Why are there no instantons in the gauge group $U(1)$? I am working my way through Srednicki's QFT book and am in chapter 93.  Near the end, Srednicki says "If the gauge group is $U(1)$, there are no instantons, and hence no vacuum angle."
I'm curious to know more about why this is.
I have a couple vague ideas:

*

*From these notes (https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/8/91170/files/2020/06/Yanjun-He-QCD-vacuum-and-Instantons.pdf), the author says that there is no winding number for a $U(1)$ abelian group (check top of page 11).


*Earlier in Srednicki's chapter 93, there is a discussion that (at least for scalar fields--not sure if and how it would change for vector fields) the tunneling amplitude between different classical field configurations is zero in the infinite space limit.
Can anyone help me get some clarity on the issue?
 A: Instantons are, by definition, solutions to the equations of motion, with finite, non-zero action. For the case of a $\mathrm{U}(1)$ gauge theory, the Euclidean action, over a manifold $X$ is $\newcommand{\d}{\mathrm{d}}$
$$ S[a] = \frac{1}{2g^2}\int_X f\wedge \star f , $$
where $f=f_{\mu\nu}\ \d x^\mu\wedge\d x^\nu$ is the curvature of the gauge field, $a=a_\mu \ \d{x^\mu}$. The equations of motion are simply the vacuum Maxwell equations
$$  \d f = 0 \qquad\text{&}\qquad \d\star f=0,$$
or, defining the codifferential $\d^\dagger := -\star\d\star$,
$$  \d f = 0 \qquad\text{&}\qquad \d^\dagger f=0.$$
But these are exactly the defining equations of a harmonic 2-form. By Hodge's theorem, the space of harmonic $p$-forms on $X$ is precisely the $p$-th cohomology group $\mathrm{H}^p(X)$. Therefore, instantons in a $\mathrm{U}(1)$ gauge theory are counted by $\mathrm{H}^2(X)$.
If $X=\mathbb{R}^4$, which is the case in Srednicki, it is obviously topologically trivial, hence $\mathrm{H}^2(X) = 0$, and thus there are no instantons.
On topologically interesting manifolds, however, you can have $\mathrm{U}(1)$ instantons. For example, on $X = \mathbb{S}^2\times\mathbb{S}^2$, $\mathrm{H}^2(X;\mathbb{R})=\mathbb{R}^2$, so there are instantons labelled by two independent integers$^{(*)}$.
As such, on topologically interesting manifolds you can add a $\theta$ angle term,
$$S_\theta = \frac{i\theta}{8\pi^2}\int_X f\wedge f,$$
as the Pontryagin index is, in this case, non-zero. Note, that this is equivalent to the fact that $f$ cannot be written everywhere as $\d{a}$. The obstruction to write $f$ as $\d{a}$ lies, again, in the non-trivial cohomology. These are all different manifestations of the same thing.

$^{(*)}$the fact that they're labelled by integers rather than reals comes from the fact that flux is quantised and thus it is $\mathrm{H}^2(X;\mathbb{Z})$ counting the instantons rather than $\mathrm{H}^2(X;\mathbb{R})$
A: I'm not sure what it is you are after; the notes you cite are more than adequate.
They remind you the boundary at infinity map to your gauge group
$S^3\mapsto U(2)\sim S^1$ is trivial, $\pi_3(S^1)=0$, which you can check naively by computing the topological invariant (Pontryagin index) (19.184),
$$
Q= {1\over 8\pi^2} \int_{S^3}\!\!d^4x ~~\epsilon^{\mu\nu\lambda\rho} \partial_\mu  A_\nu \partial_\lambda  A_\rho \\  =
{1\over 8\pi^2} \int_{S^3}\!\!d^4x ~~\epsilon^{\mu\nu\lambda\rho} \partial_\lambda ( A_\rho \partial_\mu  A_\nu )=  0
$$
as a total divergence with fields vanishing at infinity.
Photon fields commute.
