Deligne Products and Categorical Eilenberg–Watts — Examples

The algebra $k$ is finite-dimensional and unital over the field $k$ [F1], and the multiplication of [F2] on $k{\otimes }_{k}k$ satisfies $(a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}$; the $k$-linear map $k\to k{\otimes }_{k}k$, $a\mapsto a\otimes 1$, has the multiplication map $k{\otimes }_{k}k\to k$, ${\sum }_i{a}_i\otimes b_i\mapsto {\sum }_i{a}_i{b}_i$, as a two-sided inverse, so $k{\otimes }_{k}k\cong k$ as $k$-algebras and $(k{\otimes }_{k}k)\text{-}\mathrm{m}\mathrm{o}\mathrm{d}=\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}$ [F1].

By [F3] with $R=S=k$, the category $(k{\otimes }_{k}k)\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ together with the tensor bifunctor $(X,Y)\mapsto X{\otimes }_{k}Y$ is a Deligne product of $k\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ and $k\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$; under the algebra isomorphism of step 1.1, a module over $k{\otimes }_{k}k$ is a $k$-vector space, so $(k{\otimes }_{k}k)\text{-}\mathrm{m}\mathrm{o}\mathrm{d}=\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}=k\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$, and the universal bifunctor is the ordinary tensor product ${\otimes }_k$ on $\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}$ (k-linear categories and k-linear functors).

Hence $\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}⊠\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}$ is identified with $\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}$, the universal bifunctor being ${\otimes }_k:\mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}\times \mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}\to \mathbf{v}\mathbf{e}\mathbf{c}\mathbf{t}$, and its universal property is exactly the defining property of a Deligne product of finite $k$-linear categories [F3]; since Deligne products are unique up to an equivalence respecting the universal bifunctors, the identification respects the universal bifunctors up to canonical natural isomorphism (Equivalence, quasi-inverse, and adjoint equivalence of categories, The Deligne product of finite linear categories).

The algebra $A_0$ is well defined: the $k$-linear assignment $e_1\mapsto \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$, $e_2\mapsto \left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$, $u\mapsto \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ identifies $A_0$ with the algebra of upper triangular $2\times 2$ matrices, in which the listed products hold and multiplication is associative; consequently $e_1,e_2$ are orthogonal idempotents summing to $1$, $e_{1}u=u=u{e}_2$, $u{e}_1=0$, $e_{2}u=0$ and $u^2=0$.

The left ideal $A_0{e}_1$ equals the $k$-span of $e_1$: from $e_1{e}_1=e_1$, $e_2{e}_1=0$ and $u{e}_1=0$ one gets $a{e}_1=\kappa (a)e_1$ for the coefficient functional $\kappa$ of $e_1$, so $A_0{e}_1=\mathrm{span}\{e_1\}$ has ${\dim }_k{A}_0{e}_1=1$. It is moreover projective in the lifting sense of Projective modules and the lifting property: by [F1] it is a direct summand of $A_0$ with projection $\pi :a\mapsto a{e}_1$, the free module $A_0$ has the lifting property because a $A_0$-linear map out of $A_0$ is determined by its value at the generator $1$, which can be lifted along any epimorphism, and restricting a lift of $f\circ \pi$ to $A_0{e}_1$ lifts a given $f:A_0{e}_1\to M$.

The image $A_0^{\ast }{e}_1$ has dimension $2$: for $\lambda \in A_0^{\ast }$ one computes $(\lambda \cdot e_1)(x)=\lambda (e_{1}x)$, so $\lambda \cdot e_1=\lambda \circ \phi$ for the $k$-linear map $\phi :A_0\to A_0$, $\phi (x)=e_{1}x$, whose image is $e_1{A}_0=\mathrm{span}\{e_1,u\}$ of dimension $2$ by step 1.1; the restriction map $A_0^{\ast }\to (e_1{A}_0{)}^{\ast }$, $\lambda \mapsto \lambda {\mathrm{\mid }}_{e_1{A}_0}$, is surjective since a functional on the direct summand $e_1{A}_0$ extends by zero on $\mathrm{span}\{e_2\}$, and composition with the surjection $\phi$ is injective, so $A_0^{\ast }{e}_1=\{\mu \circ \phi :\mu \in (e_1{A}_0{)}^{\ast }\}\cong (e_1{A}_0{)}^{\ast }$ has dimension $2$.

