# indepence of Galois orbits on a product?

Let's take $X$ some fixed variety of a fixed base field $k$, which is assumed to be of char.zero for simplicity. Write $\Gamma_k$ for the Galois group of $k$. Given a point $x\in X(\bar{k})$, write $O(x)$ for the $\Gamma_k$-orbit $\{\sigma x:\sigma\in\Gamma_k\}$, and $ord(x)$ to be the cardinality of $O(x)$. The question is aimed to understand how independent the action of $\Gamma_k$ on different points could be.

So consider a pair $(x,y)\in X\times X(\bar{k})$. I wonder if the size of $O(x,y)$ in $X\times X$ would be close to the product $ord(x)\times ord(y)$. This seems impossible in general, but what if one considers a sequence and take limit? Say $(x_n,y_n)$ is a sequence in $X\times X(\bar{k})$ which is generic, in the sense that for any closed subvariety $Y\subsetneq X\times X$ defined over $\bar{k}$, $(x_n,y_n)$ is not in $Y$ for $n$ large. is there any results known of the form $$\lim_n\dfrac{ord(x_n,y_n)}{ord(x)\times ord(y)}=1?$$

The motivation of the question can be found in some equidistribution theorems in arithmetic geometry, see e.g. the works of Szpiro, Ullmo, Zhang, Yuan, etc. In their case no special properties are needed on the product structure, and only the genericity of the sequence of points is assumed. then they embed the base field into the complex number field, and compare the averaged Dirac measure with some canonical measure on the complex locus. But I wonder if the product phenomenon mentioned above happens to be true in some cases. Also for the independece, does it seem to have some probability theoretic interpretations?

-
It seems like you get equality whenever the stabilizers of $x$ and $y$ in the Galois group have intersection that is sufficiently small. Even you fix $x$, this should hold for generic $y$ (assuming you're not in a special case like $k$ real closed). –  S. Carnahan Nov 12 '10 at 8:01
Unless I misunderstood something: if $X=\mathbb{A}^1_k$, your quotient is just the degree of the extension $k(x_n)\cap k(y_n)$ over $k$. –  Chris Wuthrich Nov 12 '10 at 9:38