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\hypersetup{pdfauthor={ChatGPT 6 Pro},pdftitle={A fourteen-dimensional biperfect Lie bialgebra}}

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\title{A fourteen-dimensional biperfect Lie bialgebra}
\author{ChatGPT 6 Pro}
\date{October 8, 2026}

\begin{document}
\maketitle

\begin{abstract}
We construct a fourteen-dimensional Lie bialgebra for which both the
underlying Lie algebra and its dual are perfect. The construction is
defined over $\Q$ and is self-dual over $\C$. It gives an affirmative
answer to Question~4.1(3) of Etingof--Gelaki--Guralnick--Saxl.
Its twenty-eight-dimensional Drinfeld double is quasitriangular with
injective cobracket, also answering Question~4.1(4).
\end{abstract}

\section*{AI disclaimer}
\noindent\textbf{This text has been produced by ChatGPT at a request by Pavel Etingof.
It was checked by an AI agent and revised to its satisfaction. Pavel's prompt is included at the end. 
Feel free to use at your discretion.  If you rewrite it or build on it, you may say that it was inspired by this note.}

\section{Statement and construction}

A Lie algebra $\g$ is \emph{perfect} if $[\g,\g]=\g$.
A finite-dimensional Lie bialgebra is \emph{biperfect} if both $\g$ and
its dual Lie algebra $\g^*$ are perfect. The existence of a nonzero
example over $\C$ was asked in \cite[Question~4.1(3)]{EGGS}.

\begin{theorem}\label{thm:main}
There exists a fourteen-dimensional Lie bialgebra over $\Q$ such that
both $\g$ and $\g^*$ are perfect. After extension of scalars to $\C$,
it is isomorphic to its dual as a Lie bialgebra. Both underlying Lie
algebras have Levi factor $\slie_2$ and an eleven-dimensional
two-step nilpotent radical with three-dimensional center.
\end{theorem}

We construct two complementary perfect Lagrangian subalgebras of a
quadratic Lie algebra. The Manin-triple correspondence then supplies
the required Lie bialgebra \cite[\S3, pp.~802--803]{Drinfeld}.
All vector spaces below are over $\Q$ until the discussion of
self-duality.

\subsection{The ambient quadratic Lie algebra}

Let $A$ and $B$ be two-dimensional vector spaces with bases
$a_0,a_1$ and $b_0,b_1$, respectively. Equip them with alternating
forms $\varepsilon_A,\varepsilon_B$ normalized by
\[
\varepsilon_A(a_0,a_1)=\varepsilon_B(b_0,b_1)=1.
\]
Let $M=\Q^4$ with the symmetric form $(u,v)=u^tv$, and put
\[
\s=\slie(A),\qquad \tlie=\slie(B),\qquad
\kk=\s\oplus\tlie,\qquad V=A\otimes B\otimes M.
\]
The natural action of $\kk$ on $V$ preserves the nondegenerate
\emph{symmetric} form
\begin{equation}\label{eq:formV}
q(a\otimes b\otimes u,a'\otimes b'\otimes v)
=\varepsilon_A(a,a')\varepsilon_B(b,b')u^tv.
\end{equation}
Define $\mu:\wedge^2V\longrightarrow\kk^*$ by
\begin{equation}\label{eq:moment}
\mu(v,w)(x)=q(xv,w),\qquad x\in\kk.
\end{equation}
This is alternating because the action preserves $q$.

On $\dlie=\kk\oplus V\oplus\kk^*$ define
\begin{equation}\label{eq:doublebracket}
[(x,v,\alpha),(y,w,\beta)]
=\bigl([x,y],\,xw-yv,\,
\ad_x^*\beta-\ad_y^*\alpha+\mu(v,w)\bigr),
\end{equation}
where $(\ad_x^*\alpha)(y)=-\alpha([x,y])$.
Thus $V\oplus\kk^*$ is two-step nilpotent, with
$[V,V]\subseteq\kk^*$ and $[V,\kk^*]=[\kk^*,\kk^*]=0$.
The form
\begin{equation}\label{eq:formD}
\langle(x,v,\alpha),(y,w,\beta)\rangle
=q(v,w)+\alpha(y)+\beta(x)
\end{equation}
is nondegenerate, symmetric, and invariant.

