Projectivization and stereographic projection, or: Why nonlinear + nonlinear = linear?

Question

This previous question had me thinking about something I've taken for granted. Consider $\mathrm{SL}_2(\mathbb{C})$ acting on $\mathbb{C}^2$, which descends to an action on the complex projective line $\mathbb{CP}^1$, which we may think of as the Riemann sphere $\widehat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$. It acts by Möbius transformations. By stereographic projection, $\widehat{\mathbb{C}}$ may be identified with a literal sphere $\mathbb{S}^2\subset \mathbb{R}^3$. So we get a handful of different spaces:

$$\Bbb C^2 \quad\longleftrightarrow\quad \widehat{\Bbb C} \quad\longleftrightarrow\quad \mathbb{S}^2 \quad\longleftrightarrow\quad \mathbb{R}^3 $$

One may restrict to the action of the special unitary group $\mathrm{SU}(2)$, and then clearly transport the action over to $\mathbb{S}^2$. But the acts of projectivizing and stereographic projection are very, well, nonlinear, so why in the world should we expect the result to be something that extends from $\mathbb{S}^2$ to a linear action on $\mathbb{R}^3$?

(I am placing $\mathbb{S}^2$ in $\mathbb{R}^3$ so that its intersection with the $xy$-plane is the unit circle.)

In order to verify that it does, presumably one could do a horrendous calculation, but it's easier to consider three one-parameter subgroups of $\mathrm{SU}(2)$ (corresponding to the standard basis of $\mathfrak{su}(2)$) correspond to coordinate plane rotations, which reduces to showing $\mathrm{SO}(2)\subset\mathrm{SU}(2)$ corresponds to a rotation around the $y$-axis. But even that seems like it would require some calculation I find myself unwilling to finish off.

More generally, $\mathrm{SL}_2(\Bbb C)$ acts by moving $\mathbb{S}^2$ around inside $\mathbb{R}^2$ between stereographic projection from and to the Riemann sphere. A nice, short video about this is Mobius transformations revealed. But what exactly is the connection between elements of $\mathrm{SL}_2(\Bbb C)$ preserving the complex inner product on $\mathbb{C}^2$, which strikes me as more algebraic than geometric, and extending to a linear map on $\mathbb{R}^3$?

Note: I believe something similar happens for $\mathrm{SO}(2)$ inducing an action on the real projective line $\mathbb{RP}^1\simeq\mathbb{S}^1\subset\mathbb{R}^2$ and $\mathrm{Sp}(2)$ inducing an action on the quaternionic projective line $\mathbb{HP}^1\simeq\mathbb{S}^4\subset\mathbb{R}^5$, where $\mathbb{H}$ denotes the quaternions. I suspect there will even be an octonionic version.

One guess I have is that the answer may come from groups attached to spaces with indefinite signature forms, in particular Minkowski spacetimes, but I haven't headed in this direction yet.

Answer 1

$O(3)$ is the isometry group of $S^2$ with its usual round metric, and it acts linearly on $\mathbb{R}^3$ for geometric reasons. So the question is why $SU(2)$ acts on the Riemann sphere by isometries for this metric. The answer is that $SU(2)$ preserves the Fubini-Study metric on the Riemann sphere, and moreover the stabilizer of a point acts transitively on the unit tangent space at that point.

I suspect this fact together with the fact that the action of $SU(2)$ is transitive implies that the Fubini-Study metric has constant curvature (necessarily positive by the Gauss-Bonnet theorem) and hence is the round metric up to a constant, but I'm not confident.

Note that when $n \ge 2$, complex projective space $\mathbb{CP}^n$ cannot possess a metric of constant curvature (by the Killing-Hopf theorem), and in particular the Fubini-Study metric is not such a metric, so whatever's going on here really is special to the case $n = 1$.

Answer 2

Consider $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{H}$ and vector spaces over these skew fields with scalars coming from the right and linear transformations from the left. There is a smooth correspondence between the collection of $\mathbb{K}$-lines in $\mathbb{K}^2$ (or equivalently the points of $\mathbb{KP}^1$) and the right $\mathbb{K}$-linear projection maps onto said $\mathbb{K}$-lines;

$$\{\mathbb{K}\mathrm{-lines}~\ell\subset\mathbb{K}^2\} \quad\longleftrightarrow\quad \{\mathrm{projection~maps~} p_\ell\in\mathrm{End}(\mathbb{K}^2)\}.$$

