# When can a finite subgroup of $GL(2n,\mathbb{R})$ be viewed as a subgroup of $GL(n,\mathbb{C})$?

A finite group acting on a complex vector space of dimension $n$ can be seen as acting on a real vector space of dimension $2n$ just by forgetting the complex structure of the space. My question is, if I am handed a real vector space $V$ of dimension $2n$, and a group $G$ acting on it, is there a test I can perform to determine if the action arose from a complex action in this way?

Sometimes it is easy to rule out: for example, if $G$ contains anything orientation-reversing, then clearly it doesn't arise in this way. Or if one knows enough about $G$ (abstractly as a group) to know it doesn't have any faithful $n$-dimensional representation. But I would like an if-and-only-if criterion:

Is there a test I can perform on the pair $G,V$ to determine whether the action of $G$ can be obtained by beginning with a complex $n$-dimensional representation and forgetting the complex structure of the vector space?

To make a little more precise what I mean by "can be obtained": if there is an element $J\in GL(V)$ that commutes with the action of $G$ and satisfies $J^2 = - I$, then the action of $G$ "can be obtained from a complex $n$-dimensional action by forgetting the complex structure", since one can regard $V$ as a complex vector space via the action of $J$. So the question is, if I am handed $G$ and $V$, is there a test for the existence of such a $J$?

• If you were looking for a practical algorithm, then you would need more information on the input. Are the entries in the matrices given as algebraic numbers? Is the group sufficiently small that you can readily compute with it, such as finding representatives of its conjugacy classes? – Derek Holt Jul 9 '16 at 8:30
• @DerekHolt - the question was aimed both at being able to do explicit calculations in small cases and also at being able to reason about it nonexplicitly in the general case. Qiaochu's answer (and your comment on it) have given me everything I need. – benblumsmith Jul 9 '16 at 15:41

## 2 Answers

It's cleaner to ask about an arbitrary finite-dimensional real representation $V$ of a finite group $G$; the hypothesis that $V$ is faithful isn't particularly helpful. $V$ has a decomposition $\bigoplus_i n_i V_i$ into irreducible components with multiplicities, and so its endomorphism algebra takes the form

$$\text{End}(V) \cong \prod_i M_{n_i}(D_i)$$

where $D_i = \text{End}(V_i)$ are division algebras over $\mathbb{R}$ by Schur's lemma, so either $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. The question is when there is a morphism (necessarily a monomorphism) $\mathbb{C} \to \text{End}(V)$ of $\mathbb{R}$-algebras, and the answer is iff there is such a morphism into each $M_{n_i}(D_i)$, hence for each $i$ either

• $D_i = \mathbb{R}$ and $n_i$ is even, or
• $D_i = \mathbb{C}$ or $\mathbb{H}$.

We can test for this as follows. If $W$ is an irreducible real representation, then $\text{End}(W \otimes \mathbb{C}) \cong \text{End}(W) \otimes \mathbb{C}$ (all tensor products here and below taken over $\mathbb{R}$), and so exactly one of three things happens:

• $\text{End}(W) \cong \mathbb{R}$, so $\text{End}(W \otimes \mathbb{C}) \cong \mathbb{C}$, meaning that $W \otimes \mathbb{C}$ remains irreducible.
• $\text{End}(W) \cong \mathbb{C}$, so $\text{End}(W \otimes \mathbb{C}) \cong \mathbb{C} \otimes \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$, meaning that $W \otimes \mathbb{C}$ is a direct sum of two complex conjugate and nonisomorphic irreducibles.
• $\text{End}(W) \cong \mathbb{H}$, so $\text{End}(W \otimes \mathbb{C}) \cong \mathbb{H} \otimes \mathbb{C} \cong M_2(\mathbb{C})$, meaning that $W \otimes \mathbb{C}$ is a direct sum of two isomorphic irreducibles.

These three cases can be distinguished by the value of

$$\langle W, W \rangle = \frac{1}{|G|} \sum_{g \in G} \chi_W(g)^2$$

as Claudio says; it takes the values $1, 2, 4$ in the above three cases. With this modification to the orthogonality relations you can try to figure out the decomposition of $V$ into real irreducible representations and then compute the $n_i$ and the $D_i$ using the above test.

See also the Frobenius-Schur indicator for some discussion of how to classify the real irreducible representations given knowledge of the complex irreducible representations.

• So isn't the simple answer that you calculate its (complex) character, decompose it into irreducibles, and check whether it is the sum of pairs of conjugate characters? – Derek Holt Jul 9 '16 at 8:28
• @Derek: you're right. So in fact it's unnecessary to know anything about the irreducibles over $\mathbb{R}$. – Qiaochu Yuan Jul 9 '16 at 9:30
• +1 This whole discussion is just what I needed. – benblumsmith Jul 9 '16 at 15:37

Note that, since $G$ is a finite group, there is an invariant inner product on $V$. The results we need can be derived from the Schur orthogonality relations.

In the irreducible case, $V$ has an invariant complex structure iff the character function has $L_2$-norm equal to $\sqrt2$ or $2$ (otherwise it is equal to $1$). In fact, $V\otimes \mathbb C=U\oplus \bar U$ for a complex representation $U$ and $\chi_V=\chi_{V\otimes\mathbb C}=\chi_U+\chi_{\bar U}$ where $\chi_U$ and $\chi_{\bar U}$ are orthogonal (unit) vectors in case $U$ and $\bar U$ are inequivalent, or $U$ and $\bar U$ are equivalent and then $V$ admits even a quaternionic structure.

In the general case, the criterion is that the irreducible components that are not as above must occur in (equivalent) pairs $(W,W)$, as Qiaochu wrote. On $W\oplus W$ we have $\left(\begin{array}{cc}0&-\mathrm{id}\\\mathrm{id}&0\end{array}\right)$ as invariant complex structure.

• What's $W$? Do you mean $V$? – benblumsmith Jul 9 '16 at 4:03