# User:YohanN7/Gamma matrices

## Discussion from Quondums Talk page

Hi Q!

You removed my note on γ_{5} forming a Clifford algebra together with the other gammas for five spacetime dimensions. Why?

I think it is relevant, and I can source the removed statement verbatim.

Cheers, YohanN7 (talk) 16:28, 6 November 2013 (UTC)

- Hi. No offence intended. For context, this is the bit I removed after you added it:
*But it is a particularly appropriate name because the set {γ*^{0},γ^{1},γ^{2},γ^{3},γ_{5}} provide a set of gamma matrices in five spacetime dimensions.

- Even as a quote from a source, it appears to be the opinion of the author you're quoting, and (if you'll forgive the wikilawyering) thus would have to be reported as such, even if said opinion was notable. However, quoting the opinions of individual authors is not what the article is about. And the article does not deal with generalizations to other dimensions at all, aside from the statement "It is also possible to define higher-dimensional gamma matrices" in the lead (duly linked to in another article). Its presentation was only a point of incidental interest. Relevance would only be shown if the name was actually notably used in this particular way, not simply some source saying that it forms a Clifford algebra.
- The "appropriateness" of the name is that authors (Weinberg, The Quantum Theory of Fields vol 1) opinion of course (and should go out anyhow). But {γ
^{0},γ^{1},γ^{2},γ^{3},γ_{5}} giving gamma matrices in 5-d is just a simple fact, notable or not. I think we mention somewhere that the Pauli matrices are 3-d gamma matrices. Why not mention the 5-d? As for spacetime in various dimensions, it is treated in plenty of advanced QFT texts, not to mention string theory. The fact that the article doesn't treat arbitrary dimensions doesn't mean it shouldn't.

- The "appropriateness" of the name is that authors (Weinberg, The Quantum Theory of Fields vol 1) opinion of course (and should go out anyhow). But {γ
- You'd also have had to indicate that it would be a
*real*Clifford algebra. There is already enough confusion about whether the Dirac algebra is Cl_{1,3}(**R**) or its complexification Cl_{4}(**C**). This algebra is Cl_{2,3}(**R**), which we shouldn't consider as being of a 5-dimensional "spacetime", since this term would normally be intended to mean having a Lorentzian metric. This is a slightly scatty reply. but I think you get the message that I think it is inappropriate to include this. —*Quondum*06:39, 7 November 2013 (UTC)- I don't follow you exactly. I need to read up on this, but from where do you get that the algebra is Cl
_{2,3}(**R**)? I'd say the algebra is Cl_{4,1}(**R**) (or Cl_{1,4}(**R**)). YohanN7 (talk) 14:41, 7 November 2013 (UTC)- If the five matrices span the grade 1 space of a real Clifford algebra and also anti-commute, then their squares necessarily give the signature of the space. And here we have that two of the matrices square to +1, and three to −1. As a check on this, we already know that the algebra is isomorphic to M
_{4}(**C**). As a real Clifford algebra over five dimensions, Classification of Clifford algebras#Classification gives three candidates with this ring isomorphism: Cl_{4,1}(**R**), Cl_{2,3}(**R**), Cl_{0,5}(**R**). So while you could argue that though starting with a (1,*n*−1) convention, we've had to flip to a (*n*−1,1) convention to match your assertion. But to do so, we've had to replace at least two of the original matrices (e.g. by multiplying 2 or 4 of them by*i*), to get the signature to be (+ + + + −). That is to say, the five matrices as listed do not span any grade 1 subspace of this Clifford algebra. Which disqualifies them as what we mean by "gammas", even if we leave the rest as a hidden puzzle for the reader to disentangle. - But ignoring the maths, the statement still does not qualify for inclusion. You included it as a comment on the name γ
_{5}, which it fails to do in a coherent fashion. The mathematical argument is actually a diversion from this point. —*Quondum*15:27, 7 November 2013 (UTC)- I included it mostly for its mathematical message, the name was just an odd twist. Forget that.
- On the math: Note that it says {γ
^{0},γ^{1},γ^{2},γ^{3},γ_{5}}, not {γ^{0},γ^{1},γ^{2},γ^{3},γ^{5}}. The Weinberg books are famous for containing extremely few typos, and of course no such gross errors as you hint above. YohanN7 (talk) 16:18, 7 November 2013 (UTC)- I noticed the subscript. But raising the index is equivalent to flipping the sign only if the corresponding element squares to −
*I*. But since (γ_{5})^{2}= +*I*, we have γ_{5}= +γ^{5}, and it makes no difference whether the index is lower or upper. C'mon, what do you read into the sequence ((γ^{0})^{2},(γ^{1})^{2},(γ^{2})^{2},(γ^{3})^{2},(γ_{5})^{2}) = (+*I*, –*I*, –*I*, –*I*, +*I*)?^{squaring added — Quondum 02:14, 8 November 2013 (UTC)} - Since I have now gone into details of the isomorphism, it is clear that I am not contradicting the statement that this can be used as a 5d "spacetime" Clifford algebra (by minor changes to the matrices it gives you a signature (+ + + + −), so depending on how it was presented, there is no deep contradiction. Starting with
*different*gamma matrices that have a (− + + +) signature (just multiply each by*i*), you get a (− + + + +) signature with the exact construction you give. But it remains esoteric, since this construction is probably confined to only a few cases (i.e. choice of number of dimensions and signature). Add to this that at some level Clifford algebras of opposite signature are equivalent for this use, and you might find that Weinberg et al are simply glossing over this detail? Or perhaps they just use the opposite sign convention? —*Quondum*19:13, 7 November 2013 (UTC)

- I noticed the subscript. But raising the index is equivalent to flipping the sign only if the corresponding element squares to −

- If the five matrices span the grade 1 space of a real Clifford algebra and also anti-commute, then their squares necessarily give the signature of the space. And here we have that two of the matrices square to +1, and three to −1. As a check on this, we already know that the algebra is isomorphic to M

