(Euclideanized) QFT on $S^d$ vs $S^{d-1}\times S^1$ 
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*Broadly I would like to understand what is the difference in the physical interpretation of a (Euclideanized) QFT which is on space-time $S^d$ and which is on a space-time $S^{d-1}\times S^1$.
In the later case I am comfortable thinking of it as being a theory actually on a Lorentzian space-time where the spatial manifold is $S^{d-1}$ but the theory is heated to a temperature equal to the circumference of the $S^1$ factor. But for the first case what is the interpretation?


*Specifically consider the action of a conformally coupled Euclidean scalar on $S^d$ space-time as, $S = \frac{1}{2} \int_{S^d} d^dx \sqrt{G} \left [ (\nabla \phi )^2 + \frac{d-2 }{4(d-1) }R \phi^2 \right ]$. (where $R$ on $S_d$ is $\frac{d(d-1) }{a^2 }$)
Knowing this is it obvious as to how to write down the action for the same conformally coupled scalar theory on a spatial $S^{d-1}$ at a finite temperature?
 A: The analytic continuation of $S^d$ is de Sitter space, often denoted as $dS_d$. Euclidean QFT on $S^d$ then corresponds to Lorentzian QFT on $dS_d$. This can be seen a number of ways, but the quickest is to simply note that the sphere is the maximally symmetric Euclidean signature space with positive curvature, and de Sitter is the maximally symmetric Lorentzian signature space with positive curvature. For a rather famous paper which goes into both the continuation and de Sitter QFT in detail, see http://journals.aps.org/prd/abstract/10.1103/PhysRevD.32.3136.
Just a few additional comments: Euclidean QFT on $S^{d-1}\times S^1$ can be continued to Lorentzian QFT on $S^{d-1}$ at finite temperature, or it could be continued to be Lorentzian QFT on $dS_{d-1} \times S^1$, depending on how the continuation is done. Also, the action for a conformally coupled scalar does not depend on the spacetime when written covariantly as you wrote it above--therefore the formula you have is general and applies to any background manifold.
