Talk:Spinor/Archive 2

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Spinors and Pauli Matrices/Dirac Matrices

Okay, I typed in "Spinors of the Pauli Spin Matrices" back in June and no one deleted it, I'm going to tempt fate and type in "Spinors of the Dirac Algebra" and see what happens. I also added the general solutions for spin in the (a,b,c) direction for spin-1/2 Pauli particles and also for Dirac particles and antiparticles.

As an aside, there are several versions of spinors based on operator theory that probably belong somewhere on wikipedia. The basic idea is to define spinors entirely through the projection operators themselves. The Cambridge geometry group calls this "density operator" theory: http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/hd_density.html

These methods allow all the usual spinor calculations without having to specify a representation. It's far more elegant than the usual methods.

Carl BrannenCarl Brannen 09:15, 25 January 2007 (UTC)

Sorry, but I moved your edits out to spinors in three dimensions and dirac spinor, respectively. The Dirac spinors edit needs to be adjusted to fit in that context, I think, but it looks to me as though it belongs there more than here. Besides, that page needs all the help it can get. Please have a look, and see if you approve. Regards, Silly rabbit 15:56, 17 April 2007 (UTC)

Article too large already

From the discussion on the talk page it appears that at least some people want to further expand this article. Maybe my comments will go against the flow, but I feel very strongly that this article is already way too big and (a related issue) diluted. I think it can be improved a lot by removing much of inessential material, then adding some important points that are not adequately addressed (eg Dirac equation or spinor fields). If I had not known what the spinors were already, I would never even come close to finding my way to their definition through the infinite ramblings of this article! This may be one of those cases, where giving a definition and a couple of examples straight up would serve the reader much better than vague and convoluted preliminary comments about their "meaning". (Curiously, the stub of an article spinor bundle seems to be better in this regard.) There should be a paragraph close to the beginning to the effect that "Spinors are elements of the spinor representation of orthogonal group G, which is constructed from the Clifford algebra as follows...", then the example of SO(3) worked out with as little notation as possible. Normally, I would have done it myself, but given the size of the article and apparently enormous amount of effort invested in it, I wanted to discuss it first. Also, there are a couple of wrong or, at least, misleading statements right at the beginning of the article, which raise the issues of the consistency, hence not easily fixable.

• Quantization — this is not the most common use of the quantization, and certainly not the best way to define spinors (aside from the role of spinors in the Dirac equation, which indeed may be said to quantize something).
• The introduction specifically refers to inner products and Spin(n), making the impression that the space is real and the form is positively-definite. But then the article almost immediately jumps into the discussion of indefinite orthogonal groups and later, isotropic vectors and various reality issues. Arcfrk 13:08, 10 March 2007 (UTC)

I basically agree. And the 'to do' comments, with more physics planned, will only make matters worse. Material should be ruthlessly moved into specialised sub-articles. By the way, tensor is no better, after being multiply forked. Charles Matthews 12:27, 20 March 2007 (UTC)

Good job cleaning it up, Charles! There is obviously still a lot of room for improving this article, but giving it more structure is an excellent first step. Are you going to move out "higher dimensional spinors" as well? At the moment, its length and multitude of details contrasts rather sharply with the minimalistic nature of "spinors in three dimensions". It also seems to me that "history" and several sections that immediately follow should be moved closer to the beginning. I am going to wait a bit until you stabilize the article, then make my own contributions. Arcfrk 21:58, 20 March 2007 (UTC)

Ugh, big job. We always (on the physics/mathematics boundary) run into these pedagogical issues, in other words 'what is X?' versus 'how to introduce X?'. The Bourbaki-flavoured mathematician can deal with the first quite well. The theoretical physics input can deal with the second, but is inevitably multi-valued. Charles Matthews 08:11, 22 March 2007 (UTC)

Right, the article is now more in the 'concentric' style, with an Overview covering all main aspects, and more algorithmic discussion separate. Charles Matthews 08:52, 22 March 2007 (UTC)

Phantastic job! A+. Arcfrk 09:52, 22 March 2007 (UTC)

Two ways to think about ψ M ψ^~

It occurs to me that there are two ways to think about the quantity ψ M ψ^~ -- two viewpoints which can clash, unless the article explains them. But at the moment it doesn't say a word.

The first way to think about ψ M ψ^~ is to think of M as a bilinear operator, and ψ as "half a vector" (or half a multivector) -- so it and its twin ψ^~ are just the passive objects being operated on.

Thus one can write

$\psi M{\tilde {\psi }}=\phi {\tilde {\phi }}\,$

and then

$\phi {\tilde {\phi }}\,=N$

because, if you like,

$M^{1/2}:{\tilde {\psi }}\rightarrow {\tilde {\phi }}$

The other way is to think of ψ as the operator, and M as the passive object on the receiving end -- ie so that between them ψ and ψ~ define a transformation, which has the effect on M of mapping it to N:

$\psi M{\tilde {\psi }}=N\,$

because, if you like,

$\psi :M^{1/2}\rightarrow N^{1/2}$

Now the first way is the point of view from which much of the article is written. (Indeed it seems to correspond to both of the approaches described in "Two basic approaches"!). It's also the way most of us probably first get conditioned to thinking about spinors via the wavefunction |ψ> in quantum mechanics.

The second way, with the spinor as the operator, is eg the way spinors are used to do rotations in computer algebra. It's the point of view now taken in the article by the sections on spinors in 2D and spinors in 3D. It's also possible to think of the wavefunction in this way, something that boosts an initial pure state |e_j><e_j| into a density matrix |ψ>|e_j><e_j|<ψ|, and any operator M into the corresponding matrix |ψ>M<ψ|.

Unless some sort of bridge is added, at the moment I fear there is a real dissonance between the different sections. Jheald 23:07, 26 March 2007 (UTC)

ψ is easier to see as an operator when it looks like a matrix. It is more difficult to think of ψ acting like an operator when it looks like a column.

When ψ looks like a column, the way it does in the last sections of the article, what is the hidden machinery in the formalism that allows it to act like an operator? Jheald 23:13, 26 March 2007 (UTC)

Aaagh

'My' lead section was completely redone, with physics buzzwords to excess. Charles Matthews 21:13, 18 April 2007 (UTC)

Ahh... I found your version of the intro in the edit history. Someone had gone in and mangled it, and I think I must have missed yours. Yours was very good. I'll see if I can merge it with the new one. As for the 'buzzwords', I don't see any. It would be shameful not to mention the physical applications of spinors, and the discussion is quite minimal. Silly rabbit 21:25, 18 April 2007 (UTC)

Thanks for the edits. The pain has gone. Charles Matthews 09:42, 20 April 2007 (UTC)

Aaagh from me, too

It's not just the lead section. There are whole paragraphs full of buzz words, such as 'polarization' used left and right without explaining anything.

Once upon a time, after Charles had gone over the article and cleaned it up, for a while, it was actually possible to find out what spinors *are*, and have a good view of the topic: first a short overview, then the details. Now a long 'motivation' section replaced the overview, the new overview is just a rambling. The old overview survives in the commented out form with a cryptic remark:

already covered somewhat above

I am sorry if it sounds a bit harsh, since it looks like someone invested a lot of effort in editing the article, but while certain parts have been somewhat improved, the overall quality of article has gone way down. I definitely appreciate the time and effort, but it has to lead us in the right direction, I think.

