Bounded Bimodule Complexes and Derived Tensor — Examples

Since the only nonzero pairs $(p,q)$ have $p,q\in \{0,1\}$, their total degrees are $0,1,1,2$, giving exactly the four displayed summands. Formula [L1] sends $f^0\otimes x^0$ to $(u(f)\otimes x,f\otimes v(x))$, which is the displayed $d_T^0$.

On $F^1{\otimes }_A{X}^0$ the second-factor term in [L1] has sign $(-1{)}^1=-1$, while on $F^0{\otimes }_A{X}^1$ the first-factor term has sign $+1$; hence $d_T^1$ is the displayed row. The maps are well-defined on the balanced tensor and preserve the outer $B,C$-actions and internal grading by [L2].

For an elementary tensor $f\otimes x\in F^0{\otimes }_A{X}^0$, the two paths give $-(u(f)\otimes v(x))$ and $u(f)\otimes v(x)$, respectively, because $u{\otimes }_{A}1$ and $1{\otimes }_{A}v$ act on separate factors. Therefore $d_T^1{d}_T^0(f\otimes x)=0$, and additivity proves $d_T^1{d}_T^0=0$ on all of $T^0$.

In the stated integer example the displayed maps have matrices $\left(\begin{array}{c}2\\ 3\end{array}\right)$ and $\left(\begin{array}{cc}-3& 2\end{array}\right)$, whose product is $-6+6=0$. If $f$ has internal degree $r$ and $x$ has internal degree $s$, every nonzero matrix entry preserves degree $r+s$; the sign is determined only by $p$. With $F$ concentrated in degree $0$ the surviving tensor differential is $1\otimes v$, and with $F$ concentrated in degree $1$ it is $-1\otimes v$, as [L1] prescribes.

The regular left $A$-module is $A\{0\}$, generated by its degree-zero unit; [L10] makes it finite graded projective. Thus the sole nonzero term of $F$ meets the left projectivity hypothesis.

By [L2, L6] with one and two indeterminates, $A=k[x]$ and $R:=k[x,y]$ are domains, hence commutative rings by [L3]. Define the enveloping action by $(a\otimes b^{\mathrm{o}\mathrm{p}})\cdot m=amb$. The map $R\to A{\otimes }_k{A}^{\mathrm{o}\mathrm{p}}$ sending $x^i{y}^j$ to $x^i\otimes (x^j{)}^{\mathrm{o}\mathrm{p}}$ is multiplicative because $A$ is commutative; its inverse sends $f(x)\otimes g(x{)}^{\mathrm{o}\mathrm{p}}$ to $f(x)g(y)$. This inverse is balanced over $k$, and the maps are inverse on the monomials and elementary tensors that span their modules by [L4, L5, L7, L8]. Thus $A^e$ is isomorphic to the commutative ring $R$.

The underlying right regular module is projective: given a surjection $q:E\twoheadrightarrow M$ of right $A$-modules and a map $g:A\to M$, choose $e\in E$ with $q(e)=g(1)$ and define $\tilde{g}(a)=ea$; then $q\tilde{g}=g$. Since the complex $F$ is concentrated in degree zero, $A{\otimes }_{A}X\to X$, $a\otimes z\mapsto az$, is a natural chain isomorphism for every bounded left $A$-complex $X$, with inverse $z\mapsto 1\otimes z$. Under the standing size convention in [L11], its derived tensor functor is represented by this ordinary tensor operation as well.

In these coordinates the action on $A$ is induced by the surjective ring map $\mu :R\to A$, $f(x,y)\mapsto f(x,x)$; it is surjective since every $g(x)\in A$ is $\mu (g(x))$. For a monomial $x^i{y}^j$ with $j\ge 1$, $x^i{y}^j-x^{i+j}=x^i(y-x){\sum }_{r=0}^{j-1}{y}^{j-1-r}{x}^r$, while the difference is zero for $j=0$; by finite support [L4], $f-f(x,x)\in (x-y)$ for every $f$. Hence $\ker \mu =(x-y)$. This ideal is nonzero because the distinct monomials $x$ and $y$ have nonzero coefficients, and proper because $\mu (1)=1\mathrm{\ne }0$.

Suppose for contradiction that $A$ is projective as a left $R$-module. Since $\mu$ is a surjection, [L9] lifts ${\mathrm{id}}_A$ to an $R$-linear map $s:A\to R$ with $\mu s={\mathrm{id}}_A$. Then $p={\mathrm{id}}_R-s\mu$ is an $R$-linear projection onto $I:=\ker \mu$: $p(R)\subseteq I$ and $p(i)=i$ for $i\in I$. Every $R$-linear map $p:R\to R$ is multiplication by $e:=p(1)$, so $I=Re$; also $p^2=p$ gives $e^2=e$.

