## Why do flag manifolds, in the P(V_rho) embedding, look like products of P^1s?

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 2010 at 15:11
"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 2010 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 2010 at 16:34
Perhaps because of the Bott-Samelson resolution $X\to G/B$, which is an iterated $\mathbb P^1$-fibration? – VA Feb 9 2010 at 20:59
Thanks. I'd feel terrible if anyone misattributed this question to some other guy Bertrand Kostant. – Allen Knutson Mar 4 2010 at 23:39

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 2010 at 2:41 I fixed some sign errors, which had no effect on the answer. – David Speyer Mar 3 2010 at 16:54