Algebra, Modules, and Tensor Symmetry Interfaces: Examples and Counterexamples

The matrices $E,F,H$ are linearly independent and every trace-zero matrix is $aH+bE+cF$ for unique $a,b,c\in \mathbb{C}$. Direct multiplication gives $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. By bilinearity the span is closed under commutators, and it inherits alternation and Jacobi from ${\mathrm{End}}_{\mathbb{C}}(V)$; hence it is the stated Lie algebra.

The inclusion $\rho :{\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})\hookrightarrow {\mathrm{End}}_{\mathbb{C}}(V)$ is linear and preserves the bracket because both brackets are the same matrix commutator. Thus $V$ is a Lie-algebra representation.

By [F3], this representation extends uniquely to a scalar-compatible unital left $U({\mathfrak{s}\mathfrak{l}}_2(\mathbb{C}))$-module on $V$. This uses neither PBW nor injectivity of the canonical map into the enveloping algebra.

Define ${\rho }_V(1)={\rho }_W(1)=1$, ${\rho }_V(s)=1$, and ${\rho }_W(s)=-1$, and extend linearly from the basis $\{1,s\}$. These maps preserve products because the only nontrivial basis relation is $s^2=1$ and both $1^2=(-1{)}^2=1$. Thus they are unital representations by [F3], hence left $H$-modules by [F4].

By [F1]–[F2], $H$ is a bialgebra with $\mathrm{\Delta }(1)=1\otimes 1$ and $\mathrm{\Delta }(s)=s\otimes s$. The action in [F5] sends $1$ to the identity and sends $s$ to $-{\mathrm{id}}_{V\otimes W}$, since $s\cdot (v\otimes w)=(s\cdot v)\otimes (s\cdot w)=v\otimes (-w)$. This operator squares to the identity, so the algebra homomorphism $H\to {\mathrm{End}}_{\mathbb{C}}(V\otimes W)$ determined on the basis $1,s$ gives a unital module action and agrees with the coproduct formula on that basis, hence on all of $H$.

The vector-space flip $\tau (v\otimes w)=w\otimes v$ is an isomorphism by [F6]. It is $H$-linear: for $s$, both $\tau (s\cdot (v\otimes w))$ and $s\cdot \tau (v\otimes w)$ equal $-w\otimes v$, while $1$ acts identically. Linearity and the basis $\{1,s\}$ give the intertwining identity for every $h\in H$. This is a symmetric swap for this example, not a claim that every bialgebra-module category is symmetric.

The identity map $\rho :\mathbb{C}\to {\mathrm{End}}_{\mathbb{C}}(\mathbb{C})$, $a\mapsto a\mathrm{id}$, is a unital representation. Thus $M$ and $N$ are unital left $A$-modules by [F1]–[F2].

Put $u=1\otimes 1\in \mathbb{C}{\otimes }_{\mathbb{C}}\mathbb{C}$. By [F3], multiplication sends $u$ to $1$, so $u\mathrm{\ne }0$. The proposed formula sends $u$ to $a^{2}u$ for $a\in \mathbb{C}$.

If this formula were an $A$-module action, it would be additive in $a$. But it gives $2\cdot u=4u$ from the formula for $a=2$, whereas additivity and $2=1+1$ require $2\cdot u=1\cdot u+1\cdot u=2u$. Since $u\mathrm{\ne }0$ and $4u\mathrm{\ne }2u$ over $\mathbb{C}$, the formula fails the module axiom. The coproduct-based formula in [F4] is a distinct construction requiring additional bialgebra data.