I'm studying Atiyah-Hirzebruch spectral sequence from some slides, a bit hard to follow as they are a collage of pages from various books and papers, filled with writings and deletions made by hand by my teacher.
In particular, I'm trying to piece together the calculation of $\tilde{K^i}(\mathbb CP^n)$.
A theorem (that holds replacing $\mathbb CP^n$ with any space) says that the groups $E_2^{p,q}$ in the second page of the AHSS for $\tilde{K^*}(\mathbb CP^n)$ are described as: $$E_2^{p,q}=\tilde{H^p}(\mathbb CP^n,\tilde{K^q}(S^0)).$$
Moreover I know that:
- $\tilde{K^q}(S^0)=\mathbb Z$ if $q=2k$ for an integer $k$, otherwise $\tilde{K^q}(S^0) =0$;
- $\tilde{H^p}(\mathbb CP^n, G)=G$ if $p=2k$ for a (strictly) positive integer $k$, and $\tilde{H^p}(\mathbb CP^n, G)=0$ otherwise.
Hence at the second page the situation should be the following:
- $E^{p,q}_2=\mathbb Z$ if $p=2k$, for a (strictly) positive integer $k$, and $q=2k'$, for an integer $k'$;
- $E^{p,q}_2=0$ otherwise.
At this point, with methods I cannot replicate, it is computed the page at infinity of our spectral sequence, described by these drawings.
First doubt. In the image above, the first drawing depicts exactly the situation I described for the second page, in the latter two points, but the rightmost drawing suggests that the page at infinity is a bounded version of the second page: in particular, $E_\infty^{p,q}=0$ if $|q|\gt 2n$, and there seems to be an upper bound also on $p$.
Now, I know that the subsequent pages of the spectral sequence are computed in this way. Let $E_1:=\bigoplus_{p,q} E_1^{p,q}$ and $d_1:E_1\to E_1$ be the differential homomorphism. Then $E_2:=ker(d_1)/im(d_1)$, and on this new (bigraded) abelian group we can define a differential $d_2:E_2\to E_2$, and so on until the situation stabilizes. Plus I know that, in the case of the AHSS, the $p,q$-component of the differential $d_{n+1}$ is a homomorphism $d_{n+1}^{p,q}:E_{n+1}^{p,q}\to E_{n+1}^{p+n+1,q-n}$; hence I think that $E_2^{p,q}$ should be equal to $ker(d_{n+1}^{p,q})/im(d^{p-n-1,q+n}_{n+1})$. However looking at the grid drawn above it is clear that, for any $n\gt 0$, if $E_{n+1}^{p,q}=\mathbb Z$ then both the domain of $d^{p-n-1,q+n}_{n+1}$ and the codomain of $d_{n+1}^{p,q}$ are zero, so that $E_{n+2}^{p,q}=E_{n+1}^{p,q}$.
Therefore my doubt is: how come the page at infinity is not equal to the second page, but is bounded? Since by the reasoning above, I would say that the pages stabilize already from the second one.
Second doubt. If the page at infinity was bounded like in the rightmost drawing above, it would mean that any group $\tilde{K^i}(\mathbb CP^n)$ has a filtration of the form: $$\cdots\to 0\to \mathbb Z\to\cdots\to\mathbb Z\to 0\to\cdots,$$ where the central dots substitute (say $N_i$) copies of $\mathbb Z$, and the external dots substitute zeros. What do I know from such a filtration? Can I say that $\tilde{K^i}(\mathbb CP^n)\cong \mathbb Z^{N_i}$? In the slides it seems so, but there are vague mentions to extension problems that are not explained.
I hope I have been clear in stating my issues, thanks for your patience.
PS: the uncut version of the image above, with computation of the $K$-theory.
First, the cohomology groups $H^k(\mathbb{C}P^n, \mathbb{Z})$ are $\mathbb{Z}$ for $k = 0, 2, 4, \ldots, 2n$ and are $0$ otherwise. So, the $E_2$ page is bounded.
Next, the degree of the differential on the $E_k$ page is $(k, k - 1)$. Since $E_2^{p, q}$ is only nonzero when both $p$ and $q$ are even, the differential will always miss the nonzero groups. That's because either $p + k$ or $q + k - 1$ will be odd for each $k$.
Thus the $E_\infty$ page is the same as the $E_2$ page.
Also, you'll find that $K^0(\mathbb{C}P^n) = \mathbb{Z}^{n + 1}$, which certainly has a filtration of the type you mentioned.