The groups $\mathrm{SL}(n) := \mathrm{SL}(n,\mathbb{C})$ and $\mathrm{GL}(n) = \mathrm{GL}(n,\mathbb{C})$ acts on $\mathbb{C}^n$ by multiplication from the left. This induces the diagonal action on $\bigotimes^M \mathbb{C}^n$. Consider two spaces:
- $A = \mathrm{HWV}^{\mathrm{GL}(n)}_{n \times k}\bigotimes^{nk} \mathbb{C}^n$ be the highest weight space of (rectangular) weight $n\times k$ with respect to the group $\mathrm{GL}(n)$.
- $B = \left(\bigotimes^{nk}\mathbb{C}^n\right)^{\mathrm{SL}(n)}$ be the space of $SL(n)$-invariants in the tensor space.
How to prove that $A = B$?
I know constructive proof using Schur-Weyl duality (for both groups). This envolves construction of a spanning sets for each space $A$ and $B$. These spanning sets coincide so the proof follows. But this proof is somewhat indirect and therefore unsatisfactory.
$SL(n)$ is normal in $GL(n)$ so an irreducible $GL(n)$ representation contains an $SL(n)$-invariant vector iff the whole thing is $SL(n)$-invariant and the representation factors through the determinant map $GL(n) \to GL(n) / SL(n) \cong \mathbb{C}^*$. Irreducible (holomorphic) representations of $\mathbb{C}^*$ are just powers of the standard one, so irreducible representations of $GL(n)$ with $SL(n)$ invariants are just powers of the determinant -- which in terms of weights are exactly these $n \times k$ rectangular ones.