[proofplan] We adapt the unitary version of Maschke's theorem to the compact-group setting. Given a representation $(\rho, V)$ of a compact group $G$ and a $G$-invariant subspace $W \leq V$, Weyl's unitary trick provides a $G$-invariant Hermitian inner product $\langle \cdot, \cdot \rangle$ on $V$. The orthogonal complement $W^\perp$ with respect to this inner product is then automatically $G$-invariant, and $V = W \oplus W^\perp$. Iterating this on $W$ and $W^\perp$ until each summand is irreducible — the iteration terminates because $\dim V < \infty$ — exhibits $V$ as a direct sum of irreducible subrepresentations. [/proofplan]

[step:Reduce to producing a $G$-invariant complement of any $G$-invariant subspace]Let $G$ be a compact group and $(\rho, V)$ a finite-dimensional continuous representation. We first show: [claim:Every $G$-invariant subspace $W \leq V$ admits a $G$-invariant complement $W' \leq V$ with $V = W \oplus W'$] [proof] By [Weyl's Unitary Trick for Compact Groups](/theorems/2472), there exists a $G$-invariant Hermitian inner product \begin{align*} \langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}. \end{align*} Define \begin{align*} W^\perp := \{\mathbf{v} \in V : \langle \mathbf{v}, \mathbf{w} \rangle = 0 \text{ for all } \mathbf{w} \in W\} \subseteq V. \end{align*} *$W^\perp$ is a subspace.* The set $W^\perp$ is the intersection over $\mathbf{w} \in W$ of the kernels of the linear functionals $\mathbf{v} \mapsto \langle \mathbf{v}, \mathbf{w} \rangle$, hence is a linear subspace. *$V = W \oplus W^\perp$.* Standard finite-dimensional Hermitian inner-product theory: positive-definiteness of $\langle \cdot, \cdot \rangle$ implies $W \cap W^\perp = \{\mathbf{0}\}$ (if $\mathbf{w} \in W \cap W^\perp$, then $\langle \mathbf{w}, \mathbf{w} \rangle = 0$, forcing $\mathbf{w} = \mathbf{0}$), and $\dim W^\perp = \dim V - \dim W$ (the orthogonal projection $V \to W$ along $\langle \cdot, \cdot \rangle$ has kernel $W^\perp$ and image $W$, so the rank-nullity theorem gives the dimension count). Therefore $V = W \oplus W^\perp$. *$W^\perp$ is $G$-invariant.* Let $\mathbf{v} \in W^\perp$ and $g \in G$; we show $\rho(g)\mathbf{v} \in W^\perp$. For any $\mathbf{w} \in W$, we use $G$-invariance of $\langle \cdot, \cdot \rangle$ in the form $\langle \rho(g)\mathbf{x}, \rho(g)\mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{y} \rangle$ for all $\mathbf{x}, \mathbf{y} \in V$ and $g \in G$, applied with $\mathbf{x} = \mathbf{v}$ and $\mathbf{y} = \rho(g^{-1})\mathbf{w}$: \begin{align*} \langle \rho(g)\mathbf{v}, \mathbf{w} \rangle = \langle \rho(g)\mathbf{v}, \rho(g)\rho(g^{-1})\mathbf{w} \rangle = \langle \mathbf{v}, \rho(g^{-1})\mathbf{w} \rangle. \end{align*} Since $W$ is $G$-invariant and $g^{-1} \in G$, we have $\rho(g^{-1})\mathbf{w} \in W$. Combined with $\mathbf{v} \in W^\perp$, this gives $\langle \mathbf{v}, \rho(g^{-1})\mathbf{w} \rangle = 0$. Therefore $\langle \rho(g)\mathbf{v}, \mathbf{w} \rangle = 0$ for all $\mathbf{w} \in W$, i.e., $\rho(g)\mathbf{v} \in W^\perp$. Setting $W' := W^\perp$ gives the required $G$-invariant complement. [/proof] [/claim][/step]

