(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 $\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n$?

  • $\begingroup$ Your $V_{\left[1,1,\ldots,1\right]}$ is the Schur functor corresponding to the partition $\left(1,1,\ldots,1\right)$ (evaluated at $\mathbb{C}^n$), right? That is, it's $\wedge^n \mathbb{C}^n$ ? $\endgroup$ Commented Dec 5, 2016 at 18:47
  • $\begingroup$ @darijgrinberg no, $V_{[1,1,...,1]}$ is the irreducible representation of ${\sf SL}_{n}$ corresponding to the weight $[1,1,...,1]$. I use the same notation as the LiE program (wwwmathlabo.univ-poitiers.fr/~maavl/LiEman/manual.pdf): e.g., $\mathbb{C}^n$ is $V_{[1,0,...,0]}$, $\wedge^2\mathbb{C}^n$ is $V_{[0,1,...,0]}$, and so on, $\wedge^{n-1}\mathbb{C}^n$ is $V_{[0,0,...,0,1]}$ and $\wedge^{n}\mathbb{C}^n$ is $V_{[0,0,...,0,0]}$. $\endgroup$ Commented Dec 5, 2016 at 19:57
  • $\begingroup$ Ah, so you're using weight vector notation. And the partition probably is $\left(n-1,n-2,\ldots,1\right)$, right? $\endgroup$ Commented Dec 5, 2016 at 19:58
  • $\begingroup$ If I was able to translate $V_{[1,1,...,1]}$ so easily into a tableaux, probably I would already have answered my own question! But yes, your guess seems accurate: you're suggesting an upper-triangular tableaux, which is consistent with Ben Webster's answer below. $\endgroup$ Commented Dec 5, 2016 at 20:06

1 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.

  • 1
    $\begingroup$ Never heard of this skew Howe duality: what is the standard reference for it? Anyway, it looks to me that you've found the equations describing $\otimes_{i=1}^n\Lambda^i\mathbb{C}^n$ as a subspace of $\Lambda^{{n+1\choose 2}}(\mathbb{C}^n\otimes\mathbb{C}^n)$... how does this help in finding the equations describing $V_{[1,1,...,1]}$ as a subspace of $\otimes_{i=1}^n\Lambda^i\mathbb{C}^n$? $\endgroup$ Commented Dec 5, 2016 at 15:40

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