No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the divergence of the magnetic field does no vanish (it has a source/sink there!), and hence the equation that allows us to define the gauge field, namely $\mathrm{d}F = 0$, is not fulfilled.
It is valid everywhere else though, and so we consider the gauge theory on space with a single point removed. But $\mathbb{R}^3 - \{0\}$ is homotopically equivalent to the sphere $S^2$, which is topologically non-trivial - you cannot contract the sphere or $\mathbb{R}^3-\{0\}$ smoothly to a point because the monopole is "in the way". This is not a true defect on spacetime, but simply a consequence of us "rescuing" the gauge description although the monopole prohibits us from doing so globally. It is a defect in the gauge theory.
The local description on the sphere $S^2$ is most easily achieved by just taking two local gauge fields that are defined on the hemisphere overlapping at the equator, and we get a globally consistent solution of we define a gauge transformation on the overlap that glues the local solutions together1, which is just a circle - $\mathrm{U}(1)$ again. So, we must give a smooth map $\mathrm{U}(1)\to\mathrm{U}(1)$ from the circle to itself. Such maps are just given by winding the circle $n$ times around itself, if you write $\mathrm{U}(1) = \{\mathrm{e}^{\mathrm{i}\phi} \vert \phi\in[0,2\pi)\}$, then the transition function is $\mathrm{e}^{\mathrm{i}\phi}\mapsto\mathrm{e}^{n\mathrm{i}\phi}$. This transition function fully characterizes the bundle, hence there are now $\mathbb{Z}$ different bundle structures that can occur. The $n\in\mathbb{Z}$ just characterizes the monopole as having magnetic charge $\frac{2\pi}{e}n$.
Now, the "Dirac string" is merely an artifact that occurs if we try to force a global solution out of the local ones. If you take the solution on one of the hemispheres and extend it as far as you can, you find that you can extend it to everywhere but the opposite pole. If we "morph back" the sphere into $\mathbb{R}^3-\{0\}$, then the pole (like all points) morphs into a ray starting at the location of the monopole. On this ray which corresponds to the pole, the solution is not defined - this is the famous Dirac string. But recall that we had another local solution - if we glue them again (or switch between them as we need to), we get a description on the whole of $\mathbb{R}^3 - \{0\}$, and the Dirac string vanishes. The quantization as $\frac{2\pi}{e}$ for the magnetic charge arises from the requirement that this artifact must be undetectable in the local solutions, and hence is not allowed to cause a physical consequence, and only for these magnetic charges, the Aharonov-Bohm effect from such a ray vanishes.
The gluing of the local solutions in fact is a construction of the principal bundle without us explicitly saying so known as the construction of a bundle by cocycles. If we have a global solution, then we need no gluing, and we choose $n = 0$, which corresponds to the trivial bundle, and no magnetic monopole, since the solution is then, in principle, extendable to all of $\mathbb{R}^3$ and we made a whole fuzz about nothing.
1The gluing technically works like this - take the gluing transformation $\chi : S^1 \to \mathrm{U}(1)$ and the two local solutions $A_1,A_2$ and set $A_2 = A_1 + \mathrm{d}\chi$. Or rather, look at your local solutions and find $\chi$ such that this works.
