Graded Quiver Algebras and Derived Tensor Functors — Examples

The nine basis paths and their degrees. Specialising [F1] to $m=2$ lists the generators of $I_2$ as $(0\mathrm{\mid }1\mathrm{\mid }2)$, $(2\mathrm{\mid }1\mathrm{\mid }0)$, $(1\mathrm{\mid }2\mathrm{\mid }1)-(1\mathrm{\mid }0\mathrm{\mid }1)$ and $(0\mathrm{\mid }1\mathrm{\mid }0)$, and by [L2] the classes of the vertices $(0),(1),(2)$, the ascending arrows $(0\mathrm{\mid }1),(1\mathrm{\mid }2)$, the descending arrows $(1\mathrm{\mid }0),(2\mathrm{\mid }1)$ and the returns $(1\mathrm{\mid }0\mathrm{\mid }1),(2\mathrm{\mid }1\mathrm{\mid }2)$ form a $\mathbb{Z}$-basis, nine classes in all; the degrees are $0$ for the vertices and ascending arrows and $1$ for the descending arrows by [F1], and $\mathrm{deg}(1\mathrm{\mid }0\mathrm{\mid }1)=\mathrm{deg}(1\mathrm{\mid }0)+\mathrm{deg}(0\mathrm{\mid }1)=1+0=1$, $\mathrm{deg}(2\mathrm{\mid }1\mathrm{\mid }2)=1+0=1$ by the additivity of the degree over concatenation.

The vanishing and identification rules at $m=2$. By [F1] the monotone length-two paths $(0\mathrm{\mid }1\mathrm{\mid }2)$ and $(2\mathrm{\mid }1\mathrm{\mid }0)$ have class $0$, while by [L2] every path of length at least three has class $0$; the return $(0\mathrm{\mid }1\mathrm{\mid }0)$ has class $0$; and the two returns at the interior vertex $1$ satisfy $(1\mathrm{\mid }2\mathrm{\mid }1)=(1\mathrm{\mid }0\mathrm{\mid }1)$ with no further relation, because the only interior vertex is $1$.

The multiplication table. Products of non-composable paths vanish and products of composable paths are concatenations by [L3]. For length two: $(1\mathrm{\mid }0)(0\mathrm{\mid }1)=(1\mathrm{\mid }0\mathrm{\mid }1)$ and $(1\mathrm{\mid }2)(2\mathrm{\mid }1)=(1\mathrm{\mid }2\mathrm{\mid }1)=(1\mathrm{\mid }0\mathrm{\mid }1)$ by step 1.2, $(2\mathrm{\mid }1)(1\mathrm{\mid }2)=(2\mathrm{\mid }1\mathrm{\mid }2)$, and $(0\mathrm{\mid }1)(1\mathrm{\mid }0)=(0\mathrm{\mid }1\mathrm{\mid }0)=0$ by step 1.2; every product of two classes that are not composable, and every product of three or more nonvanishing arrows, is either a non-composable product or a path of length at least three and hence $0$. In particular $A_2$ is not commutative, since $(1\mathrm{\mid }0)(0\mathrm{\mid }1)=(1\mathrm{\mid }0\mathrm{\mid }1)\mathrm{\ne }0$ while $(0\mathrm{\mid }1)(1\mathrm{\mid }0)=0$.

The quotient by the arrows is ${\mathbb{Z}}^3$. The two-sided ideal generated by the arrows consists of the $\mathbb{Z}$-linear combinations of the basis elements that are not vertices, and the vertex idempotents multiply by $e_j{e}_k={\delta }_{jk}{e}_j$ by [L3]; hence the quotient $A_2\mathrm{/}(\text{arrows})$ is free of rank $3$ on the classes of $e_0,e_1,e_2$ and is isomorphic to ${\mathbb{Z}}^3$ as a ring with coordinatewise multiplication.