- I don't follow you exactly. I need to read up on this, but from where do you get that the algebra is Cl

- You are right. Weinberg uses the (− + + +) convention. What we should say (if anything) is that {γ
^{0},γ^{1},γ^{2},γ^{3},iγ^{5}} gives a 5-d Clifford algebra. B t w, I checked another reference (that uses (+ - - -)): http://www.damtp.cam.ac.uk/user/dt281/qft/four.pdf. Respected author, freely available online, page 93. He states that the reason for the terminology is, in fact, that {γ^{0},γ^{1},γ^{2},γ^{3},iγ^{5}} is a set of gamma matrices for five spacetime dimensions. YohanN7 (talk) 20:03, 7 November 2013 (UTC)

- You are right. Weinberg uses the (− + + +) convention. What we should say (if anything) is that {γ

- With the Weinberg sign convention, I guess someone might have used this as a "reason", but it doesn't quite work for me as a reason with the added
*i*. The source you gave a reference to seems to do as you say: use a (+ − − −) convention, and then adds*i*γ^{5}in an attempt to make a five dimensional (+ − − − −) Clifford algebra. The anticommutation relations given in the source and required for a Clifford algebra do hold, but this is a crucial: they are not*sufficient*. As you can check for yourself, {γ^{0},γ^{1},γ^{2},γ^{3},*i*γ^{5}} does not work for five dimensions at all: we have the identity*i*γ^{5}= −γ^{0}γ^{1}γ^{2}γ^{3}}, which is not**R**-linearly independent of the space that is generated by the other matrices; it is already present in the Clifford algebra. This collapses the 5-d Clifford algebra to a 4-d one. Weird, but it looks to me to be a genuine mistake to call this a "five-dimensional Clifford algebra". —*Quondum*20:47, 7 November 2013 (UTC)

- With the Weinberg sign convention, I guess someone might have used this as a "reason", but it doesn't quite work for me as a reason with the added

- There is nothing about linear independence of the gammas from
*everything generated by the gammas*in the defining condition. In odd spacetime dimension d the totally antisymmetric (AS) tensors of rank n are linearly related to the AS tensors of rank d - n. [The AS tensors are obtained by antisymmetrizing the products of the gammas. This is something I'll put into the Dirac algebra article.] There is a formula expressing this in the Weinberg book (with a million of indices and subscripts/superscripts, too lazy to write it here). This formula implies the identity*i*γ^{5}= −γ^{0}γ^{1}γ^{2}γ^{3}, so everything is as it should be. YohanN7 (talk) 10:35, 8 November 2013 (UTC)

- There is nothing about linear independence of the gammas from

- You seem to be mixing up the two cases. You cannot transfer Weinberg's mathematical details to the other case. Weinberg uses different gammas from the ones in the article (this must be the case to get the different signature). Also, take care not to confuse
**R**-linear with**C**-linear. {1,*i*} is**R**-linear independent, but**C**-linear dependent. Remember that the two cases use different matrices throughout, so they must be dealt with independently. For the moment I'm going to use another name (γ^{4}) for one of the matrices, to try to keep things straight.- Weinberg's (− + + +) convention, and his proposal (γ
^{0},γ^{1},γ^{2},γ^{3},γ^{4}) with γ^{4}= γ^{5}=*i*γ^{0}γ^{1}γ^{2}γ^{3}as a 1-vector basis for the generating space of a real 5-dimensional Clifford algebra: Here we have the relation γ^{4}=*i*γ^{0}γ^{1}γ^{2}γ^{3}, so γ^{4}is**R**-linearlyof the product γ**independent**^{0}γ^{1}γ^{2}γ^{3}. We also get the desired squaring to (− + + + +). - The (+ − − −) convention, and the proposal (γ
^{0},γ^{1},γ^{2},γ^{3},γ^{4}) with γ^{4}=*i*γ^{5}= −γ^{0}γ^{1}γ^{2}γ^{3}as a 1-vector basis for the generating space of a real Clifford algebra: Here we have the relation γ^{4}= −γ^{0}γ^{1}γ^{2}γ^{3}, so γ^{4}is**R**-linearlyon the product γ**dependent**^{0}γ^{1}γ^{2}γ^{3}. We get~~an awkward squaring to (+ − − − +)~~__apparent squaring to (+ − − − −), which is not the case for any Clifford algebra isomorphic to M__._{4}(**C**), denoted**C**(4) in the classification article

- Weinberg's (− + + +) convention, and his proposal (γ
- The confusion possibly comes in because previously we did not have a symbol for the fifth basis element, and were using the an expression in its place. One of the properties of a Clifford algebra of a vector space of dimension
*n*over a field*F*(often referred to as an*n*-dimensional Clifford algebra, even though it has a higher dimension over*F*as a vector space in its own right) is that its dimension is 2^{n}over*F*. That is to say, a basis of the algebra has 2^{n}*F*-linearly independent elements. If this is not what we find, we know something went wrong. —*Quondum*15:34, 8 November 2013 (UTC)- Weinberg writes that there are 2
^{n}independent elements in even spacetime dimensions, and 2^{n-1}in odd spacetime dimensions. Edit: No, he doesn't exactly. He is talking about antisymmetrized products, and I can't say immeduatly that it is the same thing. That aside, I find it hard to believe that the metric signature (exactly one + or exactly one -) has any real significance for anything we are discussing. YohanN7 (talk) 16:43, 8 November 2013 (UTC) - If we have a set of (linearly independent!) gamma matrices squaring and anticommuting the right way, I'd say we have a Clifford algebra. The article Clifford algebra says that the dimension of the algebra is 2
^{n}(no exceptions mentioned), that's true. But the definition of a Clifford algebra doesn't say anything about dimension. There is a problem somewhere. YohanN7 (talk) 17:31, 8 November 2013 (UTC) - Perhaps we could continue the discussion here: User:YohanN7/Gamma matrices. I wrote a program to confirm anticommutation/squaring. The matrices with signature (+ − − −) certainly behave as advertised. YohanN7 (talk) 19:14, 8 November 2013 (UTC)

- Weinberg writes that there are 2

- You seem to be mixing up the two cases. You cannot transfer Weinberg's mathematical details to the other case. Weinberg uses different gammas from the ones in the article (this must be the case to get the different signature). Also, take care not to confuse

Template:Od (ec: posting this here, but will continue discussion on your page when I get back to it later.)

