| Version 17 |
Version 16 |
| The Pfaffian is an analog of the determinant that is defined only for a $2n\times 2n$ antisymmetric matrix. It is a polynomial of polynomial ring $n$ in elements of the matrix, such that its square is equal to the determinant of the matrix. |
The Pfaffian is an analog of the determinant that is defined only for a $2n\times 2n$ antisymmetric matrix. It is a polynomial of polynomial ring $n$ in elements of the matrix, such that its square is equal to the determinant of the matrix. |
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| The Pfaffian is applied in the generalized Gauss-Bonnet theorem. |
The Pfaffian is applied in the generalized Gauss-Bonnet theorem. |
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| {\bf Examples} |
{\bf Examples} |
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| $\mbox{Pf}\begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}=a,$ |
$\mbox{Pf}\begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}=a,$ |
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| $\mbox{Pf}\begin{bmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0& f \\-c & -e & -f & 0 \end{bmatrix}=af-be+dc.$ |
$\mbox{Pf}\begin{bmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0& f \\-c & -e & -f & 0 \end{bmatrix}=af-be+dc.$ |
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| {\bf Standard definition} |
{\bf Standard definition} |
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| Let |
Let |
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| $$A=\begin{bmatrix} 0 & a_{1,2} & \ldots & a_{1,2n} \\ -a_{1,2} & 0 & \ldots & a_{2,2n} \\ \vdots & \vdots & \vdots & \vdots \\-a_{2n,1} & -a_{2n,2} & \ldots & 0 \end{bmatrix}.$$ |
$$A=\begin{bmatrix} 0 & a_{1,2} & \ldots & a_{1,2n} \\ -a_{1,2} & 0 & \ldots & a_{2,2n} \\ \vdots & \vdots & \vdots & \vdots \\-a_{2n,1} & -a_{2n,2} & \ldots & 0 \end{bmatrix}.$$ |
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| Let $\Pi^{}_{}$ be the set of all partition of $\{1,2, \ldots ,2n\}$ into pairs of elements $\alpha\in \Pi^{}_{}$, can be represented as |
Let $\Pi^{}_{}$ be the set of all partition of $\{1,2, \ldots ,2n\}$ into pairs of elements $\alpha\in \Pi^{}_{}$, can be represented as |
| $$\alpha^{}_{}=\{(i_1,j_1),(i_2,j_2), \ldots ,(i_n,j_n)\} $$ |
$$\alpha^{}_{}=\{(i_1,j_1),(i_2,j_2), \ldots ,(i_n,j_n)\} $$ |
| with $i_k<j_k$ and $i_1 < i_2 < \cdots < i_n$, let |
with $i_k<j_k$ and $i_1 < i_2 < \cdots < i_n$, let |
| $$\pi=\begin{bmatrix} 1 & 2 & 3 & 4 & \ldots & 2n \\ i_1 & j_1 & i_2 & j_2 & \ldots & j_{n} \end{bmatrix}$$ |
$$\pi=\begin{bmatrix} 1 & 2 & 3 & 4 & \ldots & 2n \\ i_1 & j_1 & i_2 & j_2 & \ldots & j_{n} \end{bmatrix}$$ |
| be a corresponding permutation and let us define |
be a corresponding permutation and let us define |
| $sgn(\alpha)$ to be the signature of a permutation $\pi^{}_{}$; clearly it depends only on the partition $\alpha$ and not on the particular choice of $\pi^{}_{}$. |
$sgn(\alpha)$ to be the signature of a permutation $\pi^{}_{}$; clearly it depends only on the partition $\alpha$ and not on the particular choice of $\pi^{}_{}$. |
| Given a partition $\alpha^{}_{}$ as above let us set |
Given a partition $\alpha^{}_{}$ as above let us set |
| $a_\alpha =a_{i_1,j_1}a_{i_2,j_2} \ldots a_{i_n,j_n},$ |
$a_\alpha =a_{i_1,j_1}a_{i_2,j_2} \ldots a_{i_n,j_n},$ |
| then we can define the \emph{Pfaffian} of $A$ as |
then we can define the \emph{Pfaffian} of $A$ as |
| $$\mbox{Pf}(A)=\sum_{\alpha\in \Pi} sgn(\alpha)a_\alpha.$$ |
$$\mbox{Pf}(A)=\sum_{\alpha\in \Pi} sgn(\alpha)a_\alpha.$$ |
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| {\bf Alternative definition} |
{\bf Alternative definition} |
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| One can associate to any antisymmetric $2n\times 2n$matrix $A=\{a_{ij}\}$ |
One can associate to any antisymmetric $2n\times 2n$matrix $A=\{a_{ij}\}$ |
| a bivector |
a bivector |
| :$\omega=\sum_{i<j} a_{ij} e_i\wedge e_j$ |
:$\omega=\sum_{i<j} a_{ij} e_i\wedge e_j$ |
| in a basis |
in a basis |
| $\{e_1,e_2, \ldots ,e_{2n}\}$ of $\mathbb{R}^{2n}$, then |
$\{e_1,e_2, \ldots ,e_{2n}\}$ of $\mathbb{R}^{2n}$, then |
| $$\omega^n= n!\mbox{Pf}(A)e_1\wedge e_2\wedge \cdots \wedge e_{2n},$$ |
$$\omega^n= n!\mbox{Pf}(A)e_1\wedge e_2\wedge \cdots \wedge e_{2n},$$ |
| where $\omega^n_{}$ denotes exterior product of $n$ copies of $\omega^{}_{}$. |
where $\omega^n_{}$ denotes exterior product of $n$ copies of $\omega^{}_{}$. |
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| {\bf Identities} |
{\bf Identities} |
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| For any antisymmetric $2n\times 2n$ matrix $A$' and any $2n\times 2n$ matrix $B$ |
For any antisymmetric $2n\times 2n$ matrix $A$' and any $2n\times 2n$ matrix $B$ |
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$$Pf(A)^2 = \det(A)$$
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$$\text{Pf}(A)^2 = \det(A)$$
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$$Pf(BAB^T)= \det(B)Pf(A)$$
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$$\text{Pf}(BAB^T)= \det(B)\text{Pf}(A)$$
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