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Given a finite group $G$ and a field $K$ such that the characteristic of $K$ does not divide $|G|$, Maschke's Theorem states that if $V$ is a $KG$-module and $V_1 \subseteq V$ is a submodule, then there exists a complementary submodule $V_2 = \ker q \subseteq V$, where $q \in \text{Hom}_{KG}(V,V)$ is given as $q(v) := \frac{1}{|G|} \sum_{g\in G} g(\pi(g^{-1}v)) $, and $\pi \in \text{Hom}_K (V,V_1)$ is a $K$-linear projection. My question is related to actually writing down (a $K$-basis for) $V_2$. Is it fair to say that so long as the hypotheses are satisfied I can think of the problem as if it were just a $K$-vector space? In other words, if I can write down $V_1, V_2$ as $K$-vector spaces satisfying $V_1 \oplus V_2 = V$ as a $K$-vector space, then $V_2$ as a $KG$-module has the same basis as $V_2$ as a $K$-vector space?

For context, the concrete problem is to have $\mathbb{Q}S_3$ acting on $\mathbb{Q}\{x_1,x_2,x_3\}$ in the natural way. Then I have a known submodule $V_1$, and I want to write down the complementary submodule given by Maschke's Theorem. My thought process is to just forget the $\mathbb{Q}S_3$ structure of $V_1$ and deal with it as just as a three-dimensional $\mathbb{Q}$-vector space. Then I can find a basis for $V_2$ satisfying $V_1 \cap V_2 = \emptyset$ and $V_1 \oplus V_2 = \mathbb{Q}^3$. Then I checked if that basis generates a $\mathbb{Q}S_3$-submodule with the necessary properties.

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  • $\begingroup$ No, there's no reason $V_2$ should be invariant. $\endgroup$ – Qiaochu Yuan Jan 25 at 18:59
  • $\begingroup$ @QiaochuYuan is the existence of $G$-invariant $V_2$ not the guarantee of Maschke's Theorem? $\endgroup$ – perpetuallyconfused Jan 25 at 19:38
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    $\begingroup$ Yes, but what you aren't guaranteed is that every complement is $G$-invariant, and most of them won't be in general. $\endgroup$ – Qiaochu Yuan Jan 25 at 20:32
  • $\begingroup$ @QiaochuYuan thanks! $\endgroup$ – perpetuallyconfused Jan 25 at 21:31
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Yes. You're right that this is Maschke's Theorem. No. You're not quite right that we can take $V_2$ itself to be invariant.

Consider $G = \mathbb{Z} / 2$ acting on $k^2$ by reflection, for instance. Then obviously the axis of reflection is a sub-$kG$-module, and any other (one dimensional) subspace is a complement of vector spaces. However, it's clear that the only choice of complement that is itself a sub-$kG$-module is the line orthogonal to the axis of reflection (do you see why?).

What Maschke's Theorem buys us is the existence of a complement that's also a $kG$-module, but it might not be the one we started with. When we perform the "averaging" maneuver, we will almost certainly get back a different complement.


So, to answer your original question: How do we get the basis of a $kG$-complement? It may be easy to guess one, but it would be nice to know how to do it in general.

Let $U \subseteq V$ be a sub-$kG$-module. All the usual caveats apply ($G$ should be finite and its order should be coprime to the characteristic of $k$). Then if $\pi : V \to U$ is any old projection of vector spaces, we can average it to get

$$\pi_G = \frac{1}{|G|} \sum_{g \in G} g \pi g^{-1}$$

Then (for a certain definition of "easy to see") it's easy to see that $\pi_G$ is still a $k$-linear map from $V$ onto $U$ (indeed, it still splits the inclusion map), but now it has the bonus property of being a $kG$-linear map (do you see why?). So $\text{ker}(\pi_G)$ (which is itself a $kG$-module) is the desired complement.

Now we're in the home stretch, though, because we can regard $\pi_G$ as an ordinary $k$-linear map. So finding a basis for its kernel is "just linear algebra" in the sense that any of us could do it (or at least, we should be able to... I, for one, am probably rusty). But a computer algebra system like sage can absolutely slaughter this problem. As a (fun?) exercise, you might write up this algorithm in sage. It's not so bad, and it'll convince you that you understand it.

Of course, for many problems there's an "obvious" complement to your subspace. In practice, the guess-and-check method that you employed with the obvious complement is a good way to go. If worse comes to worse, though, your only recourse might be some honest computation:

  1. Fix any old $k$-linear projection $\pi : V \to U$
  2. Average it to $\pi_G : V \to U$
  3. Compute a basis for $\text{ker}(\pi_G)$

I hope this helps ^_^

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