Fact 3.4.

1. (1) $D_Q(\pi )\geqslant \lambda $ , $f\in \operatorname {Aut}(\mathfrak {C})\ \Rightarrow \ D_{f(Q)}(f(\pi ))\geqslant \lambda $ .

2. (2) If $\pi '\subseteq \pi $ then $D_Q(\pi )\leqslant D_{Q}(\pi ')$ .

3. (3) $D_Q(\pi )\leqslant \oplus _{j<n}D_{((\varphi _j,q_j))}(\pi )$ , where $\oplus $ denotes the Cantor sum.

Proof Items (1) and (2) follow by the definition. Item (3) follows from the following fact, which can be proven by a straightforward induction on $\alpha $ : for any ordinal $\alpha =A_0\dot \cup A_1\dot \cup \cdots \dot \cup A_{n-1}$ , if $\alpha _i$ is the order type of $A_i$ for $i<n$ , then $\alpha \leqslant \alpha _0\oplus \alpha _1\oplus \dots \oplus \alpha _{n-1}$ .

Proof The implication right-to-left is straightforward and holds even without the assumption about NSOP $_1$ . Let us show the implication left-to-right.

By Remark 3.2, there is a configuration satisfying conditions (1)–(7) from Definition 3.1 such that each $M^{\alpha +1}$ is $|M^{\alpha }|^+$ -saturated. By Kim’s lemma (Fact 2.7), we can replace the condition (7) from Definition 3.1, by: for each $\alpha <\lambda $ there exists $\bar {c}^{\alpha }=(c^{\alpha }_i)_{i<\omega }\models q^{\otimes \omega }_{\eta (\alpha )}|_{M^{\alpha }}$ such that $c^{\alpha }_0=b^{\alpha }$ and $\{\varphi _{\eta (\alpha )}(x;c^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent. By saturation of $M^{\alpha +1}$ , we find $(b^{\alpha }_i)_{i<\omega }\subseteq M^{\alpha +1}$ such that $(b^{\alpha }_i)_{i<\omega }\equiv _{M^{\alpha }b^{\alpha }}(c^{\alpha }_i)_{i<\omega }$ which is the desired sequence to finish the proof.

Proof Let us deal first with the case when $Q=((\varphi (x;y),q(y)))$ . Assume that $D_Q(\pi )\geqslant \lambda>0$ and let $(b^{\alpha },M^{\alpha })_{\alpha <\lambda }$ be as in Definition 3.1 ( $\eta $ is constant, so we skip it).

We have that $b^{\alpha }\models q|_{M^{\alpha }}$ and since $b^{\alpha }M^{\alpha }\subseteq M^{\alpha +1}$ , it is $b^{\alpha }\models q|_{M^0b^{<\alpha }}$ . We know also that q is $M^0$ -invariant and that $\{\varphi (x;b^{\alpha })\;|\;\alpha <\lambda \}$ is consistent. On the other hand, by Lemma 3.9, there exists a Morley sequence $(b^0_i)_{i<\omega }\models q^{\otimes \omega }|_{M^0}$ such that $b^0_0=b^0$ and $\{\varphi (x;b^0_i)\;|\;i<\omega \}$ is inconsistent. Therefore $C_{\varphi ,q}<\omega $ and so $\lambda \leqslant C_{\varphi ,q}<\omega $ .

We switch to the general case where

$$ \begin{align*}Q=\big((\varphi_0(x;y_0),q_0(y_0)),\ldots,(\varphi_{n-1}(x;y_{n-1}),q_{n-1}(y_{n-1}) \big).\end{align*} $$

Now, the function $\eta \in n^{\lambda }$ becomes important. We have that $\lambda =\bigcup _{j< n}\eta ^{-1}[j]$ , so we are done if we can show that $|\eta ^{-1}[j]|=0$ or $|\eta ^{-1}[j]|\leqslant C_{\varphi _j,k_j}<\omega $ for each $j< n$ . Fix some $j< n$ and assume that $|\eta ^{-1}[j]|>0$ . Let $\alpha <\lambda $ be the first index such that $\eta (\alpha )=j$ . Repeating the first part of this proof for $q_j$ we obtain what we need.