As I've said before, linear independence and the commutation relations of a basis for a vector space is not a sufficient condition for producing a full Clifford algebra. If this were the case, we could construct higher-dimensional Clifford algebras with very limited representations. Take for example the quaternions **H** ≈ Cℓ_{0,2}(**R**), which is a "two-dimensional" Clifford algebra, with the algebra having 2^{2} = 4 dimensions of the algebra. (If you want a strictly real full matrix representation example, we could use Cℓ_{1,1}(**R**) ≈ M_{2}(**R**) instead, which can be used to illustrate the same point.) Now take the three elements **i**, **j**, **k**, which are **R**-linearly independent, and which satisfy the necessary commutation relations. Yet, no amount of finding products is going to build a Clifford algebra over three dimensions (i.e. is going to construct Cℓ_{0,3}(**R**) with an algebra dimension of 8), for the simple reason that we have a predefined relation **ij** = **k**, which would normally be a new, linearly independent basis element of the algebra in the construction of a Clifford algebra. The relation **ij** = **k** is an additional equivalence that collapses the attempted construction back to **H**. This is exactly the problem that is occurring in our example with gamma matrices. Or, to put it more succinctly, M_{4}(**C**) (or any subalgebra of it) is not algebra isomorphic with Cℓ_{1,4}(**R**); it simply cannot be used as a representation of the Clifford algebra Cℓ_{1,4}(**R**), even though it is a representation of Cℓ_{4,1}(**R**).

Your difficulty in accepting that switching the signature should have any effect is shared by many people, but it is a fact of Clifford algebras: this switch often produces a completely nonisomorphic algebra. However, in physics we are usually interested only in specific subspaces of the algebra, for example scalars, 1-vectors, bivectors, spinors/rotors, pseudovectors, etc., and the full algebra never comes into play. Because of this, the sign convention (and the fact that we would be working in nonequivalent algebras) has no effect: we always get the same result, regardless of the sign convention. Which will hopefully allay the feeling that this contradicts your intuition. But it does not change my assertion that the source got it wrong. — *Quondum* 19:24, 8 November 2013 (UTC)

- If you two don't mind me cutting in, I just found a very nice pdf: [1] on the Gamma matrices. It looks relevant and useful for RQM articles. Unfortunately, the pdf alone is not reliable as a source, and there could be errors in it, but at least it has a clear presentation style.
**M∧***Ŝ**c*^{2}*ħ*ε*И*20:45, 8 November 2013 (UTC)_{τlk}

- You're welcome to join. I've scanned through the pdf, though not in detail. It does not seem to focus on constraints. From the start it seems to use the imaginary unit
*i*to handle the signature so that the Kronecker delta can be used for convenience. This sort of forces one to use whatever subspace of M_{r}(**C**) happens to work without identifying this subspace, and it seems to say nothing about the size*r*of the gamma matrix representations needed. —*Quondum*04:09, 9 November 2013 (UTC)

- You're welcome to join. I've scanned through the pdf, though not in detail. It does not seem to focus on constraints. From the start it seems to use the imaginary unit

## Continued discussion

I am beginning to (finally) get the point. I don't have a problem with Cℓ_{1,4}(**R**) and Cℓ_{4,1}(**R**) not being isomorphic. [This is a sort of thing that may or may not happen in group rep theory with a rep and its dual.] But it is truly surprising that one of them fits into M_{4}(**C**), and the other one doesn't. Do I get the definition below right now? YohanN7 (talk) 21:24, 8 November 2013 (UTC)

- The additional bold seems to be the same as "freest". Think about it this way: both Clifford algebras have 32 real dimensions, which is exactly the same as the real dimensions of the matrix algebra M
_{4}(**C**). Since we have no surplus dimensions in the representation, the full matrix algebra must be exactly isomorphic to what it represents (if we had extra dimensions, we'd have freedom to select a subspace). Since the two Clifford algebras are not algebra isomorphic, it follows that we will not be able to "fit" both of them into M_{4}(**C**). —*Quondum*04:17, 9 November 2013 (UTC)- That makes perfect sense. A last(?) question on the dependence on metric convention. Suppose we redefine what a Clifford algebra is (using a minus sign on the r h s in the defining equation). This would swap the roles of Cℓ
_{1,4}(**R**) and Cℓ_{4,1}(**R**)? Now Cℓ_{4,1}(**R**) fits into M_{4}(**C**), but not Cℓ_{1,4}(**R**)? I haven't done the numbers, but I guess (and hope) that this is true. If it is true, then it truly is inconsequential which kind of metric one uses. [One has to be pretty damned careful with ones definitions, but that's another story.] The physicist may then safely regard his 4-d spacetime algebra as M_{4}(**C**) regardless of metric. YohanN7 (talk) 12:04, 9 November 2013 (UTC)

- That makes perfect sense. A last(?) question on the dependence on metric convention. Suppose we redefine what a Clifford algebra is (using a minus sign on the r h s in the defining equation). This would swap the roles of Cℓ

At any rate, these issues are quite subtle. I'm sure that our articles can become much more clear on these matters. To be sure, in the physics literature, they don't even care to say whether they consider complex Clifford algebras or not. In reality they seem to regard the real Clifford algebra as a *complex algebra* over some spacetime, but not *complex Clifford algebras* over spacetime. YohanN7 (talk) 12:28, 9 November 2013 (UTC)

