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For Minkowski space $M^4=\mathbb{R}^{1,3}$ the Clifford algebra $Cl_{1,3}$ (Dirac algebra) with $\{\gamma^\mu, \gamma^\nu \}=2 g^{\mu \nu}$ is sometimes called "spacetime algebra". What is its significance for spacetime?

I know that the generators of the Lorentz group in the $(0,1/2) \oplus (1/2,0)$ representation can be written as $S^{\mu \nu}=i[\gamma^\mu,\gamma^\nu]/4$ and that the Dirac gamma-matrices $\gamma^\mu$ somehow act as reflections in the space of Dirac spinors (and combinations like $\gamma^\mu \gamma^\nu$ as rotations). I also understand that they enabled Dirac to find a "square root" of the d'Alembertian $\square=(\partial_\mu\gamma^\mu)^2$, the Dirac operator. So to me it seems like maybe all possible operators on Dirac spinors can be written as linear combinations of products of those matrices (is that right ?).

But I do not really understand the role this algebra plays in general for physics. Is it just used because it allow us to construct a basis for the operator algebra in the space of Dirac spinors? Does it have some deeper meaning?

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  • $\begingroup$ " So to me it seems like maybe all possible operators on Dirac spinors can be written as linear combinations of products of those matrices (is that right ?)." Yes. See my post here: physics.stackexchange.com/questions/237908/… $\endgroup$ – Lewis Miller Mar 5 '16 at 17:06
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The Clifford algebra $\mathrm{Cl}(\mathbb{R}^{1,3})$ is the algebra generated by endowing the vectors in $\mathbb{R}^{1,3}$ with a free algebra multiplication and then imposing the constraint given by $v \circ v = \eta^{\mu\nu}v_\mu v_\nu$ with $\eta$ as the Minkowski metric on $\mathbb{R}^{1,3}$.

Any Clifford algebra has a natural connection to the exterior algebra because you can just give a vector space isomorphism by mapping $e_i \wedge e_j$ to $e_i \circ e_j$, and the exterior algebra has the geometric meaning of describing $k$-planes in $\mathbb{R}^n$, see also my answer about vectors and pseudovectors.

In fact, the exterior algebra is the Clifford algebra with $\eta = 0$. One can thus view the Clifford algebra $\mathrm{Cl}(\mathbb{R}^4,\eta)$ as a deformation (or quantization) of the exterior algebra, with the quadratic form $\eta$ measuring the failure of $\circ$ to be anticommutative.

Now, what has this to do with the Lorentz group? The Lorentz group $\mathrm{SO}(1,3)$ is the group of (origin-preserving) isometries of Minkowski space, and the trick is that by construction the Clifford algebra with $\eta$ carries representations of the isometries preserving $\eta$:

The Clifford algebra naturally decomposes into degrees, just like the exterior algebra does:

$$\mathrm{Cl}(\mathbb{R}^{1,3}) = \mathrm{Cl}_0 \oplus \mathrm{Cl}_1 \oplus \mathrm{Cl}_2 \oplus \mathrm{Cl}_3 \oplus \mathrm{Cl}_4$$

where $\mathrm{Cl}_n$ is the subvectorspace generated by $n$-fold products of vectors from $\mathbb{R}^{1,3}$. The second degree $\mathrm{Cl}_2$ of this is six-dimensional.

Let us first note that the Clifford algebra is a representation space of the Lorentz group with the representation simply given by linearly extending the action on the first degree, which is just $\mathbb{R}^{1,3}$ to all other degrees. Furthermore, the degrees are proper subrepresentation spaces, and one can convince oneself that the Clifford algebra really is nothing but

$$ \text{scalar} \oplus \text{vector} \oplus (1,1)\text{-tensor} \oplus \text{pseudovector} \oplus \text{pseudoscalar}$$

which is not surprising since it is isomorphic to the exterior algebra as a vector space, after all. The interesting thing is that there is another representation "hidden" inside it:

If we take an "orthonormal" basis $\gamma^i$ of $\mathbb{R}^{1,3}$ with $\eta(\gamma^i,\gamma^j) = \eta^{ij}$, then the second degree is spanned by the six $\sigma^{ij} := \frac{1}{4}(\gamma^i \circ \gamma^j - \gamma^j \circ \gamma^i) = \frac{1}{4}[\gamma^i,\gamma^j]$. The commutators of these lie now in the fourth degree, which is one-dimensional, and hence only a number. Computing the commutators of the $\sigma^{ij}$ shows that they are exactly the commutators of $\mathfrak{so}(1,3)$, hence, there is a representation of the Lorentz algebra sitting in the middle degree of this Clifford algebra!1

And this representation is neither a vector nor a tensor representation, it is a true spinor representation acting on $\mathrm{Cl}_1$, which you can see since the action of $\mathrm{Cl}_2$ takes the vectors in $\mathrm{Cl}_1$ to pseudovectors in $\mathrm{Cl}_3$, meaning it is not a proper linear representation, but a projective representation. By further inspection, one can see that $\mathrm{Cl}_2$ on $\mathrm{Cl}_1$2 is indeed the usual Dirac spinor representation, but even without any computation you can see it is spinorial since it does not behave properly w.r.t. reflections.3


1Not of the Lorentz group though - as a spinorial representation, this is only a proper linear representation of the cover of the Lorentz group.

2To get the $\mathrm{Cl}_2$ action to actually yield elements in $\mathrm{Cl}_1$ instead of $\mathrm{Cl}_3$, use the Hodge duality between $\mathrm{Cl}_1$ and $\mathrm{Cl}_3$.

3It is very interesting to note that nowhere here a complexification has happened - this version of getting a spinorial representation does not actually need complex numbers, though the quantum theory of course needs them.

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  • $\begingroup$ OK thanks, this is a very nice exposition of the mathematical aspects of the Clifford Algebra. What is its physical role? Does it just serve for the purpose of a spinor rep or are there further implications for fundamental physics? What is its significance as the so-called "spacetime algebra"? $\endgroup$ – quantumorsch Feb 18 '15 at 15:45
  • $\begingroup$ @quantumorsch: It's significance lies in that it is the deformation of the exterior algebra - and the exterior algebra very physically is just the algebra describing all subplanes of $\mathbb{R}^n$. The "deformation" makes this description compatible with the metric $\eta$ on the space, and hence automatically gives how "geometric objects" in the space transform under the group which leaves that metric invariant - the Lorentz group in the case of the Minkowski metric. $\endgroup$ – ACuriousMind Feb 18 '15 at 15:50

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