MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space. A free cohomology ring space is a space $Y$ and a map $X \to Y$ such that the $\mathbb Z/2$ cohomology of $Y$ is a polynomial ring with generators $a_1,...,a_n$, and the pullbacks of the generators along the maps form a basis for all the cohomology groups of $X$.

This definition may seem kind of, or extremely, strange, which is perhaps why I had to use a word salad title. The motivating example that interests me is the map $G_n^1 \to G_n^\infty$, for real or complex Grassmanians. (The map is induced by an embedding $\mathbb R^{n+1}\to \mathbb R^{\infty}$.) In either case, the latter is a free cohomology ring space of the former.

What I would like to know is if such a relationship could be made natural, that is, that there is a functor that takes a space $X$ to a space $Y$ and a map with this property that forms the appropriate commutative diagram. So far I have been unable to find one, and proving that a functor does not exist is probably beyond my command of category theory or topology.

share|cite|improve this question
What would be more likely is that you can get a space/spectrum that is free as a module over the steenrod algebra that relates to your space. – Sean Tilson Jan 24 '12 at 3:26
Note that while it doesn't work over $Z_2$ by Peter May's and Tom Goodwillie's answers it does work over $\mathbb Q$ if instead of an actual map $X\to Y$ you are willing to settle for its homotopy class. just take $Y$ to be the product of $K(H^n(X,\mathbb Q),n)$ with natural map $X\to Y$ coming from $H^n(X,H^n(X,\mathbb Q))=[X, K(H^n(X,\mathbb Q),n)]$. This correspondence is clearly functorial in $X$. – Vitali Kapovitch Jan 24 '12 at 4:38
For connected $X$ with a choice of basepoint, you can take $Y$ to be the infinite symmetric product $SP^\infty X$, and then the natural inclusion $X \to SP^\infty X$ gives a functorial construction with gives you the desired map on rational cohomology back. If, instead of the infinite symmetric product (the free abelian monoid on $X$) you take the free $\mathbb{Q}$-vector space (which has a canonical choice of topology), you get the same result and it provides a functorial point-set version of Vitali Kapovitch's construction. – Tyler Lawson Jan 24 '12 at 12:35
If, instead of either of these, you construct the free $\mathbb{Z}/2$-vector space on $X$, you get a space $Y$ where the map back on mod-2 cohomology has the following property. There exists a lift of the generators of the cohomology of $X$ to $Y$, and given any lift $\{a_i\}$ the cohomology of $Y$ is free, in the category of algebras with Steenrod operations satisfying the instability relations, on the generators $\{y_i\}$. – Tyler Lawson Jan 24 '12 at 12:39
(In both of these I'm assuming that the homology of $X$ is finitely generated in each degree. Otherwise, you have to be a lot more careful.) – Tyler Lawson Jan 24 '12 at 12:40
up vote 14 down vote accepted

Not all polynomial algebras over $\mathbf{Z}/2$ on generators of chosen degrees are realizable as the mod $2$ cohomology of a space, but any set of generators of any such (connected) polynomial algebra forms the basis for the mod $2$ cohomology of a space $X$ (it can be chosen to be a wedge of spheres). Therefore there can be no such functor.

share|cite|improve this answer
Which ones aren't, and why? A reference would of course work as well. – Will Sawin Jan 24 '12 at 2:01
A polynomial ring on one generator in dimension $3$ is impossible. In fact, the squaring map $H^3\to H^6$ in mod $2$ cohomology must be zero if $H^5=0$ because (1) for $x\in H^n$ with mod $2$ coefficients the Steenrod square $Sq^nx$ is the cup product $x\cup x$ and (2) $Sq^3=Sq^1\circ Sq^2$. – Tom Goodwillie Jan 24 '12 at 2:23
The problem of exactly which polynomial rings can be realized was known as the Steenrod problem, and has a complete solution. It turns out that e.g., for algebras over Z/2 the only ones possible are those arising form classifying spaces of compact Lie groups, and one extra one with generators in degree 15, 14, 12, and 8 related to the exotic 2-compact group DI(4). See "The Steenrod problem of realizing polynomial cohomology rings. J. Topology 1 (2008), 747-760" and "The classification of 2-compact groups. J. Amer. Math. Soc. 22 (2009), 387-436" for much more info. – Jesper Grodal Jan 30 '12 at 10:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.