multiplicative spectral sequence


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A spectral sequence is called multiplicative or a spectral ring if there is a bi-graded algebra structure on each page such that the differentials act as graded derivations of total degree 1.

For example the Serre-Atiyah-Hirzebruch spectral sequence with coefficients in a ring spectrum.


From spectral products on Cartan-Eilenberg systems

The following gives sufficient conditions for a Cartan-Eilenberg spectral sequence to be multiplicative. This is due to (Douady 58). The following is taken from (Goette 15a).


Let (H,η,)(H,\eta,\partial), (H,η,)(H',\eta',\partial') und (H,η,)(H'',\eta'',\partial'') be Cartan-Eilenberg systems. A spectral product μ:(H,)×(H,)(H,)\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial) is a sequence of homomorphisms

μ r:H(m,m+r)H(n,n+r)H(m+n,m+n+r) \mu_r\colon H'(m,m+r)\otimes H''(n,n+r)\to H(m+n,m+n+r)

such that for all mm, nn, r1r\ge 1, the following two diagrams commute:

H(m,m+r)H(n,n+r) μ r H(m+n,m+n+r) ηη η H(m,m+1)H(n,n+1) μ 1 H(m+n,m+n+1) \array{ H'(m, m+r) \otimes H''(n, n+r) &\stackrel{\mu_r}{\longrightarrow}& H(m+n, m+n+r) \\ \downarrow^{\mathrlap{\eta' \oplus \eta''}} && \downarrow^{\mathrlap{\eta}} \\ H'(m, m+1) \otimes H''(n, n+1) &\stackrel{\mu_1}{\longrightarrow}& H(m+n, m+n+1) }


H(m,m+r)H(n,n+r) μ r H(m+n,m+n+r) ηη H(m+r,m+r+1)H(n,n+1) μ 1+μ 1 H p+q1(m+n+r,m+n+r+1) H(m,m+1)H(n+r,n+r+1). \array{ H'(m, m+r) \otimes H''(n, n+r) &\stackrel{\mu_r}{\longrightarrow}& H(m+n,m+n+r) \\ \downarrow^{\mathrlap{\partial' \otimes \eta'' \oplus \eta' \otimes \partial''}} && \downarrow^{\mathrlap{\partial}} \\ H'(m+r, m+r+1) \otimes H''(n,n+1) \\ \oplus &\stackrel{\mu_1 + \mu_1}{\longrightarrow}& H_{p+q-1}(m+n+r, m+n+r+1) \\ H'(m,m+1) \otimes H''(n+r, n+r+1) } \,.

The first diagram in def. 1 is weaker than in (Douady 58). The second may be read as a Leibniz rule.

Write EE for the Cartan-Eilenberg spectral sequence induced from the Cartan-Eilenberg system HH.


A spectral product μ:(H,)×(H,)(H,)\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial) as in def. 1 induces products

μ r:E m rE n rE m+n r, \mu^r\colon E^{\prime r}_m\otimes E^{\prime\prime r}_n\to E^r_{m+n}\;,

such that

  1. μ 1=μ 1\mu^1=\mu_1

  2. d m+n rμ r=μ r(d m rid)±μ r(idd n r)d^r_{m+n}\circ\mu^r=\mu^r\circ(d^{\prime r}_m\otimes\mathrm{id})\pm\mu^r\circ(\mathrm{id}\circ d^{\prime\prime r}_n),

  3. μ r+1\mu^{r+1} is induced by μ r\mu^r.

(Goette 15a, following Douady 58, theorem II).


Assume by induction that μ r\mu^r is induced by μ 1\mu_1. In particular,

Z m rZ n rμ 1Z m+n r,Z^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to Z^r_{m+n}\;,
B m rZ n rμ 1B m+n r,B^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;,
Z m rB n rμ 1B m+n r.Z^{\prime r}_m\otimes B^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;.

This is clear for r=1r=1 if we put μ 1=μ 1\mu^1=\mu_1 because E p 1=Z p 1=H(p,p+1)E^1_p=Z^1_p=H(p,p+1) and B p 1=0B^1_p=0.

