Talk:Pfaffian

Introduction
The intro says: "The Pfaffian is nonvanishing only for 2n × 2n skew-symmetric matrices". This is not true. As a counter example, take some skew-symmetric matrix A, which has some nonzero pfaffian Pf(A). All of the diagonal elements $$A_{ii}$$ must be zero by assumption of skew-symmetry. Now change $$A_{11}$$ to one. A is no longer skew symmetric, but it still has the same Pfaffian because the definition of the Pfaffian does not "see" the diagonal elements. It is only a function of matrix elements like $$A_{P(a)P(2)}$$ which are permutations of distinct indices. Ltgoff —Preceding undated comment added 22:52, 2 February 2010 (UTC). Indeed - the definition of the Pfaffian projects out the symmetric part of the matrix, but there is no harm in starting with a general matrix.

The attribution of $$\operatorname{pf}^2(A) = \operatorname{det}(A)$$ to Muir is likely incorrect. Contributors to the FKT algorithm article attribute this to Caley, and give ref Cayley, Arthur (1847). "Sur les determinants gauches" [On skew determinants]. Crelle's Journal 38: 93–96. — Preceding unsigned comment added by 70.72.144.66 (talk) 00:20, 19 June 2015 (UTC)

About time this was corrected (I'm doing it now)! I've read Cayley's paper from 1847, and while he doesn't call them Pfaffians these are the point of the paper. What is interesting is he attributes the fact that $$\operatorname{pf}(A)$$ divides $$\operatorname{det}(A)$$ to Jacobi, but says Jacobi is apparently unaware that the quotient is also $$\operatorname{pf}(A)$$. The later name Pfaffian seems to be due to the fact that Jacobi was using it for studying Pfaffian systems of ODEs! (Seems a bit tenuous to me. Perhaps he should have called it the Jacobian of the matrix!) James Montaldi 14:08, 25 February 2018 (UTC)  — Preceding unsigned comment added by Jamontaldi (talk • contribs)

I don't see any problem in the introduction. Pfaffian is only defined for 2n × 2n skew-symmetric matrices. Applying Pfaffian to a non-skew-symmetric matrix is illegal and not defined. For skew-symmetric matrices, it is always true that the Pfaffian squares to the determinant.Everett (talk) 23:46, 7 April 2016 (UTC) It can be easily defined using the sum over permutations, even if it is not obviously useful to do so. N Geraint Jones (talk) 14:57, 8 April 2016 (UTC)

Derivation from Determinant
Although we have


 * $$C_{ij}^2 = \det(A_{ij,ij})\det(A), \,$$

It doesn't directly imply the sign of $$C_{ij}$$ as $$(-1)^{i+j}$$ by simply taking the square root. May I know how to reach this step in detail?

--221.127.68.176 (talk) 16:26, 28 July 2009 (UTC)

I do not think one should try hard to change \times to x both ways it looks ugly, but hopefully with Mathml the first one will look better... Tosha 15:29, 23 May 2004 (UTC)

Johann Friedrich Pfaff
Is this the guy? Tosha

Do you know the story? Gauss's mother asked a mathematician 'who is the best mathematician in Germany?' and the answer was 'Pfaff'. 'What about my son?' - 'Oh, he's the best mathematician in Europe!'. Pfaff was Gauss's teacher, in a sense.

Charles Matthews 09:04, 9 Sep 2004 (UTC)

Nice story, but is Johann Friedrich Pfaff is one who responsible for Pfaffian?


 * This web page says so: http://members.aol.com/jeff570/p.html. He had a brother JW Pfaff, who was also a mathematician, which I hadn't realised. Charles Matthews 18:55, 9 Sep 2004 (UTC)

Thank you Tosha

The Pfaffian in coordinates
(moved from generalized Gauss-Bonnet theorem)

Suppose given a coordinate coframe $$dx^i$$. The curvature 2-form admits an expression $${\Omega_i}^j$$ relative to this coframe. Define the alternating Kronecker tensor
 * $$\delta_{j_1j_2\dots j_k}^{i_1i_2\dots i_k}$$

to be the identity endomorphism of $$\bigwedge^kT^*M$$. Then


 * $$Pf(\Omega)=\delta_{j_1j_2\dots j_n}^{i_1i_2\dots i_n}{\Omega_{i_1}}^{j_1}\wedge{\Omega_{i_2}}^{j_2}\wedge\dots\wedge{\Omega_{i_n}}^{j_n}$$

where repeated indices are summed.

