This is a bit of a tricky question, we define the real Quaternions as: $$H=\left\{ a+bi+cj+dk\mid a,b,c,d\in\mathbb{R}\right\}$$ With the rule that: $$ij=-ji=k\:,\: jk=-kj=i\:,\: ki=-ik=\, j\;,\: i^{2}=j^{2}=k^{2}=-1$$ We expand the multiplication to the rest of the set by linearity and distributivity and it's worth mentioning that the inverse comes out: $$\left(a+bi+ci+jk\right)^{-1}=\frac{a-bi-cj-dk}{a^{2}+b^{2}+c^{2}+d^{2}}$$ We also define the conjugate in $H$ to be $\overline{a+bi+cj+dk}=a-bi-cj-dk$ and we define a norm on $H$ by $\left|z\right|=\sqrt{z\cdot\overline{z}}$ (which is multiplicative and real). Using this norm we can define the unit sphere in $H$ to be $$S^{3}=\left\{ z\in H\mid\left|z\right|=1\right\}$$ Now I've shown the following two things:

  1. There is a ring-isomorphism between $H$ and the matrix ring $M_{2}\left(\mathbb{C}\right)$.

  2. The unit sphere is a multiplicative group.

I need to show that if we identify $H$ with $\mathbb{R}^{4}$ by $a+bi+cj+dk\mapsto\left(a,b,c,d\right)$ and give $H$ the standard topology of $\mathbb{R}^{4}$ then the multiplication mapping $S^{3}\times S^{3}\to S^{3}$ defined by $\left(z_{1},z_{2}\right)\mapsto z_{1}\cdot z_{2}$ and the inverse mapping $S^{3}\to S^{3}$ defined by $z\mapsto z^{-1}$ are continuous.

I've tried doing this directly but it comes out being quite a pain so I assume there might be some nifty trick that would simplify the whole thing.

  • $\begingroup$ If it makes you happy, there is a mapping to 4 by 4 real matrices that works; $1$ becomes $I,$ and $i,j,k$ become matrices with only four nonzero entries each. Matrix multiplication is then continuous, and inverse of a matrix with determinant $1$ has polynomial entries from the original (just the conjugate, as you note) and continuous again. $\endgroup$ – Will Jagy Oct 20 '13 at 19:11
  • $\begingroup$ Hmm ye I suppose that would work better as it solves it without dealing at all with the topology on the sphere itself. Thanks $\endgroup$ – Serpahimz Oct 20 '13 at 19:24
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    $\begingroup$ Multiplication $\mathbb{H}\times\mathbb{H} \to \mathbb{H}$ produces coordinates that are polynomials in the factors' coordinates. Hence it is continuous. The restriction to $S^3\times S^3 \to S^3$ is hence also continuous. The conjugation is continuous on $\mathbb{H}$. On $S^3$, the inversion $z\mapsto z^{-1}$ is the restriction of conjugation, hence continuous. $\endgroup$ – Daniel Fischer Oct 20 '13 at 20:43

If you look at the multiplication mapping on $H\times H\simeq\mathbb R^4\times \mathbb R^4$ with values in $H\simeq \mathbb R^4$, each of its components is a simple polynomial function, hence they're continuous and so is the multiplication.

Hence the multiplication is continuous when restricted to $S^3\times S^3$.

For the inverse, you gave yourself the expression of the inverse of a quaternion and it's not hard to see that the inverse mapping is just the conjugation of quaternions on $S^3$ (clearly another continuous function).


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