Cohomology groups of an intersection - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:18:17Z http://mathoverflow.net/feeds/question/40936 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40936/cohomology-groups-of-an-intersection Cohomology groups of an intersection Charlie Frohman 2010-10-03T14:31:47Z 2010-10-04T05:15:48Z <p>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.</p> <p>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 $H^{|*}(X)$. </p> <p>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 $H^{*}(P\cap Q)=H^{*}(P)\otimes_{H^{*}(X)} H^{*}(Q)?$</p> <p>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 $S^2$.</p> <p>With a little work you can see that</p> <p>$H^{*}(X)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2)$, </p> <p>that is, integer polynomials in $4$ variables where the square of any variable is zero.<br> 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 $a-d,b-c$. </p> <p>Its easy to check that </p> <p>$$H^{*}(P)\otimes_{H^{*}(X)} H^{*}(Q)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2,a-b,b-c,c-d)$$ </p> <p>which is just $\mathbb{Z}[x]/(x^2)$ which is the cohomology group of the sphere.</p> http://mathoverflow.net/questions/40936/cohomology-groups-of-an-intersection/40985#40985 Answer by Torsten Ekedahl for Cohomology groups of an intersection Torsten Ekedahl 2010-10-04T05:15:48Z 2010-10-04T05:15:48Z <p>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 $H^\ast(Y,\mathbb Z)$.</p> <p>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 $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)}H^\ast(Q,\mathbb Z)$.</p> <p>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. </p>