[proofplan] A subspace $U \leq V$ is $G$-invariant when $gu \in U$ for every $g \in G$ and $u \in U$. We must show: for any $g \in G$ and $v \in W^\perp$, the vector $gv$ is again in $W^\perp$ — i.e. orthogonal to every $w' \in W$. The trick is to write the test vector $w'$ as $gw$ for some $w \in W$ — possible because $W$ is $G$-invariant and $G$ acts by bijections, so $g$ restricts to a bijection $W \to W$. Then $G$-invariance of the inner product converts $\langle gv, gw \rangle$ to $\langle v, w \rangle$, which vanishes since $v \in W^\perp$. [/proofplan] [step:Show that the action of $g$ restricts to a bijection $W \to W$] Let $g \in G$. Since $W$ is $G$-invariant, the map $\rho(g): V \to V$ satisfies $\rho(g)(W) \subseteq W$. The same applies to $\rho(g^{-1})$, giving $\rho(g^{-1})(W) \subseteq W$. Consider the restriction \begin{align*} \rho(g)|_W: W &\to W \\ w &\mapsto gw. \end{align*} This is well-defined by the inclusion $\rho(g)(W) \subseteq W$ above. It is invertible: $\rho(g^{-1})|_W$ is its two-sided inverse, since for any $w \in W$, \begin{align*} \rho(g^{-1})(\rho(g)w) = \rho(g^{-1}g)w = \rho(1)w = w, \end{align*} and similarly $\rho(g)(\rho(g^{-1})w) = w$, both using the homomorphism property of $\rho$. In particular, $\rho(g)|_W: W \to W$ is **surjective**. [guided] A small but technically necessary observation. To prove $gv \in W^\perp$, we must check $\langle gv, w'\rangle = 0$ for every $w' \in W$. The strategy will be to write $w' = gw$ for some $w \in W$ and then exploit the $G$-invariance of the inner product. For this strategy to work, **every** $w' \in W$ must be expressible as $gw$ for some $w \in W$ — that is, the map $w \mapsto gw$ must be surjective from $W$ to itself. Let us verify this. The hypothesis "$W$ is $G$-invariant" gives $\rho(g)(W) \subseteq W$ for every $g \in G$. Applied to $g^{-1}$, this also gives $\rho(g^{-1})(W) \subseteq W$. So the restriction \begin{align*} \rho(g)|_W: W &\to W \\ w &\mapsto gw \end{align*} is well-defined as a map $W \to W$. We claim it is bijective; the candidate inverse is $\rho(g^{-1})|_W$. For any $w \in W$, \begin{align*} \rho(g^{-1})(\rho(g)w) = \rho(g^{-1}g)w = \rho(1)w = w, \end{align*} where the first equality uses the homomorphism property $\rho(ab) = \rho(a)\rho(b)$ and the second uses $\rho(1) = \mathrm{id}_V$. The reverse composition is similar. So $\rho(g)|_W$ is a bijection $W \to W$ with inverse $\rho(g^{-1})|_W$. In particular, **every** $w' \in W$ has a unique preimage $w := \rho(g^{-1})w' \in W$ satisfying $\rho(g)w = w'$. This is the bijectivity we will use in the next step. [/guided] [/step] [step:Verify $gv \perp w'$ for an arbitrary $w' \in W$] Fix $v \in W^\perp$ and $g \in G$. Let $w' \in W$ be arbitrary. By the previous step, there exists $w \in W$ such that $\rho(g)w = w'$; explicitly, $w = \rho(g^{-1})w' \in W$. Then: \begin{align*} \langle gv, w' \rangle = \langle gv, gw \rangle = \langle v, w \rangle = 0, \end{align*} where: - the first equality substitutes $w' = gw$; - the second equality uses the **$G$-invariance of the inner product**, which says $\langle gx, gy\rangle = \langle x, y\rangle$ for all $x, y \in V$ and $g \in G$; - the third equality uses $v \in W^\perp$ and $w \in W$, so the inner product vanishes by definition of the orthogonal complement. [guided] Now we are ready to prove that $gv \in W^\perp$. By definition, \begin{align*} W^\perp = \{u \in V : \langle u, w'\rangle = 0 \text{ for all } w' \in W\}, \end{align*} so we must show $\langle gv, w'\rangle = 0$ for **every** $w' \in W$. Let $w' \in W$ be arbitrary. Here is where the previous step pays off: since $\rho(g)|_W$ is bijective, we can write $w'$ as $gw$ for some $w \in W$. Explicitly, take $w := g^{-1}w'$; this belongs to $W$ because $W$ is $G$-invariant (which gives $g^{-1}w' \in W$). Then $gw = g(g^{-1}w') = w'$. Now compute: \begin{align*} \langle gv, w'\rangle &= \langle gv, gw\rangle && \text{(rewrote } w' \text{ as } gw\text{)} \\ &= \langle v, w\rangle && \text{(} G\text{-invariance of the inner product)} \\ &= 0 && \text{(} v \in W^\perp \text{ and } w \in W\text{).} \end{align*} The middle step is the heart of the proof. The hypothesis "$V$ carries a $G$-invariant inner product" means precisely $\langle gx, gy\rangle = \langle x, y\rangle$ for every $g \in G$ and $x, y \in V$ — and this is exactly what we just used. Why did we need to introduce $w = g^{-1}w'$? Because the $G$-invariance of the inner product compares $\langle gx, gy\rangle$ with $\langle x, y\rangle$ — both factors must be hit by $g$. We have $gv$ on the left, but $w'$ on the right is "naked." The substitution $w' = gw$ brings the right side into the "$g$ acts on both" form, after which $G$-invariance applies and removes the $g$ from both sides simultaneously. [/guided] [/step] [step:Conclude that $W^\perp$ is $G$-invariant] Since $w' \in W$ was arbitrary in the previous step, $\langle gv, w'\rangle = 0$ for **every** $w' \in W$. By the definition of orthogonal complement, $gv \in W^\perp$. Since $v \in W^\perp$ and $g \in G$ were arbitrary, this shows $\rho(g)(W^\perp) \subseteq W^\perp$ for every $g \in G$. Therefore $W^\perp$ is $G$-invariant, completing the proof. [/step]