The vertex projectives. By [L3] the module $P_j=A_2{e}_j$ is spanned over $\mathbb{Z}$ by the nine basis paths that end at $j$, and by step 1.1 those are, for $j=0$: $(0)$ and $(1\mathrm{\mid }0)$; for $j=1$: $(1)$, $(0\mathrm{\mid }1)$, $(2\mathrm{\mid }1)$ and $(1\mathrm{\mid }0\mathrm{\mid }1)$; for $j=2$: $(2)$, $(1\mathrm{\mid }2)$ and $(2\mathrm{\mid }1\mathrm{\mid }2)$. Since the nine classes are $\mathbb{Z}$-linearly independent by step 1.1, each of these lists is a $\mathbb{Z}$-basis, so the ranks are $2,4,3$ as displayed, and their degrees read off from step 1.1 give the graded ranks $(1,1)$, $(2,2)$ and $(2,1)$ in internal degrees $0$ and $1$ by [L4].

The right modules ${}_{j}P$. Symmetrically, ${}_{j}P=e_j{A}_2$ is spanned by the basis paths beginning at $j$, namely $(0),(0\mathrm{\mid }1)$ for $j=0$; $(1),(1\mathrm{\mid }0),(1\mathrm{\mid }2),(1\mathrm{\mid }0\mathrm{\mid }1)$ for $j=1$; and $(2),(2\mathrm{\mid }1),(2\mathrm{\mid }1\mathrm{\mid }2)$ for $j=2$, the same lists with the roles of the arrow directions exchanged; these are bases by step 1.1, so the ranks are again $2,4,3$.

Conclusion. At $m=2$ the algebra $A_2$ has the nine-element path basis of step 1.1 with the multiplication of step 2.1 and the vanishing rules of step 1.2; the vertex projectives $P_0,P_1,P_2$ are free of ranks $2,4,3$ on the paths ending at $0,1,2$ with graded ranks $(1,1),(2,2),(2,1)$, and the right modules ${}_{j}P$ are free of the same ranks on the paths beginning at $j$; the quotient by the arrow ideal is ${\mathbb{Z}}^3$ by step 2.2. All computations are finite lists of the nine basis paths, and no choice principle is used.

The map $P_0\to P_1$. By [L4] right multiplication by $(0\mathrm{\mid }1)$ is the degree-zero $A_2$-linear map $P_0\to P_1$ with $(0)\mapsto (0)(0\mathrm{\mid }1)=(0\mathrm{\mid }1)$ and $(1\mathrm{\mid }0)\mapsto (1\mathrm{\mid }0)(0\mathrm{\mid }1)=(1\mathrm{\mid }0\mathrm{\mid }1)$ by [L1]; both images are basis elements of $P_1$ by [L1], the degrees are preserved because $(0)$ and $(0\mathrm{\mid }1)$ have degree $0$ and $(1\mathrm{\mid }0)$ and $(1\mathrm{\mid }0\mathrm{\mid }1)$ have degree $1$, and the map is injective because it carries a basis to a linearly independent set.

The map $P_1\to P_2$. Right multiplication by $(1\mathrm{\mid }2)$ is the degree-zero $A_2$-linear map $P_1\to P_2$ with $(1)\mapsto (1\mathrm{\mid }2)$, $(2\mathrm{\mid }1)\mapsto (2\mathrm{\mid }1)(1\mathrm{\mid }2)=(2\mathrm{\mid }1\mathrm{\mid }2)$, $(0\mathrm{\mid }1)\mapsto (0\mathrm{\mid }1)(1\mathrm{\mid }2)=(0\mathrm{\mid }1\mathrm{\mid }2)=0$ and $(1\mathrm{\mid }0\mathrm{\mid }1)\mapsto (1\mathrm{\mid }0\mathrm{\mid }1)(1\mathrm{\mid }2)=(1\mathrm{\mid }0\mathrm{\mid }1\mathrm{\mid }2)=0$, the last two being respectively a monotone length-two path and a length-three path; again $(1)$ and $(1\mathrm{\mid }2)$ have degree $0$ while $(2\mathrm{\mid }1)$ and $(2\mathrm{\mid }1\mathrm{\mid }2)$ have degree $1$.

