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I've heard $x^2 + y^2 = 1,\;$ where $x, y$ are complex numbers, is supposed to be a sphere with two points removed, or also a cylinder.
The problem is I've been trying to wrap my head around this for a while now, and I just don't see it. If it is a sphere, is it like a sphere in $\mathbb{R}^4,\;$ i.e. a $3-$sphere?

I think I see how the stereographic projection works for the extended complex numbers, but if anyone could help me see this complex curve as a sphere, that'd be great.

user376343
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Ryker
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    Did you mean $|x|^2+|y|^2 = 1$? This would mean $(\text{Re}(x))^2+(\text{Im}(x))^2+(\text{Re}(y))^2+(\text{Im}(y))^2 = 1$, which is basically a sphere in $\mathbb{R}^4$. – copper.hat Nov 07 '12 at 03:55
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    Thanks for the reply, but I actually don't. I'm sure $x^2 + y^2$ = 1 is correct. – Ryker Nov 07 '12 at 04:00
  • You used the [tag:general-topology] tag -- are you asking whether this set is topologically a sphere? – joriki Nov 07 '12 at 04:31
  • joriki, I'm new here, so I wasn't sure what to tag it as. I guess I meant it as it falling under the "general shape of things" umbrella, which I somewhat loosely associate with topology. – Ryker Nov 07 '12 at 04:43
  • Hint: consider the map $(x,y)\mapsto x$ and compute Euler char of the surface in question. – Grigory M Nov 07 '12 at 08:37

2 Answers2

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If $x =a (1+i) + b(1-i)$ and $y = c (1+i) + d(1-i)$ with $a,b,c,d$ real, then the real and the imaginary parts of the equation $x^2 + y^2 = 1$ is $ab + cd = 1/4$ and $a^2 + c^2 = b^2 + d^2$. Now we can't have $a^2 + c^2 = 0$, as then the left-hand side of the first equation is $0$. So with $r = \sqrt{a^2 + c^2}$ we have $a = r \cos(\theta)$, $c = r \sin(\theta)$, $b = r \cos(\phi)$, $d = r \sin(\phi)$, for some angles $\theta$, $\phi$, and the first equation says $r^2 \cos(\theta - \phi) = 1/4$. Thus we must have $\cos(\theta - \phi) > 0$. We may take as parameters $u = \theta - \phi \in (-\pi/2, \pi/2)$ and $v = \phi \in [0, 2 \pi)$ with $v = 0$ and $v = 2 \pi$ identified, and thus

$$ \eqalign{a &=\frac{\cos(u+v)}{2 \sqrt{\cos(u)}}\cr b &= \frac{\cos(v)}{2 \sqrt{\cos(u)}}\cr c &= \frac{\sin(u+v)}{2 \sqrt{\cos(u)}}\cr d &= \frac{\sin(v)}{2 \sqrt{\cos(u)}}\cr}$$

This gives us a homeomorphism from the cylinder $(-\pi/2, \pi/2) \times ({\mathbb R} \mod 2\pi)$ to the solution set of $x^2 + y^2 = 1$ in ${\mathbb C}^2$.

The cylinder is also homeomorphic to the $2$-sphere $S^2$ with two points removed.

Bilal Aytekin
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Robert Israel
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  • Thanks for the reply! I'm still to fully wrap my head around the answer, but what is the motivation for having $x = a(1+i) + b(1-i)$ rather than, say, $x = a + bi$? – Ryker Nov 07 '12 at 09:09
  • Oh, and also, you say it's a 2- and not a 3-sphere then? – Ryker Nov 07 '12 at 09:55
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    Doing it this way (which introduces a 45 degree rotation in the $a-b$ plane, thus a 90 degree rotation of the squares) is nicer because $a^2 + c^2 = b^2 + d^2$ has a simple geometric meaning. – Robert Israel Nov 07 '12 at 21:24
  • $S^2$ is called the $2$-sphere because it is a $2$-dimensional manifold. An ordinary sphere in $3$-dimensional space is a $2$-sphere. – Robert Israel Nov 07 '12 at 21:29
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It is a standard exercise that the solution to the equation $z_{1}^{2} + \ldots + z_{n}^{2} = 1$ is diffeomorphic to the $TS^{n-1}$, the tangent bundle of the $n$-sphere.

In your case $n = 2$ and we are looking at $TS^{1}$, the tangent bundle of the circle. The circle is parallelizable, being a Lie group, therefore this bundle is trivial and we have $TS^{1} \simeq T \times \mathbb{R}$, a cylinder like you suggested.

Piotr Pstrągowski
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