Maschke's theorem

In

.

Formulations

Maschke's theorem addresses the question: when is a general (finite-dimensional) representation built from irreducible

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

${\displaystyle G}$ has a natural ${\displaystyle G}$-invariant for the case of representations over ${\displaystyle \mathbb {C} }$ by constructing ${\displaystyle U}$ as the orthogonal complement of ${\displaystyle W}$ under this inner product.

Module-theoretic

One of the approaches to representations of finite groups is through

. Representations of a group ${\displaystyle G}$ are replaced by modules over its group algebra ${\displaystyle K[G]}$ (to be precise, there is an isomorphism of categories between ${\displaystyle K[G]{\text{-Mod}}}$ and ${\displaystyle \operatorname {Rep} _{G}}$, the category of representations of ${\displaystyle G}$). 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 ${\displaystyle K}$ is the field of complex numbers, this shows that the algebra ${\displaystyle K[G]}$ is a product of several copies of complex matrix algebras, one for each irreducible representation.^[10] If the field ${\displaystyle K}$ has characteristic zero, but is not

, for example if ${\displaystyle K}$ is the field of real or rational numbers, then a somewhat more complicated statement holds: the group algebra ${\displaystyle K[G]}$ is a product of matrix algebras over division rings over ${\displaystyle K}$. The summands correspond to irreducible representations of ${\displaystyle G}$ over ${\displaystyle K}$.^[11]

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 ${\displaystyle p_{0}:V\to W}$ be the projection function, i.e., ${\displaystyle p_{0}(w+u)=w}$ for any ${\displaystyle u\in U,w\in W}$.

Define ${\textstyle p(x)={\frac {1}{\#G}}\sum _{g\in G}g\cdot p_{0}\cdot g^{-1}(x)}$, where ${\displaystyle g\cdot p_{0}\cdot g^{-1}}$ is an abbreviation of ${\displaystyle \rho _{W}{g}\cdot p_{0}\cdot \rho _{V}{g^{-1}}}$, with ${\displaystyle \rho _{W}{g},\rho _{V}{g^{-1}}}$ being the representation of G on W and V. Then, ${\displaystyle \ker p}$ is preserved by G under representation ${\displaystyle \rho _{V}}$: for any ${\displaystyle w'\in \ker p,h\in G}$,

so ${\displaystyle w'\in \ker p}$ implies that ${\displaystyle hw'\in \ker p}$. So the restriction of ${\displaystyle \rho _{V}}$ on ${\displaystyle \ker p}$ is also a representation.

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

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

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

so φ is in fact K[G]-linear. By the splitting lemma, ${\displaystyle K[G]=V\oplus \ker \varphi }$. 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.^[12]

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

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

Notes

References

• Zbl 0984.00001
  .
• Zbl 0355.20006
  .
• OCLC 246650103
  .