Maschke's theorem

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: Then the corollary is 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: 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. 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. Category-theoretic Reformulated in the language of semi-simple categories, Maschke's theorem states == 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. 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. == Notes ==