Iterated barycentric subdivision cofinal in system of subdivisions?

Question

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, 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.

Answer 1

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 $d$.