If $\eta$ is a Grassmann variable, due to invariance under translations we get that,

$$\int d\eta\ \eta = 1 \tag1$$

Nevertheless, for being Grassmann's, $\eta$ satisfies $\eta^2 = 0$. Differentiating this condition you get,

$$d(\eta^2) = 2\eta d\eta \equiv 0 \Rightarrow \int d\eta\ \eta = 0 \tag2$$

So, Eq. (2) obtained just via definition of Grassmann variable goes against Eq. (1) that comes out from translation invariance. But I've seen the use of Eq. (1) in all books about fermions' path integral, so what is the thing that I'm misunderstanding?


I am not sure that $d(\eta^2)$ is defined at all. But if it is, then, in my opinion, you should write it in this way $$ d(\eta^2) = d(\eta\eta) = d\eta\ \eta + \eta\ d\eta $$ So you get not $\eta\ d\eta = 0$, but natural anticommutation of $\eta$ and $d\eta$: $d\eta\ \eta + \eta\ d\eta = 0$. I think the latter equality is usual for Grassmann integrals.

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  1. For Grassmann-odd Berezin integration, the integration measure $d\theta$ (called an integration form in Ref. 1) is not$^1$ a 1-form/differential form $\mathrm{d}\theta$!

  2. For Grassmann-odd Berezin integration there is no naive analogue of Stokes' theorem.

  3. Also note the curious fact that there is no top-form in the Grassmann-odd sector. E.g. the 2-form $\mathrm{d}\theta\wedge\mathrm{d}\theta\neq 0$, because there is a minus sign associated with permuting the exterior derivative $\mathrm{d}$ ("the wedge") and there is a minus sign associated with permuting the Grassmann-odd variable $\theta$. Similarly, the 3-form $\mathrm{d}\theta\wedge\mathrm{d}\theta\wedge\mathrm{d}\theta\neq 0$, and so forth.

  4. Instead the Berezin integration $\int\! d\theta=\frac{\partial}{\partial \theta}$ is the same as differentiation! For motivation of eq. (1), see e.g. this Phys.SE post.


  1. Th. Voronov. Geometric integration theory on supermanifolds, Sov. Sci. Rev. C. Math. Phys., Vol. 9, pp.1–138. Harwood Academic Publ. (1992).

  2. Th. Voronov, Supermanifold Forms and Integration. A Dual Theory, arXiv:dg-ga/9603009.


$^1$ Quoting from the introduction of Ref. 2: The naive definition of differential forms in super case has nothing to do with the Berezin integration!

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