Maximizing a skew-symmetric 4D cross product How do I find two orthonormal 4D vectors, $(x_0,x_1,x_2,x_3)$ and $(y_0,y_1,y_2,y_3)$, which maximize this function:
$-19x_1y_0 - 33x_2y_0 + 11x_3y_0 + 19x_0y_1 - 21x_2y_1 - 33x_3y_1 + 33x_0y_2 + 21x_1y_2 - 30x_3y_2 - 11x_0y_3 + 33x_1y_3 + 30x_2y_3$
Notice that the coefficients are skew-symmetric in case that helps.  The solution space should be rotationally symmetric in some 2D plane within the 4D space, so I believe $x_0$ can be set to zero without loss of generality.  I really want to know where the highest global absolute value of the above function is.  I tried direct constraint substitution and Lagrange multipliers, but things just got ugly.

This is actually an unanswered question on Math SE which solves the example in an unanswered question on Physics SE.  I find it hard to believe that finding the maximum 4D angular momentum reference frame can be so difficult and suspect that I'm missing an easy way.  But, even a numerical method would be appreciated.
 A: Too long to comment. See if this helps.
Since the matrix (defining the quadratic) is skew-symmetric, we can first decompose it into $Q\Sigma Q^\top$, where $Q$ is orthogonal, and $\Sigma$ is a block diagonal matrix (see Wiki on Skew symmetric matrix). Doing a change of variables from $x$ to $x_q=Q x$, and from $y$ to $y_q = Q y$, we can rewrite the problem as:
$$
\max \lambda_1(x_q[0]y_q[1] - x_q[1]y_q[0]) + \lambda_2(x_q[2]y_q[3] - x_q[3]y_q[2])\\
\mbox{subject to}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
x_q ~~\mbox{and}~~ y_q  ~~\mbox{being orthonormal}.
$$
Suppose $\lambda_1>\lambda_2$. By AM-GM inequality,
$$
\lambda_1(x_q[0]y_q[1] - x_q[1]y_q[0]) + \lambda_2(x_q[2]y_q[3] - x_q[3]y_q[2]) \leq \frac{\lambda_1}{2}(x^2_q[0]+y^2_q[1] + x^2_q[1] + y^2_q[0]) + \frac{\lambda_2}{2}(x^2_q[2] + y^2_q[3] + x^2_q[3] + y^2_q[2]).
$$
It is obvious that the maximum is achieved when $x_q[0]=1$ and $y_q[1]=1$, and the optimal value is $\lambda_1$. A similar argument can be made if $\lambda_2\geq \lambda_1$. Finally, work backwards to get $x$ and $y$.
