# Examples for p-form gauge fields

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I don't completely understand the notion "p-form". Can you give me examples of 1-form, 2-form and 3-form gauge fields? What kind of p-form is e.g. the Higgs field, the electromagnetic four-potential, etc.?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
Higgs, photons, W-Bosons are charge carriers, not gauge fields. Their description in QFT is achieved by means of gauge theories on fibre bundles, which in turn make use of tensor calculus on manifold, where $p$-form formalism takes place.

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user Gennaro Tedesco
Sorry, I meant not the gauge bosons but the corresponding gauge fields such as the electromagnetic four-potential. Why do we have a 3-form for Higgs and not a 2-form?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF

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One way to understand it is that the "reason" why ordinary gauge fields are given by 1-forms is that the term in the action of a charged point particle that gives, under variation, the Lorentz force of a background electromagnetic field on the particle, is simply the integral of that 1-form over the worldline of the particle.

Now as a particle = 0-brane is replaced by a p-brane, then the direct analog of the gauge coupling term is the integral of a (p+1)-form field over the worldvolume of the p-brane.

A famous example of a 2-form gauge field is the Kalb-Ramond B-field that the fundamental string (a 1-brane) is charged under. Then there is the 3-form C-field.under which the M2-brane is charged.

For more see also at Higher Prequantum Geometry I: The need for prequantization.

answered Oct 7, 2015 by (6,095 points)
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Not sure whether I understood the intention of your question correctly.

In electrodynamics you usually use a 1-form $A=A_{\mu}dx^{\mu}$ to write down an action \begin{equation*} S = \int F \wedge \star F \end{equation*} with the 2-form field strength $F=dA$, which the gives the known (vacuum) Maxwell equations.

Scalar fields $\phi$ (such as the Higgs) are described by 0-forms or alternatively by their dual $(d-2)$-forms in $d$ dimensions.

In string theory one finds cousins of the electromagnetic field $A$, but they turn out to be $p$-forms, e.g. in type IIB we have the R-R-fields $C_0, C_2,C_4$,...

psm

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
answered Oct 2, 2015 by (55 points)
Thanks a lot! However, unfortunately I am not used to your notations. Your action is equivalent to $S=\int F_{\mu \nu} F^{\mu \nu}$ or not? What exactly means wedge product followed by Hodge operator? And $dA$ means $d_{\mu}A_{\nu}-d_{\nu}A_{\mu}$?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
And what do you mean with "dual field", what is that physically?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
Yes, up to some factor $F \wedge \star F \sim d^dx F_{\mu\nu}F^{\mu\nu}$. In $d$ dimensions, $\star F$ is by definition a $(d-2)$-form if $F$ is a 2-form. Then, the wedge product can be applied as in the case of wedging a p- and a q-form. And $dA= \partial_{\mu}A_{\nu}dx^{\mu}\wedge dx^{\nu}$. (See definition of exterior derivative.)

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
Concerning the dual field: Take a p-form potential $C_p$. Its field strength tensor is then $F=dC_p$, a $(p+1)$-form. Thus, in $d$ dimensions $\tilde{F} \equiv \star dC_p$ is a $(d-p-1)$-form, and locally there exists a $(d-p-2)$-form $\tilde{C}_{d-p-2}$ s.t. $\tilde{F}=d \tilde{C}$. We call $\tilde{C}$ the dual field to $C$. Physically they should be equivalent...

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
Thanks again! I am just wondering where these dual fields come from, because up to know I never met them in QFT. When the dual field is a (d-2)-form if the field is a 2-form, how can they then be physically equivalent?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
You're welcome! Careful, the dual field of a 2-form is a $(d-4)$-form. One has to figure this out by dualising the field strength tensor. (See my last comment.) To see, why these should be physically equivalent: i) One can rewrite the action $S[C_p]$ into $S[\tilde{C}_{d-p-2}]$ and they're structurally identical. In particular, the Bianchi identity & the eq. of motion for $C$ become the e.o.m. & Bianchi id. for $\tilde{C}$. ii) The on-shell degrees of freedom for a $p$-form and a $(d-p-2)$-form gauge field are identical.

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
Notice however, that I'm not sure whether the physical equivalence holds true in all field theories involving $p$-form gauge fields. But it should be true at least for abelian gauge fields with Lagrangians involving kinetic terms as above. Maybe someone else can provide more details...

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm

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