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Maschke's theorem tells us that any representation of a finite group $G$ can be decomposed into a direct sum of irreducible representations. The proof does make intuitive sense to me (Intuition behind Maschke's theorem), but my question is really about why we should expect it to be true in the first place.

Sure, we can do some explicit examples (perhaps a most obvious starting point is the standard permutation representation of $S_n$, which has an obvious invariant subspace $\{(x,x,\ldots, x) \in \mathbb C^n: x\in \mathbb C\}$, and almost as obvious complement subspace $\{(x_1,\ldots, x_n) \in \mathbb C^n: \sum x_i = 0\}$), and perhaps one has already seen the result for abelian groups over $\mathbb C$ in the guise of linear algebra: Matrices commute if and only if they share a common basis of eigenvectors?.

However I wonder if there is any other point of view that makes Maschke's theorem feel "inevitable"...since right now, it just seems absurdly powerful/magical.

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    $\begingroup$ You should expect it to be true by induction as long as subreps have complements. So that's really the thing you should be asking about. And that's explained well by the accepted answer to the first question you linked about intuition. So I consider your question already answered. It's less about "expecting" it to be true and more about "following your nose" (assuming you have the olfactory prerequisites). $\endgroup$
    – runway44
    May 27 at 6:25
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    $\begingroup$ Honestly, I don't understand this question. What do you mean by "expect it to be true"? It either is or is not. The intuition here is that finite groups behave similarly to the trivial group (meaning representations are just vector spaces, and every vector space decomposes as a product irreducible vector space). $\endgroup$
    – freakish
    May 27 at 6:47
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    $\begingroup$ @freakish there are some results in math that one "expects" to be true, for example differentiable implies continuity. There are some results in math that seem completely miraculous, e.g. "a locally uniform limit of complex differentiable functions is complex differentiable". The latter fact can become more "expected" if one knows that complex differentiability has an equivalent formulation in terms of an integral, and integrals behave well under locally uniform convergence. $\endgroup$
    – D.R.
    May 27 at 7:35
  • $\begingroup$ @runway44 yes it's true that there's a certain amount of "just follow your nose" to it, but I still think there's something unreasonably miraculous about it. Like in the case of groups, it's not true that every group can be decomposed as the "direct sum" (Cartesian product) of simple groups. Of course we have that in general modules over semisimple rings have this nice decomposition property, but how did group rings "get so lucky" as to have the honor of being ordained semisimple? $\endgroup$
    – D.R.
    May 27 at 7:45
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    $\begingroup$ @D.R. let me put it differently: whether something is "expected", "unexpected" or "miraculous" is a personal opinion. It depends on your expectations. It is not a mathematical question. And so how are we supposed to answer that? $\endgroup$
    – freakish
    May 27 at 7:56

4 Answers 4

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If $\rho \colon G \to \operatorname{GL}(V)$ is a representation of a finite group $G$ on a finite-dimensional $\mathbb{C}$-vector space $V$, then $V$ can be equipped with an inner product $\langle -, - \rangle \colon V \times V \to \mathbb{C}$ such that every element of $G$ acts by a unitary matrix. This inner product is not mysterious either: take any old inner product $(-, -) \colon V \times V \to \mathbb{C}$ and average it over the group: $$ \langle u, v \rangle := \frac{1}{|G|} \sum_{g \in G} (\rho(g) u, \rho(g) v).$$ (The same is true over $\mathbb{R}$ if we replace "unitary" with "orthogonal"). The slogan is: we can always find a basis such that every $\rho(g)$ is a unitary matrix.

This makes Maschke's theorem completely inevitable: if a subspace $W \subseteq V$ is invariant under a unitary operator $\rho(g)$, then its orthogonal complement $W^\perp$ is also invariant under $\rho(g)$. So every subrepresentation has a complement, and moreover, doing standard orthogonal projection stuff is an effective way to decompose a vector into its components in subrepresentations.

Of course, Maschke's theorem is true in more generality than just $\mathbb{R}$ and $\mathbb{C}$, and this inner product argument really only works over them, but I think it is illustrative of why it is completely inevitable in certain contexts, and many of the other settings end up being related to these settings anyway.

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    $\begingroup$ FWIW, this is also my point of view concerning the inevitability of Maschke's theorem (in characteristic not dividing the order of the group). It displays its important geometric aspect. $\endgroup$ May 27 at 10:28
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    $\begingroup$ Ahh, so the "right" question to ask is not whether a group has a representation, but rather whether a group has a unitary representation (which I think fits the geometric intuition for group representation more anyways; like ). What you wrote tells us that any representation can be made into a unitary representation. I guess we can now generalize my question to "why should unitary representations be so nice" ... e.g. in the general case of locally compact groups $G$, the Gelfand-Raikov theorem tells us that there are many unitary representations, enough to separate points! $\endgroup$
    – D.R.
    May 27 at 19:21
  • $\begingroup$ Also relevant: math.stackexchange.com/questions/717000/… $\endgroup$
    – D.R.
    May 27 at 19:22
  • $\begingroup$ @D.R. Is there still something in your initial question that was not sufficiently answered by Joppy? $\endgroup$
    – M. Winter
    Jun 13 at 17:54
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Maschke's theorem holds for all fields, not just characteristic $0$, and including characteristics dividing the order of the group. You just have to be more careful about how you state it.

If $G$ is a finite group with subgroup $H$, $k$ is a field, a $kG$-module $V$ is relatively $H$-projective if, whenever $U$ is a $kG$-submodule of $V$, and there is a complement to $U$ as a $kH$-module, there is a $kG$-module complement.

