The 4m+1 path basis

The relations in path notation. Write $u_j:=(j\mathrm{\mid }j+1)$ and $d_j:=(j+1\mathrm{\mid }j)$ for $0\le j\le m-1$. The generators of $I_m$ read, in this notation, $u_j{u}_{j+1}(0\le j\le m-2),d_{j+1}{d}_j(0\le j\le m-2),u_i{d}_i-d_{i-1}{u}_{i-1}(0<i<m),u_0{d}_0\mathrm{.}$ Indeed $(j\mathrm{\mid }j+1\mathrm{\mid }j+2)=u_j{u}_{j+1}$, $(j+2\mathrm{\mid }j+1\mathrm{\mid }j)=d_{j+1}{d}_j$, $(i\mathrm{\mid }i+1\mathrm{\mid }i)=u_i{d}_i$ and $(i\mathrm{\mid }i-1\mathrm{\mid }i)=d_{i-1}{u}_{i-1}$. Each of these elements has class $0$ in $A_m$ by [L4]; in particular $u_i{d}_i$ and $d_{i-1}{u}_{i-1}$ have the same class for $0<i<m$, and $u_0{d}_0$ has class $0$.

A separating functional on the path ring. Let $F$ be the free abelian group with basis $e_0,\dots ,e_m,u_0,\dots ,u_{m-1},d_0,\dots ,d_{m-1},r_1,\dots ,r_m,$ a total of $4m+1$ basis elements. By [L3] there is a unique $\mathbb{Z}$-linear map $\rho :\mathbb{Z}{\mathrm{\Gamma }}_m\to F$ with $\rho ((i))=e_i,\rho ((i\mathrm{\mid }i+1))=u_i,\rho ((i+1\mathrm{\mid }i))=d_i,$ $\rho ((i\mathrm{\mid }i-1\mathrm{\mid }i))=r_i(1\le i\le m),\rho ((i\mathrm{\mid }i+1\mathrm{\mid }i))=r_i(1\le i<m),\rho (p)=0\text{ for every other path }p\mathrm{.}$ Here "other" covers $(0\mathrm{\mid }1\mathrm{\mid }0)$, the monotone length-two paths not of return form, and all paths of length at least three; the two returns at an interior vertex are both sent to $r_i$, and $(m\mathrm{\mid }m-1\mathrm{\mid }m)$, the only return at $m$, is sent to $r_m$.

Length-two paths in $A_m$. A length-two path $(i_1\mathrm{\mid }{i}_2\mathrm{\mid }{i}_3)$ has consecutive differences $\pm 1$ and is of one of four forms. If both steps go up it is $(j\mathrm{\mid }j+1\mathrm{\mid }j+2)=u_j{u}_{j+1}$ with $0\le j\le m-2$, and if both go down it is $(j+2\mathrm{\mid }j+1\mathrm{\mid }j)=d_{j+1}{d}_j$ with $0\le j\le m-2$; both have class $0$ in $A_m$ by step 1.1. If the steps are up-then-down it is the return $(j\mathrm{\mid }j+1\mathrm{\mid }j)=u_j{d}_j$ at $j$, for $0\le j\le m-1$: its class is $0$ when $j=0$ by step 1.1 and otherwise equals the class of $d_{j-1}{u}_{j-1}=(j\mathrm{\mid }j-1\mathrm{\mid }j)$. If the steps are down-then-up it is the return $(j+1\mathrm{\mid }j\mathrm{\mid }j+1)=d_j{u}_j$ at $j+1$, for $0\le j\le m-1$, whose class equals that of $u_{j+1}{d}_{j+1}$ when $j+1<m$ by step 1.1 and is the class of the unique return at $m$ when $j+1=m$. Consequently the classes of length-two paths are $0$ or one of $r_1,\dots ,r_m$, and each of the two returns at an interior vertex represents the same class.

