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This question is motivated by the following homework problem. I'm trying to explicitly compute the homeomorphism $f:S^2 \rightarrow \mathbb{CP}^1$ by using stereographic projection and considering $\mathbb{CP}^1 = \mathbb{C}\cup {\infty}$. I'll want to prove that this is an isometry, where $S^2$ has the standard angle metric and $\mathbb{CP}^1$ has the Fubini-Study metric given by $d(\overline{x},\overline{y})=2\cos^{-1}|(x,y)|$, where $x,y\in \mathbb{C}^2$ are unit vectors (and presumably $(-,-)$ is the usual Hermitian inner product). Later, I'll use this to explicitly compute the Lie group homomorphism $U(2)\rightarrow SO(3)$.

My stereographic projection is from the north pole, takes the equator to the unit circle, and puts the south pole at the origin. What I've gotten so far is that for $z\not= 1$, \begin{equation*} f(x,y,z)=\left( \frac{x}{1-z} , \frac{y}{1-z} \right) = \frac{x+iy}{1-z} = [x+iy : 1-z ], \end{equation*} where these are coordinates in $\mathbb{R}^2$, $\mathbb{C}$, and $\mathbb{C}\subseteq \mathbb{CP}^1$ respectively. This is troublesome, because philosophically I'd expect that I should be able to define this for $(x,y,z)\not= (0,0,1)$ and then end up with a function to projective space that extends continuously over the north pole; that's sort of the point of projective space, to make $\infty$ into just another point. However, it is not immediately obvious that this works, although luckily \begin{equation*} \left| \frac{x+iy}{1-z} \right| = \sqrt{ \frac{|x+iy|^2}{(1-z)^2} } = \sqrt{ \frac{1-z^2}{(1-z)^2}}, \end{equation*} and the limit of this expression as $z\rightarrow 1^-$ is indeed $\infty$.

So, fair enough. This ends up extending to a continuous function after all. But: Am I wrong in my philosophical understanding of projective space?

(For what it's worth, I tried using my calculations to verify that $f$ is an isometry, and it didn't look like it was going to work out. So maybe I really am just doing something wrong.)

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I am wondering - why are you working with $\mathbb{C}$ explicitly? Essentially, $\mathbb{C} = \mathbb{R}^2$ as metric spaces. The only difference lies in their algebraic structure. So, the explicit use of $\mathbb{C}$ only clutters things unnecessarily. Use $\mathbb{R}^2$ instead, and stereographic projection of $S^2$ onto the $(x,y)$ plane in $\mathbb{R}^3$. This might make things more transparent. I should also add that, technically, what you're doing is one-point compactification, which is always possible for locally compact Hausdorff spaces - projective space is an overkill. – William Sep 27 '10 at 6:33
See also here: Like I said, you're essentially doing one-point compactification, as far as the topological structure is concerned (see here:, but after re-reading your post, I have a feeling that you want to carry the analytic structure, too. In that case, refer to the Riemann sphere (wiki link above). – William Sep 27 '10 at 6:40
Well, the Fubini-Study metric is defined on CP^1. This also seems like the right setting for my further goals (Lie group calculations). In any case, I think that the complex numbers are kind of nice! And even though this is related to a homework problem, I really am interested in whether there's something wrong with the way I think about projective space. – Aaron Mazel-Gee Sep 27 '10 at 7:58
No, this sort of thing is not appropriate at all. – anon Sep 27 '10 at 9:00
@muad: Are you referring to the (former) title of this question? I'm sorry if it offended you. I just joined this site today, and I was browsing around and kept seeing things I wanted to comment on but I couldn't. It seemed like the easiest way to go about it -- I'm not trying to be a reputation whore... – Aaron Mazel-Gee Sep 27 '10 at 15:57
up vote 4 down vote accepted

One complication in your situation is that you are mixing real and complex coordinates.
If you were considering a map from a complex curve to $\mathbb{CP}^1$, then the kind of computation you are trying to make would work out more straightforwardly.

Because you are looking at a map of real analytic manifolds, not complex analytic ones (concretely, you are working with the variables $x,y,z$, which are real coordinates), the point of view you have adopted is perhaps not quite as natural. Nevertheless, it can be made to work, as follows: $$[x+iy:1-z] \text{ (which is where you finished) } = [x^2 + y^2: (1-z)(x - i y)]$$ $$ = [ 1 - z^2: (1-z)(x-iy)] = [1+z:(x-iy)].$$ This rewriting of your map to $\mathbb{CP}^1$ is now well-defined in a neighbourhood of $(0,0,1)$ on the sphere. (The fact that I introduced a complex conjugate of $x + i y$ to facilitate the computation is related to the real vs. complex issue mentioned above. This is also essentially the same computation you made to check that your map tends to $\infty$ as $z \to 1$, just rewritten in homogeneous coordinates.)

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I see. This is exactly what I was looking for. Thanks. Is there a different way I should've approached this problem that's more natural? It did actually ask me to look at "the isometry from $S^2$ to $\mathbb{CP}^1$ defined by stereographic projection from $S^2$ to $\mathbb{C}$", that wasn't my idea... – Aaron Mazel-Gee Sep 27 '10 at 15:46
@Aaron: Dear Aaron, I think this approach is fine; it would also be fine just to work in $\mathbb C$, and then use the usual transformation $w \mapsto 1/w$ to see what is happening at infinity. My comments about real vs. complex above are probably more heavy-handed than they should be, but I made them just to emphasize that you are working in the category of real analytic spaces (or algebraic varieties, if you prefer), whereas the magical properties of $\mathbb{CP}^1$ that you are thinking of are most clearly seen when looking at a map from a one-dimensional complex space. – Matt E Sep 27 '10 at 15:53

"Am I wrong in my philosophical understanding of projective space?"
I think your understanding is incomplete in some ways.
E.g. your remark "resolve singularities by replacing a point with a copy of projective space" doesn't make much sense to me.

Edit: I think I didn't understand the question correctly. But I couldn't ask any clarification in comments to the question like others do, because of lack of reputation!

It may help in your understanding of "projective spaces" to realize that their resolvement of $\infty$ singularities uses homogene coordinates, e.g. in $\mathbb{CP}^1 = \mathbb{C}\cup {\infty}$ one uses coordiantes $\rho\cdot (z, 1)$ for elements in $\mathbb{C}$ and $\rho (1, 0)$ for ${\infty}$. Then the mapping should involve both coordinates and account for the free factor $\rho \neq 0$.

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What the OP is referring to is resolving singularities of a map by blowing up the target. – Matt E Sep 27 '10 at 15:30
That was silly. I've heard of blowups, and that's about it -- I thought maybe that was what was going on here, but the more I think about it the more it seems clear that it's not. – Aaron Mazel-Gee Sep 27 '10 at 15:38
I am using homogeneous coordinates, by the way... – Aaron Mazel-Gee Sep 27 '10 at 15:39
"In mathematics, blowing up or blowup is a type of geometric transformation which replaces a subspace of given space with all the directions pointing out of that subspace [i.e., a copy of projective space]." -- -- this is what I was referring to. – Aaron Mazel-Gee Sep 27 '10 at 15:53

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