2
$\begingroup$

Consider the following question: If two nodes collide what do you get? First of all it can not be a strict $A_2$ node, because the delta invariant of that is $1$. So it has to be more singular than an $A_2$ node. It can be an $A_3$ node because the delta invariant of that is $2$.

Is there any simple argument to show that if an $A_2$ node and an $A_1$ node collide, then we can not get a strict $A_3$ node? The delta invariant doesn't help. Is there some other invariant that can answer this question? Note that I am NOT asking what do we actually get when an $A_2$ node and an $A_1$ node collide. I merely want to show that we can not get a strict $A_3$ node.

$\endgroup$

1 Answer 1

5
$\begingroup$

I am not sure if the following counts as a simple argument:

A miniversal deformation of the $A_3$ singularity is given by the family $y^2 = a + bx + cx^2 + x^4$. There is no member in this family with nodes of type $A_1$ and $A_2$ so it follows that we cannot get an $A_3$ node from collisions of two nodes of type $A_1$ and $A_2$.

$\endgroup$
4
  • $\begingroup$ Thank you. Is the miniversal deformation of any ADE singularity known? I wanted to show that two nodes and one cusp can not collide to a strict A_4 node. And is the miniversal deformation of E_7 node known? $\endgroup$
    – Ritwik
    Nov 17, 2011 at 14:53
  • $\begingroup$ By the way, how do we know that the family you wrote has no singularities of type A_1 and A_2? $\endgroup$
    – Ritwik
    Nov 17, 2011 at 14:56
  • $\begingroup$ The miniversal deformation of any planar singularity is easy to write down. See for example Hartshorne's "Deformation theory", section 14. $\endgroup$
    – naf
    Nov 17, 2011 at 15:12
  • $\begingroup$ The singularities in the family correspond to multiple roots of the polynomial $a + bx + cx^2 + x^4$. Since this is of degree $4$, it can never have both a double root and a triple root. $\endgroup$
    – naf
    Nov 17, 2011 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.