By [F2] the multiplication map $A_0^{\ast }{\otimes }_{A_0}{A}_0{e}_1\to A_0^{\ast }{e}_1$, $\lambda \otimes x\mapsto \lambda \cdot x$, is a well-defined surjection, and it is injective with inverse $\nu \mapsto \nu \otimes e_1$: indeed $(\lambda \cdot e_1)\otimes e_1=\lambda \otimes (e_1{e}_1)=\lambda \otimes e_1$, and for $x=c{e}_1\in A_0{e}_1$ one has $\lambda \cdot x=(\lambda \cdot c)\cdot e_1$ and $((\lambda \cdot c)\cdot e_1)\otimes e_1=(\lambda \cdot c)\otimes e_1=\lambda \otimes c{e}_1$. Hence $A_0^{\ast }{\otimes }_{A_0}{A}_0{e}_1\cong A_0^{\ast }{e}_1$, and since $N^r\cong A_0^{\ast }{\otimes }_{A_0}-$ by the given data, step 2.2 gives ${\dim }_k{N}^r(A_0{e}_1)={\dim }_k{A}_0^{\ast }{e}_1=2$.

Since ${\dim }_k{N}^r(A_0{e}_1)=2$ while ${\dim }_k{A}_0{e}_1=1$, the vector spaces $N^r(A_0{e}_1)$ and $A_0{e}_1$ are not isomorphic, so by [F4] there is no natural isomorphism $N^r\cong 1_{\mathcal{A}}$; equivalently $N^r$ is not naturally isomorphic to the identity functor. By [F4] the equivalence ${\mathrm{\Gamma }}^{rl}$ sends the identity functor, regarded as left exact, to $N^r$, so it does not send the identity to a functor naturally isomorphic to the identity, and the equivalence does not preserve the identity object.

The end is the regular bimodule: apply [F1] with $\mathcal{A}=\mathcal{B}$ and $M=A$, the regular $(A,A)$-bimodule. Then $G={\mathrm{\Phi }}^r(A)\cong 1_{\mathcal{A}}$ by [F2], so the end ${\int }_a\bar{a}⊠1(a)$ of the identity diagram is the end of the diagram $a\mapsto G(a){\otimes }_k{a}^{\ast }$ and equals the (co)end object $M=A$ of [F1], with universal wedge ${\omega }_a(m)(x)=m\otimes x$.

For $A_0$ one has $A_0{e}_1=\mathrm{span}\{e_1\}$ of dimension $1$, because $e_1{e}_1=e_1$, $e_2{e}_1=0$ and $u{e}_1=0$; and $A_0^{\ast }{e}_1$ has dimension $2$, because $(\lambda \cdot e_1)(x)=\lambda (e_{1}x)$ expresses $\lambda \cdot e_1$ as the composite of $\lambda$ with the map $x\mapsto e_{1}x$, whose image is $e_1{A}_0=\mathrm{span}\{e_1,u\}$ of dimension $2$, and every functional on that direct summand of $A_0$ extends to $A_0$.

The coend is the co-regular bimodule: apply [F1] with $\mathcal{A}=\mathcal{B}$ and $M=A^{\ast }$. Then $F={\mathrm{\Phi }}^l(A^{\ast })\cong 1_{\mathcal{A}}$ by [F2], so the coend ${\int }^a\bar{a}⊠1(a)$ is the coend of the diagram $a\mapsto F(a){\otimes }_k{a}^{\ast }$ and equals the (co)end object $M=A^{\ast }$ of [F1], with universal cowedge ${\rho }_a(f\otimes \lambda )=\lambda \circ f$; under the identification ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}⊠\mathcal{A}\simeq (A,A)\text{-}\mathrm{b}\mathrm{i}\mathrm{m}\mathrm{o}\mathrm{d}$ the object $A$ is the regular and $A^{\ast }$ the co-regular bimodule.

Consequently $A_0^{\ast }{\otimes }_{A_0}{A}_0{e}_1\cong A_0^{\ast }{e}_1$ has dimension $2$, by the multiplication isomorphism $\lambda \otimes x\mapsto \lambda \cdot x$ with inverse $\nu \mapsto \nu \otimes e_1$, while $A_0{\otimes }_{A_0}{A}_0{e}_1\cong A_0{e}_1$ has dimension $1$ by the unit isomorphism of [F3].

