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Is there a simple characterization of the big Bruhat cell of the universal (simply-connected) Chevalley groups over $\mathbb C$?

For example, it is known that the Borel subgroup of $\mathrm{SL}_n(\mathbb C)$ are the upper triangular matrices and the longest Weyl element $w_0$ is the permutation matrix for $(1 2.. n)$.
Furthermore if for every $1 ≤ i \neq j ≤ n$ let $E_{i,j}$ be the subgroup of $\mathrm{SL}(n, \mathbb C)$ consisting of the matrices with $1$ on the diagonal and zero on the $(r, s)$-entry whenever $1 ≤ r \neq s ≤ n$ and $(r, s) \neq (i, j)$ and define $U_{i,j} := \langle E_{r,s} \mid j < s\,or\,j = s\,\mbox{ and }\,r ≤ i\rangle$.
Than for every $x$ in the big Bruhat cell we can write $x=bw_0u$ for $b \in B$ and $u \in U_{n-1,n}$.

Is there something similar for the general case?

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    $\begingroup$ Exactly the same is true for any Chevellay group: one chooses any lifting of the longest Weyl group element $w_0$ to $G$ and then the big cell is $Bw_0B$ (or $Bw_0U$ if you wish, where $U$ is the unipotent radical of $B$). This is the maximal dimension cell of the Bruhat decomposition. $\endgroup$
    – Victor Petrov
    Commented Jul 21, 2019 at 19:00
  • $\begingroup$ In the simply connected case one can show that the big cell $U^{-} B$ is a principal open set, see Lemma 4.5 in "Properties and linear representations of Chevalley Groups" by Borel (Seminar on algebraic groups). The original reference is "Certains schémas de groupes semi-simples" by Chevalley, pp. 226-228. In your example for $G = SL_n(\mathbb{C})$, the big cell corresponds to matrices $(g_{ij})$ for which the $n$ leading principal minors are nonzero. $\endgroup$
    – spin
    Commented Jul 21, 2019 at 19:27
  • $\begingroup$ Yes, but how do I find the longest Weyl group element, the Borel subgroup and it's unipotent radical? $\endgroup$
    – Ami
    Commented Jul 21, 2019 at 19:49
  • $\begingroup$ @Ami: I think the point is that there are solid proofs for existene of a Borel subgroup (corresponding to any fixed choice of positive roots, say), and similarly for the longest element of $W$. To describe these explicitly in the case of exceptional simple algebraic groups may be difficult, of course. But the root systems are well-studied. Anyway, some of your restrictions on the big group are unnecessary here (for example, simple connectedness). $\endgroup$
    – Jim Humphreys
    Commented Jul 22, 2019 at 0:26
  • $\begingroup$ By the way, I came here from one of your more recent questions, and want to mention also here that $(1\ 2\ \ldots\ n)$ is not the long element of $A_{n - 1}$; it is a Coxeter element, and has length $n$. The long element is $(1\ n)(2\ n - 1)(3\ n - 2)\dotsb$, and has length $n(n - 1)/2$. $\endgroup$
    – LSpice
    Commented Aug 14, 2019 at 16:09

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