ISOMETRIC EMBEDDINGS OF POLAR GRASSMANNIANS AND METRIC CHARACTERIZATIONS OF THEIR APARTMENTS

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Problem: Reconstructed conjecture: for every polar space $\Pi$ of type $\mathsf D_n$, with $n>4$ odd, and every half-spin Grassmann graph $\Gamma_\delta(\Pi)$, there exists an isometric embedding

whose image is not an apartment. This is the natural reading because the cited paper fixes $\Pi$ before stating the conjecture.

Result: The conjecture is false as stated.

Take $\Pi=Q^+(9,2)$, the split orthogonal polar space of type $\mathsf D_5$ over $\mathbb F_2$. Let $\delta\in\{+,-\}$. By the reduction in Pankov, Section 6, any isometric embedding of $\tfrac12H_5$ into $\Gamma_\delta(\Pi)$ is either an apartment, or its image is contained in a point-residue $[q\rangle_\delta$, which is a half-spin Grassmann graph of type $\mathsf D_4(2)$.

So it is enough to show that $\tfrac12H_5$ does not embed isometrically into a half-spin graph of $\mathsf D_4(2)$. Model $\mathsf D_4(2)$ on an $8$-dimensional split quadratic space over $\mathbb F_2$, and fix a maximal singular subspace $E$. In $\tfrac12H_5$, a vertex has five vertices at maximal distance, and these five are pairwise adjacent. Their images would therefore be five maximal singular subspaces disjoint from $E$, pairwise meeting in dimension $2$.

Every maximal singular subspace disjoint from $E$ is the graph of an invertible alternating $4\times4$ matrix $A$ over $\mathbb F_2$. Pairwise adjacency gives

For alternating $4\times4$ matrices write

Then $q(A)=1$ iff $A$ is invertible, and $q(A)=0$ for nonzero rank-$2$ matrices. The polar form of $q$ is nondegenerate symplectic on the $6$-dimensional space of alternating matrices. Thus the five $A_i$ would be pairwise orthogonal nonsingular vectors. Their span is totally isotropic, hence has dimension at most $3$; on it $q$ is linear, so at most $2^{3-1}=4$ vectors can satisfy $q=1$. Contradiction.

Therefore no non-apartment isometric embedding exists in $\Gamma_\delta(Q^+(9,2))$. Since $5$ is odd and $>4$, this disproves the literal conjecture.

A natural repair is to add a field-size hypothesis. Over fields with more than two elements, one can choose a singular vector nonorthogonal to every point of a frame and project an apartment of type $\mathsf D_n$ from that vector into a residue of type $\mathsf D_{n-1}$; this gives the desired non-apartment embedding. The obstruction above is genuinely the missing $\mathbb F_2$ boundary case.

Citation: Pankov, “Isometric embeddings of half-cube graphs in half-spin Grassmannians,” arXiv:1106.5435, especially Lemma 5.3 and Section 6.