Definition of a period using cohomology.

Question

In algebraic geometry, a period is an integral of an algebraic function over an algebraic domain (for instance, $2i\pi=\int_{\partial D(0,1)}\frac{dz}{z}$). It is said in my notes that such a period can be seen as an entry of the matrix in rational bases of the de Rham isomorphism : $$ \mathbb{C}\otimes_{\mathbb{Q}} H_{{\rm sing}}^i(X(\mathbb{C}),\mathbb{Q})\simeq\mathbb{C}\otimes_K H_{{\rm dR}}^i(X) $$ where $X$ is an algebraic variety defined over a number field $K$, $H_{{\rm sing}}^i$ denotes the singular cohomology and $H_{{\rm dR}}^i$ the algebraic de Rham cohomology. I don't get why this is true, especially why the singular cohomology is used here and why tensorizing by $K$ in the right hand side. Thanks for your help.

Answer 1

What notes are these? Results of this type are generally called "comparison theorems" and often have hard proofs. As for your other questions: singular cohomology could just as easily be any other ordinary cohomology theory defined on topological spaces; the distinction between is between ordinary cohomology and something fancier, in this case algebraic de Rham cohomology.

The point of the tensor products to formalize the statement and make it functorial. The basic fact is that the cohomology groups on either side of the $\simeq$ are vector spaces of the same dimension, over $\mathbb Q$ and $K$ respectively. The tensor product just allows that equality of dimensions to be upgraded to an isom. of complex vector spaces.

So then you might object that $\mathbb Q \subset K \subset \mathbb C$, so why don't we just tensor up with $K$ and call it a day? This is true, but I'm fairly certain that there is no way to extend scalars on the cohomology of $X(\mathbb C)$ to $K$ in a meaningful way (that is, in a way that $\operatorname{Gal}(K/\mathbb Q)$ acts geometrically on the cohomology), i.e. taking $\mathbb C$-points forgets some/all of the $K$-structure on $X$. So while I cannot pinpoint where it would happen, somewhere functoriality would break down.