Is the big cell a principal open set?

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.

• Is this an intrinsic property? Does not it depend on how the affine algebraic group is realized as a linear algebraic group? – P Vanchinathan Apr 12 '14 at 8:41
• 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? – Jesko Hüttenhain Apr 12 '14 at 8:58
• @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. – Jim Humphreys Apr 12 '14 at 14:10

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.
• 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. – Jesko Hüttenhain Apr 12 '14 at 9:28
• 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 $GL(n)$. – Allen Knutson Apr 12 '14 at 15:08