In wikipedia is stated that a complex-orientable cohomology theory is a multiplicative cohomology theory $E$ such that the restriction map $E^2(\mathbb{C}\mathbf{P}^\infty) \to E^2(\mathbb{C}\mathbf{P}^1) = \mathbb{Z}$ is surjective. An element $t \in E^2(\mathbb{C}\mathbf{P}^{\infty})$ that restricts to the canonical generator of the reduced theory $\widetilde{E}^2(\mathbb{C}\mathbf{P}^1)$ is called a ''complex orientation''.

Subsequently there is an explanation how $t$ gives rise to formal group law on $E^*(\mathbb{C}\mathbf{P}^\infty) = \varprojlim E^*(\mathbb{C}\mathbf{P}^n)$:

Let $m$ be the multiplication

$$ \mathbb{C}\mathbf{P}^\infty \times \mathbb{C}\mathbf{P}^\infty \to \mathbb{C}\mathbf{P}^\infty, ([x], [y]) \mapsto [xy] $$

where $[x]$ denotes a line passing through $x$ in the underlying vector space $\mathbb{C}[t]$ of $\mathbb{C}\mathbf{P}^\infty$.

There are several issues in this approach that are not clear at all. What does the notation "$xy$" resp. $[xy]$ mean? Is it tacitly exploited that $\mathbb{C}\mathbf{P}^\infty$ has a natural multiplicative structure?

Moreover why is this structure as suggested in last sentence identical to ring of polynomials $\mathbb{C}[t]$?


1 Answer 1


Here $\mathbb{CP}^\infty$ is identified with the projectivization of the polynomial ring $\mathbb{C}[t]$ as a vector space over $\mathbb{C}$. So, a point of $\mathbb{CP}^\infty$ is a 1-dimensional subspace of $\mathbb{C}[t]$, or equivalently a nonzero element up to scalar multiples. This then induces a multiplication map on $\mathbb{CP}^\infty$: given two points of $\mathbb{CP}^\infty$, represented by elements $x,y\in\mathbb{C}[t]$, you can multiply the polynomials $x$ and $y$ to get another nonzero polynomial $xy$ which represents another element of $\mathbb{CP}^\infty$. Changing $x$ or $y$ by a scalar multiple will change $xy$ by a scalar multiple, so this gives a well-defined map $\mathbb{CP}^\infty\times\mathbb{CP}^\infty\to\mathbb{CP}^\infty$.

(The Wikipedia article seems to make the very unfortunate choice of using "$t$" to denote both the complex orientation and the variable in the polynomial ring $\mathbb{C}[t]$. These are two totally different things which should be given different names.)

  • $\begingroup$ I see, that's very enlightening, thank you. What I not see if this projectivization is in some way compatible with truncations of of $\mathbb{C}[t]$? Indeed $\mathbb{CP}^\infty$ is naturally given as colimit $\bigcup_n \mathbb{CP}^n $, while $\mathbb{C}[t]$ comes which natural truncations to $\mathbb{C}[t]/t^{n+1}$. Does the projectivization of $\mathbb{C}[t]$ in your answer prove any interesting natural relation between the finite proj spaces a $\mathbb{CP}^n $ and the $\mathbb{C}[t]/t^{n+1}$? $\endgroup$
    – user267839
    Jul 8, 2022 at 17:14
  • 2
    $\begingroup$ I'm not sure what you're asking. The quotient map $\mathbb{C}[t]\to\mathbb{C}[t]/t^{n+1}$ does not induce a map on the projectivizations, since it maps some nonzero elements to zero. The more natural thing to consider is instead that $\mathbb{CP}^n$ is the projectivization of the vector subspace of $\mathbb{C}[t]$ consisting of polynomials of degree at most $n$. $\endgroup$ Jul 8, 2022 at 18:37

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