Why are spinors interesting, from a Clifford algebra perspective

[ComeInSpinor]

Hi, I'm trying to understand spinors better, and I seem to be getting stuck on understanding the reason they're said to transform differently from vectors, and I'd appreciate any help with a justification for that. I'm sure I'm missing something pretty simple, but here goes;

Here's what I've been thinking. Consider the Clifford algebra Cl_(2,0)(R), with basis elements 1, e₁, e₂, e₁₂. Linear combinations of the even-graded elements give the complex numbers, since e₁₂e₁₂=-1. These play the role of spinors, and the linear combinations (ae₁+be₂) are just vectors. I know that this two-dimensional case is pretty simple, but I think (correct me if I'm wrong) that what I'm going to say next generalizes to higher dimensions. So if we act on a vector with an even-graded element $$\gamma$$, the vectors conjugate-commute with $$\gamma$$. Thus a vector transforms as
v'= $$\gamma$$ v $$\gamma$$*,
and in two dimensions we can get away with rewriting this as $$\gamma$$$$\gamma$$ v (by pulling the $$\gamma$$* through the v) so the vector transforms through double the angle defined by $$\gamma$$. Now a spinor just commutes, rather than conjugate-commuting, so the transformation
s' = $$\gamma$$ s $$\gamma$$ *
would just collapse down to
s' = s
Okay, so that's no good. The core of my question then, is why is it "natural" to define the transformation of a spinor as $$\gamma$$ s, rather than say $$\gamma$$ s $$\gamma$$ (which looks more like the transformation of a vector)? If you act on an object with one copy of a transformation, and another object with two copies, it seems obvious that the second object with transform twice as much. So it seems disingenuous to say that a spinor transforms through half the angle that a vector does when acted on by a given transformation, since you're actually acting on the vector with double the transformation that acted upon the spinor. Of course the spinor is going to rotate through half the angle. That doesn't seem to me like a natural, inevitable choice that is forced upon us by the maths. So what am I missing? It would seem more natural if there was a way to associate a (unique?) spinor with any vector, such that the spinor has to turn through $$\theta$$/2 when the vector turns through $$\theta$$. Is there?

Thanks in advance

 

[arkajad]

You have your algebra, so every algebra has what is being called "modules" - representations of the algebra in vector spaces. They exist, they can be studied. Their study proves to be of importance. In a general module it does not make sense to act on its element from both sides.

You may not like these modules, but they are there. In the Clifford algebra case their elements are called spinors.

It is like with Hilbert space operators. They act on states only from one side. Of course one can always discuss Hilbert algebras, that is a part of what is being called Tomita-Takesaki theory, and it is very interesting. But Hilbert spaces that are not algebras but just linear spaces on which algebra can act from one side only are nevertheless the basis of QM.

 

[ComeInSpinor]

arkajad, thank you for the reply, however I might elaborate a bit more on my question to get closer to an answer that will (hopefully) satisfy me.

So it's not a matter of whether or not I "like" modules. I have nothing against them :-) In fact I'm pretty comfortable with the idea of saying that the even-graded elements of a Clifford algebra transform (most simply) when acted upon on only one side, whereas the vectors have to be acted upon on both sides.

So here's the thing; I realize that if you have a vector (x,y,z) and a spinor ($$\psi_1$$, $$\psi_2$$) you can act upon the spinor with an SU(2) matrix with components $$\left(\begin{array}{cc}a & b \\ -b^* & a*\end{array}\right)$$ to get a new spinor, and it's possible to write out a 3 x 3 matrix, whose components are functions of $$a,\,a^*,\,b,\,b^*$$, which acts upon the vector (x,y,z), so it's easy to see there's a homomorphism between SU(2) and SO(3) in this context. But what I'm really trying to get at (and the reason for the title of this thread) is how this fits into the geometric viewpoint that comes with Clifford algebras. Vectors are directed line segments, bivectors are directed areas, etc. How does one see that there's unambiguously a homomorphism between transformations of directed areas and transformations of directed line segments, and that this homomorphism is 2-to-1?

And while we're at it, you can also say that spinors transform as $$\zeta\rightarrow U\zeta$$, or $$\eta^\dagger \rightarrow \eta^\dagger U^\dagger$$. And a vector can be written as $$v=\zeta\eta^\dagger$$ so that a vector transforms as $$v\rightarrow U v U^\dagger$$. However in the notation of Clifford algebras, the spinors are linear combinations of the even-graded elements like $$e_{ij}$$, right? So how can they multiply together to give odd-graded (vector) results?