Some specific comments:

• Sexy pictures of coffee cups are fine and dandy, but this is an article about spinors, not the fact that the special orthogonal group SO(3) isn't simply-connected. Pare down, or summarize and move out the bulk to a separate article.
• Far from everyone can grasp the meaning of 'entanglement' (if indeed there is any precise meaning to it at all). This can be very annoying.
• It's ridiculous to have to scroll down 3 pages to find a first approximation to a definition of a spinor.
• In the positive-definite case, there are no isotropic subspaces, therefore, all talk about polarizations isn't helpful. And this is one of the most important cases for applications!
• The new overview section is largely pseudo-scientific. Here is how it starts:

In higher dimensions, spinors can also be introduced as geometrical objects generalizing the notion of a vector in order to allow for the orientation entanglement of rotations. These spinors must allow for the fact that orientation entanglement occurs not just for vectors, but also for linear subspaces (such as lines, planes, and so on). Hence a spinor defines a sort of polarized multivector.[9]

There are several approaches for defining and describing the properties of spinors. In one approach, one looks closely at the notion of polarization. By factorizing each vector into isotropic components, we achieve a polarization of the space by consistently selecting one isotropic component of every vector. The spinors are formed by taking exterior products of this isotropic component of different vectors.[10] Usually, the exterior product can be thought of as a way to encode the subspace spanned by a collection of vectors, and hence spinors are (roughly speaking) polarized isotropic subspaces.[11]

This is (1) vague, (2) not particularly insightful (unless one happens to already possess suitable intuition), (3) mathematically meaningless, for the most part. For example, if spinors are geometrical objects, what can I do with them? Can I add them? Do they have length? How about angles? What does 'factorizing a vector' mean? How can a multivector be 'polarized'? And so on, and so forth. The next few paragraphs about things acting on each other are particularly unclear. In addition, some of the links, such as

polarized multivectors

are rather confusing (wrong link, in this case? I coudn't even be sure!).

Some suggestions:

• Restore the basic structure of the article, with short summaries and detailed explanations elsewhere, at least for the first half.
• State things more precisely, so that at least people who know mathematics, but may not know what spinors are or what they could be used for, could follow the flow of the article.
• Try not to talk about many possible vague interpretations of something that has not even been defined (Ok, the last one is just out of utter frustration).
• If you feel like you can say a lot about spinors using your favourite approach, then write a wikibook! As a reference for many other articles dealing with spinors, such as spin structure, it would be preferable to have a more precise and concise version of this article.

Arcfrk 01:22, 28 April 2007 (UTC)

Noted. The page has been reverted. Cheers. Silly rabbit 01:41, 28 April 2007 (UTC)

Ok, I've calmed down somewhat now, and can respond in more detail to your comment.

• I'm not married to the idea of spinors being polarized multivectors. In direct response to your criticisms: Yes, it was the right link. No it isn't mathematical nonsense as you claim. Yes, I see what you mean about buzzwords. But that's all a moot point, since as you point out I probably should avoid talking about things which haven't yet been formally introduced (at least not until much later in the article).
• When I returned to the article, it was in need of some global improvements. Firstly, about 1/3 of the article seemed to be of the opinion that spinors were elements of the Clifford algebra, rather than its representations. Further, the introduction had been mangled, and now restored and expanded per Charles Matthew's indication.
• The article (even Charles Matthew's version) still failed to define spinors correctly. Apparently subsequent editors realized this, but incorporated their views in inappropriate places (such as the introduction).
• Not a lot of motivation was presented, and I think that many readers would find a bit of elementary (sexy coffee cups) discussion helpful. Since most of the important content (largely written by myself) had been moved (by C.M.) out of the article, I was under the impression that he wanted this to be a light introduction to spinors with the real content contained elsewhere.

So, I'm sorry if I've made you angry. Let me assure you that my edits were done in good faith, with the honest intention of improving the article. Silly rabbit 03:13, 28 April 2007 (UTC)

I've made some changes to correspond approximately to your suggestions. The points you made were all good ones. You don't need to be antagonistic to get your point across. Silly rabbit 04:22, 28 April 2007 (UTC)

I've been told that my criticism, while intended to be constructive, often appears antagonistic. I should learn to express myself better, since my comments were also in good faith. Arcfrk 07:00, 28 April 2007 (UTC)

What is a spinor

To give some flavour of why definitions of spinors are often next to useless and at best highly confusing, consider the following from the current edit. I read: "a spinor must belong to a representation of the double cover of the rotation group SO(n,R)". So I think, aha, I know what a spinor is if I know what a representation is. Reading representation, I learn that a representation is a homomorphism from a group G to an automorphism group. So what that means is that a spinor is a group homomorphism?

No. That's doesn't sound right. Ah, the problem must be this phrase "belong to", what does "belonging to a representation" mean. No idea. No definition of *that* anywhere.

A similar problem occurs later in "In this view, a spinor is an element of the fundamental representation of the Clifford algebra Cℓn(C) over the complex numbers". What is an "element" of a representation?

When I was at college I was taught a representation was just as wikipedia defines it (a hom) but now I suspect that maybe we are talking about a module over the algebra and the spinor is a vector in that module. Is that right? Who can say?

I completely, 100% understand what a clifford algebra is -- and clifford algebra is pretty clear and lucid. I *think* I know what a representation is, and I could do character theory at college. I feel that I ought to understand or be able to deduce what a spinor is. But I'm afraid this page leaves me as confused as ever. Will someone help?

By the way, its no use giving an example. I'm not sure if my efforts to generalise an example will work. Eg, I am quite familiar with (say) the first few chapters of spinors and space-time by Penrose and Rindler. Great fun, but no definition of what a spinor is in general. I can manipulate different kinds of objects (called spinors) when calculating interaction cross-sections from Feynman diagrams in QED. I have seen examples. What *exactly* am I looking at? Anyone care to help? Francis Davey 21:34, 24 July 2007 (UTC)

I'm going to restore Adam's previous answer, which I thought was also quite revealing: -- Jheald (talk) 08:18, 18 December 2007 (UTC)

Good question, Francis, and I hope to be able to answer you (and the rest of the WP community) in a helpful manner, but first I'd like to ask what is meant by the question "What is an X?" I'll take as an example "What is a vector?"

Vague: A vector is an object with magnitude and direction. Good for intuition, persuades you that they are useful things, certainly that they are "independent of basis", whatever that means at this point, but not much use for maths. Actually, you can formalise this and use it to do maths, and Euclid got remarkably far, but it's painful.

Example: This arrow thingy is a vector, and so is this 3d arrow, and also this function from the reals to the reals, in a weird infinite dimensional way. Often helpful if done in the right way. Rarely definitive enough to be useful as the only definition.

Representation: A vector is an n-tuple of reals (or maybe members of some field K). The physicist might add that the components transform in a certain way under "change of basis", which is painful to a mathematician because a basis is a set of vectors.

Intrinsic: If you add two vectors, you get a vector, and if you multiply a vector by an element of the field you get a vector. The computer science term for this kind of definition is duck typing. This is based on the idea that these seem to be the only properties of vectors that you ever really need, so anything which also quacks like this we'll agree to call a vector as well. If we needed anything else, we'd be quite happy to add it in. See Hilbert space for example. Notice that this actually define a vector space, and a vector is just an element of this.

