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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)?

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    $\begingroup$ 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)! $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '11 at 16:44
  • $\begingroup$ @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? $\endgroup$ – Qiaochu Yuan Jan 31 '11 at 16:49
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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.

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  • $\begingroup$ Hrm. Is it really necessary to bring in such technology? $\endgroup$ – Qiaochu Yuan Jan 31 '11 at 17:19
  • $\begingroup$ 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, ... $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '11 at 17:34
  • $\begingroup$ (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... $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '11 at 17:35
  • $\begingroup$ (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 :) ) $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '11 at 17:39
  • $\begingroup$ 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... $\endgroup$ – Qiaochu Yuan Jan 31 '11 at 19:07
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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).

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There are of course many possible diffeomorphisms $\mathbb{CP}^1\simeq S^2$, and of course the transported action of $\mathrm{SU}(2)$ on $S^2$ will not be by rotations with respect to all of them. A few canonical diffeomorphisms are given by stereographic projection. (I say "a few" because one could perform the projection regardless of the size of the sphere or where it sits inside three-space $\mathbb{R}\times\mathbb{C}$.)

Every element of $\mathbb{CP}^1$ is a $\mathbb{C}$-line in $\mathbb{C}^2$, say $\mathbb{C}[\begin{smallmatrix} \alpha \\ \beta\end{smallmatrix}]$ (where $[\begin{smallmatrix} \alpha \\ \beta\end{smallmatrix}]$ is a unit vector), which corresponds to a projection map $p\in M_2(\mathbb{C})$ given by $p(x)=[\begin{smallmatrix} \alpha \\ \beta\end{smallmatrix}]\langle [\begin{smallmatrix} \alpha \\ \beta\end{smallmatrix}],x\rangle$ (assuming your inner product is conjugate-linear in the first argument), which equals

$$ p=\begin{bmatrix} \alpha \\ \beta\end{bmatrix}\begin{bmatrix} \alpha \\ \beta\end{bmatrix}^\dagger=\begin{bmatrix} \alpha\overline{\alpha} & \alpha\overline{\beta} \\ \beta\overline{\alpha} & \beta\overline{\beta}\end{bmatrix}. \tag{1}$$

This is a hermitian matrix with trace $|\alpha|^2+|\beta|^2=1$. If we double and subtract $I$ from it we get a traceless hermitian matrix. What is the effect of this mapping from $\mathbb{CP}^1$ to $M_2(\mathbb{C})$? Write

$$\begin{bmatrix}\alpha\\ \beta\end{bmatrix}=\frac{1}{\sqrt{1+|x|^2}}\begin{bmatrix}1 \\ x\end{bmatrix} \quad\Rightarrow\quad 2p-I=\begin{bmatrix} \displaystyle \frac{1-|x|^2}{1+|x|^2} & \displaystyle \frac{2\overline{x}}{1+|x|^2} \\ \displaystyle \frac{2x}{1+|x|^2} & \displaystyle -\frac{1-|x|^2}{1+|x|^2}\end{bmatrix}. \tag{2}$$

Denote by $\mathfrak{h}_2'(\mathbb{C})$ the vector space of traceless $2\times 2$ complex matrices. The embedding of $\mathbb{CP}^1$ is therefore the stereographic projection $\mathbb{C}\cup\{\infty\}\to\mathbb{R}\times\mathbb{C}$ composed with the obvious isomorphism $\mathbb{R}\times\mathbb{C}\cong \mathfrak{h}_2'(\mathbb{C})$. With respect to the Frobenius norm, the image of this map is the usual round sphere defined by the metric.

Observe $\mathrm{SU}(2)$ acts on $\mathfrak{h}_2'(\mathbb{C})$ by conjugation. This preserves the norm

$$ \|X\|_{\small F}^2=\sum_{i,j}|x_{ij}|^2=\sum \|\mathrm{row}\|^2=\sum\|\mathrm{column}\|^2 \tag{3}$$

because left multiplication by $g\in\mathrm{SU}(2)$ preserves column norms and right multiplication preserves row norms. Therefore it acts by rotations. Alternatively we could have used cycling:

$$\|gXg^{-1}\|_{\small F}^2=\mathrm{tr}((gXg^{-1})(gXg^{-1})^\dagger)=\mathrm{tr}(gXX^\dagger g^{-1})=\mathrm{tr}(g^{-1}gXX^\dagger)=\|X\|_{\small F}^2. \tag{4}$$

The map $\mathbb{CP}^1\to\mathfrak{h}_2'(\mathbb{C})$ is $\mathrm{SU}(2)$-equivariant because the induced action is

$$g\cdot p=\left(g \begin{bmatrix}\alpha \\ \beta\end{bmatrix}\right)\left(g \begin{bmatrix}\alpha \\ \beta\end{bmatrix}\right)^\dagger =gpg^{-1}. \tag{5}$$

Therefore, the action of $\mathrm{SU}(2)$ on $\mathbb{CP}^1$ transported via stereographic projection to $S^2$ is indeed by rotations, and we have a natural map $\mathrm{SU}(2)\to\mathrm{SO}(\mathfrak{h}'_2(\mathbb{C}))$ (essentially $\mathrm{SO}(3)$). The kernel consists of matrices whose associated linear fractional transformations are trivial, so $\{\pm I\}$.

The three nicest values $0,1,i\in\widehat{\mathbb{C}}$ correspond to the Pauli matrices:

$$0\leftrightarrow\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}, \quad 1\leftrightarrow \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad i\leftrightarrow \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \tag{6}$$

Their stabilizers correspond to rotations around the axes of an orthonormal basis (which yields surjectivity too I think if that's of interest). This should be intuitively acceptable by drawing the corresponding vector flows around the three points in $\mathbb{C}$ and then imagining their effect after stereographic projection onto the unit-radius origin-centered sphere in $\mathbb{R}\times\mathbb{C}$.

If we interpret $\mathrm{SU}(2,\mathbb{R})$ as $\mathrm{SO}(2)$ and $\mathrm{SU}(2,\mathbb{H})$ as $\mathrm{Sp}(2)$, and treat $\mathbb{H}^2$ as a right vector space over the quaternions $\mathbb{H}$, the above argument can be generalized to yield more onto $2$-to-$1$ Lie group homomorphisms (i.e. spin maps). According to Baez this can be done for octonions $\mathbb{O}$ too by suitably interpreting $\mathrm{SU}(2,\mathbb{O})$. Thus we have

$$\begin{array}{|l|} \hline \mathrm{SU}(2,\mathbb{R})\to\mathrm{SO}(2) \\ \hline \mathrm{SU}(2,\mathbb{C})\to \mathrm{SO}(3) \\ \hline \mathrm{SU}(2,\mathbb{H})\to\mathrm{SO}(5) \\ \hline \mathrm{SU}(2,\mathbb{O})\to\mathrm{SO}(9) \\ \hline \end{array} \tag{7} $$

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    $\begingroup$ That's the right argument! :-) Here is a polished-up rendering of it: twitter.com/SchreiberUrs/status/1292547211314831363 This immediately generalizes to the quaternionic case, to show that the canonical $\mathrm{Sp}(2)$-action on $\mathbb{H}P^1$ is the canonical $\mathrm{Spin}(5)$-action on $S^4$. $\endgroup$ – Urs Schreiber Aug 19 '20 at 11:36

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