Forking but not dividing

Question

Definition. A formula $\phi(x,a)$ divides over a set $B$ if there are $k<\mathbb{N}$ and a sequence $(a_i)_{i<\omega}$ such that

(1) $\text{tp}(a/B)=\text{tp}(a_i/B)$, for all $i<\omega$;

(2) $\{\phi(x,a_i)\}_{i<\omega}$ is $k$-inconsistent.

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Definition. A formula $\phi(x,a)$ forks over a set $B$ if there are $n\in\mathbb{N}$ and formulas $\psi(x,b_1),\dots, \psi(x,b_n)$ such that

(1) for each $i=1,\dots, n$, the formula $\psi_i(x,b_i)$ divides over $B$;

(2) $\phi(x,a)\models \bigvee_{i=1}^{n} \psi_i(x,b_i)$.

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It is clear that dividing implies forking.

Question. Does forking always imply dividing? If no, is there a formula that forks but does not divide?

Answer 1

No, forking does not always imply dividing. In simple theories dividing and forking are the same, but they are not the same in general. Look at the following example.

Example. Let $\mathcal{L}=\{ R^{(3)} \}$ be a language which consists of a ternary relation. Consider the $\mathcal{L}$-structure $\mathcal{M}=\big(\mathbb{S}^1, R \big)$ where $\mathbb{S}^1$ is the unit circle around the origin on the plane and $R(x,y,z)$ holds if and only if $y$ lies on the shorter arc between $x$ and $z$, ordered clock-wise, including the endpoints. Now, let $a,b$ and $c$ be three equidistant points on $\mathbb{S}^1$. Then

$$\mathcal{M}\models \forall x\bigg( R(a,x,b)\vee R(b,x,c) \vee R(c,x,a) \bigg)$$

Claim(1). Each of $R(a,x,b), R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$.

Proof of Claim(1).

We will show, for instance, the formula $R(a,x,b)$ 2-divides over $\emptyset$. Let $a, a_0,b_0, a_1,b_1,\dots, b$ be a sequence of consecutive points on $\mathbb{S}^1$. Then $(a_ib_i)_{i<\omega}$ is a sequence such that

i) $\text{tp}(a_i,b_i)=\text{tp}(a,b)$ for each $i<\omega$;

ii) $\big\{ R(a_i,x,b_i) \big\}_{i<\omega}$ is 2-inconsistent.

Therefore $R(a,x,b)$ 2-divides over $\emptyset$. Similarly we can prove that $R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$ as well.

Claim(2). The formula $x=x$ forks over $\emptyset$ but does not divide over $\emptyset$.

Proof of Claim(2).

Since $x=x\models R(a,x,b)\vee R(b,x,c)\vee R(c,x,a)$ and by Claim(1) each of $R(a,x,b), R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$, $x=x$ forks over $\emptyset$. But $x=x$ does not divide over $\emptyset$, because if $x=x$ divides over $\emptyset$, then there are $k<\omega$ and an $\emptyset$-indiscernible sequence $(a_i)_{i<\omega}$ such that $\big\{ x=x \big\}_{i<\omega}$ is $k$-inconsistent which is a contradiction.

Therefore the formula $x=x$ forks but does not divide.

Answer 2

I hope you accept Mostafa's answer, since he described the canonical example of a formula which forks but does not divide ($x=x$ in the circular order). I believe this example is originally due to Byunghan Kim, in his thesis on simple theories.

But since you asked for more examples in the comments, I'll just add some more examples and references.

• Consider the two-sorted structure $(X,\mathcal{P}(X); \in)$ where $X$ is an infinite set, $\mathcal{P}(X)$ is its powerset, and the only symbol in the language is the membership relation $\in$ between the two sorts. Let $A$ be an infinite and coinfinite subset of $X$, and let $B$ be its complement. Then $x = x$ implies $(x\in A)\lor (x\in B)$, and both $(x\in A)$ and $(x\in B)$ divide over $\emptyset$. Why? Any two infinite and coinfinite subsets of $X$ have the same type over the empty set (they are even conjugate by an automorphism), and $X$ can be partitioned into infinitely many pairwise disjoint infinite sets. So $x = x$ forks but does not divide over $\emptyset$.

