One could try to apply the Eilenberg-Moore spectral sequence to the pullback diagram • → X ← •, obtaining a spectral sequence TorH(X, R)(R, R) => H(ΩX, R), but could there be differentials or extension problems which differ for different spaces X with the same cohomology ring?


Tyler's comment to my earlier answer seems to give a solution; he suggests comparing the space $T=(S^3\vee S^3)\cup\_{[x,[x,y]]} e^8$ with a wedge $S^3\vee S^3\vee S^8$. It's probably easier to think about homology with the Pontryagin product. Homology of loops on on the wedge will be a tensor algebra on classes in 2,2,7 (since it's loops of a suspension). The homology of loops on Tyler-space $T$ should differ in dimension 6: the homology class [x,[x,y]] will be 0 (where x,y are now the homology generators in dimension 2), "killed" by the new attaching map. So H_6 (and thus H^6) of the two spaces have different rank.

To make this explicit, we have $S^7 \xrightarrow{f} X \rightarrow T$, where X is the wedge of two 3-spheres. The restriction of $\Omega f: \Omega S^7 \to \Omega X$ is a map $S^6 \to \Omega X$ adjoint to f, and on homology this hits the homology class corresponding to the [x,[x,y]]. The result follows because $\Omega S^7 \to \Omega X \to \Omega T$ is null homotopic. (I'm basically using the Hilton-Milnor theorem to understand $\Omega X$.)


Well, simply connected Lie groups tend to have cohomology which is an exterior algebra. For instance, SU and X = S3 x S5 x S7 x ... have the same cohomology ring. But ΩSU=BU, and ΩX isn't. Since each HΩSn is a divided power algebra, while HBU is a polynomial algebra, this should provide a counterexample, I think. I'm guessing this shows up in the Eilenberg-Moore spectral sequence as a non-trivial extension problem.

  • $\begingroup$ Nice, too bad I've reached my daily vote limit! $\endgroup$ – Ilya Nikokoshev Oct 22 '09 at 22:25
  • $\begingroup$ This doesn't give an example for what Reid asked, though because the cohomology groups are still the same. $\endgroup$ – Eric Wofsey Oct 22 '09 at 23:50
  • $\begingroup$ Oh okay. I interpreted his remark about extension problems as being about multiplicative extensions, rather than group extensions. $\endgroup$ – Charles Rezk Oct 23 '09 at 0:08
  • $\begingroup$ Yes, I intentionally worded my question the way I did, but it wasn't because I knew of an example for the version with "cohomology ring" in both places (it was because I expected the answer to the question in the title to be no even as stated here). This example is very nice and memorable, thanks! $\endgroup$ – Reid Barton Oct 23 '09 at 6:38

Complementing the other answers in this thread: while the cohomology ring of a simply connected space does not determine the cohomology of the loop space, the rational cohomology viewed as an $A_\infty$-algebra does.

Namely, the cohomology of any $A_\infty$-algebra $A$ over $\mathbf{Q}$ (in particular, of any differential graded algebra) carries an $A_\infty$-structure such that there is an $A_\infty$ map $H^\ast(A)\to A$ inducing the identity in cohomology; this $A_\infty$ structure is unique up to a non-unique isomorphism. See e.g. Keller, Introduction to A-infinity algebras and modules, 3.3 and references therein. By taking $A$ to be the rational singular cochains of a topological space $X$ we get an $A_\infty$-structure on $H^\ast(X,\mathbf{Q})$.

To each $A_\infty$ algebra $H$ there corresponds a bar construction, which is a free differential coalgebra on $H$ shifted by 1 to the left (see e.g. 3.6 of Keller's paper mentioned above). It is an old result of Kadeishvili (see MR0580645) the that if $H$ is the cohomology of a simply-connected space $X$ with the above $A_\infty$-structure, then the cohomology of the bar construction is the cohomology of $\Omega (X)$.

This also explains why we should expect a negative answer to the question as it is stated: all components of the $A_\infty$ structure on the cohomology participate in the bar construction, and not just the product.


It seems hard to come up with an example in which the cohomology groups of the loop spaces differ. I imagine something like the following should produce an example, in the context of rational homotopy:

Find two rational commutative dgas A and B whose cohomology algebras are the same as rings, but which have different massey product structures. Then I think it's likely that the derived tensor products Q ⊗A Q and Q ⊗B Q will have different cohomology. I can't find an example small enough that I'd like to compute it, though.

  • 3
    $\begingroup$ You should be able to attach an 8-cell to S^3 wedge S^3 to kill a Whitehead product [x,[x,y]] of the generators x,y of pi_3 in rational homotopy. The cohomology algebra is then the same as S^3 wedge S^3 wedge S^8 ignoring secondary operations, but the underlying DGA is different. $\endgroup$ – Tyler Lawson Oct 23 '09 at 4:49

My feeling is that Charles is on the right track with the answer above. But rather than looking for a counterexample, I think we should have a go at correcting the original question. Now I'm not quite sure over which rings the next statements work, possibly only over rings over a field of char 0. Perhaps someone knows the details better than I, but to make it work will probably require working with simplicial algebras as these carry a model structure over any ring.

The cochains of X carry a dg-algebra structure A. Since ΩX is the homotopy pullback of • → X ← • and taking cochains should preserve the relevant (co)limits (can someone help me here), then the cochains ring of ΩX is the homotopy pushout of k ← A → k, that is, the derived tensor product. We can then take cohomology.

For the next bit we probably do need characteristic 0. The cochains ring will be rather large, so to keep track of things we could take the cohomology, but remember the higher operations. Then as an infinity ring the cohomology H(A) will be quasi-isomorphic to A (which isn't necessarily true if we don't remember the higher operations). Then with that in mind we can calculate the derived form of k ⊗_H(A) k. Its cohomology should be the cohomology of the loop space.

It would be nice to have a counterexample though, how about complements of links, the cohomology rings aren't so bad to calculate (only depending on the number of links over the rationals at least). What about the loop spaces?


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