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?
  • 5
    $\begingroup$ For the opposite algebra $A^{opp}$ see wikipedia. $\endgroup$ Aug 6 '20 at 15:28
  • 1
    $\begingroup$ They're not related. $\endgroup$ Aug 6 '20 at 21:58
  • $\begingroup$ @DietrichBurde Thanks. I think I understand now: We (my lecture notes) are interested in the enveloping algebra $A^e$ as defined above because it gives a correspondence between the category of $A$-bimodules and left $A^e$-modules. That allows us to relate the notion of separable algebra to the notion of projective module (we say: Call a $k$-algebra $A$ separable if it is projective as an $A^e$-module). Now, we define $A^e:=A\otimes A^{opp}$ and not $A^e:=A\otimes A$ because otherwise the correspondence wouldn't exist. That is, given… $\endgroup$
    – M.C.
    Aug 7 '20 at 18:23
  • $\begingroup$ … an $A$-bimodule $B$ the "action" of $A^e$ on $B$ given by $(a_1 \otimes a_2).b:= (a_1.b).a_2$ wouldn't actually define an action/an $A^e$-module structure on $B$. That is, the compatibility between multiplication an action - i.e. $(\alpha \cdot \beta).m=\alpha . (\beta .m)$ - wouldn't be satisfied (easy to check). Analogously, defining $A^e:=A^{opp} \otimes A$ would give a correspondence between the category of $A$-bimodules and right $A^e$-modules. Is my understanding correct? $\endgroup$
    – M.C.
    Aug 7 '20 at 18:23

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 $m \in M$.


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