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## Main Question or Discussion Point

I'm reading section 2.7 of Flanders' book about

Let ##\lambda## be a ##p##-vector in ##\bigwedge^p V## and let ##\sigma^1,\ldots,\sigma^n## be a basis of ##V##. There's a unique ##*\lambda## such that, for all ##\mu\in \bigwedge^{n-p}##,$$

\lambda \wedge \mu = (*\lambda, \mu)\sigma,

$$ where ##\sigma = \sigma^1 \wedge \cdots \wedge \sigma^n##.

Flanders says it's enough to consider ##\lambda=\sigma^1\wedge\cdots\wedge\sigma^p## because of linearity, but what about ##\lambda=\sigma^H## where ##h_1<\cdots <h_p##?

In that case I get $$

*\lambda = (-1)^s(\sigma^\bar{H},\sigma^\bar{H})\sigma^\bar{H},

$$ where ##H\sqcup\bar{H}=\{1,\ldots,n\}## and ##s## is the permutation needed to permute ##H\bar{H}=(h_1,\ldots,h_p,\bar{h}_1,\ldots,\bar{h}_{n-p})## into ##(1,\ldots,n)##.

I suspect this is not correct... or is it?

*differential forms*, but I have some doubts.Let ##\lambda## be a ##p##-vector in ##\bigwedge^p V## and let ##\sigma^1,\ldots,\sigma^n## be a basis of ##V##. There's a unique ##*\lambda## such that, for all ##\mu\in \bigwedge^{n-p}##,$$

\lambda \wedge \mu = (*\lambda, \mu)\sigma,

$$ where ##\sigma = \sigma^1 \wedge \cdots \wedge \sigma^n##.

Flanders says it's enough to consider ##\lambda=\sigma^1\wedge\cdots\wedge\sigma^p## because of linearity, but what about ##\lambda=\sigma^H## where ##h_1<\cdots <h_p##?

In that case I get $$

*\lambda = (-1)^s(\sigma^\bar{H},\sigma^\bar{H})\sigma^\bar{H},

$$ where ##H\sqcup\bar{H}=\{1,\ldots,n\}## and ##s## is the permutation needed to permute ##H\bar{H}=(h_1,\ldots,h_p,\bar{h}_1,\ldots,\bar{h}_{n-p})## into ##(1,\ldots,n)##.

I suspect this is not correct... or is it?