The Groove of Math

or in other words, the Devil is in the Details

Filling the Details #1: “Generalised Cohomology of the projective space via Atiyah-Hirzebruch Spectral Sequence”

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So the aim of this article will be to fill out the details of the computation of \mathbb{E}^*(\mathbb{C}P^n) one can find in several books like “Adams’ Stable Homotopy and Generalised Homology” Lemma 2.52.5 page 39 or “Kochman’s Bordism, Stable Homotopy, and Adams Spectral Sequence Proposition 4.3.2 page 132. I feel the amount of details one usually found in these references are quite scarce. It could be a problem, since the topic is rather technical instead. Here is my attempt in filling the details.

Here is the Proposition :

Let E be an oriented spectrum with orientation x. Then:

  1. E^*(\mathbb{C} P^n)\cong (\pi_*E [i_n^*(x)])/((i_n^*(x))^{n+1}) where i_n\colon \mathbb{C} P^n\to \mathbb{C} P^{\infty} is the inclusion map.
  2.  E^*(\mathbb{C} P^{\infty})\cong \pi_*E[[x]]
  3. E_*(\mathbb{C} P^{n})=\pi_*E\{\alpha_0, \dots , \alpha_n\} where \alpha_k is the dual basis element of x^k under the pairing E^*(\mathbb{C}P^n)\otimes E_*(\mathbb{C} P^n)\to \pi_*E.
  4. E_*(\mathbb{C} P^{\infty})=\pi_*E\{\alpha_k\mid k\geq 0\} where \alpha_k is the dual basis element of x^k under the pairing E^*(\mathbb{C}P^{\infty})\otimes E_*(\mathbb{C} P^{\infty})\to \pi_*E.
  5.  E^*(\mathbb{C} P^{\infty}\times \mathbb{C} P^{\infty})\cong \pi_*E[[x_1,x_2]]
  6. E_*(\mathbb{C} P^{\infty}\times \mathbb{C} P^{\infty})\cong \pi_*E\{\alpha_j\otimes \alpha_k \mid j,k\geq 0\}

Before jumping to the proof, let us recall the concept of oriented spectrum.

Def: Let i \colon \mathbb{C}P^1 \to \mathbb{C}P^{\infty} denote the inclusion map. A ring spectrum (E,\mu,\imath) is called oriented spectrum if the following conditions are satisfied:

  • \pi_*E  is bounded below
  • there is a class x^E \in \widetilde{E}^2(\mathbb{C}P^{\infty}), called orientation class, such that i^*(x^E)=\Sigma^2 \imath, the canonical generator of  \widetilde{E}^2(\mathbb{C}P^1)

 

Let me fix some notation: i_n \colon \mathbb{C}P^n \to \mathbb{C}P^{\infty} is the usual inclusion, and y_E is the image of the orientation class in the unreduced second cohomology group of \mathbb{C}P^{\infty}.

Clearly the Eilenberg-Maclane Spectrum HR is oriented for any ring R. The Thom spectrum MU is oriented too (check on Kochman’s book, chapter 4.3)

I will prove only point 1 of the proposition for now. Maybe later I will update this article in order to include the proof of the other points, but really, there is nothing special as soon you know how to prove the first. 

So let me prove the first point for you:

Proof of (1): The first step is to identify a certain element in the AHSS for \mathbb{C}P^1 with the orientation i_1^* y_E. Recall that H^*(\mathbb{C}P^n)\cong \mathbb{Z}[y]/(y^{n+1})

Claim 1: The element y \otimes \imath \in E^{2,0} represents the orientation class i_1^*(y_E).

Proof of Claim 1: Consider the AHSS for \mathbb{C}P^1:

Screen Shot 2016-06-17 at 18.09.12

Since the edge homomorphism for the AHSS is always surjective, we have that the only possible non-zero differentials (the differentials of the 2nd page starting from the zero-th column to the second one) are trivial. This implies that the spectral sequence collapses. Since the second page is generated multiplicatively by y\otimes \imath \in E^{2,0}_2 and it is a free graded \pi_*E-module (i.e. the extension problem is trivial) the isomorphism E^2(\mathbb{C} P^1)\xrightarrow{\cong} E^{0,2}_2 \oplus E^{2,0}_2 maps i_1^*(y_E) \mapsto y \otimes \imath ( E_2^{0,2} =E^2(\ast)  and the other is the reduced cohomology group).

