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The topic of Whitney stratifications came up in a lecture, and the general procedure in the examples was to decompose the singular locus of the variety into the strata starting with the "worst" ones. On the Whitney Umbrella for example (defined by $x^2+yz^2$ say), one would want the origin, the punctured y-axis, and the smooth locus as the Whitney strata (as opposed to just the y-axis and the smooth locus).

I know that algebraic varieties have a Whitney stratification. I was wondering if the following idea works for providing a Whitney stratification for algebraic varieties over an algebraically closed field in characteristic 0:

Using the desingularization invariant from Bierstone and Milman in http://arxiv.org/pdf/math/0702375.pdf for example, we can define a tuple $inv(x) = (\nu_1(x),s_1(x),...,\nu_k(x),s_k(x),\nu_{k+1}(x))$ at a point $x$ in our variety $X$. Since the computation is in year 0 (as per the language of the literature on the topic), all the $s_i(a)=0$ since they count exceptional divisors from blowings-up during the desingularization process. The point is that the invariant measures (in a sense) the worst singularities. If we define the strata by $X_a=\{x\in X: inv(x)=a\}$ for each possible value $a$ of the invariant, does this collection define a Whitney stratification? The fact that the invariant is upper semi-continuous takes care of the closure conditions for being a stratification, but I am not sure if Whitney's condition B (which implies A) is satisfied.

In our example of the Whitney Umbrella, the invariant at the origin is $(2,0,3/2,0,1,0,\infty)$ as opposed to $(2,0,1,0,\infty)$ along the rest of the y-axis, so the origin, the punctured y-axis, and the smooth locus do give the correct result in this case.

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  • $\begingroup$ Is it possible to draw a picture for the collection $X_a$ in the case of the Whitney umbrella? $\endgroup$
    – Saurabh T
    Commented Jan 30, 2015 at 13:24
  • $\begingroup$ Well you can draw the Whitney Umbrella using the coordinates I gave, and then you can colour the origin (the pinch point) one colour, the $y$-axis minus the origin (the singular normal crossing points) another colour, and then you can colour the remaining points (all the smooth ones) with a final colour. This would be your Whitney stratification. Of course when working over $\mathbb{C}^3$ you have to draw your picture in $\mathbb{R}^3$ like this en.wikipedia.org/wiki/Whitney_umbrella $\endgroup$
    – Sergio Da Silva
    Commented Jan 30, 2015 at 14:38
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    $\begingroup$ You just described a Whitney b-stratification of the Whitney umbrella. I know that. I wanted to know if it is possible to picture $X_a$'s in this case, ($X_a$'s as in your definition). Or, do you mean $X_a$'s are indeed the strata of this b-stratification of the Whitney umbrella. $\endgroup$
    – Saurabh T
    Commented Jan 30, 2015 at 14:48
  • $\begingroup$ Yes I am referring to the $X_a$'s as the strata for the $b$-stratification (by which I assume you mean a strata that satisfies Whitney condition $b$). For example, you can see fairly easily that if you had a sequence in the smooth locus and a sequence in the $y$-axis converging to a point away from the origin, then secant lines converging to $L$ would certainly be contained in the limit of the tangent planes of the smooth points. You should also convince yourself that this does not happen at the origin if I instead chose my strata to be the entire $y$-axis and the smooth locus. $\endgroup$
    – Sergio Da Silva
    Commented Jan 30, 2015 at 16:13
  • $\begingroup$ Okay, so in the case of Whitney umbrella, the method you described gave a Whitney b-regular stratification. What about the following variety $y^2=z^2x^2+x^3$. Do you still get a Whitney b-stratification? $\endgroup$
    – Saurabh T
    Commented Jan 31, 2015 at 0:04

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