# Knots and Singularities

These next few posts are about some cool things I found a book by Frances Kirwan (Complex Algebraic Curves)

### Singularities and Knots

This comes roughly from page 11 of the book.  So $C$ is a complex algebraic curve, say sitting in $\mathbb{P}^2$.  Look at an affine patch; that is, the zero locus of some $f(x,y) \in \mathbb{C}[x,y]$ in $\mathbb{C}^2$.

Say $f$ has a singularity at $(0,0)$. Look at a small $3$-sphere near the origin: $S^3 := \{|x|^2 + |y|^2 = e^2\}$ for small $e$.  It turns out the intersection is a knot.  Forget proving this, but consider how to extract the knot from equation $f$.

Here are the steps involved.  We want to get $S^3 \cap Z(f)$ as a point set in $\mathbb{R}^3$So first defined the (stereo graphic projection) $S^3 \to \mathbb{R} \cup \infty$.

The projection is from the point $(0, e) \in S_3$.  Meaning this point maps to $\infty$.  For all other points $(a,b) \in S_3$ let $l(a,b) = (a,b) + t(-a, e -b)$ be the real line $(t \in \mathbb{R})$ connecting $(a,b)$ and $(0,e)$.  It intersects $\{(c,d)| re(d) = 0\}\cong \mathbb{R}^3$ in a unique point.  Its when $re(b) + t_b(e - re(b)) = 0$, or $t_b = \frac{- re(b)}{e - re(b)}$.  Then the explicit formula is (for $(a,b) \ne (0,e)$):

$(a,b) = (a_1 + ia_2, b_1 + ib_2) \mapsto t_b(a_1,a_2,b_2)$

Now about getting the inverse to this map
.  Say $(u,v,w) \in \mathbb{R}^3$.  We want to produce two complex numbers.  Based on the forwards map, the first should be a (real) scalar multiple of $u + iv$ and the second be roughly $x + iw$.  All in all we can write the point as

$e \cdot (su +siv, x +isw)$

you have two conditions: the value of $t_b$ above, and also that this point should have norm $e^2$, these in principle should determine $s,x$ but the equations seem to get pretty ugly (the answer is in the book).

A concrete example
$f = xy – y^2 = y(x – y)$.  Then it is the union of the horizontal line given by $y$ and the diagonal line given by $y = x$.  Its not hard to check that under the sterographic projection, the horizontal line maps to a circle

$\{(a,b,c) \in \mathbb{R}^3| a = 0, b^2 + c^2 = e^2\}$

and the diagonal line maps to the ellipse

$\{(a,b,c) \in \mathbb{R}^3| b = c, (a - e)^2 + 2b^2 = 2e^2\}$

and it happens that these are closed paths in $\mathbb{R}^3$ that are linked.  You could do this with $y^2 = x^3$ and you would get the trefoil wrapped around the torus; you get the $(2,3)$ torus knot but the equations are more cumbersome.  I wonder if you would get the $(n,m)$ torus knot if you look at $y^n = x^m$?