Let $f \colon (\mathbb{C}^{n},\mathbf{0}) \to (\mathbb{C},0)$ be a complex analytic function with isolated critical point at the origin. Define the singular hypersurface $V_{f, \kappa} = f^{-1}(\kappa)$ for small $\kappa > 0$. Let $\pi \colon \tilde{V} \to V$ be a resolution of $V = V_{f,0}$, and let $p_{g} = \dim H^{n-2}(\tilde{V}, \mathcal{O})$ denote the corresponding geometric genus of $V$, which is independent of the chosen resolution. (This is the same geometric genus in the Yau-Durfee Theorem: $\mu \geq n! \ p_{g}$ for $n > 2$, See Y93 or Y06.)
Define the map $\varphi_{f} = f/\| f \| \colon S_{\epsilon}^{2n-1} \setminus V_{f, \kappa} \to S^{1}$, where $\epsilon > 0$ is sufficiently small. The intersection of $V_{\kappa}$ with a small sphere $S_{\epsilon}^{2n-1}$ is an algebraic link.
Consider the Brieskorn-Pham polynomial $f = \sum_{i = 1}^{n} z_{i}^{a_{i}}$ with $a_{i} \geq 2$. In particular, for $f = z_{1}^{p} + z_{2}^{q}$, the intersection above is a torus knot $T_{p,q}$ if $p$ and $q$ are coprime. In 1968, Milnor conjectured that the unknotting number $u(T_{p,q})$ is related to the dimension of the local algebra $A_{f} = \mathbb{C}[x,y] / \langle x^{p-1}, y^{q-1} \rangle$, which is also the rank of the middle homology group $H_{1}(F_{f,0}; \mathbb{Z})$ of the corresponding fiber $F_{f,0} = \varphi_{f}^{-1}(e^{i \theta})$.
In 1928, Brauner proved $\mu(f) = \text{rank} \ H_{1}(F_{f,0}, \mathbb{Z}) = (p-1)(q-1)$ for $f = x^{p} + y^{q}$. In 1994, Kronheimer and Mrowka proved the Thom conjecture, and as a consequence, the Milnor conjecture followed.
(Edited) Question(s): Is the genus $g$ of the fiber $F_{f,0}$ related to the geometric genus $p_{g}$ of $V$ for $n > 2$? Is there a geometric genus-like object for $n = 2$? If so, how might it relate to invariants of the torus knots?
Thanks!
Best Answer
It is a bit unclear what you are asking. The thing you call the geometric genus is certainly not the geometric genus; the literal object you wrote is $\infty$ and if you first compactified the curve as I assume you want to do, then it is always $1$. The correct definition of geometric genus would yield $p_g(V) = 0$ in this case, as the map $t \to (t^q, -t^p)$ gives a 1-1 parameterization of your curve, showing that its normalization is $\mathbb{A}^1$.
I suspect the question you mean to ask is the following.
The answer to this is given by the formula*
$\mathrm{h}_1(f^{-1} (\epsilon) ) =: \mu = 2 \delta + 1 - b$.
I do not know the history of this formula, but it certainly appears in Milnor's book on hypersurface singularities. Here $\delta$ is as above the difference between the arithmetic and geometric genera of a projective curve containing this singularity and smooth elsewhere, and $b$ is the number of analytic local branches, here 1.
*in general, you should intersect $f^{-1}(\epsilon)$ with a small ball, but in the case of $x^p + y^q$ it is unnecessary.