Let $V$ be a $m$ dimensional vector space over a field $\mathbb{F}$ and $1\leq l<m.$ Then consider the wedge product $\bigwedge^lV$ and $\bigwedge^{m-l}V.$ Fix a basis $\{e_1,\ldots,e_m\}$ of $V$ and cosider $I(l,m)=\{(\alpha_1,\ldots,\alpha_l)\in \mathbb{Z}^l:1\leq \alpha_1 <\alpha_2\cdots<\alpha_l \leq m\}.$ We know that $\{e_{\alpha}: \alpha \in I(l,m)\}$ where $e_{\alpha}=e_{\alpha_1} \wedge \cdots \wedge e_{\alpha_l}$ forms a basis for $\bigwedge^lV$. Now for $\alpha \in I(l,m)$ let $\alpha ^c=(\alpha_{1}^c,\ldots, \alpha_{m-l}^c)$ denote the unique element of $I(m-l,m)$ such that $\{\alpha_1,\ldots, \alpha_l\}\cup\{\alpha_{1}^c,\ldots, \alpha_{m-l}^c\}=\{1,\ldots,m\}.$ So $\{e_{\alpha}^c:\alpha \in I(l,m)\}$ gives a basis for $\bigwedge^{m-l}V.$ Now we define well known Hodge star operator $h: \bigwedge^lV \to \bigwedge^{m-l}V$ by defining on the basis above as following. Define $h(e_{\alpha})=(-1)^{\alpha_1 + \cdots+\alpha_l+l(l+1)/2}e_{\alpha^c}$ for $\alpha \in I(l,m).$
How do I show that $h$ is independent of choice of basis ?
I don't know how to approach for it. Please help me. Many thanks.
It is not independent of the choice of basis (or even of the order of the basis)! For a very simple example, consider $m=2$ and $l=1$, so $h(e_1)=e_2$ and $h(e_2)=-e_1$. If you instead used the basis $e_1'=e_2$ and $e_2'=e_1$, the Hodge star map would instead be $h'(e_1')=e_2'$ and $h'(e_2')=-e_1'$, or $h'(e_1)=-e_2$ and $h'(e_2)=e_1$ (so $h'=-h$). Or if you used the basis $e_1''=e_1+e_2$ and $e_2''=e_2$, you would have $h''(e_1)=h''(e_1''-e_2')=e_2''+e_1''=e_1+2e_2$ and $h''(e_2)=h''(e_2'')=-e_1''=-e_1-e_2$.