Is there a deep mathematical reason for why bosons should be in the adjoint representation of the gauge group rather than any other representation?
[Physics] bosons be in the adjoint representation of the gauge group
gauge-theoryparticle-physicsquantum mechanicsquantum-field-theory
Related Solutions
It's a lot of questions but they have pretty easy answers, so here they are:
Gauss's law is just $\mbox{div }\vec D=\rho$ in electrodynamics. Note that it contains no time derivatives so it's not really an equation describing evolution: it's an equation restricting the allowed initial conditions. More generally, it's the equation of motion that you get by varying the Lagrangian with respect to $A_0$, the time component of the gauge field, so the corresponding equation of motion counts the divergence of the electric field minus the electric sources (charge density) that must be equal to it. This difference is nothing than the generator of the overall $U(1)$ group, or any other group, if you consider more general theories, so the classical equation above is promoted to the quantum equation in which $(\mbox{div }\vec D-\rho)|\psi\rangle=0$ which just means that the state $|\psi\rangle$ is gauge-invariant.
The traces you are encountering here - in non-Abelian theories - are just traces over fundamental (or, less frequently, adjoint) indices of the Yang-Mills group. They're different than traces over the Hilbert space. You must distinguish different kinds of indices. Tracing over some color indices doesn't change the fact that you still have operators.
"Adjoint of a gauge group" is clearly the same thing as "Adjoint representation of the Lie group that is used for the gauge group."
It's not true that all compact spaces eliminate all massless fields - for example, the Wilson line of a gauge field remains a perfectly massless scalar field on toroidal compactifications in supersymmetric theories - but in most other, generic cases, it's true that compactification destroys the masslessness of all fields. All the Fourier (or non-zero normal) components of the fields that nontrivially depend on the extra dimensions - the Kaluza-Klein modes - become massive because of the extra momentum in the extra dimensions. But even the "zero modes" become massive in general theories because of the Casimir-like potentials resulting from the compactification.
Basic excitations are not "the same thing" as one-particle states. In fact, we want to use the word "excitation" exactly in the context when the goal is to describe arbitrary multi-particle states. But the basic excitations are just the creation operators (and the corresponding annihilation operators) constructed by Fourier-transforming the fields that appear in the Lagrangian, or that are elementary in any similar way.
You haven't constructed any "specific states above" so I can't tell you how some other states you haven't described are related. In this respect, your question remained vague. All of them are some states with particles on the Hilbert space - but pretty much all states may be classified in this way.
A mode of a quantum field is the term in some kind of Fourier decomposition, or - for more general compactifications and backgrounds - another term (such as the spherical harmonic) that is an eigenstate of energy i.e. that evolves as $\exp(E_n t/i\hbar)$ with time. So for example, a field $X(\sigma)$ for a periodic $\sigma$ may be written as the sum below. The individual terms for a fixed $n$ - or the factor $X_n$ or the function that multiplies it (this terminology depends on the context a bit) - are called the modes. $$\sum_{n\in Z} X_n \,\exp(in\sigma - i|n|\tau)$$
Gauge symmetry has to commute with the Hamiltonian because we want to ban the gauge-non-invariant states, and by banning them in the initial state, they have to be absent in the final state, too. So it has to be a symmetry. On the other hand, the representation theory is trivial because, as I said at the very beginning, we require physical states to be gauge-invariant; that was the comment about Gauss's law: states have to be annihilated by all operators of the type $\mbox{div }\vec D-\rho$ which are just generators of the gauge group at various points. In other words, all of the physical states have to be singlets under the gauge group. That's why the term "symmetry" is somewhat misleading: some people prefer to call it "gauge redundancy". So the Hilbert space is surely completely reducible - to an arbitrary number of singlets. You may reduce it to the smallest pieces that exist in linear algebra - one-dimensional spaces.
I just explained you again why all the physical states are singlets under the gauge group. The fact that $n$-particle states with identical particles are completely symmetric or completely antisymmetric reduces to the basic insight that in quantum field theory, particles are identical and their wave function has to be symmetric or antisymmetric (for bosons and fermions) because the corresponding creation operators commute (or anticommute) with each other.
I am not sure if I know the correct answer (as I am a student my self), but I will try (and if I am wrong, someone please correct me).