- I believe you'd be making a mistake with that logic: playing with the sign convention of the definition of a Clifford algebra to select the opposite sign convention is simply a way of playing with conventions: it would be effectively mapping your sign convention onto the most convenient representation, without showing equivalence. You are not showing that it is immaterial which of the two Clifford algebras we use is immaterial. That said, I'm convinced that the
*relevant structures*within the two "tilts" of a Clifford algebra are indeed isomorphic. To demonstrate this would be nontrivial, and we'd need to define the "relevant structures", but I think that it comes down to that any scale-independent statements made in terms of 1-vectors will be equivalent in both tilts. By "scale independent", I mean in effect that the choice of units of measurement does not change the statement. This should be a way of achieving your "hope". And yes, these issues are subtle, but they do not seem to be addressed very much in the literature (that I'm aware of, but I'm poorly read). Lounesto does take a gander at it (*Clifford Algebras and Spinors*, 2nd ed, ch 13). The articles could be clearer, but making them so would take someone with an in-depth understanding of these subtleties. —*Quondum*18:10, 9 November 2013 (UTC)- The operation I think about is mentioned in Gamma matrices. Either multiply with i or switch sign in definition. Don't think equivalence is hard to show (if it is true).
- Clifford algebras isn't the only area where definitions aren't totally symmetric. Take complex analysis. There a (continuously? forgot details, not important) differentiable function of is analytic, whereas a function of never is. Roles could be reversed. YohanN7 (talk) 20:40, 9 November 2013 (UTC)
- The "operation"? You've lost me. But your description suggests you mean a switch between tilts (and is mentioned by Lounesto). Multiplying by
*i*can work by treating the two Clifford algebras as real algebras embedded in their complexification, which would then typically be the same algebra. This requires that the algebra have the necessary freedom (e.g. you can't complexify the complex numbers regarded as a real algebra by putting complex numbers into each real component: you need a fresh imaginary unit). So again, you have conditions to satisfy before this'll work. Working the other way around is easy: start with a Clifford algebra as a complex algebra, then you can reduce it to a real algebra with whatever signature you choose. So you can fit any real four-dimensional Clifford algebra into M_{4}(**C**) ≈ Cℓ_{4}(**C**), but you need bigger matrices to fit Cℓ_{5}(**C**) as a starting point to get Cℓ_{4,1}(**R**) and Cℓ_{1,4}(**R**). —*Quondum*23:45, 9 November 2013 (UTC)- Just got home from the pub with a couple of beers in my system, so don't take me too seriously right now. All I mean is that with suitable redefinitions of trivialities (or conventions if you put it that way) it should be possible to make either Cℓ
_{4,1}(**R**) or Cℓ_{1,4}(**R**), but not both using the same convention, sit inside M_{4}(**C**). The reason I believe this is that the choice of metric*is*a triviality - as far as I know in every other circumstance. Think about it. Either differs by a factor of minus one from the other. I can't see any*significance*of Cℓ_{1,4}(**R**) sitting in M_{4}(**C**), but not Cℓ_{4,1}(**R**) (or was it the other way around?). Using the normal conventions, then there are R-linear dependencies for one of them when attempting to make a Clifford algebra with that γ_{5}, but not for the other one. Switch a sign somewhere in the definitions, and the roles ought to be interchanged. If not, I'm mystified. It seems impossible that one of them would be a more complicated object (needing a bigger house) than the other. It*is*true, as you have shown above, if you adhere to standard definitions (which contains a natural, but arbitrary choice of sign). Can we have~~16~~25 R-linearly independent matrices in M_{4}(**C**), four of which square to 1, and one of which square to -1, that anticommute among the latter five? (The answer is yes if we replace -1 by 1.) I suspect the answer is yes. Now do you see what I mean? YohanN7 (talk) 02:28, 10 November 2013 (UTC)- Try the 1-dimensional case, which behaves the same. Take Cℓ
_{0,1}(**R**) (complex numbers) and Cℓ_{1,0}(**R**) (split-complex numbers), and squeeze them into M_{1}(**C**) ≈**C**. The first one fits in using the gamma [*i*]. The second one, differing only by a flipped signature, thus the gamma*i*[*i*] = [−1]. You'll note that the vector space has collapsed into the scalar space ([1], now a real scalar multiple (of the pseudoscalar, which here also happens to be the 1-d 1-vector space)), and all you're left with is the real numbers, Cℓ_{0,0}(**R**) instead of Cℓ_{1,0}(**R**). Just like its bigger brother example, you simply cannot squeeze Cℓ_{1,0}(**R**) into**C**, even though its signature-flipped version does fit. Run your program to find the pseudoscalar (the product of the five gammas), and you'll see the same. —*Quondum*03:45, 10 November 2013 (UTC)- I have ran the program to find the exactly what you predict. This is something we have agreed upon, and I have understood this. But read on.
- From the article:
*Note that the other sign convention for the metric, (− + + +) necessitates either a change in the defining equation:**or a multiplication of all gamma matrices by i, which of course changes their hermiticity properties detailed below. Under the alternative sign convention for the metric the covariant gamma matrices are then defined by*.- It appears as if the "multiply by " trick doesn't work in general. I suspect that it works in even spacetime dimensions only.

