Stereographic projection of "unit sphere" in a Banach space So I was wondering, if it is possible to extend the idea that a sphere (minus one point) is isomorphic to a plane to the case of an infinite dimensional Banach (or Hilbert if needed) space.
Suppose we have a $L^2$ space, then consider the unitary sphere $U$ , namely the family of function in $L^2$ that have unitary norm. Is there an analog of the Stereographic projection in this case?
Thanks in advance!
 A: Hilbert seems to be needed.  Consider a unit vector $n$ and let $S$ be the sphere with center $n$ and radius $1$.  The straight line through the north pole $2n$  and a point $f$ on the sphere is given by
$$x=2n+t(f-2n).$$
To calculate the intersection of the straight line and the orthogonal complement of $\{n\}$ solve $\langle x,n\rangle=0$ for $t$.
Hence the stereographic projection $P$ from $S\setminus\{2n\}$ to the orthogonal complement of $\{n\}$ is given by
$$
P(f)=\frac{2}{2-\langle n,f\rangle}(f-\langle n,f\rangle n).
$$
Furthermore, the inverse mapping is easily computed to
$$
P^{-1}(f)=\frac{2}{4+\|f\|^2}(\|f\|^2n+2f)
$$
again using a straight line through $2n$ and $f$; this time we assume $n\perp f$.
Works in any Hilbert-space.
A: The answer is affirmative for the unit sphere of a Hilbert space. The point I would like to make is that the stereographic projection is essentially a one-dimensional mapping, which only involves one polar coordinate. Thus, dimensionality does not really matter.
Remark. On the other hand, for a general Banach space, non-Hilbert, the stereographic projection is not generally available. I have in mind the example of $\mathbb R^2$ with the $\lVert\cdot\rVert_\infty$ norm, for which the unit sphere is the unit square. I don’t see how to stereographically project that; the straight faces of the square are surely going to pose problems. These straight faces are a manifestation of the lack of strict convexity of $(\mathbb R^2, \lVert\cdot\rVert_\infty)$.

Consider the usual stereographic projection of $P\in \mathbb S^2$ onto $\mathbf x\in \mathbb R^2$, but let us reason in polar coordinates, with the physicist’s convention that $\theta\in [0, \pi]$ and $\phi\in [0, 2\pi)$. The point
$$
P=(\cos\phi\sin \theta, \sin \phi \sin \theta, \cos \theta)$$
is mapped onto the point
$$
\mathbf x=(r\cos \phi, r \sin \phi, 0),\quad r\in [0,\infty),$$
where
$$\tag{1}
\tan\frac\theta2 = r.$$
See picture.

Writing the stereographic projection in this way, it becomes manifest that the mapping is essentially one-dimensional.

So let us replace $\mathbb S^2$ with the unit sphere $S$ of a Hilbert space $H$, where we choose a unit vector $e_1$, which will play the role of $(0,0,1)$ in the picture above.
We define a mapping of $P\in S$ onto $\mathbf x\in e_1^\bot$ by mimicking the above; for
$$
\cos \theta := \langle P, e_1\rangle, $$
we let
$$
r= \tan\frac{\theta}2\in [0, \infty), $$
and we denote
$$\tag{2}
\mathbf x = r\frac{P’}{\lVert P’\rVert},\quad \text{where }P’=P-\langle P, e_1\rangle e_1.$$
To compare this mapping with Michael Hoppe’s one, we remark that
$$
\tan \frac\theta 2 = \frac{\sin \theta}{1+\cos \theta},$$
(this trig formula can be proved by inspection of the picture above), and that
$$
\lVert P\rVert^2 = 1 = \cos^2 \theta + \lVert P’\rVert^2 \quad \Rightarrow \lVert P’\rVert =\sin \theta.$$
Thus, (2) reads
$$\tag{3}
\mathbf x = \frac{P’}{1+\langle P, e_1\rangle}, $$
where Michael’s formula reads, in our notation,
$$
\tag{4}
\mathbf x = \frac{2P’}{2-\langle P, e_1\rangle}.$$
The two mappings are not exactly the same, which corresponds to the different choices of the projection pole. Here we projected from the South Pole of $S$, which is the point $-e_1$. Micheal projected from $2e_1$.
