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Bert Kostant mentioned an odd fact to me some time ago. As usual (with such statements), fix a complex, connected, reductive) Lie group $G$, with maximal torus $T$, and Weyl vector $\rho$ equal to half the sum of the positive roots. Let $L_\beta := T \cdot $ the root $SL_2$ subgroup corresponding to the positive root $\beta$.

Quoth Bert: the character of the irrep $V_{n\rho}$ is $T$-isomorphic to the tensor product over all positive roots of the $L_\beta$-irrep with highest weight $n\beta$. (The latter is a $T$-representation by sticking $T$ diagonally into $\prod_{\Delta_+} L_\beta$.) Once someone tells you, it's very easy to prove from the Weyl character formula.

Geometrically, this says the following. Inside ${\mathbb P}^* (V^G_\rho)$, we have a copy of the flag manifold $G/B$ as the orbit of the highest weight vector. (Indeed, this is the smallest embedding by a complete linear series.) Identifying

$V^G_\rho \cong \bigotimes_{\Delta_+} V^{L_\beta}_\beta$

as $T$-representations, we also have in this projective space a Segre-embedded $\prod_{\Delta_+} {\mathbb P}^* (V^{L_\beta}_\beta),$ a product of ${\mathbb P}^1$s. Kostant's observation is that these two subvarieties have the same $T$-equivariant Hilbert series.

  1. "Why" is the flag manifold masquerading as a product of ${\mathbb P}^1$s?

  2. More concretely, we know that the two varieties lie in the same connected component of the Hilbert scheme of this projective space, by Hartshorne's thesis. Can one connect them without breaking the $T$-action? (As far as I know, Hartshorne connectivity doesn't hold in general if you keep track of the multigrading, not just the single grading.)

  3. Since the two varieties are both smooth, (EDIT:) and have different topology, there won't be a flat family over an irreducible base in which one is a general fiber, one the special. Do they have a common degeneration? Hartshorne's thesis just guarantees that we can degenerate, deform, degenerate, deform, ... to get from one to the other, not that they will be on adjacent components of the Hilbert scheme.

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Hartshorne connectivity does not hold when you keep track of the multigrading. See arxiv.org/abs/math/0201271 . I don't know about any of the more interesting questions you pose. –  David Speyer Feb 9 '10 at 15:11
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"Since the two varieties are both smooth, there won't be a flat family over an irreducible base in which one is a general fiber, one the special." Are you sure about this? I thought there was a flat family whose general fibers were $P^1 \times P^1$ and whose special fiber was $\Sigma_2$. Namely, set $E:=\mathrm{Ext}_{P^1}(\mathcal{O}(2), \mathcal{O}(0))$. Form the universal rank two vector bundle on $E \times P^1$; this is $\mathcal{O}(2) \oplus \mathcal{O}(0)$ over the origin of $E$ and $\mathcal{O}(1) \oplus \mathcal{O}(1)$ everywhere else. Projectivize to get the claimed example. –  David Speyer Feb 9 '10 at 15:51
    
You're right, $F_0$ degenerates to $F_2$, in Hirzebruch-surface notation. (It even does so $T^1$-equivariantly.) But those manifolds, unlike mine, are diffeomorphic. Question edited to reflect that. –  Allen Knutson Feb 9 '10 at 16:34
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Perhaps because of the Bott-Samelson resolution $X\to G/B$, which is an iterated $\mathbb P^1$-fibration? –  VA. Feb 9 '10 at 20:59
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Thanks. I'd feel terrible if anyone misattributed this question to some other guy Bertrand Kostant. –  Allen Knutson Mar 4 '10 at 23:39

2 Answers 2

I'm hoping someone will check this computation. I get that the Bott-Samelson over $SL_3/B$ is not homeomorphic to $(\mathbb{P}^1)^3$. That is a major obstacle for the approach that Allen and VA discuss above.

There are two reduced words for the longest element of $S_3$, but the automorphism of the Dynkin diagram exchanges them. So, as an abstract variety, it makes sense to talk about the Bott-Samelson for $SL_3$. We'll call it $X$.

Let $F$ be $\mathbb{P}^2$ blown up at a point, with $\pi: F \to \mathbb{P}^2$ the blowdown. Then $X$ is a $\mathbb{P}^1$ bundle over $F$. Specifically, let $Q$ be the tautological quotient over $\mathbb{P}^2$. I believe that $X$ is $\mathbb{P}(\pi^* Q)$.