Proof The left-to-right implication is straightforward: by Lemma 3.9 there are proper $\eta \in n^\lambda $ and $\big ((b^{\alpha }_i)_{i<\omega },M^{\alpha }\big )_{\alpha <\lambda }$ , we set $M:=M^0$ and reuse $(b^{\alpha }_i)_{\alpha <\lambda ,i<\omega }$ .

Let us take care of the right-to-left implication. We will recursively define a sequence $((e^{\alpha }_i)_{i<\omega },M^{\alpha })_{\alpha <\lambda }$ satisfying all the six conditions from Lemma 3.9. Because we do not want to get lost in a notational madness, instead of introducing new subscripts, we will sketch a few first steps.

We set $M^0:=M$ and take $M^1\preceq \mathfrak {C}$ which is a $|M^0|^+$ -saturated, and which contains $M^0$ and $(b^0_i)_{i<\omega }$ . Consider a Morley sequence

$$ \begin{align*}(c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^1_i)_{i<\omega}\models q_{\eta(\lambda-1)}^{\otimes\omega}\otimes\cdots\otimes q_{\eta(1)}^{\otimes\omega}|_{M^1}.\end{align*} $$

We have

$$ \begin{align*}(b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^1_i)_{i<\omega}\equiv_{M^0(b^0_i)_{i<\omega}} (c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^1_i)_{i<\omega}.\end{align*} $$

Now, let $M^2\preceq \mathfrak {C}$ be $|M^1|^+$ -saturated such that $M^1(c^1_i)_{i<\omega }\subseteq M^2$ and let us choose one more Morley sequence

$$ \begin{align*}(d^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(d^2_i)_{i<\omega}\models q_{\eta(\lambda-1)}^{\otimes\omega}\otimes\cdots\otimes q_{\eta(2)}^{\otimes\omega}|_{M^2}.\end{align*} $$

We see that

$$ \begin{align*}(d^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(d^2_i)_{i<\omega}\equiv_{M^1(c^1_i)_{i<\omega}}(c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^2_i)_{i<\omega},\end{align*} $$

and so

$$ \begin{align*} (d^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(d^2_i)_{i<\omega}^\smallfrown(c^1_i)_{i<\omega}^\smallfrown(b^0_i)_{i<\omega} &\equiv_{M^0} (c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^1_i)_{i<\omega}^\smallfrown(b^0_i)_{i<\omega} \\ &\equiv_{M^0} (b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^0_i)_{i<\omega}. \end{align*} $$

Set $(e^2_i)_{i<\omega }:=(d^2_i)_{i<\omega }$ , $(e^1_i)_{i<\omega }:=(c^1_i)_{i<\omega }$ and $(e^0_i)_{i<\omega }:=(b^0_i)_{i<\omega }$ .

Continuing this process we will obtain a sequence of models $M^0\preceq M^1\preceq \cdots \preceq M^{\lambda -1}$ and Morley sequences $(e^{\alpha }_i)_{i<\omega }\models q_{\eta (\alpha )}^{\otimes \omega }|_{M^{\alpha }}$ , where $\alpha <\lambda $ , such that $M^{\alpha }(e^{\alpha }_i)_{i<\omega }\subseteq M^{\alpha +1}$ and

$$ \begin{align*}(e^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(e^0_i)_{i<\omega}\equiv_M (b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^0_i)_{i<\omega}.\end{align*} $$

Since $\operatorname {dom}(\pi (x))\subseteq M=M^0$ , we have also that $\pi (x)\,\cup \,\{\varphi _{\eta (\alpha )}(x;e^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent and that $\{\varphi _{\eta (\alpha )}(x;e^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent for each $\alpha <\lambda $ . Therefore all the conditions of Lemma 3.9 are satisfied for $((e^{\alpha }_i)_{i<\omega },M^{\alpha })_{\alpha <\lambda }$ .

