The first ring seems to be what one learns first: the underlying group is the cohomology of the total singular cochain complex $C^*(\mathbf{C} P^\infty)$, which is defined as $\oplus C^n(\mathbf{C} P^\infty)$, so we can't possibly get the power series ring.

On the other hand, eventually one learns that cohomology is actually a representable functor in the stable category. Since $\mathbf{C} P^\infty=\varinjlim \mathbf{C} P^n$, perhaps we should have $H^*(\mathbf{C} P^\infty)=\varprojlim H^*(\mathbf CP^n)=R[[X]]$. At first glance this is a logical leap from $H^n$ being representable as a functor to $R$-modules to $H^*$ being representable as a functor to $R$-algebras, but I think this actually follows by abstract nonsense by the naturality of the ring structure on homs coming from a ring object.

Of course this could all be generalized to any infinite-dimensional CW complex, or further, and to more general cohomology theories $E$. I bring it up because the distinction seems in this case seems to matter for complex bordism theory and more generally for complex oriented cohomology theories. Specifically, a complex orientation gives rise to a formal group law in $E^*(\mathbf{C}P^\infty\times\mathbf{C} P^\infty)$, which only makes sense if the latter is a power series ring, not a polynomial ring. But ordinary cohomology is complex oriented, so we see again $H^*(\mathbf{C}P^\infty)$ should be the power series ring.

It seems like the argument in favor of $R[[X]]$ is more solid. But this would imply that the singular chain complex doesn't actually model the cohomology of a space, and would seem to cause problems with modeling spaces in general via $A^\infty$-algebras, which are in particular graded algebras, that is, direct sums of their terms...and don't we want to be able to do this? Maybe we actually want different rings in different situations? Maybe I'm missing something important in the second paragraph? Color me confused! Thanks in advance to anyone who can clarify this.

EDIT: Lurie seems to write down both possibilities in the same lecture notes: in the second paragraph here, the polynomial ring, whereas here, Example 5 showing that ordinary cohomology is complex orientable combined with Example 8 would give the power series ring.

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    $\begingroup$ Your first example where Lurie writes down the polynomial ring seems to be a typo. It's certainly inconsistent with his definition of complex orientability. $\endgroup$ Nov 3, 2014 at 1:36
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    $\begingroup$ Jacob Lurie makes this issue explicit in remark 1.1 of "A survey of elliptic cohomolgy" math.harvard.edu/~lurie/papers/survey.pdf $\endgroup$ May 24, 2016 at 10:40

2 Answers 2


It depends on what you want to do. It's worth noting that you can think about $H^{\bullet}(X, \mathbb{Z})$ (say) solely as a graded ring, by which I mean a sequence $R_0, R_1, \dots$ of abelian groups together with an operation $R_n \times R_m \to R_{n+m}$ satisfying etc. In particular, this definition does not require that I commit to a choice of forgetful functor from graded rings to ungraded rings. There are at least two such functors I could write down:

  • $\bigoplus_i R_i$, which on $\mathbb{CP}^{\infty}$ gets me $\mathbb{Z}[x]$, or
  • $\prod_i R_i$, which on $\mathbb{CP}^{\infty}$ gets me $\mathbb{Z}[[x]]$.

There are several arguments in favor of $\prod_i H^i(X, \mathbb{Z})$ being the thing you actually want. The basic one is that there are many situations in which you want to write down sums of cohomology classes (e.g. characteristic classes) living in the bigger ring rather than the smaller ring. The story about complex cobordism and formal group laws is a big one, but there are simpler ones: for example, the Chern character of a vector bundle lives in this bigger ring (tensored with $\mathbb{Q}$). More abstractly, $\prod_i H^i(X, \mathbb{Z})$ has the benefit of being representable by a space, namely the corresponding product $\prod_i B^i \mathbb{Z}$ of Eilenberg-MacLane spaces.

(By the way, this is not what people mean when they say that cohomology is representable in the stable category. What that means is that there is a spectrum $H \mathbb{Z}$ such that the spectrum of maps from (the suspension spectrum of) a space $X$ into $H \mathbb{Z}$ is a spectrum whose homotopy groups reproduce the cohomology groups $H^i(X, \mathbb{Z})$. The thing that is representable is a functor that outputs spectra, not graded abelian groups.)

But this would imply that the singular chain complex doesn't actually model the cohomology of a space

It implies nothing of the sort; singular cohomology gives you a graded ring in the above sense, and as mentioned you don't need to commit yourself to a choice of forgetful functor down to ungraded rings to say this.

and would seem to cause problems with modeling spaces in general via $A^{\infty}$-algebras

Do you mean thinking of a space $X$ in terms of cochains $C^{\bullet}(X, \mathbb{Z})$ on $X$? Again, this is a graded object and you don't need to commit yourself to a choice of forgetful functor down to ungraded objects.

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    $\begingroup$ This is very helpful-I had been conflating graded rings with their image under the direct sum forgetful functor. $\endgroup$ Nov 3, 2014 at 3:19
  • $\begingroup$ @Kevin: I forgot to mention a more explicit reason to prefer the product, related to two other things I said above, which is that the Chern character isomorphism for a space with cohomology in infinitely many degrees really furnishes an isomorphism between, say, $K^0(X) \otimes \mathbb{Q}$ and the product $\prod_i H^{2i}(X, \mathbb{Q})$ rather than the sum. One way to think about this is that after rationalization the representing space of K-theory, namely $\mathbb{Z} \times BU$, becomes a product of Eilenberg-MacLane spaces $\prod_i B^{2i} \mathbb{Q}$. $\endgroup$ Nov 7, 2014 at 2:28
  • $\begingroup$ This is also related to the formal group story in that this isomorphism can morally be thought of as coming from the isomorphism between the additive formal group and multiplicative formal group given over $\mathbb{Q}$ by the exponential and the logarithm respectively. But this whole formal group story relates the additive formal group to $2$-periodic ordinary cohomology, and the way in which the even and odd cohomologies are collapsed is again precisely by products as above rather than by sums (which must be the case in order for this story to fit together the way it does). $\endgroup$ Nov 7, 2014 at 2:28
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    $\begingroup$ @Riccardo a graded do is a sequence of rings with some structure. There's no one right answer as to whether it should be realized as a sum or a product. $\endgroup$ May 24, 2016 at 14:27
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    $\begingroup$ @Riccardo: I think this is a bad convention, and this discussion is my explanation why. In general, it's possible to write down a meaningful notion of "graded X-object" even when X-objects don't have a meaningful notion of direct sum. For example, a graded set is a sequence of sets. $\endgroup$ May 24, 2016 at 18:28

The limit would be taken in the category of graded rings. $R[[X]]$ is not a graded ring.

  • $\begingroup$ I suppose this raises the question of why the limit is taken in the category of graded rings? $\endgroup$
    – curious
    Nov 2, 2014 at 22:50
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    $\begingroup$ Indeed, Lurie goes ahead and takes the filtered limit in math.harvard.edu/~lurie/252xnotes/Lecture4.pdf, Example 8. $\endgroup$ Nov 2, 2014 at 22:52
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    $\begingroup$ Dear @Matt Samuel, I hope this is not a stupid question, but why $R[[x]]$ is not a graded ring? I mean, every ring is a graded ring with the trivial decomposition (everything concentrated in degree $0$) and therefore a fortiori $R[[x]]$ is a graded ring. I apologise for commenting such an old answer, but I'm really interested in this question $\endgroup$
    – Riccardo
    May 24, 2016 at 9:24

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