No path of length three survives. Let $w=(v_0\mathrm{\mid }{v}_1\mathrm{\mid }{v}_2\mathrm{\mid }{v}_3)$ be a path of length three and put $s_j:=v_j-v_{j-1}\in \{\pm 1\}$ for $j=1,2,3$, so that $w$ is the product of its three arrows by [F1]. If $s_1=s_2$, then the subpath $(v_0\mathrm{\mid }{v}_1\mathrm{\mid }{v}_2)$ is monotone with interior vertex $v_1$ satisfying $0<v_1<m$, hence is one of the generators of $I_m$ listed in step 1.1 and $w=(v_0\mathrm{\mid }{v}_1\mathrm{\mid }{v}_2)(v_2\mathrm{\mid }{v}_3)=0$ in $A_m$ by [L4]. If $s_2=s_3$, the same argument applies to $(v_1\mathrm{\mid }{v}_2\mathrm{\mid }{v}_3)$ and $w=(v_0\mathrm{\mid }{v}_1)(v_1\mathrm{\mid }{v}_2\mathrm{\mid }{v}_3)=0$. Otherwise $s_1\mathrm{\ne }{s}_2$ and $s_2\mathrm{\ne }{s}_3$, so $v_0=v_2$ and $v_1=v_3$: the path oscillates on the edge $\{j,j+1\}$ with $j=\min (v_0,v_1)$, and there are two cases. If $v_0=j$, then $w=u_j{d}_j{u}_j=(u_j{d}_j)u_j$; for $j=0$ the factor $u_0{d}_0$ is the zero return at $0$ by step 1.1, while for $1\le j\le m-1$ the relation of step 1.1 rewrites $u_j{d}_j$ as $d_{j-1}{u}_{j-1}$ and $w=d_{j-1}{u}_{j-1}{u}_j=d_{j-1}\left(u_{j-1}{u}_j\right)=0,$ the inner product being the generator $(j-1\mathrm{\mid }j\mathrm{\mid }j+1)$ of $I_m$ with $0<j<m$. If $v_0=j+1$, then $w=d_j{u}_j{d}_j$ and $j+1\ge 1$. If $m=1$ then $j=0$ and $w=d_0\left(u_0{d}_0\right)=0$, the factor $u_0{d}_0$ being the zero return at $0$ by step 1.1. If $m\ge 2$, the two possibilities are exhaustive: for $j+1<m$ the relation at the interior vertex $j+1$ writes the middle return as $d_j{u}_j=u_{j+1}{d}_{j+1}$, so $w=u_{j+1}\left(d_{j+1}{d}_j\right)=0$, the inner product being the generator $(j+2\mathrm{\mid }j+1\mathrm{\mid }j)$ with $0<j+1<m$; and for $j+1=m$ the relation at $m-1$, an interior vertex because $m\ge 2$, writes $u_{m-1}{d}_{m-1}=d_{m-2}{u}_{m-2}$, so $w=d_{m-1}\left(d_{m-2}{u}_{m-2}\right)=\left(d_{m-1}{d}_{m-2}\right)u_{m-2}=0$, the inner product now being the generator $(m\mathrm{\mid }m-1\mathrm{\mid }m-2)$ with $0<m-1<m$. All cases are exhausted, so $w$ has class $0$ in $A_m$.

$\rho$ vanishes on the relation ideal. The two-sided ideal $I_m$ consists of finite sums $asb$, where $s$ is a relation generator from step 1.1 and $a,b\in \mathbb{Z}{\mathrm{\Gamma }}_m$: such sums form an ideal containing the generators and are contained in every such ideal. By bilinearity it suffices to consider basis paths $a,b$. Every path term in $asb$, if nonzero, has length $\mathrm{\mid }a\mathrm{\mid }+2+\mathrm{\mid }b\mathrm{\mid }$. If this length is at least three, each term has $\rho$-image zero by step 1.2. Otherwise $a,b$ are vertex paths. For a monotone generator, $asb$ is either zero or the same monotone path, which $\rho$ sends to zero. The same holds for $s=u_0{d}_0$. For $s=u_i{d}_i-d_{i-1}{u}_{i-1}$, both summands start and end at $i$: multiplication by vertex paths either kills both summands or retains both, and in the latter case their images are $r_i-r_i=0$. Thus $\rho (asb)=0$ for every relation generator, and linearity gives $\rho (I_m)=0$.

