Why are Grassman integration and differentiation equivalent? The only justification of this definition I have ever scene is "Well, how else could it work?" Indeed, I don't have any other suggestions, but I'd like something a bit more rigorous or formal, or at least more detailed.

In particular, I am interested in this as it applies to fermionic systems in physics, since that is the primary context that I am familiar with. Explanations from different perspectives are welcome as well.

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    $\begingroup$ In what context are you looking for an answer? For example, my contact with Grassman numbers is their relevance in physics for describing systems of fermionic particles. But that's not necessarily the kind of answer you want. $\endgroup$ Jul 18, 2014 at 16:30
  • $\begingroup$ That is also precisely the context in which I have encountered them. I don't really know where else they are important, but something more abstract and general would also be welcome. $\endgroup$ Jul 18, 2014 at 16:32
  • $\begingroup$ Then you might consider asking this over at Physics Stack Exchange as well. If you do keep it here, your question would probably benefit from making clear the context. $\endgroup$ Jul 18, 2014 at 16:35
  • $\begingroup$ You might also tag this a bit differently: "physics", "mathematical physics", and "quantum field theory" would be appropriate additions. $\endgroup$ Jul 18, 2014 at 16:48

1 Answer 1


One way to define Grassmann integration is via axioms. See for instance this answer on Physics.SE.

An equivalent way of thinking about it that I find very natural is the following: Grassmann integration is the integration of differential forms.

To see this, suppose you're working with $n$ Grassmann variables $\theta_1, \ldots, \theta_n$. That is, the $\theta_i$ are generators of an abstract Grassmann algebra, i.e. the polynomials in $\theta_i$ modulo the anti-commutativity relation that $\theta_i^2 = 0$ for each $i$.

If you're familiar with differential forms, it's easier to recast the above in that language. That is, just rename $\theta_i$ to $dx_i$ and think of the "abstract" Grassmann algebra above as the "concrete" exterior algebra (the algebra of differential forms) over $\mathbb{R}^n$. Then the anti-commutativity relation simply means that we multiply the forms via the wedge product (as usual). However, let's suppress that in the notation so that $dx_i dx_j$ means $dx_i \wedge dx_j$.

By anti-commutativity, we need only concern ourselves with polynomials $f(dx_1, \ldots, dx_n)$ in the $dx_i$ (analytic functions of forms can be defined by Taylor expansion, which necessarily terminates after a finite number of terms), which themselves are differential forms.

Now the Grassmann integral $\int f(\theta_1, \ldots, \theta_n) d\theta_1 \ldots d\theta_n$ is simply the integral $\int_{\mathbb{R}^n} f(dx_1, \ldots, dx_n)$ of the form $f(dx_1, \ldots, dx_n)$ over $\mathbb{R}^n$ (I didn't include any "extra" differentials in the second integral because the integrand is already a differential form). What's crucial is that we are (essentially) only integrating the top-degree part of this form (because the integral of a $k$-form over a space of dimension $n > k$ is $0$ by definition). This reflects the fact that, e.g. $\int \theta_1 d\theta_1 d\theta_2 = 0$ in Grassmann integration.

It's a good exercise to see that integrating forms in this way respects the axioms of Grassmann integration.

  • $\begingroup$ This is an excellent answer. It sounds so obvious framed this way. $\endgroup$ Jan 7, 2015 at 5:06
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    $\begingroup$ @ZachMcDargh: Thanks! I learned this while reading this paper of Brydges, Imbrie, and Slade. There is a self-contained introduction to this perspective on the Grassmann integral in section 4. They credit two papers of Le Jan but I have not read those myself. $\endgroup$
    – Ben
    Jan 7, 2015 at 5:33

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