Whitney Conditions vs Equisingularity

Question

In studying singular spaces, it is often important to pick an appropriate stratification which encodes the singularity structure. One class of such stratifications are called "Whitney stratifications" after the work of H. Whitney (with many following contributions by Thom, Mather..) . A Whitney stratification obeys certain conditions that are meant to intuitively capture the idea that every point on a fixed stratum looks roughly the same.

Now, there appears to be a somewhat separate theory of "equisingularity" due to O. Zariski . My question is :

• How closely are these two theories related ? In particular, is it possible to have an "equisingular stratification" (I am not sure if this the right technical term) that is not a Whitney stratification ?

I have seen weakened versions of Whitney conditions appearing at various places in the literature, but I don't know of a place where these are compared with Zariski's notion of "equisingularity".

Answer 1

I am not sure what you mean by "equisingular" stratification. But I guess you would like to say that $X$ is equisingular along $Y$ if the local rings $\mathcal{O}_{X,x}$ have constant multiplicity for all $x \in Y$. The equisingular stratification would be a stratification with respect to the condition being equisingular along a subvariety.

In that case, equisingular stratifications and Whitney stratifications are not the same thing (though a Whitney stratification is always an equisingular stratification). You can construct a counter-example as follows.

First of all, if I remember correctly, if $Y \subset X$ is a smooth Whitney stratum, then $X$ is normally flat along $Y$. In particular, if $Y$ is closed in $X$, then the scheme-theoretic exceptional divisor of the blow-up of $X$ along $Y$ is flat over $Y$.

Now, let $Z$ be a generic hyperplane section of $\mathbb{P}^2 \times \mathbb{P}^1 \subset \mathbb{P}^5$. The surface $Z$ is a ruled cubic surface. I denote by $L$ the principal axis of the ruling.

Let us denote by $X \subset (\mathbb{P}^4)^*$ the projective dual of $Z$. It is easily checked that $X$ is a cubic hypersurface which is not a cone. In particular, all singular points of $X$ have exactly multiplicty $2$ in $X$.

Let $H$ be a hyperplane containing $L$. Since $H$ cuts $Z$ in a curve of degree $3$ (Bezout's Theorem), we have $H \cap Z = L \cup \mathcal{C}$, were $\mathcal{C}$ is a curve of degree $2$ on $Z$. In fact, one proves easily that $\mathcal{C}$ must be the union of two lines ruling $Z$, say $l_H$ and $l'_H$.

If $l_H \neq l'_H$, then $H$ is tangent to $Z$ along the two points $l_H \cap L$ and $l'_H \cap L$. Thus, $H^{\perp}$ is a singular point of $X$ and the tangency locus of $H$ along $Z$ is two distinct points.

If $l_H = l'_H$, then $H$ is tangent to $Z$ along $l_H$. Again $H$ is a singular point of $X$.

This shows that $L^{\perp}$ is in the singular locus of $X$. In fact, one proves that the singular locus of $X$ is exactly $L^{\perp}$. Hence, the singular locus of $X$ is a $\mathbb{P}^2$ (in particular it is smooth) and $X$ has multiplicity exactly $2$ along all the points of $H^{\perp}$. Hence $X - L^{\perp} ; L^{\perp}$ is an equi-singular stratification of $X$.

Let me prove that this stratification is not a Whitney stratification. By the preliminary remark, I only have to prove that $X$ is not normally flat along $L^{\perp}$.

Let $I_Z =\{(z,H^{\perp}) \in Z \times (\mathbb{P}^{N})^*, \textrm{such that} \, T_{Z,z} \subset H \}$ be the conormal variety of $Z$. Its projection on the second factor maps onto the projective dual of $Z$ (which is $X$).

If $Z$ is smooth and its projective dual is a hypersurface, it is a classical result of Nash that $I_Z$ is the blow-up of $X$ along its singular locus (which is $L^{\perp}$).

As a consequence, we see that the exceptional locus of this blow-up can not be flat over $L^{\perp}$. Indeed, if $p$ is the projection:

$$ p : I_Z \longrightarrow X$$

then the fiber of $p$ over $H^{\perp} \in L^{\perp}$ is:

_ two points when $l_H \neq l'_H$

_a line when $l_H = l'_H$.

This proves that $X$ is not normally flat along $L^{\perp}$ and so $X-L^{\perp}, L^{\perp}$ is not a Whitney stratification of $X$.

Hope for Whitney = equisingular. Teissier and his collaborators introduced in the 70's a much finer notion of multiplicity (which basically takes into account all the multiplicities of the local rings of all the generic polars passing through a fixed point of $X$) and he proves in that context that Whitney = equisingular. The paper where he proves this result is called "Variétés polaires II : Mulitplicités polaires, sections planes et conditions de Whitney". You can find it here : https://webusers.imj-prg.fr/~bernard.teissier/documents/VarPol2.pdf

Answer 2

Here is a result of Trotman relevant to your question. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ra}{\rightarrow}$ $\DeclareMathOperator{\cl}{cl}$ $\DeclareMathOperator{\dist}{dist}$

Suppose that $(X,Y)$ is a pair of $C^1$ submanifolds of the $\bR^N$, $\dim X= m$. Assume $X\subset\cl(Y)\setminus Y$. Then the pair $(X,Y)$ satisfies the Whitney regularity condition (b) along $X$ if and only if, for any open set $U\subset \bR^N$, and any $C^1$-diffeomorphism $\Psi : U\ra V$, where $V$ is an open subset of $\bR^N$, such that $$ \Psi(U\cap X) \subset \bR^m\oplus 0 \subset \bR^N, $$ the map $$ \Psi(Y\cap U)\ra \bR^m\times (0,\infty),\;\;y\longmapsto \bigl( \,{\rm proj}\,(y)\,, \,\dist(y,\bR^m)^2\,\bigr), $$ is a submersion, where ${\rm proj}:\bR^N\ra \bR^m$ denotes the canonical orthogonal projection

The submersion condition is a form of equisingularity. The fiber of the above submersion is the normal link of $X$ in $Y$. For details see D. Trotman's original paper

Geometric versions of Whitney regularity conditions, Ann. Scien. Ec. Norm. Sup., ´ 4(1979), 453-463.