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

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 κ}\{x^\kappa\}, {x α}\{x^\alpha\}, respectively, adapated to an isometric embedding

4,1 ι 5 5,1 κ= 0,1,2,3,4, α= 0,1,2,3,4, 5 \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 3H_3 on 5,1\mathbb{R}^{5,1} decomposes as

(1)H 3=F^dx 5+H^ H_3 \;=\; \widehat{F} \wedge d x^{5} + \widehat{H}

for unique differential forms of the form

F^=12F^ κ 1κ 2(x κ,x 5)dx κ 1dx κ 2 \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

H^=13!H^ κ 1κ 2κ 3(x κ,x 5)dx κ 1dx κ 2dx κ 3. \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 3H_3 has vanishing Lie derivative along the x 5x^5-direction,

(2) 5H 3=0 \mathcal{L}_5 H_3 \;=\; 0

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

In terms of this decomposition, the 6d Hodge dual of H 3H_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) 6H 3=( 5H^)dx 5 5F^ \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.)

6 6H 3=+H 3 \star_6 \star_6 H_3 \;=\; + H_3

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

(H 3= 6H 3)H 3=F^dx 5+H^(H^= 5F^). \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 5x^5-dependence (2) then the condition that H 3H_3 be a closed and self-dual 3-form is equivalent to its 5d components F^\widehat{F} (H^\widehat{H}) being the (dual) field strength/Faraday tensor satisfying the Maxwell equations of D=5 Maxwell theory (without source current):

dH 3=0 6H 3=H 3}AD=6 self-dual 3-form theory 5H 3=0H 3=F^dx 5+{ dF^=0 d 5F^=0AD=5 Maxwell theory \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.

Last revised on May 4, 2020 at 11:54:52. See the history of this page for a list of all contributions to it.