Vrbljanovwu

2022-09-06

Let $\alpha \in {\mathrm{\Lambda }}^{p}L$, which is p-th power of L, where L is linear space of dimension equal to n. Let us consider the following map ${f}_{\alpha }:L\to {\mathrm{\Lambda }}^{p+1}L$ given with the formula ${f}_{\alpha }\left(\sigma \right)=\alpha \wedge \sigma$, where $\wedge$ is a wedge product.
Prove that if $\alpha ,\beta \in {\mathrm{\Lambda }}^{p}L$ and p${f}_{\alpha }={f}_{\beta }\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\alpha =\beta .$
The part where we assume $\alpha =\beta$ is easy, but what about another implication? Does anything come your minds? This was the first thing I thought about:
${\alpha }_{1}\wedge \cdots \wedge {\alpha }_{p}\wedge \sigma ={\beta }_{1}\wedge \cdots \wedge {\beta }_{p}\wedge \sigma \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left({\alpha }_{1}\wedge \cdots \wedge {\alpha }_{p}-{\beta }_{1}\wedge \cdots \wedge {\beta }_{p}\right)\wedge \sigma =0.$

Bestvinajw

Let ${e}_{i}$ be a basis for L, and assume $\alpha =\sum _{I}{a}_{I}{e}_{I}$ where and ${e}_{\left\{{i}_{1},\dots ,{i}_{p}\right\}}={e}_{{i}_{1}}\wedge \cdots \wedge {e}_{{i}_{p}}$ with ${i}_{1}<\cdots <{i}_{p}$
We have to prove ${f}_{\alpha }=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\alpha =0$
Assume some ${a}_{I}\ne 0$, then since $|I|=p, there's an index $j\notin I$, and hence ${a}_{I}\left({e}_{I}\wedge {e}_{j}\right)$ will be a nonzero term in ${f}_{\alpha }\left({e}_{j}\right)$

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