Mathematica code
Here's a Mathematica function for computing the Pfaffian of a matrix. I don't suspect this belongs on the main page, so I'll just post it here and let the Wikipedia experts decide:

Pf[A_] := If[Length[A] == 0, 1, Module[{L, A1, MatrixDelete}, MatrixDelete[M_, i_] := Delete[#, i] & /@ Delete[M, i]; L = Length[A]; A1 = MatrixDelete[A, 1]; Sum[(-1)^i (A1i Pf[MatrixDelete[A1, i - 1]]), {i, 2, L}]]]


 * The above recursive algorithm calculates the pfaffian for the general n×n matrix in O(n!) time. However for a skew symmetric matrix, the pfaffian can be calculated more efficiently in O(n3) time by generalized Gaussian elimination. The Mathematica package implements such algorithm is originally written by R. W. Cherng, which is available here: http://physics.harvard.edu/~cherng/programs.html. However the package is written in a C/FORTRAN manner, which runs slowly on Mathematica. I modify the code by vectorizing loops to avoid explicit part references, which allows Mathematica to optimize the performance. The modified package code is posted as bellow. To use the package: copy and past to Mathematica notebook, and evaluate the cell once, then can use Pf[A] to obtain the pfaffian of the skew symmetric matrix A. Everett (talk) 08:35, 27 November 2010 (UTC)

BeginPackage["Pfaffian`"]; Pf::usage="Pf[A] gives the pfaffian of the skew symmetric A."; Begin["`Private`"]; Pf[A_] := Switch[Length[A], 0, 1, _?OddQ, 0, _?EvenQ, xPf[A, 1]]; xPf[A_, p0_] := Module[{A0, n, pivot, sign = 1, A1, p1}, n = Length[A]/2; If[n != 1, A0 = A;   pivot = First[Ordering[Normal[Abs[A02 n - 1, All]], -1]]; If[pivot != 2 n, A0{pivot, 2 n}, All = A0{2 n, pivot}, All; A0All, {pivot, 2 n} = A0All, {2 n, pivot}; sign = -1;]; p1 = A02 n - 1, 2 n; A1 = p1 A01 ;; 2 n - 2, 1 ;; 2 n - 2; A1 += (# - Transpose[#]) &@ Outer[Times, A01 ;; 2 n - 2, 2 n, A01 ;; 2 n - 2, 2 n - 1]; A1 /= p0; sign xPf[A1, p1], A1, 2]]; End[]; EndPackage[];

Applications
It would be nice to see a qualified writer add to the applications section to include minimal free resolutions of codimension three Gorenstein quotients (a la Buchsbaum-Eisenbud in the 70s). Courtney Gibbons 24 Mar 2011 — Preceding unsigned comment added by Virtualcourtney (talk • contribs) 22:47, 24 March 2011 (UTC)

While unifying the format of the references, I noticed that most of the references in this article where about an application that was removed in the revision made on 14:27, 2 February 2011. I personally agree with the removal (at least it should have been clarified where the Pfaffian enters there!), and would also favor removing the corresponding references.

However, before doing so, I wanted to start a general discussion about the Applications section. The two mathematical contexts in which Pfaffians arise seem fine to me. However, the physics and especially the machine learning applications mentioned there (with references) seem somewhat arbitrary and biased to me - at least I don't see why these applications should be so outstanding that they are mentioned as sole examples in a wikipedia article.

In my opinion, one should probably extend the Applications section by more examples, to make it more biased again. Alternatively, one could restrict the whole Applications section to the pure mathematical level, but that would mean removing more examples and references, and I'm not a particular fan of that.

I know a lot of physics examples (with references), but that alone would be biased again. Any other ideas?

Wmmrmchl (talk) 13:32, 8 April 2011 (UTC)

Simple proof of equation $$ \operatorname{pf||O^T A O||} = \operatorname{det||O||} \operatorname{pf||A||}$$
The definition of Pfaffian:
 * $$\mathrm{pf}\left\| A\right\| =\frac{1}{2^{n}n!}\epsilon

_{i_{1}i_{2}...i_{2n}}A_{i_{1}i_{2}}A_{i_{3}i_{4}}...A_{i_{2n-1}i_{2n}}.$$ Here ε is the totally antisymmetric tensor, ε12...2n = 1. One has
 * $$\mathrm{pf}\left\| O^{T}AO\right\| = \frac{1}{2^{n}n!}\epsilon

_{i_{1}i_{2}...i_{2n}}\left( O_{i_{1}j_{1}}O_{i_{2}j_{2}}A_{j_{1}j_{2}}\right) \left( O_{i_{3}j_{3}}O_{i_{4}j_{4}}A_{j_{3}j_{4}}\right) ...\left( O_{i_{2n-1}j_{2n-1}}O_{i_{2n}j_{2n}}A_{j_{2n-1}j_{2n}}\right)$$
 * $$ =\frac{1}{2^{n}n!}\det \left\| O\right\| \epsilon

_{j_{1}j_{2}...j_{2n}}A_{j_{1}j_{2}}A_{j_{3}j_{4}}...A_{j_{2n-1}j_{2n}} = \det\left\| O\right\| \mathrm{pf}\left\| A\right\|. $$