All paths of length at least three have class $0$. We prove by induction on $l\ge 3$ that every path of length $l$ has class $0$ in $A_m$. The case $l=3$ is step 2.2. For $l\ge 4$, write $w=a{w}^{\mathrm{\prime }}$ where $a$ is the first arrow of $w$ and $w^{\mathrm{\prime }}$ is the suffix of length $l-1$; by the induction hypothesis $w^{\mathrm{\prime }}$ has class $0$, hence $w=a{w}^{\mathrm{\prime }}$ has class $0$ by [L4], the product of the class of $a$ with the class of $w^{\mathrm{\prime }}$ being the class of their concatenation by the ring structure of the quotient.

The functional descends. By step 2.3 the $\mathbb{Z}$-linear map $\rho$ vanishes on the additive subgroup $I_m\subseteq \mathbb{Z}{\mathrm{\Gamma }}_m$. The quotient map $\mathbb{Z}{\mathrm{\Gamma }}_m\to A_m=\mathbb{Z}{\mathrm{\Gamma }}_m\mathrm{/}{I}_m$ is a surjective homomorphism of abelian groups with kernel $I_m$ by [L4], so $\rho$ factors through it: there is a $\mathbb{Z}$-linear map $\bar{\rho }:A_m\to F$ with $\bar{\rho }(x+I_m)=\rho (x)$ for every $x\in \mathbb{Z}{\mathrm{\Gamma }}_m$. It is surjective, because each basis element $e_i$, $u_i$, $d_i$, $r_i$ of $F$ is the image of the class of the corresponding displayed path by step 1.2.

The class map $\pi$ is surjective. By step 3.1 and step 2.1, the images in $A_m$ of the $4m+1$ displayed paths span $A_m$ as an abelian group: every path of length at least three is $0$, every length-two path is $0$ or one of the returns, and length-zero and length-one paths are the vertices and arrows. Let $\pi :{\mathbb{Z}}^{4m+1}\to A_m$ be the $\mathbb{Z}$-linear map sending the standard basis elements to these $4m+1$ classes in the displayed order; it is surjective.

The composite of $\pi$ and $\bar{\rho }$. Let $F={\mathbb{Z}}^{4m+1}$ be identified with the free group of step 1.2 in the displayed basis. For each displayed path $p$ one has $\rho (p)$ equal to the corresponding basis element of $F$; hence, for the standard basis element ${ϵ}_p$ attached to $p$, the composite $\bar{\rho }\circ \pi$ satisfies $\bar{\rho }\pi ({ϵ}_p)=\bar{\rho }(\text{class of }p)=\rho (p)={ϵ}_p$. Since both sides are $\mathbb{Z}$-linear and agree on a basis, $\bar{\rho }\circ \pi ={\mathrm{i}\mathrm{d}}_F$ by [L3].

Conclusion: a basis. From $\bar{\rho }\circ \pi ={\mathrm{i}\mathrm{d}}_F$ of step 5.1, $\pi$ is injective: if $\pi (x)=0$ then $x=\bar{\rho }\pi (x)=0$. Since $\pi$ is surjective by step 4.1 it is an isomorphism, so $A_m\cong {\mathbb{Z}}^{4m+1}$ and the images under $\pi$ of the standard basis, namely the classes of the $4m+1$ displayed paths, form a $\mathbb{Z}$-basis of $A_m$. In particular those classes are $\mathbb{Z}$-linearly independent, the rank is $4m+1$, and the vanishing and identification statements for paths of length at least two are exactly those of steps 2.1 and 3.1. The internal degrees of the basis elements are $0$ for the vertices and ascending arrows and $1$ for the descending arrows and the returns, by the degree convention of [F2], so the basis is graded as displayed.