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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

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

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You probably mean that (f(t),x(t),(t)) is an analytic (?) germ of curve which for $t\ne 0$ lies in $A^1_4$. No? – quim Oct 4 '11 at 14:50
The limit is indeed a D_7 singularity. I know how to prove it using the "Horace method", and will try to post an answer, but do you even know what the Horace method is? – quim Oct 4 '11 at 15:01
I do not know what the horace method is. But I would appreciate a reply based on that. If you can give me a reference to look up Horace method, it would be nice. – Ritwik Oct 4 '11 at 16:48
What I was wondering is, isn't there some simple way to prove this? It seems to me that some clever'' combination like $$f_{02} yf+\frac{1}{4} x^2 y f_{12}f_{x}=0$$ should magically'' yield the expression $$ \frac{-f_{31}^2}{24} + \frac{f_{50} f_{12}}{40}$$ and cancel some things earlier. I am not sure what the combination is but something like that. Does that seem to be a promising thing to try? – Ritwik Oct 4 '11 at 16:55
Évain computed a few "collisions" using the (differential) Horace method in his thesis and in the paper Compactifications des espaces de configuration dans les schémas de Hilbert Bull. Soc. Math. France 133 (2005), no. 4, 497--539. With his method it is not hard to show that a node colliding transversely with an ordinary cusp gives as a limit singularity a D5 singularity (which can probably be proved in other ways) and then blow up the origin (as explained in my answer below) gives what you want. – quim Oct 5 '11 at 7:47
up vote 4 down vote accepted

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.

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This approach is nice, but Ritwik is asking about a specific nongeneric collision, not about all possible collisions. More precisely, his node approaches along a smooth curve which has intersection exactly 2 with the smooth curve with maximal contact with the $A_4$ – quim Oct 4 '11 at 16:18
Yes exactly! I am asking about a SPECIFIC collision. – Ritwik Oct 4 '11 at 17:35
Well, the local Bezout theorem is of some help: this smooth curve along which the points collide intersects the singularities with the total multiplicity 4+2. So, it must intersect the resulting singularity with multiplicity at least 6. This rules out $E_6$. One can use some other very special tricks (of very limited use). The general approach is just to take an honest flat limit of your family. I guess this can be done in many computer systems (Singular etc.) – Dmitry Kerner Oct 4 '11 at 18:56
@qui-vadis: (1) Ritwik already knows it is at least a $D_6$, so $E_6$ and $A_6$ are ruled out, and it follows it is at least a $D_7$ (using your GBL criterion). (2) Indeed your "local Bézout" argument proves that in addition to being a $D_7$ it has maximal contact 6 with the curve $x^2=L_1^2y$, so one can in fact prove two more "relations between derivatives", not just one. – quim Oct 5 '11 at 12:42
@qui-vadis:(3) The initial family $A^1_4$ is not linear (the equation setting $A_4$ is quadratic) and this means it is not given by an ideal (or subscheme) which would be the natural setting for the computation of a flat limit. We would need a second parameter u to account for the variation of $f_{40}/f_{21}$, compute the flat limit for each u and show that it is in fact independent of u. – quim Oct 5 '11 at 12:47

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.

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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.

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