How Do Unit Quaternions Relate to Spheres and Even-Grade Subalgebras?

[mnb96]

Hello,
I read somewhere that the set of unit quaternions identifies the \mathcal{S}^3 sphere.
This makes sense; however, what happens if we consider instead a quaternion as an element of the even-grade subalgebra \mathcal{C}\ell^+_{3,0} ?

Now a unit quaternion is represented as a scalar-plus-bivector p+\mathbf{B}q which can be written in the form cos(\alpha)+\mathbf{B}sin(\alpha) where \alpha is an angle on the plane B.

So why can´t we consider a quaternion as an element of \mathcal{S}^2 instead?

 

[arkajad]

This is exactly how it is nowadays. This way quaternions can rotate both vectors and planes (or axial vectors) by just one simple formula. In other words: quaternions are Hodge dual to vectors:
\mathbf{B}=i\mathbf{b}=\star \mathbf{b}, where i=e_1e_2e_3 is in the center of the algebra with i^2=-1.

 

[mnb96]

Thanks arkajad!
if I understood correctly, here the interesting observation is that in 3 dimensional space (only!) the dual of a vector is indeed a bi-vector (and vice-versa) => there is a one-to-one correspondence between planes and their normal-vectors.

However, let's say that now, we want to define a distance between unit-quaternions.
In the case of unit complex numbers this is easy, because we know that each unit complex number identifies one point on the \mathcal{S}^1 sphere: the unit circle.

When dealing with unit-quaternions I guess we are forced to consider them as elements of the \mathcal{S}^3 sphere and use (for example) the shortest-arc on the 3-sphere as distance measure. This is because there is not necessarily a one-to-one corrspondence between unit-quaternions and points on the \mathcal{S}^2 sphere.

Is this correct?

 

[arkajad]

Well, you can take the distance from the 4-dimensional Euclidean space.

 

[mnb96]

thanks arkajad!
so, summarizing: the equivalent of the unit-circle for complex numbers is the 3-sphere for unit-quaternions.

It is true that we can indeed use the euclidean distance.
I wouldn't want to go too much off-topic, but is there a "mathematical definition" to express the relationship between the shortest arc on the sphere, and the euclidean distance between point on the sphere?

 

[arkajad]

I wrote originally:

You can look at it in a different way. Each unit quaternion is just four real numbers with squares adding to 1. So, you can associate with it a quantum vector state of a qubit. If |q>,|q'> are such states then, assuming q and q' are not too far from each other, the geodesic distance D(q,q') is given by the formula

\cos^2(D(q,q'))=|<q|q'>|^2

To relate it to physics - see, for instance, "http://arxiv.org/abs/quant-ph/0509017" ", Eq. (13).

But no, that was wrong!

From my calculations the geodesic distance between two unit quaternions q,q' is given by:

D(q,q')=\arccos (|1-\frac12||q-q'||^2|)

See the attached extract from Hanson, "Visualizing quaternions".

 

[mnb96]

Uhm...
I am bit confused now.

The formula that is shown on the scanned page is different than yours: Hanson basically computes the scalar product between two (unit?) quaternions and says that it is the cosine of the angle between them. He uses the ordinary "euclidean inner product".

You instead seem to use the inner-product commonly associated with the conformal model of geometric algebra \mathcal{C}\ell_{4,1}, which is: -\frac12(q-p)^2. Then you say that the cosine of the angle is "1 minus the inner product".

What is the difference/advantages/disadvantages between the two?
Personally, I was even thinking of using the bare Euclidean distance between points on the 3-sphere. it works for both the unit circle in 2D and the unit sphere in the 3D.

 

[arkajad]

Simple calculation in the 4d Euclidean space :

||q-p||^2=(q-p,q-p)=||q||^2+||p||^2-2(p,q)=2(1-(p,q))

You can calculate |(p,q)| from this.

 

[mnb96]

Ops...Sorry!
I should have not missed that!
I am evidently tired at this time in the evening :)
You compute the d(q,q') which is a chord-length and then simply retrieve the arc-length.
Now everything is clear.
Thanks a lot!


One very last thing: when we consider the two aforementioned distances, chord-length and arc-length (respectively d_c and d_a), are these distances "said to be something"?
In other words, is there a definition in the literature to denote two distances which have the property d_c(x,y)\leq d_c(x,z) \Leftrightarrow d_a (x,y)\leq d_a (x,z) for all x,y,z

Thanks again, you fully answered my original question.

 

[arkajad]

Well, look at the graph of the function f(d)=\arccos(1-d^2/2),\quad 0<d<1

You see that d_1<d_2 iff f(d_1)<f(d_2) - the derivative being positive!. So at least there you have what you need.

 

[mnb96]

Yes, my question was essentially:
- is there a name in literature to call two distances that satisfy that property?

Perhaps, equivalent distances ? I'm just guessing.

 

[arkajad]

I don't know that. Sorry. Or, I would say: one is locally majorized by the other.

 

[mnb96]

No problem,
it was not important. I was just curious. As I mentioned you already fully answered the main question.

Thanks!