Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f \colon (\mathbb{C}^{2}, \mathbf{0}) \to (\mathbb{C}, 0)$ is a weighted homogeneous polynomial with isolated critical point at the origin and with weights $\omega_1$ and $\omega_2$. This means that $f$ satisfies the identity \begin{align} \lambda f(x,y) = f(\lambda^{\omega_{1}} x, \lambda^{\omega_{2}} y) \qquad \lambda \in \mathbb{C} \end{align} and the local algebra $\mathbb{C} \{ x , y \} / \langle \partial_x f, \partial_y f \rangle$ is finite dimensional. Consider the isolated singularity at the origin. The Milnor-Jung formula states \begin{align} \mu = 2 \delta - r + 1 \end{align} where $\delta$ is the delta invariant, $r$ is the number of branches and $\mu$ is the dimension of the local algebra, the so called Milnor number of $f$. In general, \begin{align} \mu = (\tfrac{1}{\omega_1} - 1)(\tfrac{1}{\omega_2} - 1). \end{align}

Easy Question: If $f = x^p + y^q$, then $\mu = (p-1)(q-1)$ and $r = \text{gcd}(p,q)$. How does one compute $\delta$ in terms of $p$ and $q$ independently of the Milnor-Jung formula?

Harder Question: How does one actually compute $\delta$ and $r$ for a general weighted homogeneous curve $f$ in terms of $\omega_1$ and $\omega_2$? Are closed formulas known in the literature?

share|improve this question
The identity given only has one weight in it. –  Qiaochu Yuan May 1 '11 at 16:44
Thanks. Typo corrected. –  user02138 Jun 30 '11 at 6:25

1 Answer 1

Have a look at this paper:




share|improve this answer
Hi Jorge, if you like my question please feel free to up-vote it. –  user02138 Aug 2 '12 at 18:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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