# Computing Invariants of Weighted Homogeneous Plane Curves

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?

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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