Real representations of twisted group algebras

Question

Let $A$ be a finite abelian group, and $\sigma$ a non-trivial action of $A$ on $\mathbb{C}$ by real algebra automorphisms. In particular, using $\sigma$, we can view $\mathbb{C}^{\times}$ as a module over $A$. Let $\xi$ be a 2-cocycle representing a class in $H^2(A;\mathbb{C}^{\times}_{\sigma})$. We can consider the real algebra $\mathbb{C}^{\xi}_{\sigma}\lbrack A\rbrack$ with multiplication given by $$z_1a_1\cdot z_2a_2 = (\xi(a_1,a_2)z_1\sigma(a_1)(z_2)) a_1a_2.$$

How many real irreducible representations does this algebra have? Alternatively, one could try to count the quaternionic irreducible representations? (The center of this algebra is not too hard to describe explicitly.)

EDIT: By a real irreducible representation, I mean a representation $V$ of $\mathbb{C}^{\xi}_{\sigma}\lbrack A\rbrack$ over $\mathbb{R}$ such that $End_{\mathbb{C}^{\xi}_{\sigma}\lbrack A\rbrack}(V)\cong \mathbb{R}$.

Answer 1

The algebra $A:=\mathbb{C}_\sigma^\xi[G]$ is the universal $\mathbb{R}$-algebra that

1. contains $\mathbb{C}$ and symbols $\{u_g \mid g\in G\}$ such that
2. $u_g u_h = \xi(g,h) u_{gh}$ and $u_g z u_g^{-1} = {^g z}$ holds, where I use ${^g z}$ as a shorthand for $\sigma(g)(z)$.

This is an example of a "crossed" $G$-graded $\mathbb{R}$-algebra, i.e. an algebra that is equipped with a decomposition $A=\bigoplus_g A_g$ such that $A_g A_h\subseteq A_{gh}$ (that "$G$-graded") and $A_g\cap A^\times\neq\emptyset$ (that's "crossed"). An ordinary group algebra $K[G]$ is an example, a twisted group algebra $K^\xi[G]$ is also an example. Your example is a little bit different than those two, because the degree-1-piece is 2-dimensional over the field, not the field itself. One particularly easy example is the $C_2$-graded crossed $\mathbb{R}$-algebra $\mathbb{H}=\mathbb{C}\oplus\mathbb{C}j$ (which is also a special case of your assumptions for $G=C_2$)

Crossed, $G$-graded algebras are nice to have, because you can do Clifford theory with them: The degree-1-piece behaves very much like $K[N]$ behaves inside $K[G]$ for any normal subgroup $N\unlhd G$. In fact: That's one example - Take any $G$-graded algebra and define a $Q:=G/N$-grading by setting $A_{gN} := \sum_{h\in N} A_{gh}$. If you do that with $A=K[G]$, then the degree-1-piece in the $Q$-grading is exactly $K[N]$.

In your example, let's have a look at the normal subgroup $N:=\ker(\sigma)$ of index 2. The degree-1-piece of the $Q$-grading on $A$ is the twisted algebra $A_N:=\mathbb{C}^\xi[N]$ (just considered as an $\mathbb{R}$-algebra). This is one complication fewer and assuming you understand such algebras well enough, one can count the representations of $A$ by using Clifford theory.

The grading group acts by conjugation on modules of the degree-1-piece. The appropriate generalisation of Clifford's theorem now tells you that for any simple $A$-module $V$, the restriction $V_{|N}$ of $V$ to $A_N$ is semisimple and the occurring simple constituents are a single $Q$-orbit that all occur with the same multiplicity $e=e_V$.

This leaves us with only three cases

1. $U=V_{|N}$ is itself simple.

2. $V_{|N} \cong U\oplus U$ for some simple $U$ and conjugation switches the two copies of $U$.

3. $V_{|N} \cong U_1\oplus U_2$ for two non-isomorphic, but conjugated simple modules $U_1, U_2$.

   In that case $I_{U_i} = 1$ and $V$ is obtained by induction from either constituent, i.e. $V=\operatorname{Ind}_1^Q(U_i) = A \otimes_{A_N} U_i$. Conversely: The induction of every simple $A_N$-module $U$ with $I_U=1$ is a simple $A$-module whose restriction back to $A_N$ splits into $U$ and its conjugate.

To count real and quaternionic representation, note that $Q$ also acts via conjugation on $\operatorname{End}_{A_N}(X)$ for all $A_N$-modules $X$ and that $\operatorname{End}_A(X)$ is precisely the space of $Q$-fixed points of this action.

Consider in particular $X=V_{|N}$ and note that $\operatorname{End}_{A_N}(U) = \mathbb{C}$, because $\mathbb{C}\subseteq Z(A_N)$ so that every module is automatically a $\mathbb{C}$-vector space and the $A_N$-action is by $\mathbb{C}$-linear maps, so that we can apply Schur's lemma for finite-dimensional algebras over the complex numbers.

Let's look at $X:=V_{|N}$ in the three cases:

1.) Then $\operatorname{End}_{A_N}(X) = \mathbb{C}$. The only possible actions of $Q$ are the trivial action and complex conjugation. $V$ cannot be quaternionic and $V$ is real iff it is the latter.

2.) Then $\operatorname{End}_{A_N}(X) = \mathbb{C}^{2\times 2}$. If $Q$ acts $\mathbb{C}$-linearly, then it must act by conjugation by Skolem-Noether. If it acts semilinearly, then it acts by conjugation followed by complex conjugation. Because it switches the two copies of $U$, the conjugating matrix must be of the form $X=\begin{pmatrix}0&\ast\\\ast&0\end{pmatrix}$. The linear case gives $\mathbb{C}$, the semilinear case $\mathbb{H}$ as fixed point space.

3.) Then $\operatorname{End}_{A_N}(X) =\operatorname{End}_{A_N}(U_1)\times\operatorname{End}_{A_N}(U_2)=\mathbb{C} \times \mathbb{C}$. The conjugation permutes $U_1$ and $U_2$ and so must flip the two copies of $\mathbb{C}$. Again, because there are only two $\mathbb{R}$-automorphisms of $\mathbb{C}$, we only have a limited number of ways, this can happen: $$(z,w) \mapsto \begin{cases} (w,z) \\ (\overline{w},z) \\ (w,\overline{z}) \\ (\overline{w},\overline{z}) \end{cases}$$ with fixed point space $$\begin{cases} \{(z,z) \mid z\in\mathbb{C}\} \\ \{(r,r) \mid r\in\mathbb{R}\} \\ \{(r,r) \mid r\in\mathbb{R}\} \\ \{(z,\overline{z}) \mid z\in\mathbb{C}\} \end{cases} $$ In the first and fourth case, $V$ is neither real nor quaternionic. In the second and third case, $V$ is real.

To conclude: The number of real representations of $A$ is the number of simple $A_N$-modules that are $G/N$-invariant, extend to $A$-modules with the above $Q$-action, plus the number of simple $A_N$-modules that are not $G/N$-invariant and have the right $Q$-action, divided by two