- Try the 1-dimensional case, which behaves the same. Take Cℓ

- Just got home from the pub with a couple of beers in my system, so don't take me too seriously right now. All I mean is that with suitable redefinitions of trivialities (or conventions if you put it that way) it should be possible to make either Cℓ

- The "operation"? You've lost me. But your description suggests you mean a switch between tilts (and is mentioned by Lounesto). Multiplying by

- Now, if I opt to change the definition of a Clifford algebra from to , it seems very much equivalent to changing signature in the metric (by the last equality). By relabeling the basis vectors, the equivalence becomes an equality. Now, in this new Clifford algebra, henceforth called a
**Maschen algebra**, Mℓ_{4,1}(**R**) ≈ M_{4}(**C**), do you agree? - A Maschen algebra is every bit as relevant as a Clifford algebra, it is the "dual" of a Clifford algebra in a very precise sense (equations above).
- The apparent differences between Cℓ
_{4,1}(**R**) and Cℓ_{1,4}(**R**) in complexity ,*apart*from them being "dual" rather than isomorphic is a mere consequence of arbitrary choices of sign. (Complexity means here "needs bigger house" or "needs bigger matrices" if you want) This, and nothing else, is what I have been trying to say the last few posts. - This also provides
*some*justification for the physicists claim that γ^{0},γ^{1},γ^{2},γ^{3},iγ^{5}forms a Clifford algebra in five spacetime dimensions. The remaining error is that it isn't a Clifford algebra at all, it is a Maschen algebra. - An important point: The whole idea of mine is
*not*to go about redefining things. From the point this part of the thread began, I have understood what you patiently were trying to explain to me before. I am after*conceptual*understanding whether the choice of metric signature has any physical or mathematical*meaning*at all. It does in the sense that you will have to make your choices and then stick to them consistently - or else you will screw up in calculations and/or rigor of statements (like Tong did in his pdf). But is has no*significant*meaning in the sense I ascribe to the word. - Please ask me what I mean if it isn't clear. You and I have a fairly long history of talking past each other. It can be avoided. YohanN7 (talk) 14:06, 10 November 2013 (UTC)

- Now, if I opt to change the definition of a Clifford algebra from to , it seems very much equivalent to changing signature in the metric (by the last equality). By relabeling the basis vectors, the equivalence becomes an equality. Now, in this new Clifford algebra, henceforth called a

- I note that you changed the number of
**R**-linearly independent matrices in M_{4}(**C**) from 16 to 25. There are, as you'd expect, 32 such matrices. - The trick of multiplying of all gamma matrices by
*i*: The condition for this trick to work is that*iI*is_{n}**R**-linearly independent of all the 2^{n}products of the gammas. - I'm afraid your "Maschen algebra"
*is*a Clifford algebra. - On "complexity", you need a very specific definition of this for you statement to make any sense. In particular, you have to be interested in the smallest
*n*in the representation of the algebra in M_{n}(**C**). One might argue that split-complex numbers are*simpler*that complex numbers, because they have a representation for which multiplication is simpler: (*a*,*b*)(*c*,*d*) = (*ac*,*bd*). If you called it "convenience" instead, I might agree. - As you say, we should focus on the conceptual meaning, or significance. I've already made this point: we can arbitrarily choose the sign convention (e.g. by convenience), and although the "shape" of the full algebra changes depending this choice, it makes no difference to any results of physical significance. In a sense, the
*part*of the algebras that we are interested in is isomorphic between the two. This is a far more subtle issue than tricks with sign conventions and tricks with*i*, but I am pretty certain that once the constraints of application of the algebra have been adequately identified, they answer will be: there is no physical significance in the choice because the constraints apply, but there is a mathematical significance except in the narrow confines when the constraints are applied. Thus, with your physicist hat on you should be unconcerned about the impact of choice of convention (and hence algebra) or representation, just don't make silly mistakes by confusing formulae from two different conventions. That is to say, if I'm correct, you can do your physics calculations in two nonisomorphic algebras, and be guaranteed that the results will be identical. In fact, if they differ, you've almost certainly made a mistake somewhere. - On this last note, you may be interested that some people have tried to attribute physical significance to the choice of algebra. In particular, I've seen a paper suggesting that it would make a difference to whether neutrinoless double beta decay is possible. I suspect that they got their maths twisted, though. —
*Quondum*18:40, 10 November 2013 (UTC) - An example of a nontrivial consequence of what I say: since rotors should have the same effect on any vector regardless of the choice of metric and hence algebra, we would expect the Clifford group to be isomorphic to when we "tilt" to the opposite metric. This amounts to saying that for the indefinite orthogonal groups, we must always have O(
*p*,*q*) ≈ O(*q*,*p*). The article confirms this isomorphism. —*Quondum*20:20, 10 November 2013 (UTC)

- I note that you changed the number of

## Continued discussion continued

I note that you changed the number of **R**-linearly independent matrices in M_{4}(**C**) from 16 to 25. There are, as you'd expect, 32 such matrices.

The trick of multiplying of all gamma matrices by *i*: The condition for this trick to work is that *iI _{n}* is

**R**-linearly independent of all the 2

^{n}products of the gammas.

- Which provably doesn't work in dimensions 1 and 5. The Maschen trick is not equivalent to the multiply by
*i*trick.YohanN7 (talk) 20:45, 10 November 2013 (UTC)- Agreed. —
*Quondum*01:07, 11 November 2013 (UTC)

- Agreed. —

I'm afraid your "Maschen algebra" *is* a Clifford algebra.

Every Maschen algebra**Yes, it is!***is*a Clifford algebra, and vice versa. But and Mℓ_{p,q}(**R**) ≠ Cℓ_{p,q}(**R**), instead Mℓ_{p,q}(**R**) ≈ Cℓ_{q,p}(**R**) holds. So, they are not the same Clifford algebra. YohanN7 (talk) 20:45, 10 November 2013 (UTC)- *squeek?* I was merely reacting to your statement
*The remaining error is that it isn't a Clifford algebra at all, it is a Maschen algebra.*Do I have to point out the logical contradiction? ;-)- You need to read between the lines sometimes.
*This also provides some justification for the physicists claim that γ0,γ1,γ2,γ3,iγ5 forms a Clifford algebra in five spacetime dimensions [for metric signature (+ - - - -)]. The remaining error is that it isn't a Clifford algebra at all [for metric signature (+ - - - -)], it is a Maschen algebra for [for metric signature (+ - - - -), i.e. a Clifford Algebra for metric signature (- + + + +)].*Better? (Yes I understand how hilarious it must have seemed.)- —
*Quondum*06:56, 11 November 2013 (UTC)

- —

- You need to read between the lines sometimes.

