Let $V$ be a smooth projective variety defined over $\mathbf{Q}$ and denote by
$$
\omega: H\_{dR}^\*(V,\mathbf{Q})){\otimes\_{\mathbf{Q}}}\mathbf{C}\rightarrow H\_{B}^\*(V,\mathbf{Q})\otimes\_{\mathbf{Q}}\mathbf{C},
$$
Grothendieck's comparison isomorphism between algebraic De Rham cohomology and Betti cohomology. Choosing $\mathbf{Q}$-rational basis on both sides we may think of $\omega$ as given by a square matrix. 


In many places in the literature it is said that algebraic cycles (defined over $\mathbf{Q}$) on the $n$-iterated product of $V$, namely $V^n$, give rise to polynomial relations in the entries of the matrix $\omega$.


**Q**: How does one obtain such polynomial relations?



On Grothendieck's period relations
Suppose the cycle class of an algebraic cycle has co-ordinates $(a\_1,…,a\_n)$ on the de Rham side and $(b\_1,…,b\_n)$ on the Betti side. Since the comparison isomorphism has to preserve cycle classes, we get the relation $\sum\_j\omega\_{ij}a\_j=b\_i$. One can also think about it as follows: If there were no constraints on $\omega$, it's just a random isomorphism between two complex vector spaces. The space of such isomorphisms is a torsor under $GL(n,\mathbb{C})$, and so the co-ordinates of a generic point of the space will generate an extension of $\mathbb{Q}$ of transcendence degree $n^2$. But the fact that cycle classes have to be preserved means that the isomorphism is actually a trivialization of a torsor under a much smaller group (the 'motivic galois group' for $V$), which means that you should have non-trivial relations between the co-ordinates. Grothendieck conjectured that these are in fact all the relations. Combined with the Hodge conjecture this would imply that the transcendence degree of the extension generated by the co-ordinates is the dimension of the Mumford-Tate group for $V$.


