I am trying to get a better understanding of the concept "decomposable" element in an exterior algebra. Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. For which tuples $(k,d)$, $\bigwedge^k V $ contains non-decomposable elements?

Here is a partial answer:

  1. First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k \le d/2$.

  2. For even $k\le d/2$, there are always non-decomposable elements:

Set $\sigma = e_1 \wedge \dots \wedge e_k + e_{k+1} \wedge \dots \wedge e_{2k} $, where $e_1,\dots,e_d$ form a basis for $V$.

Then $\sigma \wedge \sigma=(1+(-1)^{k^2})e_1 \wedge \dots \wedge e_{2k}=(1+(-1)^k)e_1 \wedge \dots \wedge e_{2k}$, which for even $k$ becomes $\sigma \wedge \sigma=2e_1 \wedge \dots \wedge e_{2k} \neq 0$, so $\sigma$ must be non-decomposable.

  1. For $(1,d)$, (and hence also for $(d-1,d)$) every element is decomposable.

I am not sure what happens for odd $k \le d/2$.


1 Answer 1


The set of decomposable elements forms a smooth projective subvariety of $\Bbb{P}(\bigwedge^kV)$ of dimension $k(d-k)$, called the Grassmannian. In particular there are non-decomposable elements if and only if $$k(d-k)<\dim\Bbb{P}(\bigwedge^kV)=\binom{d}{k}-1,$$ i.e. if and only if $1<k<d-1$.

  • $\begingroup$ Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach... $\endgroup$ Commented Jan 22, 2019 at 13:23
  • $\begingroup$ I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though. $\endgroup$
    – Servaes
    Commented Jan 22, 2019 at 14:53
  • $\begingroup$ Dear Servaes, in your ingenious argument how do you prove the arithmetical equivalence stated in the last two lines? $\endgroup$ Commented Dec 7, 2020 at 21:24

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