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My question is in the title. Let's say there are two quaternions $\mathbf q_1$ and $\mathbf q_2$, and both of them reside in the upper hemisphere of $\mathbb S^3$.

Is the result of their multiplication $\mathbf q_3 = \mathbf q_1 * \mathbf q_2$ also guaranteed to lie in the upper hemisphere or are there any corner cases when $\mathbf q_3$ will move to the lower hemisphere?

For example, are there any issues when the quaternions lie on the equator?

jordi
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  • Just to be clear, can you define "upper hemisphere" and "lower hemisphere"? – Brian Tung Sep 10 '23 at 01:43
  • @BrianTung I'm not sure if I can define it, I took this terminology from this website, where the author mentions "northern hemisphere" and "upper half". My guess is that the "upper hemisphere" is the set of points that lie closer to the $\mathbf 1$ unit quaternion than to the $\mathbf {-1}$ unit quaternion (I forgot to mention that I'm only considering unit quaternions in this question), but I am not sure how to categorize the points lying on the equator. – jordi Sep 10 '23 at 01:49
  • This looks like a “just wondering” question, not showing a lot of thought beforehnd. I’d certainly want a precise definition of “upper hemisphere”. Indeed, can you give us an example of a quaternion that’s on it? – Lubin Sep 10 '23 at 03:14
  • @Lubin yes, it's a "just wondering" question (as in: I didn't fully understand the article linked in my previous comment, so I created this question, hoping to learn more about quaternions). Assuming unit quaternions lie on the surface of a 4D hypersphere, let's define the "north pole" of the hypersphere as the point $\mathbf q_N = 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$ and the "south pole" as the point $\mathbf q_S = -1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$. If $\mathbf q_N \cdot \mathbf q > 0$, then the unit quaternion $\mathbf q$ is said to lie on the "upper hemisphere". 1/2 – jordi Sep 10 '23 at 03:39
  • My motivation is to learn how to compute the "correct" spherical average of multiple unit quaternions (= average of multiple quaternions on the surface of the 4D hypersphere). The "correct" spherical average would be along the shorter geodesics. I though that by "flipping" multiple unit quaternions (which I want to average) so that they reside on a common hemisphere, I could achieve spherical average over the shorter geodesics. 2/2 – jordi Sep 10 '23 at 03:54
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    The linked article defines "upper hemisphere" as $a > 0$ (i.e. scalar part greater than zero), which is the same as the definition you gave. – Nicholas Todoroff Sep 10 '23 at 15:55
  • I did not give you the -1, even though your question angered me greatly. It looked as if you did not look at so many as just one example. You asked about what happens on the “equator”, which evidently means scalar part equaling zero, and didn’t try that case out at all. But if you square a unit quaternion with no scalar part, you necessarily get a result of $-1$, and if you had tried that out, considerations of continuity would have led you to realize that your guess was way off. – Lubin Sep 11 '23 at 00:20
  • @Lubin You mistaken my lack of intelligence for lack of trying. If something is obvious to you, it doesn't mean it's obvious to other people, I apologize if this angered you. I hoped this was a welcoming place where even a beginner could ask a question and learn. I know that points lying on the equator have zero real component (I didn't ask about that). My problem was that I didn't realize I could play with the quaternion algebra to build geometric intuition of quaternion multiplication, which is what Nicholas Todoroff's answer beautifully reminded to me. And I wanted an authoritative answer. – jordi Sep 12 '23 at 00:12
  • BTW, any nbhd of the identity in a connected group generates the whole group (so can't be closed under multiplication). – coiso Sep 18 '23 at 15:29

1 Answers1

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No.

Consider scalars $a, b$ and imaginary quaternions $v, w$ (which we can think of as 3D vectors). Then $$ (a+v)(b+w) = [ab - v\cdot w] + [aw + bv + v\times w]. $$ These quaternions $a+v$ and $b+w$ are unit quaterions if $$ a^2 + |v|^2 = 1,\quad b^2 + |w|^2 = 1 $$ so because $a, b > 0$ we can write the scalar term of the product as $$ \sqrt{1-|v|^2}\sqrt{1-|w|^2} - v\cdot w. $$ We can choose any $|v|,|w| < 1$; choose $v = w$ to get $$ 1 - 2|v|^2. $$ Clearly if $|v| \geq 1/\sqrt2$ then our resultant quaterinion is outside the northern hemisphere.

To summarize, we've shown that for any $q$ in the northern hemisphere with scalar part $\geq1/\sqrt2$, it's square $q^2$ is outside the northern hemisphere.

From a higher level perspective, the group of unit quaternions $\mathbb S^3$ naturally gives us a double cover of $\mathrm{SO}(3)$, explicitly $$ \rho : \mathbb S^3 \to \mathrm{SO}(3),\quad \rho_q(v) = qv\bar q $$ where $v \in \mathbb R^3$ is thought of as an imaginary quaternion and $\bar q = q^{-1}$ is the conjugate. We have $\ker\rho = \{\pm1\}$, meaning that both $q$ and $-q$ represent the same rotation. If the upper hemisphere $\mathbb S^3_+$ were closed under under multiplication then it would be a subgroup of $\mathbb S^3$ which is isomorphic to the "subgroup" $$ \mathrm{SO}(3)\setminus\{180^\circ\text{ rotations}\} $$ but the set of all $180^\circ$ rotations is not a subgroup. Equivalently, the equator of $\mathbb S^3$ is not a subgroup: if $v,w$ are imaginary then $$ vw = -v\cdot w + v\times w $$ which need not be imaginary.

Nicholas Todoroff
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  • Excellent answer, thank you! – jordi Sep 10 '23 at 21:14
  • A fine answer, and I gave it the +1. But I just took $\frac1{\sqrt{901}}(1+10i+20j+20k)$ and squared it to get $\frac1{901}(-899+20i+40j+40k)$, unless I made a ghastly computational error. – Lubin Sep 11 '23 at 00:09
  • @Lubin What's the issue? $-899/901 < 0$ so it's not in the northern hemisphere. – Nicholas Todoroff Sep 11 '23 at 06:29
  • The issue is that a simple example is sufficient to show that OP’s conjecture is far, far, off. – Lubin Sep 13 '23 at 18:58
  • @Lubin Ok, I thought you were saying something was wrong with my answer. Even so, such a counterexample is not enlightening. What my answer does is demonstrate two different lines of reasoning that someone can follow, one from a computational perspective and one from a more abstract perspective, to figure out and understand for themselves why the product of quaternions isn't closed in the northern hemisphere. – Nicholas Todoroff Sep 13 '23 at 19:42
  • Right, and that’s why I gave you the +1. – Lubin Sep 13 '23 at 22:05