Let $\ p:X\to B\ $ be a fibration of spaces with fiber $F$. Then there are homological and cohomological Leray-Serre spectral sequences $E^{r}_{pq}$ and $E_r^{pq}$ that converge to $H_*(X)$ and $H^*(X)$ respectively. It is well known that the cohomological spectral sequence $E^{pq}_r$ is multiplicative (spectral sequence of algebras) with the product induced by the cup product $$\smile\: :E_r^{pq}\times E_r^{p'q'}\longrightarrow E_r^{p+p',q+q'}.$$ Moreover, the product on $E_\infty^{**}$ is induced by the cup product on $H^*(X)$ and the product on $E_2^{**}$ given by the 'double' cup product on $H^*(B,H^*(F)).$

Question: Is it possible to define the cap product on the spectral sequences $$\frown\: :E_{pq}^r\times E_r^{p'q'} \longrightarrow E^{r}_{p-p',q-q'} $$ with the similar properties? I believe it is possible but I need a reference.

  • $\begingroup$ I would think so but I don't know a reference. I would first lift the cap product to the level of chains and cochains and then look at the filtration of the total space gotten by using the lifting property of fibrations applied to the cellular filtration of the base space. Does the cap product interact well with that? I don't know. But this would be how one might go about proving it. $\endgroup$ Commented Feb 15, 2016 at 13:03

1 Answer 1


I am sorry that I don't know a reference, so I will try to give a proof instead. We follow Douady's approach using Cartan-Eilenberg systems, see here.

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 $(\tilde h_\bullet,\partial,\wedge)$ be a reduced multiplicative generalised homology theory, and let $(\tilde h^\bullet,\delta,\wedge)$ be the corresponding cohomology theory. We define Cartan-Eilenberg systems $(H,\eta,\partial)$, $(H',\eta',\partial')$ by $$H(p,q)=\tilde h_\bullet(X^{-p},X^{-q})\qquad\text{and}\qquad H'(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})$$ for~$p\le q$ with the obvious maps $\eta\colon H(p',q')\to H(p,q)$, $\eta'\colon H'(p',q')\to H'(p,q)$ for $p\le p'$, $q\le q'$. The maps $\partial\colon H(p,q)\to H(q,r)$, $\partial'\colon H'(p,q)\to H'(q,r)$ come from exact sequences of triples. Sorry for the weird indexing, but this makes it the easiest to use Douady's result. The spectral sequence will not end up in the first quadrant, but you can surely bring it back into its rightful place. 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}/(X\wedge X)^{m+n-r} \cong\bigcup_{a+b=m+n}(X^a\wedge X^b)\Bigm/ \bigcup_{c+d=m+n-r}(X^c\wedge X^d)\\ \begin{aligned} \twoheadrightarrow\mathord{}&\bigcup_{a+b=m+n}(X^a\wedge X^b)\Bigm/ \Bigl(\bigcup_{a=0}^{m-1}(X^a\wedge X^{m+n-a}) \cup\bigcup_{b=0}^{n-r}(X^{m+n-b}\wedge X^b)\\ \cong\mathord{}&\bigcup_{a=0}^{r-1}(X^{m+a}\wedge X^{n-a})\Bigm/ \bigl(X^{m-1}\wedge X^n\cup X^{m+r-1}\wedge X^{n-r}\bigr)\\ \hookrightarrow\mathord{}& X^{m+r-1}\wedge X^n\bigm/ (X^{m-1}\wedge X^n\cup X^{m+r-1}\wedge X^{n-r})\\ \cong\mathord{}&(X^{m+r-1}/X^{m-1})\wedge(X^{n}/X^{n-r})\;. \end{aligned} \end{multline*} Together with the diagonal map $\Delta$, for $r\ge 1$, we define \begin{multline*} \mu_r\colon H(-n-m,r-n-m)\otimes H'(m,m+r)\\ \begin{aligned} &\cong\tilde h_\bullet(X^{m+n}/X^{m+n-r})\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\ &\stackrel{\Delta_X^*\otimes\mathrm{id}}\longrightarrow\tilde h_\bullet\bigl((X\wedge X)^{m+n}/(X\wedge X)^{m+n-r}\bigr)\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\ &\stackrel{F_{m,n,r,*}\otimes\mathrm{id}}\longrightarrow\tilde h_\bullet((X^{m+r-1}/X^{m-1})\wedge(X^{n}/X^{n-r}))\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\ &\stackrel{/}\longrightarrow\tilde h_\bullet(X^{n}/X^{n-r})=H(-n,r-n)\;. \end{aligned} \end{multline*}

Proposition For all $m$, $n$, $r\ge 1$, the following diagram commutes $\require{AMScd}$ \begin{CD} H(-m-n,1-m-n)\otimes H'(m,m+1)@>\mu_1>>H(-n,1-n)\\ @A\eta'\oplus A\eta''A@AA\eta A\\ H(-m-n,r-m-n)\otimes H'(m,m+r)@>\mu_r>>H(-n,r-n)\\ @V\partial\otimes\eta'\oplus V\eta\otimes\partial'V@VV\partial V\\ {\begin{matrix}H(r-m-n,r+1-m-n)\otimes H'(m,m+1)\\\oplus\\H(-m-n,1-m-n)\otimes H'(m+r,m+r+1)\end{matrix}}@>\mu_1\pm\mu_1>>H_{p+q-1}(r-n,r+1-n)\rlap{\;.} \end{CD}

As explained here, this Proposition allows us to define a multiplicative structure on the associated spectral sequence. In this setting, it will be the cap product you ask for: $$\frown\colon E^r_{m+n}\otimes E_r^m\to E^r_n\;.$$

Proof. The upper square commutes because the maps $F_{m,n,r}$ are defined sufficiently naturally. For the lower square, we use the Leibniz rule for the slant product and continue as here

  • $\begingroup$ Sorry for linking two other answers of mine. But otherwise, this would have become much too long. $\endgroup$ Commented Feb 15, 2016 at 20:41

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