# Stereographic projection for a link/knot

I've been trying to understand the topological "link" between algebraic varieties and their associated knots/links and to this end I've been reading F. Kirwan's book, "Complex algebraic Curves". The example on pg. 11 is, I think, a perfect example of how an algebraic (complex) curve can be associated to a (Hopf?) link.

Anyway, I've actually constructed mappings and their inverses for stereographic projections from the "north" and "south" poles of a sphere before so I have no problem with that part of Ex. 1.1, though the choice of setting $\Re(q)=0$ is still somewhat unclear - my problem is in the irred. part $y=x$ and its intersection with the sphere $S_{\epsilon}$.

Here is what I've done to replicate the example.

Have the two ingredients. For the sphere, know that it is defined by $S_{\epsilon}:=\{(x,y) \in \mathbb{C}\times\mathbb{C}\,:\,\vert x-0 \vert^2 + \vert y-0 \vert^2 = \epsilon^2\}$ (assuming $\vert\vert\cdot\vert\vert_2$)

The alg. set $V$ defined by $x\,y=y^2$, which is reducible to $y=x$ and $y=0$, and so will be done in two parts.

The mapping under stereographic projection is given by,

$$(x,y) \mapsto \left\{ \begin{array}{cc} \left(\frac{\epsilon\,\Re(x)}{\epsilon-\Re(y)},\frac{\epsilon\,\Im(x)}{\epsilon-\Re(y)},\frac{\epsilon\,\Im(y)}{\epsilon-\Re(y)}\right), & \Re(y)\neq \epsilon \\ \infty, & \Re(y)=\epsilon \end{array} \right.$$

First part

Points in $\mathbf{C}\times\mathbf{C}$ map to $\mathbf{R}^3$, for this variety, as

$$(x,y)_{|_{y=0}} \mapsto \left(\Re(x),\Im(x),0\right) \equiv (u',v',0),$$

and for the sphere,

$$(x,y)_{|_{S_{\epsilon}}} \mapsto \left( \frac{\epsilon\,u'}{\epsilon-\Re(y)}, \frac{\epsilon\,v'}{\epsilon-\Re(y)}, \frac{\epsilon\,w'}{\epsilon-\Re(y)} \right)$$

Combining them (intersect them) means substituting the variety into the form for a sphere,

$$\left(\left\lvert \frac{\epsilon\,u'}{\epsilon-\Re(y)} \right\rvert^2 + \left\lvert \frac{\epsilon\,v'}{\epsilon-\Re(y)} \right\rvert^2 + \left\lvert \frac{\epsilon\,w'}{\epsilon-\Re(y)} \right\rvert^2\right)_{|_{w=0}} = \epsilon^2,$$

which upon simplification results in (the choice of projection that I'm not fully appreciative of simplifies it further via $\Re(y)=0$?)

$$\frac{|u|^2+|v|^2}{|\epsilon-\Re(y)|^2} = 1 \overset{\Re(y)=0}{\longrightarrow} |u|^2 + |v|^2 = \epsilon^2 \sim S^1 \subset \mathbb{R}^3$$

Second part

Need to transform the points for the variety $y=x$ under stereoprojection as,

$$(x,y)_{|_{y=x}} \mapsto \left( \frac{\epsilon\,u'}{\epsilon-\Re(x)}, \frac{\epsilon\,v'}{\epsilon-\Re(x)}, \frac{\epsilon\,v'}{\epsilon-\Re(x)} \right),$$

which is then put into the form for the sphere,

$$\left\lvert \frac{\epsilon\,u'}{\epsilon-\Re(x)} \right\rvert^2 + \left\lvert \frac{\epsilon\,v'}{\epsilon-\Re(x)} \right\rvert^2 + \left\lvert \frac{\epsilon\,v'}{\epsilon-\Re(x)} \right\rvert^2 = \epsilon^2,$$

$$\frac{|\epsilon\,u'|^2}{|\epsilon-\Re(x)|^2} + \frac{|\epsilon\,v'|^2}{|\epsilon-\Re(x)|^2} + \frac{|\epsilon\,v'|^2}{|\epsilon-\Re(x)|^2} = \epsilon^2,$$

$$\frac{|\epsilon\,u'|^2}{|\epsilon-u'|^2} + \frac{|\epsilon\,v'|^2}{|\epsilon-u'|^2} + \frac{|\epsilon\,v'|^2}{|\epsilon-u'|^2} = \epsilon^2,$$

$$\frac{|\epsilon\,u'|^2+|\epsilon\,v'|^2+|\epsilon\,v'|^2}{|\epsilon-u'|^2} = \epsilon^2,$$

$$\frac{|u'|^2+|v'|^2+|v'|^2}{|\epsilon-u'|^2} = 1,$$

$$\frac{|u'|^2+2|v'|^2}{|\epsilon-u'|^2} = 1,$$

$$\sim u'^2 + 2\,v'^2 = (\epsilon-u')^2$$

$$\neq \; 2\epsilon^2 - 2\,v^2 = (u-\epsilon)^2$$

So this is the part that I am having a problem with - this last equation is what is given in the book. I hope it is just something silly I've been overlooking. Also, if anyone has any insight into why the projection chose to set $\Re(y)=0$, other than reducing the dimension, that would be helpful to know as well.

• I have a full answer to this, ready to be posted, but your strange tactics on this infamous previous question are making me reluctant to post it. What should I do, oh, what should I do? Please enlighten me... – Did Sep 2 '16 at 7:12