Pfaffians and Determinants
José de Jesús Martı́nez
Department of Mathematics, Kobe University,
Rokko, Kobe 657-8501, Japan
Contents
1 Introduction
1
2 Pfaffians and determinantal
2.1 Definition . . . . . . . . .
2.2 Exterior Algebra . . . . .
2.3 Wronskians . . . . . . . .
2.4 General Determinants . .
expressions
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1
1
2
3
3
3 Laplace Expansion
5
4 Plücker Relations
6
5 Determinant Formulae
5.1 Cofactors . . . . . . . .
5.2 The Jacobi Identity . . .
5.3 Perfect Square Formula
5.4 Bordered Determinants
6 Pfaffian Formulae
6.1 Pfaffian Identities . . .
6.2 Expansion Formula for
6.3 The Addition Formula
6.4 Pfaffian Derivatives . .
1
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6
6
7
9
12
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(a1 , a2 , 1, 2, . . . , 2n) .
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14
14
16
18
20
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Introduction
In this report we discuss those fundamental properties of pfaffians and determinants which will eventually be
used to comprehend the structure of soliton solutions. Some of the topics to study are pfaffians identities,
plücker relations, pfaffian and determinantal expressions as well as a formulae to compute pfaffian derivatives.
We focus on the necessary tools to understand the latter. To a lesser extend we also consider the meaning of
Maya diagrams and their connection to plücker relations and bilinear soliton equations.
2
2.1
Pfaffians and determinantal expressions
Definition
Let A be the determinant of an m × m antisymmetric matrix where aj,k = −ak,j . Then,
A = det(aj,k )1≤j,k≤m .
It is well known that if m is odd, then A = 0. This is called the Jacobi Theorem. If m is even, then A is the
square of a pfaffian according to a theorem of Thomas Muir. This pfaffian is of order n such that m = 2n. We
denote this pfaffian as,
1
(1, 2, . . . , 2n) .
We also define the entries as,
(j, k) = aj,k , (j, k) = −(k, j) .
In general, a pfaffian may be expanded with respect to the first index as follows,
(1, 2, . . . , 2n) =
2n
X
(−1)j (1, j)(2, 3, . . . , b
j, . . . , 2n) .
(1)
j=2
It is also possible to be expanded with respect to the last index,
=
2n−1
X
(−1)j−1 (1, 2, . . . , b
j, . . . , 2n − 1)(j, 2n) ,
j=1
where b
j means to that the index j is omitted.
To expand even further, there is a formula similar to that of the determinant which gives the sum of the
products of first-order pfaffians,
X0
(1, 2, . . . , 2n) =
(−1)P (i1 , i2 )(i3 , i4 ) . . . (i2n−1 , i2n ) .
(2)
P
From now on, the first-order pfaffian (i, j), is called a pfaffian entry.
combinations of pairs selected from the set {1, 2, . . . , 2n} such that
P0
means the sum of all possible
i1 < i2 , i3 < i4 , i5 < i6 , . . . i2n−1 < i2n ,
i1 < i3 < i5 < · · · < i2n−1 .
(−1)P takes a value of +1 or −1 if the sequence i1 , i2 , . . . i2n is an even or odd permutation.
2.2
Exterior Algebra
Now we wish to define the nth -order determinant and nth -order pfaffian using Exterior Algebra.
The nth -order determinant
Consider the one-form,
wi =
n
X
ai,j xj
(i = 1, 2, . . . , 2n) .
j=1
Then the following skew symmetric commutation relations hold arbitrarily for xj ,
xi ∧ xj = −xj ∧ xi ,
xi ∧ x i = 0 .
Note that ai,j is an arbitrary complex function.
The determinant det(ai,j )1≤i,j≤n is defined in terms of the exterior product of n one-forms,
ω1 ∧ ω2 ∧ · · · ∧ ωn ≡ det(ai,j )1≤i,j≤n x1 ∧ x2 ∧ · · · ∧ xn .
2
nth -order pfaffian
Consider the two-form or bilinear form,
Ω=
X
bi,j xi ∧ xj .
1≤i<j≤2n
Here we let (i, j) = bi,j . The nth -order pfaffian with entries (i, j) is defined by the n-copy wedge product of
two-forms,
1
2
2n
{z· · · ∧ Ω} ≡ n! (1, 2, . . . , 2n) x ∧ x ∧ · · · ∧ x .
|Ω ∧ Ω ∧
n times
2.3
Wronskians
Next we briefly show how one may express a wronskian determinant as a pfaffian. This will come in useful
later on. It was discovered by Junkichi Satsuma that we can express soliton solutions, such as that of the KdV
equation , using wronskian determinants. It is also known thanks to Freeman and Nimmo that the KP bilinear
equation is a determinantal identity when we express its solutions as wronskian determinants.
The nth order wronskian is defined as,
Wr(f1 , f2 , f3 , . . . , fn ) ≡ det
∂ j−1
fi
∂xj−1
1≤i,j≤n
(0)
f
1
(0)
f2
= .
..
(0)
fn
f1
(1)
f2
..
.
(1)
...
