# Multiplicative Structure of the Atiyah-Hirzebruch/Leray-Serre spectral sequence

This is related to this question (edit: now answered, see below). Is there a nice explanation of the multiplicative structure on the higher pages of that spectral sequence? I want to assume that $h$ is some multiplicative extraordinary cohomology theory (satisfting the wedge axiom), and I assume that $X\to B$ is a Serre fibration over a CW complex with typical fibre $F$. Then consider the Leray-Serre / Atiyah-Hirzebruch / Whitehead spectral sequence $$E^{p,q}_2=H^p(B;h^q(F))\Rightarrow h^{p+q}(X)\;.$$ Several books state that there is a cup product on each page $E_k$ such that $d_k$ satisfies a Leipniz rule, and the cup product on $E_{k+1}$ is the induced one. However, I only found a proof in G. W. Whitehead's "elements of homotopy theory", which looks rather scary. Is there a more accessible account?

Edit (again) The answers to the question mentioned above name two papers: one by Massey and one by Douady. None of these contains an actual proof, so I am still looking for a nice reference.

• Hatcher's book has a proof. Also, I don't know a reference but I think the proof is easier to follow if you work with the unrolled exact couple (like in Boardman's paper "Conditionally convergent spectral sequences" which unfortunately does not deal with multiplicative structures) – Denis Nardin Dec 8 '15 at 15:22
• Hmm... Indeed I have an old version. I'll try to sketch a proof using exact couples in an answer, since I'm unable to provide a reference. – Denis Nardin Dec 8 '15 at 16:12
• Switzer agrees with your assessment: Remark 4 of Chapter 15 outlines what you'd want to show and how you'd go about showing it, then says "Checking the details of all these statements is extremely tedious." – Eric Peterson Dec 8 '15 at 18:31
• Once a filtration is shown to be multiplicative then the associated spectral sequence is multiplicative by use of universal examples. Have a look at what Dugger does in his unpublished notes. – Sean Tilson Dec 11 '15 at 12:03
• @SeanTilson Are you referring to arXiv:0305173 and arXiv:0305187? I will have a look. – Sebastian Goette Dec 11 '15 at 14:46

Let $B$ be a CW complex and $\pi\colon X\to B$ a Serre fibration. Put $X^k=\pi^{-1}(B^k)$. A cellular approximation~$\Delta_B\colon B\to B\times B$ of the diagonal can be lifted to an approximation $\Delta\colon X\to X\times X$ of the diagonal such that $$X^k\stackrel\Delta\longrightarrow\bigcup_{m+n=k}X^m\wedge X^n\;.$$
Let $(\tilde h^\bullet,\delta,\wedge)$ be a reduced multiplicative generalised cohomology theory. We define a Cartan-Eilenberg system $(H,\eta,\partial)$ by $$H(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})$$ for~$p\le q$ with the obvious maps $\eta\colon H(p',q')\to H(p,q)$ for $p\le p'$, $q\le q'$. The corresponding exact sequences take the form $$\cdots\to\tilde h^\bullet(X^{r-1},X^{q-1})\to\tilde h^\bullet(X^{r-1},X^{p-1}) \to\tilde h^\bullet(X^{q-1},X^{p-1})\stackrel\delta\to \tilde h^\bullet(X^{r-1},X^{q-1})\to\cdots$$ We ignore the grading; it is easy to fill in.
To define a spectral product $\mu\colon(H,\eta,\partial)\times(H,\eta,\partial)\to(H,\eta,\partial)$ we consider the map \begin{multline*} F_{m,n,r}\colon(X\wedge X)^{m+n+r-1}/(X\wedge X)^{m+n-1} \cong\bigcup_{a+b=m+n+r-1}(X^a\wedge X^b)\Bigm/ \bigcup_{c+d=m+n-1}(X^c\wedge X^d)\\ \begin{aligned} \twoheadrightarrow\mathord{}&\bigcup_{a+b=m+n+r-1}(X^a\wedge X^b)\Bigm/ \Bigl(\bigcup_{a=0}^m(X^{a-1}\wedge X^{m+n+r-a}) \cup\bigcup_{b=0}^n(X^{m+n+r-b}\wedge X^{b-1})\\ \cong\mathord{}&\bigcup_{a=m+1}^{m+r}(X^{a-1}\wedge X^{m+n+r-a})\Bigm/ \bigl(X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1}\bigr)\\ \hookrightarrow\mathord{}& X^{m+r-1}\wedge X^{n+r-1}\bigm/ (X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1})\\ \cong\mathord{}&(X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\;. \end{aligned} \end{multline*}Together with the diagonal map $\Delta$, for $r\ge 1$, we define \begin{multline*} \mu_r\colon H(m,m+r)\otimes H(n,n+r) \cong\tilde h(X^{m+r-1}/X^{m-1})\otimes\tilde h(X^{n+r-1}/X^{n-1})\\ \begin{aligned} &\stackrel\wedge\longrightarrow\tilde h\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\ &\stackrel{F_{m,n,r}^*}\longrightarrow\tilde h\bigl((X\wedge X)^{m+n+r-1}/(X\wedge X)^{m+n-1}\bigr)\\ &\stackrel{\Delta_X^*}\longrightarrow\tilde h(X^{m+n+r-1}/X^{m+n-1})=H(m+n,m+n+r)\;. \end{aligned} \end{multline*}
Proposition For all $m$, $n$, $r\ge 1$, the following diagram commutes $\require{AMScd}$ \begin{CD} H(m,m+1)\otimes H(n,n+1)@>\mu_1>>H(m+n,m+n+1)\\ @A\eta\oplus A\eta A@AA\eta A\\ H(m,m+r)\otimes H(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\ @V\partial\otimes\eta\oplus V\eta\otimes\partial V@VV\partial V\\ {\begin{matrix}H(m+r,m+r+1)\otimes H(n,n+1)\\\oplus\\H(m,m+1)\otimes H(n+r,n+r+1)\end{matrix}}@>\mu_1\pm\mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)\rlap{;,} \end{CD}
Proof. The upper square commutes because the maps~$F_{m,n,r}$ are defined sufficiently naturally. For the lower square, we consider the boundary morphism $\delta$ of the triple $$(X^{m+r}\wedge X^{n+r-1}\cup X^{m+r-1}\wedge X^{n+r}, X^{m+r}\wedge X^{n-1}\cup X^{m+r-1}\wedge X^{n+r-1}\cup X^{m-1}\wedge X^{n+r},\\ X^{m+r}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r})\;.$$ The following diagram commutes: \begin{CD} \tilde h^{-p}(X^{m+r-1}/X^{m-1})\otimes\tilde h^{-q}(X^{n+r-1}/X^{n-1}) @>\wedge>> \tilde h^{-p-q}\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\ @V\delta\wedge\mathrm{id}\oplus V\mathrm{id}\wedge\delta V @VV\delta V\\ {\begin{matrix} \tilde h^{1-p}(X^{m+r}/X^{m+r-1})\otimes\tilde h^{-q}(X^{n+r-1}/X^{n-1})\\ \oplus\\ \tilde h^{-p}(X^{m+r-1}/X^{m-1})\otimes\tilde h^{1-q}(X^{n+r}/X^{n+r-1}) \end{matrix}} @>\wedge\oplus\wedge>> {\begin{matrix} \tilde h^{1-p-q}\bigl((X^{m+r}/X^{m+r-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\ \oplus\\ \tilde h^{1-p-q}\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r}/X^{n+r-1})\bigr) \end{matrix}} \end{CD} We extend this diagram to the right using the maps $F_{m,n,r}$ and conclude that the lower square also commutes.