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$(e_1\wedge e_2 + e_3\wedge e_4)$ is well-known to be of rank 2 (can't be decomposed). On the other hand, $\omega \wedge \omega = e_1\wedge e_1 \wedge ... = 0$.

According to the wikipedia article section on the rank of a $k$-vector, if $\wedge^p\omega \not = 0$, but $\wedge^{p+1}\omega = 0$, then the rank of $\omega$ is $p$.

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  • $\begingroup$ Um, how do you reason from $\omega\wedge\omega=0$ to "it must be decomposable"? The Wikipedia section you links to seems to claim no such thing. $\endgroup$ – Henning Makholm Nov 24 '14 at 21:15
  • $\begingroup$ See this question math.stackexchange.com/questions/341540/… $\endgroup$ – Daniel Nov 24 '14 at 22:38
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I believe this theorem that you mentioned only works for 2-vectors. If $w$ is a 2-vector then $w$ has rank $p$, if $\wedge^pw\neq 0$ and $\wedge^{p+1}w= 0$.

Actually, let us prove that $w=e_0\wedge(e_1\wedge e_2+e_3\wedge e_4)$ is not decomposable, if $e_0,e_1,e_2,e_3,e_4$ are linear independent vectors.

First, $v_1,\ldots,v_n$ are linear independent vectors of a vector space $V$ iff $0\neq v_1\wedge\ldots\wedge v_n\in\bigwedge^nV$. Thus, if $v_1,\ldots,v_n$ are linear independent vectors of $V$ and $v\in V$ then $0=v\wedge v_1\wedge\ldots\wedge v_n\in\bigwedge^{n+1}V$ iff $v\in\text{span}\{v_1,\ldots,v_n\}$.

Remark: If $v_1,v_2,v_3$ are linear independent in $V$ then the kernel of $f:V\rightarrow\bigwedge^4V$, $f(v)=v\wedge v_1\wedge v_2\wedge v_3$, is 3 dimensional.

Let us prove that the kernel of $T_w:V\rightarrow\bigwedge^4V$, $T_w(v)=v\wedge w$, is generated by $e_0$. Thus, $w=e_0\wedge(e_1\wedge e_2+e_3\wedge e_4)$ can not be decomposable, otherwise the kernel of $T$ would be 3-dimensional, by the remark above.

Assume $T(v)=0$, for some fixed $v\in V$. Let $w_1=v\wedge e_0\wedge e_1\wedge e_2$ and $w_2=v\wedge e_0\wedge e_3\wedge e_4$. Thus, $0=T(v)=w_1+w_2$ and $w_1=-w_2$.

Define $T_{w_i}:V\rightarrow \bigwedge^{5}V$, $i=1,2$, $T_{w_i}(v')=v' \wedge w_i$. Notice that $T_{w_1}=-T_{w_2}$.

If $w_1\neq 0$ then $w_2\neq 0$ and $\{v,e_0,e_1,e_2\}$ is a linear independent set and $\{v,e_0,e_3,e_4\}$ is linear independet too.

Thus, $\text{span}\{v,e_0,e_1,e_2\}=\ker(T_{w_1})=\ker(T_{w_2})=\text{span}\{v,e_0,e_3,e_4\}$.

Therefore, $\{e_0,e_1,e_2,e_3,e_4\}$ is a linear independent set inside the 4-dimensional vector space $\ker(T_{w_1})$. This is a contradiction. Thus, $w_1=w_2=0$.

So if $T(v)=0$ then $v\wedge e_0\wedge e_1\wedge e_2=v\wedge e_0\wedge e_3\wedge e_4=0$. This implies that $v\in \text{span}\{e_0,e_1,e_2\}\cap \text{span}\{e_0,e_3,e_4\}$. Thus, $\ker(T_w)$ is generated by $e_0$.

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