Let $V$ be a Euclidean vector space and let $V^{\mathbb{C}} = V \oplus V$ be its complexification, with complex structure
$$J = \begin{pmatrix} 0 & -\mathrm{id}\\ \mathrm{id} & 0 \end{pmatrix}.$$
Of course, we can regard $V^{\mathbb{C}}$ also as a real vector space with a canonical orientation (for a basis $v\_1, \dots, v\_n$ of $V$, the basis $(v\_1, 0), (0, v\_1), \dots, (v\_n, 0), (0, v\_n)$ is positively oriented). Now let $A$ be an anti-symmetric endomorphism of $V$ and consider the anti-symmetric endomorphism 
$$\tilde{A} = \begin{pmatrix} A& -\mathrm{id}\\ \mathrm{id} & A\end{pmatrix}$$
of $V^{\mathbb{C}}$. Then $\tilde{A}$ has a well-defined Pfaffian. On the other hand, we have $\tilde{A} = A + i$. Is it true that
$$\mathrm{Pf}(\tilde A) = \det(A+i),$$
at least if $V$ is even-dimensional? What about the odd-dimensional case?



Pfaffian equals complex determinant?
The answer is YES in every dimension, up to a sign. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore
$${\rm Pf}(\tilde A)^2=\det(I+iA)\det(I-iA).$$
On the other hand
$$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^\*=\det(I+iA),$$
yields
$${\rm Pf}(\tilde A)^2=(\det(I+iA))^2.$$
Let us remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We deduce
$${\rm Pf}(\tilde A)=\epsilon\det(I+iA)$$
for some constant $\epsilon=\pm1$. Taking $A=0\_n$ gives $\epsilon=1$.


Remark that ${\rm Pf}(\tilde A)$, as a polynomial in the entries of $A$, is even. In particular, if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$. For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then
$${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$


