---- Let $G$ be a finite group. > [!definition] Definition. ([[averaging over a group]]) > Let $G$ be a finite [[group]] [[group action|acting on]] a [[vector space]] $V$ via [[group representation|representation]] $\rho$ and let $v \in V$. The **average of $v$ over $G$** is defined $\tilde{v}:= \frac{1}{|G|}\sum_{g \in G} g \cdot v=\frac{1}{|G|} \sum_{g \in G} \rho_{g}(v).$ > Crucially, $\tilde{v}$ is $G$-invariant: $\rho_{g}(\tilde{v})=\tilde{v}$ for all $g \in G$. \ We can also average [[linear map|linear maps]] over $G$. If $(\rho^{(1)}, V_{1})$ and $(\rho^{(2)},V_{2})$ are two [[group representation|representations]] of $G$, then the **average of $L \in \hom(V_{1},V_{2})$** over $G$ is defined $\tilde{L}:= \frac{1}{|G|} \sum_{g \in G}\rho^{(2)}_{g^{-1}} \circ L \circ \rho^{(1)}_{g}.$ Crucially, $\tilde{L}$ is $G$-[[group-equivariant map|equivariant]]: $\tilde{L}=\rho^{(2)^{-1}} \tilde{L} \rho^{(1)}$. In the special case where $(\rho_{1}, V_{1})=(\rho_{2}, V_{2})$ so that $L \in \hom(V,V)$, and the representation is [[irreducible group representation|irreducible]], we can say more: $\tilde{L}=\frac{\text{tr }L}{\dim V} \textcolor{Thistle}{\id}.$ > [!basicproperties] > - [[the average of a vector over G is G-invariant]] > - [[the average of a linear map over G is G-equivariant]] > [!proof] Proof of Basic Properties. > For the second property: let > $|G| \tilde{L}:= \sum_{g \in G} \rho^{(2)}_{g^{-1}} \circ L \circ \rho^{(1)}_{g}$; we have $\begin{align} \rho^{(2)^{-1}}_{h}|G|\tilde{L}\rho^{(1)}_{h} = & \rho^{(2)^{}}_{h^{-1}} \left( \sum_{g \in G} \rho^{(2)}_{g^{-1}} \circ L \circ \rho^{(1)} _{g} \right) \rho^{(1)}_{h}\\ = & \sum_{g \in G} \rho_{g^{-1}h^{-1}}^{(2)} \circ L \circ \rho_{gh}^{(1)} \\ = & \sum_{g_{'} \in G} \rho^{(2)}_{g'^{-1}} \circ L \circ \rho_{g'}^{(1)} \\ = & |G|\tilde{L}. \end{align}$ \ Now consider the special case of [[morphism of group representations|isomorphic representations]] $(\rho^{(1)}, V_{1}) \cong (\rho^{(2)},V_{2}):= (\rho, V)$. Since $\tilde{L}$ is $G$-[[group-equivariant map|equivariant]], by [[Schur's lemma for groups]] it has the form $\tilde{L}= cI$ for some $c \in \mathbb{C}$. Take the [[trace of a linear operator|trace]] on both sides: $\text{tr }\tilde{L}= \text{tr }cI = c \dim V.$ Compute $\begin{align} \text{tr }\tilde{L}= & \text{tr } \frac{1}{|G|} \sum_{g \in G} \rho_{g^{-1}} L \rho_{g} \\ = & \frac{1}{|G|} \sum_{g \in G} \text{tr }L \ \ (\text{tr inv. under conj.}) \\ = & \text{tr }L. \end{align}$ This gives $c = \frac{1}{\dim V}\text{tr }L$. ---- #### ---- #### 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 > ```