15
$\begingroup$

Let $A\subset X$ be CW-complexes (or even manifolds). In cohomology with coefficients in a commutative ring $R$, we have a long exact sequence $$\cdots \rightarrow H^p(X,A)\rightarrow H^p(X)\rightarrow H^p(A)\stackrel{\partial_p }{\rightarrow }H^{p+1}(X,A)\rightarrow \cdots$$ Let $\alpha $ in $H^p(A)$, $\beta$ in $H^q(A)$. Is there a formula for $\partial_{p+q} (\alpha \smile\beta )$?

$\endgroup$
1
  • 4
    $\begingroup$ Have you tried writing it out by definition and get stuck somewhere? A quick glance at the formulas suggests that it obeys the Leibniz rule, the argument provided analogously by Lemma 3.3.6 of Hatcher's Algebraic Topology (i.e. computing the ordinary co-differential on cochains). Here I note that cup product is defined on mixed relative cohomologies $H^p(X,A_1)\times H^q(X,A_2)\to H^{p+q}(X,A_1\cup A_2)$ and that relative cochains can be regarded as absolute cochains which vanish on the appropriate subcomplexes. $\endgroup$ Feb 22, 2015 at 21:12

2 Answers 2

2
$\begingroup$

I will assume $(X,A)$ is a "good pair", i.e. that $A$ is a CW-subcomplex of $X$. Cup product works on mixed relative cohomologies, $H^p(X,A_1)\times H^q(X,A_2)\to H^{p+q}(X,A_1\cup A_2)$, so we can take $A_1=A_2=A$ or $A_1=A_2=\varnothing$ or $A_1=A$ and $A_2=\varnothing$ and vice versa. The coboundary map $H^n(A)\to H^{n+1}(X,A)$ is obtained by taking co-chains on $A$ and viewing them as co-chains on $X$ which vanish on $X-A$ and then pre-composing with the differential $C_{n+1}(X)\to C_n(X)$, i.e. it is obtained directly from the co-differential. If you work out the formula for the co-differential of the cup product of co-chains (which is the Leibniz rule), this should respect the values of the relative co-chains... but now I'm stuck, and the best I can say is the following:

If $\beta=i^\ast\eta$ where $i:A\hookrightarrow X$ and $\eta\in H^\ast(X)$, then the desired formula is $$\partial_{p+q}(\alpha\smile i^\ast\eta)=\partial_p\alpha\smile\eta$$

This is a "stability" result found in chapter VII section 8 of Dold's Lectures on Algebraic Topology. Exercise #3 of that section asks for a generalization of this result, which mimics the corresponding result for cross products (given in section 7 and section 2 of the same chapter). But while the cross product makes sense for general pairs $(X,A)$ and $(Y,B)$, the cup product needs $X=Y$ so that we may apply the diagonal map $\Delta:X\to X\times X$ (and appropriate relative versions). So I think Dold's desired "generalization of stability" only considers larger spaces such as $(X\times Y,A\times Y)$.

I originally wrote down a "Leibniz rule", but it's not defined (see the comments). Though for the cross product it is the case that $\partial_{p+q}(\alpha\times\beta)=\partial_p\alpha\times\beta=(-1)^p\alpha\times\partial_q\beta$.

