Consider two irreducible, unitary projective representations $\rho$ and $\tau$ of a finite group G onto the same complex matrix space. If these representations are projectively equivalent, ie. $\rho (g)=c(g)U\tau (g)U^{-1}$ for all $g$ in $G$ and for some unitary matrix $U$ and a scalar function $c:G\rightarrow \mathbb{C}$, then the factor systems of $\rho$ and $\tau$ are equivalent modulo a coboundary.

Now I want to know the converse: if the factor systems of $\rho$ and $\tau$ are known to be equivalent, under which conditions does that imply that $\rho$ and $\tau$ must be projectively equivalent?

EDIT: I will make my question more explicit. Consider a fixed cocycle $\omega:G\times G\rightarrow \mathbb{C}^\times$. It is clear that, if $\rho (g)$ is an irrep with factor system $\omega$, then so is $\tau (g)=\chi (g)U\rho (g)U^{-1}$ for a unitary $U$ and a 1D rep $\chi$ (the fact that $\chi$ is a representation means that it does not change the cocycle $\omega$.)

My question is, under what conditions on $G$ and $\omega$ is it true that all irreps with factor system $\omega$ can be related to $\rho$ in this way? By Qiaochu's answer, this is true if $\omega$ is "nondegenerate", and in this case $\chi(g)$ is always trivial since all such irreps are linearly equivalent. Since we only require projective equivalence, can we loosen the non-degenerate condition?

For example, the dihedral group $D_4$ satisfies my condition, but it does not admit a non-degenerate cocycle since it is not a group of central type. The symmetric group $S_4$, on the other hand, seems to have both 2D and 4D irreps with the same cocycle, so my property cannot be satisfied.


I apologize if this is a trivial question, I am new to this field.


1 Answer 1


Projective representations with a fixed 2-cocycle $c : G \times G \to \mathbb{C}^{\times}$ (meaning that $\rho(g) \rho(h) = c(g, h) \rho(gh)$) correspond to modules over the twisted group algebra $\mathbb{C} \rtimes_c G$, which is $\mathbb{C}[G]$ with the modified multiplication

$$g \cdot h = c(g, h) gh.$$

Like the group algebra, the twisted group algebra is semisimple, so the number of isomorphism classes of irreducible projective representations with 2-cocycle $c$ is the dimension of the center. Now we compute: $z = \sum z_g g$ is central in the twisted group algebra iff

$$h \cdot z = \sum z_g c(h, g) hg = z \cdot h = \sum z_g c(g, h) gh$$

and making the substitution $g \mapsto hgh^{-1}$ in the second sum gives that this is true iff for every $g, h \in G$ we have

$$z_g c(h, g) = z_{hgh^{-1}} c(hgh^{-1}, h).$$

This means that $z_g$ determines $z_{g'}$ for any $g'$ conjugate to $g$, but in addition it also means that if $g = hgh^{-1}$, or equivalently $h$ lies in the centralizer $Z_G(g)$, then

$$z_g c(h, g) = z_g c(g, h)$$

so either $z_g = 0$ or $c(h, g) = c(g, h)$ for all $h \in Z_G(g)$. So the dimension of the center is the number of conjugacy classes of $G$ with this property. The only conjugacy class that obviously has this property is the one containing the identity, although sometimes (e.g. if $c(g, h) = 1$ is the trivial 2-cocycle) every conjugacy class will. I think this property has a name but I don't remember it; you can look up "twisted group algebra" and probably find some references.

  • $\begingroup$ Great answer, thank you. So if I want my claim to be true (same cocycle implies projective equivalence) then I need that the conjugacy class containing the identity is the only class which satisfies your property, correct? That way, there is only 1 isomorphism class so the representations must be equivalent. I believe that, for abelian groups, this is only possible if G is a symmetric group (see Characters of Finite Groups Pt. 1, Berkovich and Zhmud Chp. 6). I wonder about the non-abelian case? $\endgroup$
    – David T
    Feb 24, 2016 at 2:18
  • $\begingroup$ @Sheepiedog: yes, that's the condition. I don't understand what you mean by "for abelian groups... a symmetric group." $\endgroup$ Feb 24, 2016 at 3:31
  • $\begingroup$ Sorry for being unclear. The result which I am trying to quote says that, if G is abelian, then a cocycle on G which satisfies my condition exists iff G is symmetric, ie. $G=G_1\times G_2$ where $G_1\cong G_2$. However I may have misinterpreted their claim. $\endgroup$
    – David T
    Feb 24, 2016 at 9:05
  • $\begingroup$ I see. I was confused because "symmetric group" usually means the groups $S_n$. Anyway, that sounds right; those should be the finite abelian groups admitting symplectic forms. $\endgroup$ Feb 24, 2016 at 16:01
  • $\begingroup$ After some reading, I believe that the property of 2-cocyles which you mentioned is called "nondegenerate", and a group which admits a nondegenerate cocycle is called "of central type". However it turns out that this is more restrictive than I need. This implies that all irreps with a given cocyle are linearly equivalent, but I require them only to be projectively equivalent, ie. isomorphic modulo a 1D representation (which is a 1-cocyle, so it does not change the factor system at all). So the problem is a bit different than I stated. I will edit the question to reflect this. $\endgroup$
    – David T
    Feb 25, 2016 at 23:43

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