As we know, for $V$ vectoral space and a orientation $\mathcal{O}$ on $V$ and $e_{1},...,e_{n}$, the hodge star operator $\ast:\wedge V^*\rightarrow\wedge V^*$ is defined for

$\ast(e_{1}\wedge...\wedge e_{n})=\pm 1$(acoording to the orientation preserve/invert for the basis $\{e_{i}\}$)

$\ast(1)=\pm e_{1}\wedge...\wedge e_{n}$(again $+$ if $[\{e_{i}\}]=\mathcal{O}$ and $-$ if $[\{e_{i}\}]=-\mathcal{O}$)

$e_{1}\wedge...\wedge e_{k}=\pm e_{k+1}\wedge...\wedge e_{n}$

Already prove that $\ast\ast=(-1)^{n(n-k)}\cdot$ and $<T,S>=\ast(T\wedge\ast S)=\ast(S\wedge \ast T)$ (where $T=\eta_{1}\wedge...\wedge \eta_{1}$, $S=\omega_{1}\wedge...\wedge \omega_{1}$ and $<T,S>=det(<\omega_i,\eta_{j}>)$)

For any $\xi\in V$ denote $\gamma:\ast:\wedge^{k+1} V^*\rightarrow \wedge^{k} V^*$ the adjoint of left exterior multiplication by $\xi$ $$<\gamma(T),S>=<T,\xi\wedge S>$$ for all $T\in\wedge^{k+1} V^*$ and $S\in\wedge^{k} V^*$.

I need to prove that $\gamma(T)=(-1)^{n-k}\ast(\xi\wedge\ast T)$ but I'm a little confused, how prove that $\gamma$ has this expression from this property?


First notice that there exists unique $\gamma:\wedge^{k+1}V^*\rightarrow \wedge^k V^*$ such that $$<\gamma(T),S>=<T,\xi\wedge S>.$$ This is due to the nondegeneracy of inner product $<,>.$ In fact, assume that there are two such $\gamma_1,\gamma_2.$ Fix arbitrary $T\in\wedge^{k+1}V^*.$ Now you get that for every $S\in\wedge^{k}V^*$ $$<\gamma_1(T),S>=<T,\xi\wedge S>=<\gamma_2(T),S>.$$ And from nondegeneracy of $<,>$ you have that $\gamma_1(T)=\gamma_2(T).$ $T$ was arbitrary, hence $\gamma_1=\gamma_2.$ Since there is unique $\gamma:\wedge^{k+1}V^*\rightarrow \wedge^k V^*$ such that $$<\gamma(T),S>=<T,\xi\wedge S>,$$ all you need to do, is to check, whether $\gamma(T)=(-1)^{n-k}\ast(\xi\wedge\ast T)$ has that property. This should be straightforward. $$<\gamma(T),S>=*(S\wedge *\gamma(T))=*(S\wedge (-1)^{n-k}\ast\ast(\xi\wedge\ast T))=\\=(-1)^{n-k}(-1)^{nk}\ast(S\wedge\xi\wedge\ast T)=(-1)^{n-k}(-1)^{nk}(-1)^{k}\ast(\xi\wedge S\wedge \ast T)=\\=(-1)^{n-k}(-1)^{(n+1)k}\ast((\xi\wedge S)\wedge\ast T)=(-1)^{n(k+1)}<T,\xi\wedge S>.$$ Unfortunately I am left with factor $(-1)^{n(k+1)}.$


I've found that you have little flaws in your previous calculation. In fact from Warner's book "Foundations of differentiable manifolds and Lie groups" we know that

Additonaly there is different formula for $\gamma$

Now the calculations go flawles.

$$<\gamma(T),S>=*(S\wedge *\gamma(T))=*(S\wedge (-1)^{nk}\ast\ast(\xi\wedge\ast T))=\\=(-1)^{nk}(-1)^{k(n-k)}\ast(S\wedge\xi\wedge\ast T)=(-1)^{nk}(-1)^{k(n-k)}(-1)^{k}\ast(\xi\wedge S\wedge \ast T)=\\=(-1)^{2nk-k^2+k}\ast((\xi\wedge S)\wedge\ast T)=(-1)^{-k(k-1)}<T,\xi\wedge S>=<T,\xi\wedge S>.$$

  • $\begingroup$ I made a mistake, the formula for your answer is the correct, thanks $\endgroup$ – Donyarley May 30 '15 at 19:18

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