A one-node coarse space misses a high-conductance channel

The grounded two-node graph with unit conductance from each node to ground and conductance $\alpha$ between the nodes has matrix $A_{\alpha }=I_2+\alpha \left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\mathrm{.}$ For $x=(x_1,x_2)$, $x^{\mathsf{T}}{A}_{\alpha }x=x_1^2+x_2^2+\alpha (x_1-x_2{)}^2$, so $A_{\alpha }$ is SPD. The coarse and local injections are $R_0=e_1$ and $R_1=e_2$, hence $A_0=A_1=1+\alpha$. The additive preconditioner from [F1] is therefore $B_{\alpha }=(1+\alpha {)}^{-1}{I}_2$. There is one local subspace, so its pairwise overlap matrix is the fixed $1\times 1$ matrix $[1]$, independent of $\alpha$.

The vectors $(1,1)$ and $(1,-1)$ are eigenvectors of $A_{\alpha }$ with eigenvalues $1$ and $1+2\alpha$. Thus $B_{\alpha }{A}_{\alpha }$ has eigenvalues $(1+\alpha {)}^{-1}$ and $(1+2\alpha )\mathrm{/}(1+\alpha )$, so ${\kappa }_A(B_{\alpha }{A}_{\alpha })=1+2\alpha \to \mathrm{\infty }$. The low-energy channel vector $(1,1)$ is not in the one-node coarse space $V_0$, and the growing condition number disproves the claim.