# The Hodge star operator

• A
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?

fresh_42
Mentor
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".

kiuhnm
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.