=0$. Choose a $\mathbb{C}$-linear projection $\pi :M\to W$, which exists by [A1] applied to a basis of $W$ extended to a basis of $M$. Averaging its conjugates, define $\bar{\pi }:=\frac{1}{\mathrm{\mid }Q\mathrm{\mid }}{\sum }_{q\in Q}P(q)\pi P(q{)}^{-1}$.

Steps 1.1 and 1.2 make $A$ a unital associative ring, and $\lambda \mapsto \lambda u_1$ is a unital ring homomorphism $\mathbb{C}\to A$ with central image, so $A$ is a $\mathbb{C}$-algebra in the sense of [F4]; by [F1] and [F3] each $u_q$ has the two-sided inverse $\alpha (q,q^{-1}{)}^{-1}{u}_{q^{-1}}$.

Each $P(q)\pi P(q{)}^{-1}$ maps $M$ into $W$, because $W$ is $A$-stable and $P(q)$ is invertible; hence $\bar{\pi }(M)\subseteq W$, and $\bar{\pi }$ fixes $W$ pointwise because $P(q)\pi P(q{)}^{-1}w=P(q)P(q{)}^{-1}w=w$ for $w\in W$.

For every $r\in Q$, $P(r)\bar{\pi }P(r{)}^{-1}=\frac{1}{\mathrm{\mid }Q\mathrm{\mid }}{\sum }_{q}P(r)P(q)\pi P(q{)}^{-1}P(r{)}^{-1}=\frac{1}{\mathrm{\mid }Q\mathrm{\mid }}{\sum }_q\alpha (r,q)\alpha (r,q{)}^{-1}P(rq)\pi P(rq{)}^{-1}=\frac{1}{\mathrm{\mid }Q\mathrm{\mid }}{\sum }_{q^{\mathrm{\prime }}}P(q^{\mathrm{\prime }})\pi P(q^{\mathrm{\prime }}{)}^{-1}=\bar{\pi }$ by step 2.1 and reindexing $q^{\mathrm{\prime }}=rq$, so $\bar{\pi }$ commutes with every $P(r)$ and hence with the action of $A$.

Let $M\mathrm{\ne }0$ be a finite-dimensional left $A$-module and let $P(q)$ be the action of $u_q$ on $M$, which is a $\mathbb{C}$-linear map by [F5]. Then $P(1)={\mathrm{id}}_M$ by [F5] and [F1], and $P(q)P(r)=\alpha (q,r)P(qr)$ because the action respects products and $u_q{u}_r=\alpha (q,r)u_{qr}$; each $P(q)$ has inverse $\alpha (q,q^{-1}{)}^{-1}P(q^{-1})$ by step 2.2, so $P(q)\in \mathrm{GL}(M)$. Thus every nonzero finite-dimensional $A$-module gives a normalized projective representation with factor set $\alpha$.

Steps 3.1 and 3.2 show that $M=W\oplus \ker \bar{\pi }$: the average is a surjection onto $W$ fixing $W$, so it is a module map with image $W$ and kernel a complement. Therefore every submodule of a finite-dimensional $A$-module has a complementary submodule.

The two constructions of steps 3.3 and 1.3 are mutually inverse: starting from $M$, the module built from $P$ acts as $u_q\cdot m=P(q)m$, which is the original action, and starting from $P$, the representation of the built module sends $q$ to the action of $u_q$, which is $P(q)$. Hence nonzero finite-dimensional left $A$-modules correspond bijectively to normalized projective $Q$-representations with factor set $\alpha$.

A finite-dimensional $A$-module has finite length: a strictly increasing chain of submodules has strictly increasing complex dimensions, so no chain of submodules can be longer than $1+{\dim }_{\mathbb{C}}M$ terms, and a maximal chain is a composition series in the sense of [F6]. By [F9], the complement property of step 4.1 makes $M$ a direct sum of simple submodules, that is, semisimple in the sense of [F7].

The zero space carries the unique $A$-module structure, all operators being zero, and it is the zero object of the category of left $A$-modules: the zero map is the only map from it and the only map to it, and both are module maps. It is therefore a module for every coefficient cocycle $\alpha$, while it determines no factor set, which is why step 4.2 is stated for nonzero modules.

Applying step 5.1 to the left regular module ${}_{A}A$, which is finite-dimensional of dimension $\mathrm{\mid }Q\mathrm{\mid }$, shows that $A$ is a semisimple ring in the sense of [F8], and step 5.1 shows in addition that every finite-dimensional left $A$-module is semisimple.

For $t\in T$ the conjugate ${}^t\rho$ is a representation of $tN{t}^{-1}=N$ whose character is ${}^t\theta =\theta ={\chi }_{\rho }$ by [F1] and [F2]; hence ${}^t\rho \cong \rho$ by [F3]. Choose $T_t\in \mathrm{GL}(S)$ with $T_t\rho (n)=\rho (tn{t}^{-1})T_t$ for every $n\in N$, and set $T_1:={\mathrm{id}}_S$, which satisfies this relation.

