The following is the definition of the neat reduct of a cylindric algebra:
For ordinals $\alpha < \beta$, If $\mathfrak{A}$ is a $\beta$-dimensional cylindric algebra then the neat-$\alpha$ reduct of $\mathfrak{A}$ is the $\alpha$-dimensional cylindric algebra $Nr_\alpha \mathfrak{A}$ whose domain is the set of all $\alpha$-dimensional elements of $\mathfrak{A}$, i.e., $$ \operatorname{\mathit{Nr}_\alpha} A = \{a \in A | \mathop{c_i} a = a \text{ for all } \alpha \leq i < \beta\},$$ with the standard operations of an $\alpha$-dimensional cylindric algebra agreeing with those of $\mathfrak{A}$ restricted to $\operatorname{\mathit{Nr}_\alpha} A$.
I think I get that, but I get confused when trying to apply the notion to cylindric set algebras. Now making $\mathfrak{A}$ a $\beta$-dimensional cylindric set algebra, it seems the most straightforward translation of the definition of $\operatorname{\mathit{Nr}_\alpha} A$ would be the following: $$ \operatorname{\mathit{Nr}_\alpha} A = \{X \in A | \operatorname{C_i} X = X \text{ for all } \alpha \leq i < \beta\}.$$ But what confuses me about this is that the resulting cylindric set algebra would still be $\beta$-dimensional -- not $\alpha$-dimensional.
Is this correct? Or is there some way I'm meant to be truncating the $\beta$-tuples in $A$ to render the elements of $\operatorname{\mathit{Nr}_\alpha} A$ sets of $\alpha$-tuples? If $\operatorname{\mathit{Nr}_\alpha} \mathfrak{A}$ is still $\beta$-dimensional, is there a general relationship (e.g., isomorphism) between $\operatorname{\mathit{Nr}_\alpha} \mathfrak{A}$ and an $\alpha$-dimensional cylindric set algebra?