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'Pfaffian'
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| Title of object: |
Pfaffian |
| Canonical Name: |
Pfafian |
| Type: |
Definition |
| Created on: |
2004-05-14 00:46:54 |
| Modified on: |
2006-09-25 10:40:27 |
| Classification: |
msc:15A15 |
| Keywords: |
antisymmetric matrix |
Revision comment (for changes between this and next version):
| Changes for correction #10174 ('has an unnecessay n'). |
Preamble:
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\usepackage{amssymb}
\usepackage{amsmath}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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Content:
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 applied in the generalized Gauss-Bonnet theorem.
{\bf Examples}
$\mbox{Pf}\begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}=a,$
$\mbox{Pf}\begin{bmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0& f \\-c & -e & -f & 0 \end{bmatrix}=af-be+dc.$
{\bf Standard definition}
Let
$$A=\begin{bmatrix} 0 & a_{1,2} & \ldots & a_{1,2n} \\ -a_{1,2} & 0 & \ldots & a_{2,2n} \\ \vdots & \vdots & \vdots & \vdots \\-a_{2n,1} & -a_{2n,2} & \ldots & 0 \end{bmatrix}.$$
Let $\Pi^{}_{}$ be the set of all partition of $\{1,2, \ldots ,2n\}$ into pairs of elements $\alpha\in \Pi^{}_{}$, can be represented as
$$\alpha^{}_{}=\{(i_1,j_1),(i_2,j_2), \ldots ,(i_n,j_n)\} $$
with $i_k<j_k$ and $i_1 < i_2 < \cdots < i_n$, let
$$\pi=\begin{bmatrix} 1 & 2 & 3 & 4 & \ldots & 2n \\ i_1 & j_1 & i_2 & j_2 & \ldots & j_{n} \end{bmatrix}$$
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 $\pi^{}_{}$.
Given a partition $\alpha^{}_{}$ as above let us set
$a_\alpha =a_{i_1,j_1}a_{i_2,j_2} \ldots a_{i_n,j_n},$
then we can define the \emph{Pfaffian} of $A$ as
$$\mbox{Pf}(A)=\sum_{\alpha\in \Pi} sgn(\alpha)a_\alpha.$$
{\bf Alternative definition}
One can associate to any antisymmetric $2n\times 2n$matrix $A=\{a_{ij}\}$
a bivector
:$\omega=\sum_{i<j} a_{ij} e_i\wedge e_j$
in a basis
$\{e_1,e_2, \ldots ,e_{2n}\}$ of $\mathbb{R}^{2n}$, then
$$\omega^n= n!\mbox{Pf}(A)e_1\wedge e_2\wedge \cdots \wedge e_{2n},$$
where $\omega^n_{}$ denotes exterior product of $n$ copies of $\omega^{}_{}$.
{\bf Identities}
For any antisymmetric $2n\times 2n$ matrix $A$' and any $2n\times 2n$ matrix $B$
$$Pf(A)^2 = \det(A)$$
$$Pf(BAB^T)= \det(B)Pf(A)$$ |
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