Some intuition on the $SL_n$-module $V_{[1,1,...,1]}$

Question

(This question highly overlaps with this and also this.)

The irreducible ${\sf SL}_{n-1}$-module $V_{[1,1,\ldots,1]}$ is the one providing the minimal projective embedding $\mathbb{P}(V_{[1,1,\ldots,1]})$ for the complete flag variety $\mathrm{Fl}(\mathbb{C}^n)$.

However one does not need representation theory in order to realise that $\mathrm{Fl}(\mathbb{C}^n)$ is projective. First, embed $$ \mathrm{Fl}(\mathbb{C}^n)\subset\mathrm{Gr}(1,\mathbb{C}^n)\times\mathrm{Gr}(2,\mathbb{C}^n)\times\cdots\times\mathrm{Gr}(n-1,\mathbb{C}^n)\, , $$ then regard each Grassmannian $\mathrm{Gr}(i,\mathbb{C}^n)$ as a projective variety in $\mathbb{P}\bigwedge^i\mathbb{C}^n$, and finally use the Segre embedding: $$ \mathrm{Fl}(\mathbb{C}^n)\subset\mathbb{P}\left(\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n \right)\, . $$

QUESTION. Is there any "evident map" defined on $\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n$ whose kernel is $V_{[1,1,\ldots,1]}$?

By "evident" I mean definable in terms of elementary operations between tensors, like skew-symmetrisation. For instance, $V_{[1,1,\ldots,1]}$ lies in the common kernel of all the skew-symmetrisations $$ \bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n \stackrel{s_{ab}}{\longrightarrow} \bigwedge^{a+b}\mathbb{C}^n\otimes\bigotimes_{i\neq a,b}\bigwedge^i\mathbb{C}^n \, , $$ where $$ s_{ab}(\cdots\otimes\omega_a\otimes\cdots\otimes\omega_b\otimes\cdots):= (\omega_a\wedge\omega_b)\otimes\cdots\, , $$ and I suspect that $V_{[1,1,\ldots,1]}$ is exactly equal to $\bigcap_{a,b}\ker s_{ab}$, though I cannot prove it!

QUESTION (reformulated). Is there an "evident way" of regarding $V_{[1,1,\ldots,1]}$ as a submodule of $\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n$?

Answer 1

You can understand this using skew Howe duality. You can regard the tensor product $\bigotimes_{i=1}^{n}\bigwedge{}^{\!i} \mathbb{C}^n$ as a subrepresentation of $\bigwedge{}^{\!\binom{n+1}{2}}(\mathbb{C}^n\otimes \mathbb{C}^n)$ sending $$v_a\otimes (v_{b_1}\wedge v_{b_2})\otimes \cdots \mapsto (v_a\otimes v_1)\wedge (v_{b_1}\otimes v_2)\wedge (v_{b_2}\otimes v_2)\wedge \cdots $$ Here $\mathfrak{sl}_n$ is acting on the left copy of $\mathbb{C}^n$ and trivially on the right one. Which means that we have a commuting copy of $\mathfrak{sl}_n$ acting on the right copy; through some usual manipulations, we see that the images of $U(\mathfrak{sl}_n)$ acting via the left and right actions are maximal mutually commuting subalgebras (anything commuting with one lies in the other). In fact, as a module over $\mathfrak{sl}_n\times \mathfrak{sl}_n$, the module $\bigwedge{}^{\!\binom{n+1}{2}}(\mathbb{C}^n\otimes \mathbb{C}^n)$ breaks up as a sum $V_{\lambda}\otimes V_{\lambda^t}$ of the rep for a partition with $\binom{n+1}{2}$ boxes which fits inside an $n\times n$ square (you can check this by finding the common highest weight vectors). The copy of $\bigotimes_{i=1}^{n}\bigwedge{}^{\!i} \mathbb{C}^n$ we've embedded is exactly the lowest weight vectors of weight $-\omega_1-\cdots -\omega_n$ for the right action, so it's the common kernels of all the maps corresponding to lowering operators. These aren't the full skew symmetrizations, but rather the partial ones, where one maps $\bigwedge{}^{\!k}\mathbb{C}^n\otimes \bigwedge{}^{\!k+1}\mathbb{C}^n\to \bigwedge{}^{\!k-1}\mathbb{C}^n\otimes \bigwedge{}^{\!k+2}\mathbb{C}^n$ by $$ (v_{a_1}\wedge \cdots \wedge v_{a_k})\otimes (v_{b_1}\wedge \cdots \wedge v_{b_{k+1}})\mapsto\\ \sum_{i=1}^k (-1)^{k-1}(v_{a_1}\wedge \cdots v_{a_{i-1}}\wedge v_{a_{i+1}}\wedge \cdots\wedge v_{a_k})\otimes (v_{a_i}\wedge v_{b_1}\wedge \cdots \wedge v_{b_{k+1}})$$ in each consecutive pair of terms.