Complex K-AHSS for spheres

By [A3] the $E_2$ page has $E_2^{0,q}=K^q(\ast )$ and $E_2^{n,q}=K^q(\ast )$ for every even $q$, and vanishes in all other positions.

For $n$ even, consider a differential $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ with source in an even coefficient row $q$ at a nonzero column $p\in \{0,n\}$. If $r$ is even then the target row $q-r+1$ is odd, so the target lies in a vanishing coefficient row. If $r$ is odd then the target column $p+r$ is odd, hence is neither $0$ nor $n$ when $p=0$ (the number $n$ being even) and exceeds $n$ when $p=n$; in both cases the target column lies outside the support $\{0,n\}$ of $H^{\ast }(S^n;\mathbb{Z})$, and the parity of the target row is irrelevant. In either case the target vanishes, so every differential is zero.

Suppose $n$ is odd. If $n=1$, every differential has $r\ge 2$ by [A1], so a source in column $0$ or $1$ has target column $p+r>1$; hence every target is zero. Now let $n\ge 3$. The only possibly nonzero differentials are $d_n:E_n^{0,q}\to E_n^{n,q-n+1}$ on even rows $q$. There is no incoming differential at column $0$, so the surviving subgroup there is $\ker d_n$. On the other hand, the edge quotient $E_{\mathrm{\infty }}^{0,q}=F^0{K}^q(S^n)\mathrm{/}{F}^1{K}^q(S^n)$ identifies with the image of restriction to the basepoint. For even $q$, [A4] and Bott periodicity give $F^1={\tilde{K}}^q(S^n)=0$, while pullback along $S^n\to \ast$ splits restriction, so that image is all of $K^q(\ast )=\mathbb{Z}$. Thus $\ker d_n=\mathbb{Z}$ inside its source $\mathbb{Z}$, forcing $d_n=0$.

Steps 2.1 and 2.2 show that all differentials vanish and $E_2=E_{\mathrm{\infty }}$. If $n$ is odd, each total-degree diagonal has only one nonzero term, so there is no extension problem. If $n$ is even, an even total degree $t$ has two graded pieces, at $p=0$ and $p=n$, and the filtration gives $0\to F^n{K}^t(S^n)\cong \mathbb{Z}\to K^t(S^n)\to K^t(\ast )\cong \mathbb{Z}\to 0$. The quotient map is restriction to the basepoint and is split by pullback along $S^n\to \ast$, since the composite $\ast \to S^n\to \ast$ is the identity. Thus $K^t(S^n)\cong \mathbb{Z}\oplus \mathbb{Z}$ in even degree, with the reduced summand $F^n\cong {\tilde{K}}^t(S^n)\cong \mathbb{Z}$ from [A4]; odd total degrees vanish. Hence the only two-piece extension is split, while the odd-dimensional cases have a single graded piece.

Steps 2.1, 2.2 and 3.1 verify the displayed reduced groups and show that the sphere $K$-AHSS has no nonzero differential and no nontrivial extension.