11
$\begingroup$

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?

$\endgroup$
15
$\begingroup$

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 $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.