how to show the orthogonal projection?

Question

Let be $(G,\circ )$ a finite group and

$ \pi : G \rightarrow U(H) $ a group homomorphism .

Set

$V:= \{ x\in H: \pi(g)x=x, \forall g \in G \} $

Then the orthogonal projection on $V$ is given as :

$$ \frac{1}{|G|} \sum_{ g \in G} \pi(g) $$

How can I verify that? Or how can start? any help very appreciated ,as I am very stuck

Answer 1

Set

$P = \displaystyle \dfrac{1}{\vert G \vert} \sum_{g \in G} \pi(g); \tag 1$

for

$y \in H, \tag 2$

we have

$P^2y = \displaystyle \dfrac{1}{\vert G \vert} \sum_{h \in G} \pi(h) \left ( \displaystyle \dfrac{1}{\vert G \vert} \sum_{g \in G} \pi(g) y \right )$ $= \dfrac{1}{\vert G \vert^2} \displaystyle \sum_{h \in G} \sum_{g \in G} \pi(h) \pi(g)y = \displaystyle \dfrac{1}{\vert G \vert^2} \sum_{h \in G} \sum_{g \in G} \pi(hg)y; \tag 3$

now for any fixed $h$ in the rightmost sum of (3),

$\displaystyle \sum_{g \in G} \pi(hg)y = \sum_{g \in G} \pi(g)y, \tag 4$

since left multiplication by $h$ merely permutes the elements of $G$ amongst themselves; it follows that this rightmost sum of (3) is comprised of $\vert G \vert$ copies of $\sum_{g \in G} \pi(g)y$; hence

$P^2y = \dfrac{1}{\vert G \vert^2} \vert G \vert \displaystyle \sum_{g \in G} \pi(g)y = \dfrac{1}{\vert G \vert} \displaystyle \sum_{g \in G} \pi(g)y = Py \tag 5$

which shows that $P$ is a projection. $P$ is orthogonal if

$P^\dagger = P; \tag 6$

we have

$P^\dagger = \dfrac{1}{\vert G \vert} \displaystyle \sum_{g \in G} \pi(g)^\dagger; \tag 7$

since

$\pi(g) \in U(H), \tag 8$

that is, $\pi(g)$ is unitary,

$\pi(g)^\dagger = \pi(g)^{-1} = \pi(g^{-1}); \tag 9$

thus

$P^\dagger = \dfrac{1}{\vert G \vert} \displaystyle \sum_{g \in G} \pi(g^{-1}) = \dfrac{1}{\vert G \vert} \sum_{g \in G} \pi(g) = P, \tag{10}$

and $P$ is an orthogonal projection.