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Yang-Mills theory with discontinuous connection

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Suppose I have a local gauge group $g$ that is a $C^0$-function over Minkowski spacetime. This implies that the corresponding gauge connection $A_\mu$ is not continuous, i.e. it can have value infinity. More concrete, an interesting case would be the case where $A_\mu$ can have the following values:

Either zero or $\infty$.

If $A_\mu$ is equal to $\infty$, the Yang-Mills action $S$ vanishes. One can define a boundary of the spacetime manifold (assumed as flat) $\partial M$ that separates the singularity of the connection. Would these assumptions lead to the theory

$\sum_{k_1, \dots, k_n=0}^1  \prod_{i=1}^n (\int \mathcal{D}[All. other. fields]e^{iS})_{k_i}$?

Here, the indices $k_i$ indicate from which points the expression $(\int \mathcal{D}[All. other. fields]e^{iS})_{k_i}$ will be separated. If for point number $i$ the value of $k_i$ is zero than the action is not defined on this region, otherwise it is defined. This would induce a sum over all manifold topologies.

Can such a theory be formulated?

asked Jan 31 in Theoretical Physics by anonymous [ no revision ]

1 Answer

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One can indeed make operator insertions in gauge theory which amount to prescribing the singularity of the gauge field near some submanifold. One proceeds as you do by removing a small neighborhood of the submanifold and thinking about it as a boundary condition. One can describe magnetic monopoles this way. Look up 't Hooft operator.

Giving these objects dynamics in any sense is a delicate thing, and usually one needs a UV completion of the theory where these singularities are smoothed. In the case of the monopole these are the 't Hooft-Polyakov monopoles one obtains by including the gauge group U(1) inside of SU(2), where the U(1) loop becomes contractible. Then one imagines there is a Higgs scale below which the monopoles look like singularities in the U(1) gauge field but above which are smooth configurations of an SU(2) gauge field which obtain mass proportional to the Higgs scale.

answered Feb 1 by Ryan Thorngren (1,455 points) [ no revision ]

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