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Consider the following construction: for a vector space $V$, define $W \subseteq \bigwedge^2 V \otimes V$ by

$W = \langle\ \alpha \otimes v : v \in \text{Span}(\alpha) \ \rangle$,

that is, $W$ is spanned by "flag tensors", as in a 2-plane containing a line. Of course, we can make similar constructions for arbitrary shapes of flags.

How is $W$ related to the Schur functor $\mathbb{S}^{2,1}(V)$?

(Edit: I have answered this particular instance of the question. See below.)

Note that the generators of $W$ satisfy the same "exchange relations" used to define the Schur functor (e.g. in Fulton's Young Tableaux): for example, if $\alpha = x \wedge y$, and $v = ax + by$, then by playing around with the tensors one can show

$x \wedge y \otimes v = v \wedge y \otimes x + x \wedge v \otimes y$,

which is the defining relation used to construct $\mathbb{S}^{2,1}(V)$ as a quotient of $\bigwedge^2 V \otimes V$ (by modding out by it). This works for all the other exchange relations for other shapes of flag as well.

The quotient picture is nice to analyze (e.g. finding a basis takes some relatively straightforward combinatorics), and is very natural for algebraic geometry, since it corresponds to the surjection

$H^0(\mathbb{P}(\bigwedge^2 V) \times \mathbb{P}(V),\mathcal{O}(1,1)) \to H^0(Fl^{2,1}(V),\mathcal{O}_{Fl^{2,1}(V)}(1,1)),$

coming from the Plücker embedding of the (2,1)-flag variety. (It's a general fact that the Schur functors give the multigraded components of the flag variety's Plücker coordinate ring in this way.)

On the other hand, the setup above with "flag tensors" is appealingly simple and I'd like to understand it. For example, the definition of $W$ is clearly functorial and a $GL$-subrepresentation (the condition $v \in \text{Span}(\alpha)$ is $GL$-invariant).

Is it actually (or almost) the same space? Is it simply dual to the quotient picture somehow? (Perhaps my $W$ is just $\mathbb{S}^\lambda(V^*)$ or something.)

Or is it entirely different? I got confused when I tried to work this out, particularly since (a) there are other ways of constructing Schur functors as subspaces rather than quotients, (b) I'm not sure how to define $W$ for a partition $\lambda$ with repeated column lengths. For instance, $\lambda = (2,2,2,1)$ should come from $Sym^3(\bigwedge^2V) \otimes V$. Should the corresponding "flag tensors" be of the form $\alpha_1 \alpha_2 \alpha_3 \otimes v$, where $v \in \text{Span}(\alpha_i)$ for each $i$?

Thanks!

Edit: A quick way to see that the $W$ given initially is isomorphic to $\mathbb{S}^{2,1}(V)$ is to use the Pieri rule, which in this case says that $\bigwedge^2V \otimes V \cong \mathbb{S}^{2,1}(V) \oplus \bigwedge^3 V$. And the definition guaranteed that

$W = \ker\left( \bigwedge^2V \otimes V \to \bigwedge^3 V\right).$

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  • $\begingroup$ More generally, for $k\geq 2$, the span of all terms of the form $\alpha \otimes v \in \wedge^{k-1} V \otimes V$ with $\alpha = a_1 \wedge a_2 \wedge ... \wedge a_{k-1} \in \wedge^{k-1} V$ and $a_1$, $a_2$, ..., $a_{k-1}$ in $V$ satisfying $v \in \left< a_1,a_2,...,a_{k-1} \right>$ is the kernel of the surjection $ \wedge^{k-1} V \otimes V \to \wedge^k V$ and is isomorphic to $\mathbb S^{2 1^{k-2}}V$. $\endgroup$ – darij grinberg Dec 15 '13 at 7:52
  • $\begingroup$ @darijgrinberg: Yes, again by the Pieri rule. I thought about this some more and have several thoughts, but haven't worked out all the details. Should I post them as a comment or an answer? (To be clear, the thoughts I had were sufficient for me to be convinced that the functors I've described are indeed the desired Schur functors.) $\endgroup$ – Jake Levinson Jan 14 '14 at 4:36
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This is an incomplete answer.

I think the "flag tensor space" $W$ described above is the Schur functor $\mathbb{S}^\lambda(V)$, as hoped. It's worth highlighting the fact that, unlike either of the two standard constructions from $V^{\otimes \lambda}$ (quotienting by exchange relations, or generating a subspace using Young symmetrizers), the space $W$ is neither a quotient nor a subspace, but actually a subquotient of $V^{\otimes \lambda}$: we first quotiented to get the exterior powers, then passed to a subspace.

In any case, to see that $W$ is the correct space, let $\lambda$ be a partition with column lengths $\mu_1, \ldots, \mu_k$, and for simplicity assume they are all distinct. First observe that the standard map

$\phi: \bigwedge^{\mu_1} V \otimes \cdots \otimes \bigwedge^{\mu_k} V \to \mathbb{S}^\lambda(V),$

defined by imposing the exchange relations, is nonzero when restricted to the "flag tensor subspace" $W$. For example, it takes the flag whose $i$-th component is $e_1 \wedge \cdots \wedge e_{\mu_i}$ to the highest weight vector of $\mathbb{S}^\lambda(V)$. In the tableau basis, this is the SSYT whose $i$-th row consists only of $i$'s. In particular, $W$ contains a(n isomorphic copy of) $\mathbb{S}^\lambda(V)$.

Now the idea is to show that $W$ is an irreducible representation, i.e. has a unique highest weight vector: namely, the one I just described. This isn't totally trivial - in fact, it highlights a weakness in this approach to defining the representation, which is that the decomposition into weight spaces is not particularly obvious. (Most of them come from, essentially, Young symmetrizers, though note that the alternating signs for the columns of $\lambda$ are automatically built-in, since we are working with wedges of vectors. Our Young symmetrizers need only be summed over permutations of the rows of $\lambda$.)

We can imagine expanding a flag tensor multilinearly into a large sum of tensors of wedges of the basis elements $e_1, \ldots, e_n$, and grouping terms by weight vectors using the obvious weight vectors on $\bigwedge^{\mu_1} V \otimes \cdots \otimes \bigwedge^{\mu_k} V$.

An alternate, handwavy argument is to use an upper-triangular matrix to "shift" a given flag tensor closer to the highest-weight flag tensor given above. Assuming this can be done, it follows that $W$ is irreducible.

If there's a clean way to complete the details here, I'd be happy to hear it, but I'm more or less convinced enough for my own satisfaction. If there are no further responses in 2-3 days, I'll accept my answer.

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