Let $\mathfrak{g}$ finite dimentional semisimple Lie algebra and $\sigma$ the usual chevalley anti-involution that fixes the Cartan subalgebra $\mathfrak{h}$ sends the weight space $\mathfrak{g}_\alpha$ in $\mathfrak{g}_{-\alpha}$ for $\alpha \in \mathfrak{h}^*$ (better, if $x_1 \dots x_n,h_1 \dots h_l,y_1\dots y_n$ is a chevalley basis, then there exists a isomophism $\sigma$ of $\mathfrak{g}$, as a vector space, that sends $x_i$ in $y_i$ for all $i$ and fixes the $h_i$'s). I would want to know if $\sigma$ lifts to an antihomorphism of the universal enveloping algebra, i.e. if there exists and application $\tilde \sigma \colon \mathcal{U}(\mathfrak{g}) \to \mathcal{U}(\mathfrak{g})$ such that $\tilde \sigma (xy)=\tilde \sigma(y) \tilde \sigma(x)$. Seems quite natural to me, but unfortunately I loose myself in technical details.

Motivation: I have to prove that the Shapovalov form is symmetric, I am following "Lie algebras with triangular decompositions" of Moody-Pianzola and seems that they used the fact above for proving it.


I solved my problem simply defining the universal property for anti-homomorphims. Let $A$ algebra, $\mathfrak{g}$ Lie algebra and $\sigma \colon \mathfrak{g} \to A$ anti-homomorphism of Lie algebras. Then this application

\begin{gather} \sigma_n \colon \mathfrak{g} \times \ldots \times \mathfrak{g} \to A \\ (x_1 \dots x_n) \mapsto \sigma(x_n) \ldots \sigma(x_1) \end{gather} is multilinear, so we can define \begin{gather} \sigma_n \colon \mathfrak{g}^{\otimes n} \to A \\ x_1\otimes \dots \otimes x_n \mapsto \sigma(x_n) \ldots \sigma(x_1) \end{gather} Now glue together all the $\sigma_n$'s on the tensor algebra to obtain a map \begin{gather} \tilde\sigma \colon T:= \bigoplus_{n \in \mathbb{N}}\mathfrak{g}^{\otimes n} \to A \end{gather} Note that by contruction $\tilde \sigma(x\otimes y)=\tilde \sigma(y)\tilde \sigma(x)$ (indeed this can be directly checked on $x=x_1 \otimes \dots \otimes x_n$ and $y=y_1 \otimes \dots \otimes y_m$). Here comes the surprising part:

\begin{multline} \tilde \sigma([x,y])=\sigma([x,y])=[\sigma (y),\sigma (x)]=\\=\sigma (y) \sigma (x)- \sigma (x)\sigma (y)=\tilde \sigma (y)\tilde \sigma (x)-\tilde \sigma (x)\tilde \sigma (y)=\\ =\tilde \sigma(x\otimes y) -\tilde \sigma(y\otimes x)=\tilde \sigma(x\otimes y-y\otimes x). \end{multline} All the elements of the form $[x,y]-(x\otimes y-y\otimes x)$ are in the kernel and then we can define

\begin{gather} \tau \colon \mathcal{U}(\mathfrak{g}) \to A \end{gather} Note that for $x,y \in \mathcal{U}(\mathfrak{g})$ we have $\tau(xy)=\tau(y)\tau(x)$. This is the universal property of the universal enveloping algebra with respect to the anti-homomorphisms.

Now consider $\sigma$ the Chevalley anti-homomorphism and compose it with the inclusion $i\colon\mathfrak{g} \to \mathcal{U}(\mathfrak{g})$, $i\tilde \sigma\colon\mathfrak{g} \to \mathcal{U}(\mathfrak{g})$, note that this last map is an anti-homomorphism. By the universal property described above $i\tilde \sigma $ lifts to a map $\tau\colon \mathcal{U}(\mathfrak{g}) \to\mathcal{U}(\mathfrak{g})$ such that by contruction $\tau(xy)=\tau(y)\tau(x)$.


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