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Usually we write the Yang-Mills theory with gauge group $G$ as $$\frac{1}{g^2} \mathrm{Tr}_{R} (F\wedge \star F)$$ But here we need to choose what $R$ is. There are several cases one may expect:

  1. $R$ has to be the rep with minimal dimension. Different choice of $R$ gives different gauge theories. For example, for $\rm SU(2)$ $R$ is 2d rep where the three generators are represented by three Pauli matrices. For $\rm SO(3)$, $R$ is 3d rep.
  2. $R$ has to take all over all possible rep of the gauge group. ($d=2,3,...$ for $\rm SU(2)$.
  3. The action does not depend on what $R$ is.

My question is: which of the above is correct? and why?

Qmechanic
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user34104
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    It is true that $\text{tr}R[\cdots] = T_R \text{tr}\text{adj}[\cdots]$. $T_R$ is known as the index of the representation. The dependence on the representation is an overall scale which one could absorb into $g$. – Prahar Jul 21 '20 at 05:27
  • Maybe I am wrong, but don't the gauge fields $A$ transform in the adjoint representation? Then there shouldn't be any freedom to choose the representation in the YM term. (However I am not 100% sure of this, so this is a comment and not an answer; if I'm wrong I'd be interested in learning more) – Andrew Jul 21 '20 at 07:01
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    @Andrew The gauge field is an element of the Lie algebra, but it can be represented in any representation you want. You can take $A_\mu = A_\mu^a T^a$ where $T^a$ are the generators in your favorite representation. The point is that the trace only depends on the representation by an overall factor. The precise one use is convention. For instance in the study of $SU(N)$ gauge theories, it is convention to compute the trace in the fundamental representation. – Prahar Jul 21 '20 at 08:32
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    @Andrew You could have worked in any other representation - the only difference between your conventions and someone else's is a rescaling of the coupling constant. – Prahar Jul 21 '20 at 08:32
  • @Prahar I see, I hadn't appreciated this point, but it makes a lot of sense. Thanks for clarifying! For posterity, this question/answer (especially the answer by tparker) goes into more detail about this point as well: https://physics.stackexchange.com/q/275499/ – Andrew Jul 21 '20 at 09:21
  • @Prahar, thanks for the comments. I agree that different representations gives a different overall rescaling factor. But for certain theories, like Chern simons theories, the overall scale is fixed to be an integer, which means there is a canonical choice of R, which is the minimal rep of G. – user34104 Jul 21 '20 at 13:58
  • I may wonder for different choice of R, we may call them different theories. For example, when R is 2d rep, the theory is SU(2) gauge theory, when R is 3d rep, the theory is SO(3) gauge theory. Am I correct? – user34104 Jul 21 '20 at 14:00
  • It is fixed to be an integer only when the trace is properly normalized or “taken in the correct representation”. There is no canonical choice. It is just a matter of convention and we always use the one that is most convenient for that problem. In CS theories, you can still use whatever representation you want, you would have to adjust the constant so that it is quantized appropriately (which means not necessarily integer, but integer quantized). – Prahar Jul 21 '20 at 14:01
  • You may not. The group that appears in the name of the gauge theory is simply the group associated to that theory. The representations of that group which appear in the Hilbert space is a totally separate matter and can be used as an addendum. For instance “SU(2) gauge theory with fundamental matter”. The representation used to describe the action is simply a matter convention. Different textbooks use different ones based on what suits them. – Prahar Jul 21 '20 at 14:03
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    @Prahar, if we consider the Wilson line. $Tr_{R'} exp(i \oint A) $. Here I, on purpose, used a different notation $R'$ to distinguish it from $R$ in the Lagrangian. If $R$ in the Langrangian is 2d of SU(2), then $R'$ in the Wilson line be any rep, because any $R'$ can be constructed from the minimal representation $R$. But if we choose $R'$ to be 3d of SU(2), then I doubt one can have $R'$ to be a 2d rep. Am I right – user34104 Jul 21 '20 at 14:06
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    I am really not following your confusion. The input that goes into the definition of a Wilson line is the representation and the path. This representation and path has nothing to do with the representation R that is used to write down the action. As I have mentioned before, the gauge field does not live in any representation. It is simply an element of the Lie algebra and no representation is required to define it. Representations are required to define other quantities built from it such as the Wilson line, etc. – Prahar Jul 21 '20 at 14:09
  • Let us continue this discussion in chat. – Prahar Jul 21 '20 at 14:12
  • Tr denotes an invariant bilinear form on the Lie algebra of G, a multiple of the Cartan-Killing form. Your Tr$_R$ may only differ by a normalization. – wonderich Jul 21 '20 at 19:58

1 Answers1

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The action should be invariant under the action of gauge group, therefore, you have to construct something, that when considering the action of gauge group remains unchanged.

Assume that the field stregth tensor $F$ transform under some representation of $G$: $$ F \Rightarrow U^{i_1 \ldots i_n}_{j_1 \ldots j_m} F $$ To make this more clear, let us list several simple examples: $$ \phi^i \Rightarrow U_{j}^{i} \phi^j - \text{fundamental representation} $$ $$ \phi_i \Rightarrow U_{i}^{j} \phi_j - \text{antifundamental representation} $$ $$ \phi_i^{j} \Rightarrow U_{i}^{a} \phi_a^{b} U_{b}^{j} - \text{adjoint representation} $$ After multiplying $F$ and $F^{*}$ and taking of the trace, there has to remain and object without free indices. The operation of multiplication and $\text{Tr}$ contracts equal number of lower and upper indices, so the only consistent way to make singlet is to have $m = n$. $$ \text{Tr} \ F \wedge F^{*} = \text{Tr} \ U^{i_1 \ldots i_m}_{j_1 \ldots j_m} U^{j_1 \ldots j_m}_{k_1 \ldots k_m} \ F \wedge F^{*} = \text{Tr} \ U^{i_1 \ldots i_m}_{j_1 \ldots j_m} U^{j_1 \ldots j_m}_{i_1 \ldots i_m} \ F \wedge F^{*} $$ The written by far expression by far doesn't correspond to irreducible representation in general - i.e, one can symmetrize (antisymmetrize) over $i_a$ or $i_j$. I may be wrong, but it seems that the general action of the aforementioned form is some irreducible representation of tensor product of several adjoint representations.

spiridon_the_sun_rotator
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