These formulas are the double extension of the abelian quadratic
Lie algebra $V$ by $\kk$ \cite[\S2.2]{MR}. For completeness, the Jacobi
identity follows from the representation property of the actions of
$\kk$ and the equivariance
\[
\ad_x^*\mu(v,w)=\mu(xv,w)+\mu(v,xw).
\]
Indeed, evaluation on $y\in\kk$ gives on the right
$q(yxv,w)+q(yv,xw)=q([y,x]v,w)$, which is the left-hand side.
Jacobi identities involving three elements of $V\oplus\kk^*$ are
immediate. Invariance of \eqref{eq:formD} follows from invariance of
$q$, the definition of the coadjoint action, and
$\langle[v,w],x\rangle=q(xv,w)$.

\subsection{Two complementary Lagrangian subspaces}

Choose
\begin{equation}\label{eq:ST}
S=\begin{pmatrix}
1&0&0&0\\0&-1&0&0\\0&0&2&0\\0&0&0&-2
\end{pmatrix},\qquad
T=\begin{pmatrix}
0&1&0&0\\1&0&0&0\\0&0&0&2\\0&0&2&0
\end{pmatrix}.
\end{equation}
Both matrices are symmetric and invertible, and
\begin{equation}\label{eq:matrixproperties}
ST=-TS,\qquad
\operatorname{Spec}(S)=\operatorname{Spec}(T)=\{1,-1,2,-2\}.
\end{equation}
In particular, each of the triples $I,S,S^2$ and $I,T,T^2$ is
linearly independent.

Define linear embeddings
\begin{align}
\iota_T:A\otimes M&\longrightarrow V,&
\iota_T(a,u)&=a\otimes b_0\otimes u+a\otimes b_1\otimes Tu,
\label{eq:iota}\\
\jmath_S:B\otimes M&\longrightarrow V,&
\jmath_S(b,u)&=a_0\otimes b\otimes u+a_1\otimes b\otimes Su,
\label{eq:jmath}
\end{align}
where the notation on pure tensors is extended linearly. Set
$U=\operatorname{im}\iota_T$ and $W=\operatorname{im}\jmath_S$.
Then $U$ is $\s$-stable and $W$ is $\tlie$-stable.

Symmetry of $T$ gives
\[
q(\iota_T(a,u),\iota_T(a',v))
=\varepsilon_A(a,a')\bigl(u^tTv-(Tu)^tv\bigr)=0.
\]
Likewise $q(W,W)=0$. Since $\dim U=\dim W=8$ and $\dim V=16$,
both are Lagrangian.

To see that they are complementary, write $z\in U$ uniquely as
\[
z=a_0\otimes b_0\otimes u+a_0\otimes b_1\otimes Tu
  +a_1\otimes b_0\otimes v+a_1\otimes b_1\otimes Tv.
\]
Membership in $W$ requires $v=Su$ and $Tv=STu$, hence
$(TS-ST)u=0$. But $TS-ST=2TS$ is invertible by
\eqref{eq:matrixproperties}. Thus $u=v=0$, proving
\begin{equation}\label{eq:VW}
V=U\oplus W.
\end{equation}

\subsection{The Manin triple}

Identify $\kk^*=\s^*\oplus\tlie^*$ in the natural way, and set
\begin{equation}\label{eq:gh}
\g=\s\oplus U\oplus\tlie^*,\qquad
\h=\tlie\oplus W\oplus\s^*.
\end{equation}
These are Lie subalgebras of $\dlie$. For example, if $u,u'\in U$
and $x\in\s$, then
\[
\mu(u,u')(x)=q(xu,u')=0,
\]
because $U$ is $\s$-stable and isotropic. Consequently
$[U,U]\subseteq\tlie^*$. Also $[\s,U]\subseteq U$ and
$\tlie^*$ is central in $\g$. The argument for $\h$ is identical.

The form \eqref{eq:formD} vanishes on each of $\g$ and $\h$:
the only pairings between $\kk$ and $\kk^*$ are
$\s$ with $\s^*$ and $\tlie$ with $\tlie^*$.
Equation~\eqref{eq:VW} gives
\[
\dlie=\g\oplus\h,
\qquad \dim\g=\dim\h=3+8+3=14.
\]
Thus $(\dlie,\g,\h)$ is a Manin triple.

