Maschke's theorem#Formulations

{{short description|Concerns the decomposition of representations of a finite group into irreducible pieces}}

In mathematics, Maschke's theorem,{{cite journal |last=Maschke |first=Heinrich |date=1898-07-22 |title=Ueber den arithmetischen Charakter der Coefficienten der Substitutionen endlicher linearer Substitutionsgruppen |language=German |trans-title=On the arithmetical character of the coefficients of the substitutions of finite linear substitution groups |journal=Math. Ann. |volume=50 |issue=4 |pages=492–498 |url=http://resolver.sub.uni-goettingen.de/purl?GDZPPN002256975 |jfm=29.0114.03 | mr = 1511011 |doi=10.1007/BF01444297 }}{{cite journal |last=Maschke |first=Heinrich |date=1899-07-27 |title=Beweis des Satzes, dass diejenigen endlichen linearen Substitutionsgruppen, in welchen einige durchgehends verschwindende Coefficienten auftreten, intransitiv sind |language=German |trans-title=Proof of the theorem that those finite linear substitution groups, in which some everywhere vanishing coefficients appear, are intransitive |journal=Math. Ann. |volume=52 |issue=2–3 |pages=363–368 |url=http://resolver.sub.uni-goettingen.de/purl?GDZPPN002257599 |jfm=30.0131.01 | mr = 1511061 |doi=10.1007/BF01476165 }} named after Heinrich Maschke,{{MacTutor|id=Maschke|title=Heinrich Maschke}} is a theorem in group representation theory that concerns the decomposition of representations of a finite group into irreducible pieces. Maschke's theorem allows one to make general conclusions about representations of a finite group G without actually computing them. It reduces the task of classifying all representations to a more manageable task of classifying irreducible representations, since when the theorem applies, any representation is a direct sum of irreducible pieces (constituents). Moreover, it follows from the Jordan–Hölder theorem that, while the decomposition into a direct sum of irreducible subrepresentations may not be unique, the irreducible pieces have well-defined multiplicities. In particular, a representation of a finite group over a field of characteristic zero is determined up to isomorphism by its character.

Formulations

Maschke's theorem addresses the question: when is a general (finite-dimensional) representation built from irreducible subrepresentations using the direct sum operation? This question (and its answer) are formulated differently for different perspectives on group representation theory.

= Group-theoretic =

Maschke's theorem is commonly formulated as a corollary to the following result:

{{math theorem|math_statement= $V$ is a representation of a finite group $G$ over a field $\mathbb{F}$ with characteristic not dividing the order of $G$ . If $V$ has a subrepresentation $W$ , then it has another subrepresentation $U$ such that $V=W\oplus U$ .{{sfn|Fulton|Harris|1991|loc=Proposition 1.5}}{{sfn|Serre|1977|loc=Theorem 1}}}}

Then the corollary is

{{math theorem|name=Corollary (Maschke's theorem)|math_statement= Every representation of a finite group $G$ over a field $\mathbb{F}$ with characteristic not dividing the order of $G$ is a direct sum of irreducible representations.{{sfn|Fulton|Harris|1991|loc=Corollary 1.6}}{{sfn|Serre|1977|loc=Theorem 2}}}}

The vector space of complex-valued class functions of a group $G$ has a natural $G$ -invariant inner product structure, described in the article Schur orthogonality relations. Maschke's theorem was originally proved for the case of representations over $\Complex$ by constructing $U$ as the orthogonal complement of $W$ under this inner product.

= Module-theoretic =

One of the approaches to representations of finite groups is through module theory. Representations of a group $G$ are replaced by modules over its group algebra $K[G]$ (to be precise, there is an isomorphism of categories between $K[G]\text{-Mod}$ and $\operatorname{Rep}_{G}$ , the category of representations of $G$ ). Irreducible representations correspond to simple modules. In the module-theoretic language, Maschke's theorem asks: is an arbitrary module semisimple? In this context, the theorem can be reformulated as follows:

{{math theorem|name=Maschke's Theorem|math_statement= Let $G$ be a finite group and $K$ a field whose characteristic does not divide the order of $G$ . Then $K[G]$ , the group algebra of $G$ , is semisimple.It follows that every module over $K[G]$ is a semisimple module.The converse statement also holds: if the characteristic of the field divides the order of the group (the modular case), then the group algebra is not semisimple.}}

The importance of this result stems from the well developed theory of semisimple rings, in particular, their classification as given by the Wedderburn–Artin theorem. When $K$ is the field of complex numbers, this shows that the algebra $K[G]$ is a product of several copies of complex matrix algebras, one for each irreducible representation.The number of the summands can be computed, and turns out to be equal to the number of the conjugacy classes of the group. If the field $K$ has characteristic zero, but is not algebraically closed, for example if $K$ is the field of real or rational numbers, then a somewhat more complicated statement holds: the group algebra $K[G]$ is a product of matrix algebras over division rings over $K$ . The summands correspond to irreducible representations of $G$ over $K$ .One must be careful, since a representation may decompose differently over different fields: a representation may be irreducible over the real numbers but not over the complex numbers.

= Category-theoretic =

Reformulated in the language of semi-simple categories, Maschke's theorem states

{{math theorem|name=Maschke's theorem|math_statement= If {{var|G}} is a group and {{var|F}} is a field with characteristic not dividing the order of {{var|G}}, then the category of representations of {{var|G}} over {{var|F}} is semi-simple.}}

Proofs

= Group-theoretic =

Let U be a subspace of V complement of W. Let $p_0 : V \to W$ be the projection function, i.e., $p_0(w + u) = w$ for any $u \in U, w \in W$ .

