"A global type which is A-invariant, ie invariant under all $\alpha\in Aut(\mathfrak{C}/A)$, does not fork over $A$.
Forking is new to me and I am not sure where to start on this. Let $p$ be the global type. If $p$ is $A$-invariant I think this means, for any $\phi(x,b)$ in $p$, we have $\phi(x,\alpha(b))$ in $p$ for all $\alpha\in Aut(\mathfrak{C}/A)$. How can we relate this to the forking condition?
Many thanks.
Well, if $p$ forks over $A$, by definition this means there are formulas $\phi_i(x,c_i)$ such that (i) each $\phi_i(x,c_i)$ divides over $A$, and (ii) $p\models\bigvee_{i=1}^n\phi_i(x,c_i)$. On the other hand, since $p$ is a global type, ie a complete type over all of $\mathfrak{C}$, it must contain at least one of the $\phi_i(x,c_i)$. (In particular, note that a global type forks if and only if it divides.)
Thus let $i\leqslant n$ be such that $\phi_i(x,c_i)\in p$, and for notational convenience denote $\phi=\phi_i$ and $c=c_i$. Now, by definition of dividing, there exists $(c_i)_{i\in\omega}$ a family of $A$-conjugates of $c$ such that the collection $(\phi(x,c_i))_{i\in\omega}$ is $k$-inconsistent for some $k\in\omega$; ie, the conjunction of any $k$ of the $\phi(x,c_i)$ is inconsistent. Now we use the hypothesis that $p$ is $A$-invariant: by definition, for each $i\in\omega$, there is an automorphism $\alpha_i\in\operatorname{Aut}(\mathfrak{C}/A)$ taking $c$ to $c_i$. Since $\phi(x,c)\in p$ and $p$ is $A$-invariant, this means that $\phi(x,c_i)=\phi(x,\alpha_i(c))\in p$ as well. So $p$ contains the entire family $(\phi(x,c_i))_{i\in\omega}$, and is thus inconsistent, a contradiction.