The map $P_2\to S_2$. Define $\epsilon$ to be the $A_2$-linear map with $\epsilon (e_2)=1$ and $\epsilon ((1\mathrm{\mid }2))=\epsilon ((2\mathrm{\mid }1\mathrm{\mid }2))=0$, which is well defined by the formula $\epsilon (a{e}_2)=a\cdot 1$: if $a{e}_2=0$, then $a\cdot 1=(a{e}_2)\cdot 1=0$ since $e_2\cdot 1=1$. This formula is $A_2$-linear by the module action in [F2]; it is surjective because $S_2=\mathbb{Z}\cdot 1$ and it is degree zero because $e_2$ has degree $0$.

Every composite in the sequence is zero. The composite $P_0\to P_1\to P_2$ is right multiplication by $(0\mathrm{\mid }1)(1\mathrm{\mid }2)=(0\mathrm{\mid }1\mathrm{\mid }2)=0$ by step 1.1, step 1.2 and [L1]; the composite $P_1\to P_2\to S_2$ kills the image of the second map, namely the basis elements $(1\mathrm{\mid }2)$ and $(2\mathrm{\mid }1\mathrm{\mid }2)$, which are sent to $0$ by step 1.3.

Exactness at $P_1$. By step 1.2 the kernel of right multiplication by $(1\mathrm{\mid }2)$ is the span of the two basis elements that are killed, $(0\mathrm{\mid }1)$ and $(1\mathrm{\mid }0\mathrm{\mid }1)$, and by step 1.1 the image of right multiplication by $(0\mathrm{\mid }1)$ is exactly the span of $(0\mathrm{\mid }1)$ and $(1\mathrm{\mid }0\mathrm{\mid }1)$; the two submodules of $P_1$ are therefore equal, so $\ker (P_1\to P_2)=\mathrm{i}\mathrm{m}(P_0\to P_1)$.

Exactness at $P_2$. A basis element of $P_2$ is in the kernel of $\epsilon$ exactly when it is not $(2)$, since $\epsilon ((2))=1$, so $\ker \epsilon =\mathbb{Z}(1\mathrm{\mid }2)\oplus \mathbb{Z}(2\mathrm{\mid }1\mathrm{\mid }2)$ by step 1.3; by step 1.2 that span is exactly the image of right multiplication by $(1\mathrm{\mid }2)$, so $\ker \epsilon =\mathrm{i}\mathrm{m}(P_1\to P_2)$.

Conclusion. The displayed sequence $0\to P_0\to P_1\to P_2\to S_2\to 0$ has finitely generated graded projective resolution terms $P_0,P_1,P_2$, all maps are degree-zero $A_2$-linear maps by steps 1.1, 1.2 and 1.3, the first map is injective by step 1.1, the last is surjective by step 1.3, every composite is zero by step 2.1 and the sequence is exact at $P_1$ and at $P_2$ by steps 2.2 and 2.3; hence it is a finite graded projective resolution of the vertex module $S_2$, of length $2$, as in [L3]. The two inner differentials are right multiplications by the degree-zero arrows $(0\mathrm{\mid }1)$ and $(1\mathrm{\mid }2)$, exactly the arrows ascending toward the vertex $2$; the module $S_2$ is a rank-one $\mathbb{Z}$-module and is therefore not a simple module over $A_2$ in the ungraded sense, the word "simple" belonging to the inherited identifier only.