By [A1] write each $i\in I$ as $i=n_i{t}_i$ with $n_i\in N$, $t_i\in T$, and set $P(i):=\rho (n_i)T_{t_i}$. This is well defined because the pair $(n_i,t_i)$ is unique, and $P(1)=\rho (1)T_1={\mathrm{id}}_S$, while $P(n)=\rho (n)T_1=\rho (n)$ for $n\in N$; each $P(i)$ is invertible, with $P(i{)}^{-1}=T_{t_i}^{-1}\rho (n_i{)}^{-1}$.

For $x\in N$ and $i\in I$ one has $P(xi)=\rho (x)P(i)$ and $P(ix)=P(i)\rho (x)$. Indeed $xi=(x{n}_i)t_i$ and $ix=n_i(t_{i}x{t}_i^{-1})t_i$ are the decompositions of [A1], so $P(xi)=\rho (x{n}_i)T_{t_i}=\rho (x)P(i)$, and $P(ix)=\rho (n_i)\rho (t_{i}x{t}_i^{-1})T_{t_i}=\rho (n_i)T_{t_i}\rho (x)=P(i)\rho (x)$ using the intertwining relation of step 1.1.

For every $i\in I$ and $m\in N$ one has $P(i)\rho (m)P(i{)}^{-1}=\rho (im{i}^{-1})$. Indeed $P(i)\rho (m)P(i{)}^{-1}=\rho (n_i)T_{t_i}\rho (m)T_{t_i}^{-1}\rho (n_i{)}^{-1}=\rho (n_i)\rho (t_{i}m{t}_i^{-1})\rho (n_i{)}^{-1}=\rho (im{i}^{-1})$, using $i=n_i{t}_i$ and step 1.1.

For $i,j\in I$ the operator $A(i,j):=P(i)P(j)P(ij{)}^{-1}$ is invertible and satisfies $A(i,j)\rho (m)=P(i)P(j)\rho ((ij{)}^{-1}m(ij))P(ij{)}^{-1}=\rho (ij(ij{)}^{-1}m(ij)j^{-1}{i}^{-1})A(i,j)=\rho (m)A(i,j)$ for every $m\in N$, by step 3.2 applied to $i$, $j$ and $ij$; since $\rho$ is irreducible over the algebraically closed field $\mathbb{C}$, [F4] gives a unique $\alpha (i,j)\in {\mathbb{C}}^{\times }$ with $A(i,j)=\alpha (i,j){\mathrm{id}}_S$, the scalar being nonzero because $A(i,j)$ is invertible, so $P(i)P(j)=\alpha (i,j)P(ij)$ for all $i,j\in I$.

The scalar $\alpha (i,j)$ of step 4.1 depends only on the cosets $iN$ and $jN$: the general element of $iN$ is $ix$ and the general element of $jN$ is $yj$ with $x,y\in N$, so it suffices to compare $A(ix,j)$ and $A(i,yj)$ with $A(i,j)$. For the first, $ixj=(ij)(j^{-1}xj)$ with $j^{-1}xj\in N$, so $P(ixj)=P(ij)\rho (j^{-1}xj)$ by step 3.1 and $\rho (j^{-1}xj)=P(j{)}^{-1}\rho (x)P(j)$ by step 3.2, whence $A(ix,j)=P(i)\rho (x)P(j)\rho (j^{-1}xj{)}^{-1}P(ij{)}^{-1}=P(i)\rho (x)\rho (x{)}^{-1}P(j)P(ij{)}^{-1}=A(i,j)$. For the second, $yj=y\cdot j$ and $iyj=(iy{i}^{-1})(ij)$ with $iy{i}^{-1}\in N$, so $P(yj)=\rho (y)P(j)$ and $P(iyj)=\rho (iy{i}^{-1})P(ij)$ by step 3.1, while $P(i)\rho (y)P(i{)}^{-1}=\rho (iy{i}^{-1})$ by step 3.2, hence $A(i,yj)=P(i)\rho (y)P(j)P(ij{)}^{-1}\rho (iy{i}^{-1}{)}^{-1}=\rho (iy{i}^{-1})A(i,j)\rho (iy{i}^{-1}{)}^{-1}=A(i,j)$, the last equality because $A(i,j)$ is a scalar.

Write $\alpha (iN,jN):=\alpha (i,j)$, which is well defined by step 5.1, so that $P(i)P(j)=\alpha (iN,jN)P(ij)$ for all $i,j\in I$; and $\alpha (N,qN)=\alpha (qN,N)=1$ for every $q\in I$, since $P(1)P(j)=P(j)$ and $P(i)P(1)=P(i)$ with $P(1)={\mathrm{id}}_S$ by step 2.1 force $A(1,j)={\mathrm{id}}_S=A(i,1)$.

The function $\alpha :Q\times Q\to {\mathbb{C}}^{\times }$ is a normalized two-cocycle on $Q=I\mathrm{/}N$ in the sense of [F6]: the normalization is step 6.1, and associativity of composition in $\mathrm{GL}(S)$ gives $\alpha (i,j)\alpha (ij,k)P(ijk)=(P(i)P(j))P(k)=P(i)(P(j)P(k))=\alpha (j,k)\alpha (i,jk)P(ijk)$, while $P(ijk)$ is invertible by step 2.1, so $\alpha (iN,jN)\alpha (ijN,kN)=\alpha (jN,kN)\alpha (iN,jkN)$; passing to cosets, this is the cocycle identity of [F6] on $Q$.