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}(\sigma )=\alpha \wedge \sigma $, where $\wedge $ is a wedge product.

Prove that if $\alpha ,\beta \in {\mathrm{\Lambda}}^{p}L$ and p${f}_{\alpha}={f}_{\beta}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\alpha =\beta .$

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:

${\alpha}_{1}\wedge \cdots \wedge {\alpha}_{p}\wedge \sigma ={\beta}_{1}\wedge \cdots \wedge {\beta}_{p}\wedge \sigma \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({\alpha}_{1}\wedge \cdots \wedge {\alpha}_{p}-{\beta}_{1}\wedge \cdots \wedge {\beta}_{p})\wedge \sigma =0.$

I am stuck here, but maybe I don't see something very obvious about this.

Prove that if $\alpha ,\beta \in {\mathrm{\Lambda}}^{p}L$ and p

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:

${\alpha}_{1}\wedge \cdots \wedge {\alpha}_{p}\wedge \sigma ={\beta}_{1}\wedge \cdots \wedge {\beta}_{p}\wedge \sigma \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({\alpha}_{1}\wedge \cdots \wedge {\alpha}_{p}-{\beta}_{1}\wedge \cdots \wedge {\beta}_{p})\wedge \sigma =0.$

I am stuck here, but maybe I don't see something very obvious about this.

Bestvinajw

Beginner2022-09-07Added 15 answers

Let ${e}_{i}$ be a basis for L, and assume $\alpha =\sum _{I}{a}_{I}{e}_{I}$ where $I\subseteq \{1,2,\dots ,n\},\text{}|I|=p$ and ${e}_{\{{i}_{1},\dots ,{i}_{p}\}}={e}_{{i}_{1}}\wedge \cdots \wedge {e}_{{i}_{p}}$ with ${i}_{1}<\cdots <{i}_{p}$

We have to prove ${f}_{\alpha}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\alpha =0$

Assume some ${a}_{I}\ne 0$, then since $|I|=p<n$, there's an index $j\notin I$, and hence ${a}_{I}({e}_{I}\wedge {e}_{j})$ will be a nonzero term in ${f}_{\alpha}({e}_{j})$

We have to prove ${f}_{\alpha}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\alpha =0$

Assume some ${a}_{I}\ne 0$, then since $|I|=p<n$, there's an index $j\notin I$, and hence ${a}_{I}({e}_{I}\wedge {e}_{j})$ will be a nonzero term in ${f}_{\alpha}({e}_{j})$

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