Why is the gauge field $A_\mu$ real for $\mathrm{U}(N)$ symmetry?

Question

From my notes I have that

The transformation law, $$A_\mu\to MA_\mu M+\frac{i}{g}\left(\partial_\mu M\right)M^\dagger\tag{1}$$ can be realised if $A_\mu$ is an element of the Lie algebra. It can therefore be written in terms of the generators, $t^a$ as $A_\mu=A_\mu^at^a$, for $a \in \{1,...,D \}$ with real coefficients $A_\mu^a$. Therefore we can think of the gauge field as a $D$-component $\color{red}{real}$ vector field. When considering the covariant derivative $D_\mu \phi$ of a field $\phi$ in a particular representation, then the gauge field would be written as a linear combination of the generators of that representation, $A_\mu = A^a_\mu T^a$.

I have a problem with the word I marked in red, put simply, I don't understand how $A_\mu$ can be real given that it has complex generators. By this I mean matrices with complex entries.

To put this problem into context consider the following question:

The Lagrangian for the $\mathrm{U}(N)$ Higgs model is $$\mathcal{L}=-\frac12\mathrm{Tr}F_{\mu\nu} F^{\mu\nu}+\left(D_\mu\phi\right)^\dagger D^\mu\phi-m^2\phi^\dagger\phi-\frac12\lambda\left(\phi^\dagger\phi\right)^2$$ where $\phi$ is a complex $N$-component Lorentz scalar field, $D_\mu=\partial_\mu+igA_\mu$ and $F_{\mu\nu}=\partial_\mu A_\nu -\partial_\nu A_\mu+ig\left[A_\mu, A_\nu\right]$, and $A_\mu=A_\mu^at^a$ is the gauge field. Show that the equation of motion for the scalar field $\phi$ is $$D_\mu D^\mu\phi+m^2\phi + \lambda\left(\phi^\dagger\phi\right)\phi=0\tag{2}$$

I'm not interested in the derivation of $(2)$, but in the beginning of the solution it is stated that

First, write the Lagrangian in terms of the components of $\phi$, $$\mathcal{L}=-\frac12\mathrm{Tr}F_{\mu\nu} F^{\mu\nu}+\color{red}{\left(\partial_\mu\phi^\dagger-ig\phi^\dagger A_\mu\right)}\left(\partial^\mu\phi+igA^\mu\phi\right)\tag{3}$$ $$-m^2\phi^\dagger\phi-\frac12\lambda\left(\phi^\dagger\phi\right)^2$$ $$=-\frac12\mathrm{Tr}F_{\mu\nu} F^{\mu\nu}+\partial_\mu\phi^\dagger\partial^\mu\phi+ig\left(\partial_\mu\phi^\dagger\right)A^\mu\phi-ig\phi^\dagger A_\mu\partial^\mu\phi $$ $$+g^2\phi^\dagger A_\mu A^\mu\phi-m^2\phi^\dagger\phi-\frac12\lambda\left(\phi^\dagger\phi\right)^2$$ $$=-\frac12\mathrm{Tr}F_{\mu\nu} F^{\mu\nu}+\partial_\mu\phi_i^\ast\partial^\mu\phi_i+igA^{a\mu}\left(\partial_\mu\phi_i^\ast\right)t_{ij}^a\phi_j-igA_\mu^a\phi_i^\ast t_{ij}^a\partial^\mu\phi_j$$ $$+g^2A_\mu^aA^{b\mu}\phi_i^\ast t_{ij}^a t_{jk}^b\phi_k-m^2\phi_i^\ast\phi_i-\frac12\lambda\phi_i^\ast\phi_i\phi_j^\ast\phi_j$$

Not shown, but the step after the final equality above involves using the Euler-Lagrange equation, $$\frac{\partial\mathcal{L}}{\partial\phi_i^\ast}-\partial_\mu\frac{\partial{\mathcal{L}}}{\partial\left(\partial_\mu\phi_i^\ast\right)}=0\tag{4}$$

In my notes, it is written that for ${\rm U}(N)$ matrix group, the generators are Hermitian $N\times N$ matrices, with dimension $D=N^2$.

So this means that $(t_{ij}^a)^\ast=t_{ji}^a$ and since the coefficients, $A_\mu^a$ are real then $(A_\mu^a)^\ast=A_\mu^a$.

It's not relevant to my question to obtain eqn. $(2)$ using $(4)$, and my concern is the term in red of equation $(3)$, specifically why is that bracket marked red not being written as $\left(\partial_\mu\phi^\dagger-ig\phi^\dagger {(A_\mu)}^\dagger\right)$?

My reason for doubting this is because the generators for $\mathrm{U}(N)$ are complex matrices, so it should follow that since the coefficients are real as stated above; $A_\mu^a \in \mathbb{R}$, then since $A_\mu=A_\mu^at^a$, it must be the case that the generators of $\mathrm{U}(N)$ are complex, and because of this $A_\mu$ must also be complex.

To simplify this assertion consider the $\mathrm{U}(1)$ group, which are simply complex numbers (a $1\times 1$ matrix), if this is true and $A_\mu=A_\mu^at^a$, then it ought to follow that $A_\mu$ must be a complex number since the coefficients are real and $$A_\mu=A_\mu^ae^{i\theta a},\tag{5}$$ where $e^{i\theta a}$ is a complex number (or phase). $\theta$ here is a matrix that is an element of the Lie algebra.