I think the problem is that we have a vague definition - lots of them in fact, depending on how much intuition you derive from words like "quantum" and "polarised" - and we have two different representations, but we have no intrinsic definition. The two representations are (a) an element of Spin(p,q) which is a subset of a clifford algebra and (b) a member of a set of explicit matrices and rules for manipulating them which are derived from specific complex representations of specific spin groups, and totally rely on the classification of real and complex clifford algebras, or at least the representations of Spin(1,3). The fact that there are 2 competing representations and no consensus means that some people (mostly mathematicians) think that a spinor 'is' (a) and some people (mostly physicists) think that a spinor 'is' (b). Also, for spinors, tensors, and many other mathematical objects, there's another kind of definition:

Formal: A spinor is an element of the connected double cover of SO(n) for n at least 2. The classic example here is that a tensor space is the universal object for bilinear maps. At least the spinor formal definition is constructive, whereas the tensor isn't even that.

These are all useful types of definition to have in an encyclopedia, and we should have all of them for the important subjects, including spinors.

So, what are you looking for?

Adam1729 10:53, 24 September 2007 (UTC)

I've just realised that there's another and more important confusion with spinors, so I've removed my previous reply. If anyone objects to me editting history in this way, I can easily put it back.

Maybe the natural mathematical definition is an element of Spin(p,q), but the natural physical definition is not the image of this element under some representation, it's a vector from the vector space on which the representation acts. This really is a vector in the mathematician's sense, but we need a new word, because it's not like the (p+q)-dimensional space-time vectors. For SO(p,q), there's a really obvious representation to use, so vectors are the obvious thing. For Spin(p,q), this is not true, so we end up with a variety of different things being called spinors. The thing that they have in common is that they are elements of vector spaces on which a representation of Spin(p,q) acts. The word 'spinor' is not strictly meaningful to apply to a single element, but it implies a set in which the spinor lives. This set is a normed vector space, and it has an extra property, namely that there is an (irreducible? faithful?) action of Spin(p,q) (for a particular p,q of interest) on this space.

Does this sentence (in the article) already cover it?

One can remove this sign ambiguity by regarding the space of spinors as a (linear) group representation of the spin group Spin(n).

Adam1729 10:51, 25 September 2007 (UTC)

Lack of definition

This article lacks easily findable definition (which should be backed with motivation, informal explanation, etc...).

Somebody fix this, please! Thank you!--83.131.0.38 (talk) 14:37, 26 November 2007 (UTC)

The article more lacks a simple understandable definition. This from the lead: "spinors can be defined as geometrical objects constructed from a given vector space endowed with a quadratic form by means of an algebraic[1] or quantization[2] procedure". is a pretty good definition. It's just not so simple, but I don't quite know what to do about that. Martijn Hoekstra (talk) 08:23, 18 December 2007 (UTC)

I disagree that the above sentence is a good definition as it doesn't say how to define a spinor but just that is can be defined in some way.

Shouldn't the definition be something like: A vector of an vectorspace V is called a Spinor, if there is an irreducible representation (rho, V) of a Spin-group. Meaning, like a position vector is a vector of a vectorspace V, that is part of a representation of SO(n), a Spinor is a vector of a vectorspace representing this further information encoded in the larger group Spin(n). -- JanCK (talk) 10:54, 18 December 2007 (UTC)

Yes, and this definition is already (effectively) in the article:

These missing representations are then labeled the spin representations, and their constituents spinors. In this view, a spinor must belong to a representation of the double cover of the rotation group SO(n,R), or more generally of the generalized special orthogonal group SO(p, q,R) on spaces with metric signature (p,q).

There are also two different explicit constructions of the spin representations, which could be called "definitions" in the sense that they refer to more or less concrete things. There is also a definition using Clifford algebras. I believe that one difficulty is that it is very difficult to define spinors in complete generality without bringing in a whole bunch of other things in the process. Either you need to be willing to accept the existence of certain representations of a simply connected double-cover of the special orthogonal group, or you need to be willing to get your hands dirty with Clifford algebras. The article currently follows a "pyramid" design: first starting out with vague statements which should (hopefully) be at least understandable to most people, and then ending with the details that make everything work, with various levels of detail in between. The grandparent poster was, apparently, not willing to go past the first few paragraphs in the search for a suitable definition. 72.95.241.69 (talk) 16:08, 22 December 2007 (UTC)

Why not geometric algebra ?

Probably very few readers of these pages have heard of geometric algebra. That fascinating and highly efficient mathematical tool originated 150 years ago in the works of Hamilton, Grassmann, Clifford. It was revived 40 years ago by David Hestenes, an american mathematician and physicist, who reinterpreted geometrically some abstract Clifford algebras, particularly those applied to real euclidian space and to Minkowski space. Among his most active followers are some astronomers and physicists of the Cambridge university (UK).

Geometric algebra (GA) has stunning simplifying effects on the manipulation and interpretation of the mathematics applied to quantum physics. By the way GA is more efficient and easier to work with than the hamiltonian quaternions, which are a subalgebra of GA.

It seems to me that everybody agrees with the fact that the mathematical object called spinor is not at all easy to understand and manipulate. But in geometric algebra the equivalent object is quite simple : in GA a spinor is an even multivector - nothing mysterious -, whose action when applied to a vector (v) by ψvψ~ means geometrically a rotation combined or not with a boost and a dilatation. In the same manner the effect of a rotation, passive or active, on a spinor is simply encoded by Rψ or ψR , where R is a rotor. Even a beginner in GA can quickly grasp these ideas ; there is no need to be an experienced algebraist.

Unfortunately todays only a few privileged students or lucky readers of internet stumbling on it when browsing, have the opportunity to study GA. Why ... ?? —Preceding unsigned comment added by Chessfan (talk • contribs) 10:30, 24 August 2008 (UTC)

The object you describe in your post is not what modern mathematicians or physicists would call a spinor. It is instead an element of the spin group, which is somewhat easier to describe than a spinor. A spinor is an element of a certain kind of representation of the spin group (called a spin representation). If you like, you can think of it this way: the spin group does the "spinning" and a spinor is the thing which "spins". There is no natural way of regarding a spinor as a multivector. A more mathematical way to think of it is that if you encode the objects you described as matrices then the spinors are not the matrices, but rather the column vectors that those matrices act upon. siℓℓy rabbit (talk) 12:15, 10 September 2008 (UTC)

Chessfan: I had the same misunderstanding as you, when I first came to this article; and in fact I made some edits based on that misunderstanding (spinors in 2D and 3D) that are still in the article, and quite probably are not be helpful or even correct.

So let me try to expand on what SR has written, and try to put it over in a more GA-orientated way.

What it boils down to, I think is this: when you write in GA a spinor is an even multivector, that's not the whole story. On the one hand, yes, a spinor ψ can always be represented in the GA as an even multivector. But (here's the but) what's missing is that not every even multivector represents a spinor. The spinors are a much smaller, more select part of the algebra.

