Miguel Marco

Plymouth, December 2018

- Thank you very much for the invitation
- Please ask questions

The set of lines through a point.

Parametrized by the points in a line.

Except for one.

Solution: add one point to the line

- The projective line is the line completed with one point at infinity.

A line cuts a circle in two points:

A line cuts a circle in two points:

A line cuts a circle in two points... or not:

\[(x-2)^2+y^2-2 = 0\]

\[y=\lambda\cdot x\]

\[(x-2)^2 + (\lambda \cdot x)^2 -2 = 0\]

\[(\lambda^2+1)x^2 - 4x + 2 = 0\]

\[x = \frac{2\pm\sqrt{2-2\lambda^2}}{\lambda^2+1}\]

- \(\lambda \leq 1\): real solutions
- \(\lambda \gt 1\): complex solutions

\(y = \lambda x\)

\(y = \lambda x\)

Is homeomorphic to the \(x\) axis: the complex plane.

\(y = \sqrt{x}\)

- \(x = e^{2\pi i\theta}\)
- \(y = \pm e^{\pi i\theta}\)
- As \(x\) makes the full turn, \(y\) makes half a turn

\(y^2 = x^3-x\)

\(y^2 = x^3-x\)

\(y^2 = x^3-x\)

**Theorem**: a smooth complex projective curve of degree \(d\) has the topology of a surface of genus \(\frac{(d-1)(d-2)}{2}\)

\(x^3-y^2=0\)

\(x^3-y^2=0\)

\(x^3-y^2=0\)

- It is the topological cone over the intersection with the boundary of the Milnor ball
- So... what is this intersection?

- The boundary of the Milnor ball is \(\mathbb{S}^3\)
- The part of the complex curve inside the disk is a piece of surface
- The intersection with the boundary is the boundary of a piece of surface:
- a curve without boundary.

- With one component:
**knots** - In general:
**links** - Can be seen in \(\mathbb{R}^3\) by stereographic projection

**Theorem**: The singular curve is the result of collapsing some circles (*vanishing cycles*) on the Milnor fibre.

Questions?