The Hodge star operator

  • #1
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1

Main Question or Discussion Point

I'm reading section 2.7 of Flanders' book about 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?
 

Answers and Replies

  • #2
12,653
9,181
I'm reading section 2.7 of Flanders' book about 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?
I don't understand your question. If you have a permutation of ##\sigma_i## you get another ##*\lambda##, with an appropriate sign. So if we have ##*\sigma##, then we get all ##*\lambda## by linear extension and sign corrections. So the only minor neglect was to say by "because of linearity" instead of "because of alternating linearity".
 
  • #3
66
1
I don't understand your question. If you have a permutation of ##\sigma_i## you get another ##*\lambda##, with an appropriate sign. So if we have ##*\sigma##, then we get all ##*\lambda## by linear extension and sign corrections. So the only minor neglect was to say by "because of linearity" instead of "because of alternating linearity".
I wasn't sure whether my formula for ##*\lambda## was correct. I had doubts because I got a wrong sign in example 1 on page 16. That example says that if the base of ##V## is ##(dx^1,dx^2,dx^3,dt)## where ##(dx^i,dx^i)=1## and ##(dt,dt)=-1##, then $$
*(dx^i dt) = dx^j dx^k,
$$ where "##(i,j,k)## is cyclic order". I didn't take into account the cyclicity of the order and so got a wrong sign. In particular, $$
*(dx^2 dt) = -(dx^1 dx^3) = dx^3 dx^1.
$$ I missed the very last step.
 

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