1. Context
In my lecture notes we defined the term enveloping algebra:
Let $k$ be a field. Let $(A, \mu_A, \eta_A)$ be a unital, associative algebra. We call the algebra $A \otimes A^{opp}$ its enveloping algebra.
Further, there is the notion of a universal enveloping algebra of a Lie algebra:
Let $\mathfrak{g}$ be a Lie algebra. Its universal enveloping algebra is the the quotient $T(\mathfrak{g})/I (\mathfrak{g})$ of the tensor algebra by the two-sided ideal $I(\mathfrak{g})$ generated by all elements of the form $x \otimes y – y \otimes x -[x,y]$ where $x,y \in \mathfrak{g}$.
2. Questions
- What does $(-)^{opp}$ stand for? Why not leave it away?
- Is the following correct? The multiplication in $A \otimes A^{opp}$ is given by $\mu_{A \otimes A^{opp}}: (\mu_A \otimes \mu_{A^{opp}})\circ(id_A \otimes \tau \otimes id_{A^{opp}}).$ The unit is given by $\eta_{A \otimes A^{opp}}: (id_A \otimes \tau \otimes id_{A^{opp}})\circ (\eta_A \otimes \eta_{A^{opp}})$. Here we used the identification $k \cong k \otimes k.$ Further, the morphism $\tau: A^{opp} \otimes A \rightarrow A \otimes A^{opp}; v \otimes w \mapsto w \otimes v$ is the twist map.
- (How) are enveloping and universal enveloping algebra related?
Best Answer
Let $A$ be an $k$-algebra. The opposite algebra $A^{\mathrm{opp}}$ (or $A^{\mathrm{op}}$) is given as follows. The underlying vector space of $A^{\mathrm{opp}}$ is the same as the underlying vector space of $A$. Let us denote for every element $a$ of $A$ by $a^{\mathrm{opp}}$ the corresponding (i.e. the same) element of $A^{\mathrm{opp}}$. The multiplication in $A^{\mathrm{opp}}$ is given in this notation by $$ a^{\mathrm{opp}} \cdot b^{\mathrm{opp}} = (b \cdot a)^{\mathrm{opp}} $$ for all $a, b \in A$. The unit of $A^{\mathrm{opp}}$ is then given by $1_{A^{\mathrm{opp}}} = 1_A^{\mathrm{opp}}$. Abstractly speaking, this means that $$ \mu_{A^{\mathrm{opp}}} = \mu_A \circ \tau \,, \quad \eta_{A^{\mathrm{opp}}} = \eta_A $$ where $\tau$ denotes the twist map from $A \otimes A$ to $A \otimes A$.
Given any two $k$-algebras $A$ and $B$ we can make the tensor product $A \otimes B$ again into a $k$-algebra, with multiplication given by $$ (a_1 \otimes b_1) \cdot (a_2 \otimes b_2) = (a_1 a_2) \otimes (b_1 b_2) $$ for all $a_1, a_2 \in A$ and $b_1, b_2 \in B$. The unit of $A \otimes B$ is then given by $$ 1_{A \otimes B} = 1_A \otimes 1_B \,. $$ The multiplication of $A \otimes B$ is thus abstractly given by $$ \mu_{A \otimes B} = (\mu_A \otimes \mu_B) \circ (\mathrm{id}_A \otimes \tau \otimes \mathrm{id}_B) \,, $$ where $\tau$ denotes the twist map from $B \otimes A$ to $A \otimes B$, and the unit of $A \otimes B$ is given by $$ \eta_{A \otimes B} = (\eta_A \otimes \eta_B) \circ \lambda $$ where $\lambda$ is the isomorphism of vector spaces $$ \lambda \colon k \to k \otimes k \,, \quad 1 \mapsto 1 \otimes 1 \,. $$
If we take $B = A^{\mathrm{opp}}$ then the above formula for the multiplication $\mu_{A \otimes A^{\mathrm{opp}}}$ agrees with the one proposed in the question. However, the proposed formula for $\eta_{A \otimes A^{\mathrm{opp}}}$ doesn’t make sense. The map $\eta_A \otimes \eta_{A^{\mathrm{opp}}}$ goes to $A \otimes A^{\mathrm{opp}}$, so we can’t apply $\mathrm{id}_A \otimes \tau \circ \mathrm{id}_{A^{\mathrm{opp}}}$ after that.
I don’t know if there is any connection between the universal enveloping algebra of a Lie algebra and the enveloping algebra of an associative, unitial algebra.
Regarding the comments under your question: Yes, an $A$-bimodule is “the same“ as a left $A^{\mathrm{e}}$-module. More precisely, if $M$ is an $A$-bimodule, then the corresponding left $A^{\mathrm{e}}$-module structure on $M$ is given by $$ (a \otimes b^{\mathrm{opp}}) \cdot m = a \cdot m \cdot b $$ for all $a, b \in A$, $m \in M$. If we would would instead use the definition ${}^{\mathrm{e}} \! A = A^{\mathrm{opp}} \otimes A$ then $A$-bimodules would be the same as right ${}^{\mathrm{e}} \! A$-modules. More precisely, if $M$ is an $A$-bimodule, then the corresponding right ${}^{\mathrm{e}} \! A$-module structure on $M$ is given by $$ m \cdot (a^{\mathrm{opp}} \otimes b) = a \cdot m \cdot b $$ for all $a, b \in A$, $m \in M$
This can also be explained in more general terms: It holds for every $k$-algebra $B$ that right $B$-modules are the same as left $B^{\mathrm{opp}}$-modules. If $M$ is a right $B$-module then the corresponding left $B^{\mathrm{opp}}$-module structure on $M$ is given by $$ b^{\mathrm{opp}} \cdot m = m \cdot b $$ for all $b \in B$ and $m \in M$. We have in our case $$ ( A^{\mathrm{e}} )^{\mathrm{opp}} = ( A \otimes A^{\mathrm{opp}} )^{\mathrm{opp}} = A^{\mathrm{opp}} \otimes (A^{\mathrm{opp}})^{\mathrm{opp}} = A^{\mathrm{opp}} \otimes A = {}^{\mathrm{e}} \! A \,. $$ We therefore find again that \begin{align*} \text{$A$-bimodules} = \text{left $A^{\mathrm{e}}$-modules} = \text{right $( A^{\mathrm{e}} )^{\mathrm{opp}}$-modules} = \text{right ${}^{\mathrm{e}} \! A$-modules}. \end{align*}
However, it needs to be pointed out that the enveloping algebra $A^{\mathrm{e}}$ has the interesting property that it is isomorphic to its own opposite algebra, since $$ ( A^{\mathrm{e}} )^{\mathrm{opp}} = {}^{\mathrm{e}} \! A = A^{\mathrm{opp}} \otimes A \cong A \otimes A^{\mathrm{opp}} = A^{\mathrm{e}} \,. $$ We can therefore interpret every $A$-bimodule not only as a left $A^{\mathrm{e}}$-module and a right ${}^{\mathrm{e}} \! A$-module, but also as a left ${}^{\mathrm{e}} \! A$-module and a right $A^{\mathrm{e}}$-module. For an $A$-bimodule $M$ the corresponding left ${}^{\mathrm{e}} \! A$-module structure is given by $$ (a^{\mathrm{opp}} \otimes b) \cdot m = b \cdot m \cdot a $$ for all $a, b \in A$ and $m \in M$, and the corresponding right $A^{\mathrm{e}}$-module structure is given by $$ m \cdot (a \otimes b^{\mathrm{opp}}) = b \cdot m \cdot a $$ for all $a, b \in A$ and $m \in M$.