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:



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> 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?
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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 $J$?



When can a finite subgroup of $GL(2n,\mathbb{R})$ be viewed as a subgroup of $GL(n,\mathbb{C})$?
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](https://en.wikipedia.org/wiki/Frobenius%E2%80%93Schur_indicator) for some discussion of how to classify the real irreducible representations given knowledge of the complex irreducible representations. 


