TITLE: Does $\mathsf{Ind}_\lambda^\mu(\mathsf{Ind}_\kappa^\lambda(\mathcal C)) = \mathsf{Ind}_\kappa^\mu(\mathcal C)$?
QUESTION [4 upvotes]: $\newcommand\Ind{\mathsf{Ind}}\newcommand\Ord{\mathsf{Ord}}\newcommand\Psh{\mathsf{Psh}}$For $\kappa \leq \lambda \leq \Ord$ regular cardinals and $\mathcal C$ an essentially small category, let $\Ind_\kappa^\lambda(\mathcal C)$ be the free completion of $\mathcal C$ under $\lambda$-small, $\kappa$-filtered colimits. So $\Ind_\kappa^{\Ord}(\mathcal C) = \Ind_\kappa(\mathcal C)$ is the usual free completion of $\mathcal C$ under $\kappa$-filtered colimits.
An explicit construction of $\Ind_\kappa^\lambda(\mathcal C)$ is given by taking the smallest full subcategory of the presheaf category $\Psh(\mathcal C)$ which contains the representables and is closed under $\lambda$-small, $\kappa$-filtered colimits. In other words, there is a monotonic, inflationary endofunction $I_\kappa^\lambda$ on full subcategories of $\Psh(\mathcal C)$ where $I_\kappa^\lambda(\mathcal D)$ comprises those objects which are $\lambda$-small, $\kappa$-filtered colimits of objects of $\mathcal D$; then $\Ind_\kappa^\lambda(\mathcal C) = \cup_{\alpha \in \Ord} (I_\kappa^\lambda)^\alpha(\mathcal C)$ is the least fixed-point of $I_\kappa^\lambda$ above the full subcategory of representables $\mathcal C \subseteq \Psh(\mathcal C)$, given by iterating $I_\kappa^\lambda$ transfinitely.
Questions:

For regular cardinals $\kappa \leq \lambda \leq \Ord$, when do we have $\Ind_\kappa^\lambda(\mathcal C) = I_\kappa^\lambda(\mathcal C)$? That is, when is it the case that a $\lambda$-small, $\kappa$-filtered colimit of $\lambda$-small, $\kappa$-filtered colimits of representables is already a $\lambda$-small, $\kappa$-filtered colimit of representables?

It is well-known that a sufficient criterion is that $\lambda = \Ord$. I suspect that when $\lambda < \Ord$, the answer is "not always". I believe it also suffices for $\mathcal C$ to have $\kappa$-small colimits. I'm interested in a sufficient criterion that holds for all $\mathcal C$. I suspect it may suffice to have $\kappa \triangleleft \lambda$ (that's the "sharply below" relation).

For regular cardinals $\kappa \leq \lambda \leq \mu \leq \Ord$, when do we have $\Ind_\lambda^\mu(\Ind_\kappa^\lambda(\mathcal C)) \simeq \Ind_\kappa^\mu(\mathcal C)$?

I suspect again that the answer is "not always", but that a sufficient criterion may be for the pairs $(\kappa,\lambda)$, $(\lambda,\mu)$, and $(\kappa,\mu)$ to each satisfy the sufficient criterion of (1). I'm particularly interested in the case $\kappa = \omega < \lambda < \mu = \Ord$, where I suspect that this always holds (keeping in mind that $\omega \triangleleft \lambda$ for all regular $\lambda$).
I would be interested to know a reference for the answers to (1) and (2).

REPLY [4 votes]: I asked myself the same questions while working on this paper [TAC].
It seemed to me that the way to answer questions like this is to embed $\textbf{Ind}_\kappa^\lambda (\mathcal{C})$ into the usual $\textbf{Ind}_\kappa (\mathcal{C})$, use known results for accessible categories, then restrict back to $\textbf{Ind}_\kappa^\lambda (\mathcal{C})$.
This is possible when $\kappa \trianglelefteq \lambda$ due to the following fact, which is basically proposition 3.5 in op. cit.: $\textbf{Ind}_\kappa^\lambda (\mathcal{C})$ is precisely the full subcategory of $\textbf{Ind}_\kappa (\mathcal{C})$ spanned by the $\lambda$-presentable objects.
Thus, to answer question 1: if idempotents in $\mathcal{C}$ split and $\kappa \trianglelefteq \lambda$, then every object in $\textbf{Ind}_\kappa^\lambda (\mathcal{C})$ is a colimit for a $\lambda$-small $\kappa$-filtered diagram of objects in $\mathcal{C}$, because every $\lambda$-presentable object in $\textbf{Ind}_\kappa (\mathcal{C})$ is such a colimit by a theorem of Makkai and Paré.
Similarly, to answer question 2: if idempotents in $\mathcal{C}$ split and $\kappa \trianglelefteq \lambda \trianglelefteq \mu$, then $\textbf{Ind}_\lambda^\mu (\textbf{Ind}_\kappa^\lambda (\mathcal{C})) \simeq \textbf{Ind}_\kappa^\mu (\mathcal{C})$, because we have $\textbf{Ind}_\lambda (\textbf{Ind}_\kappa^\lambda (\mathcal{C})) \simeq \textbf{Ind}_\kappa (\mathcal{C})$ by proposition 3.10(iv) in op. cit., which is proven by reducing to the case where $\kappa = \lambda$.
(A side remark: for fixed $\mathcal{C}$, $\textbf{Ind}_\kappa^\lambda (\mathcal{C})$ is contravariant in $\kappa$ and covariant in $\lambda$, so personally I would reverse the use of subscript and superscript in this notation.)
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