# How are weakly compact cardinals based on reflection principles?

It seems to me that the fact that the class of Mahlo cardinals below a weakly compact cardinal $\kappa$ is stationary in $\kappa$ is in fact a reflection property. Out of the many equivalent definitions of a weaklky compact cardinal, are some based on the reflection principles?

To me, the extension property seems to be very close, but it works the opposite way; for a weakly compact $\kappa$, instead of giving a smaller $V_\alpha$ where some property also holds, it gives us an elementary superstructure. Can this be seen as some sort of a reflection principle? Is there anyone who has written on this matter?

• Weakly compact cardinals are exactly the $\Pi^1_1$-indescribable cardinals. And this means that every second-order statement of the form $\forall X\varphi$, where $\varphi$ is a first-order statement, is reflected below $\kappa$. Even if we augment our structure $(V_\kappa,\in)$ by an additional predicate. – Asaf Karagila Aug 14 '16 at 12:30
• Sorry, I should have included this in the question. I am looking for an answer that is specific to weakly compact cardinals, since what you said is true of all $\Pi^m_n$–indescribable cardinals. I am specifically looking for a vertical reflection–style explanation'' why weakly compact cardinals don't break $L=V$, but measurables do. – mikulas Aug 14 '16 at 12:34
• Well. In the weakly compact case, the elementary end-extension is only a first-order elementary end-extension. In the measurable cardinals, it covers $V_{\kappa+1}$ as well. It is closer to a second-order property. Since a measure is never in its ultrapower, namely in the end-extension, it must be the case that $V\neq L$. And again, in the weakly compact case, we significantly shrink $V_{\kappa+1}$ when we take these end-extensions. But what is true, for example, is that if $V=L$ holds below a weakly compact $\kappa$, then it holds at $V_\kappa$ as well. – Asaf Karagila Aug 14 '16 at 12:38