On Grothendieck's period relations - MathOverflow

On Grothendieck's period relations

2012-06-16T14:01:22Z

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?

Answer by Keerthi Madapusi Pera for On Grothendieck's period relations

2012-06-17T03:59:10Z

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 $V$.

Answer by François Brunault for On Grothendieck's period relations

2012-06-17T10:17:34Z

Here is an example to flesh out Keerthi's nice answer in a very simple case.

Take $V$ to be the $0$-dimensional variety $V=\operatorname{Spec} F $ where $F/\mathbf{Q}$ is a finite Galois extension of degree $d$. The left hand side is given by $H^0_{dR}(V) = F$, while $V(\mathbf{C})$ is the finite set $\Sigma=\operatorname{Hom}(F,\mathbf{C})$, so that $H^0_B(V(\mathbf{C}),\mathbf{Q}) \cong \mathbf{Q}^{\Sigma}$. The comparison isomorphism is then just the usual isomorphism $\omega : F \otimes \mathbf{C} \xrightarrow{\cong} \mathbf{C}^\Sigma$.

Choosing a basis $(a_1,\ldots,a_d)$ of $F$ over $\mathbf{Q}$ and the canonical basis of $\mathbf{Q}^{\Sigma}$, the entries of the matrix of $\omega$ are just $\sigma(a_i)$ with $\sigma \in \Sigma$ and $1 \leq i \leq d$. Each time you have a polynomial relation $P(a_1,\ldots,a_d)=0$ with $P \in \mathbf{Q}[X_1,\ldots,X_d]$ (which you can interpret as an algebraic cycle on $V^d =\operatorname{Spec} F^{\otimes d}$), you get a corresponding relation on the $\sigma(a_i)$'s. As you have probably guessed, the motivic Galois group in this case is just $\operatorname{Gal}(F/\mathbf{Q})$ seen as a finite algebraic subgroup of $\mathrm{GL}(H^0_B) \cong \mathrm{GL}_{d/\mathbf{Q}}$. (NB : this "definition" of the Galois group is actually closer in spirit to Galois's original definition.)

Another example is given by an elliptic curve $E/\mathbf{Q}$ which has CM by an order in an imaginary quadratic field $K$. In this case $\mathrm{GL}(H^1_B(E)) \cong \mathrm{GL}_{2/\mathbf{Q}}$ . The complex multiplication provides an algebraic cycle on $E \times E$, and the motivic Galois group is the normalizer of $K^\times$ seen as a subgroup of $\mathrm{GL}_{2/\mathbf{Q}}$.