Proof Assume that $D_Q(\pi )\geqslant \lambda \in \mathbb {N}_{>0}$ , then by Lemma 3.12 there exist $\eta \in n^{\lambda }$ , $(b^{\alpha }_i)_{\alpha <\lambda ,i<\omega }$ , and $M\preceq \mathfrak {C}$ such that:

1. (1) $\operatorname {dom}(\pi )\subseteq M$ ,

2. (2) $q_0,\ldots ,q_{n-1}$ are M-invariant,

3. (3) $\pi (x)\cup \{\varphi _{\eta (\alpha )}(x;b^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent,

4. (4) $\{\varphi _{\eta (\alpha )}(x;b^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent for each $\alpha <\lambda $ ,

5. (5)

   $$ \begin{align*}(b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^0_i)_{i<\omega}\models q_{\eta(\lambda-1)}^{\otimes\omega}\otimes\dots\otimes q_{\eta(0)}^{\otimes\omega}|_{M}.\end{align*} $$

Let $N\preceq \mathfrak {C}$ contain M and $\operatorname {dom}(\pi ')$ , and let

$$ \begin{align*}(c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^0_i)_{i<\omega}\models q_{\eta(\lambda-1)}^{\otimes\omega}\otimes\dots\otimes q_{\eta(0)}^{\otimes\omega}|_{N}.\end{align*} $$

Then naturally $\operatorname {dom}(\pi ')\subseteq N$ and $q_1,\ldots ,q_n$ are N-invariant. Because

$$ \begin{align*}(b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^0_i)_{i<\omega}\equiv_M (c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^0_i)_{i<\omega},\end{align*} $$

we have also that $\{\varphi _{\eta (\alpha )}(x;c^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent for each $\alpha <\lambda $ , and that $\pi (x)\cup \{\varphi _{\eta (\alpha )}(x;c^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent. Moreover, $\pi \vdash \pi '$ implies that $\pi '(x)\cup \{\varphi _{\eta (\alpha )}(x;c^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent. Hence, Lemma 3.12 gives us $D_Q(\pi ')\geqslant \lambda $ .

Proof Because $\pi \cup \{\psi _i\}\vdash \pi \cup \{\bigvee _{j}\psi _j\}$ for each $i\leqslant m$ , Lemma 3.15 gives us that

$$ \begin{align*}\max\limits_{j\leqslant m} D_Q(\pi\cup\{\psi_j\})\leqslant D_Q\Big(\pi\cup\Big\{\bigvee\limits_{j\leqslant m}\psi_j\Big\}\Big).\end{align*} $$

Hence it is enough to show that

$$ \begin{align*}D_Q\Big(\pi\cup\Big\{\bigvee\limits_{j\leqslant m}\psi_j\Big\}\Big)\geqslant\lambda\quad\Rightarrow\quad \max\limits_{j\leqslant m} D_Q(\pi\cup\{\psi_j\})\geqslant\lambda.\end{align*} $$

Let $D_Q(\pi \cup \{\bigvee _{j}\psi _j\})\geqslant \lambda $ , i.e., there exists $\eta \in n^\lambda $ and $(b^{\alpha },M^{\alpha })_{\alpha <\lambda }$ as in Definition 3.1, in particular,

• • $\operatorname {dom}(\pi \cup \{\bigvee _{j}\psi _j\})\subseteq M^0$ ,

• • $\pi \cup \{\bigvee _j\psi _j\}\cup \{\varphi _{\eta (\alpha )}(x;b^{\alpha })\;|\;\alpha <\lambda \}$ is consistent.

