----- > [!proposition] Proposition. ([[group-invariant subspace admits group-invariant complement over C]]) > Let $G$ be a finite [[group]] [[group action|acting on]] a finite-dimensional [[complex numbers]]-[[vector space]] $V$. Let $W \subset V$ be a [[group-invariant subspace|G-invariant subspace]]. Then $W$ admits a [[group-invariant subspace|G-invariant]] [[complement of a linear subspace|complement]] $W'$: $V = W \oplus W' \text{ with } g \cdot W = W \text{ and } g \cdot W' = W'.$ > [!proposition] > If $W$ is a $G$-invariant subspace and $V$ has an inner product, with respect that inner product $W^{\perp}$ is $G$-invariant. > [!proof]- Proof with Projectors. ([[group-invariant subspace admits group-invariant complement over C]]) > We know $W$ admits *some* complement $W_{1}$, [[every linearly independent list extends to a basis|for we can extend]] any [[basis]] of $W$ to a [[basis]] of $V$. This induces a [[linear projector]] $P_{0}: V \to V$ with $\im P_{0} = W$, by [[linear projector iff induces direct sum decomposition]]. \ \ Defining the map $P:=\frac{1}{|G|}\sum_{g \in G} \rho_{g} P_{0} \rho_{g}^{-1},$ the claim is that $W':=\ker P = \im (I-P)$ is a $G$-invariant complement to $W$. $P$ is a linear-map-[[averaging over a group|average over]] $G$ and hence is [[average of a linear map over G is G-equivariant|G-equivariant]]. \ **Proof that $W'$ is a complement to $W$.** We need to show $V=W \oplus W'$. [[direct sum iff 0-vector decomposes uniquely|It suffices to show]] $W \cap W' = \{ 0 \}$ and $\dim W + \dim W' = \dim V$. We will do this by exhibiting $W$ as the image of $P:V \to V$, for the result then follows from the [[Rank-Nullity theorem]]. \ **1. $\im P \subset W$.** For any $v \in V$, $P$ [[linear operator|operates linearly on]] $V$ as $\underbrace{\rho_{g} \underbrace{P_{0} \overbrace{\rho_{g}^{-1}v}^{ \in V}}_{\in W_{}}}_{\in W_{} \text{ since } W_{} \ G\text{-invariant }}$ meaning that each summation entry belongs to $W$. $W$ is closed under addition as a [[linear subspace]]; the result follows. \ \ **2. $W \subset \im P$.** Let $w \in W$. We claim that $Pw=w$, meaning $w \in \im P$. We have $Pw = \frac{1}{|G|} \sum_{g \in G} \rho_{g} P_{0} \rho_{g}^{-1} w = \frac{1}{|G|} \sum_{g \in G} w = w,$ where we use that $\rho_{g}$ and $\rho_{g}^{-1}$ fix $w$ because $W$ is $G$-invariant and that $P_{0}$ is a [[linear projector]] (hence fixes [[vector|vectors]] already in its [[image]]). \ Thus the two-way inclusion is shown: $W= \im P$. By definition $W'=\ker P$, so that $W \cap W' = \{ 0 \}$ and $\dim W + \dim W' = \dim V$ so that $W \oplus W' = V$. \ **Proof that $W' := \im (I-P)=\ker P$ is $G$-invariant.** \ **Lemma. $P \rho_{h}=\rho_{h} P$ for all $h \in G$:** Let $h \in G$. We compute $\begin{align} \rho_{h}^{-1} P \rho_{h} = & \rho_{h}^{-1} \big( \frac{1}{|G|} \sum_{g \in G} \rho_{g} P_{0} \rho_{g}^{-1} \big) \rho_{h}\\ = & \frac{1}{|G|} \sum_{g \in G} \rho_{h^{-1}g} P_{0} \rho_{g^{-1}h} \\ = & \frac{1}{|G|} \sum_{g \in G} \rho_{h^{-1}g} P_{0} \rho_{h^{-1} g}^{-1} \\ ^{g':=h ^{-1} g}= & \frac{1}{|G|} \sum_{g \in G} \rho_{g'} P_{0} \rho_{g'}^{-1} \ (*) \\ = & P, \end{align}$ where the step $(*)$ is justified because as $g$ goes over the group $g'$ does too, just in a different order (to which the final sum result is indifferent). \ **Now to prove the result.** Let $w' \in W'$, then $w' = (I-P)v$ for some $v \in V$. Now for any $g \in G$, $\rho_{g}w' = \rho_{g}(I-P)v = (I-P) \overbrace{\rho_{g} w'}^{:= \hat{v}}=(I-P)\hat{v} \in \im (I-P) = W',$ where we used from **the lemma** that $P$ and $\rho_{g}$ commute. > [!proof] Proof with $G$-invariant Hermitian forms. > Endow $V$ with any [[inner product]] $\langle \cdot,\cdot \rangle$ and [[averaging over a group|average over it]] with respect to $\langle \cdot, \cdot \rangle$ to obtain the [[group-invariant function|Group-invariant]] [[inner product]] $\langle \langle \cdot,\cdot \rangle \rangle$. > We claim that, w.r.t. $\langle \langle \cdot,\cdot \rangle \rangle$, $W^{\perp}$ is [[group-invariant subspace|G-invariant]]. So let $v \in W^{\perp}$ and $g \in G$. Letting an arbitrary $w \in W^{}$, we have $\langle \langle \rho_{g}v, w \rangle \rangle = \langle \langle \rho_{g}v, \rho_{g} \underbrace{\rho_{g^{-1}} w}_{\in W } \rangle \rangle = \langle \langle v, \rho_{g^{-1}}w \rangle \rangle = 0, $ where >- We conclude $\rho_{g^{-1}}^{}w \in W$ because $W$ is $G$-invariant; >- The third equality uses that $\langle \langle \cdot,\cdot \rangle \rangle$ is $G$-invariant; >- The final equality uses that $v \in W^{\top}$ while $\rho_{g^{-1}}w \in W$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY 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 > ```