# The Hodge star operator

• A

## 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?

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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.