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A $\mathbb{Z}/2\mathbb{Z}$-graded algebra is said to supercommute if $xy = (-1)^{|x| |y|} yx$; in other words, odd elements anticommute. Why is this the "right" definition of supercommutativity? (Put another way, why is this the natural tensor product structure on super vector spaces?) Answers from both a categorical or physical point of view would be great.

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up vote 15 down vote accepted

The categorical answer is that (in characteristic zero) this is the only way that you can make a suitable symmetric tensor category, other than by using group representations. There is a Tannakian theorem of Deligne to this effect in the algebraic setting.

One of the physical answers is equivalent to the categorical answer. "Parastatistics" is the topic of self-consistent linear actions of the symmetric group on identical quantum-mechanical particles. The parastatistics theorem in physics (or theorems or conjectures; the level of rigor of the real point is not entirely clear) is a lot like Deligne's theorem. It says that parastatistical particles come in two kinds, parafermions and parabosons, and that they can all be modeled as fermions and bosons together with internal state spaces which are group representations.

Bosons and fermions may not look exactly the same as commutative or supercommutative algebras. But they are the same topic, because (if you apply second quantization in reverse) the values of their fields commute or anticommute.

For particles in 2D, the correct group action is the braid group, not the symmetric group. So in this case, the parastatistics theorem does not hold and you can have "anyons". Then the allowed statistics is given by a unitary ribbon tensor category. However, since the category in question is no longer symmetric, there is no clear way to define commutativity; at least, nothing that's clearly important.

Note also that isn't just that the principle of available symmetric tensor categories comes from category theory and is needed in physics. It's also needed in topology. The most traditional supercommutativity in mathematics is cohomology.

To answer Qiaochu's question below, there's nothing wrong with using the standard switching map $v \otimes w \mapsto w \otimes v$ to define commutativity. It shows up all the time. The point is that the signed switching map $v \otimes w \mapsto (-1)^{|v||w|}w \otimes v$ is another valid and inequivalent symmetric monoidal structure. (The symmetric tensor structure is interpreted as what it means to permute factors of a product, of course.) There is nothing to prevent the signed switching map from arising among topological invariants or in physics, so it does arise. The structure theorems say that all "suitable" choices for the switching map are essentially these two, possibly disguised by a restriction to tensors that are invariant under a group action.

For both good and bad reasons, I was deliberately vague about what it means for the symmetric tensor category to be suitable, in the sense that it will satisfy a structure theorem. You want some extra axioms and properties to hold, some of them related to existence of duals and traces. One version of the structure theorem, due to Deligne, is reviewed in this arXiv paper by Etingof and Gelaki. (The theorem cited as [De2] is the relevant one.)

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Could you clarify what you mean by "suitable"? I don't see anything wrong with the identification of v x w with w x v except that it's more boring; it seems to give a legitimate symmetric monoidal category. – Qiaochu Yuan Nov 7 '09 at 1:29
A followup to Greg's answer, and a more precise notion of suitable in the mathematical context can be found here: <>; (one can seek the references for the original source). Deligne's theorem says that every symmetric tensor category which is not obscenely large (meaning that lengths of Jordan-Holder series grow super-exponentially) is equivalent to representations of a super-group G, and in particular, the objects can be viewed as vector spaces (w/ G action) and the braiding given by either v x w --> w x v, or v x w --> (-1)^|v||w| w x v. – David Jordan Nov 9 '09 at 22:18
Just for the record, here are more references: – Urs Schreiber Mar 31 '15 at 12:45

This is not a satisfying answer to your question, but one observes that the exterior algebra and Clifford algebras have this kind of commutativity, so it certainly arises "naturally".

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