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

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 $c$.

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