Representations of the Dirac algebra, hermitian adjoint and traces

Question

Strictly speaking this is a math question, but since the Dirac algebra is much more important in physics than in math I thought I'd have a better chance of getting an answer here.

The Dirac algebra can be defined as the Clifford algebra associated to the Minkowski metric. In physics texts it is not always clear if the real or the complex Clifford algebra is meant, I would say that morally it's the real one, but strictly speaking the complex one. The latter is isomorphic to the algebra of $4\times 4$ complex matrices, but not in a canonical way. By the Skolem-Noether theorem (and probably in some more elementary way as well) we see that all such isomorphisms are conjugate, and even that all complex representations of the real Dirac algebra (which is simple) are conjugate.

Now in physics texts the Dirac algebra is often defined to be an algebra of $4\times 4$ matrices, but with a representation of $\Bbb R^{1,3}$ contained within it, in the form of the four (unspecified) matrices $\gamma^\mu$.

In that context, there are obvious definitions of the trace and the Hermitian adjoint, namely as the trace and the Hermitian adjoint of the matrix. By the remark above and the fact that the trace is invariant under conjugation, we see that the trace is well-defined on the abstract algebra level. Of the Hermitian conjugate that is not so clear (I don't know if it's true).

This is unsatisfactory for several reasons:

• We do really want to work with different representations, so it would be nice if we could define the trace in a way that is independent of a specific representation, or maybe some canonical representation like the regular representation or (even better) if there is some canonical 4-dimensional space on which it acts.
• It doesn't work so well for other spaces $\Bbb R^{1,d}$.
• Things like the trace identities are messy and hard to memorize.

My questions:

• Does this only work well in $\Bbb R^{1,3}$?
• Is it more correct or useful to view the Dirac algebra as the real or the complex Clifford algebra?
• Is there a canonical, or representation-independent definition of the trace? Ideally one that works for all Clifford algebras.
• Is there a canonical definition of the Hermitian adjoint, or maybe a Hermitian inner product?
• Can the Dirac equation or more a Dirac spinor field be conveniently interpreted in terms of the abstract Dirac algebra without any explicit reference to a 4-dimensional representation of it?

EDIT

It isn't hard to show that the Dirac algebra has a unique anti-involution which I'll denote $\ast$ that satisfies $\gamma^{\mu\ast} = g^\mu_{\ \ \nu}\gamma^\nu$. If we call this the Hermitian adjoint, this would say that $\gamma^0$ is Hermitian, and $\gamma^i$ is anti-Hermitian. We could then demand that in a matrix representation, these map to a Hermitian and anti-Hermitian matrices. However, this is not automatic: it doesn't follow from the algebraic structure that this must be the case, and must be seen as an additional requirement on a representation of the Dirac algebra.

To be explicit, one could consider the real Clifford algebra or $\Bbb R^{1,1}$. This can be mapped isomorphically onto $\mathcal M_2(\Bbb R)$ in such a way that $\gamma^0$ is symmetric and $\gamma^1$ is antisymmetric. It can also be mapped onto a subalgebra of $\mathcal M_2(\Bbb C)$ in such a way that $\gamma^0$ is Hermitian, and $\gamma^1$ is anti-Hermitian.

Answer 1

Actually, your claim about "different representations" is kind of misplaced: We do not need to worry about different representations of the Clifford algebra, because it has at most two non-isomorphic irreducible representations, and those have the same dimensions. See, for example, this question and this question. However, we may choose different realizations of these representations as matrices, e.g. Weyl basis vs. Majorana basis.

You cannot define a trace or a Hermitian conjugate on the algebra itself because both the trace and the Hermitian conjugate are properties of a representation. Famously, the trace in a particular representation is called the character and an important tool to distinguish representations. This is usually done for groups, but since the spin and pin groups sit within the Clifford algebra, this applies equally well here.

Likewise, the abstract Clifford algebra does not carry a Hermitian product. The proper abstract definition of the (real) Clifford algebra in dimension $d= p+q$ is as the quotient of the tensor algebra $\bigoplus_{i = 0}^\infty \left(\mathbb{R}^d\right)^{\otimes i}$ modulo the relation $v\otimes v = \eta^{\mu\nu}v_\mu v_\nu$ where $\eta$ is the metric with signature $(p,q)$. It's a non-trivial (but not so hard) fact that all its representations are "pseudo-unitarizable" in the sense that we can choose a Hermitian product on the representation such that the Hermitian adjoint of all $\rho(\gamma^\mu)$ is $\eta_{\mu\mu}\rho(\gamma_\mu)$ (no summation convention), where $\eta$ is our spacetime metric.

Lastly, the Dirac algebra is just a tool to construct spinor representations, it is not the fundamental thing under which spinors transform. The crucial thing is that the second degree of the Clifford algebra contains the Lorentz algebra, so the representations of the Dirac algebra induce representations of the Lorentz algebra, and it's easier to find the (few) representations of the Dirac algebra than those of the Lorentz algebra. In particular, the Weyl and Majorana representations that exist in certain dimensions and signatures are not representations of the Dirac algebra (the full Dirac representation of dimension $2^{\lfloor d/2\rfloor}$ is always irreducible as a representation of the Clifford algebra), but only of the Lorentz algebra. The split into Weyl subrepresentations is the split into two eigenspaces of the top degree of the Clifford algebra, which represents parity, and the split into Majorana representations is rather complicated in arbitrary signature but has to do with the existence of so-called real structures.

Therefore, to expect the spinors to transform "abstractly" under the Clifford algebra is physically misguided - what we are actually after are representations of the Lorentz algebra $\mathfrak{so}(p,q)$.