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I would like to know why the relation: R(x,y) iff x independent from y (i.e: tp(x/y) doesn't fork over the empty set) is type definable in stable theory?

Thanks to the helper.

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2 Answers

First, the relation $R(x,y)$ as you defined it is in general not type-definable (although it is definable in any $\aleph_0$-categorical theory, since it is invariant). Indeed, take any stable structure in which $X = \mathrm{acl}(\emptyset)$ is infinite, and take any sequence $(a_n) \subseteq X$ without repetitions. Then we have $R(a_n,a_n)$ for all $n$, but in an ultra-power, if $a = [a_n]$ (the class of the sequence modulo the ultra-filter) then $a \notin \mathrm{acl}(\emptyset)$, so $\neg R(a,a)$. This happens, for example, in the theory of algebraically closed fields, which is as stable as you can get.

Second, for any fixed $b$, the relation $R(x,b)$ is type-definable: $$ R(x,b) = \{ \neg \varphi(x,b) \colon \text{the formula } \varphi(x,b) \text{ forks over } \emptyset \}. $$ I stated this over the empty set, as in the question, but the same is true over an arbitrary parameter set.

The problem is that as $b$ varies, the property "$\varphi(x,b)$ forks over $\emptyset$" varies in a non type definable fashion, so you really must know $b$

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Work in a large saturated model, $\mathcal M.$ Things like $x$ and $y$ are not assumed to be singletons. They may be tuples.

First, I want to be careful about just what the question says. To say that this relation is type definable would mean that as a subset of $\mathcal M^2,$ the set of points satisfying the relation is the intersection of definable sets in $\mathcal M^2.$

Lets use the following characterization of forking - everything is over the empty set, but I want to assume that the types are stationary. $a$ forks with $b$ if $tp(a/b)$ represents a new formula. In a stable theory, types are definable, that is, fix a formula $\psi(x,y)$ and consider the set of parameters $c$ so that $b \models \psi (x,c).$ Definability of types says that this is a definable set. In fact, it is a boolean combination of instances of $\psi .$

So, for each new formula that $a$ might represent over $b,$ we can simply tell, definably. Intersecting over the set of all potential new formulas which $a$ might represent gives a type-definable subset of $\mathcal M^2.$

Maybe someone else can comment: is there a way to get around stationarity in a simple manner? Maybe one should just do things over the algebraic closure of the empty set and somehow go back down? Am I just being silly? (I am used to working over models, not over the empty set...)

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I don't understand what do you mean by 'tp(a/b) represents a new formula', but it seems to me that you fixing the variables a and b. p.s: for my purpose it suffice to assume that the empty set is Algebraically closed (so the type is stationary) –  henry Aug 23 '11 at 16:15
    
Ok, good. I was not sure about what would happen without stationarity, though some people who work with general stable theories might comment on this. Representing a new formula is a criteria for forking. In, say, Pillay's small stability theory book, this sort of notion is definitely covered. This should be in Marker's model theory book as well. Perhaps the language is the issue. This terminology goes with the old style of describing heirs, etc. –  James Freitag Aug 23 '11 at 18:11
    
About fixing the variables. You are correct technically; I suppose that I am sort of proving that the fibers of the relation are type definable. Perhaps it can be completed by varying the definition in a type definable way with the first coordinate. I would have to think for a bit to write this down carefully... –  James Freitag Aug 23 '11 at 18:13
    
The problem with this approach is not stationarity. It is that the $\varphi$-definition of a type over $\emptyset$ (or over $\mathrm{acl}(\emptyset)$ - we do want it to be stationary) is not obtained type-definably from a realisation. In the example I give in the answer below, you have a sequence of types which represnts the formula $x = y$, but no accumulation point of the sequence represents it. –  Itaï BEN YAACOV Aug 24 '11 at 15:23
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Ok, so my last comment is clearly wrong. Itai is right; it is the premise of the original question that is wrong. –  James Freitag Aug 24 '11 at 21:18
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