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!



Cohomology class of the diagonal
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.


