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first of all, I need to say that I'm a mathematician, so this question may sound a little stupid. Keeping this is mind, please, try to use coordinate-free notations.

Along this question, I will use mainly the notation from the paper http://arxiv.org/abs/1209.2530 (which is actually the source of my doubt).

Let $X$ be a 4 dimensional manifold without boundary (usually a globally hyperbolic Lorentzian manifold), $j_e \in \Omega^3_c (X, \mathfrak{g})$, $G = U(1)$, $\mathcal{U} = \{U_i\}_i$ an open covering of $X$ (if necessary $\mathcal{U}$ may have all finite intersections contractible or even diffeomorphic to the euclidean space).

In $ U(1)$-gauge theory without magnetic charges, the Lagrangian density is given by $$L[A] = \frac{-1}{2} F_A \wedge \star F_A + A \wedge j_e$$ and $d\star F_A = j_e$ for every $A = \{ A_i\}_{U_i} \in \prod \Omega^1 (U_i, \mathfrak{g})$ satisfying the usual gluing condition on the open covering $\mathcal{U}$ of $X$.

I have two questions, where the first of is the the main question.

1)Since $A$ is not globally defined as a $1$-form, what's $\int_X A \wedge j_e$ in this case?

2) The Lagrangian $L$ is invariant under the change $A \mapsto A + \alpha$ for $\alpha$ some closed $1$-form. More precisely, $$L[A + \alpha] - L[A] = - d (\alpha \wedge \star F_A)$$. So this means that the action $S [A] = \int_X L[A]$ is invariant under transformations $$A \mapsto A + \alpha$$. Therefore these transformations should be considered instead of only the gauge transformations when defining the space of field configurations as a quotient.

In the above paper, such transformations are called gauge transformations instead of the usual ones $A \mapsto A + gdg^{-1}$ (see page $5$). So my question is why people usually does not consider the classical space of field configurations as $$\Omega^1 (X)/\Omega^1_{closed} (X) $$ ?

I know that in the quantum level such transformations are not consistent with the space of field configurations, since Wilson loops would attain different values on different elements of the same class. However, classically, as I understand, the entire theory is only determined by the curvature $F_A$.

Thanks in advance.

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  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1209.2530 $\endgroup$ – Qmechanic Feb 16 '16 at 10:15
  • $\begingroup$ @Qmechanic Why linking directly to the pdf file is not recommended? $\endgroup$ – user40276 Feb 17 '16 at 5:04
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    $\begingroup$ E.g. if people have slow connection / equipment. $\endgroup$ – Qmechanic Feb 17 '16 at 6:46
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  1. As a manifold, $X$ admits partitions of unity. Integrating a gauge-invariant functional $G[A]$ over all of $X$ while $A$ is actually only defined by local fields $A_i$ on a cover $\{U_i\}$ of $X$ is simply defined by choosing a partition of unity $\{\chi_i\}$ on $\{U_i\}$ and then writing $$ \int_X G[A] = \sum_i \int_{U_i}\chi_i G[A_i]$$ where gauge invariance of the functional ensures $G[A_i] = G[A_j]$ on overlaps (due to the form of the gluing condition for the $A_i$) so that this is well-defined and independent of the chosen cover or partition.

    The object $A\wedge j_e$ is not gauge invariant, but nearly so: Under $A\mapsto A+\mathrm{d}\phi$, we get $A\wedge j_e\mapsto A\wedge j_e + \mathrm{d}(\phi\wedge j_e)$ since $\mathrm{d}j_e=0$.