Thus there is $i_0\leqslant m$ such that $\pi \cup \{\psi _{i_0}\}\cup \{\varphi _{\eta (\alpha )}(x;b^{\alpha })\;|\;\alpha <\lambda \}$ is consistent. Because $\operatorname {dom}(\pi \cup \{\psi _{i_0}\})\subseteq \operatorname {dom}(\pi \cup \{\bigvee _j\psi _j\})\subseteq M^0$ , we have that $D_Q(\pi \cup \{\psi _{i_0}\})\geqslant \lambda $ and so also $\max _{j} D_Q(\pi \cup \{\psi _j\})\geqslant \lambda $ .

Proof Because for each $\beta $ we have $D_Q(\pi _{\beta })\geqslant \lambda $ , by Lemma 3.12, for each $\beta $ there exists a function $\eta _{\beta }\in n^{\lambda }$ , a sequence of sequences $(b^{\beta ,\alpha }_i)_{\alpha <\lambda ,i<\omega }$ and a model $M^{\beta }\preceq \mathfrak {C}$ such that:

1. (1) $\operatorname {dom}(\pi _{\beta })\subseteq M^{\beta }$ ,

2. (2) q is $M^{\beta }$ -invariant,

3. (3) $\pi _{\beta }(x)\cup \{\varphi _{\eta _{\beta }(\alpha )}(x;b^{\beta ,\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent,

4. (4) $\{\varphi _{\eta _{\beta }(\alpha )}(x;b^{\beta ,\alpha }_i)\;|\;i<\omega \}$ is inconsistent for each $\alpha <\lambda $ ,

5. (5)

   $$ \begin{align*}(b^{\beta,\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^{\beta,0}_i)_{i<\omega}\models q^{\otimes\omega}\otimes\dots\otimes q^{\otimes\omega}|_{M^{\beta}}.\end{align*} $$

Let $N\preceq \mathfrak {C}$ be such that $\bigcup \limits _{\beta }M^{\beta }\subseteq N$ . Moreover, let us pick up the following Morley sequences:

$$ \begin{align*}(c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^0_i)_{i<\omega}\models q^{\otimes\omega}\otimes\dots\otimes q^{\otimes\omega}|_N,\end{align*} $$

and, without loss of generality, let $\{\varphi _{\eta _{\beta }(\alpha )}(x;y)\;|\;\beta ,\alpha <\lambda \}=\{\varphi _0(x;y),\ldots ,\varphi _{r-1} (x;y)\}$ for some $r\leqslant n$ . We introduce $\psi (x;y):=\bigvee \limits _{i<r}\varphi _i(x;y)$ , and note that

$$ \begin{align*}\pi(x)\,\cup\,\{\psi(x;c^{\alpha}_0)\;|\;\alpha<\lambda\}\end{align*} $$

is consistent (otherwise, by compactness, $\pi _{\beta }(x)\,\cup \,\{\psi (x;c^{\alpha }_0)\;|\;\alpha <\lambda \}$ is inconsistent for some $\beta $ , which is impossible, since

$$ \begin{align*}(c^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(c^0_i)_{i<\omega}\equiv_{M^{\beta}}(b^{\beta,\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^{\beta,0}_i)_{i<\omega}\end{align*} $$

and $\pi _{\beta }(x)\cup \{\varphi _{\eta _{\beta }(\alpha )}(x;b^{\beta ,\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent). Consider $d\models \pi (x)\,\cup \,\{\psi (x;c^{\alpha }_0)\;|\;\alpha <\lambda \}$ , then for each $\alpha <\lambda $ there is $i_{\alpha }<r$ such that $\models \varphi _{i_{\alpha }}(d,c^{\alpha }_0)$ . Let $\eta \in n^{\lambda }$ be given by $\eta :\alpha \mapsto i_{\alpha }$ .

