It is known (a theorem of Komjáth and Hajnal) that it is consistent (by adding a Cohen real to a universe where there is no Suslin line) that there exists an order type $\theta$ such that for any other order type $\eta$, there exists a bad coloring $f: [\eta]^2 \to \omega$ such that for any order preserving embedding $g: \theta \to \eta$, $f''[g''\theta]^2=\omega$, i.e. $\eta \not \rightarrow [\theta]\_\omega^2$, in particular $\eta \not \rightarrow (\theta)\_\omega^2$ (there are more results but enough for now). 


I'm wondering if there are references about similar phenomenon for polarized partition relation, more concretely: do there exist order types $\theta\_0, \theta\_1$ and cardinal $\gamma$ such that for any order types $\eta\_0, \eta\_1$, there exists $f: \eta\_0\times \eta\_1 \to \gamma$ such that for all order-preserving embeddings $g\_0: \theta\_0 \to \eta\_0, g\_1: \theta\_1\to \eta\_1$, $f''[g\_0''\theta\_0\times g\_1''\theta\_1]=\omega$ (or just not constant), i.e. ${\eta\_0 \choose \eta\_1} \not \rightarrow {\theta\_0\choose \theta\_1}\_\gamma^{1,1}$. Any pointers would be appreciated. Thanks in advance. 



Strongly non-Ramsey order type in polarized partition problems
We can't find such pair of bad order types, aka there indeed is some Erdös-Rado phenomenon in the polarized partitions with respect to linear orderings. Indeed given $\gamma$ number of colors and $\theta$, $\psi$ we can find large saturated dense linear orders $\alpha, \beta$ such that for any $f: \alpha\times \beta \to \gamma$, there exists $X\subset \alpha, Y\subset \beta$ such that $tp(X)=tp(\alpha), tp(Y)=tp(\beta)$ such that $f''X\times Y$ is constant.


To elaborate: Given $\gamma$ the number of colors and order types $\theta, \psi$, given let $\lambda$ be regular a lot larger than $|\theta|, |\psi|, \gamma$, let $\beta$ be a $\lambda$-saturated dense linear order. Let $\lambda'$ be a lot larger than $2^{|\beta|}, \gamma$ and let $\alpha$ be $\lambda'$-saturated dense linear ordering, we claim that ${\alpha\choose \beta}\to {\theta\choose \psi}^{1,1}\_{\gamma}$. Given a coloring $f: \alpha \times \beta\to \gamma$, for each $i\in \alpha$, consider $f\_i: \beta\to \gamma$. 



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> There exists a monochromatic under $f\_i$ sub order of $\beta$ that is $\lambda$-saturated.
>  Why? Suppose not, then induct on $i\in \gamma$ to build a cut. Stage 0, there exists $\eta\_0,\eta\_1<\lambda$ and $A\_0=\{a\_j: j<\eta\_0\}<B\_0=\{b\_j: j<\eta\_1\}$ such that there is no $c\in \beta$ that has color 0 such that $A\_0<c<B\_0$. Note that $(A\_0,B\_0)=\{c\in \beta: A\_0<c<B\_0\}$ is still $\lambda$-saturated. Inductively at stage $\alpha$, suppose we have $A\_0<A\_1<\cdots<A\_\beta<\cdots < B\_\beta<\cdots <B\_0$ (i.e. $\forall \beta<\gamma<\alpha \ A\_\beta<A\_\gamma<B\_\gamma<B\_\beta$). Since $\beta$ is $\lambda$-saturated, let $A'\_\alpha= \bigcup\_{i<\alpha}A\_i, B'\_\alpha=\bigcup\_{i<\alpha}B\_i$, we know $(A'\_\alpha, B'\_\alpha)$ is still $\lambda$-saturated. By hypothesis, we can find $A\_\alpha, B\_\alpha$ such that $A'\_\alpha<A\_\alpha<B\_\alpha<B'\_\alpha$ such that no element in $(A\_\alpha, B\_\alpha)$ has color $\alpha$. This can proceed in $\gamma$ steps since $\gamma<\lambda$ but absurdity arises since $\beta$ is saturated and every element gets colored.
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Denote the monochromatic $\lambda$-saturated suborder of $\beta$ under $f\_i$ $\beta\_i$, and the color be $\gamma\_i$. Now define a coloring $g: \alpha\to 2^{\beta}\times \gamma$ such that $g(i)=(\beta\_i,\gamma\_i)$. By the similar argument above we can get a suborder of $\alpha$, say $\alpha'$ that is monochromatic and $\lambda'$-saturated. Let $(\beta',\gamma')$ be the color $g''\alpha'$, then $f''\alpha'\times \beta'=\gamma'$, as $\alpha'$ and $\beta'$ are sufficiently saturated we can embed $\theta$ and $\psi$ into them respectively.


