Let $L$ be the Lazard's universal ring, and $R=\mathbb{Z}[b_1,b_2,\cdots,b_n,\cdots]$, regarded as a graded ring with the degree of $b_i$ equal to $2i$. Let $\theta: L\rightarrow R$ be the homomorphism carrying the universal formal group law $\mu^L$ to the formal group law $$\mu^R(x_1,x_2)=\exp(\log(x_1)+\log(x_2)),$$ where the power series $$\exp(x)=x+\sum_{i\geq 1}b_ix^{i+1},$$ and $\log(x)$ its inverse, denoted as $$\log(x)=x+\sum_{i\geq 1}m_ix^{i+1}.$$ Let $MU$ be the complex cobordism spectrum, and by Quillen's theorem we have the following commutative diagram $\require{AMScd}$ \begin{CD} L @>\theta>> R\\ @V \cong V V @VV \cong V\\ \pi_*(MU) @>>h> H_*(MU;\mathbb{Z}) \end{CD}

where $h$ is the Hurewicz homomorphism.

In Section 9, Part II of

Adams, J. F., Stable homotopy and generalised homology, Chicago Lectures in Mathematics. Chicago - London: The University of Chicago Press. X, 373 p. 3.00 (1974). ZBL0309.55016.**

it is stated that the class $[\mathbb{C} P^n]\in\pi_*(MU)$ is sent to $(n+1)m_n\in H_*(MU;\mathbb{Z})$ by $h$, and it is indicated there that the argument is a Chern number computation, but I am not seeing the argument.**

I would greatly appreciate your help if you could sketch the proof or point out a reference containing a proof. Thank you!

  • 3
    $\begingroup$ I think the idea to make this a characteristic class computation is that the image of $[\mathbb{C}P^n]$ under the Hurewicz homomorphism is the same as pushing forward the fundamental class into $BU$ via classifying the tangent bundle, then using the Thom isomorphism to get a homology class in $MU$. $\endgroup$ Mar 19 '20 at 17:31
  • $\begingroup$ @Connor Malin Thank you for your helpful comment. $\endgroup$
    – Xing Gu
    Mar 20 '20 at 15:53

It is a key result that the composite $$ MU_* \xrightarrow{h} H_*(MU;\mathbb Z) \xrightarrow[\sim]{\Phi^{\vee}}H_*(BU;\mathbb Z),$$

where $\Phi^{\vee}$ is the dual of the Thom isomorphism $\Phi$, agrees with evaluating on normal Chern numbers.

In other words, $\langle \Phi(c), h([M])\rangle = \bar c(M)$ for all $c \in H^*(BU)$ and for all $[M] \in MU_*$.

(A reference in the real case is the diagram on page 228 of A concise course in algebraic topology by J.P.May. The complex case is identical.)

So the assertion is just a Chern number calculation:

$h([\mathbb CP^n]) = (n+1)m_n$ if and only if, for all $c \in H^{2n}(BU;\mathbb Z)$, $$\langle \Phi(c), (n+1)m_n\rangle = \bar c(\mathbb CP^n).$$

  • $\begingroup$ Thank you for your answer and reference! They are very helpful. $\endgroup$
    – Xing Gu
    Mar 20 '20 at 15:54

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