# Does a graded vector space isomorphism between the homology of two loop spaces imply the existence of an algebra isomorphism?

Consider ordinary homology with coefficients in a field. For $X$ a path-connected pointed space, the graded vector space $\bigoplus_{q\ge 0} H_q(\Omega X)$ has the structure of an algebra with the multiplication induced as follows: $$H_p(\Omega X) \otimes H_q(\Omega X) \rightarrowtail H_{p+q}(\Omega X \times \Omega X) \to H_{p+q}(\Omega X)$$ Here, the first map is a monomorphism by Kunneth's theorem. The second map is induced by the concatanetion of loops $\Omega X \times \Omega X \to \Omega X$.

Let two path-connected pointed spaces $X$ and $Y$ be given. Suppose that, for each $q\ge 0$, the two vector spaces $H_q(\Omega X)$ and $H_q(\Omega Y)$ have the same dimension. Hence $\bigoplus_{q\ge 0} H_q(\Omega X)$ and $\bigoplus_{q\ge 0} H_q(\Omega Y)$ are isomorphic as graded vector spaces. Is it necessary that $\bigoplus_{q\ge 0} H_q(\Omega X)$ and $\bigoplus_{q\ge 0} H_q(\Omega Y)$ are isomorphic as algebras?

• This doesn't seem like such a natural question... why would you hope for this ? To build a concrete counterexample, I think it might be useful to know that rationally, a theorem of Moore-Milnor says that if $X$ is simply connected $H_*(\Omega X) \cong UL$, the universal enveloping algebra of some graded Lie algebra. Just take an abelian graded Lie algebra and a non-abelian one on the same vector space. – Daniel Pomerleano Nov 4 '14 at 13:38
• Ah, in the example I have in mind, in fact $Y= \Sigma Y'$ is the suspension of some other space, and hence is simply connected. Noting Tyler's answer, I now realize that this added condition of $Y$ being a suspension space is crucial. – user2529 Nov 4 '14 at 13:58
• Right, but I'm explaining that this doesn't actually help even if you assume your space is simply connected. – Daniel Pomerleano Nov 4 '14 at 13:59

Let $X = B\Bbb Z/4$ and $Y = B(\Bbb Z/2 \times \Bbb Z/2)$ be classifying spaces for the two groups of order four. Then, as loop spaces, $\Omega X$ and $\Omega Y$ are homotopy equivalent to the discrete spaces $\Bbb Z/4$ and $\Bbb Z/2 \times \Bbb Z/2$ respectively.
Their rational homology groups are the same, because the underlying spaces $\Omega X$ and $\Omega Y$ are homotopy equivalent: both of them are homotopy equivalent to a discrete set with four points. However, $H_*(\Omega X)$ is the group algebra $\Bbb Q[\Bbb Z/4]$ and $H_* (\Omega Y)$ is the group algebra $\Bbb Q[\Bbb Z/2 \times \Bbb Z/2]$. Those are nonisomorphic rings: one has three prime ideals, and the other has four.
Here's a partial positive answer, in response to your comment. Suppose $X=\Sigma X'$ and $Y=\Sigma Y'$ are suspensions of connected spaces whose homology is finitely generated in each degree. Then if $H_*(\Omega X)$ and $H_*(\Omega Y)$ have the same dimension in each degree, they are isomorphic as algebras. Indeed, a theorem of Bott and Samelson says that $H_*(\Omega\Sigma X')$ is naturally isomorphic to the tensor algebra on the reduced homology $\tilde{H}_*(X')$. Under the assumed finiteness conditions, the dimensions of $\tilde{H}_*(X')$ can be recovered from the dimensions of $H_*(\Omega \Sigma X')$, and so the algebra structure on the latter is uniquely determined (up to isomorphism) by its dimensions.
• Thank you Eric for your answer. You answer together with Tyler's answer leaves just one case, where we know only that one of $X$ or $Y$ is the suspension of a connected space. – user2529 Nov 4 '14 at 14:58
• Of course, for simplicity, we can assume that both $X$ and $Y$ has homology which is finitely generated in each degree. – user2529 Nov 4 '14 at 14:59