Let $V$ be a finite dimensional vector space over a field $k$ of characteristic $0$. Let $G$ be a finite group and $\rho : G\to GL(V)$ be a group homomorphism (so $V$ is a linear representation of $G$). Let $W$ be a $1$-dimensional $G$-invariant subspace of $V$. Then by Maschke's theorem, we can write $V=W\oplus U$ for some $G$-invariant subspace $U$ of $V$. My question is, is the existence of $U$ unique? i.e. if $U_1, U_2$ are $G$-invariant subspaces with $V=W\oplus U_1=W \oplus U_2$, then is $U_1=U_2$ ? I can show that $U_1,U_2$ are $G$-equivariant isomorphic, but I am not sure if they have to be exactly equal or not.
I am especially interested in the case where $G=S_n$ is the permutation group, $V=k^n$ and $\rho$ is the permutation representation i.e. $\rho(g)$ is the invertible matrix whose $ij$-th entry is $1$ when $j=g(i)$ and $0$ otherwise. And $W$ is the subspace spanned by the vector $(1,…,1)$ (all entries are $1$).
Best Answer
Consider for example a 2-dimensional vector space with a trivial action of $G$. Then any pair of distinct 1-dimensional subspaces gives a decomposition into irreducible $G$ subspaces, so the complement is far from unique. This same behavior happens whenever you have multiple copies of the same irreducible representation inside $V$.
However for each irreducible representation $W$ of $G$ its isotypic component in a representation $V$ is the subspace $V_W$ spanned by all copies of $W$ in $V$. $V$ always decomposes as a direct sum of its isotypic components $V_W$ for different irreducible representations $W$, and moreover this decomposition is unique. In fact this isotypic decomposition is the common eigenspace decomposition for the action of the center of $\mathbb{C}[G]$ on $V$, and the character table tells you the eigenvalues.
In the case you are interested in for $S_n$, since $V$ decomposes as a direct sum of two non-isomorphic irreducible subspaces then indeed this decomposition is unique as each of the two factors is its own isotypic component.