Simple proof of equation $$ \operatorname{pf^2||A||}=\operatorname{det||A||}$$
Matrix A is antisymmetric. It defines a symplectic structure. According to the Darboux theorem, there exists 2n × 2n matrix O that brings A into the canonical form
 * $$O^TAO = J \equiv \begin{bmatrix}

\begin{matrix} 0 & 1\\ -1 & 0\end{matrix} & 0 & \cdots & 0 \\ 0 & \begin{matrix}0 & 1\\ -1 & 0\end{matrix} & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & \begin{matrix}0 & 1\\ -1 & 0\end{matrix} \end{bmatrix}.$$ One can check that det||J|| = pf||J|| = 1, and so
 * $$ \operatorname{det||O^TAO||}=\operatorname{det^2||O||} \operatorname{det||A||} = 1,$$
 * $$ \operatorname{pf||O^TAO||}=\operatorname{det||O||} \operatorname{pf||A||} = 1.$$

These two lines being combined give
 * $$ \operatorname{pf^2||A||}=\operatorname{det||A||}.$$

Trompedo (talk) 08:21, 16 January 2013 (UTC)

Simple proof of equation $$ \operatorname{pf||A^{2m + 1}||}= (-1)^{nm}\operatorname{pf||A||}^{2m + 1}$$
that caused difficulties for Ms. Pendleton. The following equations are obvious
 * $$\operatorname{pf}(A^{m}A(A^{T})^{m})=\operatorname{pf}(A^{m}A(-A)^{m})=\operatorname{pf}(A^{m}A(-)^{m}A^{m})=(-)^{nm}\operatorname{pf}(A^{2m+1}),$$

and
 * $$\operatorname{pf}(A^{m}A(A^{T})^{m})=\det (A^{m})\operatorname{pf}(A)

=(\operatorname{det}(A))^{m}\operatorname{pf}(A) =(\operatorname{pf}(A)^{2})^{m}\operatorname{pf}(A)=\operatorname{pf}(A)^{2m+1}.$$ Comparing these two lines, we obtain
 * $$\operatorname{pf}(A^{2m+1})=(-)^{nm}\operatorname{pf}(A)^{2m+1}.$$

In the special case m = - 1, we arrive at the equation in the text. Trompedo (talk) 06:30, 21 February 2013 (UTC)

Reference to Wells
There's a reference to Wells' Curious and Interesting Numbers in the References section. I have this work (metaphorically) open in front of me now, and I can't see anything on p.182 that seems remotely relevant. Can anyone help explain what that's all about? --Matt Westwood 07:21, 24 June 2018 (UTC)

Is it the same as Pfaffinant?
WaterWaterWaterLooLooLoo (talk) 22:11, 21 July 2019 (UTC)

References and clarification for the trace formula
The trace log identity for the product of pfaffians, and the resulting numerical method lacks references. https://arxiv.org/pdf/1605.00447.pdf appears to be a possible relevant reference, but the range of applicability of the formula is not clear from there either. The trace identity is stated for the case when A^T B is positive definite, but there is no mention of this restriction in the numerical methods section. The trace identity tr(log(AB)) = tr(log(A)) + tr(log(B)) is not true in general, only modulo 2 pi i because of the ambiguity of choosing the branch cut of the log. — Preceding unsigned comment added by Dvarjas (talk • contribs) 21:08, 27 February 2020 (UTC)

PARI/gp code
I saw someone contributing Mmca code in an earlier section, so I allow myself to contribute PARI code: pf(A)=if(#A>1,sum(j=2,#A,(-1)^j*A[1,j]*pf(A[^j,^j][^1,^1])),!#A) This is not optimized but certainly much more efficient than to factor the determinant which already for a 6x6 antisymmetric matrix is a homogeneous polynomial of degree 6 with about 200 terms in the 10 independent components. &mdash; MFH:Talk 01:41, 28 March 2020 (UTC)

Problem in Examples section
The section Examples ends with this sentence:

"(Note that any skew-symmetric matrix can be reduced to this form with all $$b_i$$ equal to zero; see Spectral theory of a skew-symmetric matrix.)"

This makes no sense at all, however, because the symbol "$$b_i$$" has not been defined. 2601:200:C000:1A0:3932:3D49:97C7:A86A (talk) 17:21, 21 September 2021 (UTC)

Bad sentence
The section Examples ends with this sentence (in parentheses):

"Note that any skew-symmetric matrix can be reduced to this form; see Spectral theory of a skew-symmetric matrix."

But nowhere in this section or earlier in the article is anything defined as a bi for any value of i.

I hope someone can fix this or else delete the otherwise meaningless sentence.


 * Never mind; I took care of it by removing the unneeded and confusing phrase "with all bi equal to zero".