Any $\mathbb{K}$-line in $\mathbb{K}^2$ looks like $\ell=[\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}]\mathbb{K}$ for some (without loss of generality) unit vector $[\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}]$. If we write down an abritrary vector $v=[\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}]c+[\begin{smallmatrix}\gamma\\ \delta\end{smallmatrix}]d$ for an orthogonal vector $[\begin{smallmatrix}\gamma\\ \delta\end{smallmatrix}]$ (i.e.$\langle [\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}],[\begin{smallmatrix}\gamma\\ \delta\end{smallmatrix}] \rangle=\overline{\alpha}\gamma+\overline{\beta}\delta=0$) and some scalars $c,d\in\mathbb{K}$, then the projection map $p_\ell$ should extract the first component. Observe this is

$$p_\ell(v)=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}\left\langle\begin{bmatrix}\alpha \\ \beta\end{bmatrix},v\right\rangle=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}\begin{bmatrix}\alpha \\ \beta\end{bmatrix}^\dagger v=\begin{bmatrix} \alpha\overline{\alpha} & \alpha\overline{\beta} \\ \beta\overline{\alpha} & \beta\overline{\beta} \end{bmatrix} v \tag{$\circ$}$$

This gives an explicit formula for $p_\ell$. Notice that right multiplying $[\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}]$ by a unit size scalar does not change the projection map $p_\ell$, as should be expected.

These projection maps have trace $\mathrm{tr}(p_\ell)=|\alpha|^2+|\beta|^2=1$. Then $\mathcal{S}=\{p_\ell-\frac{1}{2}\mathrm{Id}\}$ is a subset of the real vector space of traceless hermitian matrices $\mathfrak{th}_2(\mathbb{K})$. There is a norm on this vector space coming from the trace form, $\|A\|^2:=\mathrm{tr}(A^\dagger A)=\sum_{i,j}|a_{ij}|^2=\sum \|\mathrm{row}\|^2=\sum\|\mathrm{column}\|^2$, and we may then compute that $\|p_\ell-\frac{1}{2}\mathrm{id}\|^2=\frac{1}{2}$. In fact, $\mathcal{S}$ is the sphere of dimension $n=\dim\mathbb{K}$ and radius $\frac{1}{\sqrt{2}}$ centered at the origin in $\mathfrak{th}_2(\mathbb{K})$.

Transformations in $\mathrm{GL}_2(\mathbb{K})$ act on lines in the obvious way by acting on their representatives. The effect on the projection maps is $p_{A\ell}=(A[\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}])(A[\begin{smallmatrix}\alpha\\\beta\end{smallmatrix}])^\dagger=Ap_\ell A^\dagger$. Then if $A\in\mathrm{SU}(2,\mathbb{K})$ (interpret appropriately for different $\mathbb{K}$), $A^\dagger=A^{-1}$ so then $\mathrm{SU}(2,\mathbb{K})$ acts by conjugation. As $\mathrm{Id}$ is central, we might as well displace this to be an action of $\mathrm{SU}(2,\mathbb{K})$ on $\mathcal{S}\subset\mathfrak{th}_2(\mathbb{K})$. Indeed, there's nothing stopping us from defining conjugation to be an $\mathbb{R}$-linear action on all of $\mathfrak{th}_2(\mathbb{K})$. Note that left multiplication by transformations $A\in\mathrm{SU}(2,\mathbb{K})$ acts on each individual column of a matrix, preserving the square norm of the columns, hence preserving the trace form, and similarly for right multiplication, so in fact $\mathrm{SU}(2,\mathbb{K})$ acts by isometries, and therefore we have homomorphisms $\mathrm{SU}(2,\mathbb{K})\to\mathrm{O}(\mathfrak{th}_2(\mathbb{K}))$.

In fact this yields the spin homomorphisms $\mathrm{SO}(2)\to\mathrm{SO}(2)$, $\mathrm{SU}(2)\to\mathrm{SO}(3)$ and $\mathrm{Sp}(2)\to\mathrm{SO}(5)$, and I believe it even yields $\mathrm{SU}(2,\mathbb{O})\to\mathrm{SO}(9)$ suitably interpreted.

The connection to stereographic projection can be seen by writing $\ell=\begin{bmatrix} x \\ 1 \end{bmatrix}\mathbb{K}$ and computing

$$p_\ell-\frac{1}{2}\mathrm{Id}=\frac{1}{1+|x|^2}\begin{bmatrix} |x|^2 & x \\ \overline{x} & 1\end{bmatrix}-\frac{1}{2}\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=\frac{1}{2}\begin{bmatrix}\frac{|x|^2-1}{|x|^2+1} & \frac{2x}{|x|^2+1} \\ \frac{2\overline{x}}{1+|x|^2} & \frac{1-|x|^2}{1+|x|^2}\end{bmatrix}.$$

Up to scaling, then, the sphere $\mathcal{S}$ is none other than the equivariant image of $\widehat{\mathbb{K}}=\mathbb{K}\cup\{\infty\}$ (acted on by Mobius transformations) under stereographic projection $\widehat{\mathbb{K}}\to \mathbb{R}\oplus\mathbb{K}$ postcomposed with the obvious isometric isomorphism $\mathbb{R}\oplus\mathbb{K}\to\mathfrak{th}_2(\mathbb{K})$.