While I'm still trying to understand the issues raised on my previous question, I decided to first address the Clifford algebra used on formulating the famous Dirac equation. In this context, what is found in physics books is the following. There are four $n\times n$ matrices (let's take $n=4$ for simplicity) $\gamma^{\mu}$, $\mu = 0,1,2,3$ satisfying: $$\{\gamma^{\mu},\gamma^{\nu}\} := \gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu} = 2g_{\mu\nu}I$$ where $g = \mbox{diag}(1,-1,-1,-1)$ is the Minkowski metric. These relations should define a Clifford algebra. Moreover, a copy of $\mathbb{C}^{4}$ on which the Dirac matrices $\gamma^{\mu}$ act is called a spinor space, and its elements are called (Dirac) spinors.

As I mentioned in my previous question, the mathematical definition (I know) of a Clifford algebra is the following.

Definition: Let $V$ be a $\mathbb{K}$-vector space, $\varphi: V \times V \to \mathbb{K}$ a symmetric bilinear map and $\Phi: V \to \mathbb{K}$ the quadratic form associated to $\varphi$. A Clifford algebra $\mathcal{Cl}(V, \Phi)$ associated to $V$ and $\Phi$ is a $\mathbb{K}$-associative algebra with unit together with a linear map $i_{\Phi}:V \to \mathcal{Cl}(V,\Phi)$ such that:

(a) $(i_{\Phi}(v))^{2}=\Phi(v)\cdot 1$, $\forall v \in V$,

(b) (Universal Property) For every $\mathbb{K}$-algebra $A$ and every linear map $f: V \to A$ such that $(f(v))^{2}=\Phi(v)\cdot 1_{A}$ ($\forall v \in V$), there exists a unique $\mathbb{K}$-homomorphism $\bar{f}: \mathcal{Cl}(V,\Phi)\to A$ such that $f = \bar{f}\circ i_{\Phi}$.

Property (a) can be rephrased in an equivalent form: $$i_{\Phi}(v)i_{\Phi}(w) + i_{\Phi}(w)i_{\Phi}(v) = 2\varphi(v,w)\cdot 1$$

I'm trying to relate the physicist approach to the above definition. When it comes to Clifford algebras, there are a lot of information out in the internet and it is really difficult to focus on what's important if you have no background on the subject. According to Wikipedia, the Dirac algebra should be $\mathcal{Cl}_{4}(\mathbb{C})$ or $\mathcal{Cl}_{1,3}(\mathbb{C})$ which, frankly, I don't know exactly what it means.

My guess is to take $V = \mathbb{R}^{4}$ the Minkowski space and $\varphi$ the Minkowski inner product: $$\varphi(x,y) = x_{0}y_{0}-x_{1}y_{1}-x_{2}y_{2}-x_{3}y_{3}$$ If $\gamma^{\mu}$, $\mu = 0,1,2,3$ are complex $4\times 4$ known matrices, we can define $i_{\Phi}$ by sending each element $e_{\mu}$ of the canonical basis of $\mathbb{R}^{4}$ to its associated $\gamma^{\mu}$ and extend $i_{\Phi}$ by linearity. However:

(1) If all this reasoning is correct, I don't know how to prove that the universal property is satisfied.

(2) Wikipedia says this construction should be $\mathcal{Cl}_{4}(\mathbb{C})$ and $\mathcal{Cl}_{1,3}(\mathbb{C})$ and I'm taking $\mathbb{R}^{4}$ instead of $\mathbb{C}$, so I don't know what is the connection between these approaches.

Can someone help me with these problems?

