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

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

`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