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Suppose $G$ is a semi-simple group of adjoint type over an algebraic closed field, and $X$ its wonderful compactification a la De Concini and Procesi. Let $P=MU$ be a parabolic subgroup in $G$, and let $Y$ be the wonderful compactification of $M$.

Question. Can you describe the closure of $P$ in $X$? In particular, is it true that the closure of $P$ fibers over $Y$?

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  • $\begingroup$ The closure in $X$ of a maximal torus $T_G$ is the toric variety $\overline{T}_G$ whose fan is the Weyl fan of $G$. So for the maximal torus $T_M$ of the Levi factor the closure of $T_M$ inside $X$ equals the closure of $T_M$ in $\overline{T}_G$. Does that toric variety admit a morphism to the toric variety $\overline{T}_M$ of $M$? That should be something about Weyl fans . . . $\endgroup$ – Jason Starr Jul 21 '16 at 21:08
  • $\begingroup$ . . . if I remember correctly. $\endgroup$ – Jason Starr Jul 21 '16 at 21:10
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    $\begingroup$ There is a problem: Since $M$ is not semisimple, let alone of adjoint type, it doesn't have a wonderful embedding. $\endgroup$ – Friedrich Knop Jul 22 '16 at 18:33
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    $\begingroup$ Let $w$ be the longest Weyl group element of $M$. Then the closure of $P$ is the closure of the $B\times B$-orbit $BwB$ in $X$. The orbit closures of arbitrary $B\times B$-orbits have been studied thoroughly in a series of papers by Springer. In particular, they are all normal with rational singularieties. $\endgroup$ – Friedrich Knop Jul 22 '16 at 18:36
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    $\begingroup$ @Ramin: $BwB$ is dense in $P$. That's it. Your question can be seen in the context of spherical varieties for which Borel group orbits have been studied to some extent. In your case, the group acting is $G\times G$ and $B\times B$ is its Borel subgroup. $\endgroup$ – Friedrich Knop Jul 23 '16 at 19:53
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I am just writing my comment above as an answer, just in the special case that the parabolic is a Borel subgroup $B$. In this case, the reductive part $M$ is a maximal torus $T$ in $G$. Denote by $W\subset N_G(T)$ a Weyl group. There is a conjugation action of $G$ on itself, and there is an induced conjugation action of $W$ on $T$. Denote by $q_T:T\to T/W$ the quotient of this finite group action. It is a "classical fact" that there is a conjugation-invariant morphism $$q_G: G \to T/W,$$ that extends $q_T$ and that is a geometric quotient of the conjugation action of $G$ on itself, cf. Section 6 of the following.

MR0180554 (31 #4788) Reviewed
Steinberg, Robert
Regular elements of semisimple algebraic groups.
Inst. Hautes Études Sci. Publ. Math. No. 25 1965 49–80.
14.50 (20.75)
http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1965__25_/PMIHES_1965__25__49_0/PMIHES_1965__25__49_0.pdf

In particular, compare the restriction of $q_G$ to $B$ with the composite morphism $$ B = T\times U \xrightarrow{\text{pr}_T} T \xrightarrow{q_T} T/W.$$ By restricting to the open dense subset of regular elements, it is clear that these two morphisms are equal.

Why is this relevant? The following is derived from an article with Xuhua He.

MR2833468 Reviewed
He, Xuhua(PRC-HKST); Starr, Jason(1-SUNYS)
Semistable locus of a group compactification. (English summary)
Represent. Theory 15 (2011), 574–583.
14L30 (14M27)
https://arxiv.org/abs/0907.0281

The Zariski closure $\overline{T}$ in $\widehat{G}$ is projective and normal. There is a maximal open subscheme $\widehat{G}^{ss}\subset \widehat{G}$ containing $G$ that is explicitly described in that article on which the morphism $q_G$ extends to a regular morphism, $$\widehat{G} \supseteq \widehat{G}^{ss} \xrightarrow{\overline{q}_G} \overline{T}/W.$$ If I am reading the article correctly, for the opposite Borel $B'$ to $B$, for the associated left-right action of $B'\times B$ on $\widehat{G}$, and for the conjugation action of a diagonal copy $G_\Delta$ in $G\times G$, the semistable locus is a disjoint union of $G_\Delta\cdot (B'\times B)$-orbits indexed by subsets of the set of simple roots (presumably one for each stratum in the boundary stratification of $\widehat{G}$).

Denote by $\overline{B}$ the normalization of the closure of $B$ (I believe the closure is already normal, but I am not completely certain). Edit. Friedrich Knop points out that Springer proved that the closure of $B$ is normal. Denote by $\overline{B}^{ss} \subset \overline{B}$ the fiber product of $\overline{B}$ and $\widehat{G}^{ss}$ over $\widehat{G}$. Denote by $\overline{B}^o$ the maximal open subscheme of $\overline{B}$ on which the following rational transformation extends to a regular morphism, $$\overline{B} \supseteq B \xrightarrow{\text{pr}_T} T \subseteq \overline{T}.$$ By Zariski's Main Theorem, $\overline{B}^o$ is also the maximal open on which this extends to a finite-to-one correspondence. In particular, since $\overline{T}\to \overline{T}/W$ is finite, $\overline{B}^o$ equals the maximal regular domain of the rational transformation, $$ \overline{B} \supseteq B \xrightarrow{q_T} T/W \subseteq \overline{T}/W.$$ Thus $\overline{B}^o$ contains $\overline{B}^{ss}$ (presumably these are equal open subsets).

Anyway, the deepest $G\times G$-orbit of $\widehat{G}$ is isomorphic to $G/B\times G/B'$. The intersection with $\overline{B}$ appears to equal $\{B/B\}\times G/B'$ union $G/B\times \{B'/B'\}$. It looks to me like the intersection with $\overline{B}^o$ equals the complement of the point $(B/B,B'/B')$. Certainly in the special case that $G=\textbf{PGL}_2$, all of this is straightforward to compute.

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  • $\begingroup$ This is helpful detail. (It may be useful for people to know that the papers referred to are freely accessible online.) $\endgroup$ – Jim Humphreys Jul 22 '16 at 17:37
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    $\begingroup$ @JimHumphreys. Thank you for the suggestion. I added web links to the papers. $\endgroup$ – Jason Starr Jul 22 '16 at 21:13
  • $\begingroup$ Here is a link to the published paper by He and you, which I guess is the same as the preprint version. ams.org/mathscinet-getitem?mr=2833468 $\endgroup$ – Jim Humphreys Jul 22 '16 at 21:57

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