Let's compute $H^*(X)$. Everything is in even degree. $H^*(\mathbb{P}^2) = \mathbb{Z}[H]/H^3$, where $H$ is the hyperplane class. The blowup $F$ has $H^*(F) = \mathbb{Z}[H,E]/\langle H^2+E^2,\ HE \rangle$ where $E$ is the class of the exceptional fiber. The chern class of $Q$ is $1+H+H^2$. So $$H^(X) = H^*(F)[Z]/\langle Z^2 + ZH + H^2 \rangle = \mathbb{Z}[H,E,Z]/\langle H^2+E^2,\ HE,\ Z^2+ZH+H^2 \rangle.$$

So $H^2(X)$ is three dimensional. We have a cubic form on $H^2(X)$ given by $\alpha \mapsto \int_X \alpha^3$. I get that this cubic form is $$\int_X \begin{pmatrix} Z^3 & & & \\ Z^2 H & Z^2 E & & \\ Z H^2 & ZHE & Z E^2 & \\ H^3 & H^2 E & H E^2 & E^3 \end{pmatrix}= \begin{pmatrix} 0 & & & \\ 1 & 0 & & \\ -1 & 0 & 1 & \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Of course, the corresponding matrix for $(\mathbb{P}^1)^3$ is $$\begin{pmatrix} 0 & & & \\ 0 & 0 & & \\ 0 & 1 & 0 & \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Now, if there were a homeomorphism $X \cong (\mathbb{P}^1)^3$, then there would be an isomorphism $ H^2(X) \otimes \mathbb{C} \cong H^2((\mathbb{P}^1)^3) \otimes \mathbb{C}$ taking one cubic to the other. But I get that the first cubic is a line times a conic, while the second is three lines.

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While I haven't fully checked your calculation, I will note that the Bott-Samelson for a Schubert divisor in $Flags({\mathbb C}^3)$ is $F_1$, not homemorphic to $F_0$, so it seems very plausible. –  Allen Knutson Feb 11 '10 at 2:41
    
I fixed some sign errors, which had no effect on the answer. –  David Speyer Mar 3 '10 at 16:54

Your first question is about two objects becoming isomorphic after quantization, and you're asking "Why?"

Here, the relevant quantum object is the spin representation of $\mathfrak g$, which is a representation of $\mathfrak g\ltimes \mathit{Cliff}(\mathfrak g)$, where $\mathit{Cliff}(\mathfrak g)$ is the Clifford algebra of (the underlying vector space of) $\mathfrak g$, with respect to some invariant inner product; a $\mathbb Z/2$-graded algebra.

Let $S$ be the unique (up to grading reversal) irreducible $\mathbb Z/2$-graded representation of $\mathit{Cliff}(\mathfrak g)$. It has a graded-commuting action of $$C:=\begin{cases}\mathbb C&\text{ if }\quad \dim(\mathfrak g) \text{ is even} \\ \mathit{Cliff}(1)&\text{ if }\quad \dim(\mathfrak g) \text{ is odd.} \end{cases} $$ Let $V$ be the category of modules of the above algebra, so that $S\in V$.
To make things a bit more canonical, one can use the graded Morita equivalence between $C$ and $\mathit{Cliff}(\mathfrak h)$ to identify $V$ with the category of $\mathit{Cliff}(\mathfrak h)$-modules.

Let $\alpha$ denote the adjoint action of $G$ on $\mathit{Cliff}(\mathfrak g)$. For any element $g\in G$, we can pre-compose the action of $\mathit{Cliff}(\mathfrak g)$ on $S$ by $\alpha_g$ to get a new, isomorphic $\mathit{Cliff}(\mathfrak g)$-module in $V$ [here, I'm using that $G$ is connected]. The map that sends $g$ to such an isomorphism is unique up to scalar, and so we get a projective representation of $G$ on $S$.

If $G$ is simply connected, this lifts to an honest action of $G$ on $S$, and so we get an action of $G\ltimes\mathit{Cliff}(\mathfrak g)$ on the object $S\in V$.

All in all, (the underlying vector space of) $S$ has actions of $G$, of $\mathit{Cliff}(\mathfrak g)$, and of $\mathit{Cliff}(\mathfrak h)$.
Now, we can "cancel" two $\mathit{Cliff}(\mathfrak h)$ actions to get a vector space with actions of $G$ and of $\mathit{Cliff}(\mathfrak g\ominus h)$. That's the irreducible $G$-rep with highest weight $\rho$.

This vector space has two descriptions:
(1) The irreducible $G$-rep with highest weight $\rho$.
(2) The irreducible $\mathit{Cliff}(\mathfrak g\ominus h)$-module.

Note however that this operation of "canceling" the two $\mathit{Cliff}(\mathfrak h)$ actions is a bit unnatural. For example, the vector space (1) is purely even, while (2) is genuinely $\mathbb Z/2$-graded. You probably see that same weirdness on the classical side of the problem when you try to identify the flag variety with a product of $\mathbb P^1$s.

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This looks pretty great, but I have two questions. (1) What about $n\rho$ for $n>1$? (2) Does this suggest that we should see the two strangely related spaces as living inside the same space ${\mathbb P}S$? –  Allen Knutson Nov 14 '13 at 18:58
    
The short answer: I don't know. To cover the $n>1$ case, one would need to replace the Clifford by some other kind of algebra... and I have no idea what that could be. As far as (2) is concerned, I would say that it doesn't make sense to try to address (2) before understanding (1). So, I guess, I have nothing to offer, sorry. –  André Henriques Nov 14 '13 at 20:00

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