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Let $G$ a connected reductive group and $S=U^{-}TU$ the open cell.

Do we have $G=\bigcup\limits_{g\in G}gSg^{-1}$?

And also if I assume that $G$ is adjoint and $\overline{G}$ is the de Concini-Procesi compactification.

We also have an open Schubert cell $S=U^{-}\overline{T}U$, do we have that: $\overline{G}=\bigcup\limits_{g\in G}gSg^{-1}$?

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    $\begingroup$ Is $G$ connected reductive (so "open schubert cell" makes sense)? If so then every $h \in G$ lies in a Borel, and every Borel is conjugate to a single Borel, so since an open cell contains a Borel we are done (for your first question). $\endgroup$
    – Marguax
    Commented Oct 6, 2013 at 19:12
  • $\begingroup$ The open Bruhat cell is the double coset $Bw_0 B$ in the Bruhat decomposition of $G$ (whereas the Schubert cells live in the flag variety isomorphic to $G/B$). Aside from this, the identity element of $G$ lies in the double coset $B$ but not in any other double coset, so your union of conjugates can't be all of $G$. The question needs some reformulation, I think. $\endgroup$
    – Jim Humphreys
    Commented Oct 6, 2013 at 19:41
  • $\begingroup$ yes sorry, I corrected it. $\endgroup$
    – prochet
    Commented Oct 6, 2013 at 19:49
  • $\begingroup$ @prochet: With this version of the open cell, which contains $B$, Marguax's comment takes care of your first question (since every element of $G$ is conjugate to an element of $B$ by standard structure theory) and probably also the second. $\endgroup$
    – Jim Humphreys
    Commented Oct 6, 2013 at 20:43

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