Let [a]Z m r[a]\in Z^{\prime r}_m, [b]Z n r[b]\in Z^{\prime\prime r}_n be represented by a=η(a 0)H(m,m+1)a=\eta'(a_0)\in H'(m,m+1), b=η(b 0)H(n,n+1)b=\eta''(b_0)\in H''(n,n+1) with a 0H(m,m+r)a_0\in H'(m,m+r), b 0H(n,n+r)b_0\in H''(n,n+r). Using the first diagram and the construction of d m+n rd^r_{m+n}, we conclude that

(d m+n rμ r)([a][b])=d m+n r[μ 1(ab)]=d m+n r[η(μ r(a 0b 0))]=(μ r)(a 0b 0). (d^r_{m+n}\circ\mu^r)([a]\otimes[b])=d^r_{m+n}[\mu_1(a\otimes b)]=d^r_{m+n}[\eta(\mu_r(a_0\otimes b_0))]=(\partial\circ\mu_r)(a_0\otimes b_0) \;.

From the second diagram, we get

(μ r)(a 0b 0)=μ 1(a 0ηb 0)±μ 1(ηa 0b 0)=μ r(d m r[a][b])±μ r([a]d n r[b]). (\partial\circ\mu_r)(a_0\otimes b_0)=\mu_1(\partial'a_0\otimes\eta''b_0)\pm\mu_1(\eta'a_0\otimes\partial''b_0)=\mu^r(d^{\prime r}_m[a]\otimes[b])\pm\mu^r([a]\otimes d^{\prime\prime r}_n[b]) \;.

This proves the Leibniz rule (2).

From the Leibniz rule and the facts that ker(d p r)=Z p r+1/B p r\ker(d^r_p)=Z^{r+1}_p/B^r_p and im(d p r)=B p r+1/B p r\mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p, we conclude that μ r\mu^r induces a product on E p r+1ker(d p r)/im(d p r)E^{r+1}_p\cong\ker(d^r_p)/\mathrm{im}(d^r_p), which proves (3). Because μ r\mu^r is induced by μ 1\mu_1, so is μ r+1\mu^{r+1}, and we can continue the induction.


AHSS for multiplicative cohomology

We discuss that the multiplicative structure on the cohomology Serre-Atiyah-Hirzebruch spectral sequence for multiplicative generalized cohomology. This is taken from (Goette 15b).


For π:XB\pi\colon X\to B a Serre fibration over a CW-complex BB. And for (h˜ ,δ,)(\tilde h^\bullet,\delta,\wedge) a multiplicative reduced generalized (Eilenberg-Steenrod) cohomology theory, define a Cartan-Eilenberg system (H,η,)(H,\eta,\partial) by

H(p,q)=h˜ (X q1/X p1) H(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})

(where X k=π 1(B k)X^k=\pi^{-1}(B^k)) for pqp\le q with the obvious maps η:H(p,q)H(p,q)\eta\colon H(p',q')\to H(p,q) for ppp\le p', qqq\le q'.

The Cartan-Eilenberg spectral sequence of this Cartan-Eilenberg system is the Serre-Atiyah-Hirzebruch spectral sequence.


The spectral product μ:(H,η,)×(H,η,)(H,η,)\mu\colon(H,\eta,\partial)\times(H,\eta,\partial)\to(H,\eta,\partial), def. 1, on the Cartan-Eilenberg system of def. 2 is that given by the following morphism

F m,n,r :(XX) m+n+r1/(XX) m+n1 a+b=m+n+r1(X aX b)/ c+d=m+n1(X cX d) a+b=m+n+r1(X aX b)/( a=0 m(X a1X m+n+ra) b=0 n(X m+n+rbX b1) a=m+1 m+r(X a1X m+n+ra)/(X m+r1X n1X m1X n+r1) X m+r1X n+r1/(X m+r1X n1X m1X n+r1) (X m+r1/X m1)(X n+r1/X n1). \begin{aligned} 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) / \bigcup_{c+d=m+n-1}(X^c\wedge X^d) \\ & \twoheadrightarrow \bigcup_{a+b=m+n+r-1} (X^a\wedge X^b)/(\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 \bigcup_{a=m+1}^{m+r}(X^{a-1}\wedge X^{m+n+r-a}) / \bigl(X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1}\bigr) \\ & \hookrightarrow X^{m+r-1}\wedge X^{n+r-1}/(X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1}) \\ & \cong (X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1}) \;. \end{aligned}