(@SR: Spinors of course tend to be presented as objects being acted on, rather than objects doing the acting. But in Clifford Algebra based approaches like GA all objects are part of the algebra, so as well as being objects being acted on, they also in themselves represent the different actions)

Now SR says the set of even elements of the Clifford Algebra represents a spin group. I'm not sure that's right. I think Cℓ⁺(n) actually represents an even bigger group than a spin group.

Yes you're right. It seemed that there was an additional requirement being applied: that φ must give a rotation+boost, which would make it an element of the spin-group. siℓℓy rabbit (talk) 11:09, 11 September 2008 (UTC)

But each particular type of spinors, as represented in GA, is only a subset of that even subalgebra -- one which corresponds to one or another of the irreducible representations of the group structure of the whole thing. A representation of the group is a set of objects which is closed under the action of the whole group. An irreducible representation is set which contains no such smaller subsets apart from one element which is invariant to everything.

matrix and column representations

Now, as SR says, one way to go is the matrix route. If you're doing GA calculations in a computer, it's fairly straightforward to map each element of the GA to a matrix. (This may not be the most computationally efficient way to do things, but it's one way to go). So you end up with a set of matrices that represent all the even multivectors, and combine by matrix multiplication, ψφ -> ψ'. The full action of the algebra corresponds to φ being able to represent any different even multivector in the algebra. By construction the algebra is closed, so choosing the set {ψ} to be the set of all even multivectors does indeed give a representation of the group of the whole algebra.

But it's not an irreducible representation, because one can identify the action of the group on ψ as a whole as composed of the action of the group (the matrix product) on each column of ψ individually.

In matrix terms, columns can be isolated by right multiplication by matrices φ = $\left({\begin{smallmatrix}1&0&...&0\\0&0&&\vdots \\\vdots &&\ddots &0\\0&...&0&0\end{smallmatrix}}\right)$ for the first column,$\left({\begin{smallmatrix}0&0&...&0\\0&1&&\vdots \\\vdots &&\ddots &0\\0&...&0&0\end{smallmatrix}}\right)$, for the second, etc.

Post-multiplying a matrix ψ by any of these generates a new matrix ψ' in which all the columns but one have been zeroed, and which remain zeroed no matter what other matrix φ' it is subsequently post-multiplied by. So this fits our requirement, that a representation of a group is a set which is closed under the action of all of its elements. That may not be enough to establish that the new representation {ψ'} cannot be reduced further, but it's a step in the right direction. And it explains why spinors can be represented by column-like things.

more abstract representations

Of course, one of the mantras of the GA crowd is to see explicit matrix representations as a last resort, and wherever possible to try to understand things in a more representation-free geometrical way.

So what's going on here?

The matrix calculation above corresponds to obtaining left ideals ψ' of the Clifford algebra by right multiplication of the elements of the algebra by idempotent elements φ_i, which act to project the whole set of (multivector) elements of the algebra down into a (multivector) subspace.

The Clifford algebra as a whole is spanned by basis elements made up of exterior products of the basis elements of the underlying vector space {v}. One can show that the left ideal Δ of elements of Cℓ_n(C) φ is spanned by basis elements made up of exterior products of the basis elements of the projected space {w} = {v} φ.

... to be continued. (Lodging this before I get a computer crash). Jheald (talk) 10:56, 11 September 2008 (UTC)

Sorry, coming back to this wiki discussion I see that I gave unadvertently a false idea of the possibilities offered by GA. Of course it is an easy task in GA to establish the restricting conditions we must impose on even multivectors to define the true equivalence of classical spinors. One must read Hestenes and/or the Cambridge Group : " They have ears but don't want to hear ... ". Jheald : thanks for your explanations, but looking for matrix calculations is indeed not in the spirit of GA. Chessfan (talk) 11:44, 4 February 2010 (UTC)

A last word about something funny : One might say in 1912 Cartan anticipated Picasso ; he accidentally cut a Hamilton quaternion in pieces, put the first and last term together, and did the same with the second and third term, piled them up in a column matrix and labelled them (Pauli) spinor . It took more than 60 years to recognize the remnants ! And still the vast majority of Clifford specialists and of quantum physicists (for different reasons) refuse that simple truth. See Roger Boudet Annales de la Fondation Louis de Broglie n°26 special 2001 . Chessfan (talk) 14:29, 4 February 2010 (UTC)

Spinors in the Geometric Algebra literature

Chessfan: Sorry for the above, I was still trying to work out exactly what it was that made spinors what they are. And I didn't get it entirely right.

In the conventional literature, spinor is the name given to what is most often represented as a column of complex numbers.

Such a quantity can be translated most literally over into the Geometric Algebra context, if we consider that the column of complex numbers is isomorphic to a corresponding matrix with all but one column zeroed off, which in turn can be identified as a representation of a left ideal of a Clifford algebra -- the image of the whole multivector algebra when projected down by an appropriate idempotent.

Lounesto, Clifford Algebras and Spinors (2e, 2001), p. 60, (Google books) gives the example of the projection of Cl₃ when right-multiplied by the idempotent ½(1+e₃), to give a linear subspace spanned by:

f₀ = ½(1+e₃),

f₁ = ½(e₂₃+e₂),

f₂ = ½(e₃₁-e₁),

f₃ = ½(e₁₂+e₁₂₃),

and then shows some results with it that I still need to work through.

But this is not the whole story; because he then (p. 63 Google books) introduces Cl₃⁺ as an operator representation of the spinors. So it seems the 'columnness' is not the key requirement. (Lounesto also deals with various other spinors, later).

Note added Another version of the same material can be found in Lounesto's opening chapter of Rafał Abłamowicz, Garret Sobczyk, Lectures on Clifford (geometric) algebras and applications (2004). [Google books, p.22]. The text is almost word-for-word identical, bar some additional (useful) subheadings, but I found it made clearer that, at least for Cl₃⁺, an explicit recipe is given for the "operator spinors", by adding to the matrixified column spinor $\scriptstyle {\psi }$ its Clifford grade-involute $\scriptstyle {\hat {\psi }}$, to give an element of the even subalgebra Cl₃⁺ which still has the original column-spinor as one column of its matrix representation.

It would be interesting to know whether this also holds (or something like it) in other dimensional spaces, so that the conventional notion and the Hestenes ("operator") notion below of how to define a spinor can more generally be related together. It may be covered later in Lounesto's book -- I'll have to check. Jheald (talk) 15:11, 24 February 2010 (UTC)

The point is explicitly made in this explanation in Lasenby, Doran and Gull (1998), "Gravity, gauge theories and geometric algebra", Phil. Trans. R. Soc. Lond. A vol. 356 no. 1737 (15 March 1998), 487-582 doi:10.1098/rsta.1998.0178 section 3(c) -- page 20 of this pdf -- echoing as it does the definition given by Geometry Guy (comment of 22:11, 11 September 2008, below). (Emphasis added)

"[R]otation of a multivector is performed by the double-sided application of a rotor. The elements of a linear space which is closed under single-sided action of a representation of the rotor group are called spinors. In conventional developments a matrix representation for the Clifford algebra of spacetime is introduced and the space of column vectors on which these matrices act defnes the spin-space. But there is no need to adopt such a construction. For example the even subalgebra of the STA forms a vector space which is closed under single-sided application of the rotor group. The even subalgebra is also an eight-dimensional vector space, the same number of real dimensions as a Dirac spinor, and so it is not surprising that a one-to-one map between Dirac spinors and the even subalgebra can be constructed." (as they do, in their Appendix A).

