Studding Character theory, and been bouncing back and forth between reading Dummit and Foote, and Character theory of finite groups by Martin Issacs.

In section 18.1 of Dummit and Foote, we are given a Bijection: $$\{V: V \text{ is an } FG-\text{module}\} \leftrightarrow \{ (\phi, V): \text{is an } F-\text{ represention from } G \rightarrow GL(V)\}$$

at the start of chapter 2 of Character theory of finite groups, It states that it is possible to have fields $F < E$, and an irreducible $F$-representation $X$, such that $X$ is reducible as an $E$-representation.

I don’t understand how $X$ can be irreducible as an $F$-representation and reducible as an $E$-representation. Since if $X$ was a representation with corresponding module $V$, and if $\phi$ and $\psi$ were $E$-representations with corresponding $E$-modules $W$ and $U$, such that $X = \phi + \psi$. Then shouldn’t $W$ and $U$ also be $F$-modules and $V = W + U$, so there would have the be $F$-representations that sum up to $X$?

Somone gave me the example: "$X : C_3 \rightarrow GL_2(\mathbb{R})$, where $C_3 = \langle g \rangle$ is cyclic of order $3$ and $X(g)$ is the matrix whose first row is $[ 0 \ \ 1 ]$ and whose second row is $[ -1 \ \ -1 ]$. The eigenvalues of this matrix are not real (they are the two primitive cube roots of $1$). Therefore, $X$ is irreducible as an $\mathbb{R}$-representation of $C_3$. Since $X(g)$ is diagonalizable over the complex numbers $\mathbb{C}$, it follows that as a representation, $X : C_3 \rightarrow GL_2(\mathbb{C})$ reduces. So, over $\mathbb{C}$, the representation is equivalent to a block diagonal sum of two linear representations of $C_3$

I understand the example, but I'm still not seeing where and why my logic fails?

  • 4
    $\begingroup$ The point is that you start with a representation $V$ over $F$, not over $E$. Then reducibility over $E$ means that the corresponding representation $G\to \operatorname{GL}(V\otimes_F E)$ is reducible, that is you get an invariant submodule $W \subset V\otimes_F E$. Such a thing need not be coming from an invariant submodule of $V$, as the example shows. $\endgroup$
    – Alex K
    Jun 11, 2022 at 21:20
  • $\begingroup$ I think my issue was that I was taking the $F$-module $V$ corresponding to the $F$-representation $X$, as the $E$ module corresponding to $X$ when viewed as an $E$-representation. It makes sense that the modules shouldn't be the same, since every $F$ module is not necessarily an $E$-module $\endgroup$ Jun 11, 2022 at 22:07

3 Answers 3


I think this is based in a misunderstanding of how we can convert between $E$ and $F$ vector spaces. For concreteness, let's work with $\mathbb{C}$ and $\mathbb{R}$, but you'll see that the same idea works for any field extension.

You're entirely right that, if we have a decomposition of $\mathbb{C}G$ modules $V = U \oplus W$, then we can apply restriction of scalars to get a decomposition of $\mathbb{R}G$ modules $V = U \oplus W$.

But here's the key thing! When we pass from $\mathbb{C}$ vector spaces to $\mathbb{R}$ vector spaces, the dimension doubles. If $V$ used to be $n$-dimensional as a $\mathbb{C}$-space, it's now $2n$-dimensional as an $\mathbb{R}$-space. Why is this important? Let's look at your example:

Given a representation of $C_3 = \langle g \rangle$ where $g \mapsto \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$ in $\mathsf{GL}_2(\mathbb{C})$, then (as Dylan says in their answer) $V = \mathbb{C}^2$ decomposes as

$$ \mathbb{C}^2 = \left \langle \begin{pmatrix} \omega \\ 1 \end{pmatrix} \right \rangle \oplus \left \langle \begin{pmatrix} \omega^2 \\ 1 \end{pmatrix} \right \rangle $$

but now, if we apply restriction of scalars, what representations do we see?

$$ \mathbb{R}^4 = \left \langle \begin{pmatrix} - \frac{1}{2} \\ \frac{\sqrt{3}}{2} \\ 1 \\ 0 \end{pmatrix}, \ \begin{pmatrix} - \frac{\sqrt{3}}{2} \\ - \frac{1}{2} \\ 0 \\ 1 \end{pmatrix} \right \rangle \oplus \left \langle \begin{pmatrix} - \frac{1}{2} \\ -\frac{\sqrt{3}}{2} \\ 1 \\ 0 \end{pmatrix}, \ \begin{pmatrix} \frac{\sqrt{3}}{2} \\ - \frac{1}{2} \\ 0 \\ 1 \end{pmatrix} \right \rangle $$

here we're using the well known fact that $\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$, and restriction of scalars turns us from a 2D $\mathbb{C}$-vector space into a 4D $\mathbb{R}$-vector space (where $i$ acts in the obvious way).

