Hodge self-duality in Minkowski spacetime

Question

I was computing the dual map for $k$-forms in Minkowski spacetime, and I found that any $2$-form is either self-dual or anti-self-dual if and only if it is the null form. Does this result make any sense to you? Here are the calculations.

Let the metric on Minkowski spacetime be

$$ (\eta_{\mu\nu})= \begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix} $$

and let $\omega$ be the $2$-form

$$ \omega=\omega_{\mu\nu}\ dx^{\mu}\wedge dx^{\nu}=(\omega_{\mu\nu}-\omega_{\nu\mu})\ dx^{\mu}\otimes dx^{\nu} $$

The $\eta$-raised components of $\omega$ are

$$ (\omega)^{\mu\nu}=\eta^{\mu\lambda}\eta^{\nu\tau}\,(\omega_{\lambda\tau}-\omega_{\tau\lambda}) $$

We have

$$ \eta^{\mu\lambda}\eta^{\nu\tau}\omega_{\lambda\tau}=(\eta\omega\eta)^{\mu\nu} $$

Thus, if

$$ \omega_{\lambda\tau}= \begin{pmatrix} 0&\omega_{01}&\omega_{02}&\omega_{03}\\ \omega_{10}&0&\omega_{12}&\omega_{13}\\ \omega_{20}&\omega_{21}&0&\omega_{23}\\ \omega_{30}&\omega_{31}&\omega_{32}&0 \end{pmatrix} $$

$\eta\omega\eta$ has components

$$ (\eta\omega\eta)^{\mu\nu}= \begin{pmatrix} 0&-\omega_{01}&-\omega_{02}&-\omega_{03}\\ -\omega_{10}&0&\omega_{12}&\omega_{13}\\ -\omega_{20}&\omega_{21}&0&\omega_{23}\\ -\omega_{30}&\omega_{31}&\omega_{32}&0 \end{pmatrix} $$

The $\eta$-raised components of $\omega$ are thus

$$ (\omega)^{\mu\nu}= \begin{pmatrix} 0&\omega_{10}-\omega_{01}&\omega_{20}-\omega_{02}&\omega_{30}-\omega_{03}\\ \omega_{10}-\omega_{10}&0&\omega_{12}-\omega_{21}&\omega_{13}-\omega_{31}\\ \omega_{20}-\omega_{20}&\omega_{21}-\omega_{12}&0&\omega_{23}-\omega_{32}\\ \omega_{30}-\omega_{30}&\omega_{31}-\omega_{13}&\omega_{32}-\omega_{23}&0 \end{pmatrix} $$

Let $*\,\omega$ denote the Hodge-dual of $\omega$, such that (as $\sqrt{|\det(\eta)|}=1$)

$$ (*\,\omega)_{\tau\lambda}=\frac{1}{2}\ (\omega)^{\mu\nu}\,\epsilon_{\mu\nu\tau\lambda} $$

Then, for example

$$ (*\,\omega)_{01}=\frac{1}{2}\ (\omega)^{\mu\nu}\,\epsilon_{\mu\nu 01}=\frac{1}{2}\,(\omega^{23}-\omega^{32})=\omega_{23}-\omega_{32} $$ and

$$ (*\,\omega)_{23}=\frac{1}{2}\ (\omega)^{\mu\nu}\,\epsilon_{\mu\nu 23}=\frac{1}{2}\,(\omega^{01}-\omega^{10})=\omega_{10}-\omega_{01} $$ Thus

$$ \begin{cases} (*\,\omega)_{23}=(\omega)_{23}\\ \\ (*\,\omega)_{01}=(\omega)_{01} \end{cases} $$ implies

$$ \begin{cases} \omega_{01}=\omega_{10}\\ \\ \omega_{23}=\omega_{32} \end{cases} $$ and the same goes for the other indices. The result is the same provided that

$$ \begin{cases} (*\,\omega)_{23}=-(\omega)_{23}\\ \\ (*\,\omega)_{01}=-(\omega)_{01} \end{cases} $$

Answer 1

Yes, this is so. A simpler way to see this is to use clifford algebra. Suppose $\omega$ is (anti-)self-dual. Then the clifford product of $\omega$ and the unit pseudoscalar $\epsilon$ is

$$\epsilon \omega = \pm \omega$$

$\epsilon$ is invertible under the clifford product, yielding the equation

$$\omega= \pm \epsilon^{-1} \omega \implies \epsilon^{-1} \omega = \pm \omega$$

But $\epsilon^{-1} = -\epsilon$ for Minkowski spacetime, yielding

$$\epsilon^{-1} \omega = \pm \omega \implies -\epsilon \omega = \pm \omega \implies \epsilon \omega = \mp \omega$$

which is a contradiction from what we initially supposed.

In 4d Euclidean space, there is no contradiction, as $\epsilon^{-1} = \epsilon$ there.

Edit: in notation only using Hodge star, this is basically the same result as saying that, in Minkowski spacetime,

$$\star \star \omega = -\omega$$

So if $\star \omega = \pm \omega$, then $\star (\pm \omega) = + \omega$ on the one hand, but also $-\omega$ on the other.

Answer 2

Similarly to the second part of Muphrid answer, let $* \omega=\pm\omega$. Then applying $*$ again \begin{equation} *^2\omega=* (\pm\omega) =\omega \end{equation} on the other hand on a Lorentzian $n$-manifold $*^2=-(-1)^{p(n-p)}$ where $p$ is the degree of the form on which $*$ is acting. In this case $n=4,p=2$ hence we get $*^2\omega = - \omega$, therefore \begin{equation} \omega = -\omega \Rightarrow \omega = 0. \end{equation}