Best Answer
I) Here we will give an explanation at the physics level of rigor. We are doing QM (as opposed to QFT) with a 1D position target space. Coleman et. al. in Ref. 1 are ultimately interested in the Minkowskian partition function/path integral
$$ \begin{align}Z^M~=~& \langle x_f | \exp\left[-\frac{iH \Delta t^M e^{-i\epsilon}}{\hbar} \right] | x_i \rangle \cr ~=~&N \int [dx] \exp\left[\frac{iS^M[x]}{\hbar} \right], \end{align}\tag{A} $$
with Minkowskian action
$$ S^M[x]~=~\int_{t^M_i}^{t^M_f} \! dt^M \left[ \frac{e^{i\epsilon}}{2} \left(\frac{dx}{dt^M}\right)^2-e^{-i\epsilon}V(x)\right],\tag{B} $$
because this connects most easily to physics, such as e.g., unitarity, optical theorem, and decay rates (as opposed to Euclidean signature). We have included Feynman's $i\epsilon$-prescription in order to make the argument of the exponential (A) have an infinitesimal positive real part to help convergence. The corresponding Euclidean partition function/path integral is
$$ \begin{align} Z^E~=~& \langle x_f | \exp\left[-\frac{H \Delta t^E e^{i\epsilon}}{\hbar}\right] | x_i \rangle \cr ~=~&N \int [dx] \exp\left[-\frac{S^E[x]}{\hbar} \right], \end{align}\tag{C} $$
with Euclidean action
$$ S^E[x]~=~\int_{t^E_i}^{t^E_f} \! dt^E \left[ \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\right)^2+e^{i\epsilon}V(x)\right].\tag{D}$$
The Minkowskian and Euclidean formulations are connected via a Wick rotation
$$ t^E e^{i\epsilon}~=~e^{i\frac{\pi}{2}} t^M e^{-i\epsilon}. \tag{E} $$
In anticipation that we might hit branch cuts & singularities at the imaginary and real time axes, we have shorten the $\frac{\pi}{2}$ Wick rotation with an infinitesimal angle $\epsilon$ at both ends of the Wick rotation. In other words, we have inserted an $i\epsilon$-prescription in the Euclidean partition function (C) as well. If we didn't do this, the Euclidean partition function (C) would be manifestly positive (possibly infinite), and it would be impossible to derive the main complex result of Ref. 1,
$$ {\rm Im}(Z^E)_{\text{one bounce}} ~\approx~\frac{Nz_1e^{-\frac{S^E[\bar{x}]}{\hbar}}}{2\sqrt{|\det^{\prime}A|}},\tag{2.23} $$
where $A$ and $z_1$ are defined in eqs. (I) and (L) below, respectively. The prime in eq. (2.23) means that zero-modes should be excluded.$^1$
II) The evaluation of the Euclidean path integral (C) uses the method of steepest descent (MSD), where $\hbar$ is treated as a small parameter. It is an Euclidean version of the WKB approximation. The steepest descent formula explicitly displays a quadratic approximation to the Euclidean action (D) around saddle points. The MSD integration contour should pass through a saddle point in the direction of steepest descent. One should realize that the higher orders of the action (D) implicitly enter in the justification of MSD approximation, cf. e.g. Section VII below.
III) It would be an interesting exercise to implement the $i\epsilon$-prescription from Section I consistently in what follows. However, here we shall just use it for the evaluation of (what naively appears to be) an unbounded Gaussian integral (if one ignores non-quadratic contributions):
$$\begin{align} \int_{\mathbb{R}} & \!\frac{dc_0}{\sqrt{2\pi\hbar}} \exp\left[-\frac{e^{i\epsilon}V(c_0)~\Delta t^E}{\hbar}\right] \cr ~=~&\int_{\mathbb{R}} \!\frac{dc_0}{\sqrt{2\pi\hbar}}\exp\left[\frac{|\lambda_0|}{2\hbar} \left(e^{i\frac{\epsilon}{2}}c_0\right)^2+\text{non-Gaussian terms}\right] \cr ~\stackrel{z=e^{i\frac{\epsilon}{2}}c_0}{=}&~ e^{-i\frac{\epsilon}{2}} \int_{-e^{i\frac{\epsilon}{2}}\infty}^{e^{i\frac{\epsilon}{2}}\infty} \! \frac{dz}{\sqrt{2\pi\hbar}}\exp\left[\frac{|\lambda_0|z^2}{2\hbar} +\text{non-Gaussian terms}\right]\cr ~\stackrel{\text{MSD}}{=}&~ e^{-i\frac{\epsilon}{2}} \int_{-e^{i\frac{\pi}{2}}\infty}^{e^{i\frac{\pi}{2}}\infty} \! \frac{dz}{\sqrt{2\pi\hbar}}\exp\left[\frac{|\lambda_0|z^2}{2\hbar} \right] \cr ~\stackrel{z=iy}{=}&~ie^{-i\frac{\epsilon}{2}} \int_{\mathbb{R}} \! \frac{dy}{\sqrt{2\pi\hbar}}\exp\left[-\frac{|\lambda_0|y^2}{2\hbar} \right]\cr ~=~&\frac{ie^{-i\frac{\epsilon}{2}}}{\sqrt{|\lambda_0|}} ~\approx~\frac{i}{\sqrt{|\lambda_0|}}. \end{align}\tag{F} $$
The upshot is that the MSD naively instructs us to integrate along the imaginary $c_0$-axis from $-i\infty$ to $+i\infty$ (as opposed to the other direction).
