# Decomposable elements of $\Lambda^k(V)$

I have a conjecture. I have a problem proving or disproving it.

Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if $w$ is decomposable.

For example, for $u=e_1\wedge e_2 + e_3 \wedge e_4$ we have $W_u = 0$.

• Your conjecture is correct and your example is nice:+1 – Georges Elencwajg Mar 26 '13 at 9:20
• May I ask why the down vote? – tom Mar 26 '13 at 23:08
• Dear tom, you may ask but experience shows that you will not be told! Anyway, don't worry: there is absolutely no reason to believe that the downvoter has any valid criticism of your question nor even competence on the subject. – Georges Elencwajg Mar 26 '13 at 23:28

## 3 Answers

A $k$-vector $w \in \bigwedge^kV$ is $m$-decomposable if there is a linearly independent set $\{e_1, \dots, e_m\}$ of $V$ and $\alpha \in \bigwedge^{k-m}V$ such that $w = e_1\wedge\dots\wedge e_m\wedge\alpha$; note that $k$-decomposable is what is normally called decomposable. Furthermore, $w$ is strictly $m$-decomposable if it is $m$-decomposable but not $(m+1)$-decomposable.

Suppose that $w$ is $m$-decomposable, i.e. $w = e_1\wedge\dots\wedge e_m\wedge\alpha$ for some linearly independent set $\{e_1, \dots, e_m\}$ in $V$ and $\alpha \in \bigwedge^{k-m}V$. Clearly $e_1, \dots, e_m \in W_w$ so $W_w$ is at least $m$-dimensional.

Now suppose $w \in \bigwedge^kV$ and $\{e_1, \dots, e_m\}$ is a linearly independent subset of $W_w$. We can extend this to a basis $\{e_1, \dots, e_n\}$ for $V$. There is an induced basis for $\bigwedge^kV$ given by $\{e_{i_1}\wedge\dots\wedge e_{i_k} \mid 1 \leq i_1 < \dots < i_k \leq n\}$, so we can write

$$w = \sum_{1 \leq i_1 < \dots < i_k \leq n}a_{i_1\dots i_k}e_{i_1}\wedge\dots\wedge e_{i_k}$$

for some coefficients $a_{i_1\dots i_k}$. Now note that

$$e_1\wedge w = \sum_{1 \leq i_1 < \dots < i_k \leq n}a_{i_1\dots i_k}e_1\wedge e_{i_1}\wedge\dots\wedge e_{i_k} = \sum_{1 < i_1 < \dots < i_k \leq n}a_{i_1\dots i_k}e_1\wedge e_{i_1}\wedge\dots\wedge e_{i_k}.$$

As $\{e_1\wedge e_{i_1}\wedge\dots\wedge e_{i_k} \mid 1 < i_1 < \dots < i_k \leq n\}$ is a linearly independent subset of $\bigwedge^{k+1}V$, $e_1\wedge w = 0$ implies that $a_{i_1\dots i_k} = 0$ for $1 < i_1 < \dots < i_k \leq n$. Therefore,

$$w = \sum_{1 = i_1 < \dots < i_k \leq n}a_{i_1\dots i_k}e_{i_1}\wedge\dots\wedge e_{i_k} = \sum_{1 < i_2 < \dots < i_k \leq n}a_{1i_2\dots i_k}e_1\wedge e_{i_2}\dots\wedge e_{i_k}.$$

Proceeding in the same fashion, the conditions $e_2\wedge w = 0, \dots, e_m\wedge w = 0$ imply that we have

\begin{align*} w &= \sum_{m < i_{m+1} < \dots < i_k \leq n}a_{1\dots mi_{m+1}\dots i_k}e_1\wedge\dots\wedge e_m\wedge e_{i_{m+1}}\wedge\dots\wedge e_{i_k}\qquad (\star)\\ &= e_1\wedge\dots\wedge e_m\wedge\left(\sum_{m < i_{m+1} < \dots < i_k \leq n}a_{1\dots mi_{m+1}\dots i_k}e_{i_{m+1}}\wedge\dots\wedge e_{i_k}\right)\\ &= e_1\wedge\dots\wedge e_m\wedge\alpha \end{align*}

where $\alpha \in \bigwedge^{k-m}V$ is given by

$$\alpha = \sum_{m < i_{m+1} < \dots < i_k \leq n}a_{1\dots mi_{m+1}\dots i_k}e_{i_{m+1}}\wedge\dots\wedge e_{i_k}.$$

So $w$ is $m$-decomposable.

