Skip to main content
added 7 characters in body
Source Link
Denis Serre
  • 52.3k
  • 10
  • 146
  • 300

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 $$\mathrm{Pf}(A) = \det(A+i),$$$$\mathrm{Pf}(\tilde A) = \det(A+i),$$ at least if $V$ is even-dimensional? What about the odd-dimensional case?

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 $$\mathrm{Pf}(A) = \det(A+i),$$ at least if $V$ is even-dimensional? What about the odd-dimensional case?

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?

Source Link
Matthias Ludewig
  • 9.9k
  • 1
  • 30
  • 71

Pfaffian equals complex determinant?

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 $$\mathrm{Pf}(A) = \det(A+i),$$ at least if $V$ is even-dimensional? What about the odd-dimensional case?