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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?

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Suppose the cycle class of an algebraic cycle has co-ordinates $(a_1,\ldots,a_n)$ on the de Rham side and $(b_1,\ldots,b_n)$ 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 –  Keerthi Madapusi Pera Jun 16 '12 at 14:33
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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 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 $V$. –  Keerthi Madapusi Pera Jun 16 '12 at 14:38
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@Keerthi: MO has an answer box for answers and a comment box for comments. :-) –  Martin Brandenburg Jun 16 '12 at 14:42
    
Sorry, I initially meant it to be a comment and then it sort of expanded. –  Keerthi Madapusi Pera Jun 17 '12 at 3:57

2 Answers 2

up vote 12 down vote accepted

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 $V$.

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Thanks Keerthi, so the key point is that the cycle class map commutes with $\omega$. –  Hugo Chapdelaine Jun 17 '12 at 15:38
    
Hugo--Yes, and it's believed that this is the only constraint on $\omega$, in a suitable sense. –  Keerthi Madapusi Pera Jun 17 '12 at 18:34

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 $\mathrm{GL}_{2/\mathbf{Q}}$.

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Thanks Francois for the simple but instructive examples. So are you suggesting that in general the motivic galois group should act on the ring generated by the periods obtained from the comparison isomorphism? –  Hugo Chapdelaine Jun 17 '12 at 15:45
    
Hugo--Yes. In fact, you would expect the ring generated by the periods to be the ring of functions of a torsor under the motivic Galois group. –  Keerthi Madapusi Pera Jun 17 '12 at 18:34

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