Relationship between determinants. Given an orthogonal matrix $O$ with dimensions $4n \times 4n$ and $\det O = -1$, how to prove that
$\det[O_{11} - O_{22} + i (O_{12} + O_{21})] = 0$?
Here $O$ is a block matrix $[[O_{11}, O_{12}], [O_{21}, O_{22}]]$, and all blocks have equal size.
An equivalent statement is the following: if $\det O = -1$, then $\det(\Omega - O \Omega O^T)=0$, with $\Omega = [[0, I],[-I, 0]]$. The reason why I believe this is correct is due to the physical meaning of it (it's about presence of bound states in a certain system), and numerical checks. I am not sure why odd $n$ doesn't work, but it makes sense for physics reasons.
Example of a matrix for which it works is below:
$$\left(\begin{array}{cccc}
0.44090815& -0.71206204& -0.44576549&  0.31600755 \\\
0.12767731&  0.35584235&  0.19884606&  0.90417641 \\\
0.88038152&  0.229104  &  0.30519084& -0.28159952 \\\
0.11927654&  0.56022784& -0.81768693& -0.05749749
\end{array}\right)$$
Edit(Will Jagy):In particular, this fails for $(4n - 2) \times (4n - 2),$ as in the  $ 2 \times 2$
$$  O \; \; = \; \;   
\left(  \begin{array}{cc}
  \cos t &  \sin t\\\
  \sin t &  - \cos t  
\end{array} 
  \right)  , $$
with the relevant  $ 1 \times 1$ matrix of complex numbers
$$ \left( \cos t - ( - \cos t) + i ( \sin t +  \sin t )\right)  = \left( 2 e^{it} \right) . $$
Edit(Anton Akhmerov):
Yet another statement which could solve the problem is the following:
Prove that any $4n\times 4n$ orthogonal matrix $O$ can be brought to a block-diagonal form by a transformation $O\rightarrow S_1 O S_2$ with $S_1$ and $S_2$ symplectic matrices. If this was correct, the rest follows immediately.
 A: What you are seeing is that the orthogonal matrices of determinant $-1$ swap the two spin representations. The first several parts of this argument will be valid for $(4n+2) \times (4n+2)$ matrices as well, so I'll rename the size of the matrix to $2m$ and specialize to $m$ even when it becomes relevant.
In more detail: Let $S$ be the $2m \times 2m$ change of basis matrix
$$\frac{1}{\sqrt{2}} \begin{pmatrix} \mathrm{Id} & i \cdot \mathrm{Id} \\ \mathrm{Id} & -i \cdot \mathrm{Id} \end{pmatrix}$$
Note that $O$ is orthogonal if and only if $SOS^{-1}$ preserves the split bilinear form $x_1 x_{m+1} + x_2 x_{m+2} + \cdots + x_{m} x_{2m}$. Explicitly,
$$S O S^{-1} = \frac{1}{2} \begin{pmatrix} O_{11} - i O_{12} + i O_{21} + O_{22} & O_{11} + i O_{12} + i O_{21} - O_{22} \\ O_{11} - i O_{12} - i O_{21} - O_{22} & O_{11} + i O_{12} - i O_{21} + O_{22} \end{pmatrix}$$
We want to show that, for $m$ even: If $U$ preserves the split bilinear form, and has determinant $-1$, then the upper right $m \times m$ block of $U$ is singular.
Consider the action of $U$ on $\bigwedge^{m} \mathbb{C}^{2m}$; this is a big $\binom{2m}{m} \times \binom{2m}{m}$ matrix which we will write $\bigwedge^m U$. The entries of $\bigwedge^m U$ are $m \times m$ minors of $U$. In particular, the entry we care about is the coefficient with which $e_1 \wedge e_2 \wedge \cdots \wedge e_m$ is taken to $e_{m+1} \wedge e_{m+2} \wedge \cdots e_{2m}$. 
