Serre spectral sequence of the quaternionic Hopf fibration

For $U_j=\{[z]:z_j\mathrm{\ne }0\}$ and unit $z$, right multiplication by $z_j^{-1}\mathrm{\mid }{z}_j\mathrm{\mid }$ gives the unique unit representative $s_j([z])$ whose $j$th coordinate is positive real. The formulas $z⟼([z],z_j\mathrm{/}\mathrm{\mid }{z}_j\mathrm{\mid }),([z],\lambda )⟼s_j([z])\lambda$ are inverse bundle charts $p^{-1}(U_j)\cong U_j\times S^3$. Their order is essential: all scalar multiplication is on the right. With $a=1\mathrm{/}(2(n+1))$, the normalized functions ${\rho }_j([z])=\frac{\max \{\mathrm{\mid }{z}_j{\mathrm{\mid }}^2-a,0\}}{{\sum }_i\max \{\mathrm{\mid }{z}_i{\mathrm{\mid }}^2-a,0\}}$ are defined because some norm square is at least $1\mathrm{/}(n+1)$, and their supports lie in $U_j$. Thus the finite bundle is numerable and [F1] makes it a Serre fibration.

Put $P^r={\mathbb{H}\mathbb{P}}^r$. The map $D^{4r}⟶P^r,w⟼[w_0:\cdots :w_{r-1}:\sqrt{1-\mathrm{\parallel }w{\mathrm{\parallel }}^2}]$ attaches one $4r$-cell to $P^{r-1}$: the boundary lands in $P^{r-1}$, and right normalization of the last nonzero coordinate gives a unique positive-real representative on the complement. Compact-to-Hausdorff quotient descent proves the attachment map is a homeomorphism. Starting at a point gives one cell in dimensions $0,4,\dots ,4n$. Every cellular differential is zero, since adjacent cellular dimensions are never both occupied. Hence [F2, F3] give $H^k(P^n;\mathbb{Z})=\{\begin{array}{ll}\mathbb{Z},& k=0,4,\dots ,4n,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$

The CW structure has one vertex and no one-cells, so the base is simply connected and the fiber system is constant. Its groups and the base groups are free, so the multiplicative sequence has $E_2^{p,q}=H^p(P^n;\mathbb{Z})\otimes H^q(S^3;\mathbb{Z}),$ with rows only at $q=0,3$ and columns only at $p=0,4,\dots ,4n$. Thus the only possible differential is $d_4:E_4^{p,3}\to E_4^{p+4,0}$. The total sphere has cohomology only in degrees $0$ and $4n+3$ by [F3], so every $d_4$ with $0\le p<4n$ is an isomorphism of infinite cyclic groups; the upper-right class at $(4n,3)$ survives.

Put $x=d_4(u)$. The first isomorphism makes $x$ a primitive generator of $H^4(P^n;\mathbb{Z})$. Since $d_4$ is zero on the bottom row and $\mathrm{\mid }x\mathrm{\mid }$ is even, the derivation rule gives $d_4(u{x}^k)=x^{k+1}(0\le k<n)\mathrm{.}$ It follows inductively that $x^k$ generates $H^{4k}(P^n;\mathbb{Z})$ for $0\le k\le n$. Dimension gives $x^{n+1}=0$, so evaluation defines a surjection $\mathbb{Z}[X]\mathrm{/}(X^{n+1})\to H^{\ast }(P^n;\mathbb{Z})$. It is injective degree by degree because every $x^k$ in the allowed range has infinite order.

The survivor $u{x}^n$ has no target because column $4n+4$ is absent and is the unique class accounting for $H^{4n+3}(S^{4n+3})$. This also eliminates any hidden extension: the actual base cup powers were computed and each relevant degree has one cyclic group. For $n=1$, $d_4(u)=x$ is the sole differential and $ux$ is the top survivor. The unit, zero groups, first and last columns, both rows, both differential endpoints, both ring-map directions, and reversal of the orientation generator are explicit. AC occurs only through [F1], [F3], and [F4]; the finite quaternionic charts and cell argument make no choices. There is no converse or splitting claim.