...
(1)
...
fn
(n−1) f1
(n−1) f2
.. .
. (n−1) fn
Next we define the pfaffian entries as follows
(j)
(dj , i) ≡ fi
, (dj , dk ) ≡ 0 ,
where (i = 1, 2, . . . , n) and (j, k = 0, 1, . . . , n − 1).
Thus, we express the nth order wronskian as an nth order pfaffian,
Wr(f1 , f2 , f3 , . . . , fn ) = (d0 , d1 , d2 , . . . , dn−1 , n, . . . , 3, 2, 1) .
2.4
General Determinants
We begin by expressing the nth order determinant as an nth order pfaffian.
Consider the nth order determinant,
B ≡ det(bj,k )1≤j,k≤n .
Then we say that B can be expressed as an nth order pfaffian,
B = (1, 2, . . . , n, n∗ , . . . , 2∗ , 1∗ ) .
3
Here the pfaffian entries, (j, k), (j , k ∗ ) and (j ∗ , k ∗ ), satisfy the following properties
(j, k) = 0 , (j ∗ , k ∗ ) = 0 , (j , k ∗ ) = bj,k .
Next we consider a 2nth order determinant as an nth order pfaffian. We recall the one-form,
ωi =
2n
X
ai,j xj
,
i = 1, 2, . . . , 2n .
j=1
By definition, the determinant det(ai,j )1≤i,j≤2n is expressed as,
ω1 ∧ ω2 ∧ · · · ∧ ω2n ≡ det(ai,j )1≤i,j≤2n x1 ∧ x2 ∧ · · · ∧ x2n .
We also use the two-form
X
Ω=
(i, j) xi ∧ xj .
(3)
1≤i<j≤2n
With the expression above we define the nth order pfaffian
1
2
2n
|Ω ∧ Ω ∧
{z· · · ∧ Ω} ≡ n! (1, 2, . . . , 2n) x ∧ x ∧ · · · ∧ x .
(4)
n times
Now consider a different two-form,
Ω ≡ ω1 ∧ ω2 + ω3 ∧ ω4 + · · · + ω2n−1 ∧ ω2n .
(5)
From here we see that,
|Ω ∧ Ω ∧
{z· · · ∧ Ω} = n! ω1 ∧ ω2 ∧ . . . ω2n .
n times
Now recall that the one-form,
ωi =
2n
X
ai,j xj
,
i = 1, 2, . . . , 2n .
(6)
j=1
gives the definition of the nth order determinant,
ω1 ∧ ω2 ∧ · · · ∧ ω2n ≡ det(ai,j )1≤i,j≤2n x1 ∧ x2 ∧ · · · ∧ x2n .
Thus, we can see that the following expression holds,
1
2
2n
|Ω ∧ Ω ∧
{z· · · ∧ Ω} = n! det(ai,j )1≤i,j≤2n x ∧ x ∧ · · · ∧ x .
n times
4
(7)
At this point we must show that the two-forms (Ω) in (4) and (7) are the same. Substituting one-form (6)
into two-form (5) gives,
Ω=
X
bi,j xi ∧ xj ,
1≤i<j≤2n
bi,j =
n
X
(a2m−1,i a2m,j − a2m−1,j a2m,i ) .
m=1
The Ω above is of the same form as (3) where (i, j) ≡ bi,j . By comparing (4) and (7) we see that,
det(ai,j )1≤i,j≤2n = (1, 2, . . . , 2n) .
3
Laplace Expansion
An n-order determinant An = det(ai,j )1≤i,j≤n can be expressed as a sum of products of r and (n − r)-order
determinants.
Choose indices
i1 < i2 < · · · < ir , ir+1 < ir+2 < · · · < in .
from the set {1, 2, . . . , n} for indices 1 to r and assign the remaining to indices r + 1 to n. We do the same for
the indices of j.
The Laplace Expansion Formula is then written as follows,
X
i
P
An =
(−1) ∆ 1
j1
ir
i
∆ r+1
jr
jr+1
...
...
...
...
in
jn
,
where P = i1 + · · · + ir + j1 + · · · + jr .
The expression,
∆
i1
j1
i2
j2
...
...
ir
jr
,
denotes the r-order determinant of the matrix (aip ,jq )1≤p,q≤r .
Similarly,
i
∆ r+1
jr+1
ir+2
jr+2
. . . in
. . . jn
,
denotes
P the r-order determinant of the matrix (aip ,jq )r+1≤p,q≤n .
denotes the sum over all possible such choices n Cr = n!/[r!(n − r)!] in number of j1 , . . . , jr . Now we wish
to express the Laplace Expansion Formula using exterior algebra. Consider the one-form,
ωi =
n
X
j=1
5
ai,j xj .
By definition, the n-order determinant is given by,
ω1 ∧ · · · ∧ ωn ≡ det(ai,j )1≤i,j≤n x1 ∧ x2 ∧ · · · ∧ xn .
We then see that the l.h.s. may be written as,
(ω1 ∧ ω2 ∧ · · · ∧ ωr ) ∧ (ωr+1 ∧ ωr+2 ∧ · · · ∧ ωn ) ,
which can be arranged as a sum of products of r and (n − r) -order determinants.
