Let $p \in \mathbb{S}^{n}$, then the stereogaphic projection is a diffeomorpshim $h:\mathbb{S}^{n} \setminus \{p\} \to \mathbb{R}^{n-1}$.

Suppose that $p$ is the 'north pole' ($p = (0,0,..,1)$), then $h$ is easily computable with the formula given by $[h(x)]_{i} = \frac{[x]_{i}}{1 - [x]_{n}}$, for $x \in \mathbb{S}^{n}$ and $1 \leq i \leq n-1$.

However, if $p$ is some arbitrary point, then $h$ seems somewhat more elusive. Is there a general way to express $h$ in such a case?

Obviously, one could rotate the sphere such that $p$ is at the pole. But, that doesn't seem very efficent.

Stereographic projection when the "North/South Pole" is not given by $(0,...,\pm 1)$? Gives an answer, but the image of the function is embedded in the hyperplane orthgonal to $p$ and not in $\mathbb{R}^{n-1}$ itself.

  • $\begingroup$ The projection I think you're describing loses some of the nice properties of the standard stereographic projection; for example, it does not map every circle to either a circle or a straight line. I don' t know what the name of the projection is but I think it's something other than "stereographic". $\endgroup$
    – David K
    Mar 26, 2015 at 18:27
  • $\begingroup$ There is nothing special about the north pole, all points are equivalent on the sphere. So it shouldn't matter what point we choose to project with respect to. $\endgroup$
    – Cain
    Mar 26, 2015 at 18:29
  • 1
    $\begingroup$ You can make a stereographic projection from any point. That is true. The projection is onto a plane parallel to the tangent plane at that point, however, that is, it is exactly what is described in math.stackexchange.com/a/607434/139123. Since you rejected that answer, I conclude you mean a different projection. $\endgroup$
    – David K
    Mar 26, 2015 at 18:32
  • $\begingroup$ The answer described there is correct, but it's given as a vector in $\mathbb{R}^{n}$, while i wish to present it as a vector in $R^{n-1}$. $\endgroup$
    – Cain
    Mar 26, 2015 at 18:35
  • $\begingroup$ OK, I misunderstood your objection. You can construct an orthonormal basis for the hyperplane and then identify it with $\mathbb R^{n-1}$. That is, you can rotate the sphere and do the projection as you first described, or you can do the projection onto a hyperplane orthogonal to $p$ and then rotate the image. I do not think there is a simpler way. $\endgroup$
    – David K
    Mar 26, 2015 at 18:40

1 Answer 1


You could do a stereographic projection from the north pole, followed by a Möbius transformation. A Möbius transformation is determined by the images of three points. You'd want to map the image of $p$ to the point at infinity, the image of the antipode of $p$ to the origin, and any point on the equator polar to $p$ to the unit circle. All of this has the same effect as performing the rotation in space, but depending on your background you might consider it easier.

If you don't care for changes due to translations or scaling, then it would be enough to map the image of $p$ whcih I'll call $p'$ to the point at infinity, e.g. using $z\mapsto\frac1{z-p'}$.


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