7
$\begingroup$

Let $G$ be a complex affine reductive algebraic group, $B\subseteq G$ a Borel with maximal torus $T$ and unipotent radical $U$. Let $w\in\operatorname N_G(T)$ be a representative of the longest Weyl element. I am wondering whether the big open Bruhat cell $BwB\subseteq G$ is a principal open set, i.e. whether there is a regular function $f\in\mathbb C[G]$ with $BwB=\{ g\in G\mid f(g)\ne 0 \}$.

This is true for $G=\operatorname{GL}_n(\mathbb C)$, because the open Bruhat cell is the set of all invertible matrices with nonvanishing principal minors, in this case $f$ would be the product of those.

A little more generally, this is true when $\mathbb C[G]$ is factorial: The complement of any affine in a noetherian, normal and separated scheme is pure of codimension one. Since algebraic groups are smooth and the open cell is isomorphic to the affine variety $B\times U$, its complement is pure of codimension one. Each codimension one subvariety of $G$ will be the vanishing set of a single regular function because $\mathbb C[G]$ is a UFD. Hence, the product of these functions will cut out the complement of the open cell set-theoretically.

I do not see how I would go about proving the statement in the general case, though - and I am not sure if is correct at all.

$\endgroup$
3
  • 1
    $\begingroup$ Is this an intrinsic property? Does not it depend on how the affine algebraic group is realized as a linear algebraic group? $\endgroup$ Apr 12, 2014 at 8:41
  • 1
    $\begingroup$ I am asking whether the sections $\mathcal O_G(BwB)$ are equal to a localization of $\mathcal O_G(G)=\mathbb C[G]$, so I think this is intrinsic? $\endgroup$ Apr 12, 2014 at 8:58
  • 1
    $\begingroup$ @Jesko: It's worth emphasizing that the picture is basically the same for all (connected) reductive groups in all characteristics. The reference by Knop et al. to a 1976 Advances in Math. paper by Birger Iversen is most relevant, I think. $\endgroup$ Apr 12, 2014 at 14:10

2 Answers 2

15
$\begingroup$

This is true if $G$ is (semi-simple) simply-connected, because then $\mathrm{Pic}(G)=(0)$, which means that $\mathbb{C}[G]$ is factorial; however, it is false for the simplest non simply-connected example, namely $G=\mathrm{PGL}(2)$. Indeed $G$ is the complement of the quadric $ad-bc=0$ in $\mathbb{P}^3$; this implies that $\mathrm{Pic}(G)=\mathbb{Z}/2$, generated by the line bundle $\mathcal{O}_{\mathbb{P}^3}(1)$ restricted to $G$. The complement of the big cell is the divisor $a=0$, whose class in $\mathrm{Pic}(G)$ is the nonzero element; hence it is not principal.

$\endgroup$
6
  • $\begingroup$ Dear abx, thanks a lot! Can you name a reference for the fact that simply connected groups have trivial Picard group? Also, is it important to require $G$ semi-simple for this to hold? The case of $\operatorname{GL}_n$ suggests that reductive might be enough. $\endgroup$ Apr 12, 2014 at 9:28
  • $\begingroup$ No, semi-simple is not important. I suggest §4 of this paper. Of course the results are much older, but they are nicely put together, and the authors give many references to original work. $\endgroup$
    – abx
    Apr 12, 2014 at 9:48
  • $\begingroup$ That's a wonderful reference. Thanks again, this is a good start to the weekend =). $\endgroup$ Apr 12, 2014 at 9:56
  • $\begingroup$ @abx: The reference you include is most helpful, even though they limit their discussion for convenience to characteristic 0 (while pointing out the general case). Maybe it's helpful to add the source of the artcile: Algebraische Transformationsgruppen und Invariantentheorie, 63–75, DMV Sem., 13, Birkhäuser, Basel, 1989. $\endgroup$ Apr 12, 2014 at 14:13
  • 8
    $\begingroup$ This is missing perhaps the most concrete and interesting part, which is what the function is. The answer is $g \mapsto \prod_\omega \langle \vec v^\omega, g \cdot \vec v_\omega \rangle$ where $\omega$ ranges over the fundamental representations. The property you want of $G$ is that you have all these fundamental representations available. Note that this function generalizes the one you mentioned for $GL(n)$. $\endgroup$ Apr 12, 2014 at 15:08
4
$\begingroup$

Actually the answer is YES and it does not depend on whether G is simply connected. The big cell is the non-vanishing locus of the regular function $f(g)=\langle \rho(g) v, w\rangle$, where $(\rho,V)$ can be taken to be any irreducible representation of $G$ whose highest weight $\chi$ lies inside the interior of the Weyl chamber. Here $v$ is a highest weight vector of $\rho$ and $w$ is a lowest weight vector of the contragredient representation $(\rho^*,V^*)$.

The point is that $u:=\rho(w_0)v$ is the lowest weight vector of $\rho$ and hence, for any $b \in B$, the vector $\rho(bw_0)v \in \rho(B)u$ has a non-zero component on $u$ modulo the sum of the weight spaces of weight $\neq w_0\chi$. This shows that $f$ does not vanish on $Bw_0B$. On the other hand if $g$ belongs to any other Bruhat cell $BwB$ for $w \neq w_0$, then $\rho(g)v \in \rho(B)\rho(w)v$. But $\rho(w)v$ is a weight vector of weight $w\chi$, hence $\rho(B)\rho(w)v$ is contained in the sum of weight spaces with weight $\ge w\chi$, hence distinct from the lowest weight $w_0\chi$.

So $PGL_2$ is perfectly fine. In the answer given by @abx the divisor $a=0$ is principal, because it is the zero locus of $a^2/(ad-bc)$, a regular function on $PGL_2$.

And by the way one must distinguish the big Bruhat cell $Bw_0B$ from the big Gauss cell $w_0Bw_0B$. It is for the latter that one has the characterization in terms of non-vanishing principal minors when $G$ is $GL_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.