The four summands and their degrees. Since $R_i$ has its two terms in degrees $-1$ and $0$ by [F2] and $X$ has its two terms in degrees $-1$ and $0$ by [F3], the index pairs $(p,q)$ with both $R_i^p$ and $X^q$ nonzero are $(-1,-1),(-1,0),(0,-1),(0,0)$, and the diagonal ${⨁}_{p+q=n}$ of [L1] collects them as $p+q=-2$ for $(-1,-1)$, as $p+q=-1$ for $(-1,0)$ and $(0,-1)$, and as $p+q=0$ for $(0,0)$; this gives the three displayed degrees with the four tensor summands $U_i{\otimes }_{A_m}{X}^{-1}$, $U_i{\otimes }_{A_m}{X}^0$, $A_m{\otimes }_{A_m}{X}^{-1}$, $A_m{\otimes }_{A_m}{X}^0$.

The differentials. By [L1] the differential on $U_i{\otimes }_{A_m}{X}^{-1}$ is $d_R\otimes 1+(-1{)}^{-1}1\otimes d_X={\beta }_i\otimes 1-1\otimes d$, which on an elementary tensor is $r\otimes x\mapsto {\beta }_i(r)\otimes x-r\otimes dx$, and the differential on $U_i{\otimes }_{A_m}{X}^0$ is ${\beta }_i\otimes 1+(-1{)}^{-1}1\otimes d_X={\beta }_i\otimes 1$, since $d_X=0$ on $X^0$ by [F3]; the differential on $A_m{\otimes }_{A_m}{X}^{-1}$ is $0+(-1{)}^{0}1\otimes d=1\otimes d$, that is $a\otimes x\mapsto a\otimes dx$, and on $A_m{\otimes }_{A_m}{X}^0$ it is $0$ because $R_i^1=0$ and $d_X^0=0$. No internal degree enters any sign, and all four maps preserve the total internal degree because ${\beta }_i$ and $d$ are degree-zero maps by [F2] and [F3].

The two routes cancel. For $r\otimes x\in U_i{\otimes }_{A_m}{X}^{-1}$ step 1.2 gives $d^{-2}(r\otimes x)={\beta }_i(r)\otimes x-r\otimes dx$, an element of the two summands of degree $-1$; applying $d^{-1}$ to the two pieces separately gives $d^{-1}({\beta }_i(r)\otimes x)={\beta }_i(r)\otimes dx$ in the first summand and $d^{-1}(-r\otimes dx)=-{\beta }_i(r)\otimes dx$ in the second, the latter because $d^{-1}$ on $U_i{\otimes }_{A_m}{X}^0$ is ${\beta }_i\otimes 1$ and $d_X(dx)=0$; the two results are negatives of one another, so $d^{-1}{d}^{-2}=0$ on the bottom term, and $d^0{d}^{-1}=0$ holds trivially because $d^0=0$. Hence the four displayed maps make the totalization a complex, as [L1] guarantees in general.

Unit form of the two upper summands. By [L4] the summands $A_m{\otimes }_{A_m}{X}^{-1}$ and $A_m{\otimes }_{A_m}{X}^0$ are degree-zero isomorphic to $X^{-1}$ and $X^0$ through the multiplication maps, so the middle term of the totalization may be written as $X^{-1}\oplus (U_i{\otimes }_{A_m}{X}^0)$ and the top term as $X^0$, with the differentials $x\mapsto dx$ and the $U_i$-component ${\beta }_i\otimes 1$ respectively.

Conclusion. A two-term twist complex $R_i$ and a two-term projective complex $X$ produce the totalization $U_i{\otimes }_{A_m}{X}^{-1}\to \left(U_i{\otimes }_{A_m}{X}^0\right)\oplus \left(A_m{\otimes }_{A_m}{X}^{-1}\right)\to A_m{\otimes }_{A_m}{X}^0$ with the four summands and the differentials of step 1.2, whose square vanishes by the explicit cancellation of step 2.1, and whose two $A_m$-columns may be read as $X^{-1}$ and $X^0$ by step 2.2. The sign $-1$ in the bottom differential is the Koszul sign $(-1{)}^p$ at $p=-1$, that is, it is attached to the homological degree of the first factor and not to any internal degree, which is the point of the construction.

