# Cohomology groups of an intersection

Suppose that $A$ is an affine algebraic variety and $P$, and $Q$ are subvarieties. It is easy to see that the coordinate rings $C[P]$ and $C[Q]$ of $P$ and $Q$ are modules over $C[A]$ and $C[P]\otimes_{C[A]} C[Q]$ is frequently the coordinate ring of $P\cap Q$. For instance, if the sum of the ideals of $P$ and $Q$ is its own radical.

If $X$ is a topological space, and $P$ and $Q$ are subspaces then their singular cohomology groups $H^{\*}(P)$ and $H^{\*}(Q)$ are modules over $H^{|*}(X)$.

To make it simple, assume that $X$ is a smooth manifold and $P$ and $Q$ are smooth submanifolds. Are there simple conditions that imply that $H^{\*}(P\cap Q)=H^{\*}(P)\otimes_{H^{\*}(X)} H^{\*}(Q)?$

Here is an example. Let $X=(S^2)^4$. Let $P\subset X$ of all pairs of the form $(x,x,y,y)$. Let $Q\subset X$ of all pairs of the form $(x,y,y,x)$. Notice that $P\cap Q$ is all pairs of the form $(x,x,x,x)$ so it is homeomorphic to $S^2$.

With a little work you can see that

$H^{\*}(X)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2)$,

that is, integer polynomials in $4$ variables where the square of any variable is zero.
We get $H^{\*}(P)$ is the quotient of $H^{\*}(X)$ by the ideal generated by $a-b,c-d$ and $H^{\*}(Q)$ is the quotient of $H^*(X)$ by the ideal generated by $a-d,b-c$.

Its easy to check that

$$H^{\*}(P)\otimes_{H^{\*}(X)} H^{\*}(Q)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2,a-b,b-c,c-d)$$

which is just $\mathbb{Z}[x]/(x^2)$ which is the cohomology group of the sphere.

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Take X to be the complex affine plane C^2, and P and Q are two smooth curves of high genuses, so that H^1(P) or H^1(Q) is non-trivial. Then the righthand-side is non-trivial in degree 1, but P\cap Q is discrete (assume P and Q are in general position). So I guess maybe there is no "simple" conditions to make it true. Might be true in some special cases. – shenghao Oct 3 '10 at 14:46
The cohomology of the intersection fits into a Meyer-Vietoris sequence which shows that the proposed formula is essentially never true. – Torsten Ekedahl Oct 3 '10 at 15:03
Don't we need open subsets (or sets whose interior....) for Mayer-Vietoris sequence? To Charlie: the coordinate ring can be interpreted as coherent cohomology. So maybe we don't expect to have analog for singular cohomology. – shenghao Oct 3 '10 at 20:25
When the closed subspaces are well-behaved we have a Mayer-Vietoris also for closed subspaces. This is the case for closed submanifolds. – Torsten Ekedahl Oct 3 '10 at 20:40
Excuse me, Torsten, can you tell me precisely what the Mayer-Vietoris sequence is in the case of closed submanifolds (or give a reference)? Do we use the cohom of the ambient space or the union of the two submanifolds in the sequence? Thanks. – shenghao Oct 3 '10 at 21:22

If $P$ and $Q$ are closed subspaces of $Y$ and $Y$ is their union, we have a short exact sequence of sheaves on $Y$ $0\rightarrow\mathbb Z\rightarrow i_\ast\mathbb Z\bigoplus j_\ast\mathbb Z\rightarrow k_\ast\mathbb Z\rightarrow0$, where $i$, $j$ resp. $k$ are the inclusions of $P$, $Q$ and $P\bigcap Q$ respectively. This gives a long exact sequence of Cech cohomology $$\cdots\rightarrow\check H^i(Y,\mathbb Z)\rightarrow \check H^i(P,\mathbb Z)\bigoplus \check H^i(Q,\mathbb Z)\rightarrow \check H^i(P\bigcap Q,\mathbb Z)\rightarrow\cdots$$ and when $Y$, $P$, $Q$ and $P\bigcap Q$ are well-behaved Cech cohomology coincides with singular cohomology and we get a similar sequence in singular cohomology. Hence $H^\ast(P\bigcap Q,\mathbb Z)$ is close (but not always equal to because of non-triviality of the boundary map) the additive pushout $H^\ast(P,\mathbb Z)\bigoplus_{H^\ast(Y,\mathbb Z)} H^\ast(Q,\mathbb Z)$. It seems in any case clear that anything we can hope to know about the cohomology of of $P\bigcap Q$ in terms of $P$ and $Q$ and some ambient space $X$ should be obtained through the Mayer-Vietoris sequence. If we compare that with the multiplicative pushout $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)} H^\ast(Q,\mathbb Z)$ we see essentially that it is close to the additive pushout only in very simple cases. It is true that the multiplicative pushout could hit more of $H^\ast(P\bigcap Q,\mathbb Z)$ because the image is the subring generated by the images of $H^\ast(P,\mathbb Z)$ and not just $H^\ast(Q,\mathbb Z)$ but on the other hand all multiplicative relations would only rarely be coming from $H^\ast(Y,\mathbb Z)$.
Now, replacing $Y$ by some ambient space $X$ doesn't seem to help as we have a natural surjection $$H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z) \rightarrow H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)}H^\ast(Q,\mathbb Z)$$ so if $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z)$ works then so does $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)}H^\ast(Q,\mathbb Z)$.
To be a little bit more specific, if $H^\ast(X,\mathbb Z)\rightarrow H^\ast(P,\mathbb Z)$ and $H^\ast(X,\mathbb Z)\rightarrow H^\ast(P,\mathbb Z)$ are not surjective, then it seems that $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z)\rightarrow H^\ast(P\bigcap Q,\mathbb Z)$ should not be injective except in some very trivial cases. However, even when they are we easily get into trouble: Let $P$ and $Q$ be lines in the complex projective plane $X$. Then $H^\ast(X,\mathbb Z)=\mathbb Z[x]/(x^3)$ and $H^\ast(P,\mathbb Z)=H^\ast(Q,\mathbb Z)=\mathbb Z[x]/(x^2)$ so that $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z)=Z[x]/(x^2)$ while $P\bigcap Q$ is a point.
Does the check mark mean that you determined that you answered my question? The remark at the end of the first paragraph " it is close to the additive pushout only in very simple cases" Seems to be headed towards an answer, but mostly the answer is magnifying the statement in your remark, which is that additive and multiplicative pushouts are in general different. It certainly works when $X$ is the cartesian product of $CP(1)$'s and $P$ and $Q$ are half dimensional components of the small diagonal. Is that the only time it works, or is are there families of examples where it works? – Charlie Frohman Oct 4 '10 at 10:23
I get it, the check is to indicate that I have the option of saying its the answer. Sorry about that. I would conjecture that the example I gave above is not the only one, but that there are a couple of universal constructions that should produce families of examples where its true. Maybe, suppose $X$ has no torsion in its homology and $P$ and $Q$ are components of the small diagonal of $X^n$ then its true? – Charlie Frohman Oct 4 '10 at 13:16
If you mean examples such as you gave in your question, then yes your formula is OK. I have to admit that they go beyond what I expected but I still think that they are quite artificial. Even though the multiplicative structure of the rings involved can be arbitarily complicated, their interrelations are given by tensor products of rings and ring homomorphisms of the type $R\bigotimes R\to R$ given by multiplication. Also they are verified after having computed explicitly the cohomology rings involved. – Torsten Ekedahl Oct 4 '10 at 19:15