By [L2, L3, and L6], $R$ is a commutative domain. Thus $e^2=e$ implies $e(e-1)=0$, so $e=0$ or $e=1$. Then $I=Re$ is either zero or all of $R$, contradicting Step 2.2, where $I$ was shown nonzero and proper. Therefore $A$ is not projective over $A^e$.

The zero polynomial has empty support and lies in $\ker \mu$; the kernel calculation includes empty finite sums. By [L2], $1\mathrm{\ne }0$ in $k$, so the zero algebra is excluded. In characteristic two, $x-y=x+y$ still has two distinct monomials, so $I$ remains nonzero and proper. The complex $F$ has exactly one nonzero cochain term in degree zero, with zero differential, so both endpoints are covered by the unit calculation. The only lift used is the single lift of ${\mathrm{id}}_A$ supplied by projectivity; no family of choices or AC is used. This example proves no iff statement. [L2, L4, L9, L10, step 1.1, step 2.1, step 2.2, step 4.1, given, algebra] $\mathrm{\square }$

The only nonzero differential of $F$ is the degree-zero bimodule map $d_F^{-1}={\mathrm{id}}_A$; every composite of two consecutive differentials is zero because the next differential is zero, so $F$ is a bounded complex supported at the endpoints $-1$ and $0$.

Define $k^0:F^0\to F^{-1}$ to be ${\mathrm{id}}_A$ and all other components to be zero; then $k^0{d}_F^{-1}=\mathrm{id}$ in degree $-1$ and $d_F^{-1}{k}^0=\mathrm{id}$ in degree $0$, hence $d_{F}k+k{d}_F={\mathrm{id}}_F$, with every component internal-degree preserving and bimodule-linear.

For any bounded graded left $A$-complex $X$, the signed totalization has $T^n=(A{\otimes }_A{X}^{n+1})\oplus (A{\otimes }_A{X}^n)$; on elementary tensors $(a\otimes x,b\otimes y)$, with $x\in X^{n+1}$ and $y\in X^n$, its differential is $D^n(a\otimes x,b\otimes y)=(-a\otimes d_{X}x,a\otimes x+b\otimes d_{X}y)$, where the first sign is $(-1{)}^{-1}$ and the second-factor signs are $(-1{)}^{-1}$ and $(-1{)}^0$ on the two rows, and the formula extends additively to each balanced total term.

Each nonzero term $A=A\{0\}$ is a degree-zero direct summand of itself and hence finite graded projective on the left by [L3]; as a right module it is projective because, viewed as a left $A^{op}$-module, any fixed surjection $q:E\twoheadrightarrow M$ and right-linear $f:A\to M$ admit $e$ with $q(e)=f(1)$, and $\tilde{f}(a)=ea$ is a right-linear lift by [L4], while zero terms are projective on both sides. Thus $F$ and the zero complex satisfy [L5] and [L8].

By [L2], $H^n(a\otimes x,b\otimes y)=(b\otimes y,0)$; then $D^{n-1}{H}^n(a\otimes x,b\otimes y)=(-b\otimes d_{X}y,b\otimes y)$ and $H^{n+1}{D}^n(a\otimes x,b\otimes y)=(a\otimes x+b\otimes d_{X}y,0)$, whose sum is $(a\otimes x,b\otimes y)$ because the mixed terms cancel, also in characteristic two. Thus $dH+Hd={\mathrm{id}}_T$.

Let $0$ be the zero bimodule complex and take the zero maps $u:F\to 0$, $v:0\to F$, the homotopy $-k$ for $vu-{\mathrm{id}}_F=-{\mathrm{id}}_F$, and the zero homotopy for $uv-{\mathrm{id}}_0=0$; [L9] gives natural isomorphisms of their tensor functors on the bounded projective homotopy category and on ordinary and graded bounded derived categories, while [L6] and [L7] identify the latter with derived tensor and $0{\otimes }_A-$ is zero.

For every chain map $g:X\to Y$, both composites in the naturality square for $H$ send $(a\otimes x,b\otimes y)$ to $(b\otimes g(y),0)$, so the contraction is natural on bounded complexes.

If $X$ is zero or has empty support, all terms and homotopy maps are zero; if $X$ is concentrated in one degree the same formula applies with missing rows zero; if $d_X=0$ the differential terms vanish but step 2.1 still gives $dH+Hd=\mathrm{id}$. If $X$ is supported in $[c,d]$, step 1.3 gives $T$ support $[c-1,d]$, and all terms and maps outside those bounded endpoints are zero. The contraction is explicit, and the projectivity argument in step 1.4 uses only one preimage for one fixed lifting square, so no Axiom of Choice is used; the example states no iff claim. [step 1.3, step 2.1, step 1.4, algebra] $\mathrm{\square }$