- *squeek?* I was merely reacting to your statement

On "complexity", you need a very specific definition of this for you statement to make any sense. In particular, you have to be interested in the smallest *n* in the representation of the algebra in M_{n}(**C**). One might argue that split-complex numbers are *simpler* that complex numbers, because they have a representation for which multiplication is simpler: (*a*,*b*)(*c*,*d*) = (*ac*,*bd*). If you called it "convenience" instead, I might agree.

- I
~~am~~was interested in the smallest*n*in the representation of the algebra in M_{n}(**C**). Thought it was evident. For every Clifford algebra it is the same as for the corresponding (see above) Maschen algebra. YohanN7 (talk) 20:45, 10 November 2013 (UTC)- But then, your twin might have insisted on the smallest M
_{n}(**R**). Heh. —*Quondum*01:07, 11 November 2013 (UTC)

- But then, your twin might have insisted on the smallest M

As you say, we should focus on the conceptual meaning, or significance. I've already made this point: we can arbitrarily choose the sign convention (e.g. by convenience), and although the "shape" of the full algebra changes depending this choice, it makes no difference to any results of physical significance. In a sense, the *part* of the algebras that we are interested in is isomorphic between the two. This is a far more subtle issue than tricks with sign conventions and tricks with *i*, but I am pretty certain that once the constraints of application of the algebra have been adequately identified, they answer will be: there is no physical significance in the choice because the constraints apply, but there is a mathematical significance except in the narrow confines when the constraints are applied.

- My interpretations of what we (finally) have agreed(?) on is very different. The fact that trivial tricks with minus signs or multiplication by i casts one convention into the other shows that there are no
*true*subtleties left. It is just the usual mess that different sets of notation/terminology and conventions bring along. The systems are in my mind mathematically and physically "isomorphic", where I intentionally leave out the definition of isomorphic because it's obvious enough for me that there is nothing of interest there, physically or otherwise. The asymmetry that is still there I attribute fully to the arbitrary choice of sign when defining what a Clifford (Maschen) algebra is. It can probably be mathematically explained in full as a consequence of -1 = (-1)^{2n + 1}≠ (-1)^{2n}= 1. YohanN7 (talk) 20:45, 10 November 2013 (UTC)- This argument, if it works, relies on subtleties beyond arguing that it is obvious. While our conclusions on the physics side are the same, the reasoning is completely different. So we'll have to agree to differ on the path of getting there. —
*Quondum*01:07, 11 November 2013 (UTC)

- This argument, if it works, relies on subtleties beyond arguing that it is obvious. While our conclusions on the physics side are the same, the reasoning is completely different. So we'll have to agree to differ on the path of getting there. —

Thus, with your physicist hat on you should be unconcerned about the impact of choice of convention (and hence algebra) or representation, just don't make silly mistakes by confusing formulae from two different conventions. That is to say, if I'm correct, you can do your physics calculations in two nonisomorphic algebras, and be guaranteed that the results will be identical. In fact, if they differ, you've almost certainly made a mistake somewhere.

- Good point, with which I fully agree. Also a point emphasized strongly as an
*important point*in my last post.

On this last note, you may be interested that some people have tried to attribute physical significance to the choice of algebra. In particular, I've seen a paper suggesting that it would make a difference to whether neutrinoless double beta decay is possible. I suspect that they got their maths twisted, though. — *Quondum* 18:40, 10 November 2013 (UTC)

- Nature has made choices before, for example when choosing between left handed and right handed for electrons, muons, etc. This is a choice between two inequivalent Lorentz reps (thus, of sorts, connected to reality). The choice of metric is of another nature. It is merely a choice of
*notation*that*we*make. But this is still interesting. In can't rule out that nature has made a choice between a bunch of non-isomorphic Clifford algebras in the same way that it has made a choice between left and right and choices of certain representations of some groups, but not other. (Representations of SU(3) of flavour doesn't come as free particles for the standard representaion (quarks), the gluons are elements of the adjoint representation space, etc) But this would be a completely new ingredient in physics, much corresponding to the introduction of gauge groups. (This would still leave us free to chose the metric as we please.) YohanN7 (talk) 20:45, 10 November 2013 (UTC)- We seem to agree here. But I'd suggest that in his instance nature cannot even in principle choose between nonisomorphic Clifford algebras, because the only part that is used is the part that is isomorphic between them. This is actually something I'd like to do: identify exactly this part that is shared between them, and see what the resulting algebra looks like. —
*Quondum*01:07, 11 November 2013 (UTC)- "...the only part that is used is ..." I mean something fundamentally new and undiscovered like that tachyons are governed by Cl
_{48,33}(or Ml_{33,48};) ), that is, nature might*use*something that we aren't aware of. I don't believe it, but it can't be ruled out, it is just very very unlikely. For another example, the most complicated exceptional Lie group, E8 has a decisive role in heterotic string theories.- In that sense, of course I agree. There might be some use for a Clifford algebra in which non-isomorphism is significant, because the whole algebra is used. I can even think of an example where this is the case (complex vs. split-complex). I'm simply saying that when a Clifford algebra is used to describe the behaviour of a vector space, and the rest of the algebra is just interpreted in terms of the vector space (rotating vectors, volumes, scaling, torque etc.) and then rest is just machinery to manipulate these, then either tilt will do. —
*Quondum*06:56, 11 November 2013 (UTC)