The bimodules $A_0^{\ast }$ and $A_0$ are not isomorphic: an isomorphism would by [F3] induce an isomorphism $A_0^{\ast }{\otimes }_{A_0}{A}_0{e}_1\cong A_0{\otimes }_{A_0}{A}_0{e}_1$, hence equality of dimensions, contradicting step 2.2. Hence the regular kernel $A_0$ and the co-regular kernel $A_0^{\ast }$ of steps 1.1 and 2.1 are distinguished, so an end and a coend of the same functor need not agree.

Finally, the end $A={\mathrm{\Psi }}^r(1_{\mathcal{A}})$ is the image of the identity functor regarded as an object of $\mathrm{Rex}(\mathcal{A},\mathcal{A})$ under ${\mathrm{\Psi }}^r$, and the coend $A^{\ast }={\mathrm{\Psi }}^l(1_{\mathcal{A}})$ is the image of the identity regarded as an object of $\mathrm{Lex}(\mathcal{A},\mathcal{A})$ under ${\mathrm{\Psi }}^l$, by steps 1.1 and 2.1; applying ${\mathrm{\Phi }}^r$ and ${\mathrm{\Phi }}^l$ recovers the Nakayama functors $N^r={\mathrm{\Phi }}^r{\mathrm{\Psi }}^l(1_{\mathcal{A}})\cong A^{\ast }{\otimes }_A-$ and $N^l={\mathrm{\Phi }}^l{\mathrm{\Psi }}^r(1_{\mathcal{A}})\cong {\mathrm{Hom}}_A(A^{\ast },-)$ of the Nakayama calculus (Left and right Nakayama functors by finite kernel calculus, The left-to-right exact equivalence sends the identity to the Nakayama functor).

The specified algebra is the upper triangular $2\times 2$ matrix algebra under $e_1\mapsto E_{11}$, $e_2\mapsto E_{22}$, $u\mapsto E_{12}$, so the products are associative and define a unital algebra. In $A_0$ the relations $e_1+e_2=1$, $e_1{e}_2=e_2{e}_1=0$ and $e_{1}u=u=u{e}_2$ hold with $e_1,e_2,u$ a $k$-basis of the regular bimodule, so left multiplication by $u$ sends $e_2$ to $u{e}_2=u\mathrm{\ne }0$ and right multiplication by $u$ sends $e_1$ to $e_{1}u=u\mathrm{\ne }0$, while $A_0$ has $k$-dimension $3$ [F1, F2].

On a one-dimensional left or right module, the action of $u$ is multiplication by a scalar $c\in k$. Since $u^2=0$, the module law gives $c^2=0$, hence $c=0$ because $k$ is a field. Thus $u$ acts as zero on every one-dimensional module on either side.

Suppose the regular bimodule $A_0$ were isomorphic to $b{\otimes }_k{a}^{\ast }$. By [F3] the two sides have the same $k$-dimension and the same vanishing pattern of the two multiplications, and by [F2] $3={\dim }_k{A}_0={\dim }_{k}b\cdot {\dim }_k{a}^{\ast }={\dim }_{k}b\cdot {\dim }_{k}a$, so one of the two factors is one-dimensional. If ${\dim }_{k}b=1$, then left multiplication by $u$ is zero on $b$ by step 2.1, hence zero on $b{\otimes }_k{a}^{\ast }$ because $u\cdot (x\otimes \lambda )=(ux)\otimes \lambda =0$, contradicting step 1.1, where left multiplication by $u$ sends $e_2$ to $u\mathrm{\ne }0$. If ${\dim }_{k}a=1$, then $u$ acts as zero on $a$, so $(\lambda \cdot u)(x)=\lambda (ux)=0$ for every $\lambda \in a^{\ast }$, and right multiplication by $u$ is zero on $b{\otimes }_k{a}^{\ast }$ because $(y\otimes \lambda )\cdot u=y\otimes (\lambda \cdot u)=0$, contradicting step 1.1, where right multiplication by $u$ sends $e_1$ to $u\mathrm{\ne }0$. Both alternatives contradict the assumed bimodule isomorphism, so the regular bimodule $A_0$ is not isomorphic to any external tensor factor $b{\otimes }_k{a}^{\ast }$.