Is the ideal of a closure of a Bruhat cell generated by generalized minors? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:32:44Z http://mathoverflow.net/feeds/question/77187 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77187/is-the-ideal-of-a-closure-of-a-bruhat-cell-generated-by-generalized-minors Is the ideal of a closure of a Bruhat cell generated by generalized minors? Ben Webster 2011-10-05T00:35:13Z 2011-10-06T08:50:13Z <p>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 $\overline{BwB}$. </p> <p>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 <strike>$\omega_{w'\delta,v}(g)=w'\delta(gv)$</strike> $\Delta_{w'\delta,v}(g)=w'\delta(gv)$ 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 $w'>w$. </p> <p>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:</p> <blockquote> <p>Is the ideal generated by these generalized minors already radical?</p> </blockquote> http://mathoverflow.net/questions/77187/is-the-ideal-of-a-closure-of-a-bruhat-cell-generated-by-generalized-minors/77208#77208 Answer by Alexander Braverman for Is the ideal of a closure of a Bruhat cell generated by generalized minors? Alexander Braverman 2011-10-05T06:25:37Z 2011-10-05T06:25:37Z <p>First of all, I think you need to write $w'\lambda&lt; w\lambda$ (look what happens when $w$ is 1).</p> <p>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 $\lambda$).</p> http://mathoverflow.net/questions/77187/is-the-ideal-of-a-closure-of-a-bruhat-cell-generated-by-generalized-minors/77341#77341 Answer by Allen Knutson for Is the ideal of a closure of a Bruhat cell generated by generalized minors? Allen Knutson 2011-10-06T08:50:13Z 2011-10-06T08:50:13Z <p>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</p> <p>.1.</p> <p>111</p> <p>.1.</p> <p>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.</p>