As we want to use Lemma 3.12 to show that $D_Q(\pi )\geqslant \lambda $ , and we have already defined $\eta $ , $N\preceq \mathfrak {C}$ and $(c^{\alpha }_i)_{\alpha <\lambda ,i<\omega }$ , we need to verify whether all the five conditions from Lemma 3.12 hold. Obviously, $\operatorname {dom}(\pi )\subseteq N$ and q is N-invariant, so we have the first and the second condition. The fifth condition is naturally satisfied by the choice of $(c^{\alpha }_i)_{\alpha <\lambda ,i<\omega }$ . The third condition says that $\pi (x)\,\cup \,\{\varphi _{\eta (\alpha }(x,c^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent, which is witnessed by element d. For the fourth condition, we need to note that $\{\varphi _{\eta (\alpha )}(x,c^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent for every $\alpha <\lambda $ . Because $\eta (\alpha )< r$ , there exist $\beta $ and $\alpha '<\lambda $ such that $\eta (\alpha )=\eta _{\beta }(\alpha ')$ . We know that $\{\varphi _{\eta _{\beta }(\alpha ')}(x,b^{\beta ,\alpha '}_i)\;|\;i<\omega \}$ is inconsistent, that $(b^{\beta ,\alpha '}_i)_{i<\omega }\models q^{\otimes \omega }|_{M^{\beta }}$ and that $(c^{\alpha }_i)_{i<\omega }\models q^{\otimes \omega }|_{M^{\beta }}$ . Thus also $\{\varphi _{\eta (\alpha )}(x,c^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent.

Proof The proof is completely standard, but let us sketch it anyway. Consider

$$ \begin{align*}\tilde{\pi}(x):=\pi(x)\,\cup\,\{\neg\psi(x)\in L(A)\;|\;D_Q(\pi(x)\,\cup\,\{\psi(x)\})<D_Q(\pi(x))\}.\end{align*} $$

By Lemmas 3.15 and 3.16, the set $\tilde {\pi }$ is a partial type over A. Let $p(x)\in S(A)$ be any extension of $\tilde {\pi }$ .

If $D_Q(p)<D_Q(\pi )$ then by Lemma 3.17 there exists a finite subset $p_0(x)\subseteq p(x)$ such that $D_Q(p_0)<D_Q(\pi )$ . Let $\psi (x):=\bigwedge p_0(x)\in L(A)$ , we have that

$$ \begin{align*}D_Q(\pi(x)\cup\{\psi(x)\})\leqslant D_Q(\psi(x))\leqslant D_Q(p_0(x))<D_Q(\pi(x)).\end{align*} $$

Thus $\neg \psi (x)\in \tilde {\pi }(x)\subseteq p(x)$ and we got a contradiction with $\psi (x)\in p(x)$ .

Proof As T is NSOP $_1$ , there is some $\lambda <\omega $ such that $D_Q(\pi )=\lambda $ . Let $\eta $ and $(b^\alpha ,M^\alpha )_{\alpha <\lambda }$ be as in Definition 3.1. For each $\alpha <\lambda $ there exists $k_{\alpha }<\omega $ such that $\varphi _{\eta (\alpha )}(x;b^{\alpha })$ $k_{\alpha }$ -divides over $M^{\alpha }$ . Set $K:=\max \{k_0,\ldots ,k_{\lambda -1}\}$ . Then, by definition, $D(\pi ,\{\varphi _0,\ldots ,\varphi _{n-1}\},k)\geqslant \lambda $ , provided $k\geqslant K$ .

