Consider a subspace $V$ of $\mathcal{H}\_A \otimes \mathcal{H}\_B$, with $\mathcal{H}\_A$ and $\mathcal{H}\_B$ finite-dimensional Hilbert spaces, that is $1\_A \otimes U$ invariant for all unitary operators $U$ on $\mathcal{H}\_B$. Is $V = V\_A \otimes \mathcal{H}\_B$ for some subspace $V\_A$ of $\mathcal{H}\_A$?



Is a $1_A \otimes U$ invariant subspace of $\mathcal{H}_A \otimes \mathcal{H}_B$ a product $V_A \otimes \mathcal{H}_B$?
The action of $1\_A\otimes U$ on $V$ defines the action of the unitary group $U(H\_B)$ on $V$. This action induces the action of Lie algebra $\mathfrak{u}(H\_b)$ on $V$. The action has the form $1\otimes T$ for skew-hermitian $T$. Every operator on $H\_B$ has the form $T\_1+iT\_2$ for $T\_1,T\_2$ skew-hermitian, so we deduce that the subspace $V$ closed under the action of $1\otimes S$ for any operator $S$. It follows that $V=V\_A\otimes H\_B$ for some $V\_A\subset H\_A$.


It seems like we can also directly use the fact, that unitary operators span the vector space of all operators.


