Is


$\mathrm{sd}^2 (\Delta^n) = \mathrm{sd}^2(\partial \Delta^n) \times \Delta^1 \cup\_{\mathrm{sd}^2(\partial \Delta^n) \times \{1\}} Cone(\mathrm{sd}^2(\partial \Delta^n))$


? Here $\mathrm{sd}^2$ means the second barycentric subdivision of a simplicial set / complex, and 


$Cone(\mathrm{sd}^2(\partial \Delta^n)) = \mathrm{sd}^2(\partial \Delta^n) \times \Delta^1 / (\mathrm{sd}^2(\partial \Delta^n) \times \{0\})$.


I've verified this for $n\leq 2$ by drawing pictures. Is this in fact the case for all $n$? And is it written down somewhere?


If so, then presumably there's a generalization of this description to a description of $\mathrm{sd}^k(\Delta^n)$ for higher $n$, but it's not so simple. I believe that $\mathrm{sd}^k(\Delta^n)$ will always divide into $2^{k-1} + 1$ "layers" according to the minimum number of edges one must traverse to get to the barycenter -- I think these are nested subcomplexes which are all homeomorphic to $S^{n-1}$ (except for the 0th layer at the barycenter itself), and I think that the edges connecting two adjacent layers all point in the same direction, which alternates from one layer to the next. But these layers (even excluding the central point) are not isomorphic simplicial sets -- for example in $\mathrm{sd}^3(\Delta^2)$ the layers have 0, 24, 24, 48, and 24 non-degenerate edges as one goes out from the center.


But what I really happen to care about is what's going on just for $k=2$ and arbitrary $n$. (That's what I need to compute a fibrant replacement in the Thomason model structure on $\mathsf{Cat}$ via the $Ex^2$ functor).



Geometry of the second barycentric subdivision (and Thomason-fibrant replacement)
Yes, that's true. See Remark 4.1 in [this paper](https://arxiv.org/abs/1408.2743) by Meier and Ozornova. A generalization to $k > 2$ would be interesting and I believe that everything you wrote about it is right. However, I don't know any general statement.