Together with the diagonal map Δ\Delta, for r1r\ge 1, we define

μ r :H(m,m+r)H(n,n+r) h˜(X m+r1/X m1)h˜(X n+r1/X n1) h˜((X m+r1/X m1)(X n+r1/X n1)) F m,n,r *h˜((XX) m+n+r1/(XX) m+n1) Δ X *h˜(X m+n+r1/X m+n1)=H(m+n,m+n+r). \begin{aligned} \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}) \\ &\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}

With def. 3, then for all mm, nn, r1r\ge 1, the following diagram commutes

H(m,m+1)H(n,n+1) μ 1 H(m+n,m+n+1) ηη η H(m,m+r)H(n,n+r) μ r H(m+n,m+n+r) ηη H(m+r,m+r+1)H(n,n+1) μ 1±μ 1 H(m+n+r,m+n+r+1) H(m,m+1)H(n+r,n+r+1). \array{ H(m,m+1) \otimes H(n,n+1) &\stackrel{\mu_1}{\longrightarrow}& H(m+n, m+n+1) \\ \uparrow^{\mathrlap{\eta \oplus \eta}} && \uparrow^{\mathrlap{\eta}} \\ H(m,m+r) \otimes H(n,n+r) &\stackrel{\mu_r}{\longrightarrow}& H(m+n,m+n+r) \\ \downarrow^{\mathrlap{\partial \otimes \eta \oplus \eta \otimes \partial}} && \downarrow^{\mathrlap{\partial}} \\ H(m+r, m+r+1) \otimes H(n,n+1) \\ \oplus &\stackrel{\mu_1 \pm \mu_1}{\longrightarrow}& H(m+n+r, m+n+r+1) \\ H(m,m+1) \otimes H(n+r, n+r+1) } \,.

Hence by prop. 1 the spectral product of def. 3 defines a mutliplicative structure on the Serre-WhiteheadAtiyah-Hirzebruch spectral sequence for multiplicative generalizted cohomology.


The upper square commutes because the maps F m,n,rF_{m,n,r} are natural transformations. For the lower square, we consider the boundary morphism δ\delta of the triple

( X m+rX n+r1X m+r1X n+r, X m+rX n1X m+r1X n+r1X m1X n+r, X m+rX n1X m1X n+r). \begin{aligned} (& 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}) \end{aligned} \;.

The following diagram commutes:

h˜ p(X m+r1/X m1)h˜ q(X n+r1/X n1) h˜ pq((X m+r1/X m1)(X n+r1/X n1) δididδ δ h˜(1p)(X m+r/X m+r1)h˜ q(X n+r1/X n1) h˜ 1pq((X m+r/X m+r1)(X n+r1/X n1)) h˜ p(X m+r1/X m1)h˜ 1q(X n+r/X n+r1) h˜ 1pq((X m+r1/X m1)(X n+r/X n+r1)). \array{ \tilde h^{-p}(X^{m+r-1}/ X^{m-1}) \otimes \tilde h^{-q}(X^{n+r-1}/X^{n-1}) &\stackrel{\wedge}{\longrightarrow}& \tilde h^{-p-q}((X^{m+r-1}/X^{m-1}) \wedge (X^{n+r-1}/X^{n-1}) \\ \downarrow^{\mathrlap{\delta \wedge id \oplus id \wedge \delta}} && \downarrow^{\mathrlap{\delta}} \\ \tilde h(1-p)(X^{m+r}/X^{m+r-1}) \otimes \tilde h^{-q}(X^{n+r-1}/X^{n-1}) && \tilde h^{1-p-q}((X^{m+r}/X^{m+r-1}) \wedge (X^{n+r-1}/X^{n-1})) \\ \oplus &\stackrel{\wedge \oplus \wedge}{\longrightarrow}& \\ \tilde h^{-p}(X^{m+r-1}/X^{m-1}) \otimes \tilde h^{1-q}(X^{n+r}/X^{n+r-1}) && \tilde h^{1-p-q}( (X^{m+r-1}/ X^{m-1}) \wedge (X^{n+r}/ X^{n+r-1}) ) } \,.

By extend this diagram to the right using the maps F m,n,rF_{m,n,r} once concludes that the lower square above also commutes.


Revised on May 6, 2016 12:23:53 by Urs Schreiber (