 

Claim 2: The AHSS for \mathbb{C} P^n collapses at the second page and we have that y\otimes \imath \in E^{2,0}_2 represents the orientation class i_n^*(y_E)

Proof of Claim 2:  Let us have a look at the AHSS for \mathbb{C} P^n:

Screen Shot 2016-06-17 at 18.18.46

Since the AHSS is multiplicative and thanks to the ring structure of the second page, it’s enough to show that the element y \otimes \imath \in E_2^{2,0} is an infinite cycle. In fact, one proceed inductively using the fact that any other element in the previous page (i.e. in the second page) is a \pi_*E-linear combination of powers of y\otimes \imath. Consider the inclusion i_1^n \colon \mathbb{C} P^1 \to \mathbb{C} P^n. We know that i_n^*(y_E) is sent to i_1^*(y_E) by the fact that (i_1^n)^*i_n^*=i_1^*. Now recall that i_1^n induces a map of spectral sequences. Since we know that AHSS for \mathbb{C} P^n converges a priori to E^*(\mathbb{C}P^n) ( recall we assumed \pi_*E is required to be bounded below) there is an element in the stable page E_{\infty}^{p,q}, for some p,q \in \mathbb{Z} which is a representative of i_n^*(y_E). By Claim 1, we already know that the representative of i_1^*(y_E) lies in E_{\infty}^{2,0}\cong F^2 E^2(\mathbb{C} P^1)/F^3 E^2(\mathbb{C} P^1). Now suppose that the class of i_n^*(y_E) lives in F^p E^{p+q}(\mathbb{C} P^n)/F^{p+1} E^{p+q}(\mathbb{C} P^n), since the inclusion preserves the filtration, it would send our class to an element lying in F^p E^{p+q}(\mathbb{C} P^1)/F^{p+1} E^{p+q}(\mathbb{C} P^1), but since we already established that the image of i_n^*(y_E) lies in F^2 E^2(\mathbb{C}P^1) / F^{3} E^2(\mathbb{C}P^1), it must be that p\geq 2 and q=0 (Since F^pE^2(\mathbb{C} P^n)\subseteq F^{p-1}E^2(\mathbb{C} P^n)). By definition of the filtration for the cohomological AHSS, if this element lies in F^p E^2(\mathbb{C}P^n) / F^{p+1} E^2(\mathbb{C}P^n) for p>2 in particular its representatives lie in F^2 E^2(\mathbb{C}P^n) meaning that when restricted to $\mathbb{C}P^1$ they are all zero (it is crucial here that the inclusion identifies the 2 skeleton of \mathbb{C}P^n with \mathbb{C}P^1). Since we know that the restriction to \mathbb{C}P^1 of the orientation i_n^*(y_E) is a non-zero element, it must be p=2 and q=0. This shows that we have to look for a representative for i_n^*(y_E) in E_{\infty}^{2,0}. Consider the following diagram, where E_r'^{p,q} will denote the group in position p,q, page r of the AHSS for \mathbb{C}P^n:

Screen Shot 2016-06-17 at 18.34.20

The lower map is an isomorphism since it is the map induced between the second singular cohomology groups of \mathbb{C} P^1 and \mathbb{C} P^n. The diagram implies that the unique preimage of the representative of i_1^*(y_E)\in E^2(\mathbb{C}P^1), which by Claim 1 is identified y \otimes \imath, can only be the element y \otimes \imath in E_2'^{2,0}, and therefore it has to be an element of E_{\infty}'^{2,0} i.e. an infinite cycle. This readily implies that the AHSS collapses at the second page, since it is multiplicative: the second page is generated multiplicatively by $latex y \otimes \imath$, and we just showed that it is an infinite cycle, by an easy inductive reasoning we have that the differentials in every page must be zero since they are \pi_*E-linear derivations.

Since the stable page is a free graduated \pi_*E module, the extension problem is trivial. Therefore we have an isomorphism of \pi_*E-modules E^m(\mathbb{C}P^n) \cong \bigoplus_{p+q=m}E_{2}^{p, q}.

Using the fact that AHSS is multiplicative, and the ring structure on the stable page is the one one would expect to have, by compatibility of the two product structures (by axiom of multiplicative spectral sequence)  and the above iso we have the claim. There is a nice discussion on Math.StackExchange about this proposition: I will update this answer accordingly

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Author: RicPed

Young mathematician :)

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