$CP$ Invariance of Yang-Mills Vacua in Electroweak Theory It is well know that quantum Yang-Mills theory has a periodic vacuum structure. Consider electroweak theory. For a single generation of fermions, the theory is CP invariant. I would like to know if the periodic vacua of the theory are also CP invariant.
One expects the trivial vacuum with topological charge $n=0$ to be CP invariant, where the topological charge, defined as
\begin{equation}
n= \int d^4x \mathcal{P}(x),
\end{equation}
is the integral of the Pontryagin density $\mathcal{P}(x)$ over the spacetime manifold. However, since $\mathcal{P}(x) \sim tr\left( F \tilde{F} \right)$ is odd under CP, one generally expects gauge configurations that have non-zero topological charge to be odd under CP (?), and this includes the non-trivial vacua.
On the other hand, electroweak theory is different from QCD in that there is no explicit theta angle term in the action (more precisely, the chiral anomaly always allows us to "rotate away" such a term), so it seems to me that, as opposed to QCD, there should be no way for us to physically distinguish between the vacua and therefore they should all have the same properties under C, P and T (note however that $\textit{changing}$ vacua via sphaleron and instanton processes does have physical significance, but that is not the main concern of the question).
What is the resolution of this apparent paradox? 
Will the pure gauge configurations that can be connected to the trivial vacuum via large gauge transformations (and hence have non-zero topological charge) be even under CP or odd?
 A: There is an explicit $\theta$ term in the electroweak Lagrangian.  However, the value of $\theta$ is basis-dependent: the chiral anomaly means that certain redefinitions of the fermion fields can shift $\theta$---see, e.g., here, or section 29.5 of Schwartz's Quantum Field Theory and the Standard Model.  In the electroweak theory we can choose a basis in which $\theta = 0$, and since physics is basis-independent this $\theta = 0$ theory will make the same predictions as the theory we started with.  So the theta angle term is there, it just happens to vanish in a certain choice of coordinates.
EDIT: I think I understand the question better now.  There's an argument that the EW vacuum should break CP symmetry and there's an argument that it should respect CP symmetry, and this is the paradox.  I don't think either argument quite works, though.
The first argument says that the quantum vacuum state should not be CP invariant, because the classical vacua aren't.  In general an argument like this doesn't work.  For example, the quantum ground state of a symmetric double-well potential is reflection-symmetric, while the two classical ground states are located in one or the other well and so break that symmetry.  In a gauge theory the ground state is generally a superposition of states that correspond to classical vacua.  For example, in the pure Yang--Mills case the Hamiltonian commutes with the operator that increments the topological charge, so they can be simultaneously diagonalized and a general vacuum state will be of the form
$$
  \lvert \theta \rangle = \sum_{n} e^{-in\theta}\lvert n \rangle
$$
where $\theta$ is the vacuum angle and $\lvert n \rangle$ is the state corresponding to the classical vacuum in the sector with topological charge $n$.  (Note, in particular, that you can't change vacua via instanton processes; that is, $\langle \theta' \vert \theta \rangle \propto \delta(\theta' - \theta)$.  If you could, they wouldn't be vacua.)  Whether this state is CP-symmetric depends on the value of $\theta$.  So whether or not the EW vacuum is CP-symmetric will depend on details of the superposition.  In this theory the vacuum will be CP-invariant just in case $\theta = 0$.  The argument in the question involves a theory with matter included, but it'll still be invariant under large gauge transformations and so the quantum ground states will still be superpositions of states in different topological sectors.
The second argument says that the vacuum should be CP invariant because the Lagrangian is CP invariant.  This kind of argument generally doesn't work either: it often happens that a symmetry of the laws isn't shared by the vacuum, in which case the symmetry is said to be spontaneously broken.  That said, I don't see any for the CP symmetry to be spontaneously broken in this case.  So I expect the vacuum to be CP-invariant in the theory in the question, though I don't know anywhere the details are worked out.