We shall remove all $i\epsilon$'s from now on. It will only enter the calculation to determine a sign convention for unstable Gaussian integrals a la formula (F).
IV) Next Ref. 1 considers a lopsided potential $V(x)$, cf. Fig.1.
$\uparrow$ Fig. 1. A lopsided potential $V(x)$ with a false vacuum at $x=0$ and a true vacuum at $x=\infty$.
We impose Dirichlet boundary conditions (BC)
$$ x(t^E_i)~=~ x_i~=~0~=~x_f~=~ x(t^E_f).\tag{G} $$
First we should identify the classical paths with Dirichlet BC eq. (G). There are the trivial path $x\equiv 0$, the bounce $\bar{x}$, and various (possibly repeated) combinations thereof, cf. Fig. 2.
$\uparrow$ Fig. 2. The graph in Fig. 1 turned upside down. In order to apply the stationary action principle, the Euclidean Lagrangian (D) should be of the form 'kinetic energy minus potential energy'. Hence the apparent potential becomes minus $V$. The bounce solution $t^E\mapsto \bar{x}(t^E)$ starts and ends at $x=0$ and reflects at $x=\sigma$.
Ref. 1 is interested in the contribution from precisely one bounce. The bounce solution $\bar{x}$ is determined by the fact that the 'kinetic energy plus potential energy' is conserved on-shell (and equal to zero, since that's what it was in the beginning of the bounce):
$$ \begin{align} \frac{1}{2} \dot{\bar{x}}^2-V(\bar{x})~=~&0\cr ~\Updownarrow~&\cr \dot{\bar{x}}~=~&\pm \sqrt{2V(\bar{x})} . \end{align}\tag{H} $$
We implicitly assume that $\int_0^{\sigma}\frac{dx}{\sqrt{2V(x)}} \leq \frac{\Delta t^E}{2} $, so that the bounce can be realized in the allocated time period $\Delta t^E:=t_f^E-t_i^E $. The action of the bounce becomes$^2$
$$ \begin{align} S^E[\bar{x}]~\stackrel{(D)}{=}~&\int_{t^E_i}^{t^E_f} \! dt^E \left[ \frac{1}{2}\dot{\bar{x}}^2 +V(\bar{x})\right]\cr ~\stackrel{(H)}{=}~&\int_{t^E_i}^{t^E_f} \! dt^E ~\dot{\bar{x}}^2\tag{2.13a} \cr ~\stackrel{(H)}{=}~& 2\int_0^{\sigma}\! dx \sqrt{2V(x)}. \tag{2.13b}\end{align} $$
The Euler-Lagrange (EL) eq. reads
$$ \frac{\delta S^E[\bar{x}]}{\delta \bar{x}}~=~-\ddot{\bar{x}}+V^{\prime}(\bar{x})~=~0. \tag{2.8} $$
Eq. (H) is a first integral to eq. (2.8).
