I posted the following question on MSE but I was advised to ask the question here.

Let $\mathfrak{g}$ be a lie algebra. Consider the anti-automomorphism

$$\phi: \mathfrak{g} \rightarrow \mathfrak{g} $$ such that $$\phi(X) = X^T, \quad \forall \quad X \in \mathfrak{g}$$ where $X^T$ is the usual transpose of $X$.

Observe that $$\phi([X,Y]) = \phi(XY-YX) = (XY-YX)^T = (XY)^T-(YX)^T.$$ Thus, we have that $$\phi([X,Y)] = Y^TX^T- X^TY^T = [\phi(Y), \phi(X)].$$

Note that we can extend this anti-automorphism to the universal enveloping algebra $\mathfrak{Ug}$ such that $\phi : \mathfrak{Ug} \rightarrow \mathfrak{Ug}$ still denoted by $\phi.$

My question: Show that $$\phi(a)= a \quad$$ for all $a \in Z\mathcal{g} $ where $Z\mathcal{g}$ denotes the centre of $\mathfrak{Ug}.$

I thought about using the Harish-Chandra isomorphim but I dont seem to see how this helps.

  • 5
    $\begingroup$ What's the "usual transpose" on an arbitrary Lie algebra? $\endgroup$ Apr 27, 2017 at 5:41
  • $\begingroup$ A lie algebra anti-homomorphism is a linear map such that $$\phi([X,Y]) = [\phi(Y), \phi(X)].$$In this specific case it just follows from the fact that $(AB)^T = (BA)^T$ for two matrices $A,B$ $\endgroup$
    – Jaynot
    Apr 27, 2017 at 5:48
  • 1
    $\begingroup$ What is the field? is there any assumption that your Lie algebra is semisimple? And indeed as Francois says, "transpose" is not defined. $\endgroup$
    – YCor
    Apr 27, 2017 at 6:14

1 Answer 1


$\newcommand{\g}{\mathfrak g}$ François is right that it doesn't quite make sense to talk about transpose for a general Lie algebra. Since you mention the Harish-Chandra isomorphism I assume you want $\g$ to be a simple Lie algebra. In that case there is an anti-involution $\omega$ given by $e_i\mapsto f_i$, $f_i \mapsto e_i$, $h_i \mapsto h_i$ for a choice of Chevalley generators $e_i,f_i,h_i$. If $\g$ is $\mathfrak{gl}_n$ this recovers the transpose, so I'll assume your question is about that.

I feel there should be a much simpler argument but here's an idea involving the Harish-Chandra map as you suggest. Let $\g=\mathfrak{n}^+ \oplus \mathfrak{h}\oplus \mathfrak{n}^-$ be the decomposition induced by your choice of generators. By PBW this induces a decomposition $$U(\g)= U(\mathfrak{h})\oplus (\mathfrak{n}^-U(\g)+ U(\g)\mathfrak{n}^+).$$ Let $p$ be the projection on $U(\mathfrak{h})$ relative to this decomposition.

I claim that $p\circ \omega=p$. Now, the restriction of $p$ to the center of $U(\g)$ is actually injective: this is essentially a reformulation of Harish-Chandra. The result you want follows.


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