• In both the circular order example and the powerset example, the formula which forks but doesn't divide is $x=x$, considered over $\emptyset$. But many variations on these examples can be cooked up to satisfy different properties, e.g. to show that forking $\neq$ dividing in general even over models. See Section 5 of Forking in NTP$_2$ theories by Artem Chernikov and Itay Kaplan (and note that Example 5.1, which they partially credit to Martin Ziegler, is really a combination of the circular order and the powerset examples). This paper is also the reference for the theorem mentioned by tomasz that forking $=$ dividing over models in NTP$_2$ theories.

When thinking about the difference between forking and dividing, we should be careful about whether we're talking about formulas or complete types. That is:

• "forking $=$ dividing for formulas" is the statement that for all formulas $\varphi(x,y)$, all sets $A$, and all parameters $b$, if $\varphi(x,b)$ forks over $A$, then $\varphi(x,b)$ divides over $A$.
• "forking $=$ dividing for complete types" is the statement that for every complete type $p(x)\in S(B)$ and every $A\subseteq B$, if $p(x)$ forks over $A$, then $p(x)$ divides over $A$.

Forking $=$ dividing for formulas implies forking $=$ dividing for complete types. Why? Suppose $p(x)\in S(B)$ forks over $A\subseteq B$. Then $p(x)$ contains a formula $\varphi(x,b)$ which forks over $A$. By forking $=$ dividing for formulas, $\varphi(x,b)$ divides over $A$, so $p(x)$ divides over $A$.

But the converse does not hold. Specifically, you might have a formula which forks but does not divide, but nevertheless any complete type containing it divides.

The circular order example and the powerset example above both give examples of forking $\neq$ dividing for complete types. In the circular order example, there is a unique $1$-type $p(x)$ over $\emptyset$, and in the powerset example, there is a unique $1$-type $p(x)$ in the "$X$ sort" over $\emptyset$, and in both examples $p(x)$ forks but does not divide over $\emptyset$. Every example I know in which forking $\neq$ dividing for complete types is closely related to one of these two examples. (That's not to say there aren't other quite different examples - I just don't know them.)

But I do know lots of other of examples of theories in which forking $=$ dividing for complete types, but forking $\neq$ dividing for formulas. This sort of behavior seems to be very common in theories without the strict order property which are not simple.

• In Forking and dividing in Henson graphs, Gabe Conant showed that forking $=$ dividing for complete types in the theory $T_n$ of the generic $K_n$-free graph, for $n\geq 3$, but that every such theory has a formula which forks but does not divide. (These theories are all SOP$_3$ but NSOP$_4$.) The example in $T_3$ is $\bigvee_{1\leq i<j\leq 4} (xEb_i\land xEb_j)$, where $b_1,b_2,b_3,b_4$ is any $4$-tuple with no edges between any of the $b_i$. Each formula $(xEb_i\land xEb_j)$ divides over $\emptyset$, but the disjunction of all $6$ of these formulas does not divide over $\emptyset$.

• In the generic binary function (the model companion of the empty theory in the language with a single binary function symbol $f$), the formula $\varphi(x;b_1,b_2)$ given by $(f(x,b_1) = b_2)\lor x = b_1$ forks but does not divide over $\emptyset$ whenever $b_2$ is not in the substructure $\langle b_1\rangle$ generated by $b_1$. This is because for every indiscernible sequence $(b_1^ib_2^i)_{i\in \omega}$, either the sequence $(b_1^i)$ is constant, in which case $\{x = b_1^i\mid i\in \omega\}$ is consistent, or the $b_1^i$ are pairwise distinct, in which case $\{f(x,b_1^i) = b_2^i\mid i\in \omega\}$ is consistent. Nick Ramsey and I showed that forking $=$ dividing for complete types in this theory (and that the theory is NSOP$_1$) in our paper Generic expansion and Skolemization in NSOP$_1$ theories.

• As far as I know, it could be that for all NSOP$_1$ unsimple theories, forking $=$ dividing for complete types, but there is a formula which forks but does not divide. ("As far as I know" doesn't mean "I conjecture"! I haven't even considered this question in every known example.) Certainly examples which are very similar to the previous one occur in other NSOP$_1$ unsimple theories. For another example with a similar flavor, see Proposition 4.23 in my paper Independence in generic incidence structures with Gabe Conant.