- In that sense, of course I agree. There might be some use for a Clifford algebra in which non-isomorphism is significant, because the whole algebra is used. I can even think of an example where this is the case (complex vs. split-complex). I'm simply saying that when a Clifford algebra is used to describe the behaviour of a vector space, and the rest of the algebra is just interpreted in terms of the vector space (rotating vectors, volumes, scaling, torque etc.) and then rest is just machinery to manipulate these, then either tilt will do. —
- About what you want to explore: Isn't there a formula relating Cl
_{p,q}to Cl_{q,p}? There ought to be a well defined map (not a Clifford algebra isomorphism) between them -**the Maschen transform**. YohanN7 (talk) 04:24, 11 November 2013 (UTC)- There is no simple 1-to-1 single mapping. Lounesto devotes a whole chapter to this, and his phrase "tilt to the opposite metric" seems to work nicely. At a glance, you have to separate out the type of object (vector, spinor, point derivative, whatever) and map it differently. You cannot just map the whole algebra uniformly. I guess my idea would be to keep different types of object separated from the start; that way they'd map into either tilt directly. —
*Quondum*06:56, 11 November 2013 (UTC)

- There is no simple 1-to-1 single mapping. Lounesto devotes a whole chapter to this, and his phrase "tilt to the opposite metric" seems to work nicely. At a glance, you have to separate out the type of object (vector, spinor, point derivative, whatever) and map it differently. You cannot just map the whole algebra uniformly. I guess my idea would be to keep different types of object separated from the start; that way they'd map into either tilt directly. —

- "...the only part that is used is ..." I mean something fundamentally new and undiscovered like that tachyons are governed by Cl

- We seem to agree here. But I'd suggest that in his instance nature cannot even in principle choose between nonisomorphic Clifford algebras, because the only part that is used is the part that is isomorphic between them. This is actually something I'd like to do: identify exactly this part that is shared between them, and see what the resulting algebra looks like. —

An example of a nontrivial consequence of what I say: since rotors should have the same effect on any vector regardless of the choice of metric and hence algebra, we would expect the Clifford group to be isomorphic to when we "tilt" to the opposite metric. This amounts to saying that for the indefinite orthogonal groups, we must always have O(*p*,*q*) ≈ O(*q*,*p*). The article confirms this isomorphism. — *Quondum* 20:20, 10 November 2013 (UTC)

- Why would you say that O(
*p*,*q*) ≈ O(*q*,*p*) supports your view that the "tilt" is mathematically significant? I'd happily use the same argument to support my claim that "tilt" is mathematically insignificant. - To wrap up:
- You say "I am pretty certain that once the constraints of application of the algebra have been adequately identified, they answer will be: there is no physical significance in the choice because the constraints apply, but there is a mathematical significance except in the narrow confines when the constraints are applied."
- I say "There is not even mathematical significance unless you give mathematical significance to much simpler things than I do. I attribute the mathematical differences to arbitrary choices of convention." YohanN7 (talk) 20:45, 10 November 2013 (UTC)
- This discussion looks interesting and I've been following it time to time. Unfortunately I can't really comment much on the general theory of Clifford algebras (only recently come to terms with geometric algebra). I may actually have something to ask you two later, but not for now.
- Anyway... why... the heck... have you named an algebra after me when I'm
a professional/distinguished mathematician!? (Unless of course you name after the real village Maschen in Harburg, Germany ^_^).**not****M∧***Ŝ**c*^{2}*ħ*ε*И*21:07, 10 November 2013 (UTC)_{τlk}- An algebra should have a good solid reputable name, don't you agree? Seriously, the thing needed a name, and Q needles me if I don't define things minutely. Since you are, are least sporadically, a part of this discussion, well, why not? YohanN7 (talk) 21:25, 10 November 2013 (UTC)
- Yes, it's a perfectly good a name, in this case as a way of identifying a Clifford algebra with the opposite sign convention on the defining quadratic form. Since WP has set a guideline to use of one convention under the name "Clifford algebra", a way of referring to the other is useful.
- Going back to your point: Non-isomorphism is the very epitome of mathematical significance, and in this instance is far from simple. As full Clifford algebras, the two are utterly different. E.g. how many solutions are there to
*n*^{2}− 1 = 0? In complex numbers 2 and in split-complex numbers 4. How many solutions to*n*^{2}= 0? 1 and infinity respectively. These are two simple Clifford algebras that differ in exactly the way that you are suggesting is "mathematically insignificant": a change in sign of the metric signature of the defining quadratic form. Next you'll be saying that the difference between Euclidean and Minkowski space is not "mathematically significant". Though I do not feel the need to convince you of my point, on this we do differ. —*Quondum*01:07, 11 November 2013 (UTC)

- An algebra should have a good solid reputable name, don't you agree? Seriously, the thing needed a name, and Q needles me if I don't define things minutely. Since you are, are least sporadically, a part of this discussion, well, why not? YohanN7 (talk) 21:25, 10 November 2013 (UTC)

- We seem to understand each other 100% now, and even agree about the single item we do not agree on; the level of significance of the non-isomorphism. The same phenomenon occurs in group theory. A representation and its dual representation may be non-isomorphic. But every single piece of information present in one of them is present in the other (The dual of the dual is (isomorphic to) the thing we started with). It is in this sense I regard the Clifford algebra and the Mashen algebra of the same spacetime signature not being
*significantly*different. But it is curious how differently the algebras embedd in matrix spaces. And, I admit that I haven't*proven*that there isn't any interesting difference. But, I believe that, plausibly, such a thing would have been discovered a hundred years ago.

- We seem to understand each other 100% now, and even agree about the single item we do not agree on; the level of significance of the non-isomorphism. The same phenomenon occurs in group theory. A representation and its dual representation may be non-isomorphic. But every single piece of information present in one of them is present in the other (The dual of the dual is (isomorphic to) the thing we started with). It is in this sense I regard the Clifford algebra and the Mashen algebra of the same spacetime signature not being

- But don't worry, there are days when I have my mathematical rigor hat on as well. I don't even like the mathematicians habit of
*identifying*things that*are*isomorphic in some sense, particularly not when they use equality signs between sets that aren't equal as sets. Heck, I don't even like the reference to*the*tangent space to a point in a manifold. There are 4-5 (at least) canonically equivalent definitions of such things. But they aren't*equal*since they contain very different objects. YohanN7 (talk) 04:24, 11 November 2013 (UTC)

- But don't worry, there are days when I have my mathematical rigor hat on as well. I don't even like the mathematicians habit of

## Definition

Specifically, a Clifford algebra is a unital associative algebra which contains and is generated by a vector space *V* equipped with a quadratic form *Q*. The Clifford algebra *C*ℓ(*V*, *Q*) is the "freest" algebra generated by *V* subject to the condition

where the product on the left is that of the algebra, and the 1 is its multiplicative identity.

