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}^3So 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?


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