# Hodge dual on orthonormal basis: two inconsistent answers

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.

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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.

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@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 $$\eta^{11}\Lambda^1_1\Lambda^2_1 + \eta^{22}\Lambda^1_2\Lambda^2_2 = 0$$ – Willie Wong Oct 30 '12 at 16:25
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 $(\Lambda^2_{\ 1})^2$? – jorgen Oct 30 '12 at 17:28
@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. – Muphrid Oct 30 '12 at 19:28
@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. – Willie Wong Oct 31 '12 at 8:10
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! – jorgen Oct 31 '12 at 11:13

First of all, I don't understand why you keep using the notation $\eta^{ij}$. If your bases are orthonormal, then $\eta^{ij} = \delta^{ij}$, i.e. $1$ when $i=j$ and $0$ otherwise. That's what I will assume.

Second, I'm having a hard time understanding this statement: "...so that $\Lambda \in SO(r,s)$ where $r+s=2$". What is $SO(r, s)$? This kind of notation doesn't seem to be standard, I can only guess what it means. For the problem to make sense, I would have to assume that $\Lambda \in SO(2)$, which means that $\det \Lambda = 1$ and $\Lambda$ is orthogonal, i.e. $\Lambda_i^j \Lambda_j^k = \delta_i^k$.

If we assume this problem statement, then you don't really have a problem, because for a 2x2 matrix $\Lambda$ from $SO(2)$ you automatically have $\Lambda_1^1 = \Lambda_2^2$ and $\Lambda_1^2 = - \Lambda_2^1$.

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I forgot to mention that $g$ is a pseudometric, which means that $\eta^{ij}=\pm\delta^{ij}$. In $SO(r,s)$, $r$ and $s$ are the numbers of positive and negative entries respectively along the diagonal of $\eta$. One example would be the Lorentz group, which is O(3,1). – jorgen Oct 30 '12 at 16:14
$SO(r,s)$ is the special orthogonal group acting on the vector space $V$ equipped with a non-degenerate symmetric bilinear form of signature $(r,s)$. The notation is standard. – Willie Wong Oct 30 '12 at 16:15