In summary, we have the following:

An element $w \in \bigwedge^kV$ is $m$-decomposable if and only if $\dim W_w \geq m$. In particular, $w$ is strictly $m$-decomposable if and only if $\dim W_w = m$.

Setting $m = k$, we get the desired result. That is, $\dim W_w = k$ if and only if $w$ is decomposable.

• Beautiful! After two years I got the answer :D That is what I call, necromancy done well! – tom Mar 17 '15 at 22:13

Roughly another 2 years later let me try giving a different proof.

Suppose first $w$ is a decomposable $k$-vector. Then there exist $k$ linearly independent vectors $\{e_i\}$, $i=1,\ldots k$ such that $w=e_1\wedge\cdots\wedge e_k$. Complete $\{e_i\}$ to a basis of $V$, it is then clear that $e_i \wedge w=0$ for $i=1,\ldots k$, $e_i\wedge w\neq 0$ for $i=k+1, \ldots n$, hence $\mathrm{dim}(W_w)= k$.

Suppose now $\mathrm{dim}(W_w)= k$, and let $\{e_i\}$, $i=1,\ldots k$ be a basis of $W_w$. By using the following fact:

Let $w$ be a $k$-vector, $u$ a non-vanishing vector. Then $w \wedge u =0$ if and only if $w= z\wedge u$ for some $(k-1)$-vector $z$.

one establishes that $w$ is of the form $w=c e_1 \wedge\cdots\wedge e_k$ for some constant $c$.

• The quoted fact is a theorem of Dibag from 1974 in the paper "Factorization in exterior algebras". – Russ Woodroofe May 19 '20 at 13:06
• Thanks for the reference, I've seen this result in textbooks, so I did not think of leaving one. In Ward and Wells "Twistor geometry and field theory" it is referred to as Cartan's lemma and given as an exercise, 1.3.1 page 28. – GFR May 19 '20 at 13:24
• Yes, it's basic enough that it's surely been reproved many times. If Cartan proved it, then likely that predates Dibag. – Russ Woodroofe May 19 '20 at 13:26
• What you attribute to Dibag is a particular case of the exactness of the Koszul complex studied by Koszul in the late 1940's. See this section in Wikipedia. That special case is probably due to Elie Cartan, but I don't know a reference. On a personal note I followed a course by Jacques Frisch (a student of Douady) on that subject in 1970 or 1971, where he mentioned that complex and facetiously called it the coKoszul complex, which sounds funny in French since "Coco" is what parrots are traditionally called in French comics. – Georges Elencwajg Dec 11 '20 at 16:11

Let us assume $V$ has dimension $n$. Suppose $w$ is a $k$-form which is indecomposable, hence of pure type $v_{i_{1}}\wedge v_{i_{2}}...\wedge v_{i_{k}}$. Then $W_{w}$ is comprised by linear combinations of $v_{i_{j}}$, $j=1,2,...k$. So $W_{w}$ is precisely a $k$ dimensional vector subspace of $V$.

• You seem to confuse indecomposable and decomposable in your second sentence. To tell the truth, I don't understand your answer. – Georges Elencwajg Mar 26 '13 at 9:22
• what does indecomposable, pure type mean? – tom Mar 26 '13 at 9:27
• Plus you have some wrong arguments there." In this case $u\wedge w=0$if and only if $u\wedge w_s=0$ for certain $s$". For example take $w=e_1\wedge e_2$, $u = e_1+e_2$. – tom Mar 26 '13 at 9:31
• I see. Editted. – Bombyx mori Mar 26 '13 at 10:41
• I still don't understand the first part. What does indecomposable mean? And how can you suppose that $w$ if k-form? it is k-vector. – tom Mar 26 '13 at 11:22