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My question is simple:

Given an algebraic group $G$ acting on a variety $X$ algebraically. If the orbits are of finite number then they form what is called an algebraic stratification of $X$.

Now my question is: Is this (the stratification by $G$-orbits) also a Whitney Stratification?

I think showing this by hand is quite painfull since the definition of a Whitney stratification is very unhandy.

I would appreciate any reference.

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    $\begingroup$ I'm pretty sure this is true, but don't know a precise reference. A good person to ask might be Tom Braden. $\endgroup$
    – naf
    Commented Apr 30, 2013 at 14:59
  • $\begingroup$ This is true in the case of partial flag varieties, cf the comments at the beginning of section 3.12 in this paper: arxiv.org/pdf/0809.4785.pdf $\endgroup$
    – Chuck Hague
    Commented Apr 30, 2013 at 16:54
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    $\begingroup$ yes, this is true and Braden is the perfect person to ask as ulrich says! If all you want is that the categories are preserved by the six operations then this is not difficult to see using equivariant sheaves (see related comments at the beginning of "tilting exercises" by Beilinson, Bezrukavnikov, Mirkovic). $\endgroup$
    – Geordie Williamson
    Commented Apr 30, 2013 at 18:03

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So Ulrich and Geordie were right, Tom Braden was the right person to ask and here is what he told me:

The answer is yes, in the case above $X$ is Whitney stratified.

The argument goes roughly as follows. In the paper [K] the following is shown. Let $X$ be an algebraic stratified variety and $ S$ a stratum. The set $Sing(S )$ of points in $S $ which do not fulfil Whitney's b condition has the structure of a (semi-)variety of dimension strictly lower that $\dim S $.

Now applied to our situation : $Sing(S)$ is necessarly $G$-invariant, hence empty. This answers my question.

[K] ``A Geometric Proof of Existence of Whitney Stratifications’’, Moscow Math. Journ., 5 (2005), no.1, 125—133

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  • $\begingroup$ The author of the cited paper [K] is Vadim Kaloshin. $\endgroup$
    – user44400
    Commented Mar 23, 2018 at 7:44
  • $\begingroup$ Note that there is a gap in the proof, as discussed in Nyguen-Trivedi-Trotman's paper in the Illinois Math Journal. (The singularity set that Kaloshin defines does account for every failure of the Whitney conditions). $\endgroup$
    – Phil Tosteson
    Commented Sep 27 at 14:04
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See:

MR0342523 (49 #7269) Luna, Domingo Slices étales. (French) Sur les groupes algébriques, pp. 81–105. Bull. Soc. Math. France, Paris, Memoire 33 Soc. Math. France, Paris, 1973.

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