Lie Sphere Geometry Part 3: oriented circles

 This post is a continuation of Lie Sphere Geometry Part 2: unoriented circles. We will move to oriented circles now. Here I am trying to follow the exposition of this subject as it is described in Ch. 15. 1, Oriented circles in S³ of [1]. Instead of S³ I am taking S² first. I say "I am trying to follow" instead of "I am following". The reason for that is that I am not entirely happy with my understanding of the exposition in this source. 

 I am not entirely happy with my understanding of the exposition in this source.

Of course there are other sources discussing the Lie Sphere Geometry, but each author has a different approach, different angle of approach, different main concepts, and I have chosen Jensen as best fitting to our purpose, at least for now. In this post we will discuss oriented circles on the sphere S². Let us recall the definition from the previous post.

S² = {x∈R³:  x² = 1}

For every m∈S², and for every r∈[0,π] the unoriented circle S_r(m)is defined as

S_r(m) = {x∈S²:  x·m = cos r}.

We have then

S_r(m)=S_π-r(-m).

We now extend this definition of S_r(m) any r∈[0,2π]. Of course it makes sense, but S_r(m) = S_2π-r(m). To remove this degeneracy, and following Ref. [1],  we define the oriented circle as follows:

Definition. An oriented circle in S² is a circle S_r(m), together with a choice of  continuous unit normal vector field on it.

Ref. [1] expands this definition as follows.

We can use the radius to define an orientation of S_r(m) as follows. Recall that x∈S_r(m) if and only if x·m = cos(r). Let n∈T_xS² be the unit vector satisfying the equation

m = cos(r)x + sin(r)n.         (0)

Except for the cases r = aπ for any integer a, this uniquely determines n for each point x on the locus. The cases r = aπ, a any integer, are the point spheres m and -m, and there is no condition on n in

n(x) = (m - cos(r) x)/sin(r).                (1)

We consider point spheres to be in the set of all oriented spheres, but without orientation. Thus, the oriented spheres with
center m are parametrized by their radius r satisfying 0 ≤ r < 2π.

I am having some problem with the formula (1) as r approaches 0. The reasoning above was causing me some stomach problems. Therefore I suggest a somewhat different approach. The rest of this post is my own thinking about the subject.

So, let us consider the family of circles for a fixed m. In order to easily visualize thi family let us take m = (1,0,0). I will also use the letter t instead of r. That is because I want to treat the radius as a parameter of some kind of a dynamics. You will see it below.

The parametric equations x(t,ϕ) of the family are then:

x(t,ϕ) = cos(t),
y(t,ϕ) = sin(t)cos(ϕ),
z(t,ϕ) = sin(t)sin(ϕ).

Here ϕ is a parameter along the circle. For t=0 we have the point m=(1,0,0). We notice that

x(t,ϕ)² =  x(t,ϕ)²+y(t,ϕ)²+z(t,ϕ)² ≡1,

so that the circles of constant t are indeed all on S².

Now, for a fixed ϕ calculate the partial derivative ∂x(t,ϕ)/∂t with respect to t. We will call it the "velocity vector field" and denote it n(t,ϕ):

n(t,ϕ) = ∂x(t,ϕ)/∂t,

or, explicitly

n_x(t,ϕ) = - sin(t),
n_y(t,ϕ) = cos(t)cos(ϕ),
n_z(t,ϕ) = cos(t)sin(ϕ).

We see that n(t,ϕ)² = 1, so we have a field of unit vectors. Moreover n(t,ϕ)is tangent to the sphere at x(t,ϕ), as it is tangent to the great circle of constant  ϕ.. Moreover, it is perpendicular to the circle of constant t. Let us calculate the numerator of (1) in our parametrization:

m_x - cos(t)x(t,ϕ) = 1 -cos²(t) = sin²(t) = -sin(t)n_x(t,ϕ) ,
m_y - cos(t)y(t,ϕ) = 0 - cos(t)sin(t)cos(ϕ) = - sin(t)n_y(t,ϕ) ,
m_z - cos(t)z(t,ϕ) = 0 - cos(t)sin(t)sin(ϕ). = - sin(t)n_z(t,ϕ) .

Comparing with (1) we see that our n(t,ϕ) coincides with -n(x) of (1), but it makes a perfect sense also for t=0 and t=π.