[guided]The argument has two ingredients. First, *existence* of a $G$-invariant inner product, supplied by Weyl's unitary trick on compact groups (Step provided by theorem 2472). Second, the *miracle* that the orthogonal complement of an invariant subspace is again invariant — this uses the unitarity of $\rho$ with respect to $\langle \cdot, \cdot \rangle$. Why does $G$-invariance of $W^\perp$ work this way? Because $\langle \cdot, \cdot \rangle$ being $G$-invariant means $\rho(g)$ is *unitary*: $\langle \rho(g) \mathbf{x}, \rho(g) \mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{y} \rangle$. Equivalently, $\rho(g)^* = \rho(g)^{-1} = \rho(g^{-1})$. The orthogonality computation $\langle \rho(g)\mathbf{v}, \mathbf{w} \rangle = \langle \mathbf{v}, \rho(g)^* \mathbf{w} \rangle = \langle \mathbf{v}, \rho(g^{-1})\mathbf{w} \rangle$ then reduces the question to $\rho(g^{-1})\mathbf{w} \in W$, which is just $G$-invariance of $W$. Without $G$-invariance of the inner product, this fails: with a generic inner product, $W^\perp$ has no reason to be $G$-invariant. Weyl's trick is precisely the device that produces an inner product making $\rho$ unitary.[/guided]

[step:Iterate the splitting until each summand is irreducible] We show by strong induction on $\dim V$ that every finite-dimensional continuous representation $V$ of $G$ decomposes as a direct sum of irreducible subrepresentations. *Base case ($\dim V = 0$):* The empty direct sum $V = \mathbf{0}$ is the required (vacuous) decomposition. *Inductive step ($\dim V \geq 1$):* If $V$ is irreducible, then $V = V$ is itself the required decomposition (a direct sum with one summand). If $V$ is not irreducible, there exists a $G$-invariant subspace $W \leq V$ with $0 < \dim W < \dim V$. By the claim of Step 1, $V = W \oplus W'$ where $W'$ is a $G$-invariant complement. Since $0 < \dim W < \dim V$ and $\dim W' = \dim V - \dim W$, we have $0 < \dim W' < \dim V$. The subspaces $W$ and $W'$ inherit representation structures from $V$ (the restrictions $\rho|_W: G \to \operatorname{GL}(W)$ and $\rho|_{W'}: G \to \operatorname{GL}(W')$ are continuous representations). By the inductive hypothesis applied to $W$ and to $W'$ (each of strictly smaller dimension than $V$), each decomposes as a direct sum of irreducible subrepresentations: \begin{align*} W = U_1 \oplus \cdots \oplus U_k, \qquad W' = U_{k+1} \oplus \cdots \oplus U_m \end{align*} with each $U_i$ an irreducible $G$-invariant subspace. Concatenating, \begin{align*} V = W \oplus W' = U_1 \oplus \cdots \oplus U_m \end{align*} expresses $V$ as a direct sum of irreducible subrepresentations. By induction, $V$ is completely reducible. [/step]

[step:Conclude]Combining Steps 1 and 2: every finite-dimensional continuous representation of a compact group $G$ admits a $G$-invariant Hermitian inner product (Step 1, via Weyl), and the orthogonal-complement construction together with induction on dimension produces a decomposition into irreducibles (Step 2). This is the statement of [Maschke's theorem for compact groups](/theorems/2473).[/step]

[guided]This is the natural generalisation of [Maschke's theorem](/theorems/2409) for finite groups. The finite-group proof averages an arbitrary projection onto $W$ over $G$ via $\frac{1}{|G|} \sum_{g \in G} \rho(g) \pi \rho(g)^{-1}$ to produce a $G$-equivariant projection. The compact-group proof replaces the finite sum with an integral against Haar measure: the abstract pattern is *averaging over the group action to manufacture invariance*, and Haar measure is the device that makes the average well-defined for compact groups. The compactness hypothesis is essential. For non-compact groups such as $\operatorname{SL}_2(\mathbb{R})$, complete reducibility fails: there exist representations with $G$-invariant subspaces that admit no $G$-invariant complement. The standard counterexample is the action of $\operatorname{SL}_2(\mathbb{R})$ on $\mathbb{R}^2$, restricted to the upper-triangular subgroup, where the line $\mathbb{R} \mathbf{e}_1$ is invariant but has no invariant complement. The other place compactness matters is finite-dimensionality of $V$. The induction is on $\dim V$ and uses that $\dim V < \infty$ to terminate. The infinite-dimensional analogue (continuous unitary representations of compact groups on Hilbert spaces) is also true, but the proof requires substantially more machinery: the Peter-Weyl theorem, which decomposes $L^2(G)$ as a Hilbert-space direct sum of finite-dimensional irreducibles.[/guided]