The fifth stable offset in integral Eilenberg–Mac Lane homology vanishes

First derive the integral degree-two model from the actual principal bundle in [F1]. Its total $S^{\mathrm{\infty }}$ is contractible and its fiber $S^1$ is a marked $K(\mathbb{Z},1)$. The fibration exact sequence gives ${\pi }_i({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }})\cong {\pi }_{i-1}(S^1)$ for every $i\ge 2$, while its component segment gives ${\pi }_1({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }})=0$ because $S^1$ is connected. Thus ${\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$ is a CW $K(\mathbb{Z},2)$, with its degree-two generator marked by this connecting isomorphism. Marked uniqueness in [F1] compares it with the chosen $K_2$, so the integral ring and finite-homology base case apply to that actual model. All integral homology groups of every $K_m$ are finitely generated. Start with $K_2$ in [F1], and induct on $m$ using its path fibration. Within this induction induct on the base homology degree $k$. In the homological sequence the bottom term $E_{k,0}^2=H_k(K_m)$ has no incoming differential. Every outgoing target has bidegree $(k-r,r-1)$ with smaller base degree. Its $E_2$ group is finitely generated by the coefficient UCT, the induction on $k$, and the finite generation of the fiber homology. Hence every outgoing image is finitely generated. Only finitely many such maps occur, and the surviving subgroup is zero because the total space is contractible. The successive kernels express the bottom group as a finite extension of finitely generated images. This proves the induction; all bases are simply connected, so there is no coefficient monodromy.

We construct the mod-two calculation for the integral models, rather than use a theorem about $K({\mathbb{F}}_2,m)$. For $m\ge 2$, the algebra $H^{\ast }(K_m;{\mathbb{F}}_2)$ is polynomial on $S{q}^I{\bar{\iota }}_m$ with $I$ admissible, containing no $S{q}^1$, and $e(I)<m$. The initial case is the polynomial ring on ${\bar{\iota }}_2$ from [F1]. For the induction, powers $z^{2^j}$ of each polynomial generator form a simple system: unique binary expansion of each exponent makes products of distinct such powers a basis. The following index calculation identifies these powers without a hidden counting step. If $I=(i_1,I^{\mathrm{\prime }})$ is admissible, then $i_1=e(I)+\mathrm{\mid }{I}^{\mathrm{\prime }}\mathrm{\mid }$. When $e(I)=m$, its operation on ${\bar{\iota }}_m$ is the square of $S{q}^{I^{\mathrm{\prime }}}{\bar{\iota }}_m$ by the top-square formula. Removing the first entry repeatedly therefore expresses it as a unique $2^j$-power of an admissible sequence of excess less than $m$. Conversely, prepend $m+\mathrm{\mid }I\mathrm{\mid }$ to an admissible $I$ of excess at most $m$; it remains admissible, has excess $m$, and evaluates to the square. Thus the simple system is indexed precisely by admissible sequences with no $S{q}^1$ and excess at most $m$.

We now give the entire odd-prime calculation needed, rather than import an odd-primary operation algebra. Fix an odd prime $p$ and $R={\mathbb{Z}}_{(p)}$. In all following path sequences, $x$ denotes the fundamental base class of degree $m$ and $a$ the fundamental fiber class of degree $m-1$. Their first differential is $d_{m}a=x$; replacing a marked generator by its negative fixes the sign if necessary. Leibniz gives $d_m(a^k)=k{a}^{k-1}x$ when $a$ is even. For an odd $a$, one has $a^2=0$ since $2$ is a unit. Coefficient systems in nonzero free rows are literal copies of $R$; a torsion row uses $H^s(K_m;R\mathrm{/}p)$, not an unjustified tensor-product formula. Such coefficients vanish for $0<s<m$ by connectivity.

