The A_(ell-1) path-space tower branches by adjacent vertices

For any path $(a_0,\dots ,a_n,b)$ ending at $b$, its penultimate vertex $a$ is uniquely determined and satisfies $a\sim b$; deleting $b$ gives a unique prefix in ${\mathsf{P}}_{n,a}$. Conversely appending $b$ to any path in ${\mathsf{P}}_{n,a}$ with $a\sim b$ gives a path ending at $b$. Hence the set of paths ending at $b$ is the disjoint union of the appended path sets $\{pb:p\in {\mathsf{P}}_{n,a}\}$ over all such $a$. The basis vectors therefore split as the asserted direct sum, and $E_{p,q}^{(n,a)}$ acts on each appended block as the same matrix unit while acting by zero on the other predecessor blocks.

Take matrix units $X=E_{p,q}^{(n,a)}$ and $Y=E_{r,s}^{(n,c)}$. If $a\mathrm{\ne }c$, then $XY=0$; in the product ${\iota }_n(X){\iota }_n(Y)$ a nonzero matrix-unit product would require the middle path $qb$ to equal $rd$ for some $b\sim a$ and $d\sim c$, forcing both $q=r$ as length-$n$ paths and $a=c$, a contradiction. Thus the images multiply to zero.

If $a=c$, then $XY={\mathbf{1}}_{q=r}{E}_{p,s}^{(n,a)}$. On multiplying the two sums in [F2], the only nonzero terms have the same appended vertex $b$ and equal middle path $qb=rb$, which is equivalent to $q=r$. The product is consequently ${\mathbf{1}}_{q=r}{\sum }_{b\sim a}{E}_{pb,sb}^{(n+1,b)}={\iota }_n(XY)$. Adjoints are preserved because ${\iota }_n((E_{p,q}^{(n,a)}{)}^{\ast })={\sum }_{b\sim a}{E}_{qb,pb}^{(n+1,b)}={\iota }_n(E_{p,q}^{(n,a)}{)}^{\ast }$. Bilinearity proves multiplicativity and $\ast$-preservation on all of $K_n$.

The identity of $K_n$ is ${\sum }_{a,p}{E}_{p,p}^{(n,a)}$. Its image is the sum of the diagonal units indexed by every one-step extension $pb$ of every length-$n$ path $p$; by step 1.1 each length-$(n+1)$ path occurs exactly once. This is the identity of $K_{n+1}$. For injectivity, every vertex $a$ has at least one neighbor in $A_{\mathrm{\ell }-1}$, since $\mathrm{\ell }\ge 3$. For each source unit $E_{p,q}^{(n,a)}$ choose one neighbor $b$; the coordinate $E_{pb,qb}^{(n+1,b)}$ in its image determines the source unit uniquely by deleting the last vertex. Distinct source units therefore have disjoint image coordinates, so no nonzero linear combination maps to zero. Hence ${\iota }_n$ is a unital injective $\ast$-homomorphism and step 1.1 gives the multiplicity-one branching rule.