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.


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.

  • 2
    $\begingroup$ @jorgen: To expand on Muphrid's answer: did you use the fact that $\Lambda$ preserves the bilinear form? This should give you a system of equations looking like: $$\eta^{11} \left[ (\Lambda^1_1)^2 - 1\right] + \eta^{22} (\Lambda^1_2)^2 = \eta^{11}(\Lambda^2_1)^2 + \eta^{22} \left[ (\Lambda^2_2)^2-1\right] = 0 $$ and $$ \eta^{11}\Lambda^1_1\Lambda^2_1 + \eta^{22}\Lambda^1_2\Lambda^2_2 = 0$$ $\endgroup$ Oct 30 '12 at 16:25
  • $\begingroup$ Thanks both of you for answers! Unfortunately I'm pretty slow here. Going through the example of $r=s=1$ I can see that $\eta^{11}=-\eta^{22}=1$ and $\Lambda^1_{\ 1}=\Lambda^2_{\ 2}$, but not that $\Lambda^1_{\ 2}=\Lambda^2_{\ 1}$.. @Willie: shouldn't the first line say $\Lambda^1_{\ 2}\Lambda^2_{\ 1}$ instead of both $(\Lambda^1_{\ 2})^2$ and $(\Lambda^2_{\ 1})^2$? $\endgroup$
    – jorgen
    Oct 30 '12 at 17:28
  • $\begingroup$ @jorgen: This is something you know from working with rotations and boosts. Look up Lorentz transformation matrices; you'll see the off-diagonal components are symmetric. $\endgroup$
    – Muphrid
    Oct 30 '12 at 19:28
  • $\begingroup$ @jorgen: no. What I wrote is correct. The first line says that the diagonal part of the metric $\eta$ is preserved. The second line says that the off-diagonal part of the metric vanishes (and is preserved). In other words, the first line says that $\eta^{11} = \tilde{\eta}^{11} = \langle \tilde{e}^1,\tilde{e}^1\rangle$ and we then expand the inner product in terms of the original non-tilde basis. $\endgroup$ Oct 31 '12 at 8:10
  • $\begingroup$ 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! $\endgroup$
    – jorgen
    Oct 31 '12 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.