\section{Perfectness and the Lie bialgebra structure}

Since $U\cong A\otimes M$ as an $\s$-module, we have
\begin{equation}\label{eq:semisimplebrackets}
[\s,\s]=\s,\qquad [\s,U]=U.
\end{equation}
It remains to prove $[U,U]=\tlie^*$. Let
\[
x=\begin{pmatrix}\alpha&\beta\\\gamma&-\alpha\end{pmatrix}
\in\tlie.
\]
Directly from \eqref{eq:formV} and \eqref{eq:iota},
\begin{equation}\label{eq:keycalculation}
\mu(\iota_T(a,u),\iota_T(a',v))(x)
=\varepsilon_A(a,a')\,
u^t\bigl(2\alpha T+\beta T^2-\gamma I\bigr)v.
\end{equation}
If $x$ annihilates $[U,U]$, taking $a=a_0$, $a'=a_1$ and
arbitrary $u,v\in M$ gives
\[
2\alpha T+\beta T^2-\gamma I=0.
\]
The independence of $I,T,T^2$ implies $x=0$. Hence the annihilator
of $[U,U]\subseteq\tlie^*$ is zero, so
\begin{equation}\label{eq:centralbrackets}
[U,U]=\tlie^*.
\end{equation}
Together with \eqref{eq:semisimplebrackets}, this proves
$[\g,\g]=\g$.

Interchanging $A,B$ and replacing $T$ by $S$ gives
\[
[\tlie,\tlie]=\tlie,\qquad
[\tlie,W]=W,\qquad [W,W]=\s^*.
\]
Thus $[\h,\h]=\h$ as well.

To specify the cobracket explicitly, identify $\h$ with $\g^*$
by $\xi(x)=\langle x,\xi\rangle$, and define
\begin{equation}\label{eq:cobracket}
\langle\delta(x),\xi\otimes\eta\rangle
=\langle x,[\xi,\eta]\rangle,
\qquad x\in\g,\quad \xi,\eta\in\h.
\end{equation}
Here $\wedge^2\g$ is identified with the skew tensors in
$\g\otimes\g$. The Manin-triple correspondence guarantees that
$\delta$ satisfies the co-Jacobi identity and the Lie bialgebra
cocycle identity. Its dual Lie algebra is precisely $\h$, which
we have proved perfect. Moreover,
\[
\ker\delta=[\h,\h]^\perp\cap\g=0.
\]
Every formula and every subspace used above is defined over $\Q$;
in particular the nondegenerate pairing between $\g$ and $\h$
defines \eqref{eq:cobracket} over $\Q$.

Put $\mathfrak n=U\oplus\tlie^*$. Then
\[
\operatorname{rad}(\g)=\mathfrak n,\qquad
\g/\mathfrak n\cong\s,\qquad
Z(\mathfrak n)=Z(\g)=\tlie^*.
\]
Indeed, $\mathfrak n$ is a two-step nilpotent ideal with semisimple
quotient, so it is the radical of $\g$. The component of
\eqref{eq:keycalculation} corresponding to $\gamma$ is the
nondegenerate alternating form $-\varepsilon_A\otimes(u^tv)$ on
$A\otimes M$, so no nonzero element of $U$ is central in
$\mathfrak n$. Thus $Z(\mathfrak n)=\tlie^*$. Since $\s$ has zero
center, every central element of $\g$ lies in $\mathfrak n$;
as $\tlie^*$ is central in $\g$, this also gives
$Z(\g)=\tlie^*$. The same statements hold for $\h$ with $\s$
and $\tlie$ interchanged.

\section{Self-duality and the quasitriangular double}

Over $\C$, put
\[
P=\frac1{\sqrt2}
\begin{pmatrix}
1&1&0&0\\1&-1&0&0\\0&0&1&1\\0&0&1&-1
\end{pmatrix}.
\]
Then $P^tP=P^2=I$ and
\[
PSP=T,\qquad PTP=S.
\]
Identify $A$ and $B$ through their chosen bases. Swapping their
tensor factors and applying $P$ to $M$ gives an isometry
\[
\Phi_V(a\otimes b\otimes u)=b\otimes a\otimes Pu
\]
of $V$. Together with the swaps $\s\leftrightarrow\tlie$ and
$\s^*\leftrightarrow\tlie^*$, it extends to an involutive
isometric Lie algebra automorphism $\Phi$ of $\dlie$.
The identities $PT=SP$ and $PS=TP$ show that
$\Phi(U)=W$ and $\Phi(W)=U$. Hence $\Phi(\g)=\h$.

