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\begin{center}
\begin{large}
{\bf An invariant tensor in $S^3(\g)$ for $\g=\mathfrak{sl}(n)$, ~$n\ge 3$}
\end{large}
\bigskip

Ami Haviv
\end{center}
\bigskip

In this short note we describe a simple construction of a non-zero
$\g$-invariant tensor in $S^3(\g)$, where $\g=\mathfrak{sl}(n)$, $n\ge 3$. 
We learned of this construction from Eric Weinstein (with the help of Dror 
Bar-Natan). Since elements in $S^3(\g)$ are antipodally odd and the PBW 
isomorphism $S(\g)^\g\cong U(\g)^\g$ is antipode preserving, the Weinstein's
tensor provides a counterexample to the author's unmotivated guess that the 
elements of $U(\g)^\g$, for $\g$ semisimple, are antipodally even.

Let $\g$ be a classical simple complex Lie algebra in its standard 
representation as a Lie subalgebra of $\mathfrak{gl}(m)$, for some $m$. We 
endow the latter
Lie algebra with the invariant metric $(X,Y)=\tr(XY)$. Recall that this metric
remains non-degenerate when restricted to $\g$. We use the induced metric
on $\g$ to identify $\g\cong \g^*$.

We define the following tri-linear form on $\g$:
\[
	(X,Y,Z) \longrightarrow (XY+YX,Z), \qquad X, Y, Z \in \g \, .
\]
This form is symmetric (because $\tr(ABC)=\tr(CAB)$) and $\g$-invariant.
(For example, take $G$ to be the compact Lie group corresponding to the
compact real form of $\g$ and use $\tr(gAg^{-1})=\tr(A), g\in G$, to get 
$G$-invariance, which implies $\g$-invariance.)

So we have the promised invariant symmetric tensor {\em provided} our 
tri-linear form is not identically zero. This is the case for 
$\g=\mathfrak{sl}(n)$, $n\ge 3$:
Take $X=Y$ having their upper left $2\times 2$ block equal to
$\left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$
 and all other entries are zero. Take
$Z$ to be the diagonal matrix $\diag(1, 0, \ldots, 0, -1)$. We have
$(X,Y,Z) \rightarrow 2 \neq 0$, as desired.

We now demonstrate that the construction ``fails'' for the other
classical algebras --- the symplectic and orthogonal Lie algebras (hence
also for $\mathfrak{sl}(2)\cong \mathfrak{sp}(1)$).
We only need to know that each of  these algebras is defined as the set of 
matrices $X \in \mathfrak{gl}(m)$
satisfying 
\begin{equation}
\label{eq:Def}
	X^t A+AX=0 \, ,
\end{equation}
where $A$ is a certain invertible matrix. Simple computations yield the
following two results:
\begin{itemize}
\item
If $X, Y$ satisfy~\eqref{eq:Def}, then $Z=XY+YX$ satisfies
\begin{equation}
\label{eq:Def-}
	Z^t A-AZ=0 \, ,
\end{equation}

\item
If $X$ satisfies~\eqref{eq:Def} and $Z$ satisfies~\eqref{eq:Def-},
then $(ZX)^t=-A(XZ)A^{-1}$, hence
\[ \tr(XZ)=0 \, . \]   
\end{itemize}
Clearly, these results complete our demonstration.

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