Now, importantly, this is not the representation $g \mapsto \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$ in $\mathsf{GL}_2(\mathbb{R})$! Instead, it's a representation $C_3 \to \mathsf{GL}_4(\mathbb{R})$.

Notice your proof tells us that this 4D vector space decomposes, as indeed it does! But you're trying to conclude that the 2D representation $g \mapsto \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$ in $\mathsf{GL}_2(\mathbb{R})$ decomposes. Of course, these are very different representations!

What is the relationship? And what goes wrong?

Given $V = \mathbb{R}^2$ with this $C_3$ representation, can freely turn it into a $\mathbb{C}$ representation by taking the complexification $V \otimes \mathbb{C}$. But this new space will be a $2$D $\mathbb{C}$-space. In fact, it's exactly the 2D space that we know decomposes from earlier!

The difference is that, when we take the restriction of scalars of $V \otimes \mathbb{C}$, we don't get our old representation back! Again, we find the dimension doubled!

What's the tl;dr, then? If we start with a reducible $\mathbb{C}G$ representation, you're entirely correct that it gives us a reducible $\mathbb{R}G$ representation too -- in fact, your exact proof works.

However, if we're starting with an $\mathbb{R}G$ representation, it probably isn't a $\mathbb{C}G$ representation. We can make it a $\mathbb{C}G$ representation by tensoring with $\mathbb{C}$, but this changes the space (for instance, its dimension). So if $V \otimes \mathbb{C}$ is reducible, that doesn't necessarily tell us anything about reducibility of $V$!

As an aside, we can still ask when knowledge of $V \otimes \mathbb{C}$ can tell us things about $V$ itself. This turns out to be one of the motivating problems in "descent theory", and while this is difficult, it's fairly well understood.

I hope this helps ^_^


Let's apply your logic to the example, since clearly something must go wrong.

Here $F = \mathbb{R}$, $E = \mathbb{C}$, $X: C_3 \rightarrow GL_2(\mathbb{R})$ given by $g \mapsto X(g) = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$.

Then as a $\mathbb{C}-$representation $(X,V)$ decomposes into two one dimensional representations, and the generators of these one dimensional representations are the eigenvectors of $X(g)$, you can work these out to be $\begin{pmatrix} \omega \\ 1 \end{pmatrix}$ and $\begin{pmatrix} \omega^2 \\ 1 \end{pmatrix}$ where $\omega$ is a cube root of unity.

Thus as a $\mathbb{C}-$ representation $(X,V)$ decomposes into $\big \langle \begin{pmatrix} \omega \\ 1 \end{pmatrix} \big \rangle \,\oplus \, \big \langle \begin{pmatrix} \omega^2 \\ 1 \end{pmatrix} \big \rangle $ (this is your $U$ and $W$, if you like) .

However as an $\mathbb{R}-$representation you cannot take the span of these eigenvectors since $\omega \notin \mathbb{R}$, so you no longer have an eigenvector decomposition and so the representation doesn't decompose.


Take a matrix $2\times 2$ matrix of order $4$ $$M = \left( \begin{matrix} \cos \frac{\pi}{2} & - \sin \frac{\pi}{2} \\ \sin\frac{\pi}{2} & \cos \frac{\pi}{2}\end{matrix} \right)= \left( \begin{matrix} 0 & - 1 \\ 1 & 0\end{matrix} \right)$$

Consider the representation of $\mathbb{Z}/4 \mathbb{Z}$, $$\mathbb{Z}/4 \mathbb{Z} \ni k \mapsto M^k\in GL(2, \mathbb{R})$$

Now, since the matrix $M$ has no real eigenvalues, it cannot be brought to a lower triangular form over $\mathbb{R}$. This implies in particular that the above representation is irreducible over $\mathbb{R}$.

However, the matrix $M$ is diagonalizable over $\mathbb{C}$, so the above representation ( considered as a map to $GL(2, \mathbb{C})$ ) is a direct sum of two subrepresentations of dimension $1$, so reducible.


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