# Aharonov–Bohm effect on a non-contractible manifold

The Aharonov–Bohm effect shows the vector potential $\mathbf{A}$ is more fundamental than magnetic flux density $\mathbf{B}$. However, vector potential is introduced by $$\nabla \times \mathbf{A} = \mathbf{B}$$ or $$\mathrm{d} \mathbf{A} = \mathbf{B}$$ which requires the Poincaré's lemma to ensure the existence of $\mathbf{A}$. If the manifold is not contractible, there is no guarantee that $\mathbf{A}$ exists. On a non-contractible manifold, what would happen for the Aharonov–Bohm effect?

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$\mathbf{B}$ is "magnetic flux density", not "magnetic potential"... – Mehrdad Aug 24 '14 at 21:50
Thanks. Edited. – user26143 Aug 25 '14 at 5:08
– Qmechanic Dec 13 '14 at 16:42

The necessary and sufficient condition for $dB = 0$ (more commonly written $\nabla \cdot \mathbf B = 0$) to imply $B = dA$ is the vanishing of the second de Rham cohomology $H^2 (M)$. This is guaranteed for a contractible manifold since cohomology is homotopy invariant. However this is for a 2-form $B$ defined on all of $M$. Consider instead the restriction of $B$ to some open set $U$, $B|_U$. We can take $U$ to be contractible, for instance, $U$ could be the image of a coordinate ball. Then $H^2(U)$ vanishes and we can find $A_U$ such that on $U$, $dA_u = B|_U$. Therefore there always exists a local vector potential.

The Aharonov-Bohm effect is related to the nonvanishing of $$\Phi = \oint A$$ even when $dA = 0$. When the first de Rham cohomology $H^1(M)$ vanishes, we have that $dA = 0$ is equivalent to $A = df$, and so by Stokes's theorem $\Phi$ is always $0$. In particular $\Phi$ can never be non-zero for a contractible space.

When a global $A$ cannot be found, one can still make sense of the quantity $\Phi$, but the best way to do this is with the tools of gauge theory. The book Gauge Fields, Knots, and Gravity by Baez and Muniain discusses much of the concepts mentioned in this answer.

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Can you give a non-trivial example when both $H^1$ and $H^2$ are non-vanishing? Thanks for the book btw. – firtree Aug 24 '14 at 18:16
Is a torus $T^2 = S^1 \times S^1$ non-trivial enough? – Robin Ekman Aug 24 '14 at 21:41
Torus will do, thanks :-) So, in our physical $R^3$ space this example coiuld be implemented as two deleted rings, linked as chain links, right? Is it of any interest to reproduce the Aharonov-Bohm-like experiment for such configuration of solenoids? – firtree Aug 25 '14 at 0:19

Formal accounts of EM do not need the Poincare lemma for the vector potential to exist:

Formulating a theory like electromagnetism on any manifold is best done by viewing it as the $\mathrm{U}(1)$ gauge theory on said manifold. Then, the "vector potential" is simply a connection on the circle bundle over the manifold, which exists regardless of whether Poincare holds or not. The electric and magnetic fields are then components of the curvature tensor $F = \mathrm{d}A$, just as in the covariant formulation of EM.

The Aharanov-Bohm effect essentially already takes place on a non-contractible manifold, since the usual arrangement removes an entire line from the manifold considered where the solenoid is. And $\mathbb{R}^3$ with a line removed is a cylinder, and not contractible. (see also this question)

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So the answer is, $\mathbf{A}$ is defined locally? – user26143 Aug 24 '14 at 17:53
@user26143: That is one way of interpreting it, yes. It is globally defined on the principal bundle over the manifold, but projects down onto the manifold itself only locally, and the different local $A$ are related by gauge transformations. (The procedure of gluing the local forms by gauge transformations can also be understood as defining the principal bundle) – ACuriousMind Aug 24 '14 at 17:56
Thank you for your explanation. – user26143 Aug 24 '14 at 17:57