So I have a specific real $4\times4$ matrix $\mathbf{P}$ given by \begin{align} \mathbf{P}= \begin{pmatrix} p_{11} & -p_{21} & p_{13} &-p_{23}\\ p_{21} & p_{11} & p_{23} & p_{13}\\p_{31} & -p_{41}& p_{33} & -p_{43}\\p_{41} & p_{31} & p_{43} & p_{33}. \end{pmatrix}, \end{align} and I'm confident that if this matrix is in $SO(4)$ then it must be block-diagonal OR anti-block-diagonal i.e. if $\mathbf{P}\in SO(4)$ then either $p_{11}=p_{21}=p_{33}=p_{43}=0$ or $p_{13}=p_{23}=p_{31}=p_{41}=0$, but I can't seem to show this...
I've tried picking one column and generating a set of orthonormal vectors to fill the matrix, but this just gives examples where it works and I'd like to show it generally. Is this possible? Am I just missing something trivial here?
Skip to the end for the big reveal, or read through this for the "how I got there" version.
Let's rewrite that as \begin{align} \mathbf{P}= \begin{pmatrix} a & -b & p &-q\\ b & a & q & p\\ c & -d& r & -s\\ d & c & s & r. \end{pmatrix}, \end{align}
Orthogonality of the first and third and first and 4th columns tells us that \begin{align} ap + bq + cr + ds &= 0\\ -aq + bp - cs + dr &= 0\\ \end{align} Cross-multiply to get \begin{align} apq + bq^2 + crq + dsq &= 0 \\ -apq + bp^2 - cps + dpr &= 0\\ \end{align} and sum to get \begin{align} b(p^2 + q^2) + crq - cps + dsq + dpr &= 0 \\ \end{align} Doing the same for columns 2 against 3 and 4 gives \begin{align} a(p^2 + q^2) - drq + csq + dps + cpr &= 0 \\ \end{align} Let's factor those to get \begin{align} b(p^2 + q^2) + c(rq - ps) + d(sq + pr) &= 0 \\ a(p^2 + q^2) + c(sq + pr) - d(rq - ps) &= 0 \\ \end{align} Looking at the dot product between rows 2 and 3, we see that $$ rq - ps = ad - bc $$ and similarly, for rows 2 and 4, we get $$ qs + pr = - (ac + bd) $$ so \begin{align} b(p^2 + q^2) + c(ad-bc) - d(ac + bd) &= 0 \\ a(p^2 + q^2) - c(ac + bd) - d(ad-bc) &= 0 \\ \end{align} which simplifies to \begin{align} b(p^2 + q^2) - bc^2 - bd^2 &= 0 \\ a(p^2 + q^2) - ac^2 - ad^2 &= 0 \\ \end{align} which become \begin{align} b(p^2 + q^2 - c^2 - d^2) &= 0 \\ a(p^2 + q^2 - c^2 - d^2) &= 0 \\ \end{align}
We conclude that either (1) $p^2 + q^2 = c^2 + d^2$ or (2) $a = b = 0$.
In the second case, we have that the squared norm of the first row is $p^2 + q^2$, which must be $1$, and the squared norm of the first column is $c^2 + d^2$, which must also be $1$, hence $p^2 + q^2 = c^2 + d^2$.
In other words, in all cases, $p^2 + q^2 = c^2 + d^2$.
By looking at the squared norms of the first row and the 4th column, we find that $$ a^2 + b^2 = r^2 + s^2 $$ as well.
But nothing else obvious seems to jump out...
...and so I began to wonder if it was actually true, and came up with this:
\begin{align} \mathbf{P}= \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 0 &-s\\ \frac{1}{2} & \frac{1}{2} & s & 0\\ \frac{1}{2} & -\frac{1}{2}& 0 & s\\ \frac{1}{2} & \frac{1}{2} & -s & 0 \end{pmatrix}, \end{align} where $s = \frac{1}{\sqrt{2}}$.
That seems to be a counterexample to your conjecture. So I guess the answer to your question is "Yes, you are just missing something trivial." :)