Simulation of n(t,ϕ) for varying t, and ϕ=0, π/2,π,3π/2. 

To be continued ...

References

[1] G.R. Jensen et al., Surfaces in Classical Geometries, Springer 2016.

The Quirks of Quaternions

The Spark of Curiosity

This post is inspired by a fascinating conversation I had with Igor Bayak and Bjab in the comments of my previous blog. Their input sparked my curiosity to dive deep into quaternions and vector fields on a three-dimensional sphere. 

These concepts aren’t just abstract math—they could be quite handy for my spinor studies. Plus, there’s something aesthetically satisfying about painting these mathematical fields. And let’s be honest, who doesn’t love a little visual beauty in math?

So, here’s a look at the world of quaternions through my curious eyes.

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Quaternions 101: Meet i, j, and k

Let’s start with our three quirky quaternion friends: i, j, and k. These are the building blocks of quaternions. A general quaternion, x, can be written like this:

$\mathrm{<b>x</b>}=x^1\mathrm{<b>i</b>}+x^2\mathrm{<b>j</b>}+x^3\mathrm{<b>k</b>}+x^4$

Simple enough, right? Now, it’s time to turn these quaternions into something a little more structured—a matrix representation. (Cue dramatic music.)

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Quaternion Matrix Magic

We’re going to multiply x by i, j, and k from the left and see what matrices pop out. This is where the magic happens. Let’s start with i.

When we multiply i by x, we get:

$\mathrm{<b>i</b>}\mathrm{<b>x</b>}=-x^1+x^2\mathrm{<b>k</b>}-x^3\mathrm{<b>j</b>}+x^4\mathrm{<b>i</b>}$

Now, for the matrix interpretation:

• At i, we have x^4, which gives us the first row of the matrix: {0, 0, 0, 1}.
• At j, we have -x^3, giving the second row: {0, 0, -1, 0}.
• At k, we have x^2, producing the third row: {0, 1, 0, 0}.
• Finally, at unity, we get -x^1, completing the fourth row: {-1, 0, 0, 0}.

Putting it all together, the matrix that represents multiplication by i on the left is:

$L1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}$

Matrix magic! (Applause, please.)

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Now Multiply by j and k

Using the same process, we get matrices for multiplication by j and k on the left:

For j, we get:

$L2 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}$

For k, we have:

$L3 = \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}$

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What Happens on the Right Side?

Not to be left out (pun intended), we can also multiply quaternions from the right. When we do this, we get the following matrices:

For i on the right:

$R1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}$

For j:

$R2 = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}$

For k:

$R3 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}$

So now we’ve got both the left and right multiplication matrices. Our quaternion friends are getting quite versatile.

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Quaternions, SU(2), and a Three-Dimensional Sphere

Now things get even more interesting. Quaternions with a norm of 1 form a three-dimensional sphere. And not only that, but they form a group that is isomorphic to SU(2)—fancy math speak for “this group behaves like SU(2).”

To make this more concrete, consider the action of this group on the space of all quaternions. Let’s define an action u as:

$u: x \rightarrow ux$

We now have a representation of this group acting on functions, written as:

$(T(\mathrm{<b>u</b>})f)(\mathrm{<b>x</b>})=f({\mathrm{<b>u</b>}}^{-\mathrm{<b>1</b>}}\mathrm{<b>x</b>})$

Are you still with me? Good! Let’s move on.

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Vector Fields: Expanding the Fun

Consider a one-parameter subgroup, say exp(ti). This subgroup generates a vector field, which we’ll call X1:

$X1(f)=\frac{d}{dt}f(exp(-t\mathrm{<b>i</b>})x){\mid }_{t=\mathrm{0<span\; style="font-size:\; large;"></span>}}$

When we calculate the action of X1 on the coordinate functions x^i, we get the components of X1:

$X^i (x) = - L1^i_j x^j$

In the same way, we can derive the vector fields X2 and X3. 

Here are the results:

The Takeaway

Quaternions may sound intimidating at first, but once you break them down, they’re not only manageable—they’re downright fascinating. Their matrix representations, their connection to SU(2), and the rich vector fields they generate all have deep implications in both math and physics. And who knew matrices could be so much fun?

So, the next time you’re pondering the mysteries of the universe (or trying to impress someone at a party), just casually drop some knowledge about quaternion vector fields. It’s sure to be a hit!