Csdrhrt

So I have a specific real $4times4$ 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." :)