Proof Assume that , which means that $\operatorname {tp}(a/N)$ Kim-divides over M. Let $\varphi (x,b)\in \operatorname {tp}(a/N)$ Kim-divide over M. There exists an M-invariant $q(y)\in S(\mathfrak {C})$ extending $\operatorname {tp}(b/M)$ such that $\varphi (x,b)\ q$ -divides over M. We set $Q=\big ((\varphi (x;y),q(y))\big )$ . By Lemma 3.15 and Corollary 3.11, it follows that

$$ \begin{align*}D_Q(\operatorname{tp}(a/N))\leqslant D_Q(\operatorname{tp}(a/M)\;\cup\;\{\varphi(x;b)\})=:\lambda<\omega.\end{align*} $$

Therefore there exists a sequence $(b^{\alpha },M^{\alpha })_{\alpha <\lambda }$ such that:

1. (1) $Mb\subseteq M^0$ ,

2. (2) q is $M^0$ -invariant,

3. (3) $M^{\alpha }\preceq \mathfrak {C}$ and $M^{\alpha +1}$ is $|M^{\alpha }|^+$ -saturated for each $\alpha <\lambda $ ,

4. (4) $b^{\alpha }M^{\alpha }\subseteq M^{\alpha +1}$ for each $\alpha +1<\lambda $ ,

5. (5) $b^{\alpha }\models q|_{M^{\alpha }}$ for each $\alpha <\lambda $ ,

6. (6) $\operatorname {tp}(a/M)\;\cup \;\{\varphi (x;b)\}\;\cup \;\{\varphi (x;b^{\alpha })\;|\;\alpha <\lambda \}$ is consistent,

7. (7) $\varphi (x;b^{\alpha })$ Kim-divides over $M^\alpha $ for each $\alpha <\lambda $ .

By Lemma 3.12, we can modify $M^0$ and so assume that $M^0$ is $|M|^+$ -saturated, which we do. We set $M^{-1}:=M$ and $b^{-1}:=b$ . Checking that $(b^{\alpha },M^{\alpha })_{-1\leqslant \alpha <\lambda }$ witnesses that $D_Q(\operatorname {tp}(a/M))\geqslant \lambda +1$ is routine (note that, as $\lambda <\omega $ , $(b^{\alpha },M^{\alpha })_{-1\leqslant \alpha <\lambda }$ can be naturally indexed by the elements of $\lambda +1$ ).

Proof Let $D_Q(a/M)=:\lambda <\omega $ , it means by Lemma 3.9 that there exists $\big ( (b^{\alpha }_i)_{i<\omega }, M^{\alpha }\big )_{\alpha <\lambda }$ such that:

1. (1) $M\subseteq M^0$ ,

2. (2) q is $M^0$ -invariant (which comes for free as q is M-invariant and $M\subseteq M^0$ ),

3. (3) $M^{\alpha }\preceq \mathfrak {C}$ for all $\alpha <\lambda $ , $(M^{\alpha })_{\alpha <\lambda }$ is continuous, and $M^{\alpha +1}$ is $|M^{\alpha }|^+$ -saturated for all $\alpha +1<\lambda $ ,

4. (4) $b^{\alpha }_{<\omega }M^{\alpha }\subseteq M^{\alpha +1}$ for all $\alpha +1<\lambda $ ,

5. (5) $b^{\alpha }_{<\omega }\models q^{\otimes \omega }|_{M^{\alpha }}$ for all $\alpha <\lambda $ ,

6. (6) $\operatorname {tp}(a/M)\;\cup \;\{\varphi (x;b^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent,

7. (7) $\{\varphi (x;b^{\alpha }_i)\;|\;i<\omega \}$ is inconsistent for all $\alpha <\lambda $ .

Step 1. Let $d\models \operatorname {tp}(a/M)\;\cup \;\{\varphi (x;b^{\alpha }_0)\;|\;\alpha <\lambda \}$ and let $N"\preceq \mathfrak {C}$ be such that $aN\equiv _M dN"$ . By the existence axiom for forking independence, there exists $N'\equiv _{Md}N"$ such that

Because , we have that and then also that . As $N'$ is $|M|^+$ -saturated, Proposition 5.4 from [Reference Casanovas4] assures us that and combine into . Then monotonicity of gives us .