  2. If the transformation $A\mapsto A+\alpha$ has $\alpha$ closed but not of the form $g^{-1}\mathrm{d}g$, then it is not a gauge transformation in the physical sense, as it does not arise from the actual gauge group. In particular, this doesn't work for non-Abelian groups. It is an "accident" of choosing $\mathrm{U}(1)$ as the gauge group that such transformations also leave the action invariant - and one that only shows on topologically non-trivial manifolds, since if $\alpha = \mathrm{d}\phi$ is exact, then it is of the form $g^{-1}\mathrm{d}g$ for $g= \exp(\phi)$. Here's why a generic closed form doesn't work for non-Abelian theories:

    For the local forms $A_i$, on the overlaps $U_i\cap U_j$ with transition function $f_{ij}$ we have $$ A_i = \mathrm{ad}_{f_{ij}}(A_j) + f_{ji}^\ast\theta$$ where $\theta$ is the Cartan-Maurer form on the gauge group, and is the more familiar $f_{ji}^\ast\theta = f_{ji}^{-1}\mathrm{d}f_{ji}$ for matrix groups. Now, if you do $A\mapsto A+\alpha$, you get $$ A_i + \alpha = \mathrm{ad}_{f_{ij}}(A_j + \alpha) + f_{ji}^\ast\theta$$ so we get $\alpha =\mathrm{ad}_{f_{ij}}(\alpha)$ as an additional requirement, i.e. consistency of the local expression for the gauge field would require $\alpha$ to be central, or the gauge principal bundle to be trivial.

    But there's more: the crucial point is that $A\mapsto A+\alpha$ does not induce a bundle automorphism of the gauge principal bundle. In the usual setting, we have a bundle automorphism $t : P\to P$, that is equivalently a $\mathrm{ad}$-equivariant function $g : P\to G$ with $g(ph) = \mathrm{ad}_h g(p)$, which in turn descends to local functions $g_i: U_i\to G$ with $g_i = \mathrm{ad}_{f_{ij}}(g_j)$, and all collections of those functions can be glued back together to a bundle automorphism. You can't do that with $\alpha$, but you might now think to try local forms $\alpha_i = \mathrm{ad}_{f_{ij}}(\alpha_j)$. However, no one forbids choosing the $U_i$ contractible, and then every such $\alpha_i$ is already of the form $g_i^{-1}\mathrm{d}g_i$, meaning that doesn't add anything.

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  • $\begingroup$ Thanks for your answer. However my first question regarded not the integration of the entire Lagrangian. However I've already understood its meaning. The form $j_e$ is actually distributional of compact support (although the author does not mention this). In this case, Poincaré duality applies to $A$ even if it's not a closed form. The Poincaré dual however need not be a smooth curve as long as I know. If it's smooth the integral is evaluated as a path ordered integral. However don't know how to deal the non-smooth case if it's too "wild" $\endgroup$ – user40276 Feb 17 '16 at 5:12
  • $\begingroup$ About 2, I'm not really convinced of why one does not incorporate this additional symmetry. It's weird too that something which is a symmetry fails to be a symmetry in the quantum level, but maybe this is something well know to physicists… Now, about your comment regarding non-abelian groups, are you saying that the action of Yang-Mlls functional is not invariant under this kind of transformation $A \mapsto A + \alpha$? I don't think so... $\endgroup$ – user40276 Feb 17 '16 at 5:31
  • $\begingroup$ @user40276: I updated the answer with a short remark why $A\wedge j_e$ is "gauge invariant enough" for my integral to work, and with a longer explanation why a merely closed $\alpha$ does not properly respect the structure of a general gauge theory. $\endgroup$ – ACuriousMind Feb 18 '16 at 15:58
  • $\begingroup$ Thanks for your additions to your answer. You're right about the non-abelian case. I was too dumb to assume that the curvature would not change by this kind of more general transformation. However I still not convinced that these transformations should not be applied to electromagnetism in the classical case. Not taking these transformations into account would generate a too much more bigger space of field configurations. Now, about the $A \wedge j_e$, it would probably be invariant for general gauge transformations after being exponentiated (i.e., taking $mod \mathbb{Z}$), ... $\endgroup$ – user40276 Feb 18 '16 at 18:54
  • $\begingroup$ however the problem of the Poincaré dual still there… Sorry for being too annoying :P $\endgroup$ – user40276 Feb 18 '16 at 18:54

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