I'm struggling to understand a remark in Pierre Simon's book on NIP theories. Let $T$ be a complete theory and $\mathfrak{U}\models T$ a monster model. Suppose $p(x),q(y)\in S(\mathfrak{U})$ are global types with distinct free variables, and that $p$ is $E$-invariant for some small subset $E\subset\mathfrak{U}$. Then the product type $p(x)\otimes q(y)$ in the free variables $xy$ is defined by taking $\phi(x,y,c)\in p\otimes q$ if and only if there exists some $b$ realizing $q|_{Ec}(y)$ and such that $\phi(x,b,c)\in p$; this is well-defined by $E$-invariance of $p$. Simon says that $p$ and $q$ "commute" if $p(x)\otimes q(y)=q(y)\otimes p(x)$.
Exercise 2.22 has us show that, if $T=\text{DLO}$ and $p(x)$ and $q(y)$ are $1$-types, then $p$ and $q$ commute provided that $p$ and $q$ are inequivalent. I've done this without trouble, but I'm a bit perplexed by a subsequent remark:
This is no longer true in $\text{RCF}$: if $p$ and $q$ are two invariant $1$-types which concentrate on definable cuts (either $\pm\infty$ or $a^{\pm}$), then they do not commute.
I'm struggling to see why this is the case. By quantifier elimination, a global $1$-type $p(x)\in S(\mathfrak{U})$ is uniquely determined by the $<$-cut of $\mathfrak{U}$ it represents, and the data of whether or not the equation $f=0$ lies in $p$ for each polynomial $f\in\mathfrak{U}[x]$. Since polynomials over a field have finitely many roots, if $p$ contains an equation $f=0$ then it is an algebraic type and hence has all possible realizations already in $\mathfrak{U}$. In particular, if $p(x),q(y)$ are distinct $1$-types that are not realized in $\mathfrak{U}$, then (i) they must represent different cuts in $\mathfrak{U}$, and (ii) we will have that $p(x)\otimes q(y)$ contains $f(x,y)\neq 0$ for each $f\in\mathfrak{U}[x,y]$. Furthermore, if I'm not mistaken, $p(x)\otimes q(y)\supset p(x)\cup q(y)$, so the only possible way for $p(x)\otimes q(y)$ and $q(y)\otimes p(x)$ to differ is if one of them contains $x<y$ and the other contains $x\geqslant y$. But $p(x)$ and $q(y)$ represent different cuts, so without loss of generality there exists $c\in\mathfrak{U}$ such that $x<c\in p(x)$ and $c<y\in q(y)$, and then necessarily $x<y$ lies in both $p(x)\otimes q(y)$ and $q(y)\otimes p(x)$, again since each of these types contains $p(x)\cup q(y)$. So it's unclear to me how the circumstances described by Simon can arise; does anyone have any insight?
Hint: Suppose $p(x)$ is the type at $\infty$ and $q(y)$ is the type at $-\infty$. Now consider the formula $x\leq -y$. Is this in $p(x)\otimes q(y)$? What about $q(y)\otimes p(x)$?
The point is that in RCF, all the definable cuts are non-orthogonal in the sense that any realization of one definable cut contains in its definable closure (over the model) realizations of all the other definable cuts.