Morley sequences and forking-independent sequences Let $T$ be a complete theory and $\mathfrak{U}$ a monster model, and $C$ a small parameter set. A $C$-indiscernible sequence $\mathcal{I}=(a_i)_{i\in I}$ is $C$-independent if $a_i\perp^f_C \{a_j:j<i\}$ for each $i\in I$. On the other hand, $\mathcal{I}$ is a $C$-Morley sequence if there exists a $C$-invariant global type $p(x)\in S(\mathfrak{U})$ such that $$\operatorname{tp}(a_{i_1},\dots,a_{i_n}/C)\subseteq\underbrace{p\otimes\dots\otimes p}_{n \text{ times}},$$ ie such that each $a_{i_k}$ realizes $p|_{Ca_{i_1}\dots a_{i_{k-1}}}$, for every $i_1<\dots<i_n\in I$. (In this latter case, say that $\mathcal{I}$ is "generated" by $p$.) Because a $C$-invariant type cannot fork over $C$, it is the case that any $C$-Morley sequence will also be $C$-independent. I am wondering about when the converse of this fact holds:

Question: In what circumstances is a $C$-indiscernible and $C$-independent sequence also a $C$-Morley sequence?


Here are two examples. Let $T$ be the theory of the random graph, say with the symbol $R$ for adjacency:
First, thanks to James Hanson for the following argument that the implication does hold when $\mathcal{I}$ is a sequence of elements; suppose $\mathcal{I}$ is a (non-constant) $C$-indiscernible and $C$-independent sequence of elements of $\mathfrak{U}$. Let $C_0=\{c\in C:cRa_0\}$. We have two cases; if $\mathfrak{U}\models a_0 Ra_1$, then let $p(x)$ be the unique unrealized global type containing $$\{\neg(xRc):c\in C\setminus C_0\}\cup\{xRc:c\in(\mathfrak{U}\setminus C)\cup C_0\}.$$ If instead $\mathfrak{U}\models\neg(a_0 Ra_1)$, then let $p(x)$ be the unique unrealized global type containing $$\{\neg(xRc):c\in \mathfrak{U}\setminus C_0\}\cup\{xRc:c\in C_0\}.$$ In either case, $p$ is $C$-invariant, and it generates $\mathcal{I}$, whence $\mathcal{I}$ is indeed a $C$-Morley sequence.
On the other hand, if I'm not mistaken, here is an example showing that the implication does not hold for arbitrary sequences in $T$. Let $\mathcal{I}=(a_ib_i)_{i\in\omega}$ be any indiscernible sequence of $2$-tuples such that (i) $\{a_i,b_i\}$ and $\{a_j,b_j\}$ are disjoint for each $i\neq j$, and (ii) $\mathfrak{U}\models a_{i+1}Ra_i$ and $\mathfrak{U}\models\neg(a_{i+1}Rb_i)$ for each $i\in\omega$. By condition (i), and the characterization of forking in $T$, $\mathcal{I}$ is $\varnothing$-independent. But $\mathcal{I}$ is not generated by any $\varnothing$-invariant type $p(xy)$; indeed, suppose $p(xy)$ generates $\mathcal{I}$. Since $a_1b_1$ realizes $p|_{a_0b_0}$, $p$ must contain the formulas $xRa_0$ and $\neg(xRb_0)$. But there is an automorphism of $\mathfrak{U}$ swapping $a_0$ and $b_0$, so $p$ is indeed not $\varnothing$-invariant.

