Let $E^*$ be a multiplicative generalized cohomology theory. By the suspension isomorphism we have: $$ \tilde{E^2}(S^2)\cong\tilde{E^0}(S^0)=E^0(pt) $$ So there is a special element $t\in\tilde{E^2}(S^2)$ which corresponds to $1_E\in E^0(pt)$ under the above isomorphism.

$E^*$ is said to be complex orientable if the inclusion map $i:S^2=\mathbb{C}P^1\to\mathbb{C}P^{\infty}$ induces a sujective morphism $i^*:\tilde{E^2}(\mathbb{C}P^{\infty})\to\tilde{E^2}(S^2)$. A pair $(E^*,x_E)$ of a complex orientable cohomology theory $E^*$ and a choice of an element $x_E\in\tilde{E^2}(\mathbb{C}P^{\infty})$ such that $i^*(x_E)=t$ is called an oriented cohomology theory.

It is well known that singular cohomology, complex K-theory and complex cobordism are examples of orientable cohomology theories.


According to various textbooks, for example [1, Theorem 3.10], we have the following classical result: for any complex oriented cohomology theory $(E^*,x_E)$, there is an isomorphism $$ E^*(\mathbb{C}P^{\infty})\cong E^*(pt)[[x_E]] $$

Now my question is: $E^*(\mathbb{C}P^{\infty})=\bigoplus_{n\in\mathbb{Z}}E^n(\mathbb{C}P^{\infty})$ or $E^*(\mathbb{C}P^{\infty})=\prod_{n\in\mathbb{Z}}E^n(\mathbb{C}P^{\infty})$? More generally, how is $E^*(X)$ defined for a topological space $X$?

Indeed, if I assume that $E^*(X)$ is the direct sum (as in [1, page 26]), then I found a contraddiction taking as oriented cohomology the singular cohomology with the first Chern class: $$ \mathbb{Z}[c_1]=H^*(\mathbb{C}P^{\infty},\mathbb{Z})\neq \mathbb{Z}[[c_1]] $$ Taking $E^*(X)$ to be the cartesian product I do not found any contraddiction, but it is a bit messy if compared with the standard notation in algebraic topology, where usually the cohomology ring is defined to be the direct sum. For example, if we consider K-theory, we know that $K^{2i+1}(pt)=0$ and $K^{2i}(pt)=\mathbb{Z}\beta^i$, where $\beta\in K^2(pt)$ is the Bott element. So, according to this definition $K^*(pt)=\mathbb{Z}[[\beta,\beta^{-1}]]$ and thus $$ K^*(\mathbb{C}P^{\infty})=\mathbb{Z}[[\beta,\beta^{-1}]][[x_K]] $$ where $x_K$ is the first Chern class in K-theory. This result seems a bit strange to me, since I used to know that $K^*(\mathbb{C}P^{\infty})=\mathbb{Z}[\beta,\beta^{-1}][[x_K]]$.

However, the definition of $E^*(X)$ via the cartesian productseems to be the right choice, as suggested in this previous question Is $H^*(\mathbf{C} P^\infty)=R[X]$ or $R[[X]]$?.  Someone can help? What am I missing?


[1] A.Kono, D.Tamaki-Generalized cohomology

[2] J.Lurie-Chromatic homotopy theory course http://people.math.harvard.edu/~lurie/252x.html

[3] M.Hopkins-Complex oriented chomology theories and the language of staks https://people.math.rochester.edu/faculty/doug/otherpapers/coctalos.pdf

  • $\begingroup$ I was very confused by this too when I was learning. When talking about complex oriented stuff the notation is to use the direct product. If we are considering the groups homogenously, there is no difference but once we start talking about the combinatorics, it is useful to consider them as power series. $\endgroup$ Feb 9, 2021 at 19:31
  • $\begingroup$ This seems like a duplicate of math.stackexchange.com/questions/1003299/…. In what way is your concern not answered by the answer there? Everything is resolved if you simply think of a graded object as a sequence, rather than forcing yourself to turn it into an ungraded object. $\endgroup$ Feb 9, 2021 at 19:54
  • $\begingroup$ Thank you for your reply. Yes, my question is pretty similar to the old one. I understand that a graded ring can be turned in an ungraded one using two functors: the direct sum and the Cartesian product and the best choice depends on the context. The point that is not clear to me is: what is the best choice in this case? In the result about the cohomology of the infinite projective space, E*(CP^infinity) and E*(pt) are Cartesian products or direct sums? $\endgroup$ Feb 9, 2021 at 20:50
  • $\begingroup$ You don't have to make any choice. Both sides of the statement $E^*(\mathbb{C}P^{\infty})\cong E^*(pt)[[x_E]]$ can be defined as graded rings, without turning either of them into ungraded rings. $\endgroup$ Feb 9, 2021 at 20:53
  • $\begingroup$ Thank you very much, I believe I got it! You are totally right, I think my confusions was due to my wrong habit to think graded rings as a direct sum of abelian groups with a graded multiplication instead of a family of groups with suitable maps. $\endgroup$ Feb 9, 2021 at 21:23

1 Answer 1


The correct definition is to use neither products nor direct sums, but to instead define graded rings as sequences. That is, define a $\mathbb{Z}$-graded ring $A$ to be a sequence of abelian groups $(A_n)_{n\in\mathbb{Z}}$ together with multiplication maps $A_n\times A_m\to A_{m+n}$ satisfying the usual axioms. If $A$ is a $\mathbb{Z}$-graded ring, then you can define a power series graded ring $A[[x]]$ (where $x$ has degree $d$) for which $A[[x]]_n$ is the set of formal expressions $\sum_{k=0}^\infty a_kx^k$ where $a_k\in A_{n-dk}$ for each $k$, with multiplication defined in the obvious way. The statement $$E^*(\mathbb{C}P^{\infty})\cong E^*(pt)[[x_E]]$$ is then an isomorphism of graded rings in this sense.

As you have observed, this statement is not true in general if you define both sides as rings by taking direct sums. It also does not make sense in general if you take direct products, though: $E^*(pt)$ is not even a ring in general if you take direct products! If $E$ is connective then it is a ring and you can define multiplication in the obvious way, but if $E^*(pt)$ is nontrivial in both positive and negative degrees that are arbitrarily large then there may be products that do not make sense. For instance, for $K$-theory, if you tried to define $K^*(pt)=\mathbb{Z}[[\beta,\beta^{-1}]]$ as a direct product, then how would you compute $$\left(\sum_{n\in\mathbb{Z}} \beta^n\right)^2?$$ There would be infinitely many terms that contribute to the same degree.

(I suppose you could get a statement that is always correct by defining $E^*(X)=\prod_{n\geq 0} E^n(X)\oplus \bigoplus_{n<0}E^n(X)$; i.e., you take a sort of restricted product where you can have infinitely many nonzero coordinates only in the positive direction. I don't know of any benefit to doing this instead of just leaving your graded rings as sequences, though.)


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