# Cohomology class of the diagonal

I have a pretty basic question:

given is a complex manifold of dimension n or a smooth projective variety over $k$ (char 0, algebraically closed). Then people are often speaking of "the cohomology class of the diagonal". Can you explain me how you realize this class, i.e. in which cohomology group does it land, how do you formally construct it, etc.

Thank you!

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This will depend on context. Given a reasonable cohomology theory for your variety (singular, etale, crystalline, etc) there is a way to associate classes to subvarieties. There is probably some unified abstract general "motivic" way to do it that then passes to any cohomology you want. I guess what I'm trying to say is: are these in reference to etale cohomology or something else so that an answer can be better suited to what you want. –  Matt Aug 31 '11 at 18:31
I would prefer singular cohomology theory... –  Descartes Aug 31 '11 at 20:04
I mean with coefficients in $\mathbb Z$ or $\mathbb Q$. –  Descartes Aug 31 '11 at 20:11
Thank you, Kevin, Neil and Carnahan for your diverse and always useful answers. I'll accept Kevin's simply because I actually only wanted some elementary explanation for the construction in the simplest case; but I definitely think many will profit from Neil's very general and conceptual answer, so thanks for your great comment, Neil! –  Descartes Sep 1 '11 at 6:43

Let $\Delta : M \to M \times M$ be the diagonal map. Since $M$ is a complex manifold, say of complex dimension $n$, it has a canonical orientation class $[M] \in H_{2n}(M, \mathbb{Z})$. Then you can take the pushforward in homology to get $\Delta_\ast [M] \in H_{2n}(M \times M, \mathbb{Z})$. If $M$ is compact then $M \times M$ is also compact and you can use Poincare duality to get an element in $H^{2n}(M \times M, \mathbb{Z})$. This is the cohomology class of the diagonal.

More generally, the words to look up are Thom isomorphism theorem or Gysin sequence or Gysin map. The inclusion $\Delta$ induces a Gysin map $\Delta_\ast: H^i(M) \to H^{i-(-2n)}(M \times M)$. The cohomology class of the diagonal is the image of $1 \in H^0(M)$ under this map. You can do the same kind of thing in $K$-theory, Chow groups, etc.

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I'll answer for smooth closed manifolds. I'll also assume choose Riemannian metrics everywhere. (The space of such metrics is convex and therefore contractible, so anything depending on the metric will usually be canonical up to homotopy.) Suppose we have a smooth embedding $f:N\to M$ of codimension $d$. This gives a $d$-dimensional vector bundle $\nu_f$ over $N$, whose fibre at $x$ is the orthogonal complement to $f_*(T_xN)$ in $T_{f(x)}M$. Next, for $(x,v)\in E(\nu_f)$ I define $g_1(x,v)$ to be $\alpha(1)$, where $\alpha:\mathbb{R}\to M$ is the geodesic with $\alpha(0)=f(x)$ and $\dot{\alpha}(0)=v$. This map $g_1:E(\nu_f)\to M$ will be an embedding on some neighbourhood of the zero section. Choose a smooth embedding $h:E(\nu_f)\to E(\nu_f)$ whose image is close to the zero section, and which is the identity on some even smaller neighbourhood of the zero section. Now put $g=g_1\circ h:E(\nu_f)\to M$; this is called a tubular neighbourhood of $f$. We can define a map $f^!:M_\infty\to E(\nu_f)_\infty$ of one-point compactifications by $f^!(g(x,v))=(x,v)$ and $f^!(m)=\infty$ whenever $m$ is not in the image of $g$. This is called the Pontrjagin-Thom construction. This in turn gives a map $\widetilde{H}^*(E(\nu_f)_\infty)\to\widetilde{H}^*(M_\infty)$. Now $M$ is compact, so the point $\infty\in M_\infty$ is isolated, so $\widetilde{H}^*(M_\infty)=H^*(M)$. On the other hand, $E(\nu_f)_\infty$ is otherwise known as the Thom space $N^{\nu_f}$, so (provided that $\nu_f$ is oriented) there is a canonical element $u\in \widetilde{H}^d(N^{\nu_f})$ (called the Thom class) such that $\widetilde{H}^d(N^{\nu_f})$ is freely generated by $u$ as a module over $H^*(N)$. We now have an element $v=(f^!)^*(u)\in H^d(M)$; this is called the cohomology class of $N$ (or of $f$) in $M$.

The cohomology class of the diagonal is obtained by applying this procedure to the diagonal embedding $f:M\to M^2$ defined by $f(m)=(m,m)$.

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Here is one answer for the question of Descartes in the case $X$ is compact Kahler manifold and the coefficient is $\mathbb{Q}$: Then we can work with coefficient $\mathbb{C}$ as well. This answers uses the de Rham cohomology.

If $X$ is a compact Kahler manifold of complex dimension $n$, then $\Delta_X\subset X\times X$ is a cycle of (real) dimension $2n$, hence acts on smooth $2n$ forms on $X\times X$ by integration. Also, Delta has no boundary, hence it is closed (by Stokes theorem). Thus it represents a homology class in $H_{2n}(X\times X)$. Now, $\Delta_X$ acts on a form by integration on the restriction of that form on $\Delta_X$. Since $\Delta_X$ is a complex manifold of dimension $n$, if $\varphi$ is a $(p,q)$ form on $X\times X$ with $p+q=2n$, its restriction to $\Delta_X$ is nonzero iff $p=q=n$. This shows that $\Delta_X$ is in $H_{n,n}(X)$, and so its Poincare dual represents a cohomology class in $H^{n,n}(X)$.

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Hi Carnahan, thanks for your answer. I think you forgot the second factor $X$ in the last line. –  Descartes Sep 1 '11 at 6:36