Step 2. Because $aN\equiv _M dN"\equiv _M dN'$ and $q(y)$ is M-invariant, we have that $D_Q(a/N)=D_Q(d/N')$ , so it is enough to show that $D_Q(d/N')\geqslant \lambda $ . Note that $\operatorname {tp}(d/N')\;\cup \;\{\varphi (x;b^{\alpha }_0)\;|\;\alpha <\lambda \}$ is consistent.

Step 3. Note that

$$ \begin{align*}b^{<\lambda}_{<\omega}=(b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^0_i)_{i<\omega}\models q^{\otimes\omega}\otimes\dots\otimes q^{\otimes\omega}|_M.\end{align*} $$

Since and the type $q^{\otimes \omega }\otimes \dots \otimes q^{\otimes \omega }|_M$ ( $\lambda $ -many repetitions) is stationary, we have that also

$$ \begin{align*}b^{<\lambda}_{<\omega}=(b^{\lambda-1}_i)_{i<\omega}^\smallfrown\ldots^\smallfrown(b^0_i)_{i<\omega}\models q^{\otimes\omega}\otimes\dots\otimes q^{\otimes\omega}|_{N'}.\end{align*} $$

Step 4. Now, Lemma 3.12 gives us easily $D_Q(d/N')\geqslant \lambda $ .

Fact 3.24. (Arbitrary T) Let be any invariant ternary relation satisfying symmetry [over models] and extension [over models]. Then for any parameter set [any model] M and $a,b$ , if $\operatorname {tp}(ab/M)$ is -stationary then $tp(a/M)$ is -stationary.

Proof Let $N\succeq M$ and let $a'\equiv _M a"\equiv _M a$ be such that and . By extension, symmetry, and invariance, there are $b',b"$ such that $a'b'\equiv _M a"b"\equiv _M ab$ , and . Then by -stationarity of $\operatorname {tp}(ab/M)$ we have that $a'b'\equiv _N a"b"$ , so in particular $a'\equiv _N a"$ .

Proof We start with the proof of (1). The implication from left to right follows from Fact 3.24. Assume $p|_{x_0}$ is stationary for every finite subtuple $x_0$ of x and let $q_0,q_1$ be global non-forking extensions of p. Then for every finite subtuple $x_0$ of x we have that $q_0|_{x_0}$ and $q_1|_{x_0}$ are non-forking extensions of $p|_{x_0}$ , so by stationarity of $p|_{x_0}$ they are equal, hence $q_0=q_1$ and p is stationary.

The argument for Kim-stationarity is exactly the same.

Proof We start with the proof of (1). By Remark 3.25, it is enough to show that for each $n<\omega $ , the type $q^{\otimes n}|_A$ is stationary. This can be shown inductively by the use of the following claim:

Claim. If $p(x)\in S(\mathfrak {C})$ is A-invariant, such that $p|_A$ is stationary, then $q\otimes p|_A$ is stationary.

Proof of the claim

Let $B\supseteq A$ , let $ab\models q\otimes p|_A$ be such that , and let $a'b'\models q\otimes p|_B$ . Our goal is $ab\models q\otimes p|_B$ . As , we have . Since $b'\models p|_B$ and p is A-invariant, also . Stationarity of $p|_A$ implies that there exists $f\in \operatorname {Aut}(\mathfrak {C}/B)$ such that $f(b)=b'$ , thus $b\models p|_B$ . Because $a\models q|_{Ab}$ , we have that $f(a)\models q|_{Ab'}$ . Since q is A-invariant, we obtain that . On the other hand, changes via f into , then symmetry, base monotonicity, normality and monotonicity of give us . By transitivity of , and imply that , which by the fact that $a'\models q|_{Bb'}$ (so ) and by stationarity of $q|A$ implies $f(a)\equiv _{Bb'}a'$ . As $f(a)\models q|_{Bb'}$ , we have that $a\models q|_{Bb}$ , thus $ab\models q\otimes p|_B$ and we end the proof of the claim.

