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I don't seem to understand the concept of a Grassmann-variable. When studying superspaces and superfields I am told that the coordinates being used are $$(x^\mu, \theta_\alpha, \bar{\theta}_\alpha)$$ and that the last two coordinates are Grassmann-odd. What is the meaning of a Grassmann variable and what does it mean for it to be odd?

Qmechanic
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user7077252
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    Possible duplicates: https://physics.stackexchange.com/q/9960/2451 , https://physics.stackexchange.com/q/40746/2451 and links therein. – Qmechanic Feb 11 '20 at 04:16

1 Answers1

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This is a quick and dirty route to Grassmann numbers. It's worth being more careful and thorough, but this is the general idea.

Start by considering a complex vector space $V$, which has a basis $\{\theta_1,\theta_2,\ldots\}$. That is, any $x\in V$ can be written $$x=x^\mu \theta_\mu = x^1\theta_1 + x^2\theta_2 + \ldots$$ where the components $x^\mu$ are complex numbers.

Next, define the antisymmetric wedge product $\theta_\mu\wedge\theta_\nu = - \theta_\nu \wedge\theta_\mu$. In particular, $\theta_\mu\wedge\theta_\mu = 0$. For example, if $x = x^1 \theta_1 + x^2 \theta_2$ and $y = y^1 \theta_1 + y^2 \theta_2$, then

$$x\wedge y = (x^1 \theta_1 + x^2 \theta_2)\wedge(y^1 \theta_1 + y^2\theta_2)$$ $$ = x^1y^1 \underbrace{\theta_1\wedge\theta_1}_{=0}+ x^1y^2 \theta_1 \wedge \theta_2 + x^2 y^1\underbrace{\theta_2\wedge \theta_1}_{=-\theta_1\wedge \theta_2} + x^2 y^2 \underbrace{\theta_2\wedge \theta_2}_{=0}$$ $$ = (x^1y^2 - x^2y^1) \theta_1 \wedge \theta_2$$

The set of such objects forms a vector space in its own right. We call it $\bigwedge^2V$, and it can be written

$${\bigwedge}^2 V := \left\{ \sum_{ij} c_{ij} \theta_i \wedge \theta_j \ \ : c_{ij}\in \mathbb C\right\}$$

We can extend this construction in a natural way:

$${\bigwedge}^3 V := \left\{ \sum_{ijk} c_{ijk} \theta_i \wedge \theta_j\wedge \theta_k \ \ : c_{ijk}\in \mathbb C\right\}$$

so on and so forth.


The exterior algebra over $V$ is the vector space

$$\bigwedge(V) = \mathbb C \oplus V \oplus {\bigwedge}^2V \oplus {\bigwedge}^3 V \oplus \ldots $$

equipped with the naturally defined exterior product operation and component-wise arithmetic. By way of example, let two elements of $\bigwedge(V)$ be $x= 3+2\theta_1 + 4\theta_1\wedge \theta_2$ and $y = 2 + 3\theta_2$. Then e.g.

$$x+y = 5 + 2\theta_1 + 3\theta_2 + 4\theta_1\wedge \theta_2$$ $$x\wedge y = (6+9\theta_2) + (4\theta_1+6\theta_1\wedge \theta_2) + (8\theta_1\wedge\theta_2 + 12\underbrace{\theta_1\wedge\theta_2\wedge\theta_2}_{=0\text{ via antisymmetry}})$$ $$ = 6+4\theta_1 + 9\theta_2 +14\theta_1\wedge\theta_2$$


If you understand that, then you have your answer. A Grassmann number is an element of the exterior algebra over some vector space $V$. A Grassmann variable or generator is an element of the basis set $\{\theta_\mu\}$ which you use to construct the exterior algebra. A quantity is Grassmann-odd if it is the wedge product of an odd number of generators.

As an important note, for notational purposes the wedge is usually dropped. I think it's a useful thing to write down, at least in the beginning.

J. Murray
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