2
$\begingroup$

I'm trying to learn differential geometry using Göckeler & Schücker's book and I have some problems with the hodge star. As an example, say we have two orthonormal bases $e^i$ and $\widetilde{e}^j=\Lambda^j_{\ k}e^j$ with $g(e^i,e^j)=g(\widetilde{e}^i,\widetilde{e}^j)=\eta^{ij}$ and $i,j=1,2$ of a 2-dimensional vector space, so that $\Lambda\in SO(r,s)$ where $r+s=2$.

The book defines the hodge star on an orthonormal basis as $*(e^{i_1}\wedge\cdots\wedge e^{i_p})=\epsilon_{i_1\ldots i_n}\eta^{i_1i_1}\cdots\eta^{i_ni_n}e^{i_{p+1}}\wedge\cdots\wedge e^{i_n}$ (no sum). My problem comes when I try to calculate the star of a basis form in two ways:

$*(\widetilde{e}^1)=\widetilde{\eta}^{11}\widetilde{e}^2=\eta^{11}\Lambda^2_{\ k}e^k=\eta^{11}(\Lambda^2_{\ 1}e^1+\Lambda^2_{\ 2}e^2),$

and then using the linearity of the hodge star:

$*(\widetilde{e}^1)=\Lambda^1_{\ k}*(e^k)=\Lambda^1_{\ k}\epsilon_{kl}\eta^{kk}e^l=\Lambda^1_{\ 1}\eta^{11}e^2-\Lambda^1_{\ 2}\eta^{22}e^1.$

These aren't equal, even using $\det(\Lambda)=1$. Can anyone see what I'm doing wrong?

I first thought that using the linearity I also have to use the hodge star on $\Lambda^1_{\ k}$. But since the hodge star takes $0$-forms, or scalars, to $2$-forms(?), this would be a product of a $2$-form and a $1$-form and thus zero.

1 Answers 1

3

Actually, if $\Lambda$ is a boost operator, they are equal. $\Lambda_1^1 = \Lambda_2^2$ and $\Lambda_1^2 = \Lambda_2^1$. You know that $\eta^{11} = - \eta^{22}$ also. These simplifications make it clear that the two results, while appearing different, are actually the same for the kind of linear operator used here.

Edit: I answered for the (1,1) signature case. In the (2,0) or (0,2) cases, the off-diagonal components of the $\Lambda$ matrix are no longer equal, but the diagonal terms of the $\eta$ matrix are, and this accomplishes the same result.

  • 0
    I see now, I got confused applying the transpose in $\widetilde{\eta}=\Lambda^T\eta\Lambda=\eta$ together with upper/lower indices. Thanks a lot both of you!2012-10-31