Concrete example of a gauge transformation of a vector potential

I’m reading through the book Gauge Fields, Knots, and Gravity by John Baez, and trying to make sure I have a firm grasp on gauge transformations. To that end, I’ve been looking at his concrete example of a gauge transformation of the trivial GG-bundle E=M×CE=M\times\mathbb{C} where G=U(1)G = U(1). The example goes as follows:

[NOTE: This book is a book on mathematics as it applies to physics and as such there is a lot of “sloppiness”, for example many times when dealing with representations of a group, I will write gg when really I mean ρ(g)\rho(g), so please bear with me.]

The connection DD on E can be described by the vector potential AA, an End(EE)-valued 1-form. But since we have a trivial bundle E=M×CE=M\times\mathbb{C}, End(EE) = End(C\mathbb{C}) =C= \mathbb{C}. But if we think of the connecton DD as a U(1)U(1) connection, we want the components, AμA_\mu of the vector potential to live in u(1)\mathfrak{u}(1), i.e., we want AμA_\mu to be of the form Aμ(p,v)⟶(p,dρ(x)v)A_\mu(p,v)\longrightarrow(p,d\rho(x)v) where ρ\rho is the representation of U(1)U(1) on C\mathbb{C}. Since u(1)={ix | x∈R}\mathfrak{u}(1) = \{ix\ |\ x\in\mathbb{R}\}, and ρ\rho is really just the map ρ:U(1)⟶U(1)⊂C\rho:U(1)\longrightarrow U(1)\subset\mathbb{C}, this amounts to the AμA_\mu’s just being imaginary functions.

Now, we apply a gauge transformation gg: A′μ=gAμg−1+g∂μg−1A_\mu’ = gA_\mu g^{-1} + g\partial_\mu g^{-1} But U(1)U(1) is abelian, so we get A′μ=Aμ+g∂μg−1A_\mu’ = A_\mu + g\partial_\mu g^{-1} and since gg is a gauge transformation, it lives in U(1)U(1) and so has the form g(p)(p,v)=(p,ρ(g(p))v)g(p)(p,v)=(p,\rho(g(p)) v) we can write g=e−f(p)g = e^{-f(p)} so that A′μ=Aμ+∂μfA_\mu’ = A_\mu + \partial_\mu f or A′=A+dfA’ = A + dfwhich is the usual result from electromagnetism.

Now, to make sure I understand this stuff correctly I wanted to go through the same argument on my own using a different representation of U(1)U(1), namely ρ:U(1)⟶SO(2)⊂GL(R2)\rho:U(1) \longrightarrow SO(2) \subset GL(\mathbb{R}^2). In this case, we take the GG-bundle to be the trivial bundle E=M×R2E = M\times\mathbb{R}^2 and we want AμA_\mu to live in u(1)\mathfrak{u}(1) which means it has the form Aμ(p,v)⟶(p,dρ(x)v)A_\mu(p,v)\longrightarrow(p,d\rho(x)v) where x∈u(1)x\in\mathfrak{u}(1) and dρ(x)d\rho(x) is now an element of so(2)\mathfrak{so}(2) and has the form dρ(iθ)=θ(0−110)d\rho(i\theta)=\theta\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right). So, really the AμA_\mu’s are real 2×22\times 2 matrix-valued functions on MM. We apply a gauge transformation, gg as before and obtain A′μ=Aμ+g∂μg.A_\mu’=A_\mu + g\partial_\mu g. gg lives in U(1)U(1) and so has the form g(p)(p,v)⟶(p,ρ(g(p))v)g(p)(p,v)\longrightarrow(p,\rho(g(p))v) so we can write g(p)=(cos(f(p))sin(f(p))−sin(f(p))cos(f(p)))g(p) = \left(\begin{matrix}\cos(f(p))&\sin(f(p))\\ -\sin(f(p))&\cos(f(p))\end{matrix}\right) which gives A′μ=Aμ+(0−110)∂μfA_\mu’=A_\mu + \left(\begin{matrix}0&-1\\1&0\end{matrix}\right)\partial_\mu f Now, I realize this looks messy since AμA_\mu are really matrices themselves so I have (0−(A′μ)12(A′μ)210)=(0−(Aμ)12(Aμ)210)+(0−∂μf∂μf0)\left(\begin{matrix}0&-(A_\mu’)_{12}\\(A_\mu’)_{21}&0\end{matrix}\right) = \left(\begin{matrix}0&-(A_\mu)_{12}\\(A_\mu)_{21}&0\end{matrix}\right) + \left(\begin{matrix}0&-\partial_\mu f\\\partial_\mu f&0\end{matrix}\right)

And here is where I am not sure if I am right: since A_\muA_\mu is of the form \theta\left(\begin{matrix}0&-1\\1&0\end{matrix}\right)\theta\left(\begin{matrix}0&-1\\1&0\end{matrix}\right) then it must be anti-Hermitian so that (A_\mu)_{12} = (A_\mu)_{21}(A_\mu)_{12} = (A_\mu)_{21}? If this is right, then I can think of the components of the vector potential as a real-valued functions on MM that satisfy A_\mu’ = A_\mu + \partial_\mu fA_\mu’ = A_\mu + \partial_\mu f as expected. Is this a valid argument?