It is easily shown that, for any uncountable infinite cardinal $\kappa$, $\square\_\kappa$ implies that for any stationary $S\subseteq \kappa^+$, there exists a stationary $T\subseteq S$ such that $T$ does not reflect at (i.e. is not stationary in) any $\alpha<\kappa$ of uncountable cofinality. The standard proof does not go through, however, when $\square\_\kappa$ is replaced by the weaker notion of $\square(\kappa^+)$. Is $\square(\kappa^+)$ compatible with stationary reflection? More precisely, if $\kappa$ is an uncountable infinite cardinal, is $\square(\kappa^+)$ consistent with the statement "every stationary $S\subseteq \kappa^+$ consisting of ordinals of cofinality $<\kappa$ reflects at some $\alpha<\kappa^+$"?



Square and stationary reflection
The answer is yes (at least for $\kappa = \omega\_1$): Shelah and Harrington showed that you can force that every stationary subset of $S\_{\omega}^{\omega\_2}$ reflects starting with a Mahlo cardinal. See Theorem A in [Some exact equiconsistency results is set theory](http://projecteuclid.org/euclid.ndjfl/1093870823). 


Since $\neg \square(\omega\_2)$ implies that $\omega\_2$ is weakly compact in $L$, if we start with a Mahlo cardinal $\kappa$ which is not weakly compact in $L$ and collapse it to $\omega\_2$ using the forcing of Shelah-Harrington we'll have $\square(\omega\_2)$ in the generic extension.


I don't know if this is true for $\kappa > \omega\_1$ as well.