$\endgroup$
6
  • 2
    $\begingroup$ I don't think that this makes sense. Note that $\partial_p\alpha$ lies in $H^*(X,A)$ and $\beta$ lies in $H^*(A)$, and there is no natural product $H^*(X,A)\otimes H^*(A)\to H^*(X,A)$ so $\partial_p\alpha\cup\beta$ is not defined. $\endgroup$ Feb 23, 2015 at 1:18
  • $\begingroup$ Sure there is, see chapter VII section 8 of Dold's Lectures on Algebraic Topology, where $(X,A,\varnothing)$ is an excisive triad. The "stability" property 8.10 seems highly relevant. $\endgroup$ Feb 23, 2015 at 1:34
  • 1
    $\begingroup$ @ChrisGerig In that situation, Dold's book has $A_1 = A$ and $A_2 = \emptyset$, so this diagram actually states that $\partial$ is map of $H^*(X)$-modules. (It does show that you are correct in the case where $\partial_q \beta = 0$.) $\endgroup$ Feb 23, 2015 at 1:50
  • $\begingroup$ True, but an exercise in that book gives this as a special case to a more general "stability" result, which I was hopeful would be of use here. And I now agree with Neil's comment. I was originally equating $H^\ast(X\times A,A\times A)$ with $H^\ast(X,A)$ and I no longer hold that in my mind. $\endgroup$ Feb 23, 2015 at 7:52
  • 1
    $\begingroup$ Here is a kind of an example that kind of shows we need some other structure than derivations (maybe): the coboundary map $H^n(X)\arrow H^{n+1}(CX,X)\simeq H^{n+1}(\Sigma X)$ is the suspension isomorphism, where $CX$ is the cone on $X$ and $\Sigma X$ is the suspension. Any reasonable action of the cohomology of X on cohomology of $(CX,X)$ should be up to sign the same if acted on the right or on the left. If we pick a class $x\in H^n(X)$, then we would get the $\Sigma(x\smile x)$ is $\Sigma(x)\smile x \pm x\smile \Sigma(x)$, which 0 module 2. Yuck! $\endgroup$ Feb 23, 2015 at 8:14
1
$\begingroup$

Let me give an answer for $R=\mathbf Z$, the ring of integers, and let us translate to sheaf cohomology. Your long exact sequence comes from the short exact sequence $$ 0 \rightarrow j_! \mathbf Z \rightarrow \mathbf Z \rightarrow i_\star \mathbf Z \rightarrow 0 $$ of sheaves of abelian groups on $X$, where $i$ is the inclusion of the closed subset $A$ in $X$, and $j$ is the inclusion of its complement. In the derived category $\mathcal D^+$ of sheaves of abelian groups on $X$, it gives rise to a triangle $$ j_! \mathbf Z \rightarrow \mathbf Z \rightarrow i_\star \mathbf Z \rightarrow j_! \mathbf Z[1]. $$ The last morphism $$ \delta\colon i_\star \mathbf Z \rightarrow j_! \mathbf Z[1] $$ is responsable for the coboundary map in the long exact sequence of cohomology: $$ \partial(\alpha)=\delta[p]\circ\alpha, $$ where $\alpha$ is alternatively interpreted as an element of $\mathrm H^p(A)$ and $\mathrm{Hom}(\mathbf Z,i_\star \mathbf Z[p])$. Here $\mathrm{Hom}$ means morphisms in the derived category $\mathcal D^+$. If one continues by identifying both groups with yet another, $\mathrm{Hom}( i_\star \mathbf Z, i_\star \mathbf Z[p])$, the formula $$ \partial(\alpha\cup\beta)=\delta[p+q]\circ\alpha[q]\circ\beta $$ makes sense and is valid for $\beta\in\mathrm{Hom}(\mathbf Z, i_\star \mathbf Z[q])=\mathrm H^q(A)$, since composition in the derived category coïncides with cup product.

If we go on and interpret $\delta$ as a cohomology class in the local cohomology group with support in $A$ and coefficients in $j_! \mathbf Z$ $$ \mathrm H_A^1 (X,j_! \mathbf Z)=\mathrm{Hom}(i_\star \mathbf Z, j_! \mathbf Z[1]), $$ one has $$ \partial(\alpha)=\delta[p]\circ\alpha=\delta\cup\alpha, $$ for the extra-ordinary cup product $$ \cup\colon \mathrm H_A^1(X,j_! \mathbf Z) \times \mathrm H^p(A) \rightarrow \mathrm H_A^{p+1}(X,j_! \mathbf Z). $$ The only formula for $\partial (\alpha\cup\beta)$ that I can make out of this is $$ \partial(\alpha\cup\beta)=\delta\cup(\alpha\cup\beta)=(\delta\cup\alpha)\cup\beta, $$ formula where among the $5$ cups, $2$ are ordinary, and $3$ are extra-ordinary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.