KMS condition and quasi-free states In algebraic formulation of QFT, it is known that if a state is KMS with respect to some time parameter $\tau$, then the Wightman 2-point functions must satisfy certain conditions, namely stationarity $W(\tau,\tau') = W(\tau-\tau',0) \equiv W(\Delta \tau)$, complex anti-periodicity $W(\Delta t+i\beta) = W(-\Delta t)$, and some holomorphicity condition for $W(z)$ on the complex strip $\Im(z)\in (0,\beta)$.
I am fairly sure that the converse is not true, that even if a free field theory has Wightman 2-point function that satisfy these condition, the state needs not be KMS. However, I cannot find the place that says this (or the opposite claim). For example, if one restricts to a subclass of algebraic states $\omega$, can the converse be true? Would something like Gaussian states (is it quasi-free state?) suffice to make one-to-one correspondence between the KMS state and the three-properties of the Wightman functions?
For all practical purposes I just need to know this for flat space, but if the idea works for curved spacetimes in general (or at least spacetimes in which the relevant states are defined, that's good enough). I am also somewhat curious if Kerr black hole spacetimes would have something similar, since if I remember correctly there is no Hadamard states to begin with.
 A: I think that indeed the double implication only holds for quasifree states. In my understanding, any operator in the algebra can be written as linear combinations and strong limits of Weyl operators, so one only has to check that
$\omega(\alpha_t(e^{i\phi(f)})e^{i\phi(g)})$ fulfills the KMS condition for all f and g, where $\alpha_t$ is an algebra isomorphism.  If $\alpha_t$ is an isomorphism, then
$\omega(\alpha_t(e^{i\phi(f)})e^{i\phi(g)})=\omega(e^{i\alpha_t\phi(f)}e^{i\phi(g)})$.
Finally, if the state is (even) quasifree, we get
$\omega(e^{i\alpha_t\phi(f)}e^{i\phi(g)})=e^{-\frac{1}{2}\omega(\alpha_t\phi(f)^2)-\frac{1}{2}\omega(\phi(g)^2)-\omega(\alpha_t\phi(f)\phi(g))}$.
The exponents can be written in terms of the Wightman function, which shows that if the Wightman function fulfills the conditions you stated then the KMS condition will hold.
For non-quasifree states, I would say stick to the fact that the cumulants of any distribution, even a classical one, are independent, but there may be subtleties beyond my scope.
