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 $\Omega X \to X^{S^1} \to X$.

3 hours later…
1:19 PM
not to be mistaken with the other fiber sequence, $\Omega X \to * \to X$
(which you can sometimes use to compute the cohomology of $\Omega X$)

6 hours later…
7:03 PM
@TimCampion the shadow of the Discord