7
$\begingroup$

The complex infinity is defined as a pole on the Riemann sphere, which is the result of the 1-point compactification of the complex plane. Considering that quaternions are an extension of complex numbers, and that unit quaternions are points on the 4D unit sphere, is there any analogue of this for quaternions, like quaternion infinity?

$\endgroup$
4
$\begingroup$

Unit complex numbers form a 1D sphere $S^1$ (a circle), while all complex numbers with a point at infinity adjoined form a 2D sphere $S^2$ (the "Riemann sphere" or complex projective line $\mathbb{CP}^1$). Unit quaternions from a 3D sphere $S^3$ and quaternions with a point at infinity adjoined form a 4D sphere $S^4$, the quaternionic projective line $\mathbb{HP}^1$.

One can embed $\mathbb{R}^n\hookrightarrow S^n$ using stereographic projection (interpret $S^n\subset\mathbb{R}\times\mathbb{R}^n$):

$$ x\mapsto \left(\frac{1-|x|^2}{1+|x|^2},\frac{2x}{1+|x|^2}\right). $$

See the link for the geometric interpretation from whence the formula came. This applies in particular with $\mathbb{C}\cong\mathbb{R}^2$ and $\mathbb{H}\cong\mathbb{R}^4$ and even octonions $\mathbb{O}\cong\mathbb{R}^8$. There is also an octonionic projective line $\mathbb{OP}^1\simeq S^8$ after all.

We have $SL(2,\mathbb{C})$ acting on $\mathbb{C}^2$, hence on $\mathbb{CP}^1$, hence on $S^2$. (The action is by "Lie sphere transformations," which are transformations of either spheres or Euclidean space which send hyperspheres to other hyperspheres.) If we restrict our attention to $SU(2)$, it in fact acts by rotations on $S^2$. The same thing happens with $SO(2)$ acting by rotations on $S^1$ (for $\mathbb{RP}^1$) and with $Sp(2)$ acting by rotations on $S^4$ (for $\mathbb{HP}^1$) and even a version for $\mathbb{OP}^1\simeq S^8$ suitably interpreted.

This spinorial interpretation of $SU(2)$ acting on $S^2$ can be seen here or here. In particular it explains where the Pauli matrices come from. Also, one can use stereographic projection to interpret the action of $SL(2,\mathbb{C})$ on $S^2$, see the youtube video Mobius transformations revealed.

For the usual copy of $\mathbb{C}\hookrightarrow\mathbb{CP}^1$, we have $SL(2,\mathbb{C})$ acting by Mobius transformations on $\mathbb{C}$, in which we sometimes have to divide by zero, or by infinity, or even evaluate indefinite rational expressions such as $(a\cdot\infty+b)/(c\cdot\infty+d)$. All of this applies for $SL(2,\mathbb{H})$ acting on $\mathbb{H}$ as well, with a new $\infty$ element, but you need to write $(aq+b)(cq+d)^{-1}$ and keep track of what's on the left vs. what's on the right (since $\mathbb{H}$ is noncommutative).

Exercise. Check the composition of two Mobius tranformations is also one, for quaternions.

(I suppose you could talk about $GL(2,\mathbb{H})$ to be easier. The group $SL(2,\mathbb{H})$ is generated by the exponentials of matrices in $\mathfrak{sl}(2,\mathbb{H})$, the lie algebra of $2\times 2$ tracles quaternionic matrices.)

The situation for octonions $\mathbb{O}$ is trickier and I don't know the details.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ senpai why did you leave me $\endgroup$ – Balarka Sen Jul 8 '18 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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