Rephasing the trivial projective representation of C2 by a coboundary

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 $-i$.