One of my lecturers mentioned in passing that the universal enveloping algebra of a Lie algebra is isomorphic to its opposite ring, so I wanted to prove this fact.
To this end, let $\mathfrak{g}$ be a (finite-dimensional, for convenience) Lie algebra over $\mathbb{k}$ with basis $\{x_1, \dots, x_n\}$, and let $U(\mathfrak{g}) = \mathbb{k}\langle x_1, \dots, x_n \rangle / \langle x_ix_j - x_j x_i - [x_i,x_j] \mid 1 \le i < j \le n\rangle$ be the universal enveloping algebra of $\mathfrak{g}$. To show that $U(\mathfrak{g}) \cong U(\mathfrak{g})^\text{op}$ it suffices to exhibit an anti-isomorphism of $U(\mathfrak{g})$. Since $\mathbb{k}\langle x_1, \dots, x_n \rangle$ is a free algebra, we may define an anti-homomorphism by saying what it does to the generators $x_i$, so define \begin{align*} \phi : \mathbb{k}\langle x_1, \dots, x_n \rangle \rightarrow U(\mathfrak{g}), \hspace{5pt} x_i \mapsto -\overline{x_i}. \end{align*} At this point, one could attempt to show directly that $\text{ker}(\phi) = \langle x_ix_j - x_j x_i - [x_i,x_j] \mid 1 \le i < j \le n \rangle$ and then we're done by the first isomorphism theorem. This appears difficult, so instead I argued as follows: We have that \begin{align*} \phi(x_ix_j - x_jx_i - [x_i,x_j]) &= \overline{(-x_j)(-x_i) - (-x_i)(-x_j) - (-[x_i,x_j])}\\ &= -\overline{(x_ix_j - x_jx_i - [x_i,x_j])}\\ &= \overline{0} \end{align*} so $\phi$ descends to a well-defined (surjective) anti-homomorphism \begin{align*} \psi : U(\mathfrak{g}) \rightarrow U(\mathfrak{g}), \hspace{5pt} \overline{x_i} \rightarrow -\overline{x_i}. \end{align*} It remains to show injectivity. Noting that $U(\mathfrak{g})$ has PBW bases $B_1 = \{x_{i_1}^{k_1} \dots x_{i_r}^{k_r} \mid 1 \le i_1 < \dots < i_r \le n\}$ and $B_2 = \{x_{i_1}^{k_1} \dots x_{i_r}^{k_r} \mid 1 \ge i_1 > \dots > i_r \ge n\}$ (where we use $B_1$ in the domain and $B_2$ in the codomain), suppose that $\psi(y) = 0$. We can write $y$ as a $\mathbb{k}$-linear combination of elements of the form $\lambda x_{i_1}^{k_1} \dots x_{i_r}^{k_r}$, $1 \le i_1 < \dots < i_r \le n$, and then $\psi(y)$ is simply a $\mathbb{k}$-linear combination of elements of the form $(-1)^{\sum k_j} \lambda x_{i_r}^{k_r} \dots x_{i_1}^{k_1}$. Since $B_2$ is a PBW basis and $\psi(y) = 0$, the coefficients $\lambda$ must all be zero, and so $y=0$.
My questions are as follows:
- Is this proof ok?
- Can we show directly that the kernel of $\phi$ is $\langle x_ix_j - x_j x_i - [x_i,x_j] \mid i,j \in I \rangle$?
- Is there a way to finish my proof without invoking PBW bases?
The "principal" involution on $\mathfrak{g}$ given by $x\mapsto -x$ naturally extends to an involution of $U(\mathfrak{g})$, i.e., we have an anti-automorphism $U(\mathfrak{g})\mapsto U(\mathfrak{g})^{op}$. And an involution $\psi$ of a ring is a bijection because of $\psi^2=id$.