When does $\kappa^{<\lambda}=\kappa^\lambda$?

Question

Can cardinals $\kappa$ and $\lambda$ with $$\bigcup_{\delta<\lambda}\kappa^\delta=\kappa^\lambda$$ be characterized? Or at least a sufficient condition for this equality be given. I thing that regularity of both of them is not sufficient.

Answer 1

One necessary condition is that the values $\kappa^\delta$ for $\delta<\lambda$ must stabilize. That is, there must exist some $\delta<\lambda$ such that for all $\delta'$ with $\delta\leq\delta'<\lambda$, $\kappa^{\delta'}=\kappa^\delta$. Indeed, suppose no such $\delta$ exists (note that this in particular implies $\lambda$ is a limit cardinal). Then we can pick a cofinal sequence $(\delta_\alpha)$ of the cardinals below $\lambda$ such that $\kappa^{\delta_{\alpha}}<\kappa^{\delta_{\alpha+1}}$ for each $\alpha$. We then have $$\kappa^{<\lambda}=\sum_\alpha \kappa^{\delta_{\alpha}}$$ and $$\kappa^\lambda=\kappa^{\sum_{\alpha}\delta_{\alpha+1}}=\prod_\alpha \kappa^{\delta_{\alpha+1}}$$ and so $\kappa^{<\lambda}<\kappa^\lambda$ by König's theorem.

Now assuming that $\lambda$ is a limit cardinal and the values of $\kappa^{\delta}$ for $\delta<\lambda$ do stabilize (say, with stable value $\theta$), we have $$\kappa^{<\lambda}=\theta\cdot\operatorname{cf}(\lambda)$$ and $$\kappa^{\lambda}=\theta^{\operatorname{cf}(\lambda)}.$$ Note though that clearly $\kappa^{<\lambda}\geq\lambda\geq\operatorname{cf}(\lambda)$ as long as $\kappa>1$ (and the cases $\kappa\leq 1$ are trivial), so $\kappa^{<\lambda}=\theta$. If $\lambda$ is singular, then $\operatorname{cf}(\lambda)<\lambda$ so $\theta=\kappa^\delta$ for some $\delta\geq \operatorname{cf}(\lambda)$ and so $\kappa^\lambda=\theta$. So, when $\lambda$ is singular, $\kappa^{<\lambda}=\kappa^{\lambda}$ iff the values of $\kappa^\delta$ for $\delta<\lambda$ stabilize.