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  Coulomb Branch vs. Higgs Branch (and the connection with D-branes, AdS/CFT)

+ 4 like - 0 dislike

I am confused about the difference between the Coulomb and Higgs branches of the moduli space of supersymmetric gauge theories. It's easy to find a definition for $D=4$, $\mathcal{N}=2$ supersymmetric field theories: The moduli space is locally a direct product between the two branches: $\mathcal{M} = \mathcal{M}_C \times \mathcal{M}_H$. The Coulomb branch corresponds to scalar fields in the vector multiplet getting VEV's, and the Higgs branch corresponds to scalar fields in the hypermultiplet getting VEV's.

I'm interested in this question for the case of $D=4$, $\mathcal{N}=4$ SYM, i.e. the worldvolume theory of D3-branes. Again, it's easy to find statements to the effect that the Coulomb branch corresponds to separating the D3 branes in the transverse directions, which gives mass to the some of the gauge fields and thus Higgses them. My main question is: what is the brane interpretation of the Coulomb branch?

Follow up questions are:

1) what is the rationale behind naming the branch of moduli space where the gauge fields are Higgsed as Coulomb, and the one where they are not as Higgs? Is there a universe in which this naming convention makes sense?!

2) I've read that Coulomb branch vacua cannot support finite temperature (for example in http://arxiv.org/abs/hep-th/0002160). I think this has a nice brane interpretation, at least for the case of many stacked branes so that the SUGRA description is valid. Two parallel but transversely separated stacks of branes are mutually BPS and therefore feel no force between each other. If we heat one of them up, then this no-force condition no longer holds, and the two stacks will gravitationally attract. In the SUGRA description, this is essentially the D3 brane version of the fact that extremal black holes can be superimposed to create multi-center solutions, but any finite temperature ruins this since the gravitational and electric forces are no longer balanced.

Can the Higgs branch vacua support finite temperature, and if so, is there a nice way to understand this via above sort of reasoning?

This post imported from StackExchange Physics at 2015-01-31 12:17 (UTC), posted by SE-user Surgical Commander
asked Jan 31, 2015 in Theoretical Physics by Surgical Commander (155 points) [ no revision ]
retagged Jan 31, 2015

1 Answer

+ 3 like - 0 dislike
Some remark about 1): in general, the gauge fields are Higgsed both on the Coulomb and on the Higgs branch. The difference is that at a generic point of the Higgs branch the gauge group is in general completely Higgsed whereas on the Coulomb group, the Higgsing is only partial: the maximal torus $U(1)^r$ of the gauge group is not Higgsed. The name Coulomb refers to the fact that at a generic point of the Coulomb branch, the low energy effective theory is an abelian gauge theory. The name Higgs refers to the fact that at a generic point of the Higgs branch, the full gauge group is Higgsed.

About $D=4$, $N=4$ SYM. This theory has 6 real scalars (corresponding to the 6 transverse dimensions to the D3 branes in 10 dimensional spacetime). Giving expectation values to these scalars define the moduli space $M$ of the supersymmetric vacua of the theory. For gauge group $U(n)$, one has $M = \mathbb{R}^{6n}/S_n$ where $S_n$ is the symmetric group acting on the n eigenvalues of the scalar expectation values. So $D=4$, $N=4$ SYM has a unique branch of vacua, whose brane interpretation is to separate the various branes in the transverse directions. I am not sure what is the correct name for this branch. If one wants to apply the definition of Coulomb/Higgs branches given for $N=2$ to $N=4$, one has to choose an embedding of the $N=2$ superalgebra in the $N=4$ superalgebra. If one does so, one finds that $M$ is a mixed branch as at a generic point all 6 scalar expectation values are turned on, 2 being in the $N=2$ vector multiplet and 4 being in $N=2$ hypermultiplets. So from a $N=2$ point of view, $D=4$, $N=4$ SYM has nor Coulomb or Higgs branch but only a mixed branch. But one could use another definition and say that a Coulomb branch for a $N=4$ theory is a branch with expectation values of scalars of the $N=4$ vector multiplet. With this definition, $M$ is a Coulomb branch. It seems to me that both terminologies are used in the litterature.
answered Jan 31, 2015 by 40227 (5,140 points) [ revision history ]
edited Jan 31, 2015 by 40227

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