On the face of it a quite different defining property is given by Dorst, Fontijne and Mann (2007) Geometric Algebra for Computer Science, section 7.6.4, page 195: (Google books)

In the literature of mathematical physics there are elements called spinors, traditionally associated with the description of rotations in quantum mechanics. These are closely related to rotors. It is useful to understand this link, since some of the spinor literature is relevant to geometry.

Spinors are not introduced as geometric products of vectors, but as elements that preserve grade under a sandwiching product in a Clifford algebra. Consider the set of elements S that can transform a vector x into a vector by the operation S x S⁻¹. (This is called the Clifford group.) When such elements are normalized to S S^˜ = ±1 and of even grades, they are called spinors, making up a spin group (though some authors appear to permit odd spinors as well [51 (Porteous, I. R. Clifford Algebras and the Classical Groups. Cambridge: Cambridge University Press, 1995)]).

The special spin group is the subgroup of the spin group consisting of the elements for which S S^˜ = +1. Its elements are most closely related to the rotors, but careful study shows (see e.g., [33 (Hestenes, D., and G. Sobczyk. Clifford Algebra to Geometric Calculus. Reidel, 1984.)], pg. 106) that there are some special spinors that are not rotors. They consist of the weighted sum of a rotor and its dual, but they are rare (they only occur in spaces of dimensionality 0 mod 4). So it is almost true that "special spinor" and "rotor" are equivalent terms. In summary:
All rotors are special spinors; almost all special spinors are rotors.

This is essentially explicitly the definition of an operator that is an element of a spin group, rather than a conventional definition of a spinor. According to Lasenby et al above, at least sometimes the Clifford elements that correspond to rotors can be considered spinors (or, at least, isomorphic to them) -- but are Dorst et al right to say they always can? Or is Silly Rabbit (12:15, 10 September 2008) right to distinguish spinors from elements of a spin group? Jheald (talk) 14:36, 7 February 2010 (UTC)

Modern GA's founding father is David Hestenes, so let's see what he has to say in the book with Sobczak cited by Dorst et al above: (Google books)

Our use of the word 'spinor' in reference to elements of Spin(p, q) is unusual. It is justified by our unusual general definition of 'spinor' , which is as follows: we say that an even multivector ψ in G(A_{(p, q)}) is a spinor of A_{(p, q)} if for each vector x in A_{(p, q)}, ψ x ψ^† is also a vector. Versors satisfy this definition, so the question is, how much more general can ψ be? [Algebra omitted]... Thus a spinor is always an even versor unless ¼n = ¼(p+q) is an integer, in which case a spinor can always be expressed as the sum of two even versors.

At first sight our definition appears to be quite different from the conventional definition, but the two have been proved to be equivalent in the cases of physical interest (see [H2] (D Hestenes (1967), "Real spinor fields", J. Math. Phys. 8, 798-808), [H6] (D. Hestenes (1973) "Local observables in the Dirac theory", J. Math. Phys., 14 (7), May 1973, 893-905) and [H9] (D. Hestenes (1975), "Observables, operators and complex numbers in the Dirac theory", J. Math. Phys. 16, 556-572. ).

Our definition has the advantages of simplicity in its algebraic formulation and its geometrical interpretation. Thus (8.10) shows that a spinor determines an orthogonal transformation and a dilation (by a factor ρ)

These and subsequent papers can be found on Hestenes' page on Dirac theory, here.

Evidently, this is the sense that Dorst is quoting. This may be all very well, but in my view it is highly confusing to glibly introduce spinors in this sense, without any discussion of how the use relates to the more conventional understanding of the term in physics.

Yet this is just what Hestenes does also, in New Foundations for Classical Mechanics, at page 51 (Google Books).

(Lasenby and Doran (2003), Geometric algebra for physicists is a bit more careful, glossing the issue on first acquaintance (page 66, Google Books), before presenting the correspondence of their Cl₃⁺ spinors with traditional 2-component Pauli spinors more fully in chapter 8, pp 268-275 Google Books).

But given that background, I think it's no wonder that there have been some confused people here. Jheald (talk) 16:41, 7 February 2010 (UTC)

Jheald: Thanks for that interesting and useful ltterature review. Well, I never dared to give a definition of spinors generally speaking, but restrict myself to Pauli and Dirac spinors. I come to the conclusion, to answer Silly Rabbit's critics, that every multivector in Euclidean 3 space is indeed a spinor, but that in Minkowski space one must impose additional conditions. If we follow Lasenby and Doran (p.145), to define the rotor transformation describing the proper orthochronous transformations (restricted Lorentz group), we must write :

             R=scalar + relative vector a + dual of relative vector b + pseudoscalar
             pseudoscalar = 0          a . b = 0

L and D do not mention these conditions, but they are easy to verify. Chessfan (talk) 10:40, 15 February 2010 (UTC)

Jheald: I recently delved deeper in that spinor question, with geometric algebra. As you certainly know we can define unitary matrices in SU(2) which multiplied by a column vector u=(1,0) are equal to a general spinor in SU(2); and the relation between the spinor and the matrice is bijective. Let us write (psi)= U u . We can also define a bijective relation between any vector v in R3 and an hermitian matrix M=(xi)(sigma^i). If now we rotate the vector v in R3 we must transform the matrix M with some U matrix, let us call it V, to get V M V^-1 . If you rotate in a similar way a spinor you get something like (psi)V or V(psi). It seems that gave rise to the strange idea that the spinor is something like "the square root of a vector".

Now if you transpose everything in GA, you get a much better idea of what happens and why it happens. If in R3 you rotate the basis vectors to e'=R e R^-1 nothing happens to any intrinsic vector related to the physical system you study, let's say to the spin vector s , we have s'= s . If you do the same with a vector v rigidly related to the basis vectors you get v'=R v R^-1 . But with the spinor (psi) you must write (psi)'= (psi) R^-1 !! Why ? Because the spinor itself, sandwiching any physical quantity you study (measure) represents a rotation now augmented by the additional rotation R (which is also a spinor).

Conclusion : the spinors are multivectors, but not intrinsic ones, they are basis dependent. They sit somewhere between the intrinsic properties of a physical system and the strictly basis determined tools.

Did anybody express a likewise idea in standard quantum mechanics ? Chessfan (talk) 10:38, 23 February 2010 (UTC)

I would need to get much practised at group representation theory before being able to take this further. Question: how close are the Lasenby/Doran/Gull definition ("elements of a linear space which is closed under single-sided action of a representation of the rotor group"), and the Hestenes/Sobczak one (multivectors which map a vector to a vector under the sandwich product). If one insists that the L/D/G "linear space" is a linear space of multivectors, does that make the two definitions equivalent?

Second question. If I understand correctly, in group representation theory the column vectors are particularly with investigating whether group representations are reducible -- the representation is reducible if we can apply similarity transformations such that the column vector can be split into two (or more) parts, which are not mixed under the group action. Of course, this generalises directly into the notion of a space breaking into a product of independent subspaces. But does the nature of what the vector (or not-a-vector) is made of make a difference? I seem to remember that representations which a reducible over a vector of real numbers may not be reducible over a vector of complex numbers, or a vector of quaternions... but maybe it's just that the reducibility shows itself in a different way.