Define $p(x) = \frac{1}{\#G} \sum_{g \in G} g \cdot p_0 \cdot g^{-1} (x)$ , where $g \cdot p_0 \cdot g^{-1}$ is an abbreviation of $\rho_W{g} \cdot p_0 \cdot \rho_V{g^{-1}}$ , with $\rho_W{g}, \rho_V{g^{-1}}$ being the representation of G on W and V. Then, $\ker p$ is preserved by G under representation $\rho_V$ : for any $w' \in \ker p, h \in G$ ,

$\begin{align}
p(hw') &= h \cdot h^{-1} \frac{1}{\#G} \sum_{g \in G} g \cdot p_0 \cdot g^{-1} (hw') \\
&= h \cdot \frac{1}{\#G} \sum_{g \in G} (h^{-1} \cdot g) \cdot p_0 \cdot (g^{-1} h) w' \\
&= h \cdot \frac{1}{\#G} \sum_{g \in G} g \cdot p_0 \cdot g^{-1} w' \\
&= h \cdot p(w') \\
&= 0
\end{align}$

so $w' \in \ker p$ implies that $hw' \in \ker p$ . So the restriction of $\rho_V$ on $\ker p$ is also a representation.

By the definition of $p$ , for any $w \in W$ , $p(w) = w$ , so $W \cap \ker\ p = \{0\}$ , and for any $v \in V$ , $p(p(v)) = p(v)$ . Thus, $p(v-p(v)) = 0$ , and $v - p(v) \in \ker p$ . Therefore, $V = W \oplus \ker p$ .

= Module-theoretic =

Let V be a K[G]-submodule. We will prove that V is a direct summand. Let π be any K-linear projection of K[G] onto V. Consider the map

$\begin{cases}
\varphi:K[G]\to V \\
\varphi:x \mapsto \frac{1}{\#G}\sum_{s \in G} s\cdot \pi(s^{-1} \cdot x)
\end{cases}$

Then φ is again a projection: it is clearly K-linear, maps K[G] to V, and induces the identity on V (therefore, maps K[G] onto V). Moreover we have

$\begin{align}
\varphi(t\cdot x) &= \frac{1}{\#G}\sum_{s \in G} s\cdot \pi(s^{-1}\cdot t\cdot x)\\
&= \frac{1}{\#G}\sum_{u \in G} t\cdot u\cdot \pi(u^{-1}\cdot x)\\
&= t\cdot\varphi(x),
\end{align}$

so φ is in fact K[G]-linear. By the splitting lemma, $K[G]=V \oplus \ker \varphi$ . This proves that every submodule is a direct summand, that is, K[G] is semisimple.

Converse statement

The above proof depends on the fact that #G is invertible in K. This might lead one to ask if the converse of Maschke's theorem also holds: if the characteristic of K divides the order of G, does it follow that K[G] is not semisimple? The answer is yes.{{sfn|Serre|1977|loc=Exercise 6.1}}

Proof. For $x = \sum\lambda_g g\in K[G]$ define $\epsilon(x) = \sum\lambda_g$ . Let $I=\ker\epsilon$ . Then I is a K[G]-submodule. We will prove that for every nontrivial submodule V of K[G], $I \cap V \neq 0$ . Let V be given, and let $v=\sum\mu_gg$ be any nonzero element of V. If $\epsilon(v)=0$ , the claim is immediate. Otherwise, let $s = \sum 1 g$ . Then $\epsilon(s) = \#G \cdot 1 = 0$ so $s \in I$ and

$sv = \left(\sum1g\right)\!\left(\sum\mu_gg\right) = \sum\epsilon(v)g = \epsilon(v)s$

so that $sv$ is a nonzero element of both I and V. This proves V is not a direct complement of I for all V, so K[G] is not semisimple.

Non-examples

The theorem can not apply to the case where G is infinite, or when the field K has characteristics dividing #G. For example,

Consider the infinite group $\mathbb{Z}$ and the representation $\rho: \mathbb{Z} \to \mathrm{GL}_2(\Complex)$ defined by $\rho(n) = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ . Let $W = \Complex \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ , a 1-dimensional subspace of $\Complex^2$ spanned by $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ . Then the restriction of $\rho$ on W is a trivial subrepresentation of $\mathbb{Z}$ . However, there's no U such that both W, U are subrepresentations of $\mathbb{Z}$ and $\Complex^2 = W \oplus U$ : any such U needs to be 1-dimensional, but any 1-dimensional subspace preserved by $\rho$ has to be spanned by an eigenvector for $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ , and the only eigenvector for that is $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ .
Consider a prime p, and the group $\mathbb{Z}/p\mathbb{Z}$ , field $K = \mathbb{F}_p$ , and the representation $\rho: \mathbb{Z}/p\mathbb{Z} \to \mathrm{GL}_2(\mathbb{F}_p)$ defined by $\rho(n) = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ . Simple calculations show that there is only one eigenvector for $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ here, so by the same argument, the 1-dimensional subrepresentation of $\mathbb{Z}/p\mathbb{Z}$ is unique, and $\mathbb{Z}/p\mathbb{Z}$ cannot be decomposed into the direct sum of two 1-dimensional subrepresentations.