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An element $F\in \mathbb{C}[[x,y]]$ defines a germ of plane curve. We assume $F(0,0)=0$. The multiplicity $mult$ of the germ is defined to be a minimal number $i$ such that $F\in m^i$ where $m=(x,y)$ is the maximal ideal in $\mathbb{C}[[x,y]]$. Other standard invariants of the germ are Milnor number:

$$ \mu=\dim \mathbb{C}[[x,y]]/(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}) $$

and delta invariant:

$$ \delta=\frac{\mu+r-1}{2}$$

where $r$ is number of branches of curve $F=0$ at $(0,0)$.

Question: What is the minimum of $\delta$ and $\mu$ among the germs of given multiplicity?

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3 Answers 3

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As Roy remarked in his answer, the delta invariant of the germ of plane curve singularity $f(x,y)=0$ at $p$ is equal to

$\delta(f) = \sum \frac{m_q(m_q-1)}{2}$,

where the sum is extended over all the points $q$ which are "infinitely near" to $p$ and $m_q$ denotes the multiplicity at $q$.

Then $\delta(f)$ is minimal among germs of a given multiplicity when there are no infinitely near points, in other words when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do the job among double points.

Now, since the Milnor number is equal to

$\mu(f)=2 \delta(f)-r(f)+1$,

where $r(f)$ is the number of branches, it follows that $\mu(f)$ is minimal among plane singularities of given multiplicity $n$ when $\delta(f)$ is minimal and the number of branches is maximal, in other words when $f=0$ is the "ordinary" $n$-ple point.

As an example, for the ordinary double point (node) $y^2=x^3+x^2$ we have

$(\delta, \mu)=(1,1)$,

whereas the ordinary cusp $y^2=x^3$ satisfies

$(\delta, \mu)=(1,2)$.

Summing up, the ordinary $n$-ple point is the only germ of plane curve singularity which minimizes both $\delta$ and $\mu$, and the corresponding values are

$(\delta, \mu)= (\frac{n(n-1)}{2}, (n-1)^2)$.

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I am not an expert, but I would assume, from reading Walker's plane curves for the polynomial case, that the delta invariant is the sum of the numbers $\frac{n(n-1)}{2}$, summed over all multiplicities $n$ at the given point and all "infinitely near" points. Hence it would seem this number is minimal when there are no infinitely near singular points, e.g. when the point is "ordinary" of multiplicity $n$. Thus the minimal delta invariant at a point of multiplicity $n$ would be $\frac{n(n-1)}{2}$.

This ordinary case has $r = n$ branches, the maximum number of branches, so from the formula above, this would seem also to minimize the milnor number, as $n(n-1)+1-n = (n-1)^2$.

Does this seem plausible?

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  • $\begingroup$ Yes, this is completely correct. My answer below is essentially an expanded version of yours. $\endgroup$ Nov 12, 2010 at 9:52
  • $\begingroup$ I took the liberty of fixing your LateX $\endgroup$ Nov 12, 2010 at 9:57
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I think the following should work for $\mu.$ If $F \in m^i,$ then both derivatives of $F$ live in $m^{i-1}.$ Denoting the ideal generated by the partial derivatives by $J_F,$ it follows that $J_F \subset m^{i-1}.$

Then $(\mathbb{C}[x,y]/J_F)/(m^{i-1}/J_F) \simeq \mathbb{C}[x,y]/m^{i-1}.$

This shows that $\mu = \frac{i(i-1)}{2} + \text{dim }m^{i-1}/J_F.$

I think (and this should probably be verified) that this latter dimension is minimized when the two generators of $J_F$ are two generators for $m^{i-1}.$ We could then assume, for example, that $J_F = (x^{i-1}, y^{i-1}).$ Then $\text{dim }m^{i-1}/J_F = i-2$ (with a basis given by $x^{i-2}y, \ldots, xy^{i-2}$).

Putting this together gives a minimum $\mu$ value as $\frac{i(i-1)}{2} + i-2.$

I notice that this disagrees with Roy Smith's answer above when $i>2$ (my answer is smaller). Perhaps I've made a mistake somewhere?

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  • $\begingroup$ Why doesn't $x^i + y^i$ work, for example? $\endgroup$ Nov 12, 2010 at 14:55
  • $\begingroup$ This is an ordinary $i$-uple point, and the result must be $\mu=(i-1)^2$. You probably made some mistake in your manipulation of formal series (by the way, you should write $\mathbb{C}[[x,y]]$ and not $\mathbb{C}[x,y]$. $\endgroup$ Nov 12, 2010 at 15:17

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