# Why are differential forms on a n-dimensional manifold a Grassmann algebra?

This is stated as an obvious example of a Grassmann algebra on page 32 in this tutorial I am trying to read, but to me it is unfortunately not so obvious.

So can somebody expand this comment a bit and explain this to me?

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If you just think of each $\theta_i$ as a one form, and wherever two of them are multiplied (e.g. in eq 2.1), just do the wedge product of forms, the relevant properties will appear e.g. $\theta^2=0$ because $\theta \wedge \theta = 0$ etc –  twistor59 Mar 31 at 15:26
The basic reason is dx^dy = -dy^dx. –  Prathyush Mar 31 at 15:27
I think this is just because Grassmann algebras are associative and anticommutative, which wedge products satisfy trivially. $a\wedge (b \wedge c) = (a \wedge b) \wedge c$, and $a \wedge b = -b \wedge a$ –  Chay Paterson Mar 31 at 15:33