Specifically, a Clifford algebra is a unital associative algebra which contains and is generated by a vector space *V* equipped with a quadratic form *Q*. The Clifford algebra *C*ℓ(*V*, *Q*) is the "freest" algebra generated by *V* subject to the condition

* and that condition only*, where the product on the left is that of the algebra, and the 1 is its multiplicative identity.

Perhaps "freest" captures that no more relations are allowed, even if V is endowed with it. YohanN7 (talk) 20:00, 8 November 2013 (UTC)

Then again, in a finite-dimensional matrix space, relations are unavoidable. YohanN7 (talk) 20:15, 8 November 2013 (UTC)

- Yes, I interpret "freest" to mean no more relations are allowed. In effect, it is saying that we start with a vector space
*V*(such as being "endowed" with a predefined multiplication to be co-opted!), and then we are**with no added structure***defining*the multiplication through (a) having an identity and being associative, (b) a definition of the product of pairs of elements in the subspace corresponding to the original vector space, and (c) that when a product is not forced to be in a subspace already defined through (a) and (b), it will be treated as a new dimension (I'll claim poetic license here, but this is a restatement of the crucial point). When you use matrix multiplication, you have a multiplication that is incompatible in the requirement (c). On your last point (finite-dimensional matrix spaces), for a finite-dimensional real or complex Clifford algebra, you can always find a finite-dimensional real matrix representation. In fact, for any Clifford algebra over an*n*-dimensional*F*-vector space, you can directly construct a matrix algebra using 2^{n}×2^{n}matrices over*F*, though far more compact representations generally exist. —*Quondum*03:33, 9 November 2013 (UTC)

- Yes, I think this pinpoints the problem, though this is a rather subtle point, with many seductive reasons for missing it. —
*Quondum*18:19, 9 November 2013 (UTC)

- Yes, I think this pinpoints the problem, though this is a rather subtle point, with many seductive reasons for missing it. —

## A program

int main()

{

BArray<YSquareMatrix<YComplex>,5> gammas; gammas[0] = "(1 0) (0 0) (0 0) (0 0)|"\ "(0 0) (1 0) (0 0) (0 0)|"\ "(0 0) (0 0) (-1 0) (0 0)|"\ "(0 0) (0 0) (0 0) (-1 0)";

gammas[1] = "(0 0) (0 0) (0 0) (1 0)|"\ "(0 0) (0 0) (1 0) (0 0)|"\ "(0 0) (-1 0) (0 0) (0 0)|"\ "(-1 0) (0 0) (0 0) (0 0)";

gammas[2] = "(0 0) (0 0) (0 0) (0 -1)|"\ "(0 0) (0 0) (0 1) (0 0)|"\ "(0 0) (0 1) (0 0) (0 0)|"\ "(0 -1) (0 0) (0 0) (0 0)"; gammas[3] = "(0 0) (0 0) (1 0) (0 0)|"\ "(0 0) (0 0) (0 0) (-1 0)|"\ "(-1 0) (0 0) (0 0) (0 0)|"\ "(0 0) (1 0) (0 0) (0 0)";

gammas[4] = "(0 0) (0 0) (0 1) (0 0)|"\ "(0 0) (0 0) (0 0) (0 1)|"\ "(0 1) (0 0) (0 0) (0 0)|"\ "(0 0) (0 1) (0 0) (0 0)";

for(int i = 0; i <= 4; i++) { cout << "Square i = " << i << ":" << endl << gammas[i]*gammas[i] << endl; } cout << endl; cin.get();

for(int i = 0; i <= 4; i++) { for(int j = 0; j <= 4; j++) { cout << "Anticommutator: i = " << i << " j = " << j << ":" << endl << gammas[i]*gammas[j] + gammas[j]*gammas[i] << endl; } cin.get(); cout << endl; }

}

## The output

I'm going to teach my math library to output in Latex one day, but this is at least readable.

Square i = 0:

(1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (1, 0)

Square i = 1:

(-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0)

Square i = 2:

(-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0)

Square i = 3:

(-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-1, 0)

Square i = 4:

Anticommutator: i = 0 j = 0:

(2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (2, 0)

Anticommutator: i = 0 j = 1:

(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0)

Anticommutator: i = 0 j = 2:

(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0)

Anticommutator: i = 0 j = 3:

(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0)

Anticommutator: i = 0 j = 4:

Anticommutator: i = 1 j = 0:

Anticommutator: i = 1 j = 1:

(-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0)

Anticommutator: i = 1 j = 2:

Anticommutator: i = 1 j = 3:

Anticommutator: i = 1 j = 4:

Anticommutator: i = 2 j = 0:

Anticommutator: i = 2 j = 1:

Anticommutator: i = 2 j = 2:

(-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0)

Anticommutator: i = 2 j = 3:

Anticommutator: i = 2 j = 4:

Anticommutator: i = 3 j = 0:

Anticommutator: i = 3 j = 1:

Anticommutator: i = 3 j = 2:

Anticommutator: i = 3 j = 3:

(-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0) (0, 0) (0, 0) (0, 0) (0, 0) (-2, 0)

Anticommutator: i = 3 j = 4:

Anticommutator: i = 4 j = 0:

Anticommutator: i = 4 j = 1:

Anticommutator: i = 4 j = 2:

Anticommutator: i = 4 j = 3:

Anticommutator: i = 4 j = 4: