everyone, I want to ask is there any result in the literature similar to the following:

Let $ X=\mathbb{P}^1\backslash \{0,1,\infty\}$, then $X$ is defined over $\mathbb{Z}$. Let $X_{\mathbb{Q}}$ denote the its generic fiber, and for any prime number $p$, $X_{\mathbb{Q}_p}:= X_{\mathbb{Q}}\times Spec(\mathbb{Q}_p)$ be the base change to $\mathbb{Q}_p$.

Consider the global section of log differntial form $\Omega_{X_{\mathbb{Q}}}(log(0,1,\infty))$. It has a basis $\{\frac{dz}{z},\frac{dz}{z-1}\}$ , where $z$ is the standard affine coordinate.

The result is :

For any global section of log 1-form $\omega=a\frac{dz}{z} + b\frac{dz}{z-1}$, with $a,b\in \mathbb{Z}$, one can associate a system of representation of algebraic fundamental group. More explicitly, for any prime number $p$, such that $p\nmid a$, or $p\nmid b$, then one has a nontrivial rank two $\mathbb{Z}_p$-representation of $\pi_1^{alg}(X_{\mathbb{Q}_p})$, where $\pi_1^{alg}$ denote the algebraic fundamental group, that is

$$ \rho_{\omega,p}: \pi_1^{alg}(X_{\mathbb{Q}_p})\to GL_2(\mathbb{Z}_p). $$ It has the following property:

(i): the restriction of $\rho_{\omega,p}$ to the geometry $\pi_1$ is an nontrivial extension of two trivial representation, so when restricted to geometry $\pi_1$, it will never be trivial.

(ii): If one has a $\mathbb{Q}_p$-point of $X_{\mathbb{Q}_p}$, then the induced Galois represention of $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$ is an extension of trivial representation and Tate twist $\mathbb{Z}_p(1)$.

I really want to know wether there is a trivial way to realize the above association. Thank you!

  • Is the construction supposed to have some useful property that would help characterize it? – Will Sawin Apr 10 '14 at 22:14
  • @Sawin, I have edied the problem to give more description of the representation. We have made such construction through some sophisticated method, but I suspect that there will be a trivial way to do so. – Lan Apr 11 '14 at 1:47
up vote 4 down vote accepted

Yes, I think so. The geometric $\pi_1$ is a normal subgroup of $\pi_1$, and by (1) consists of upper triangular unipotent matrices. So the whole representation must live in the normalizer, which is all upper-triangular matrices. In other words, the representation is an extension of two one-dimensional representations, $\chi_1$ and $\chi_2$. $\pi_1^{geom}$ acts trivially on these, so are just Galois representations. By (2), we see these are the trivial representation and the Tate twist $\mathbb Z_p(1)$.

In $\pi_1$-representations, extensions of the $\mathbb Z_p$ by $\mathbb Z_p(1)$ are classified by $Hom(\pi_1^{geom}, \mathbb Z_p(1)) = H^1_{et} ( X_{\overline {\mathbb Q}}, \mathbb Z_p(1))$. This is a rank-two lattice over $\mathbb Z_p$. The most natural way to view it is as triples of numbers $a,b,c$ satisfying $a+b+c=0$, where $a$ is the local contribution at $0$, $b$ is the local contribution at $1$, and $c$ is the local contribution at $\infty$. (In De Rham cohomology, we would view $a,b,c$ as measuring the integral of the $1$-form over a small loops around the respective point.)

Presumably this is your associated representation?

  • @Sawin, If I understand correctectly, $Hom(\pi_1^{geom},\mathbb{z}_p(1))=H^1_{et}(X_{\bar{\mathbb{Q}}},\mathbb{Z}_p(1))$, should be consider as isomorpshic Galois module. So if one want to extend the representaion of $\pi_1^{geom}$ to $\pi_1^{alg}$, one need to consider the Galois invariant of $H^1_{et}(X_{\bar{\mathbb{Q}}},\mathbb{Z}_p(1))$.But how do you know that it has two basis of Galois invariant. That is the Galois representation of $H^1_{et}(X_{\bar{\mathbb{Q}}_p},\mathbb{Z}_p(1))$ is isomorphic to trivial representation? After this point,I should be our associated representation. – Lan Apr 11 '14 at 11:17
  • For high genus smooth projective curve $X_0$ defined over $\bar{\mathbb{F}}_p$, such that its Jacobian is ordianry abelian variety. We can always find some nonzero 1-form over $X_0$, and some $W_2$-lifting $X_1$ of $X_0$, then for any $W(k)$-lifting $X$ of $X_1$ , we can produce a rank two nontrivial $\mathbb{F}_p$-representation of $\pi_1^{alg}(X_{\mathbb{Q}_p})$, which has the extension structure similar to the one of the representaion of $P^1\backslash \{0,1,\infty\}$. So by your method, can one say something about the Galois invariant of $H^1(X_{\bar{\mathbb{Q}}},\mathbb{F}_p)$? – Lan Apr 11 '14 at 11:37
  • 2
    Yes this is exactly correct. We can check that it is Galois invariant by observing that it is Poincare dual to $H^1_c(X_{\overline{\mathbb Q}}, \mathbb Z_p)$, which fits into the long exact sequence $H^0_c(\mathbb P^1_{\overline{\mathbb Q}}, \mathbb Z_p )\to H^0_c(\{0,1,\infty\},\mathbb Z_p) \to H^1_c(X_{\overline{\mathbb Q}}, \mathbb Z_p) \to H^1_c(\mathbb P^1_{\overline{\mathbb Q}},\mathbb Z_p)$. The first two terms are Galois-invariant and the last term is zero. – Will Sawin Apr 11 '14 at 14:08
  • @Sawin, thank you for your great help! – Lan Apr 12 '14 at 2:08

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