# Hodge self-duality in Minkowski spacetime

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

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

• I think you mean $(\star \omega)_{23} = \omega_{10}$? – Muphrid Jun 21 '15 at 20:04
• Where, exactly? – Giorgio Comitini Jun 21 '15 at 20:07
• Nevermind. By "null form" you mean the zero 2-form? – Muphrid Jun 22 '15 at 2:28
• Yes, I mean the zero 2-form – Giorgio Comitini Jun 22 '15 at 7:33

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.

• Right, thank you so much. I didn't notice that in Minkowski spacetime you need to change the sign for the bidual operator (from $(-1)^{k(n-k)}$ to $(-1)^{k(n-k)+1}$). – Giorgio Comitini Jun 22 '15 at 16:46
• Hi Muphrid. I have a question regarding the implications of the lack of anti/self-dual forms on Minkowski spacetime. Is there any chance that I could ask for your opinion in private? – Giorgio Comitini Jun 23 '15 at 9:15
• I'm open to suggestion regarding how to go about that. – Muphrid Jun 23 '15 at 14:50
• Well, I was thinking about doing it in a chatroom. It's better than nothing, and there's no need for extra privacy :-) – Giorgio Comitini Jun 23 '15 at 15:04
• Fair enough, just let me know where and when. – Muphrid Jun 23 '15 at 15:12

Similarly to the second part of Muphrid answer, let $* \omega=\pm\omega$. Then applying $*$ again $$*^2\omega=* (\pm\omega) =\omega$$ 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 $$\omega = -\omega \Rightarrow \omega = 0.$$