Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One way to get the famous double cover $\text{SU}(2) \to \text{SO}(3)$ is to note that $\text{SU}(2)$ is isomorphic to the group of unit quaternions and to let unit quaternions $q$ act on the subspace $V$ of $\mathbb{H}$ spanned by $i, j, k$ via conjugation $t \mapsto qtq^{-1}$; this preserves the norm. (Alternately, this is the adjoint action on the Lie algebra, which preserves the Killing form.)

Another way to do this is to let $\text{SU}(2)$ act on $\mathbb{P}^1(\mathbb{C})$, which is diffeomorphic to the sphere. This gives an action of $\text{SU}(2)$ by conformal automorphisms. However, I don't know how to prove that $\text{SU}(2)$ actually acts by rotations (at least, not without some explicit and unenlightening calculations).

To be more precise, if we fix an inner product on $\mathbb{C}^2$, then the space of lines in $\mathbb{C}^2$ can be given the Fubini-Study metric, which $\text{SU}(2)$ preserves. But how can I prove that the Fubini-Study metric agrees with the natural metric on the sphere (up to a constant)?

share|cite|improve this question
It is an homogenous metric, and there are not a lot of those (on a compact simply connected surface, but you probably do not need that much information)! – Mariano Suárez-Alvarez Jan 31 '11 at 16:44
@Mariano: I figured something like that was true, but I would appreciate a somewhat thorough explanation of this. Is it enough to show that SU(2) acts transitively? – Qiaochu Yuan Jan 31 '11 at 16:49
up vote 5 down vote accepted

With the F-S metric, $\mathbf P^1(\mathbb C)$ is a Riemannian surface upon which a group acts transitively. That implies that the curvature is constant. Now, the classification of space forms, for example, shows that such a thing is covered locally isometrically by $S^2$, the round sphere, $E^2$, the flat plane, or $H^2$, the hyperbolic plane. Since $\mathbf P^1(\mathbb C)$ is simply connected and compact, its only covering is the identity, and since it is compact, it must be a round sphere.

share|cite|improve this answer
Hrm. Is it really necessary to bring in such technology? – Qiaochu Yuan Jan 31 '11 at 17:19
I don't think you can get more out of the transitivity of the action than the constancy of curvature. That leaves you with a compact surface of constant curvature: that is not enough information, for there are others. If you compute the sign of the curvature, then that is enough to decide, because there is only one compact orientable surface of positive constant curvature, but you need to compute it. The other thing you know is that it simply connected, to use it you end you end up using the classification I mentioned---notice that the case of positive curvature, the only one we need, ... – Mariano Suárez-Alvarez Jan 31 '11 at 17:34
(cont.) is the simplest part of that theorem. I know many ways to characterize a round sphere when it is embedded in $\mathbb R^3$ (i.e., extrinsic characterizations) but not a lot of characterzations for an 'abstract' round sphere... – Mariano Suárez-Alvarez Jan 31 '11 at 17:35
(Maybe one can use the fact that the group acting is $\mathrm{SU}(2)$ in some way and some representation theory---bringing in technology from another toolbox :) ) – Mariano Suárez-Alvarez Jan 31 '11 at 17:39
Well, I found a proof that does not require much computation: all one has to do is verify that all the maximal tori in SU(2) act as rotations. But I was hoping a more abstract proof would be easier. Perhaps in this case it isn't... – Qiaochu Yuan Jan 31 '11 at 19:07

Okay, forget the Fubini-Study metric. I think I have an alternate solution. Let's rephrase the problem as follows. We'll pick a local coordinate $z$ and think of elements of $\text{PSL}_2(\mathbb{C})$ as fractional linear transformations $z \mapsto \frac{az + b}{cz + d}$. Adding an inner product on the underlying copy of $\mathbb{C}^2$ lets us associate to any $z$ its "orthogonal complement" $- \frac{1}{\bar{z}}$, and the right choice of stereographic projection sends orthogonal complements to antipodes.

Now, an element of $\text{PSL}_2(\mathbb{C})$ respects antipodes (equivalently, orthogonal complements) if and only if it lies in $\text{PSU}(2)$, so it remains to show that any conformal orientation-preserving antipode-preserving automorphism of the Riemann sphere is a rotation. Certainly a fractional linear transformation $g$ preserving antipodes must have two antipodal fixed points of the same type. They can't both be attractive or both repelling (that contradicts the fact that the product of the eigenvalues is $1$), so they are both neither. This ought to already be enough to conclude that $g$ preserves distances to its fixed points, which should only be possible if it's a rotation (and expanding at a local coordinate at the fixed points shows this if nothing else does).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.