# In which degrees there exist non-decomposable elements in the exterior algebra?

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 $$1. For which tuples $$(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

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.

• 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... – Asaf Shachar Jan 22 at 13:23
• 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. – Servaes Jan 22 at 14:53