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The C-field in 11-dimensional supergravity is an elusive object that is not the simple higher $\mathrm{U}(1)$-gauge field one would naively make this out to be. For an overview of possible models for this object, see, for instance, section 3 of "The M-theory 3-form and $E_8$ gauge theory" by Diaconescu, Freed and Moore (henceforth DFM).

However, it is always an object that carries with it a notion of "gauge transformation", and for naive higher $\mathrm{U}(1)$-gauge fields with a transformation law $$ C\mapsto C + \mathrm{d}\Lambda$$ for $C$ a $p$-form and $\Lambda$ a $(p-1)$-form, one can easily see that there are "gauge transformations" that actually do not change the gauge field at all - those with $\mathrm{d}\Lambda = 0$. However, objects charged under this $\mathrm{U}(1)$ would transform by $\mathrm{e}^{\mathrm{i}\Lambda}$, meaning this transformation is non-trivial on the other fields. This means that there is a global symmetry group associated to this gauge transformation law, namely given by all closed $(p-1)$-forms $C^{p-1}_\mathrm{dR}(M)$ on our spacetime manifold $M$ that does not change the gauge field (hence does not need to be quotiented out in order to "fix the gauge") and therefore remains even after quantization. If we consider that the objects charged under such a symmetry are $(p-1)$-dimensional objects, it is clear that the proper symmetry operator is $\mathrm{e}^{\mathrm{i}\int_\Sigma \Lambda}$ where $\Sigma$ is the charge object, and so the final global symmetry group is in fact $H^{p-1}(C,\mathrm{U}(1))$ since exact forms just act as the identity.1

A similar reasoning seems to be carried out in "M-Theory Dynamics On A Manifold Of $G_2$ Holonomy" by Atiyah and Witten to obtain the total and unbroken global symmetry groups associated to the C-field. However, as I mentioned in the first paragraph of this question, the C-field is not a simple higher gauge field, and its exact "gauge group" is a subtle question.

For instance, in one of DFM's models, the proper notion of a gauge transformation is that "the C-field" is a pair of objects $(A,c)$ where $A$ is an ordinary $E_8$-gauge field and $c$ a 3-form with integral periods, and the gauge transformations are given by $$ A\mapsto A+\alpha \quad c\mapsto c - \mathrm{CS}_3(A,A+\alpha) + \omega,$$ where $\mathrm{CS}_3(A,A+\alpha)$ is the relative Chern-Simons invariant 3-form between two connections given by integrating $\mathrm{tr}(F^2)$ along the straight line between $A$ and $A+\alpha$ in connection space (which is affine, so this is possible). The $\alpha$ is simply a 1-form on $\mathrm{ad}(P)$ and $\omega$ is a 3-form with integral periods.

There seems to be no evident notion of how such a transformation acts on objects charged under the C-field, nor do there seem to be global transformations in this case or indeed a straightforward relation of this transformation to the $\Lambda$ considered earlier. Section 7 of DFM defines the proper notion of charge for the $E_8$-model of the C-field, but does not consider how objects charged thusly transform as far as I can see.

What is the proper action of such a gauge transformation in the sense of DFM on charged objects? How can this be reconciled with the analysis of global symmetries of the C-field as done by Atiyah and Witten? Is there a notion of the global symmetries associated to the gauge symmetry of the C-field in the stricter formulation where it is no longer a naive higher $\mathrm{U}(1)$-gauge field?


1The clear analogy in electromagnetism is the global $\mathrm{U}(1)$ symmetry that persists even after quantization, corresponding to $H^0(\mathbb{R}^4,\mathrm{U}(1)) = \mathrm{U}(1)$.

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  • $\begingroup$ Is it obvious that there is matter "charged" under the $C$ field? $\psi\to\exp\!\left(\text{i}\Lambda\right) \psi$ does not seem to be a very reasonable transformation if $\Lambda$ is a $p$-form. So I'm not sure I'd interpret the $\text{d}\Lambda$ as some $g^{-1}\text{d}g$ with $g$ an element of some group. $\endgroup$ – Toffomat Aug 18 '17 at 10:45
  • $\begingroup$ @Toffomat I mention in the second paragraph that the proper operator is not $\exp(\mathrm{i}\Lambda)$ but $\exp(\mathrm{i}\int \Lambda)$, where the integral is over the extent of the charged object. People certainly say things like "the 2-brane is charged under the $C$-field", so some notion of being charged must exist, and Atiyah and Witten do claim that the global symmetry group is given by $H^2$, which I can only explain by observing that $\exp(\mathrm{i}\int\Lambda)$ is trivial for exact forms, since without that we would just have the group of all closed forms as symmetry. $\endgroup$ – ACuriousMind Aug 18 '17 at 11:12
  • $\begingroup$ Usually, form field couple to branes via some $\int_W C$ or more complicated terms (eg. the Chern-Simons coupling, see Eq. (3) in arxiv.org/abs/hep-th/9910053). The gauge transformations $C\to C+\text{d}\Lambda$ is still not derived from some group element, and the brane does not transform the way a field does in QFT. In that sense, I wouldn't think there is a gauge roupassociated with $C$ in the usual sense. $\endgroup$ – Toffomat Aug 19 '17 at 17:59
  • $\begingroup$ @Toffomat I'm asking for a global symmetry group, not a gauge group. (For example, in the case of an ordinary SU(N) gauge theory, the answer to "what is the global symmetry group?" is the center $\mathbb{Z}_n$, not the gauge group SU(N)). I'm willing to entertain the thought that the $\exp(\int C)$ operator is not the correct way to think about this, but any other answer to the global symmetry group needs to explain why Atiyah/Witten claim the global symmetry group is $H^2$. I also think that the $\exp(\int C)$ is compatible with arxiv.org/abs/1412.5148v2, but I'm not fully certain. $\endgroup$ – ACuriousMind Aug 19 '17 at 18:13
  • $\begingroup$ I guess Atiyah/Wittens Eqns 2.10 - 2-12 seem to indicate that $e^\int \Lambda$ or somthing like it could be a reasonable guess for a symmetry of the low-energy theory, i.e. after integrating out (part of) the extra dimensions. (But I don't see how such an operator could be a symmetry of the full theory, since its not even local.) Is that what you have in mind? $\endgroup$ – Toffomat Aug 21 '17 at 16:41

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