Let us consider two different quantum field theories in 4 dimensional Minkowski spacetime, call them theory A and theory B, with 8 supercharges. (i.e. 4D $\mathcal{N}=2$ theories). Let $G_A$ be the flavor symmetry group of A, and $G_B$ the flavor symmetry group of B.

  1. Is it true that if $G_A$ and $G_B$ are isomorphic Lie groups, then the Higgs branch of the moduli space of theory A is isomorphic (as an algebraic variety) to the Higgs branch of the moduli space of theory B?

  2. If it is not true, there exist an explicit counterexample of two theories with same flavour symmetry group and different Higgs Branches of the moduli space?

  3. Does this fact that the flavor symmetry fixes the Higgs Branch also hold in different dimensions, or with a smaller amount of supersymmetry?

PS if it is simpler, we can restrict the question to Lagrangian Theories.

  • $\begingroup$ The idea seems meaningful. Why don't you write a paper on it, instead of searching the answer here? $\endgroup$
    – Mass
    Commented Aug 30, 2016 at 17:38
  • 2
    $\begingroup$ Because I am not able to prove this fact in a clear enough way to write it in a paper. Or at least I would like first of all to know if this fact is already known or not by the community, cause a rapid search in the literature gave no result. I might be a very trivial fact known for a long time and therefore not explained in any recent paper. I think it is ok to post research level questions here. Different experts read StackExchange and often comment. If I am mistaken, please tell me and I can remove the post. $\endgroup$ Commented Aug 30, 2016 at 17:44
  • $\begingroup$ All i can say, this is a thought provoking question and is quite non trivial. $\endgroup$
    – Mass
    Commented Aug 30, 2016 at 17:48

2 Answers 2


1.) NO

2.) Consider the lagrangian theories with gauge group G= USP(2N), four fundamental hypers and one antisymmetric, all these models have flavor symmetry SU(2) x SO(8), but the higgs branch is different in each case and it is the N-instanton SO(8) moduli space. These are different Hyper-Kahler manifolds with dimensions 4 N (N+1).

3.) NO

  • $\begingroup$ Thank you very much for the clear and straightforward answer. $\endgroup$ Commented Mar 19, 2018 at 16:40

This is an interesting question. My first sense was to say this is negative. However, maybe the Higgs branch for symmetries for gauge symmetry if equal to that for the Higgs on the color symmetries of fermions of that force. This is maybe an interesting research topic. It may have been pursued, maybe within the context of technicolor. I outline a possible way this might in fact be correct.

I will start with the definition of the Higgs field on its vacuum. We know that for a standard quantum field, such as that with the Lagrangian ${\cal L}~=~\frac{1}{2}|\partial\phi|^2~-~\frac{1}{2}|\phi|^2$ has an orbit in the quadratic potential that has nonzero energy, and is at the vacuum $\langle\phi\rangle~=~0$ when the field is zero. Contrary to that with the Higgs the potentially $$ V(\phi)~=~-\mu|\phi|^2~+~\lambda|\phi|^4 $$ has a minimum, found by evaluating $\partial V(\phi)/\partial\phi~=~0$, that gives a set of vacua at the fields $|\phi|^2~=~\mu/2\lambda$. This is the same as defining a set of complex numbers that have a modulus or magnitude on a circle in the complex plane. This is a set of vacua that occur for Higgs field that are non-zero. This means the vacuum configuration of the Higgs field is nonzero, which is a condensate $$ \langle\phi\rangle~\ne~0. $$ Condensates occur with symmetry breaking or with statistical sets of degenerate states.

The field is degenerate according to $\Phi(x)~=~\phi(x)~-~C\mathbb I$, for $C$ a constant with respect to $\phi$, so that $$ \langle\Phi\rangle~=~\langle\phi\rangle~-~C\langle\mathbb I\rangle, $$ leading to $\langle\phi\rangle~=~C$

This is a bit of a sketch, but I might argue it is the case the color gauge and flavor fermion branches of the Higgs are isomorphic. Now propose an elementary scheme where the fields $\Phi(x)$ and $\phi(x)$ are related by unitarity $\Phi(x)~=~U\phi(x)U^\dagger$, where $U~=~e^{f({\cal O})(a-a^\dagger)}$, where $a$ and $a^\dagger$ are the raising and lowing operators for $\phi$ at an IR momentum $k_0$. Further, $f({\cal O})$ represents the following: $$ f({\cal O})(a-a^\dagger)~=~\epsilon A(a-a^\dagger) $$ or $$ f({\cal O})(a-a^\dagger)~=~\epsilon b^\dagger(a-a^\dagger)b. $$ The first of these reflects the gauged derivative $D_\mu\phi~=~\partial_\mu\phi~+~A\phi$, in particular the $A\phi$, and the second is a cryptic form of the Yukawa Lagrangian ${\cal L}_y~=~\bar\psi H\psi$. Now consider $\epsilon~<<~1$ and this becomes $$ \phi~=~\Phi~+~\epsilon f({\cal O})([a,~\Phi]~-~[a^\dagger,~\Phi]). $$ A Fourier expansion of the field $$ \phi~-~i\sum_k(a_ke^{-ikx}~-~a^\dagger_ke^{-ikx}). $$ leads then to $$ \phi~=~\Phi~+~2\epsilon f({\cal O})cos(k_0x), $$ where for the gauge or fermion case we have $f({\cal O})~=~A$ or $b^\dagger b$.

This means that gauge field and fermion sectors track each other. The moduli space for the gauge sector appears identical to that of the flavor sector. It might even be argued if there are Gribov ambituities with the gauge branch these carry over to the fermion branch. This is an interesting set of problems to look into.

  • $\begingroup$ I am sorry, but this answer does not answer any of the questions asked above. The question is about Higgs Branch (set of gauge inequivalent vacua parametrized by vevs of the hypermultiplets) of supersymmetric QFTs with 8 supercharges. (i.e. N=2 in d=4). Furthermore, even the conclusion you suggest is wrong. Assuming that "the gauge branch" is a non-standard name for the Coulomb Branch, and the "flavor branch" is a non-standard name for the Higgs branch, you argue that the two of them are always isomorphic. This is generically false. Overall, I am very perplexed about this answer. $\endgroup$ Commented Aug 30, 2016 at 17:33

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