# Defining the enveloping algebra

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?
• For the opposite algebra $A^{opp}$ see wikipedia. Aug 6 '20 at 15:28
• They're not related. Aug 6 '20 at 21:58
• @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…
– M.C.
Aug 7 '20 at 18:23
• … 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?
– 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 $$m \in M$$.