Perfect Complexes and Triangulated Grothendieck Groups — Examples

By [F6] the module $P$ is a degree-zero direct summand of a finite direct sum $A\{r_1\}\oplus \cdots \oplus A\{r_n\}$; since $A$ is finite dimensional over $k$, that sum and hence its summand $P$ are finite dimensional, so $P$ and $P\{2\}$ are objects of the finite-dimensional graded module category in which $K_0^{\mathrm{g}\mathrm{r}}(A)$ and $G_0^{\mathrm{g}\mathrm{r}}(A)$ are formed. Applying the invertible shift $\{2\}$ to the splitting exhibits $P\{2\}$ as a degree-zero direct summand of $A\{r_1+2\}\oplus \cdots \oplus A\{r_n+2\}$, so $P\{2\}$ is again finite graded projective by [F6]. The stalk complex $C$ with $C^0=P$ and all other terms and differentials zero is a bounded complex of finite graded projectives with zero, hence degree-zero, differentials, so $C$ and its shifts are objects of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}^{\mathrm{g}\mathrm{r}}(A)$ by [F2]. Internal shift acts termwise and does not touch cochain degrees, while the cochain shift does not touch the internal grading, so $C\{2\}=P\{2\}[0]$ as complexes, and $C[1]\{2\}=C\{2\}[1]$ is the complex with $P\{2\}$ in cohomological degree $-1$ and zero differential, i.e. the class $[P[1]\{2\}]$ is the class of $(P\{2\})[1]$ in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}^{\mathrm{g}\mathrm{r}}(A))$.

Applying the shift-sign identity of [F3] inside the essentially small triangulated category $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}^{\mathrm{g}\mathrm{r}}(A)$ to the object $X=C\{2\}$ gives $[X[1]]=-[X]$, that is $[C\{2\}[1]]=-[C\{2\}]$; by step 1.1 the left-hand class is $[P[1]\{2\}]$ and the right-hand class is $-[P\{2\}[0]]$, so $[P[1]\{2\}]=-[P\{2\}[0]]$ in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}^{\mathrm{g}\mathrm{r}}(A))$.

In $K_0^{\mathrm{g}\mathrm{r}}(A)$ the shift action gives $v^2[P]=[P\{2\}]$ by [F5], and the graded comparison of [F4] sends this class to ${\iota }_{\ast }(v^2[P])=[(P\{2\})[0]]=[P\{2\}[0]]$; independently, the inverse Euler class of [F4] evaluated on the two-term complex $P[1]\{2\}$ of step 1.1, whose only nonzero term is $P\{2\}$ in cohomological degree $-1$, equals $(-1{)}^{-1}[P\{2\}]=-v^2[P]$. Hence ${\iota }_{\ast }(v^2[P])=[P\{2\}[0]]$ and ${\iota }_{\ast }^{-1}([P[1]\{2\}])=-v^2[P]$.

Combining steps 2.1 and 2.2, $[P[1]\{2\}]=-[P\{2\}[0]]=-{\iota }_{\ast }(v^2[P])$; that is, identifying $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}^{\mathrm{g}\mathrm{r}}(A))$ with $K_0^{\mathrm{g}\mathrm{r}}(A)$ along the comparison isomorphism ${\iota }_{\ast }$ of [F4], the class of the homological-and-internal shift of the degree-zero complex is $[P[1]\{2\}]=-v^2[P]$, the sign coming from $[1]$ and the factor $v^2$ from $\{2\}$. Since the Cartan map of [F5] sends $[Q]\mapsto [Q]$ and both $K_0^{\mathrm{g}\mathrm{r}}(A)$ and $G_0^{\mathrm{g}\mathrm{r}}(A)$ have $v$ acting by internal shift, $c_A^{\mathrm{g}\mathrm{r}}(-v^2[P])=-v^2[P]$, so the image of $[P[1]\{2\}]$ in $G_0^{\mathrm{g}\mathrm{r}}(A)$ obeys the same formula $-v^2[P]$. The differentials of $P[1]\{2\}$ vanish identically because $P$ sits in a single cohomological degree, so the internal shift introduces no cochain sign; only the homological shift contributes the sign $-1$, and no finite-global-dimension or Noetherian hypothesis is used.

For $a+b\epsilon \in A$ one has $\epsilon (a+b\epsilon )=a\epsilon$, so multiplication by $\epsilon$ has $\ker (\epsilon \cdot )=(\epsilon )=\mathrm{im}(\epsilon \cdot )$: the kernel consists exactly of the multiples of $\epsilon$, and it equals the image. The quotient augmentation $\pi :A\to S$ is surjective with kernel $(\epsilon )$, equal to the image of the differential into the degree-zero copy of $A$. Thus the sequence is exact at every copy of $A$ and at $S$, and is a free resolution $Q_{\bullet }\to S$ with every term $A$ finitely generated free.

