## Friday, December 6, 2013

### A Left-for-the-reader in stability theory

Yes, yes, I know I haven't posted in forever. I've been busy proving shit.

Completely out of left field, my research has indicated relevance to stability theory, an area of model theory that I've never had the urge to learn. Well, now I feel the urge. So much for purity of intention.

Ralph and I had a little fun today thinking through a left-for-the-reader in a paper that I hope to raid for its methods. So much fun that I want to share it with you, O gentle reader.

Definition 1:
• Let $$\lambda$$ be an infinite cardinal. A theory $$T$$ is $$\lambda$$-stable if for every model $$\mathbf{M} \models T$$ and every $$C \subset M$$ with $$|C| \leq \lambda$$, the space $$S_1(T/C)$$ of 1-types with parameters from $$C$$ has cardinality no greater than $$\lambda$$.
• $$T$$ is stable if it is $$\lambda$$-stable for some $$\lambda$$, and superstable if it is $$\lambda$$-stable for all cardinals $$\lambda$$ greater than some fixed cardinal.
Let $$L$$ be the first-order language with a constant symbol $$0$$, a binary function $$+$$, and a unary function $$F$$, and let $$T$$ be the theory axiomatized by the following sentences:
• $$+$$ is associative
• $$0 = x + x$$
• $$x + 0 = 0 + x = F(x) = F(F(x))$$
• $$x + y = F(x) + F(y) = F(x+y)$$
Note that these axioms imply that $$F$$ is an endomorphism of every model of $$T$$. It is less visually obvious, but still pretty quick, to show that the kernel of $$F$$ is always a strongly abelian congruence.

Exercise: this theory is unstable. That is, it fails to be $$\lambda$$-stable for any infinite cardinal $$\lambda$$.

Proof: For every model $$\mathbf{M} \models T$$, $$\mathbf{M}/ \ker F$$ is an abelian group $$\mathbf{A}$$ of exponent 2. Conversely, for every such group and every function $$j$$ from $$A$$ into cardinals, we can produce a split extension $$\mathbf{A} \leq \mathbf{M} \models T$$ where $$| F^{-1}(a) | = j(a)$$. (Note that $$\mathbf{M}$$ will not in general be a group, since addition will not typically be surjective.)

Let $$\lambda$$ be an infinite cardinal. We must produce a model of $$T$$ and a set $$C = \{ c_i \colon i < \lambda \}$$ such that $$T$$ has more than $$\lambda$$ types with parameters in $$C$$. Let $$\mathbf{A}$$ be the free $$\mathbb{Z}_2$$-module on generators $\{ c_i \colon i < \lambda \} \cup \{ x_g \colon g \in \omega^\lambda \}$and construct the split extension $$\mathbf{M}$$ described above with $$| F^{-1}(x_g + c_i) | = g(i) + 1$$, for each $$g \in \omega^\lambda$$.

Observe that if $$g_1 \neq g_2$$, then $$\mathrm{typ}(x_{g_1}/C) \neq \mathrm{typ}(x_{g_2}/C)$$; indeed, if $$g_2(i) > g_1(i)$$, then $\mathbf{M} \models \exists^{=g_1(i)} y \quad y \neq F(y) = x_{g_1} + c_i$while $\mathbf{M} \models \exists^{>g_1(i)} y \quad y \neq F(y) = x_{g_2} + c_i$This suffices to complete the proof, since $$\omega^\lambda = 2^\lambda > \lambda$$, showing that $$T$$ has more than $$\lambda$$ types with parameters from $$C$$.