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?

  • $\begingroup$ Related: mathoverflow.net/questions/13502/… $\endgroup$ – anon May 24 '13 at 2:18
  • $\begingroup$ It would be correct to say that "every group ring of a finite group over $\Bbb R$ is semisimple by Maschke's theorem, hence by Wedderburn's theorem is a product of matrix rings over $\Bbb R,\Bbb C$ or $\Bbb H$ (not necessarily all the same dimension or division ring)." $\endgroup$ – rschwieb May 24 '13 at 2:22
  • $\begingroup$ A short rebuttal is that for $|G|>1$, the augumentation ideal is a nontrivial ideal. But all square matrix rings over $\Bbb R$ are simple, so none of them are going to be group rings... $\endgroup$ – rschwieb May 24 '13 at 19:39

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: $$\begin{array}{ccccc|c} n_1 & n_2 & n_3 & n_4 & n_5 & G \\ \hline 1 & 1 & 1 & 1 & 2 & D_8 \\ 1 & 1 & 2 & 2 & 2 & D_{14} \\ 1 & 1 & 2 & 3 & 3 & S_4 \\ 1 & 3 & 3 & 4 & 5 & A_5 \end{array}$$

  • $\begingroup$ Ah, I was being silly. I was doing a lot of hard work in the case $m = 3$ because I didn't realize that ambivalent means the number of conjugacy classes is the number of real conjugacy classes... do you know whether the number of possibilities is always finite for fixed $m$? $\endgroup$ – Qiaochu Yuan May 24 '13 at 3:25
  • $\begingroup$ Yeah, almost all of the observations are already in your answer. Miller's classification could actually be reworded to fit into this framework (cuts down on one or two cases where $[G:G'] = 3$) but this answer started out as a comment and is already a long answer :P $\endgroup$ – Jack Schmidt May 24 '13 at 3:29
  • 1
    $\begingroup$ Yes, this is a special case of Landua's theorem that there are only finitely many isomorphism classes of finite groups with a fixed number of conjugacy classes (then only take the ones with all indicators = 1). $\endgroup$ – Jack Schmidt May 24 '13 at 3:30
  • $\begingroup$ Somewhere there is a good paper on groups all of whose reps are real, but I haven't remembered the title even with google's assistance. $\endgroup$ – Jack Schmidt May 24 '13 at 3:31
  • 1
    $\begingroup$ This isn't it, but is nice anyways: $G$ must be generated by involutions, Wang–Grove (1988) ams.org/mathscinet-getitem?mr=963550 $\endgroup$ – Jack Schmidt May 24 '13 at 3:36

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.

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.

  • 1
    $\begingroup$ I added m=3 through 5 and noted that the hypothesis in the question implies the $n_i$ are exactly the character degrees and $m$ is also the number of any kind of conjugacy class. $\endgroup$ – Jack Schmidt May 24 '13 at 3:24
  • $\begingroup$ Oh, of course. That's clear after tensoring with $\mathbb{C}$ but somehow I didn't see it. $\endgroup$ – Qiaochu Yuan May 24 '13 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.