Let $\alpha_{+}$ and $\alpha_{-}$ be the highest weights of the spin reps of $\mathrm{Spin}(2m)$. Let $V_{\pm}$ be the irrep with highest weight $2 \alpha_{\pm}$. Then $V_{+}$ and $V_{-}$ both inject into $\bigwedge^{m} \mathbb{C}^{2m}$. The vector $e_1 \wedge \cdots \wedge e_m$ is a high weight vector for $V_{+}$. The vector $e_{m+1} \wedge \cdots e_{2m}$ is a low weight vector for $V_{(-1)^m}$. In particular, when $m$ is even, Both these wedge products lie in $V_{+}$. An orthogonal matrix with determinant $-1$ switches the spin representations. So this entry of $\bigwedge^m U$ is $0$. 

If you'd like a "geometric" presentation of the same argument, note that 
$$O_{11} + i O_{12} + i O_{21} - O_{22} = \begin{pmatrix} \mathrm{Id} & i \cdot \mathrm{Id} \end{pmatrix} O \begin{pmatrix} \mathrm{Id} \\ i \cdot \mathrm{Id} \end{pmatrix}.$$  
The kernel of the left hand matrix is an isotropic $m$-plane; call it $L$. The image of the right hand matrix is also $L$. So we want to show that, if $O$ has determinant $-1$, then $L \cap O L$ is nonzero. 
Let $Q$ be the quadric hypersurface $\sum x_i^2=0$ in $\mathbb{CP}^{2m-1}$. This is a complex variety of (complex) dimension $2m-2$. The projectivization $\mathbb{P} L$ is a subvariety of dimension $m-1$. It is well known that $H^{2m-2}(Q) \cong \mathbb{Z}^2$, with orthogonal matrices of determinant $1$ switching the two generators. It is less well known that the intersection form on $H^{2m-2}(Q)$ has matrix $\left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$ when $m$ is even and ` $\left( \begin{smallmatrix} 1 & 0  \\ 0 & 1 \end{smallmatrix} \right)$ when $m$ is odd. 
We are in the former case. So the cup product $[\mathbb{P}L] \cup [O \cdot \mathbb{P}L]$ is $1$ and we deduce that $L$ and $OL$ have nonzero intersection, as desired.
A: I denote your matrix $\Omega$ by $W$ for the sake of brevity. Note that $W$ is both skew-symmetric and orthogonal, i. e. it satisfies $W=-W^T$ and $W^2=-I$. (And this is all I am going to use about $W$.)
The only thing I am going to use about the matrix $O$ is that $O^TO=I$. The assumption that $O$ is a real matrix will not be needed (it could be from any field of characteristic $\neq 2$).
We are going to use notion of the Pfaffian of a skew-symmetric matrix.
Lemma 1. Let $R$ be a commutative ring with $1$. Let $A\in R^{2n\times 2n}$ be a skew-symmetric matrix, and $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, the matrix $B^TAB$ is skew-symmetric as well, and $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$.
Proof of Lemma 1. Clearly, the matrix $B^TAB$ is skew-symmetric. It remains to prove that $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Since the square of the Pfaffian of a skew-symmetric matrix is the determinant of this matrix, we have
$\left(\mathrm{Pf}\left(B^TAB\right)\right)^2=\det\left(B^TAB\right)=\underbrace{\det B^T}_{=\det B}\cdot\underbrace{\det A}_{=\left(\mathrm{Pf}A\right)^2}\cdot \det B = \left(\det B\cdot\mathrm{Pf} A\right)^2$.`
Thus, FOR EVERY $A$ and $B$, we have either $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$ OR $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$. By a Zariski-topological argument, we can interchange the words "for every" with the words "or" here, so we obtain: (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$) OR (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$). Since (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$) is wrong (take $A=W$ and $B=\mathrm{id}$), we thus must have (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$), and Lemma 1 is proven.
Lemma 2. Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ and $C\in R^{2n\times 2n}$ be arbitrary matrices. Then, $\det\left(W-BWC\right)=\det\left(W-CWB\right)$.
Proof of Lemma 2. We have $W-BWC=W\cdot\left(I+WBWC\right)$ (because $-I=W^2$), and thus
$\det\left(W-BWC\right)=\det\left(W\cdot\left(I+WBWC\right)\right)=\det W\cdot\det\left(I+WBWC\right)$
and similarly
$\det\left(W-CWB\right)=\det W\cdot\det\left(I+WCWB\right)$.