The fundamental fiber class has a relative lift transgressing to the fundamental base class. Put $B=K_{m+1}$, $E=PB$ and $F=\mathrm{\Omega }B$, for $m\ge 2$. Represent the marked generator by $a:(D^{m+1},\mathrm{\partial }{D}^{m+1})\to (B,\ast )$. Contract the disk to a chosen boundary point and apply $a$ to the contraction, read from that point to $x$; this gives a continuous lift $v:D^{m+1}\to PB$ with $pv=a$ and $v(\mathrm{\partial }{D}^{m+1})\subset F$. By the connecting marking in [F1], its boundary represents the marked generator of ${\pi }_m(F)$. Absolute Hurewicz in [F2], transported along the marked equivalence $K_m\to F$, makes that boundary the generator of $H_m(F;\mathbb{Z})$. Since $E$ is contractible, the singular pair boundary $H_{m+1}(E,F;\mathbb{Z})\to H_m(F;\mathbb{Z})$ is an isomorphism and the lower positive relative homology vanishes. Thus $v$ represents its generator, which $p$ sends to the marked generator of $H_{m+1}(B,\ast ;\mathbb{Z})$. The UCT and naturality of evaluation now give $\delta {\bar{\iota }}_m=p^{\ast }{\bar{\iota }}_{m+1}$ in the actual relative cohomology group: both evaluate as one on this generator. No CW structure on $(E,F)$ is asserted. By [F4] every composite $S{q}^I$ is relatively transgressive with representative $S{q}^I{\bar{\iota }}_{m+1}$. Apply the Borel comparison [F4] to the locally finite simple system from step 1.2; it gives the polynomial algebra on admissible sequences of excess at most $m$, equivalently less than $m+1$, containing no $S{q}^1$. This proves the induction, including linear independence.

For $m=3$, the fiber is $R[a]$, $\mathrm{\mid }a\mathrm{\mid }=2$, and $d_3(a^k)=k{a}^{k-1}x$, with $x^2=0$ since $\mathrm{\mid }x\mathrm{\mid }=3$. Through total degree nine, only $k\le 4$ occurs in a first-column source. If $p\ge 5$, all $1,2,3,4$ are units; the displayed pairs cancel, leaving only $R$ in base degrees zero and three through degree nine. To justify that no unlisted bottom class is hidden, take the first such base column. Its possible incoming sources have smaller base column, hence column zero or three. The zero-column powers and column-three products have just been paired by $d_3$, so no source survives to hit it. It would survive to the contractible total, a contradiction.

For $n\ge 8$, degree $n+6<2n$ contains no product of two positive-degree generators from step 2.1. Its basis is exactly $S{q}^6{\bar{\iota }}_n,S{q}^{4}S{q}^2{\bar{\iota }}_n$. Adem gives $S{q}^{1}S{q}^6{\bar{\iota }}_n=S{q}^7{\bar{\iota }}_n$ and $S{q}^{1}S{q}^{4}S{q}^2{\bar{\iota }}_n=S{q}^{5}S{q}^2{\bar{\iota }}_n$. These are distinct basis classes in degree $n+7<2n$. Thus $S{q}^1$ is injective on degree $n+6$. Every reduction of an integral class is in its kernel, since ${\beta }_{\mathbb{Z}}{\rho }_2=0$. The coefficient exact sequence embeds $H^{n+6}(K_n;\mathbb{Z})\mathrm{/}2$ into this kernel, so that quotient is zero. Finite generation from step 1.1 shows that $H^{n+6}(K_n;\mathbb{Z})$ has neither a free part nor any two-primary torsion.

For $p=3$ the same calculation has precisely one surviving group through that range: $R\mathrm{/}3$ at bidegree $(3,4)$, the cokernel on $a^{2}x$ of $d_3(a^3)=3a^{2}x$. The source $a^3$ itself dies, since multiplication by three on $R$ is injective; $a^4$ dies with coefficient four. The first-unlisted-column argument of step 2.2 makes the other columns between three and eight zero. The surviving $R\mathrm{/}3$ can only die by $d_5$ to $(8,0)$, so contractibility makes that differential an isomorphism and $H^8(K_3;R)=R\mathrm{/}3$. A differential into $(9,0)$ could only come from column zero, where $a^4$ has died, or column three with fiber degree five, which is zero; column eight would require $r=1$. Therefore $H^9(K_3;R)=0$. This proves that through degree nine the nonzero positive groups are $R$ in degree three and, only for $p=3$, $R\mathrm{/}3$ in degree eight.

