\begin{example}
Consider the 
the pure quadratic, orthogonally constrained problem
\beq
\label{eq:pq}
\begin{array}{ccc}
\mu^*:=
&\min & \tr AXBX^T \\
 &  \mbox{~s.t.~} &XX^T = I,
    \end{array}
\end{equation}
with  $2\times 2$ matrices
$$
A= \left( 
\begin{array}{cc}
1 & 0 \\
0 & 2 
\end{array} 
\right), ~~~~~~~~
B= \left( 
\begin{array}{cc}
3 & 0 \\
0 & 4 
\end{array} 
\right).
$$
The dual problem is
\beq
\label{eq:d2}
\begin{array}{ccc}
\mu^D := &\max & -\tr {S} \\
&  \mbox{~s.t.~}  &  (B \otimes A + I \otimes {S}) \succeq 0\\
  && S=S^T.
\end{array}
\end{equation}
Then $\mu^* = 10.$ But is the dual optimal value
 $\mu^D$ also 10? We have
$$
B \otimes A = \left( 
\begin{array}{cccc}
3 & 0 & 0 & 0 \\
0 & 6 & 0 & 0 \\
0 & 0 & 4 & 0 \\
0 & 0 & 0 & 8 \\
\end{array} 
\right).
$$
Then in order to satisfy dual feasibility, we must
have $S_{11}\geq -3$ and $S_{22}\geq -6$.
In order to maximize the dual, equality must hold. 
Therefore $-\tr {S} = 
9$ in the optimum. Thus we have a duality gap for this simple example.
\end{example}
