Representing sums of matrix algebras as group rings

Question

Let $A = M_{n_1}(\mathbb R) \oplus M_{n_2}(\mathbb R) \oplus ... \oplus M_{n_m}(\mathbb R)$ be a direct sum of real matrix algebras. Under what conditions does there exist a group ring $\mathbb R[G]$ which is isomorphic to $A$?

I know every group ring is isomorphic to some such $A$ by Wedderburn's theorem, and I want to determine the extent to which the converse holds. I know that $A$ must have a $\mathbb R$-summand, corresponding to the linear span of $\sum_{g\in G} g \in \mathbb R[G]$. Is this condition sufficient as well? If not, is there a nice criterion, at least for small numbers of summands?

Answer 1

This is similar to Qiaochu Yuan's answer (+1) but makes the claims more strongly and uses Miller's classification of groups with very few conjugacy classes.

Suppose $G$ is a finite group such that $\mathbb{R}[G]$ is the direct product of matrix rings $M_{n_i}(\mathbb{R})$ for $n_1 \leq n_2 \leq \dots \leq n_m$.

The original question claims this is true for any group, but actually this places fairly severe restrictions on the group which we can use to classify those groups with small $m$. Since all representations are real, we have that the $n_i$ are the character degrees of $G$, and so we have that $n_i$ divides $\sum n_i^2 = |G|$ and that $m$ is the number of conjugacy classes of $G$ (which must all be real). The number of the $n_i$ that are equal to 1 is the index $[G:G']$, and in particular is at least one. Since all the representations are real, we also have that $G$ contains exactly $\sum n_i$ elements of order dividing 2, and so $|G|$ is even.

If $m=1$, there is only one degree and we know it is $n_1=1$ so that $|G|=1$ and $G$ is the trivial group.

If $m=2$, then $G$ has only two conjugacy classes, so $|G|=2$ and $G$ is cyclic of order 2 with $n_1=n_2=1$.

If $m=3$, then $G$ has only three conjugacy classes, so $|G| \in \{3,6\}$, but only $G$ the nonabelian group of order 6 has only real conjugacy classes, so $n_1= n_2=1$ and $n_3=2$.

If $m=4$, then $G$ has only four conjugacy classes, so $|G| \in \{4,10,12\}$, but only $G = C_2 \times C_2$ and $G = D_{10}$ work, giving either $n_1=n_2=n_3=n_4=1$ or $n_1=n_2=1$ and $n_3=n_4=2$. $$\begin{array}{ccccc|c} n_1 & n_2 & n_3 & n_4 & G \\ \hline 1 & 1 & 1 & 1 & C_2 \times C_2 \\ 1 & 1 & 2 & 2 & D_{10} \end{array}$$

If $m=5$, then we get only four possibilities: $$\begin{array}{ccccc|c} n_1 & n_2 & n_3 & n_4 & n_5 & G \\ \hline 1 & 1 & 1 & 1 & 2 & D_8 \\ 1 & 1 & 2 & 2 & 2 & D_{14} \\ 1 & 1 & 2 & 3 & 3 & S_4 \\ 1 & 3 & 3 & 4 & 5 & A_5 \end{array}$$

Answer 2

This is in fact not what Artin-Wedderburn says over $\mathbb{R}$. Artin-Wedderburn says that $\mathbb{R}[G]$ is a product of matrix algebras over (finite-dimensional) real division algebras. Over $\mathbb{R}$ there are three such algebras, namely $\mathbb{R}, \mathbb{C}, \mathbb{H}$.

Which of these algebras appear can be computed from the character table of $G$ (see Frobenius-Schur indicator). In order for $\mathbb{R}$ to be the only division algebra that occurs, the Frobenius-Schur indicator of every irreducible representation of $G$ over $\mathbb{C}$ must be equal to $1$. In particular, every irreducible representation of $G$ over $\mathbb{C}$ must be self-dual, which is equivalent to the character table being real, which is in turn equivalent to every element of $G$ being conjugate to its inverse (ambivalent groups).

If the above condition holds, then the $n_i$ are precisely the dimensions of the real irreducible representations of $G$. There are various conditions that these dimensions need to satisfy (for example, the dimensions of the complex irreducible representations must divide $|G|$, so I think for each $i$ either $n_i$ or $\frac{n_i}{2}$ must divide $|G|$, or something like that) and I am doubtful that anything simple can be said in general.

Note that we must have $|G| = \sum n_i^2$, so if this number is particularly simple (e.g. prime) then you can hope to classify all of the finite groups of order $|G|$ and compute their character tables.

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Here are some things that can be said for very small values of $m$.

$m = 1$: Every group has the trivial representation, so one of the $n_i$ must be equal to $1$. So in the case of one summand, the only possibility is $A = \mathbb{R}$ and $G$ is the trivial group.

$m = 2$: We must have $A = \mathbb{R} \times M_n(\mathbb{R})$ for some $n$ and $G$ must be a finite group with $|G| = n^2 + 1$ such that $G$ has only one nontrivial real irreducible representation. This is a very difficult constraint to satisfy: it implies that $G$ has an irreducible complex representation of order either $n$ or $\frac{n}{2}$, hence either $n$ or $\frac{n}{2}$ divides $n^2 + 1$. The only $n$ with this property are $n = 1, 2$. In the second case $n^2 + 1 = 5$ and the cyclic group of order $5$ does not work, so we can only have $G = C_2$ and $A = \mathbb{R} \times \mathbb{R}$.

Alternatively, it's known that the number of real irreducible representations of $G$ is the number of "real conjugacy classes" of $G$, where real conjugacy is the equivalence relation generated by conjugacy and inverses. Since elements of different order can't be real conjugate, the number of real conjugacy classes is at least one more than the number of prime factors of $|G|$, hence if this number is equal to $2$ then $|G|$ must be prime and in fact must be equal to $2$ as well.