TITLE: Is the ideal of a closure of a Bruhat cell generated by generalized minors?
QUESTION [10 upvotes]: 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?

REPLY [3 votes]: 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.
111
.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.