V) We next expand the path integration variable $x$ around the $\bar{x}$ bounce solution
$$ \begin{align} x(t^E)~=~&\bar{x}(t^E) + y(t^E), \cr y(t^E)~:=~&\sum_{n=0}^{\infty}c_n x_n(t^E), \end{align}\tag{2.5} $$
where $x_n$ are real orthonormal eigenfunctions
$$\begin{align} \int_{t^E_i}^{t^E_f} \! dt^E ~x_n(t^E) x_m(t^E)~=~&\delta_{nm}, \cr x_n(t^E_i)~=~0~=~&x_n(t^E_f), \end{align}\tag{2.6} $$
and $\lambda_n$ are eigenvalues $Ax_n=\lambda_nx_n$ of the Hessian operator
$$ A~:=~-\left(\frac{d}{dt^E}\right)^2 + V^{\prime\prime}(\bar{x}). \tag{I} $$
The integral measure is defined as
$$\tag{2.7} [dx]~=~\prod_{n=0}^{\infty} \frac{dc_n}{\sqrt{2\pi\hbar}}. $$
The point spectrum consists of one negative eigenvalue $\lambda_0<0$; one zero eigenvalue $\lambda_1=0$; and positive eigenvalues $0<\lambda_2<\lambda_3< \ldots$. Differentiation of the EL eq. (2.8) wrt. $t^E$ yields that the velocity $\dot{\bar{x}}$ is a zero-mode: $A\dot{\bar{x}}=0$. The normalized zero-mode reads
$$ x_1~=~\frac{\dot{\bar{x}}}{\sqrt{S^E[\bar{x}]}}, \tag{2.18} $$
cf. eqs. (2.6) & (2.13a). The zero-mode reflects the time-translational symmetry of the bounce
$$ \begin{align} \bar{x}(t^E)+ \dot{\bar{x}}(t^E)~dt^E_0~=~&\bar{x}(t^E+dt^E_0)\cr ~=~&\bar{x}(t^E)+ x_1(t^E)~dc_1.\end{align}\tag{J} $$
In other words, we can identify the zero-mode $c_1$ with the central instant $t^E_0$ of the bounce
$$ \begin{align} dc_1~\stackrel{(2.18)+(J)}{=}&~\sqrt{S^E[\bar{x}]}~dt^E_0, \cr \bar{x}(t^E_0)~=~~&\sigma,\end{align}\tag{K} $$
up to an affine transformation. The integrated zero-mode contribution is therefore given by
$$ \sqrt{2\pi\hbar} z_1 ~:=~ \int \! dc_1 ~\stackrel{(K)}{=}~ \sqrt{S^E[\bar{x}]}~\Delta t^E. \tag{L} $$
In eq. (L) we have for simplicity assumed that the period of the bounce is much smaller than $\Delta t^E$. The velocity (2.18) has a zero/node at $x=\sigma$ because of energy conservation (H). This shows that there must be a nodeless eigenfunction $x_0$ with negative eigenvalue $\lambda_0<0$. We may assume without loss of generality that $x_0>0$ is positive.
VI) The quadratic action reads
$$ \begin{align} S^E_2[x]~=~&S^E[\bar{x}] +\frac{1}{2}\int_{t^E_i}^{t^E_f} \! dt^E ~y(t^E) Ay(t^E) \cr ~=~& S^E[\bar{x}] +\frac{1}{2} \sum_{n=0}^{\infty}\lambda_n c_n^2.\end{align}\tag{M} $$
If we naively apply the MSD to the quadratic action (M) we would get a purely imaginary number
$$ (Z^E)_{\text{one bounce}}^{\text{MSD}} ~\approx~\frac{iNz_1e^{-\frac{S^E[\bar{x}]}{\hbar}}}{\sqrt{|\det^{\prime}A|}},\tag{N} $$
which is twice the estimate (2.23). Here we have used the sign convention from eq. (F). The estimate (N) is unrealistic for various reasons. For starters, it looks phony that (N) doesn't have any real part. One would naively expect that the imaginary part could develop gradually, not just as an on-off effect.
