3
$\begingroup$

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!

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Is the construction supposed to have some useful property that would help characterize it? $\endgroup$ – Will Sawin Apr 10 '14 at 22:14
  • $\begingroup$ @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. $\endgroup$ – Lan Apr 11 '14 at 1:47
4
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – Lan Apr 11 '14 at 11:17
  • $\begingroup$ 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 $H^1(X_{\bar{\mathbb{Q}}},\mathbb{F}_p)$? $\endgroup$ – Lan Apr 11 '14 at 11:37
  • 2
    $\begingroup$ 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. $\endgroup$ – Will Sawin Apr 11 '14 at 14:08
  • $\begingroup$ @Sawin, thank you for your great help! $\endgroup$ – Lan Apr 12 '14 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.