The answers to these questions may, I suspect, be rather obvious to people with a better knowledge and familiarity with representation theory - which at the moment I have not got. So that's something for me to go away and get a bit more practiced in. But eventually, I think it is of importance for the article to consider (and explain) more general things that spinors may identify - along the lines of the L/D/G - rather than thinking of them only as vectors of complex numbers; how the properties and uses initially identified working with complex vectors translate into these different spaces; and whether the different spaces do give good new ways for thinking about our basic question in this article, namely what is a spinor? Jheald (talk) 15:54, 24 February 2010 (UTC)

Having thought on this a bit further, to answer my own questions a bit, it seems clear that anything meeting the Hestenes/Sobczak definition will also meet the L/D/G one, since the single-sided action of a rotor on a multivector will produce another multivector, which will map a vector to the sandwich product of the rotor with a vector, which will be another vector, thus satisfying the L/D/G requirement of closure; and since we can linearly map a (finite dimensional) vector space to a space of real column vectors, this also directly relates to the conventional "column vector" definition.

Going the other way, if (as you suggest) we take each coefficient in the column vector and multiply it by an appropriate basis matrix (or alternatively, by a basis element of the Clifford algebra) we clearly get to a set of representatives of Clifford algebra, which will transform the right way under the single-sided action of the spin group. Yes, there are lots of different possible sets that could be produced in this way; it would then fall to argue that the Hestenes/Sobczak choice is particularly meaningful. And I think one would need to somehow count degrees of freedom, to show (if it's true) that the Hestenes/Sobczak choice always completely exhausts all the freedom the spinor has. Plus of course there are other linear spaces the spinor may co-ordinatise (eg solutions to differential equations) -- in such contexts is the Hestenes/Sobczak way of thinking about/breaking down the spinor always the most useful; or are there other ways of thinking about it/breaking it down that are also useful, so that the "column vector" should be retained as the most neutral way of thinking about the spinor, albeit in the awareness that there are quite important natural ways to interpret what that vector may mean...

Sorry, a bit of a stream-of-consciousness, but at least for me it does take things on a bit from where I was in my last post. Jheald (talk) 21:15, 24 February 2010 (UTC)

This starts to look like simply an application of a general rule that is no doubt quite basic and well known to those who know about such things. Viz, that any group can be given a (faithful, minimal) linear representation that can be thought of as acting like a matrix on a column vector. But the group can also act on itself. Therefore one can naturally create a one-to-one linear map from any state of the column vector to any matrix representing an element of the group. So given such a vector, it becomes natural to think about it by linking it in a one-to-one way with an element of the group.

This may not be saying much more than that if I have a solution of some covariant system with coefficients (a1,a2,a3,a4), I may be able to expect to be able to relate that coefficient vector to a particular way of boosting and transforming some standard underlying template solution. (?)

Presumably there is some standard terminology and language for discussing all of this, for which this association is just one particular application. Jheald (talk) 22:43, 24 February 2010 (UTC)

Uniqueness - or not

Please note that a maximal isotropic subspace does not have a unique isotropic complement in dimension >2. For instance, in R^{2,2}, with quadratic form q(w,x,y,z) = w z + x y, the maximal isotropic subspace

{(w,x,0,0):w,x real}

has complementary isotropic subspaces of the form

{(c y, -c z, y, z): y,z real}

for any real number c. Geometry guy 14:42, 7 September 2008 (UTC)

That was obviously a mistake, for which I take full responsibility. siℓℓy rabbit (talk) 15:04, 7 September 2008 (UTC)

No worries - it's an easy one to make. I only mention it to reduce the risk that it gets added to the article again in the future! Geometry guy 15:30, 7 September 2008 (UTC)

PS. I got something out of this exercise: it is interesting that the exterior algebra construction seems to depend on both the choice of W and W', while the minimal ideal construction only depends on W. I've never been a fan of the minimal ideal approach, but this does seem to be a point in its favour.

Idempotents vs. Nilpotents

Please forgive my ignorance - I don't have Chevalley to hand - but can someone explain the following passage?

There are two variations on this theme: one can either find a primitive element ω which is a nilpotent element of the Clifford algebra, or one which is an idempotent. The construction via nilpotent elements is more fundamental in the sense that an idempotent may then be produced from it.

Thanks. Geometry guy 19:28, 9 September 2008 (UTC)

I believe there was a discussion, unfortunately split over several user talkheaders, on whether one should use orthogonal idempotents in the Clifford algebra to define the spinor space. This is the usual way that one first learns about constructing minimal ideals in an Artinian algebra: they are a collection of principal ideals each generated by one idempotent out of a family of mutually orthogonal idempotents. Some authors, Porteus for example, emphasize this approach to spinors. However, the approach of Chevalley (and so that of the article) is to use rank a two nilpotent element, called here ω, to generate the spinor space as a principal ideal. Once such an element is selected, if ω' is any element such that ωω' + ω'ω = 1, the element ω'ω is idempotent, and generates the same minimal left ideal as ω. siℓℓy rabbit (talk) 11:32, 10 September 2008 (UTC)

Ah, good, thanks for this. I guess ω' can be the guy associated to another max isotropic, as in the article, right? I think it would be good to mention this in the article anyway: it will only take two sentences, and will keep the idempotent fans happy. Geometry guy 18:26, 10 September 2008 (UTC)

This doesn't work: for my choice of ω', the expression ωω' + ω'ω will have components in form degree 4k, and it already fails to be a scalar for dim W = dim W' = 2. So how do you get an ω' which does work? Geometry guy 16:07, 13 September 2008 (UTC)

Interesting. Ok, the obvious ω' does work (in the sense that ω'ω is idempotent, and generates the same minimal ideal as ω), but the commutator identity that I incorrectly quoted appears only to hold modulo higher degree terms. One needs only that ωω'ω=ω and ω'²=0. siℓℓy rabbit (talk) 16:40, 13 September 2008 (UTC)

Ah, good, I think I get it now. Similarly ωω' generates the same ideal as ω' so they certainly don't add up to 1 unless the Clifford algebra is the direct sum of these two ideals! Geometry guy 19:32, 13 September 2008 (UTC)

Clebsch-Gordan decompositions and summary style

I notice that in the Comments field of the maths rating box above that Geometry guy suggests that the Clebsch-Gordan decomposition might be too detailed for this article. However the existing article Clebsch-Gordan decomposition appears not to be suitable as a move target as it was clearly written straight out of a quantum mechanics book (it is only applicable to SO(3), from the looks of it). Anyway, the Clebsch-Gordan decomposition is probably the most important single fact about spinors, since it encodes essentially all of the operations of interest on spinors (including triality, inner products, relationships with tensor representations of the orthogonal groups, and even the Dirac operator). While one could spend an entire book discussing each of these topics, it is all succinctly summarized in the Clebsch-Gordan decomposition. Weyl and Brauer spent a large part of their seminal paper on spinors in n dimensions discussing it. And I don't personally find it undue weight to have such a section here, but I would just as well happily move it elsewhere if there were a suitable move target. siℓℓy rabbit (talk) 11:55, 10 September 2008 (UTC)

I agree it is pretty important. It may be more natural to describe it at spin representation, as it is about the representation theory of spinors, rather than the spinors themselves. Lets see how things look once we've got spin representation into decent shape (work in progress, as I'm sure you've seen). Ultimately it would be good to have summary style links to all the nice topics you mention as applications of Clebsch-Gordan. Geometry guy 18:20, 10 September 2008 (UTC)

Definition?

