10:33 AM
@jdc I believe it's the case that if X is an H-space, then the free loop space splits as a product $X^{S^1} = X \times \Omega X$
I suppose I must be at least assuming grouplike H-space
That would give a splitting of the cohomology ring.
Even in this case, I think it may be subtle in general to compute the cohomology ring of $\Omega X$ in terms of the cohomology of $X$?
I am the wrong person to ask, though :)
In general, of course there is a fiber sequence $\Omega X \to X^{S^1} \to X$.