The Spin Chronicles (Part 3): Spin frames

What’s a Spin Axis, Anyway?

This post was inspired by a question from Bjab, who asked: "What do you mean by ‘spin axis’?" Well, buckle up because I’m about to explain that! I’m not entirely thrilled with the answer I have right now, but hey—it works (kind of). Hopefully, in the near future, I'll come up with something that satisfies both Bjab and my restless curiosity. For now we have to be happy with the standard concept of the spin structure - representing the universe as a set of interlocked rotations - described below.

Spin structure

Where Do Spinors Live?

Imagine a strange and mysterious land—let’s call it V—a two-dimensional complex vector space. It’s not just any space, though. It has "structure." Picture it equipped with a hermitian non-degenerate scalar product that we’ll denote by $(u,v)$, and a bilinear antisymmetric non-degenerate form we’ll call $\epsilon$. Sounds fancy, right? These two should be related in a specific way, and here’s how:

There exists a special kind of basis for this space, which we’ll call “special orthonormal.” It satisfies:

1. $(e_1,e_1) = (e_2,e_2) = 1$,
2. $(e_1, e_2) = 0$,
3. $ϵ(e_1,e_2)=1.$

“Special” indeed. The first natural question is: Can this magical basis even exist?

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The Magic of C²: Relax, It’s All Possible

Yes, it exists! Here’s why. Let’s choose any basis in V. Using this, we can identify V with $C^2$ (our familiar two-dimensional complex space). In $C^2$, the scalar product $(u,v)$ is defined as:

$$(u,v)=u^{\mathrm{1\ast }}{v}^1+u^{\mathrm{2\ast }}{v}^2$$

where the star (*) represents complex conjugation. As for $\epsilon$, we define it as:

$$ϵ(u,v)=u^1{v}^2-u^2{v}^1$$

Now, using a standard basis $e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, it’s easy to check that our special orthonormal basis conditions hold. So, crisis averted! We don’t need to worry about whether this basis exists—it does. Phew.

Oh, and here’s a fun fact: if $(e_1, e_2)$ is a special orthonormal basis, then $(-e_1, -e_2)$ is also a special orthonormal basis. Isn’t symmetry beautiful?

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Back to the Mysterious World of V

So, we’ve got our scalar product, our fancy $\epsilon$, and we’re cruising in V—home to the enigmatic spinors. But are there other “special orthonormal” bases? And if so, how many?

Let’s say $e'_1, e'_2$ is another basis. There exists some matrix $A$ such that:

$$e_i^{\mathrm{\prime }}=e_j{A}_i^j(i,j=1,2)$$

(Yes, that’s a summation over “j”.)

Now, here’s the neat part: $e'_1, e'_2$ will also be special orthonormal if and only if the matrix $A$ is unitary with determinant 1. Matrices like this form a group called SU(2).

By the way, SU(2) is isomorphic to the group of unit quaternions. I won’t dive into that now, but it’s a pretty cool connection between algebra and geometry.

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The Spinor Connection

Let’s denote by O(V) the space of all these special orthonormal bases. The group SU(2) acts on O(V) from the right, both transitively and freely. The elements of O(V) are what we call spinor bases, and the elements of V are simply spinors. If $u$ is a spinor, and you change the basis using an SU(2) matrix $A$, the components of the spinor change according to this formula:

$$u^{\mathrm{\prime }i}=A_j^i{u}^j$$

Got it? Good! Now, let’s connect this with something real, like the lab—where actual experiments happen (sometimes with coffee stains on the data sheets).

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What About Our Lab?

To bridge this to the physical world, we need some more algebra (don’t worry, just a little more). Quaternions are often used for handling 3D rotations, but since we’re sticking with matrices, here’s what we do in quantum mechanics:

For each point $x = (x^1, x^2, x^3)$ in $R^3$, associate the following 2x2 Hermitian matrix, $\sigma(x)$:

$$\sigma (\mathrm{<b>x</b>})=\left(\begin{array}{cc}{x}^3& x^1-i{x}^2\\ x^1+i{x}^2& -x^3\end{array}\right)$$

Now, for any SU(2) matrix $A$, it turns out:

$$A\sigma (\mathrm{<b>x</b>})A^{\ast }=\sigma (R(A)\mathrm{<b>x</b>})$$

where $R(A)$ is a 3x3 real rotation matrix, an element of the special orthogonal group SO(3). In fact, the map from $A\to R(A)is\; a\; homomorphism.\; And\; every\; element\; of\; SO(3)\; can\; be\; generated\; this\; way.\; The\; catch?\; If$$A$ and $B$ are two SU(2) elements, then $R(A) = R(B)$ if and only if $B = \pm A$.

Hence, SU(2) is a double cover of SO(3). It’s like the universe giving you two chances to rotate.

Note: With quaternions it would look more "natural". To each x we would associate a pure imaginary quaternion 

q(x) = xi+yj+zk, 

and for each unit quaternion u, we would have 

uq(x)u* = q(R(u)x), 

where R(u) is in SO(3). 

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The Spin Structure: So What About That Spin Axis?

Let’s bring this back to the real world—our lab, where we have Cartesian coordinates. Any two right-handed orthogonal systems (say, different setups in the same room) are related by a unique rotation matrix from SO(3). The set of all such frames forms a space O(L), which SO(3) acts on.

Now, here’s the kicker: nature somehow maps each element of O(V) (the spinor world) to an element of O(L) (our lab’s coordinate system) in a way that respects the rotations. We call this map the spin structure.

In other words, when you rotate an element of O(V) by an SU(2) matrix A, the corresponding lab element rotates, too, By and SO(3) matrix R(A). It’s like a cosmic interlock between spinors and real-world rotations.

Note (for experts): This is a baby version of a spin structure, good for a 3D flat space or, more generally, for a space with a distant parallelism. But this baby version is all that we need for now.

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What’s Next?

All this brings us closer to answering the question: what do I mean by a spin axis? Stay tuned, because we’ll tackle that mystery in the next post! We will also take under the loop the spin structure - the most important point here.

P.S. I am experimenting with the mathematical notation on this blog. Sometimes I mess up, then correct what can be corrected.

P.S. I have uploaded the pdf version of this post to academia.edu.

P.S. 18-10-24 9:33 Igor Bayak (1.1.6)

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!