Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
The above is a repost from math.stackexchange. I have some comments there. math.stackexchange.com/questions/9404/… –  Ryan Budney Nov 8 '10 at 16:13
add comment

1 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.

In what way is the $\delta$-invariant of the singularity -- i.e., the local contribution to the arithmetic genus -- related to the dimension of the first homology group of the Milnor fibre?

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.

share|improve this answer
    
$gcd(p,q)$ is the number of path-components in the torus link. –  Ryan Budney Nov 8 '10 at 18:31
    
@Ryan,Vivek: Rephrased: Is b related to gcd(p,q)? –  user02138 Nov 8 '10 at 18:40
    
yes, b = gcd(p,q). also your "geometric genus" is still wrong –  Vivek Shende Nov 8 '10 at 18:52
3  
Actually Kronheimer and I proved the Milnor conjecture in 1992 as consequence of the genus minimizing property of complex curves in a K3 surface. This original proof used Donaldson invariants. In 1994 we proved the Thom conjecture (genus minimizing property for complex curves in the projective plane) using the Seiberg-Witten equations. –  Tom Mrowka Jan 18 '11 at 2:24
add comment

Your Answer

 
discard

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.