Given a rational polyhedral fan $\Sigma$ in $\mathbb{R}^d$ (say with full-dimensional support), its barycentric subdivision $\mathrm{bar}(\Sigma)$ is obtained by performing star-subdivision at the barycenters
$$
\rho\_\sigma = \sum\_{\tau \in \sigma(1)} \rho\_\tau
$$
of its cones $\sigma$ (where $\rho\_\tau$ are the primitive generators of the rays $\tau$ of $\sigma$, and the star-subdivisions are performed in order of decreasing dimension on the cones $\sigma \in \Sigma$).


**Question:** Is it true that for any fan $\Sigma'$ with support $|\Sigma'|=|\Sigma|$ there exists $n \geq 0$ such that the $n$-th iterated barycentric subdivision $\mathrm{bar}^{\circ n}(\Sigma)$ is a refinement/subdivision of $\Sigma'$?


For my purposes it would also suffice to restrict to the case where $\Sigma$ consists of the the positive orthant $\sigma\_d = \mathbb{R}\_{\geq 0}^d$ and its faces.


**Example:** For $d=2$ the positive orthant is subdivided at the rays spanned by
$$
(1,1); (2,1), (1,2); (3,1), (3,2), (2,3), (1,3); \ldots
$$
in the successive barycentric subdivisions. The generators $(x,y)$ of these rays exactly run through the primitive integer vectors in the interior of $\sigma\_2$, and the ratios $x/y$ form the entries of the [Stern-Brocot tree](https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree), which enumerates all rational numbers in $(0,1)$. In particular, it is the case that any fan $\Sigma'$ with support $\sigma\_2$ will eventually be refined by some iterated barycentric subdivision of $\sigma\_2$ (since the slopes of the rays of $\Sigma'$ appear at some point in the Stern-Brocot tree).


The behaviour on the rays generalizes: for $\sigma\_d$, the ray generators of the iterated barycentric subdivisions run through the primitive vectors in $\sigma\_d$, so at least the $1$-skeleton of $\Sigma'$ will eventually be captured by this process.



Iterated barycentric subdivision cofinal in system of subdivisions?
It turns out that the answer is "No" for dimension $d \geq 3$.


Indeed, for the counter-example in dimension $3$ take $\Sigma$ given by $\sigma\_3$ and its faces, and consider the fan $\Sigma'$ obtained by subdividing $\Sigma$ along the hyperplane $H=-x\_1 + 2 x\_2 + x\_3 = 0$. The fact that $H$ has both positive and negative coefficients implies that it meets the interior of $\sigma\_3$.


Now the first barycentric subdivision of $\Sigma$ will contain the cone
$$
\overline \sigma = \left\{x\_1' \cdot \pmatrix{1\\0\\0} + x\_2'\cdot \pmatrix{1\\1\\1} + x\_3' \cdot \pmatrix{1\\1\\0} : x\_1', x\_2', x\_3' \geq 0\right\}
$$
Pulling back the equation $H$ to the coordinates $x\_1', x\_2', x\_3'$ on $\overline \sigma$ we get
$$H' = -(x\_1' + x\_2' + x\_3') +2(x\_2'+x\_3')+x\_2' = -x\_1' + 2 x\_2' + x\_3',$$
which has the *same* coefficients $(-1,2,1)$ as $H$. So we observe that $H$ meet the interior of the maximal cone $\overline \sigma$ of the first barycentric subdivision of $\Sigma$, and thus $\mathrm{bar}(\Sigma)$ does not refine $\Sigma'$. But in fact since the first barycentric subdivision contains a cone on which the equation of $H$ has not changed, in fact any iterated barycentric subdivision also contains such a cone, so none of these will refine $\Sigma'$.


Since $\sigma\_3$ appears as a face of $\sigma\_d$ for $d > 3$, it's easy to see that this also gives a counterexample in arbitrary dimension $d$.


