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Let $G$ be a reductive group acting on a smooth projective $X$. Let $P$ be a parabolic subgroup of $G$, and $Y$ a locally closed subvariety invariant under $P$. Assume in addition $Y$ is smooth. ( This all come from Kirwan's 'Cohomology of quotients ' Chapter 13). Then Kirwan used the notion $G\times_PY$, and claimed this variety is smooth.

My question: 1. What is the precise definition of $G\times_PY$, it looks like a fiber product but I could not figure out how.

2.Why is $G\times_PY$ smooth?( if this does not follow automatically from the definition)

Thanks!

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    $\begingroup$ The notation is unfortunate because of the clash with the fibre product notation. Here one defines $G\times_PY=(G\times Y)/\sim$ where $(gp,y)\sim (g,py)$ for all $p\in P$. $\endgroup$
    – Peter McNamara
    Commented Nov 14, 2019 at 3:06
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    $\begingroup$ $G \times_P Y$ is smooth because $G \times Y$ is a $P$-torsor over $G \times_P Y$. Then $G$ and $Y$ are both smooth, so $G \times Y$ is smooth. In general if $f: X \to Z$ is a smooth surjection then $X$ being smooth implies $Z$ is smooth. In fact, you only need that $f$ is fppf, though this fact is harder and the homological characterization of regularity. So smoothness of $G \times Y$ implies smoothness of $G \times_P Y$. $\endgroup$
    – Aaron Landesman
    Commented Nov 14, 2019 at 4:40
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    $\begingroup$ A commonly used notation, which avoids the confusion with fibered products, is $G \times^{P}Y$. $\endgroup$
    – Angelo
    Commented Nov 14, 2019 at 9:17
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    $\begingroup$ I believe it's called a "contracted product" in the literature. $\endgroup$
    – Minseon Shin
    Commented Nov 14, 2019 at 11:03

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