----- > [!proposition] Proposition. ([[the average of a vector over G is G-invariant]]) > Let $G$ be a [[group]] [[group action|acting on]] a [[vector space]] $V$ and let $v \in V$. > 1. The [[averaging over a group|average]] $\tilde{v}$ of $v$ over $G$ is [[group-invariant function|G-invariant]]. > 2. If $w$ is already [[group-invariant function|G-invariant]], then $\tilde{w}=w$. > [!proof]- Proof. ([[the average of a vector over G is G-invariant]]) > **1.** We want to show that $h \cdot \tilde{v}=\tilde{v}$ for all $h \in G$. We have $\begin{align} h \cdot \tilde{v} = & \rho_{h} \big( \frac{1}{|G|} \sum_{g \in G} g \cdot v \big) \\ = & \frac{1}{|G|} \sum_{g \in G} \rho_{h}(g \cdot v) \\ = & \frac{1}{|G|} \sum_{g \in G} h \cdot (g \cdot v) \\ = & \frac{1}{|G|} \sum_{g \in G} (hg) \cdot v \\ \\ ^{g':=hg}= & \frac{1}{|G|} \sum_{g \in G} g' \cdot v \ \ (*)\\ = & \tilde{v}, \end{align}$ where the step $(*)$ is justified because as $g$ runs through the [[group]], $g'$ does too, just in a different order (to which the final summation is indifferent). **2.** Let $w$ already be $G$-invariant. Then [[averaging over a group|averaging]] $w$ over $G$ simply gives $\tilde{w}= \frac{1}{|G|} \sum_{g \in G}\rho_{g}(w)= \frac{1}{|G|} \sum_{g \in G} w =w.$ ----- #### ---- #### 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 > ```