Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a smooth projective geometrically connected curve over $\mathbf{Q}$ of genus at least two. Fix an algebraic closure $\overline{\mathbf{Q}}$ of $\mathbf{Q}$ and let $G_{\mathbf{Q}}$ be the absolute Galois group of $\mathbf{Q}$. Moreover, fix a rational base point $x$ in $X(\mathbf{Q})$.

Let $\sigma:G_{\mathbf{Q}}\to \pi_1(X)$ be a section of the exact sequence of groups $$ 1\to \pi_1(X_{\overline{\mathbf{Q}}})\to \pi_1(X) \to G_{\mathbf{Q}}\to 1.$$

Let $K\subset \overline{\mathbf{Q}}$ be a finite field extension of $\mathbf{Q}$. (I don't want to assume $K$ to be Galois, but please do if this helps.)

Let $G_K$ be the absolute Galois group of $K$. Then $G_K$ is an open subgroup of $G_{\mathbf{Q}}$. Note that $\pi_1(X_K)$ (with the same base point $x$) injects into $\pi_1(X)$.

Question. Does there exist a section $\sigma^\prime:G_{\mathbf{Q}}\to \pi_1(X)$ which is a $\pi_1(X_{\overline{\mathbf{Q}}})$-conjugate of $\sigma$ such that the image of $\sigma^\prime|_{G_K}$ lies in $\pi_1(X_K)$?

Motivation. If $a\in X(\mathbf{Q})$, then $a\in X(K)$. Thus, if $\sigma$ is a section associated to $a$, then the answer to the above question is positive. My question is really about sections that a priori do not come from a rational point.

Note. I always use the base point $x$ to define the fundamental group and we can replace $\mathbf{Q}$ by any number field.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

$\pi_1(X_K)$ is just the subgroup of $\pi_1(X)$ that is the inverse image of $G_K$. So, since the map is a section, its restriction immediately factors through - we do not even have to conjugate!

share|improve this answer
    
Thanks a lot. I was clearly confused. –  Jan Hendrik Nov 3 '12 at 19:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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