The official problem description for the Yang-Mills existence and mass gap problem references "compact simple gauge groups" but does not define the terms "compact" and "simple." I know what "compact" means, but "simple" is ambiguous.

There are different definitions of "simple" for general groups and Lie groups. Presumably the Lie group version is intended, since gauge groups are Lie groups. But according to Wikipedia, "there is no universally accepted definition of a simple Lie group."

Is this an oversight of the authors of the problem? Or is there an definition of "simple gauge group" in physics that is sufficiently standard that everyone knows exactly how to interpret it? If so, what is this definition?

  • $\begingroup$ There is no official textbook on group theory to be used by physicists, so one is free to choose the definition from an "official mathematical source" such as Pontrjagin's classic. $\endgroup$
    – DanielC
    May 17, 2021 at 0:33

2 Answers 2


In high-energy physics, "simple" always means "any group whose Lie algebra is any of the algebras in the Cartan classification $\{A,B,C,D,E,F,G\}$" where $A_n=\mathfrak{su}_{n-1}$, $B_n=\mathfrak{so}_{2n+1}$, $C_n=\mathfrak{sp}_n$, $D_n=\mathfrak{so}_{2n}$ and $E_6,E_7,E_8,F_4,G_2$ are the exceptional algebras.

Therefore, in hep-th "simple" means a gauge theory over groups like the classic matrix groups $SU(N),Spin(N),Sp(N)$, or the exceptional ones $E,F,G$, or finite quotients or extensions, like $PSU(N),O(N)$, etc.

You can also specify that you want groups that are either connected or simply-connected. If so, the convention is to say it explicitly (e.g., consider "Yang-Mills over a connected, simply-connected simple group").

The most general gauge theory involves reductive groups. These are, by definition, groups whose Lie algebra is the direct sum of a finite number of simple algebras, plus a finite number of copies of $\mathfrak u_1$. So, the most general gauge theory corresponds to groups of the form $G\ltimes SU(N)\times Spin(M)\times U(1)^n\times\cdots/\Gamma$ where in the numerator you can have as many copies of simple groups as you want, and as many copies of $U(1)$ as you want, and $G,\Gamma$ denote finite groups (like $\mathbb Z_n$, etc.).

  • $\begingroup$ Thanks, this sums it up cleanly. Can you recommend any texts that cover this terminology? I'm trying to review YM theory + Lie theory, but couldn't find anything that addressed this in particular. $\endgroup$
    – WillG
    May 20, 2021 at 15:34
  • 1
    $\begingroup$ @WillG The second volume by Weinberg does a great job, here you will be able to find the explicit proof of some of my claims. Francesco et al, big yellow book on CFT, has a physics-oriented chapter on simple groups which is quite good, although they do not discuss gauge theories. There is also more specific texts, like Georgi's "Lie algebras in particle physics", which proves the classification theorem in chapter 20. $\endgroup$ May 22, 2021 at 9:04

I think there is an underlying assumption that the answer will not depend depend on which definition of "simple" is involved. In principle, a group is simple if it has no proper normal subgroup. However, since Lie groups are related to Lie algebras, we can also bring in the notion of a simple Lie algebra, which is one having no proper ideal. (For the algebras, this can be further generalized to the definition of a semisimple algebra, which has no nonzero Abelian ideal. A semisimple alegebra is then a direct sum of simple algebras.)

Sometimes, the notion of a simple Lie group means one that has a simple Lie algebra. However, this conflicts with the strict group theoretic definition—although the conflict is not actually as bad as it might look. In a small enough neighborhood of the origin, the Lie group and the Lie algebra are in one-to-one correspondence, which means that if the algebra is simple, the group also looks simple in the immediate vicinity of the origin.

This has a lot of important consequences, the key one being that it is always possible to form a quotient of a Lie group with a simple Lie algebra by a discrete group to get another Lie group with the same Lie algebra. That is, if $G_{1}$ is a Lie group whose Lie algebra $g_{1}$ is simple, the there is a discrete normal subgroup $Z$ of $G_{1}$ such that $G_{2}=G_{1}/Z$ is simple, and the Lie algebra $g_{2}$ of $G_{2}$ is isomorphic to $g_{1}$. One of the best known examples of this is that the special orthogonal group $SO(3)$ and the special unitary group have the same Lie algebra, $so(3)\cong su(2)$, and the two groups are related by the quotient $SO(3)=SU(2)/\{\pm I\}$, where $\{\pm I\}$ is the $C_{2}$ group consisting of the positive and negative $3\times3$ identity matrices.

This means, for example, that the perturbative quantum field theory with a gauge group $G_{1}$ looks almost exactly the same as the perturbative theory with the gauge group $G_{2}$.* There is more to the mass gap problem than just perturbative physics, but it is generally believed that there can be no essential difference between the behavior of Yang-Mills theories with $G_{1}$ and $G_{2}$ (or any other intermediate gauge group). In fact, it is presumably felt that the equivalence will be trivial to demonstrate, once the structure of the Yang-Mills theory is well enough understood to demonstrate the presence of a mass gap (and/or color confinement) rigorously. In fact, it seems like it ought to be very simple to generalize from a simple compact gauge group to any compact Lie group.

*The only difference is that some representations of the larger group $G_{1}$ may not correspond to representations of the smaller $G_{2}$. In the example with $SO(3)=SU(2)/\{\pm I\}$, the larger group $SU(2)$ has unitary representations of every positive integer dimension, but $SO(3)$ only has unitary representations with odd dimensions. [This is the well-known fact that the Lie algebra $so(3)$ with commutation relations $[J_{i},J_{j}]=i\hbar\epsilon_{ijk}J_{k}$ only has orbital angular momentum representations with integer angular momentum $j$—and thus odd dimension $2j+1$.] However, this difference in the representation structure only affects how the gauge field can couple to charged matter fields. It does not affect the pure Yang-Mills theory that is the subject of the Clay Millennium Problem, since that involves only the adjoint representation of the gauge group; and since the adjoint is defined by the action of the group on its own Lie algebra, it exists with the same essential structure for both $G_{1}$ and $G_{2}$.

  • $\begingroup$ Good answer, although I have a rather minor comment. Yang-Mills over simply-connected groups is believed to be gapped and have a trivial vacuum, i.e., there is a mass gap and there is a single state below the gap. If the group is not simply-connected, the theory is still believed to be gapped but the vacua are non-trivial (they carry topological degrees of freedom), i.e., there is a mass gap but there are several states below the gap, and they may show long range entanglement. So, the theories $G_1$, $G_2$ are not quite the same, and they show different behaviour below the gap. $\endgroup$ May 17, 2021 at 8:56

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