The antenna-vertex conjecture for the partition graph

Less than 1 year old · traced to 2026

Let $L_1(n)$ denote the set of vertices of local simplex dimension $1$ in the partition graph on partitions of $n$ . The two antenna vertices are the partitions $(n)$ and $(1^n)$ .

Antenna-vertex conjecture. For every $n\geq 4$ , the only vertices of local simplex dimension $1$ are the two antenna vertices:

(n),\qquad (1^n).

Equivalently,

L_1(n)=\{(n),(1^n)\}.

Computations show that $|L_1(n)|=2$ for $4\leq n\leq 30$ , with these two vertices precisely the antenna vertices. The conjecture asserts that this stable pattern persists for all $n\geq 4$ .

References

Primary source

Fedor B. Lyudogovskiy, “Simplex Stratification and Phase Boundaries in the Partition Graph”, arXiv:2603.23228 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Open

The conjecture remains open: computations support it through size 30, but no proof or counterexample was found.

The conjecture asserts that, for every $n\geq 4$ , the only partition-graph vertices with local simplex dimension $1$ are $(n)$ and $(1^n)$ . The directly relevant source presents this as a conjecture, not a theorem.

Known results

Computations verify $|L_1(n)|=2$ for $4\leq n\leq 30$ , with exactly $(n)$ and $(1^n)$ .
The two antenna vertices have local simplex dimension $1$ for every $n\geq 2$ ; uniqueness for all $n\geq 4$ is not established.

Current status (as of August 2026): The two antenna vertices are known to have local simplex dimension $1$ , and the conjecture is computationally verified through $n=30$ ; whether any other such vertices occur for $n>30$ remains open.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Proof that the antennas are exactly the one-dimensional vertices

The previously established local corner formulas in arXiv:2603.18696, Corollary 3.4, Proposition 3.7, and Theorem 7.1, are used as known input.

Write

\lambda=(d_1^{m_1},\ldots,d_k^{m_k}), \qquad d_1>\cdots>d_k>0, \qquad g_i=d_i-d_{i+1}, \qquad d_{k+1}=0.

For removable corner $i$ and addable corner $j$ , the known corner-incidence matrix is

M_{ij} = 1-\mathbf1_{\{j=i,\ m_i=1\}} -\mathbf1_{\{j=i+1,\ g_i=1\}}.

Hence the star capacities are

s_i=k+1-\mathbf1_{\{m_i=1\}}-\mathbf1_{\{g_i=1\}},

and every top capacity is at most $k$ . Consequently

\dim_{\mathrm{loc}}(\lambda)= \begin{cases} 0,&\lambda=(1),\\ k-1,&\lambda=(k,k-1,\ldots,1),\ k\ge2,\\ k+1,&m_i\ge2\text{ and }g_i\ge2\text{ for some }i,\\ k,&\text{otherwise}. \end{cases} \tag{1}

Fix $n\ge4$ . If $k\ge2$ and $\lambda$ is not a staircase, (1) gives

\dim_{\mathrm{loc}}(\lambda)\ge k\ge2.

A staircase with $k\ge3$ has dimension $k-1\ge2$ . The only remaining staircase with $k=2$ is $(2,1)$ , whose size is $3$ and therefore does not occur.

Thus a vertex of local dimension one must have $k=1$ , so

\lambda=(d^m),\qquad dm=n.

Here (1) gives dimension two exactly when both $m\ge2$ and $d\ge2$ . Hence dimension one occurs precisely when

m=1\quad\text{or}\quad d=1.

The corresponding partitions are exactly $(n)$ and $(1^n)$ . Therefore

\boxed{L_1(n)=\{(n),(1^n)\}\qquad(n\ge4).}

This proves the full antenna-vertex conjecture.