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 $Y$? 



Wonderful compactification
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.