The first thing that took me some time to figure out is what they mean by adjoint representation. In Georgi's book he defines the adjoint representation of a generator as: \begin{equation} [T_i]_{jk} \equiv -if_{ijk} \end{equation} which is equivalent to the adjoint representation of a Lie algebra. However, when discussing monopole, they actually mean the adjoint representation of a Lie group . This means that $\phi$ takes values in the Lie algebra (the vector space formed by the generators) and can be expressed in terms of the generators in an arbitrary representation: \begin{equation} \phi = \phi^a t^a \end{equation} where $t^a$ denote the generators in an arbitrary representation (and there is an implicit sum over repeated indices).
Now, let us look at the simplest example, which is the bosonic part of the $\mathrm{SU(2)}$ gauge invariant Georgi-Glashow model: \begin{equation} \mathcal{L}=\frac{1}{8} \mathrm{Tr} (F_{\mu \nu} F^{\mu \nu}) - \frac{1}{4} \mathrm{Tr}(D_\mu \phi D^\mu \phi) - \frac{\lambda}{4}(1-\phi^a \phi^a )^2 \end{equation} We can write the kinetic and potential energy, $T$ and $V$, as: \begin{equation} T=\int \left( - \frac{1}{4} \mathrm{Tr} (F_{0i}F_{0i}) - \frac{1}{4} \mathrm{Tr}(D_0 \phi D_0 \phi) \right) \mathrm{d^3}x \end{equation} and: \begin{equation} V=\int \left( - \frac{1}{8} \mathrm{Tr} (F_{ij}F_{ij}) - \frac{1}{4} \mathrm{Tr}(D_i \phi D_i \phi) + \frac{\lambda}{4}(1-\phi^a \phi^a )^2 \right) \mathrm{d^3}x \end{equation} where we used $L= \int \mathcal{L} \; \mathrm{d^3}x = T-V$. In order to get finite energy solutions we have to impose boundary conditions such that the total energy of the model vanished at spatial infinity. It should be clear that one of the requirements to ensure that the energy vanishes is: \begin{equation} \phi^a \phi^a =1 \end{equation} This implies that the Higgs vacuum corresponds to an infinite amount of degenerate vacuum values lying on the surface of a unit two-sphere in field space, which we will denote by $S^2_1$. Furthermore, by imposing the aforementioned finite energy boundary condition, this gives rise to the following map: \begin{equation} \phi : S^2_\infty \to S^2_1 \end{equation} where $S^2_\infty$ denotes the two-sphere associated with spatial infinity (in 3 dimensions). This is in fact the definition of the winding number (or degree) between two two-dimensional spheres and is therefore classified by $\pi_2(S^2)=\mathbb{Z}$ (and it is in theory possible to construct topological solitons). Now, if $\phi$ was in the fundamental representation, then I don't think it is possible to construct these topological solitons.
Best Answer
I guess by "bosons" you're referring to gauge bosons?
If so then start with some matter field $ \psi(x)$ which transforms under the gauge group. For local gauge transformations the gauge group element $g$ is spacetime dependent $g(x)$, and the transformation is $$\psi(x) \longrightarrow \psi'(x) = g(x)\psi(x).$$
Derivatives would transform as
$$\partial_{\mu}\psi(x) \longrightarrow g(x)\partial_{\mu}\psi(x)+(\partial_{\mu}g(x))\psi(x),$$
i.e. inhomogeneously. We would like a gauge covariant derivative $D_{\mu}$ which transforms homogeneously as
$$D_{\mu}\psi(x) \longrightarrow g(x)D_{\mu}\psi(x).$$
To achieve this, we define
$$D_{\mu}\psi = \partial_{\mu}\psi - A_{\mu}\psi,$$ where $A_{\mu} = \mathbf{A}_{\mu} \cdot\boldsymbol{{\tau}}$ and $\boldsymbol{\tau}$ are the generators of the Lie algebra of the gauge group and $A_{\mu}$ is our bosonic gauge field. This introduction of gauge bosons via the derivative term is sometimes referred to as minimal coupling.
In order to achieve this, $A_{\mu}$ is forced to have the transformation law
$$A_{\mu} \longrightarrow A'_{\mu} = gA_{\mu}g^{-1} + (\partial_{\mu}g)g^{-1}.$$
Just looking at how the $A_{\mu}$ are transforming under the group action (the first term), we recognize the adjoint representation.
Of course, on the global stage, the fields $\psi$ can be interpreted as bundle sections and the gauge fields as bundle connections. $A_\mu$'s transformation law will be recognisable as a transformation of connection coefficients under the action of the bundle's structure group. A good reference is Nakahara, or this link.