Now, we are moving to the proof of (2). Assume that $\bar {a}=(a_i)_{i\in I}\models q^{\otimes I}|_A$ and $\bar {b}=(b_i)_{i\in I}\models q^{\otimes I}|_A$ are such that and for some $B\supseteq A$ . We need to show that $\operatorname {tp}(\bar {a}/B)=\operatorname {tp}(\bar {b}/B)$ , which holds if and only if for each $n<\omega $ and each $i_1,\ldots ,i_n\in I$ , such that $i_1<\cdots <i_n$ , we have $\operatorname {tp}(a_{i_n}\ldots a_{i_1}/B)=\operatorname {tp}(b_{i_n}\ldots b_{i_1}/B)$ . Consider such $i_1,\ldots ,i_n\in I$ and choose any infinite sequence $I_0\subseteq I$ starting with $(i_1,\ldots ,i_n)$ which has the order type of $(\omega ,<)$ . Let $a_{I_0}:=(a_i)_{i\in I_0}$ and $b_{I_0}:=(b_i)_{i\in I_0}$ . Then $a_{I_0}\models q^{\otimes \omega }|_A$ , $b_{I_0}\models q^{\otimes \omega }|_A$ . Monotonicity of gives us that and . As $q^{\otimes \omega }|_A$ is stationary we have that $a_{I_0}\equiv _B b_{I_0}$ , so in particular $\operatorname {tp}(a_{i_n}\ldots a_{i_1}/B)=\operatorname {tp}(b_{i_n}\ldots b_{i_1}/B)$ .

The proof of (3) is similar to the proof of (2), as the only property of used was monotonicity, which also holds for .

Question 3.29. Let $M\preceq \mathfrak {C}$ and let $q(y)\in S(\mathfrak {C})$ be M-invariant such that $q|_M$ is Kim-stationary. Is the type $q^{\otimes \omega }|_M$ Kim-stationary?

Proof (1) is exactly the content of Lemma 3.22. For (2), we use Lemma 3.26 and repeat the proof of Lemma 3.22, where in Step 1, instead of the saturation assumption we use transitivity, which naturally holds in simple theories.

(3) also uses a modification of the proof of Lemma 3.22. More precisely, in Step 1, we use that implies and that transitivity (over arbitrary sets) holds in NSOP $_1$ with existence. We obtain and then . Then, Step 3 follows by the definition of strong Kim-stationarity. The rest (Step 2 and Step 4) remains the same.

Proof By compactness, we may assume that $I=\{i_1,\dots ,i_{n}\}$ is finite. Assume by induction that $bc_1\dots c_{n-1}\equiv f(b)c_1\dots c_{n-1}$ . Pick $\tilde f\in \operatorname {Aut}(\mathfrak {C})$ sending $bc_1\dots c_{n-1}$ to $f(b)c_1\dots c_{n-1}$ . Then, as $\tilde f(q)=q$ by $B/A$ -invariance, we have

$$\begin{align*}\tilde f(c_n)\models \tilde f(\operatorname{tp}(c_n/bc_1\dots c_{n-1}))=\tilde f(q)|_{\tilde f(b)\tilde f(c_1)\dots \tilde f(c_{n-1)}}=q|_{bc_1\dots c_{n-1}}=\operatorname{tp}(c_n/bc_1\dots c_{n-1}) \end{align*}$$

which shows that $f(b)c_1\dots c_n\equiv f(b)c_1\dots c_{n-1} \tilde f(c_n)=\tilde f(bc_1\dots c_n)\equiv bc_1\dots c_n$ .

Proof As $tp(b/A)=r|_A=q|_A$ , we have that $q,r\in S(A\,\cup \,\operatorname {tp}(b/A)(\mathfrak {C}))$ and we can repeat the proof of Proposition 3.15 from [Reference Kaplan and Ramsey20].