Since $A$ is commutative, the resolution of step 1.1 supplies both a left and a right projective resolution of $S$. Applying $S{\otimes }_A(-)$ to its unaugmented complex $Q_{\bullet }$ gives a complex with $S{\otimes }_{A}A\cong S$ in every nonnegative degree and induced differentials equal to multiplication by $\epsilon$ on $S$, which is zero because $\epsilon S=0$; hence its homology is $S\cong k$ in every degree $i\ge 0$. By [F4] the specified-resolution Tor is ${\mathrm{Tor}}_i^{A,Q}(S,S)\cong k$ for every $i\ge 0$; under the DC supplied by AC [F3], the balanced bifunctor [F5] identifies this with ${\mathrm{Tor}}_i^A(S,S)\cong k$, and [F7] then gives $H^{-n}(S[0]{\otimes }_A^{\mathbf{L}}S[0])\cong {\mathrm{Tor}}_n^A(S,S)\cong k$ for every $n\ge 0$, in particular $H^{-n}\mathrm{\ne }0$ for all $n\ge 0$.

Suppose $S[0]$ were perfect; then [F1] supplies a bounded cochain complex $P$ of finitely generated projective left $A$-modules together with an isomorphism $P\cong S[0]$ in $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$. Both $P$ and $S[0]$ are bounded above, so [F9] lifts this isomorphism to $D^{-}(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$. Since the bounded-above derived tensor is a bifunctor in its second variable [F6], the lifted isomorphism gives $S[0]{\otimes }_A^{\mathbf{L}}S[0]\cong S[0]{\otimes }_A^{\mathbf{L}}P$. The identity $P\stackrel{\mathrm{i}\mathrm{d}}{\to }P$ is a quasi-isomorphism from a bounded-above complex of projective modules, so it is a supplied projective replacement as required by [F6]. Thus $S[0]{\otimes }_A^{\mathbf{L}}P$ is represented by $\mathrm{Tot}(S[0]{\otimes }_{A}P)$, which by [F8] is the bounded complex $S{\otimes }_{A}P$: its differential is $1\otimes d_P$ since the first factor is in degree zero, and it vanishes outside the finite support of $P$. Therefore $H^{-n}(S[0]{\otimes }_A^{\mathbf{L}}S[0])\cong H^{-n}(S{\otimes }_{A}P)=0$ for all sufficiently large $n$, contradicting step 2.1, which gives the nonzero $k$ in every degree $n\ge 0$. Hence $S[0]$ is not perfect, and since it is a complex concentrated in degree $0$ with $H^0(S[0])=S\cong k$ finite dimensional over $k$ and all other cohomology zero, this failure of perfectness is not detected by boundedness or by finite-dimensional cohomology.

Under cochain reindexing $X^i=X_{-i}$, the chain-cone terms $D_n\oplus C_{n-1}$ of [F1] become $Q^i\oplus P^{i+1}$, agreeing with the cochain formula of [F2]. For degree-zero stalk complexes, the only nonzero terms are $Q$ in degree $0$ and $P$ in degree $-1$, with differential $f:P\to Q$. Thus $\mathrm{Cone}(f)$ is bounded with finitely generated projective terms.

The cone triangle $P[0]\to Q[0]\to \mathrm{Cone}(f)\to P[1]$ of [F2] is a distinguished triangle of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$, since all three terms are bounded complexes of finitely generated projectives; its relation and the shift sign [F3] give $[\mathrm{Cone}(f)]=[Q[0]]+[P[1]]=[Q[0]]-[P[0]]$ in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$. Independently, the Euler class formula of [F4] on the two-term complex of step 1.1 gives $\chi (\mathrm{Cone}(f))=(-1{)}^{-1}[P]+(-1{)}^0[Q]=[Q]-[P]$ in $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))$, and the comparison isomorphism of [F5] carries this class to $[Q[0]]-[P[0]]$, so the two computations agree as promised.

Right multiplication $r_a(x):=xa$ is left $A$-linear: $r_a(bx)=(bx)a=b(xa)=b{r}_a(x)$ for all $b,x\in A$. Taking $P=Q=A$ and $f=r_a$ in step 2.1 gives $[\mathrm{Cone}(r_a)]=[A[0]]-[A[0]]=0$ and $\chi (\mathrm{Cone}(r_a))=[A]-[A]=0$ for every $a\in A$, whatever the kernel $\{x:xa=0\}$ and cokernel $A\mathrm{/}Aa$ may be: the two projective terms cancel even when the cone is not acyclic, and no assertion that its cohomology modules are projective is used.