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I believe the answer to be yes, but I realize that sometimes physicists place additional constraints that might not be obvious. If superalgebras are clifford algebras, why make a literary distinction?

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up vote 3 down vote accepted

The answer is No, as Lubos Motl has already pointed out. Here I would like to make a couple of general remarks.

1) On one hand, the notion of superalgebras is a huge topic, which includes, e.g., associative superalgebras and Lie superalgebras. Important examples of Lie superalgebras are super-Poincaré algebras.

2) On the other hand, a Clifford algebra $\mathrm{Cl}(V,g)$ over a vector space $V$ satisfies

$$vw+wv=2g(v,w){\bf 1}, \qquad v,w\in V.$$

In physics applications, the vector space $V$ is often a vector space spanned by a basis of gamma matrices,

$$\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}{\bf 1}. $$

In more mathematical applications, the vector space $V$ sometimes comes with an odd grading, so that the anticommutator is a supercommutator

$$[v,w]=2g(v,w){\bf 1},$$

which can be viewed as a super Heisenberg algebra with odd grading, and which is hence an example of a superalgebra.

More generally, the vector space $V$ could be a super vector space $V=V_0\oplus V_1$ with both an even and an odd sector, leading to a notion of super Clifford algebras.

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Saying that the vector space $V$ is sometimes assigned an odd grading is tricky, insofar as a vector space does not support a multiplication. What I think you mean is either that the image of the vector space $V$ in an algebra is an odd graded subspace or that the vector space $V$ as a natural subspace of the algebra is an odd graded subspace. I see the Wikipedia page gives a Clifford algebra as an example of a superalgebra, which seems a good way to go. Super-Poincaré algebras are "important" as Mathematics/tentative theoretical Physics, not yet as empirical Physics. Useful Answer, +1. – Peter Morgan May 17 '11 at 13:33
Dear @Peter Morgan. A (super) vector space $V$ can have a $\mathbb{Z}_2$ grading even if $V$ does not support an intrinsic multiplication $V\times V \to V$. – Qmechanic May 17 '11 at 15:06
Thanks, Qmechanic. Very helpful nudge. – Peter Morgan May 17 '11 at 15:55

On a mathematics stack exchange, you could succeed with such an "isomorphism". But in physics, superalgebras are interpreted totally differently than Clifford algebras. So not only you're forbidden to identify the two terms universally; but in reality, there is not a single example in physics that could be called both a Clifford algebra and a superalgebra.

Clifford algebras are reserved for algebras of matrices that have explicit Grassmann-even i.e. bosonic entries - especially the algebra of Dirac matrices $$ \gamma_i \gamma_j + \gamma_j \gamma_i = 2 \delta_{ij} $$ while superalgebras are reserved for algebras with Grassmann-odd generators that can't be represented by Grassmann-even matrices in any natural way, e.g. $$ \{Q_a,\bar Q_{\bar b}\} := Q_a Q_{\bar b} + Q_{\bar b} Q_a = 2 p_{a\bar b}. $$ So despite the mathematical similarities in the anticommutators, they're just totally different things. The difference between Grassmann-even (bosonic) and Grassmann-odd (fermionic) objects in physics is viewed as a very important one; one can't say that the bosonic and fermionic objects are the same thing. Whether they change their sign or not after a 360-degree rotation decides about their statistics, too - by the spin-statistics theorem.

In fact, the $\{A,B\}$ brackets for the anticommutator are only "natural" in the case of the superalgebras. For example, for fermionic fields, it's the anticommutator that should vanish at spacelike separations. In the case of the Clifford algebras, the objects such as $\gamma_i$ are not generators of any physically natural groups, especially not fermionic ones, so one shouldn't talk about their anticommutators. In the Dirac matrix case, the anticommutators are sometimes used but they're just a bookkeeping device.

On the other hand, in the superalgebra case, they're real replacements for commutators.

Again, there may be mathematical analogies in the formulae defining the algebras but these similarities ignore the physical meaning that is very different.

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A Clifford algebra is a geometrically rich algebraic structure. In the case of the Dirac algebra, which is a Clifford algebra over the complex field, we introduce a map from the tangent space of Minkowski space at a point into a superalgebra. Given a vector $V\in\mathcal{V}$, we can introduce a map into an algebra, $\gamma:\mathcal{V}\rightarrow\mathcal{A};V\mapsto\gamma(V)$. The algebraic structure can be presented in a number of ways, but informally they amount to working systematically with the linear dependence $\gamma(U)\gamma(V)+\gamma(V)\gamma(U)=2(U,V)$. In the Dirac algebra case, the inner product $(U,V)$ in component terms is obviously $U_\mu g^{\mu\nu}V_\nu$. The complex structure makes no appearance in the linear dependence that I have used to informally present the algebra.

For a number of reasons, I haven't worked with superalgebras in the terms described by Luboš, so I'm scandalously uncertain of his notation, but it looks as if the two distinct conjugations in his Answer, of the $Q$'s and of the $Q$-indices, signify that the inner product in the “superalgebra in Physics” case is sesquilinear (which seems to be complex anti-linear in the second component, though that's immaterial except as a convention). I presume that the inner product presented by Luboš as $p_{a\bar{b}}$ in general has no connection to the space-time metric. There is a level of abstraction at which the sesquilinear structure makes no difference, however it's enough that as categories the two constructions are different in detail because the complex structure is now nontrivially involved in the construction of the algebra, we're not just working with the complexification of a Clifford algebra over the reals.

I made quite a few starts at writing interpretational remarks, but since they were essentially responses to aspects of Luboš's Answer, not to your Question, I've decided to leave you to draw your own conclusions.

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I recommend Qmechanic's Answer. – Peter Morgan May 17 '11 at 16:01

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