Closure of singular points - MathOverflow

Closure of singular points

2011-10-04T13:35:13Z

Let $f(x,y)$ be a complex degree $d$ polynomial that has this particular form.

$$ f = \frac{f_{02}}{2} y^2 + \frac{f_{21}}{2} x^2 y + \frac{f_{12}}{2} x y^2 + \frac{f_{03}}{6} y^3 + \frac{f_{40}}{24} x^4+ \ldots $$

This polynomial $f$ can be thought of as an element of $\mathbb{C}^{M_d}$, where $M_d = \frac{d^2+3d-10}{2}$. Note that aside from vanishing at the origin, the following derivatives at the origin also vanish $$ f_{10}, f_{01}, f_{20}, f_{11}, f_{30}=0.$$

Let us now define a subset of $$ A_4^1 \subset \mathbb{C}^M_d \times \mathbb{C}^2$$ given by

$$ A_{4}^1:= ( (f,x,y) \in \mathbb{C}^{M_d} \times \mathbb{C}^2 : f(x,y)=0, ~~f_{x} =0, ~~ f_{y} =0, ~~ (x,y) \neq (0,0), ~~ f_{02} \neq 0, $$

$$ f_{40} f_{02} - 3 f_{21}^2 =0. )$$

I have a question regarding the closure of the space $\overline{A_{4}^1}$. Suppose the curve $x(t) = L_1 t$ and $ y(t) = t^2 $, $t\neq 0$, lies in the space $A_4^1$ for all $t\neq 0 $. Further suppose that $f_{02}(t) = L_2 t^r$, for some $r > 0$. Assume that $L_1$ and $L_2$ are fixed non zero complex numbers (they don't depend on $t$).

What happens to the derivatives $f_{ij}$ in the limit as $t$ tends to zero? We basically want to see what happens in the closure when you approach it via the path $ x = L_1 t$, $y = t^2$ and $f_{02} = L_2 t^r$.

It is easy to see that $f_{21}$ will tend to zero, using the equation $f_{y}=0$. Further, using that $f_{21}$ will tend to zero and using $f_{x} =0$ we get that $f_{40}$ will go to zero. I expect another condition to come up, using the fact that $$ f_{40} f_{02} - 3 f_{21}^2 =0.$$

In fact, I expect (but can't prove) that in the limit

$$ \frac{-f_{31}^2}{24} + \frac{f_{50} f_{12}}{40} =0.$$

In any case even if that last claim is wrong, I still expect another condition to come up. The remaining coefficients can not be arbitrary is what I think. May be we get different conditions depending on what $r$ is?

This may seem like a random question, but let me explain intuitively what I am asking. Look at the form of the function $f$ that I have taken. This curve has an $A_3$ singularity (a tacnode) at the origin. What this is means is that at the origin, the first derivatives vanish, the Hessian has a Kernel ( which we have fixed to be $(1,0)$) and the third derivative along the kernel of the Hessian is zero. The condition $$ f_{40} f_{02} - 3 f_{21}^2 =0$$ is the condition for an $A_4$ singularity. Hence, the space $A_4^1$ is the space of curves with an $A_4$ singularity at the origin and one node at a point distinct form the origin. I wish to know how much more singular the curves becomes if the two points come together in the particular way I said i.e $ x = L_1 t$, $y = t^2$ and $f_{02} = L_2 t^r$. The conditions $f_{02}=0$, $f_{21} =0$ and $f_{40} =0$ imply that the curve is at least as singular as a $D_6$-node. I expect it to be as singular as a $D_7$ node which is the condition

$$ \frac{-f_{31}^2}{24} + \frac{f_{50} f_{12}}{40} =0.$$

Answer by Dmitry Kerner for Closure of singular points

2011-10-04T15:46:25Z

If I understand correctly, you ask what can be the results of the collision of two singular points, of $A_4$ and $A_1$ types. In general there does not seem to exist an ultimate effective method to treat such questions. Only in some simple cases, for example in this case.

A somewhat simpler question is: given a point of some singularity type, to which other types can it split by deformation? Of course, it is enough to classify only the "primitive splittings", i.e. those that cannot be factorized through others.

In your particular case, if one restricts to ADE types, you ask: Which types deform to $A_4+A_1$? For ADE's you can use the classical criterion (Grothendieck/Brieskorn/Lyashko) that says: a type S (one of ADE's) deforms to a bunch of types $(S_1,..,S_k)$ iff the disjoint union of Dynkin diagrams of $(S_1,..,S_k)$ is obtained from that of $S$ by erasing some vertices.

Therefore you get immediately, that $D_7$, $E_6$ and $A_6$ deform to $A_4+A_1$ while $D_6$ does not deform.

Unfortunately for higher singularities no such simple general "iff" criterion is known. In your particular case, however the deformation of any other singularity to $A_4+A_1$ factorizes through these "prime" splittings.

You can see some additional results and references in my paper.

Answer by quim for Closure of singular points

2011-10-04T16:27:08Z

Assuming you know that a node colliding transversely with an ordinary cusp gives as a limit singularity a $D_5$ singularity, the answer is easier.

Blow up the point $(0,0)$ (ie, take y=xz, $f_t$ becomes divisible by $x^2$) and the family of proper transforms $f_t(x,xz)/x^2$ of your $f_t$'s have exactly an ordinary cusp and a node approaching transversely. The limit curve has at least a $D_5$, ie, it has intersection multiplicity 3 with the exceptional. The proper transform of a point of multiplicity 2 cannot intersect with multiplicity 3, so x (the equation of the exceptional divisor) divides $f_0(x,xz)/x^2$ at least once (it is the smooth branch of the $D_5$), and the quotient (which is the actual strict transform of the limit curve, $f_0(x,xz)/x^3$) has at least an $A_2$ (ie an ordinary cusp). So the limit curve has a $D_7$ as you say.

Answer by quim for Closure of singular points

2011-10-11T15:02:13Z

Almost everything in this answer has already been said by qui-vadis or in the comments, but now I'll translate it to your notation. I'll write $f(x,y,t)$ for $f$, and $f(x,y,0)$ for the limit $f$.

First remark that $u^4(u-t)^2$ divides $f(L_1u,u^2,t)$ so $u^6$ divides $f(L_1u,u^2,0)$ (this is qui-vadis' local Bézout). Expanding this and passing to the limit, $$\frac{L_1^5}{5!}f_{50}+\frac{L_1^3}{3!}f_{31}+\frac{L_1}{2}f_{12}=0.$$

Next oberve that the limit vanishings of $f_{40}$, $f_{21}$ and $f_{02}$, together with $f_{40}(L_1t,t^2,t)f_{02}(L_1t,t^2,t)−3f^2_{21}(L_1t,t^2,t)=0$, give $$Q:=f_{t40}(0,0,0)f_{t02}(0,0,0)−3f^2_{t21}(0,0,0)=0,$$ i.e., the limit of $f_t=\partial f/ \partial t$ also has an $A_4$ at least. Now using $f_x=0, f_y=0$ and the vanishing (in the limit) of $f_{ij}$ for $i+2j\le 4$ which you already know, it is possible to write the unknowns $f_{t40}(0,0,0)$, $f_{t02}(0,0,0)$, $f_{t21}(0,0,0)$ in terms of $f_{50}$, $f_{31}$ and $f_{12}$. Substitute in $Q$, and the resulting equation is exactly what you were looking for.