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Let $G$ be your favorite complex semi-simple algebraic group, and let $B\supset T$ be your favorite Borus. For any $w\in W$, we have the Bruhat cell $BwB$, and its closure $\overline{BwB}$.

Now, it's very easy to write down some functions that cut out this variety. Let $V$ be any finite-dimensional representation of highest weight $\lambda$, and let $v$ be highest weight vector, and $\delta$ a non-zero functional killing all but the highest weight space. Then the generalized minors $\omega_{w'\delta,v}(g)=w'\delta(gv)$ $\Delta_{w'\delta,v}(g)=w'\delta(gv)$ for all $w'$ with $w'\lambda > w\lambda$ all vanish on $BwB$ (since $BwBv$ in contained in the sum of weight spaces $\leq w\lambda$), and on no Bruhat cells $Bw'B$ with $w'>w$.

That is, the radical of the ideal generated by these functions is all functions vanishing on $\overline{BwB}$. In fact, it's enough to just consider $\lambda$ fundamental to get an ideal with the correct radical. So, my question is:

Is the ideal generated by these generalized minors already radical?

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First of all, I think you need to write $w'\lambda< w\lambda$ (look what happens when $w$ is 1).

It seems to me that when $G$ is not $SL(n)$ the answer is no. For example assume that $w=1$. Then you know that the relations are generated by all matrix coefficients $\omega_{\eta,v}$ (your notations) where $\eta$ is a functional which vanishes on the lowest weight vector of $V$ and your generators correspond to $\eta$ being an extremal weight vector. But if the fundamental representations of $G$ are not minuscule I don't see how you get relations with $\eta$ not being an extremal weight vector in $V^*$ (for fundamental $V$) from those with extremal $\eta$ - this seems impossible for degree reasons (if you introduce the multigrading corresponding to $\lambda$).

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  • $\begingroup$ Not sure where my brain was when I wrote $\omega$ instead of $\Delta$. Just writing too fast. $\endgroup$
    – Ben Webster ♦
    Commented Oct 5, 2011 at 14:39
  • $\begingroup$ I'm also not sure I follow. What's the grading the corresponding to $\lambda$? Shouldn't one pick a coweight rather than a weight to get a grading? $\endgroup$
    – Ben Webster ♦
    Commented Oct 5, 2011 at 14:48
  • $\begingroup$ By the way, I think the fact I really wanted was the Theorem from Ramanathan you linked to in your deleted answer. Thanks. $\endgroup$
    – Ben Webster ♦
    Commented Oct 5, 2011 at 14:54
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    $\begingroup$ OK, good. That's what seemed to me at the first glance, but now it seems to me that the way you actually formulated the question is different. Ramanathan says that the ideal of a Schubert variety is generated by elements of degree 1 in the homogeneous ring of the flag variety, whereas in your question you are talking about some particular elements of degree 1. $\endgroup$
    – Alexander Braverman
    Commented Oct 5, 2011 at 17:09
  • $\begingroup$ Right, but one doesn't always ask the question one actually wanted to know the answer to. $\endgroup$
    – Ben Webster ♦
    Commented Oct 6, 2011 at 4:42
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Exactly as Alexander said, this will fail for $G/P$ nonminuscule. The smallest example is the closed orbit $SO(5)/P$ of $SO(5)$ acting on ${\mathbb P}({\mathbb C}^4)$. The $T$-weight diagram of this representation is

.1.

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.1.

The representation arises as the space of sections of ${\mathcal O}(1)$ on $SO(5)/P$. If we take the space of sections over the Schubert point $P/P$, we get just the top $1$. The extremal weight vectors you want to kill correspond to the left, right, and bottom $1$s. But to get the Schubert point on the nose, you have to kill the $1$ in the middle too.

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