Sorry to be annoying, but is there any chance of putting a definition in the lead? All this deep discussion is very interesting, but in the first few paragraphs there should be something like "A spinor is a...." or "Spinors are members of ....". The best that the article can manage is to tell me that spinors are group homomorphisms (because that is what representations are defined to be in wikipedia). I am pretty certain spinors aren't defined as members of a group of homomorphisms, so how about a definition, one that doesn't begin "You can think of..." or some such weasel phrase? This is a real concern for me and I am sure many others reading an article like this. It may be that if you *know* what is meant you can interpret it. At the moment what is there is neither a definition a lay person can understand nor in internal terms a pedantically correct one. Francis Davey (talk) 19:57, 10 September 2008 (UTC)

I am not aware of any definition of a spinor in the literature that is both mathematically correct and understandable by a lay-person. The article itself actually does define what a spinor is in several different ways. Some of them are reasonably explicit, but they are all rather technical and not amenable to being condensed to a lead paragraph description. The lead does say that spinors are elements of a certain vector space which carries a representation of the spin group (or a projective representation of the orthogonal group). This doesn't say which representation, and so is incomplete. I don't know if there is a compelling way to say this in the lead, other than to indicate that they are the representations which do not arise by Weyl's construction. siℓℓy rabbit (talk) 20:17, 10 September 2008 (UTC)

Sorry, you've misread what I wrote. I don't look for a definition that is both mathematically correct and understandable by a lay-person but one that falls into either category. You have improved the lead, but now I don't know what it means for a vector space to "carry" a representation. As far as I can see that is not discussed in the various representation pages on wikipedia. surely this can be explained somehow? The thing is there are lots of isomorphic vector spaces. The fact that (say) Lorentzian 2-spinors are members of a 2D complex vector space doesn't tell me much since there are many such spaces that (might) arise. A spinor is obviously a vector in a particular vectors space. Which one or ones? There must be some link between that vector space and some geometry - in the same way that we call objects contravariant and covariant vectors depending on whether they are members of the tangent or cotangent spaces, and there is a natural way of generating those spaces from a differentiable manifold, what are the ingredients I need for a spinor and how do I make spinor spaces from those ingredients? Perhaps if you explain what "carrying" means that would help. Francis Davey (talk) 22:09, 10 September 2008 (UTC)

To illustrate (please humour me). Suppose I have a (complex) representation of Spin(3,1), my understanding of what that means is that I have a homomorphism Phi: Spin(3,1) -> M where M is some algebra of nxn complex matrices. OK. Now where are the spinors for that representation? Francis Davey (talk) 22:16, 10 September 2008 (UTC)

M is an algebra of matrices, and the spinors are then the column vectors which those matrices act upon. This is what it means for a vector space to carry a representation: that the group acts as matrices on the space. siℓℓy rabbit (talk) 01:10, 11 September 2008 (UTC)

With nearly all of the ingredients: As indicated in second paragraph of the article, a spinor space (or spin representation) can be constructed from a quadratic form defined on a vector space. A spinor space is a pair (V,ρ) where ρ is a group representation of the spin group on GL(V), ρ:Spin(n)→GL(V), which does not factor through the quotient map Spin(n)→SO(n). That is, it is a representation of the spin group which doesn't come from a representation of SO(n). There are various equivalent characterizations of this condition: it is a representation which doesn't come from Weyl's construction (that is, which are not "tensors" in the classical sense); it is a representation of SO(n) (at least in the complex case) having highest weights (1/2,1/2,...,1/2) (which actually just gives one type of irreducible spin representation).

Now a spinor is an element of a spinor space. Obviously talking about a spinor only makes sense if you have a representation to work with, otherwise as you say the vector space doesn't give any interesting structure to its elements. Also, in response to the overall vibe I get of what is a spinor and how to build one, the answer is: with great difficulty. In the Explicit constructions section, you see how to build a spinor using the exterior algebra of a maximal isotropic space for the quadratic form. The particular vector space that you obtain in this way depends on an initial choice of isotropic space, but any two such spaces are isomorphic and carry isomorphic representations of the spin group.

For a spinor on a manifold, additional information is required: a spin structure. This is a topological condition on the manifold itself. siℓℓy rabbit (talk) 01:55, 11 September 2008 (UTC)

Excellent. At last a sentence with "A spinor is..." in which I can decompile all the subdefinitional elements via wikipedia! Great. That a spinor space is a pair may have been implicit and obvious to everyone else, but it isn't said in the article and now that makes a great deal of sense. From that the rest is obvious - in particular the spin structure stuff. I'll have a look through the clifford algebra/explicit construction stuff later. I'm not sure that the reason for my confusion is clear though 8-(. Francis Davey (talk) 09:30, 11 September 2008 (UTC)

So in 2 sentences, to see if I'm simplifying this correctly.

A spinor is an element of a spinor space.

A spinor space Δ is the name given to an irreducible representation ρ of a rotation group on GL(V), if ρ not a space of scalars, vectors, or tensors."

Would that be correct? Or have I oversimplified?

Wikipedia defines a representation as being a homomorphism. Taken literally (and how else should a non expert reader take it) that means a spinor is a group homomorphism. As explained above - no it isn't. I completely understand the concept of "abuse of notation" and "abuse of terminology" but it seems to me there's a lot of that going on here. Fine for an expert, but wikipedia might just not be read by people who already know the answer. So, no. that is oversimplified. I'm happy with "A spinor space .. is the name given to a pair consisting of...". That will do for me. Sorry to be pedantic but I think its reasonable to be so. Francis Davey (talk) 13:49, 11 September 2008 (UTC)

Okay, for the moment I'll go along with being pedantic about the word "representation". But then what extra do we get by calling the spinor space a pair (V,ρ), rather than just saying it's the image of the group under this representation? What am I missing, if I'm thinking of a spinor space as just a particular subspace of my geometric algebra Cℓ(n) ? Jheald (talk) 17:29, 11 September 2008 (UTC)

The one important thing I think I've missed out, above, is that the rotation group is Spin(n) "which can be constructed as a subgroup of the invertible elements in the Clifford algebra Cℓ(n)", according to its article.

Could one also write that "Δ is the name given to an irreducible representation of the invertible elements in the Clifford algebra Cℓ(n), if Δ is not a space of scalars, vectors, or tensors," or would that let in other kinds of representations that are not spinors? Jheald (talk) 12:44, 11 September 2008 (UTC)

Okay, there is often a confusion between a representation as a space on which a group acts, and a representation as a group homomorphism into the general linear group of that space. But this is no different from saying "the integers are a group under addition" instead of "the quadruple consisting of the integers, addition, zero and minus is a group". The latter is more correct, but annoyingly pedantic.