As a smaller question, does the counterexample above look right? For the main question, if there is not much to be said in general, any references that have investigated this question in a specific context would be much appreciated. I am particularly interested in whether this implication holds generally in NIP theories.
 A: The counterexample is fine, but basically the same example works for the theory of a single equivalence relation with two classes, both infinite (which is $\aleph_0$-stable, not only NIP): if $(a_n)_n$ is an injective sequence of elements in one class and $(b_n)_n$ is an injective sequence in the other class, then the sequence $(a_nb_n)_n$ is indiscernible and independent, but not a Morley sequence over $\emptyset$ for the same reason as in your example.
On the other hand, I think in this example also you have that indiscernible independent sequences over models (or any set containing $\operatorname{acl}^{\mathrm{eq}}(\emptyset)$) is a Morley sequence. The random graph example shows that this is not true even over models. So I think there is a chance of a positive result if besides tameness, you also assume that the base is either a model or $\operatorname{acl}^{\mathrm{eq}}$-closed.
A: First a comment on terminology: A $C$-indiscernible $C$-(forking)-independent sequence is usually called a Morley sequence over $C$. This was the original definition of Morley sequence, and if you just say "Morley sequence", people will often assume you mean this notion, especially in the context of a stable or simple theory. More recently, there has been a lot of theory developed around Morley sequences generated by a global invariant type $p$ (and as you note, these sequences are always Morley sequences in the non-forking sense). But to avoid confusion, unless it's clear from context, it's probably better to say "$p$-Morley sequence" or "Morley sequence in a global invariant type" or maybe just "invariant Morley sequence".
Now to your question: In general, there is not much to say - some Morley sequences are invariant Morley sequences, and some aren't. But in the setting of an NIP theory, every Morley sequence is a Morley sequence for a global Lascar invariant type! Note that I had to put Lascar's name in there - I'll get to that technicality below.
Let's start by working over a model $M$, for simplicity. The key fact is that in an NIP theory, a global type $p$ does not fork over $M$ if and only if $p$ is $M$-invariant (this is Corollary 5.22 in Simon's book A Guide to NIP Theories). If we define $a\perp_M^i b$ to mean $\mathrm{tp}(a/Mb)$ extends to a global $M$-invariant type, it follows that $a\perp_M^f b$ if and only if $a\perp_M^i b$.
Now suppose $I = (a_i)_{i\in \omega}$ is a Morley sequence over $M$ (an $M$-indiscernible $M$-independent sequence). Stretch $I$ to have length $\omega+1$, adding an element $a_\omega$ to the end of the sequence. Then $a_\omega\perp^f_M I$ (since if $a_\omega\not\perp^f_M I$, then $a_\omega\not\perp^f_M a_0\dots a_{n-1}$ for some $n$, and by $M$-indiscernibility, $a_n\not\perp^f_M a_0\dots a_{n-1}$, contradicting $M$-independence). By NIP, $a_\omega\perp^i_M I$, so $\mathrm{tp}(a_\omega/MI)$ extends to a global invariant type $p$. By indiscernibility, $\mathrm{tp}(a_n/Ma_0\dots a_{n-1}) = \mathrm{tp}(a_\omega/Ma_0\dots a_{n-1})= p|_{Ma_0\dots a_{n-1}}$, so $I$ is an invariant Morley sequence in $p$ over $M$.
If you want to work over an arbitrary base set $C$, you run into the problem that a type over $C$ may not extend to any global $C$-invariant type. Fortunately, the above argument goes through if we replace types by Lascar strong types and invariance by Lascar invariance - see Section 5.1 of A Guide to NIP Theories. The general fact is Proposition 5.21 there: in an NIP theory, a global type $p$ does not fork over a set $C$ if and only if $p$ is $\mathrm{Lstp}_C$-invariant.
To remove Lascar from the picture, you need to be working over a set $C$ such that Lascar strong types over $C$ are the same as ordinary types over $C$. Such a set is called "bdd-closed" (well, really a bdd-closed set is one in which Kim-Pillay strong types are the same as ordinary types, but in NIP Lascar and Kim-Pillay strong types are the same). Models are bdd-closed in any theory. In stable theories, a set is bdd-closed if and only if it is $\mathrm{acl}^{\text{eq}}$-closed. The counterexample in tomasz's answer appears exactly because $\emptyset$ is not $\mathrm{acl}^{\text{eq}}$-closed in the theory of an equivalence relation with two infinite classes.