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Final point:

I would just like to mention that I know I am wrong about this, but I want to know why I am wrong.

In this linked question the user that answered my question in the comment stated that the gauge field, $A_\mu$, for $\mathrm{U}(1)$ is real, but did not justify this claim.

For simplicity, in the $\mathrm{U}(1)$ case there is a complex phase multiplying the real coefficient $A_\mu^a$. So it really ought to follow that the gauge field, $A_\mu$, is complex. But this directly contradicts what I have been told.

Please help me understand this.

Answer 1

1. The first point is that the Lie group $G=U(N)$ [which consists of unitary $N\times N$ matrices] is a real Lie group, which is perhaps best explained as that the tangent spaces, or equivalently, the Lie algebra $\mathfrak{g}=u(N)$ [which consists of anti-Hermitian$^1$ $N\times N$ matrices] is a real vector space, i.e. its field is $\mathbb{F}=\mathbb{R}$.

   Note in particular that:

   • It is irrelevant for the above terminology that the matrices in $U(N)$ and $u(N)$ can have complex entries.

   • We can not impose a complex structure on $U(N)$ and/or $u(N)$, i.e. $U(N)$ is not a complex Lie group, and $u(N)$ is not a complex vector space.

2. In non-Abelian gauge theory [such as Yang-Mills theory] the gauge field $A=A_{\mu}\mathrm{d}x^{\mu}$ is a Lie algebra-valued 1-form field, $$\begin{align}A_{\mu}~=~&\sum_{a=1}^{\dim\mathfrak{g}}A^a_{\mu}t_a~\in~\mathfrak{g},\cr A^a_{\mu}~\in~&\mathbb{F},\cr \mathfrak{g}~=~&{\rm span}_{\mathbb{F}} \left\{t_a\mid a\in\{1,\ldots,\dim\mathfrak{g}\}\right\},\end{align}$$ and hence for OP's Lie algebra $\mathfrak{g}=u(N)$, the component fields $A^a_{\mu}\in\mathbb{R}$ are real.

3. Wrt. to eq. (3), the Hermitian conjugate$^1$ of the gauge-covariant derivative of a field $\phi$ in the fundamental representation of $G$ is $$ (D_{\mu}\phi)^{\dagger}~=~(\partial_{\mu}\phi+igA_{\mu} \phi)^{\dagger}~=~\partial_{\mu}\phi^{\dagger}-ig \phi^{\dagger}A_{\mu}^{\dagger} ~=~\partial_{\mu}\phi^{\dagger}-ig \phi^{\dagger}A_{\mu},$$ since $A_{\mu}^{\dagger}=A_{\mu}$ is Hermitian, cf. footnote 2.

4. Abelian case $N=1$: Electromagnetism. Then the Lie algebra $\mathfrak{g}=u(1)=\mathbb{R}$, cf. footnote 1. The Lie algebra generator is conventionally chosen to be $t_{a=1}=1$, so $A_{\mu}=A_{\mu}^{a=1}$ is a real number.

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$^1$ For more details on the Hermitian conjugate, see e.g. my related Phys.SE answer here.

$^2$ Be aware that in much of the physics literature [including OP's source] $u(N)$ is instead identified with the set of Hermitian $N\times N$ matrices, and compensating factors of the imaginary unit $i$ is inserted at pertinent places, e.g. the gauge-covariant derivative has in OP's source an explicit factor of the imaginary unit $i$: $$ D~=~\mathrm{d}+igA.$$ See also my related Phys.SE answer here.

Answer 2

First, I question the importance of this in practice. Because if you have an action where the $A_\mu^a$ are initially real and you decide that you don't like this, you can always change this by redefining the generators.

But your notes are right that one can always choose $A_\mu^a$ to be real. This is because the gauge field transforms in the adjoint which is a real representation. When the matrices representing an algebra have real entries, you can always use a similarity transformation to spoil this reality. Which is what has often been done when you see matrices for the adjoint representation with an imaginary part. But the reverse is often not possible because there are so called "complex" and "pseudo-real" representations as well.

The $U(1)$ example is too trivial to gain intuition about this because the adjoint representation of $U(1)$ is just the singlet. Also, phases $e^{i\theta}$ are elements of the group, not the algebra.

Answer 3

It's perfectly possible to have a real vector space in which the vectors have complex components. Remember that a real vector space consists of a set of vectors $V$ such that $a v_1 + b v_2 \in V$ for any $v_1,v_2\in V$ and $a,b\in \mathbb R$. In other words, a real vector space is one for which the field of scalars is $\mathbb R$.

To illustrate the point, observe that $\mathbb C$ can be understood as a real vector space of dimension $2$, with basis $\{1,i\}$, or as a complex vector space of dimension $1$, with basis $\{1\}$.

The Lie group $\mathrm{U}(2)$ has Lie algebra $\mathfrak{u}(2)$ which consists of all anti-hermitian $2\times 2$ complex matrices. A generic element of this set takes the form

$$\pmatrix{ai & b + c i \\ -b + ci & di}$$ for real numbers $a,b,c,d$. As a result, this is a 4-dimensional real vector space spanned by $$\pmatrix{i & 0 \\ 0 & 0} \qquad \pmatrix{0 & 0 \\ 0 & i} \qquad \pmatrix{0 & i \\ i & 0 } \qquad \pmatrix{0 & -1 \\ 1 & 0}$$