---- > [!theorem] Theorem. ([[Maschke's Theorem]]) > Every [[group representation|representation]] $(\rho, V)$ of a finite [[group]] $G$ on a nonzero, [[finite-dimensional vector spaces MOC|finite-dimensional]] [[vector space]] over $\mathbb{C}$ is a [[direct sum of representations|direct sum of irreducible]] [[group representation|representations]]: $\rho \cong \bigoplus_{i=1}^{h} m_{i}\rho_{i}.$ In particular, if $\chi_{i}$ is the [[character of a representation|character]] of $\rho_{i}$, we have $m_{i} = \langle \chi_{\rho}, \chi_{i} \rangle ,$ > where $\langle \cdot,\cdot \rangle$ is the usual [[inner product]] for the [[class function|space of class functions]] on $G$. > [!proof]- Proof. ([[Maschke's Theorem]]) > That $\rho \cong \bigoplus_{i=1}^{h}m_{i}\rho_{i}$ for *some* $m_{i}$ is a corollary of [[group-invariant subspace admits group-invariant complement over C]]: start a [[group representation|representation]] $(\rho,V)$ If $(\rho, V)$ is [[irreducible group representation|irreducible]] we're done. Else we have $V=W_{1} \oplus W_{2}$ for [[group-invariant subspace|G-invariant]] $W_{1}, W_{2}$. Now we repeat the process for each of $(\rho, W_{1})$ and $(\rho, W_{2})$, continuing until we have broken $\rho$ down entirely. > \ > Now, because the $\chi_{i}$ [[class function|form an]] [[orthonormal basis]] of $\text{Cl}(G)$, using [[writing a vector as a linear combination of orthonormal basis using projections|this result]] we have $\chi_{\rho}= \sum_{i=1}^{h} \langle \chi_{\rho}, \chi_{i} \rangle \chi_{i} \ \ (1)$ . From the [[block matrix]] property of the [[direct sum of representations]] it is clear that the [[character of a representation|character]] of a [[direct sum of representations]] is the sum of the [[character of a representation|characters]]: $\chi_{\rho} = \chi_{\bigoplus_{i=1}^{h} m_{i} \rho_{i}} = \sum_{i=1}^{h} m_{i} \chi_{i} \ \ (2).$ > From here we can match terms in $(1)$ and $(2)$ to get $m_{i}=\langle \chi_{\rho}, \chi_{i} \rangle$ as required. > > [!basicexample] > 1. Let $G:=C_{2}=\{ e,g \}$ [[group action|act on]] $V:=\mathbb{C}^{2}$ as $g \cdot \begin{bmatrix} x \\ y \end{bmatrix}:= \begin{bmatrix} y \\ x \end{bmatrix}.$ We claim that the consequent 2-dimensional representation $(\rho, V)$ can be written as a [[direct sum of representations|direct sum]] of two 1-dimensional [[irreducible group representation]]s. The notion of a [[direct sum of representations|(internal) direct sum of representations]] requires that $G$ [[group action|acts on]] the [[linear subspace|subspaces]] in question— i.e., that they're [[group-invariant subspace|G-invariant]]. > So what are some $G$-invariant subspaces of $V$? The line $\ell_{1}$ [[submodule generated by a subset|spanned by]] $\begin{bmatrix}x \\ x\end{bmatrix}$ is one ($g$ doesn't affect it), as is the line $\ell_{2}$ [[submodule generated by a subset|spanned by]] $\begin{bmatrix}-x \\ x\end{bmatrix}$ ($g$ 'flips it on itself'). Since $\ell_{1} \cap \ell_{2} = \{ 0 \}$, $\ell_{1} \oplus \ell _2$ [[direct sum of two subspaces|exists]]; then [[dimension of direct sum is sum of dimensions]] implies $\dim (\ell_{1} \oplus \ell_{2})=2$ and [[the (finite) dimension theorem]] yields $V \cong \ell_{1} \oplus \ell_{2}$. > > Then $\rho=\rho |_{\ell_{1}} \oplus \rho |_{\ell_{2}}$ where $\rho_{\ell_{1}}$ and $\rho_{\ell_{2}}$ are [[irreducible group representation]]s (since they're 1-dimensional). Note that $\ell_{1}$ corresponds to the [[trivial group representation]]. > >2. Let $G:=C_{3}=\{ e,g,g^{2} \}$ [[group action|act on]] $V:=\mathbb{C}^{3}$ as $g \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} z \\ x \\ y \end{bmatrix}.$ We claim that the consequent 3-dimensional representation $(\rho, V)$ is a [[direct sum of representations|direct sum]] of $1$-dimensional [[group representation|representations]], i.e., we want to write $V=\ell_{1} \oplus \ell_{2} \oplus \ell_{3}$ for lines $\ell_{1},\ell_{2},\ell_{3}$ that are [[stabilizer|stable]] under $g$. Idea: $g$ is [[circulant matrix|circulant]], hence [[diagonalizable|diagonalized]] in the [[discrete fourier basis]]. We can define $\ell_{1}, \ell_{2}, \ell_{3}$ to be the corresponding [[eigenspace|Eigenspaces]] of $\rho_{g}$ since these are [[invariant subspace|invariant]] under the operation. These are $\ell_{1} =\span \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \ell_{2}= \span \begin{bmatrix} 1 \\ \omega \\ \omega^{2} \end{bmatrix}, \ell_{3}= \span \begin{bmatrix} 1 \\ \omega^{2} \\ \omega \end{bmatrix},$ where $\omega=e^{2\pi i / 3}$ is a third [[roots of unity|root of unity]]. Then $\rho=\rho |_{\ell_{1}} \oplus \rho |_{\ell_{2}} \oplus \rho | _{\ell_{3}}$ where the $\rho_{\ell_{k}}$ are [[irreducible group representation]]s (since they're 1-dimensional). > [!basicnonexample] Warning. > We needed to use the fact that [[complex numbers]] has [[characteristic of a field|characteristic zero]] here. Indeed, the theorem is can be false otherwise. For instance, it is false when the ground [[field]]'s [[characteristic of a field|characteristic]] [[divides]] the [[order of a group|order of]] $G$: as a (counter-)example, let $k$ be a [[field]] of [[characteristic of a field|characteristic]] $p$. Let $G=C_{p}$ [[group action|act on]] the two-dimensional [[vector space]] $V:=k^{2}$ by sending a [[generating set of a group|generator]] of $G$ to the [[matrix]] $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \in \text{GL}_{2}(k)$ which has [[order of an element in a group|order]] $p$. This [[group representation|representation]] is not a [[direct sum of representations|dire[](field.md)md)of [[irreducible group representation]]s because we can find a proper nontrivial [[group-invariant subspace|G-invariant]] [[subspace]] $W \[](field.md)ld.md)th no [[group-invariant subspace|group-invariant]] [[complement of a linear subspace|complement]]. In particular, $W:= \span \begin{bmatrix}1 \\ 0\end{bmatrix}$ is $G$-invariant. Consider some [[complement of a linear subspace|complement]] $U = \span \begin{bmatrix}u_{1} \\ u_{2}\end{bmatrix}$ of $W$, where $u_{1}, u_{2} \in k$. We have $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix}= \begin{bmatrix} u_{1}+u_{2} \\ u_{2} \end{bmatrix} \notin U$ and so $U$ is not $G$-invariant. ^c0cd53 ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```