nLab relation between 5d Maxwell theory and self-dual 3-forms in 6d -- section

Relation between 5d Maxwell theory and self-dual 3-forms in 6d

Consider 5d- and 6d-dimensional Minkowski spacetime equipped with global orthonormal coordinate charts $\{x^\kappa\}$, $\{x^\alpha\}$, respectively, adapated to an isometric embedding

$$\array{ & \mathbb{R}^{4,1} &\overset{\;\;\;\iota_5\;\;\;}{\hookrightarrow}& \mathbb{R}^{5,1} \\ \kappa = & 0, 1, 2, 3, 4\phantom{,} \\ \alpha = & 0, 1, 2, 3, 4, && 5 }$$

With this notation, the pullback of differential forms along this embedding is notationally implicit.

Now any differential 3-form $H_3$ on $\mathbb{R}^{5,1}$ decomposes as

(1)$$H_3 \;=\; \widehat{F} \wedge d x^{5} + \widehat{H}$$

for unique differential forms of the form

$$\widehat F \;=\; \tfrac{1}{2}\hat F_{\kappa_1 \kappa_2}(x^\kappa, x^5) d x^{\kappa_1} \wedge d x^{\kappa_2}$$

and

$$\widehat{H} \;=\; \tfrac{1}{3!} \widehat{H}_{\kappa_1 \kappa_2 \kappa_3}(x^\kappa, x^5) d x^{\kappa_1} \wedge d x^{\kappa_2} \wedge d x^{\kappa_3} \,.$$

In the case that $H_3$ has vanishing Lie derivative along the $x^5$-direction,

(2)$$\mathcal{L}_5 H_3 \;=\; 0$$

then also these components forms do not depend on $x^5$ are actualls pullbacks of differential forms on $\mathbb{R}^{4,1}$.

In terms of this decomposition, the 6d Hodge dual of $H_3$ is equivalently given by the 5d Hodge duals of these components as (best seen by the relation to Hodge pairing according to this Prop.)

(3)$$\star_6 H_3 \;=\; \big( \star_5 \widehat{H}\big) \wedge d x^{5} - \star_5 \widehat{F}$$

Since the Hodge star operator squares to unity in the special case that it is applied to differential 3-forms on 6d Minkowski spacetime (by this Prop.)

$$\star_6 \star_6 H_3 \;=\; + H_3$$

we may ask for $H_3$ to he Hodge self-dual. By (3) this means equivalently that its 5d components are 5d Hodge duals of each other:

$$\big( H_3 \;=\; \star_{6} H_3 \big) \;\;\; \overset{ H_3 = \widehat{F} \wedge d x^5 + \widehat{H} }{ \Leftrightarrow } \;\;\; \big( \widehat{H} = \star_5 \widehat{F} \big) \,.$$

It follows that if there is no $x^5$-dependence (2) then the condition that $H_3$ be a closed and self-dual 3-form is equivalent to its 5d components $\widehat{F}$ ($\widehat{H}$) being the (dual) field strength/Faraday tensor satisfying the Maxwell equations of D=5 Maxwell theory (without source current):

$$\underset{ \color{blue} { {\phantom{A}} \atop {\text{D=6 self-dual 3-form theory}} } }{ \left. \array{ & d H_3 = 0 \\ & \star_6 H_3 = H_3 } \right\} } \;\;\; \overset{ {\mathcal{L}_{5} H_3 = 0} \atop {H_3 = \widehat{F}\wedge d x^5 + \cdots} }{ \Leftrightarrow } \;\;\; \underset{ \color{blue} { {\phantom{A}} \atop \text{D=5 Maxwell theory} } }{ \left\{ \array{ & d \widehat{F} = 0 \\ & d \star_5 \widehat{F} = 0 } \right. }$$

This may be summarized as saying that the massless part of the Kaluza-Klein reduction of self-dual 3-form theory from 6d to 5d is D=5 Maxwell theory.

Essentially this relation underlies the formulation of the M5-brane via the Perry-Schwarz Lagrangian.