Reversing the two Lagrangian subalgebras gives the dual Lie bialgebra:
$\h$ carries $\delta_{\h}$, the transpose of the bracket on $\g$.
For $x,y,z\in\g$, we have
\begin{align*}
\langle\delta_{\h}(\Phi x),y\otimes z\rangle
 &=\langle\Phi x,[y,z]\rangle\\
 &=\langle x,[\Phi y,\Phi z]\rangle\\
 &=\langle\delta(x),\Phi y\otimes\Phi z\rangle\\
 &=\langle(\Phi\otimes\Phi)\delta(x),y\otimes z\rangle.
\end{align*}
Here we used $\Phi^2=1$ and preservation of the bracket and pairing.
Thus $\delta_{\h}\Phi=(\Phi\otimes\Phi)\delta$, so
$\Phi|_{\g}:\g\to\h\cong\g^*$ is a Lie bialgebra isomorphism.
This proves the self-duality asserted in Theorem~\ref{thm:main}
and completes its proof. The canonical double cobracket restricts
to $-\delta_{\h}$ on $\h$, as in Corollary~\ref{cor:double} below.

\begin{remark}
The fourteen-dimensional example is not coboundary. Its central
subspace $\tlie^*$ has a nonzero, indeed injective, cobracket: the
restriction of \eqref{eq:cobracket} to $\tlie^*$ is the transpose
of the bracket on $\tlie$. A coboundary cobracket vanishes on the
center. The same observation applies to the dual example.
\end{remark}

\begin{corollary}\label{cor:double}
There exists a twenty-eight-dimensional quasitriangular Lie bialgebra
over $\Q$ with injective cobracket.
\end{corollary}

\begin{proof}
Equip $\dlie=D(\g)$ with its canonical quasitriangular structure.
If $e_i\in\g$ and $f_i\in\h$ are dual bases for the pairing, its
classical $r$-matrix is $r=\sum_i e_i\otimes f_i$.
The cobracket $\delta_{\dlie}(z)=[z\otimes1+1\otimes z,r]$
restricts to $\delta$ on $\g$ and to $-\delta_{\h}$ on $\h$,
where $\delta_{\h}$ is the transpose of the bracket on $\g$.
Both restrictions are injective because $\g$ and $\h$ are
perfect, and their images lie in the disjoint subspaces
$\wedge^2\g$ and $\wedge^2\h$ of $\wedge^2\dlie$.
Therefore $\delta_{\dlie}$ is injective.
\end{proof}

This is the consequence anticipated in \cite[Remark~4.2(3)]{EGGS}
and answers its Question~4.1(4).

\Needspace{28\baselineskip}
\section*{Acknowledgements}

This paper was written in response to the following prompt by Pavel Etingof:
\begin{quote}
Is there a nonzero finite-dimensional Lie bialgebra $\g$ over $\mathbb C$ for which both $\g$ and $\g^*$ are
perfect?
\end{quote}

\begin{thebibliography}{9}

\bibitem{Drinfeld}
V.~G.~Drinfeld,
\emph{Quantum groups},
Proceedings of the International Congress of Mathematicians,
Berkeley, 1986, Vol.~1, Amer. Math. Soc., Providence, RI, 1987,
pp.~798--820.
\href{https://ncatlab.org/nlab/files/Drinfeld-QuantumGroups.pdf}{Full text}.

\bibitem{EGGS}
P.~Etingof, S.~Gelaki, R.~Guralnick, and J.~Saxl,
\emph{Biperfect Hopf algebras},
arXiv:math/9912068 (1999).
\url{https://arxiv.org/abs/math/9912068}.

\bibitem{MR}
A.~Medina and P.~Revoy,
\emph{Alg\`ebres de Lie et produit scalaire invariant},
Ann. Sci. \'{E}cole Norm. Sup. (4) \textbf{18} (1985), no.~3,
553--561.
\href{https://doi.org/10.24033/asens.1496}{doi:10.24033/asens.1496}.

\end{thebibliography}
\end{document}