## Return to Answer

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As Roy remarked in his answerabove, 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 word 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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As Roy remarked in his answer above, 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 to $p$ and to all the points $q$ which are "infinitely near" to $p$.p$and$m_q$denotes the multiplicity at$q$. Then$\delta(f)$is minimal among germs of a given multiplicity when there are no infinitely near points, in other word when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do the job among double points. Now, one can prove that since the Milnor number is equal to$\mu(f)=2 \delta(f)-r(f)+1$, where$r(f)$is the number of branches(see [Gruel-Lossen-Shustin, introduction to Singularities and Deformations, Proposition 3.35]). It it follows that$\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)$. 3 added 197 characters in body; deleted 4 characters in body; added 6 characters in body; added 7 characters in body As Roy remarked in his answer above, the delta invariant of the germ of plane curve singularity$f$f(x,y)=0$ at $p$ is equal to

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

where the sum is extended to $p$ and to all the point points $q$ which are "infinitely near" to $p$ (with $p$ included, of course).p$. Then$\delta(f)$is minimal among germs of a given multiplicity when there are no "infinitely near points"points, in other word when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do (the job among double points)points. Now, one can prove that the Milnor number is equal to$\mu(f)=2 \delta(f)-r(f)+1$, where$r(f)$is the number of branches (see [Gruel-Lossen-Shustin, introduction to Singularities and Deformations, Proposition 3.35]). It follows that$\mu(f)$is minimal among plane singularities of given multiplicity$n$when$\delta$\delta(f)$ 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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