However, in the case at hand, the key feature of spinors is how they transform under the Spin group, not where they live. The Clifford/geometric algebra is a representation of the Spin group in two ways. First the spin group acts by conjugation. Second it acts by (say) left multiplication. Under the first action, the Clifford algebra decomposes into a direct sum of forms of degrees 0 to n. Under the second action the Clifford algebra is (roughly) a tensor product of a spin representation and a trivial representation. Under this second action, elements of the Clifford/geometric algebra are spinors, under the first they are not. A minimal left ideal is just an irreducible under the second action. They aren't preserved by the first action. It doesn't matter whether one considers spinors as elements in a geometric algebra or not. The key point is that the action of the spin group is not the conjugation action, but an action via the spin representation, which, in geometric algebra is left multiplication. Geometry guy 22:11, 11 September 2008 (UTC)

Examples - clifford aglebras

OK. I am still trying to work out in my mind what a spinor is (I don't think most of you have any idea how hard it is to approach this material - there are so many hidden assumptions, abuse of notation and logical leaps that its hard for someone even with a maths degree like myself to follow). In the examples we have the following:

The action of an even Clifford element γ ∈ Cℓ⁰_2,0 on vectors, regarded as 1-graded elements of Cℓ_2,0, is determined by mapping a general vector u = a₁σ₁ + a₂σ₂ to the vector

$\gamma (u)=\gamma u\gamma ^{*}\,$,

where γ^* is the conjugate of γ, and the product is Clifford multiplication. In this situation, a spinor is an ordinary complex number. The action of γ on a spinor φ is given by ordinary complex multiplication:

$\gamma (\phi )=\gamma \phi \,$.

Note that the sentence beginning "In this situation, a spinor" plucks the notion of spinor out of thin air. The example has worked through the fact that you can embed vectors into the clifford algebra (in the natural way) and that the even graded elements act on those vectors by conjugation such that they each represent rotations. Fine. But then there is a leap where the "spinor" is mentioned without being introduced. When I first read it I thought that spinors were also the vector elements of the algebra but acted on by multiplication not conjugation, but reading the 3D example made me think that what is actually happening is that the spinors are the even graded elements, acted on by left multiplication. Each spinor corresponds to a vector in a natural way, but the rotations require 4pi rotation not 2. Is that a correct understanding? I.e. that (i) in this example the spinors are the even subalgebra and (ii) that there is a 1-1 correspondence between 2D vectors and spinors? If clarification could be given, the example could be tightened up a bit to explain that. Francis Davey (talk) 17:43, 21 September 2008 (UTC)

I echo these comments, nearly 9 months later. I can't fix this article at this point, but I recognize that this section is broken. 420ftjesus (talk) 14:23, 8 June 2009 (UTC)

POTENTIAL ERROR:

I think it is not correct to say that so(V,g) is embedded as a Lie subalgebra in Cℓ(V,g) because in particular Cℓ(V,g) is not a lie algebra. Rather, one should say that one can construct a representation of so(V,g) from one of Cℓ(V,g) called the spinor representation of so(V,g), by:

$\Sigma _{ab}={\frac {1}{4}}(\Gamma _{a}\Gamma _{b}-\Gamma _{b}\Gamma _{a}),$.

where \Gamma_a span a representation of Cℓ(V,g). These representations act on vector spaces of the same dimension.

Any associative algebra can be considered as a Lie algebra with the commutator as the Lie bracket. What you wrote down above is exactly such an embedding. (RogierBrussee (talk)) —Preceding undated comment added 21:06, 11 July 2009 (UTC).

Lead

There seems to be a reversion war occurring in the lead section. This is not good to see, and reverts withour even an edit summary are a bad idea. Everyone is agreed that spinors turn out to be needed, both in mathematics and physics; but the reason why they are needed can be explained from different angles (basically analytical, algebraic and topological). No one reason should be privileged in the lead: in particular the talk about the Lie algebra is only one way of looking at it, and the talk about tensoring up representations to find less than all representations is also only one way of looking at it. Everyone should bear in mind that this is an introductory section to an encyclopedia article, and not a place for pedagogical quarrels.Charles Matthews (talk) 10:06, 11 July 2009 (UTC)

I have added one para to the lead, trying to incorporate the main points made in previous versions, removing excess bulleting, and providing a transition to the Overview. Charles Matthews (talk) 10:27, 11 July 2009 (UTC)

I prefer my version of the lead (the current version of the first paragraph as restored by User:Charles Matthews, that is) because, as you say, it doesn't make firm commitments on the style of approach to the notion of a spinor. As for the additional paragraph, the few sentences on Einstein-Cartan theory seem quite out of place to me, as this is pretty far outside of mainstream physics I think. I am removing that portion of the text. Sławomir Biały (talk) 21:32, 11 July 2009 (UTC)

Introducing spinors from the point of representation theory of the lie algebra so(V, g) is by far the most natural way to introduce spinors. Once you see that non tensorial representations of the Lie algebra exist, you have to find a way to construct them, which is where Clifford algebra's come in naturally. Now you can argue that the Clifford algebra is perfectly natural, even without its relation to so(V,g) and that spinors therefore live independent of the Lie algebra and I would agree. However, searching for vector spaces on which the infinitessimal rotations (or Lorentz trafos) act, and which are therefore behave nicely under infinitessimal coordinate transformations seems as close to a pedagogical starting point as you can get. In addition the Lie algebra point of view forces you to deal with the fact that representations of the Lie algebra are representations of the universal covering of orthogonal group SO^+(V,g) rather than representations of the group itself, i.e. topology. As for the analytical side, for analysis you need to talk about spinor _fields_ i.e. spinor valued functions (or more generally sections of a spinor bundle) and Dirac operators. But to understand spinor valued functions you better understand what spinors are and to understand Dirac operators you better understand the representation of the Clifford algebra point of view.

Now you can disagree with all that but at least defining spinors as non tensorial representations of Lie algebra's that are subspaces of algebra representations of Clifford algebra's, seems like a definite improvement over a "definition" that reads "a complex vectorspace whose extra dimensions are needed to reveal the full structure of orthogonal groups". I realise that part of this edit war comes from different expectations between mathematicians and physicists, but could we at least get a definition of spinors back that is unambiguous or as unambiguous as possible please.

If people want to improve this article then in it would be much better to go through the low dimensional examples (dim 1 up to 4 say) in the concrete and abstract definitions of spinors that are given. The way the current low dimensional examples are done are not very nice in my opinion. (RogierBrussee (talk) 22:23, 11 July 2009 (UTC))

First of all, often mathematics articles do not contain a definition in the lead paragraph. This article is further complicated by the need to accommodate people of different backgrounds, and we already have very detailed articles on spin representations and Clifford algebras. Nowhere in MOS:LEAD or WP:MTAA does it say that articles must contain precise definitions in the first paragraph. On the contrary, even our own WP:MOSMATH urges against putting precise definitions too early on in the article. Secondly, I don't see how "part of this edit war comes from different expectations between mathematicians and physicists". As far as I can tell, all parties involved are mathematicians. Sławomir Biały (talk) 22:57, 11 July 2009 (UTC)

I agree that the examples could be clearer. Sławomir Biały (talk) 23:36, 11 July 2009 (UTC)

Clifford modules

In addition to this page on spinors and the page on spin representations, there's also a small page on Clifford modules. These are closely related. But the different pages aren't linked very well. And I'm too lazy to do it. John Baez (talk) 03:18, 7 October 2009 (UTC)