Picture of the isotopy class of a degree $d$ smooth complex curve

Question

All smooth complex curves of degree $d$ in $\mathbb{C}P^2$ are isotopic. Let $C$ be such a curve. I often picture $\mathbb{C}P^2$ as a 2-dimensional disk bundle over $S^2$ (of Euler class 1) which on its boundary is $S^3$ (with the disk boundaries giving a decomposition of $S^3$ as the fibers of the Hopf fibration) together with a 4-ball slapped on. By Bezout's theorem, any such $C$ will intersect this disk bundle (picking the disks to be small enough) in $d$ disk bundle fiber disks.

So the intersection of $C$ with the 4-ball will be on the boundary $S^3$ a link $L_d$ (namely, the inverse image of d points under the Hopf fibration - I'm not sure if this link has a name) and in the rest of the 4-ball some properly embedded genus $\frac{(d-1)(d-2)}{2}$ orientable surface with $d$ boundary components.

I would like to see an explicit picture of this surface - does anyone know where I can find such a picture? I would also love to understand the derivation of the picture.

Answer 1

This might be a bit annoying to justify (see below), but I'm quite sure that the surface you're looking for is constructed as follows. (And in any case this is too long for a comment.)

Start with the $d$-braid $\beta = \sigma_1\sigma_2\dots\sigma_{d-1}$, whose closure is the unknot. The torus link $T(d,d)$ (the union of $d$ fibres of the Hopf link, which is the link you're describing) is the closure of the braid $\beta^d$. $\beta^d$ is obtained from $\beta$ by adding $(d-1)^2$ generators.

Now, start with the closure of $\beta$, take the disc it bounds in $S^3$, and push it into the 4-ball. Each addition of a generator $\sigma_i$ to your braid can be viewed as a band attachment in $S^3\times [0,1]$, which gives you a cobordism (with Euler characteristic 1) from the link before-adding and the link after-adding. Do this for each of the $(d-1)^2$ braid generators you need to add to $\beta$ to get to $\beta^d$, and this gives you a cobordism in $S^3\times [0,(d-1)^2]$ from the unknot to $T(d,d)$. Glue in the ball with the disc on the lower boundary component, and you get your desired surface.

As a sanity check: the Euler characteristic of this surface is $1-(d-1)^2 = 2d-d^2$ (the $1$ is from the disc, the $-(d-1)^2$ is from the bands). The Euler characteristic of the surface you were looking for is $(3d-d^2)-d = 2d-d^2$ (the $3d-d^2$ is from the algebraic curve, and the $-d$ is from the $d$ discs you're removing). So, ok, at least this checks.

One piece of justification for why what I'm doing is actually what you want, is that each band attachment corresponds to a plumbing of a positive Hopf band, so what I'm describing is the fibre of a fibration of $T(d,d)$ as a link in $S^3$, which arises as the boundary of a Lefschetz fibration of $B^4$. This is exactly the picture you'd expect, since $T(d,d)$ is the link of the singularity of $\{x^d + y^d = 0\}$, and the piece of algebraic curve you're looking for is a fibre of the Milnor fibration associated to this singularity.