TITLE: Minimum of Milnor number for the curve singularities of fixed multiplicity 
QUESTION [10 upvotes]: 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?

REPLY [7 votes]: 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
$(\delta, \mu)= (\frac{n(n-1)}{2}, (n-1)^2)$.