# Tag Info

## Hot answers tagged superalgebra

6

Let me just retell a part of the story in slightly different language — maybe it will become less mysterious then. The story starts with a module $M=\mathbb C[\xi]$ (free supercommutative algebra with one odd generator — aka exterior algebra on one generator) over a Clifford algebra $A=\mathbb C[\xi,\frac\partial{\partial\xi}]$ (“odd differential ...

2

The $\mathbb{Z}_2$ grading is easy enough to anticipate. Given an integer it is either even or odd. So, there's your grading. A one-form is odd. A two-form is even. Even elements commute with all other elements under the wedge product whereas the product of odd elements anticommute. All of this is plainly seen in:  \alpha \wedge \beta = (-1)^{pq} \beta \...

2

Since they are tied to groups, and groups are tied to with geometry, one would expect so. From the wiki article on supersymmetry: Supersymmetry can be reinterpreted in the language of noncommutative geometry and quantum groups. In particular, it involves a mild form of noncommutativity, namely supercommutativity. See the main article for more details.

2

Let us consider exterior (Grassmann) algebra $\Lambda\mathbb R^{2n}$ with $2n$ generators $v_k$, $k=1, \ldots, 2n$. It is $2^{2n}$-dimensional real algebra. It is possible to introduce operators (1, p.15) $M_k$ and $\delta_k$, $k=1, \ldots, 2n$, where $M_k(1) = v_k$, $M_k(\omega) = v_k \wedge \omega$ and $\delta_k$ is adjoint of $M_k$ (see 1, p.15 for ...

1

Not to be gauche, but I ended up finding a good answer to my own question: Varadarajan's Supersymmetry for Mathematicians: An Introduction

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