  • 2
    $\begingroup$ Your (mathematical) definition of Clifford algebra is (in some sense) incomplete. If I define something through a universal property, I have to give a model to show that the thing I defined actually exists. Do you know of such a model? If you have one at hand, it's much easier to answer your questions. $\endgroup$ – Peter Mar 13 at 2:54
  • $\begingroup$ @Peter I'm definitely not much trained on this subject so I don't know if I understood you correctly, but here are some comments. This is the most general definition of a Clifford algebra I know and I've seen some proofs of its existence in books. Thus, I'm considering it exists. In any case, if I succeed on showing that the Dirac algebra satisfies the properties of the definition, I'd be indirectly proving its existence again by constructing an explicit example of it. Did I answer your question? $\endgroup$ – IamWill Mar 13 at 4:39
  • $\begingroup$ It is not that hard to show, by explicit computation, that a finite-dimensional irrep of a Clifford algebra with $n$ generators $\gamma^{\mu}$ (in the Dirac theory, $n$ is the dimensionality of spacetime) must be $m$-dimensional, where $m=2^{[n/2]}$. This is mostly a matter of showing that the fundamental anticommutation relations imply that the algebra contains at least $m^2$ linearly independent elements. That a $m\times m$ matrix irrep exists is then easy. (Dan Freedman had some good lecture notes on this.) This basically covers the universality for finite-dimensional representations. $\endgroup$ – Buzz Mar 13 at 21:28

Here's a partial answer.

Heads up: that Wikipedia article seems to be written by physicists for physicists, so IMHO it's not the best source to learn these things. Greub's "Multilinear algebra" has a chapter on Clifford algebras. I believe my favorite one used to be R. Shaw's "Linear algebra and group representations", though.

Clifford algebras can be constructed as quotients of the tensor algebra, just as how it's done with the exterior algebra. Let $V$ be a vector space over $k$ endowed with an orthogonal bilinear form $B:V\times V\to k$ of signature $(p,q)$. Then, the Clifford algebra associated to it is the quotient $$ Cl_{p,q}(k)=Cl(V,B) = \left(\bigoplus_n T^nV\right) / (v\otimes w + w\otimes v=2B(v,w)1), $$ where $T^nV=V^{\otimes n}$ with $T^0V:=k$. The quotient is by the ideal generated by the written relations and $1\in T^0V$ is the unit of this ring.

This definition is equivalent to the one using quadratic forms, so I'd suggest you pick one way of doing it and stick with it. To see this you need to understand the equivalence between bilinear forms and quadratic forms.

Now, in physical applications (especially when dealing with relativistic effects) you usually want to consider a bilinear form with signature $(3,1)$ (or $(1,3)$?), i.e., the one used in the Minkowski spacetime.

Having said all this let me say a couple of things about your questions.

  1. You can show this quotient solves the universal property the same way you show the tensor algebra satisfies its universal property. This might be written in Greub's.

  2. To really understand what's going on you have to study the classification theory of Clifford algebras and their irreducible representations. From the mathematics side the gamma matrices (Dirac matrices) come from a representation of $Cl(V,B)$. I speculate that physicists complexify this algebra in order to access certain representations that are more sensible from a physical point of view. These notes seem to give a good overview of the whole situation from the mathematical perspective.

  • $\begingroup$ Your partial answer is already very useful! So, in short: I can keep my definition using quadratic forms as it is and show that the construction via quotient spaces provide a particular example where the definition holds (i.e. it proves the existence). I guess this is what Greub does. After, I have to choose $B$,$V$ and so on to construct the Dirac algebra. Is that correct? I still need to understand the details about the second part, on how to construct the Dirac algebra from the general quotient construction tho. $\endgroup$ – IamWill Mar 13 at 12:36
  • $\begingroup$ Also, If you have other references to suggest, where this construction using quotient space is done, I'd appreciate. $\endgroup$ – IamWill Mar 13 at 12:36
  • $\begingroup$ An additional comment: I just checked the construction on Greub's book and he uses the ideal $u\otimes u -B(u,u)$ apparently. Is this equivalent to your approach? $\endgroup$ – IamWill Mar 13 at 14:09
  • 1
    $\begingroup$ Why is this a partial answer? It seems to cover everything. Ah, the connection to Dirac matrices. The $\gamma$-matrix identities are precisely those needed by the universal property, so they define a representation of $Cl_{1,3}$ on $\mathbb{C}^4$. If the representation is faithful, then the $\gamma$-matrices generate an isomorphic copy of $Cl_{1,3}$. That is the case (see Trautman's note), so you could have taken it as the definition, which is what had happened historically. $\endgroup$ – Igor Khavkine Mar 13 at 15:07
  • $\begingroup$ @IamWill The two ideals agree as long as the characteristic of the underlying field is not $2$. In characteristic $2$, though, Greub's ideal that you quoted is larger and, as far as I know, is the preferred version. $\endgroup$ – Andreas Blass Mar 13 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.