Comparing singular cohomology with algebraic de Rham cohomology Let $X$ be a smooth projective variety over a number field $K$. Then there are two cohomology groups we can attach to $X$: the algebraic de Rham cohomology group
$H^k_{\text{dR}}(X/K), $
which is a finite dimensional $K$-vector space, and the singular cohomology group
$H^k_{\text{sing}}(X(\mathbf{C}), \mathbf{Q}), $
which is a finite dimensional $\mathbf{Q}$-vector space. De Rham's theorem gives us an isomorphism between these two cohomology groups:
$$\sigma: H^k_{\text{dR}}(X/K) \otimes_{K} \mathbf{C} \xrightarrow{\sim} \,H^k_{\text{sing}}(X(\mathbf{C}), \mathbf{Q}) \otimes_{\mathbf{Q}} \mathbf{C}.$$
The two groups in this isomorphism both have a rational structure. The de Rham cohomology group $H^k_{\text{dR}}(X/K) \otimes_{K} \mathbf{C}$ has a $K$-lattice inside it given by $H^k_{\text{dR}}(X/K)$. And the singular cohomology group $H^k_{\text{sing}}(X(\mathbf{C}), \mathbf{Q}) \otimes \mathbf{C}$ has a $\mathbf{Q}$ lattice inside it given by $H^k_{\text{sing}}(X(\mathbf{C}), \mathbf{Q})$.
My question is: what is the relation between these two lattices under the isomorphism $\sigma$? For example, if $K = \mathbf{Q}$, then does the $K$-lattice on the de Rham side map to the $\mathbf{Q}$-lattice on the singular cohomology side? In general, if $K \neq \mathbf{Q}$, is there any relation between these two lattices we can speak of?
 A: This is the subject of periods: recall that the de Rham isomorphism between $H^k_{\text{dR}}(X/K) \otimes_K \mathbf C = H^k_{\text{dR}}(X_{\mathbf C}/\mathbf C)$ and $H^k_{\text{sing}}(X(\mathbf C),\mathbf Q) \otimes_{\mathbf Q} \mathbf C = H^k_{\text{sing}}(X(\mathbf C),\mathbf C)$ is defined by integrating $k$-forms $\eta \in H^k_{\text{dR}}(X_{\mathbf C}/\mathbf C)$ along $k$-cycles $\Delta^k \to X(\mathbf C)$.
This shows that the two lattices should not be the same: the integral of a $\mathbf Q$-valued $k$-form along an integral $k$-cycle $\Delta^k \to X(\mathbf C)$ need not be a rational number. For instance, if $X = \mathbf G_m = \mathbf A^1 \setminus \{0\}$, then $H_1(X(\mathbf C),\mathbf Z)$ is generated by a loop around the origin, and $H^1_{\text{dR}}(X/\mathbf Q) = H^0(X,\Omega_X^1)$ is generated by $\tfrac{\text{d}z}{z}$. The integral is $2 \pi i$, which is not a rational number. Thus, the lattices do not agree. You can see an incarnation of this in Hodge theory, where Tate twists show up by multiplying a lattice by $2\pi i$.
There are similar examples in the projective case, for instance on elliptic curves (but they become harder to compute). The question is which complex numbers are periods is one that is heavily studied. A great introduction is the book Periods and Nori motives by Annette Huber and Stefan Müller-Stach, Parts III and IV (Chapters 11–16).
