# Whitney Conditions vs Equisingularity

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".

## 2 Answers

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

• My terminology was indeed imprecise. I think you have given one precise version of what can be termed a equisingular stratification. As for whether Whitney implies $X$ should be normally flat along $L^{\perp}$ (in your example), I am not sure. Would failure of this lead to a failure of just Whitney condition B or will it lead to failure of both Whitney conditions A and B ? (failure of Whitney condition A would automatically mean the failure of condition B) – Aswin Mar 9 '17 at 12:01
• @Aswin : I am almost entirely sure that Whitney implies normally flat. The result of Trotman quoted by Liviu Nicolaescu below is a $\mathcal{C}^1$ version of this. The normal cone and the normal link are somehow the same thing. I do not remember which of Whitney conditions is violated by the failure of normal flatness. Looking at the result of Trotman, I would say condition $b$, but not for sure. – Libli Mar 9 '17 at 22:44

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.

• In this case, I think the first Whitney condition (usually called condition $A$) is still obeyed.. right ? In general, does the condition of equisingularity require that atleast the first Whitney be obeyed or is even this not a necessary requirement ? – Aswin Mar 9 '17 at 11:54
• It is known that B implies A. Condition A does not guarantee equisingularity The Whintey cusp $y^2+x^3-x^2z^2=0$ satisfies $A$ along the $z$ axis, but not $B$. – Liviu Nicolaescu Mar 9 '17 at 12:15
• Yes, I am aware that B => A for Whitney conditions. I was asking a question about equisingular stratifications that are not Whitney. Do these stratifications obey condition A or do they not even obey that ? – Aswin Mar 9 '17 at 13:00
• Heres a bit of history. Initially Whitney believed that A guarantees equisingularity. He then observed that the Whitney cusp with an obvious stratification satisfies A yet it is not equisingular. That is when he improved A to B. I suggest you have of look at Section 4.2 of the notes www3.nd.edu/~lnicolae/Morse2nd.pdf It is a rather informal introduction to Whitney stratifications with pictures, examples and a discussion of equisingularity that might address some of your questions. It also contains references I found useful. – Liviu Nicolaescu Mar 9 '17 at 14:21
• Those are some very nice notes and a fantastic introduction to the Whitney theory! But, unless I have misunderstood your answer and comments.. these are in a slightly different direction from my question. I believe Zariski's definition of equisingularity is in general different from the one Whitney obtained using his two conditions (the other answer by Libli + ref to Tessier's works seems to also indicate this). – Aswin Mar 9 '17 at 14:37