The witness is nonzero. By [L2] the module $P_i=A_m{e}_i$ is a nonzero finitely generated graded projective left $A_m$-module for each $0\le i\le m$, in particular for $1\le i\le m$; let $\underset{‾}{P_i}$ denote the complex with ${\underset{‾}{P_i}}^0=P_i$ and ${\underset{‾}{P_i}}^n=0$ for $n\mathrm{\ne }0$, an object of $C_m$ by [L1].

The two shifted complexes. By [L1] the internal shift acts termwise, so $(\underset{‾}{P_i}\{1\}{)}^n={\underset{‾}{P_i}}^n\{1\}$ is nonzero exactly for $n=0$, where it equals $P_i\{1\}$, and its differentials are those of $\underset{‾}{P_i}$, all zero; the homological shift reindexes, so $(\underset{‾}{P_i}[1]{)}^n={\underset{‾}{P_i}}^{n+1}$ is nonzero exactly for $n=-1$, where it equals $P_i$, and again all differentials vanish. In particular $(\underset{‾}{P_i}\{1\}{)}^0=P_i\{1\}\mathrm{\ne }0$ by [L3] while $(\underset{‾}{P_i}[1]{)}^0=0$.

No morphism, hence no isomorphism. Let $f:\underset{‾}{P_i}\{1\}\to \underset{‾}{P_i}[1]$ be a morphism in $C_m$ represented by a chain map. By [L1] its degree-$0$ component is a degree-zero $A_m$-linear map $f^0:P_i\{1\}\to 0$, which must be the zero map; every other component of $f$ has either zero source or zero target by step 2.1, so $f=0$ and the only morphism between the two objects is the zero morphism. The identity of the one-term nonzero complex $\underset{‾}{P_i}\{1\}$ is nonzero in $C_m$: with both differentials zero, a homotopy cannot make its degree-$0$ identity map null. As its Hom group to $\underset{‾}{P_i}[1]$ is zero, the two objects are not isomorphic in $C_m$; consequently there is no natural isomorphism between the functors $\{1\}$ and $[1]$, because such a natural isomorphism would supply an isomorphism $\underset{‾}{P_i}\{1\}\cong \underset{‾}{P_i}[1]$ for this particular object.

The $K_0$ comparison. By [L4] one has $[\underset{‾}{P_i}[1]]=-[\underset{‾}{P_i}]$ and $[\underset{‾}{P_i}\{1\}]=q[\underset{‾}{P_i}]$ with $q$ an invertible endomorphism of $K_0(C_m)$; the two shift functors therefore act on the class of the witness by the operators $-1$ and $q$. The displayed classes are not asserted to be unequal: deciding that would require the additional input that $[\underset{‾}{P_i}]$ is not annihilated by $q+1$, that is, a basis computation in $K_0(C_m)$, which this counterexample does not use. The non-isomorphism of step 3.1 is a homological-support statement and needs no such computation.

Conclusion. For every $1\le i\le m$ the objects $\underset{‾}{P_i}\{1\}$ and $\underset{‾}{P_i}[1]$ of $C_m$ have nonzero terms in the distinct homological degrees $0$ and $-1$ by step 2.1, there is no nonzero morphism between them by step 3.1, and consequently the internal shift $\{1\}$ and the homological shift $[1]$ are not naturally isomorphic functors on $C_m$; the refuted claim fails already on a single vertex projective concentrated in degree $0$. The internal shift moves internal degrees and leaves homological placement fixed, the homological shift reindexes without touching internal degrees, and the two are compared in $K_0(C_m)$ by the operators $q$ and $-1$ of step 4.1 rather than identified. No choice principle is used, and the witness is the single module $P_i$ in one homological degree.