The next five calculations give a finite table, with $T=R\mathrm{/}3$ if $p=3$ and $T=0$ if $p\ge 5$: for $3\le m\le 7$, through degree $m+6$ the nonzero positive groups are $R$ in degree $m$, $T$ in degree $m+5$, and also $R$ in degree $2m$ when $m=4$ or $6$. We verify every transition, including the extra degree. For $m=4$, the free fiber generator is odd of degree three; its $d_4$ maps $a{x}^k$ to $x^{k+1}$, so the free base classes through degree ten are exactly $x,x^2$. The torsion fiber class, if present, has degree eight; its only possible nonzero differential is its transgression $d_9$ to bottom degree nine. Earlier targets with nonzero base column four or eight have fiber degrees five or one, both zero; the potential target at $(6,3)$ has zero base group in degree six. All other targets have zero base column or zero fiber row. Contractibility makes $d_9$ an isomorphism onto $T$. For bottom degree ten, the first-column source has fiber degree nine and is zero by step 3.2; columns four and eight require fiber degrees five and one, and column nine would require $r=1$. Hence degree ten is zero.

For $m=5$, the free fiber classes in the required range are $a,a^2$ in degrees four and eight. They die by $d_{5}a=x$ and $d_5{a}^2=2ax$, with two a unit; the odd base class satisfies $x^2=0$. The fiber $T$ in degree nine has no earlier nonzero target: the only candidate positive base column before its transgression is column five with fiber degree five, which is zero; other positive columns below ten vanish. Its $d_{10}$ therefore gives the sole torsion group $T$ in base degree ten. An incoming differential to degree eleven has first-column source degree ten, zero by step 4.1, or column-five source fiber degree five, also zero. Thus degree eleven is zero, and the stated table holds for $m=5$.

For $m=6$, the free fundamental fiber class has degree five. The possible $d_6:T=E_6^{0,10}\to E_6^{6,5}=Rxa$ is zero because its source is torsion and its target is free. Connectivity and the already computed lower groups leave this target equal to $R$ through page six; no earlier differential changes it. The torsion class consequently survives to its transgression $d_{11}$, which is an isomorphism onto the bottom group $T$ in degree eleven. For degree twelve the first-column source is fiber degree eleven, zero by step 5.1. The sole other source is $Rxa$ at $(6,5)$ with $d_6(xa)=x^2$. Contractibility makes this surject onto the whole degree-twelve group. Its rational rank is one by [F3], so this surjection from $R$ is also injective. Hence degree twelve is $R{x}^2$, with no torsion. This explains why the torsion row at $(0,10)$ cannot be omitted even though the odd fundamental class has square zero.

For $m=7$, the fiber $T$ in degree eleven transgresses isomorphically to the bottom group $T$ in degree twelve. All earlier possible positive-column targets have fiber degree at most five and vanish. The fiber class $a^2$ of degree twelve is killed earlier by $d_7(a^2)=2ax$, an isomorphism between the free copies of $R$ at $(0,12)$ and $(7,6)$. Thus it cannot survive to hit degree thirteen. Every other positive-column source to degree thirteen again requires fiber degree at most five, below the fiber's first positive degree six. Hence $H^{13}(K_7;R)=0$. The free and torsion groups in all the stated ranges are now completely listed. The first-unlisted-column argument from step 2.2 rules out an additional bottom group at every intermediate zero degree in these finite calculations for $m=4,5,6,7$: its surviving potential sources are exactly the free pair and the indicated single torsion transgression, already accounted for.

For $m\ge 8$, a mixed positive-column/positive-row term in total degree through $m+6$ would have degree at least $m+(m-1)=2m-1>m+6$. Thus only the axes occur in the whole relevant incoming-differential region. Every positive fiber class in degrees through $m+5$ transgresses to the next base degree, since the total space is contractible. The induction table of step 7.1 therefore shifts by one: base degree $m$ is $R$, degree $m+5$ is $T$, and degree $m+6$ is zero, with all intervening groups zero. In particular $H^{n+6}(K_n;{\mathbb{Z}}_{(p)})=0$ for every odd prime $p$ and $n\ge 8$. There is no additive extension ambiguity in these assertions about the bottom axis: its entire incoming list and its zero abutment have been specified.

Steps 3.1 and 8.1 eliminate every primary torsion subgroup of $H^{n+6}(K_n;\mathbb{Z})$; localization of cohomology agrees with cohomology with localized coefficients here by the finite-generation coefficient UCT, so these are the actual primary groups. By the cohomological UCT, its torsion subgroup is $\mathrm{Ext}(H_{n+5}(K_n;\mathbb{Z}),\mathbb{Z})$. The rational computation in [F3] gives $H^{n+5}(K_n;\mathbb{Q})=0$, since $n<n+5<2n$. Finite generation from step 1.1 then kills both the torsion and free parts of $H_{n+5}$. Thus $H_{n+5}(K_n;\mathbb{Z})=0$, as claimed.