VII) Discussion. Ref. 1 stresses that the Euclidian partition function $Z^E(g)$ can only be defined via analytic continuation of some parameter $g\in\mathbb{R}$ (which Ref. 1. calls $z$) to the complex plane. Since The Euclidean partition function $Z^E(g)$ is manifestly real, the analytical continuation hence satisfies Schwarz reflection principle
$$\begin{align} 2i{\rm Im} Z^E(g)~=~&Z^E(g+i0^+)-Z^E(g-i0^+), \cr g~\in~&\mathbb{R}.\end{align}\tag{O}$$
For instance, we can define a 1-parameter family of lopsided potentials$^3$
$$ V_g(x)~=~V(x)\{g\theta(x)+\theta(-x)\}.\tag{P}$$
The exact form of eq. (P) is not important. For $g<0$ ($g>0$) there is (is not) an instanton & zero-mode, respectively. The original potential $V(x)=V_{g=1}(x)$ corresponds to $g=1$. Let us vary $g$ infinitesimal near $g=1\in\mathbb{C}$ with fixed function basis $(x_n)_{n\in\mathbb{N}_0}$ for the path integral. We expect the $(c_n)_{n\in\mathbb{N}}$ integration contours (whose steepest-decent direction is along the real $c_n$-axes) to vary continously except for the $c_0$ integration (whose steepest-decent direction is along the imaginary $c_0$-axis). Since $x_0>0$ is positive, we expect the $c_0$ integration contour along the negative real half-axis to vary very little, while the positive real half-axis varies to counteract the effect of a complex $g$, cf. Fig. 3-4 and Refs. 2-3.
$\uparrow$ Fig. 3. For ${\rm Im}(g)<0$ the full integration contour for the $c_0$-variable in the complex $c_0$-plane is approximately the negative real $c_0$-axis combined with the positive imaginary $c_0$-axis.
$\uparrow$ Fig. 4. For ${\rm Im}(g)>0$ the full integration contour for the $c_0$-variable in the complex $c_0$-plane is approximately the negative real $c_0$-axis combined with the negative imaginary $c_0$-axis.
Note that the difference between the $c_0$-contours in Fig. 3-4 is the imaginary axis, i.e. the steepest-decent direction, cf. eq. (F). We conclude that the MSD estimate (N) should instead be identified as
$$ {\rm Im} (Z^E)_{\text{one bounce}}^{\text{MSD}} ~\approx~ - \frac{Nz_1e^{-\frac{S^E[\bar{x}]}{\hbar}}}{2\sqrt{|\det^{\prime}A|}}.\tag{Q} $$
This explains the half in the main formula (2.23) although we get the opposite sign.
References:
C.G. Callan, Jr. & S. Coleman, Fate of the false vacuum. II. First quantum corrections, Phys. Rev. D 16 (1977)1762.
J. Zinn-Justin, QFT & Critical Phenomena, 2002; chapter 39.
M. Marino, Instantons and large $N$; chapter 2.
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$^1$ Equations labelled by numbers (as opposed to letters) are taken from Ref. 1.
$^2$ There is missing a factor 2 in eq. (2.13b) of Ref. 1.
$^3$ It is tempting to define instead a 1-parameter family of lopsided potentials
$$ V_g(x)~=~V(x)\{g\theta(x-\sigma)+\theta(\sigma-x)\}.\tag{R}$$
For $g<0$ this is a non-negative double-well potential bounded from below. The instanton one-bounce $\bar{x}$ and the zero-mode $x_1$ are independent of $g$. Again since the zero-mode mode $x_1$ still has a node, there must still be a negative mode, i.e. it still an unstable decay. In other words, the analytical continuation (R) has achieved nothing!