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Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.

Under which condistions does the following hold:

If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.

For example, is this true if $G$

  • is walk-regular, (resp. vertex-transitive)
  • is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
  • is distance-regular (resp. distance-transitive), or
  • has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.

Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.

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  • $\begingroup$ Do you have any evidence/heuristics for why this should hold in some cases? Also what is special about $\theta_2$? $\endgroup$ Jul 18, 2020 at 1:50
  • $\begingroup$ @AntoineLabelle My intuition comes from spectral embeddings of $G$, where you assign points in Euclidean space to the vertices of $G$, and you do this by somehow using the eigenvectors of $G$ to some eigenvalue. It seems to be common intuition, that the $\theta_2$-embedding is "as expanded as possible" (imagine the vertices as repelling each other, but the edges keeping them together, and then the $\theta_2$-embedding is an equilibrium configuration under these condition). And to be as expanded as possible, it is plausible that antipodal vertices are embedded opposite to each other. $\endgroup$
    – M. Winter
    Jul 18, 2020 at 8:48
  • $\begingroup$ @AntoineLabelle My evidence is that it holds for all the graphs I have checken (e.g. the skeletons of regular polytopes, but also many others). $\endgroup$
    – M. Winter
    Jul 18, 2020 at 8:52

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