Maximizing a skew-symmetric 4D cross product

Question

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.

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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.

Answer 1

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 $y$.