Thus, in order to prove Lemma 2, it remains to show that $\det\left(I+WBWC\right)=\det\left(I+WCWB\right)$. This follows from the general fact that if $U$ and $V$ are two square matrices of the same size, then $\det\left(I+UV\right)=\det\left(I+VU\right)$ (for a proof of this fact, apply Corollary 2 in MathLinks post #1491761 to $X=-1$). Thus, Lemma 2 is proven.
Lemma 3. Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$.
Proof of Lemma 3. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Applying Lemma 2 to $C=B^T$, we obtain $\det\left(W-BWB^T\right)=\det\left(W-B^TWB\right)$. Since the determinant of a skew-symmetric matrix is the square of its Pfaffian, this rewrites as $\left(\mathrm{Pf}\left(W-BWB^T\right)\right)^2=\left(\mathrm{Pf}\left(W-B^TWB\right)\right)^2$. Again, a Zariski density argument like in the proof of Lemma 1 shows us that we must have $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$ (because we cannot have $\mathrm{Pf}\left(W-BWB^T\right)=-\mathrm{Pf}\left(W-B^TWB\right)$ for all $B$, since this would fail for $B=0$). This proves Lemma 3.
Now let us solve our problem:
We have $\mathrm{Pf}U=\mathrm{Pf}\left(  -U\right)  $ for every skew-symmetric $4n\times 4n$ matrix $U$ (because $\mathrm{Pf}$ is a homogeneous polynomial of degree $2n$). Now,
$\mathrm{Pf}\left(  W-OWO^{T}\right)  =\mathrm{Pf}\left(  W-O^{T}WO\right)  $ (after Lemma 3, applied to $B=O$)
$=\mathrm{Pf}\left(  -\left(  W-O^{T}WO\right)  \right)  $ (since $\mathrm{Pf}U=\mathrm{Pf}\left(  -U\right)  $ for every skew-symmetric $4n\times4n$ matrix $U$)
$=\mathrm{Pf}\left(  O^{T}WO-W\right) $
$ =\mathrm{Pf}\left(  O^{T} WO-O^{T}OWO^{T}O\right)  $ (since $O^{T}O=I$ yields $W=O^{T}OWO^{T}O$)
$=\mathrm{Pf}\left(  O^{T}\left(  W-OWO^{T}\right)  O\right) $
$=\underbrace{\det O}_{=-1}\cdot\mathrm{Pf}\left(  W-OWO^{T}\right)  $ (by Lemma 1, applied to $B=O$ and $A=W-OW^{T}O^{T}$)
$=-\mathrm{Pf}\left(  W-OWO^{T}\right)  $,
so that $2\mathrm{Pf}\left(  W-OWO^{T}\right)  =0$ and thus $\mathrm{Pf}\left(  W-OWO^{T}\right)  = 0$, so that $\det\left(  W-OWO^{T}\right) =\left(\mathrm{Pf}\left(  W-OWO^{T}\right)  \right)^2=0^2=0$, qed.
A: Might be getting there. So far, I have
$$ \begin{array}{ll}
& \left(O_{11} - O_{22} + i O_{12} + i O_{21}  \right) \cdot \left( O_{11}^t + O_{22}^t - i O_{12}^t + i O_{21}^t  \right)  = \\\
&  (O_{11} O_{22}^t - O_{22} O_{11}^t   ) +  (O_{21} O_{12}^t - O_{12} O_{21}^t   ) + \\\
& i (O_{12} O_{11}^t - O_{11} O_{12}^t   ) + i (O_{21} O_{22}^t - O_{22} O_{21}^t   )
\end{array}$$
so that the resulting real and imaginary parts are skew symmetric. However, as they are $ 2n \times 2n$ this does not imply singular for either part.
Note: there are some generally unknown but useful properties of orthogonal matrices, for example, in this case:
$$ \det O_{11} =  - \det O_{22}, $$ see
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