Is the big cell a principal open set?

Question

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.

Answer 1

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.

Answer 2

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 $GL_n$.