 

[arkajad]

However in the notation of Clifford algebras, the spinors are linear combinations of the even-graded elements like $$e_{ij}$$, right?

No. Spinors are vectors in a module. This can be an abstract module, or it can be a minimal left ideal in the Clifford algebra. It is not constructed out of the even Clifford algebra alone.

 

[ComeInSpinor]

No. Spinors are vectors in a module. This can be an abstract module, or it can be a minimal left ideal in the Clifford algebra. It is not constructed out of the even Clifford algebra alone.

Okay, thanks, that's clearly a significant point that hadn't sunk in before. Can you illustrate with an example or two, please? I know a module is a generalization of a vector space, where scalar multiplication of the vectors is by scalars from a ring rather than a field. But the most common examples of spinors (and certainly the example I used above of an SU(2) matrix acting on a 2-spinor) involves scalar multiplication by ordinary complex numbers, which in Clifford algebras are represented by the bivectors (since $$e_{ij}^2 = e_{ij}e_{ij} = - e_{ij}e_{ji} = -1$$). So it looks to me like there are simple instances of spinors, whose transformations can be written in terms of even-graded Clifford algebra elements acting on each other, and more general cases in which it's not that straightforward. I think a couple of examples would help a lot, thanks.

 

[arkajad]

I think a couple of examples would help a lot, thanks.

First of all about modules. Here they are simply representation spaces for the algebra. Clifford algebras are real, so we mainly think about real representations. But Clifford algebra Cl(3,0) has a center isomorpfic to complex numbers. Therefore any real representation space of this algebra becomes naturally a complex vector space - a module for Cl(3,0).

Now, Cl(3,0) is isomorphic (as a real algebra) to the algebra of complex 2x2 matrices with Pauli matrices as its generators. Pauli matrices multiplied by i generate, in this isomorphism, the even subalgebra.

It is easy to show that each minimal left ideal is generated by a non-zero matrix of determinant 0. The simplest such matrix is

$$P=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$

The left ideal generated by $P$ consists of matrices

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\c&0\end{pmatrix}$$

that is, essentially, of column vectors - Pauli spinors. You can easily check that left action on this ideal corresponds to the action of the algebra on the standard 2-spinors.

This way the circle closes. Of course you could choose a different minimal idempotent and a different left ideal. This way you would get an equivalent representation - thus no real difference.

P.S. Of course you can define another complex structure on the representation space of Cl(3,0), by replacing i with -i. This corresponds to taking the "complex conjugated representation".

 

[ComeInSpinor]

Arkajad, thank you very much. I disappeared for a while over the new year break, but I let your comments sink in and I think I have got it now. So allow me to paraphrase, and please correct me if I'm wrong...

In an algebra $${\cal A}$$, you have a subalgebra $${\cal L}$$ which is a left ideal, i.e. if $$U\in{\cal A}$$, and $$\zeta \in {\cal L}$$ then $$U\zeta\rightarrow\zeta'\in{\cal L}$$. This is the transformation law we expect for spinors, so the spinors are elements of the left ideal.

There could also be a right ideal, whose elements $$\eta^\dag$$ transform as $$\eta^\dag U^\dag \rightarrow \eta'^\dag$$.

The product of an element of the left ideal and the right ideal, $$v= \zeta\eta^\dag$$ will obey the transformation law for vectors $$UvU^\dag \rightarrow v'$$.

How this all relates to Clifford algebas is that depending on the particular Clifford algebra you happen to be considering, you need to construct the left ideal of the algebra. In some cases this might consist of the even-graded elements, but not in general, hence your comment that "spinors are vectors in a module... It is not constructed out of the even Clifford algebra alone."

Am I on the right track?

BTW Branson Clan, thank you for your input too, although I was already aware of the binomial triangle pattern in the Clifford algebras.

 

[arkajad]

The interesting left ideals are the minimal ideals. These are generated by primitive idempotents.

The point is that in a Clifford algebra of a quadratic form there is no canonical way of selecting such an idempotent (or an ideal). That is why many mathematicians prefer to deal with abstract modules - in order to avoid this problem. In physics however such a cross-section of the Clifford algebra bundle may result form some kind of a "spontaneous symmetry braking". Say, by selecting a particular solution of some differential equation of physical significance. However this path is not well researched so far.