# Star operator, exterior product

Let $\ast$ be the star operator $$\ast:\Lambda^p(V)\to \Lambda^{d-p}(V)$$ so that we have $$\ast(e_{i_1}\wedge...\wedge e_{i_p})=e_{j_1}...\wedge e_{j_{d-p}}$$ where $$e_{i_1},...,e_{i_p},e_{j_1},...,e_{j_{d-p}}$$ is a positive basis of $V$. For $v,w\in\Lambda^p(V)$ we have $$\langle v,w\rangle=\ast(w\wedge \ast v)=\ast(v\wedge \ast w).$$

Why then $$\ast(1)=\sqrt{\det(g_{ij})}dx^1\wedge ...\wedge dx^d$$ and also how do we use the above property to yield $$\int_M \langle\alpha,\beta\rangle\ast(1)=\int_M \alpha\wedge \ast\beta$$ for $\alpha,\beta\in \Omega^p(M)$ with compact support?

• Shouldn't $e_{i_1},...,e_{i_p},e_{j_1},...,e_{j_{d-p}}$ be a positive basis for $V^*$? – md2perpe Aug 17 '18 at 19:40
• Jurgen Jost here on the page 103 in line 9 writes for $V$. – user122424 Aug 18 '18 at 8:32
• Okay. Anyway, note that $e_1, \ldots, e_n$ is an ON-basis while $dx^1, \ldots, dx^n$ is not necessarily ON. – md2perpe Aug 18 '18 at 9:14
• OK. And can you please answer my entire question ? – user122424 Aug 18 '18 at 12:41
• Don't be distracted by the integral or the requirement for compact support. Show that $\langle\alpha,\beta\rangle\,{\star}(1)=\alpha\wedge{\star}\beta$, which almost trivially gives you the final result from what you have stated. – qman Sep 16 '18 at 23:51