Maschke's theorem, in full generality, is the following.

Theorem: Let $G$ be a finite group and let $k$ be a field of characteristic $p\geq 0$. Let $P$ be a Sylow $p$-subgroup of $G$ (where $P=1$ if $p=0$). Every $kG$-module is relatively $P$-projective.

For $p=0$ or $p\nmid G$, this says that every $kG$-module is relatively $1$-projective. Writing this out, that says that if a $kG$-submodule has a $k1$-complement, so a vector space complement (which it obviously does) then it has a $kG$-complement.

This removes the requirement that $p\nmid |G|$, and places Maschke's theorem in a more general context. The intuition here comes from the theory of projective modules, but is unfortunately too advanced for most first courses in representation theory.

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Finite groups (and more generally compact groups) have an averaging/integration operation that can be used to produce projections onto the trivial submodule. Supposing $V$ is a finite-dimensional $G$-module with $G$ a finite group, then the linear map $\pi:V\to V$ defined by $$ \pi(v) = \frac{1}{\lvert G\rvert } \sum_{g\in G} gv$$ is both a homomorphism of $G$-modules and a projection, where the subspace $\pi(V)\subseteq V$ has a trivial $G$-action. This projection, by the way, comes from the action of the element $\frac{1}{\lvert G\rvert}\sum_g g$ in the group algebra $\mathbb{C}[G]$, a sort of average over the group itself. Linearity is key here, which is why we don't see a similar sort of thing with plain $G$-actions.

An important example of what this is able to do is give a description of $\hom_G(V,W)$ where $V$ and $W$ are $G$-modules. The characterization of $f\in\hom_G(V,W)$ is that it is a linear map $f\in\hom_{\mathbb{C}}(V,W)$ with the property that $f(gv)=gf(v)$ for all $v\in V$ and $g\in G$. Or, in other words, that $gf(g^{-1}v)=f(v)$. The space $\hom_{\mathbb{C}}(V,W)$ is a $G$-module by the action defined by $(g\cdot f)(v) = gf(g^{-1}v)$ (this corresponds to the action of $G$ on $V^*\otimes W$, which is isomorphic to $\hom_{\mathbb{C}}(V,W)$), so $\hom_G(V,W)$ is nothing other than the trivial submodule of $\hom_{\mathbb{C}}(V,W)$, which by the above analysis gives us that $$\hom_{G}(V,W) = \pi(\hom_{\mathbb{C}}(V,W)).$$ Another way we can use this is to get decompositions. For example, if $W\subseteq V$ is a $G$-submodule, then the above projection lets us take any map $f:V\to W$ and get a $G$-module homomorphism $\pi(f)$. If $f$ were to have the property that $f|_W=\mathrm{id}_W$, then one can check that $\pi(f)$ does as well since $W$ is a $G$-module. Thus, by letting $f$ be any linear map extending $\mathrm{id}_W$, one gets that $\ker(\pi(f))$ is a complementary subspace to $W$. That's the key step for Maschke's theorem!

So, the inevitability of the theorem might be that finite groups have an averaging/integration operation that can be used to get the trivial part of a $G$-module, and that is very useful when paired with the fact that the trivial part of the space of linear maps is the space of $G$-module homomorphisms.

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  • $\begingroup$ I'm not very familiar with representation theory outside the finite or (very similar) compact case over $\mathbb{C}$ (e.g., modular or $L^2$ representations). Given that the main idea of Maschke's proof is the existence of a projection on $\mathbb{C}[G]$, does the representation theory of amenable groups also look like the finite case? $\endgroup$
    – anomaly
    May 27 at 12:32
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    $\begingroup$ @anomaly I'm not sure. My understanding is that you can't get this sort of averaging operation out of an amenable group's mean ($\mathbb{Z}$ is an amenable group for example), but amenable groups seem to have the notion of "almost invariant vectors" and "weak containment of representations," so maybe there's a notion of weak decomposition? $\endgroup$ May 27 at 13:39
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As sort of an anti-answer, why should we expect it to be true? Isn't it a bit surprising when you first see it?

Just look at representations of the most basic infinite group, the infinite cyclic group $\mathbb{Z}$. The representations correspond to matrices, and their decomposition corresponds to the Jordan normal form. For example, the representation $$f: (\mathbb{Z},+) \rightarrow GL_2(\mathbb{C}),\ f(x) = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$$ is not completely reducible.

So what about finite groups? Look at the simplest example, a cyclic group $G = \langle g \rangle$, say $|g| = n$. For a representation $f: G \rightarrow \operatorname{GL}_d(\mathbb{C})$, you have $f(g)^n = I_d$. So the minimal polynomial of $f(g)$ divides $x^n - 1$, which is a polynomial with no repeated roots. Hence $f(g)$ is diagonalizable, meaning after a change of basis $f(g)$ is a diagonal matrix $$f(g) = \begin{pmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_d \end{pmatrix}.$$

Hence representations of finite cyclic groups are completely reducible.

But why should we expect representations of all finite groups to be completely reducible? I do not think it is obvious at all a priori.

And also it's not true over fields of positive characteristic, here is one example: take a finite cyclic group $G = \langle g \rangle$ of prime order $p$. Then $$f: G \rightarrow \operatorname{GL}_2(\mathbb{Z} / p\mathbb{Z}), f(g^k) = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$$ is a representation which is not completely reducible.

So you somehow need to use the fact that the group is finite, and the fact that the field is of characteristic zero (or more generally, of characteristic coprime to the order of the group). As shown in the other answers, a certain averaging argument does the trick.

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