4
Plücker Relations
Consider the fourth-order determinant,
a0
b0
0
0
a1
b1
a1
b1
a2
b2
a2
b2
a3 a0
b3 b0
=
a3 0
b3 0
0
0
a1
b1
0
0
a2
b2
0 0 =0.
a3 b3 By expanding the determinants above we obtain the following expression,
a0
b0
a1 a2
b1 b2
a3 a0
−
b3 b0
a2 a1
b2 b1
a3 a0
+
b3 b0
a3 a1
b3 b1
a2 =0.
b2 By letting the column vector ~ci = (ai , bi )t , then
~c0 ~c1 ~c2 ~c3 − ~c0 ~c2 ~c1 ~c3 + ~c0 ~c3 ~c1 ~c2 = 0 .
Since only the indices are relevant, we express the above as,
(0 1)(2 3) − (0 2)(1 3) + (0 3)(1 2) = 0 .
(8)
Expression (8) is the simplest case of a plücker relation.
The importance of plücker relations is their connection to solutions to soliton equations.
Plücker
Relations
5
5.1
−−−−→
Maya
Diagram
−−−−→
Young
Diagram
Bilinear
−−−−→ Soliton
Equations
Determinant Formulae
Cofactors
Let D be the determinant of matrix A = (ai,j )1≤i,j≤n . Then the cofactor ∆i,j to A is the determinant of the
matrix obtained by eliminating the ith row and j th column from matrix A, multiplied by the signature (−1)i+j .
6
Through a special case of the orthogonality relations we obtain a formula to expand a determinant by its
cofactors.
D=
n
X
ai,j ∆ij =
i=1
n
X
ai,j ∆i,j ,
j=1
where i, j = 1, 2, . . . , n.
In the case of pfaffians the cofactor Γ(i, j) with respect to (i, j) in the n-order pfaffian (1, 2, . . . , 2n) is defined
by,
Γ(i, j) = (−1)i+j+1 (1, 2, . . . , bi, . . . , b
j, . . . , 2n) ,
Γ(j, i) = −Γ(i, j) ,
Γ(i, i) = 0 .
The formula to expand a pfaffian by its cofactors is given by,
(1, 2, . . . , 2n) =
2n
X
(i, j)Γ(i, j) ;
(i = 1, 2, . . . , 2n) .
j=1
By fixing i = 1, we obtain the expansion formula of a pfaffian with respect to the first index. In a similar
way, by fixing j = 2n, we get the expansion formula of a pfaffian with respect to the last index.
Next, consider the case where D is of even order and the entry ai,j is equal to the pfaffian entry (i, j). Then
we have that
2n
X
(i, j) ∆i,j = (1, 2, . . . , 2n)2 .
i=1
By comparing the result above with the formula to expand a pfaffian by its cofactors, we see that the
following cofactor relation holds,
∆i,j = Γ(i, j) (1, 2, . . . , 2n) .
The relation above will be used later on to derive a derivative formula for a pfaffian.
5.2
The Jacobi Identity
j
The (j, k) minor D
of the N-order determinant, D = det(ai,j )1≤i,j≤N , is obtained by eliminating the j th
k
row and k th column.
th
a1,1
.
..
j
D
= j
k
.
..
aN,1
Note that D being an N -order determinant, D
a1,2
..
.
...
k
...
..
.
a1,N
..
.
..
.
aN,2
...
...
aN,N
.
j
is an (N − 1)-order one.
k
7
The cofactor ∆jk is then defined as,
∆jk ≡ (−1)
j+k
j
D
.
k
The (N − 2)-order determinant obtained by eliminating the lth and mth columns and the j th and k th rows, is
written as,
D
j
l
k
.
m
Now we may express the Jacobi Identity as follows,
i
j
i
j
i
D
D
−D
D
=D
i
j
j
i
i
j
D.
j
Consider the following case,
n−1
n
n−1
n
n−1
D
D
−D
D
=D
n−1
n
n
n−1
n−1
n
D.
n
Recall that the n-order determinant can be expressed as an n-order pfaffian,
D = (1, 2, . . . , n − 1, n, n∗ , n − 1∗ , . . . , 2∗ , 1∗ ) ,
(j, k) = 0 , (j ∗ , k ∗ ) = 0 , (j, k ∗ ) = aj,k .
Thus,
D
j
l
k
m
...
c∗ , . . . , lb∗ , . . . , 1∗ ) .
= (1, . . . , b
j, . . . , b
k, . . . , n , n∗ , . . . , m
...
Therefore, the Jacobi Identity can be written as a pfaffian expression,
(1, . . . , n − 2, n, n∗ , n − 2∗ , . . . , 1∗ )
× (1, . . . , n − 1, n − 1∗ , . . . , 1∗ )
−(1, . . . , n − 2, n, , n − 1∗ , . . . , 1∗ )
× (1, . . . , n − 1, n∗ , n − 2∗ , . . . , 1∗ )
=(1, . . . , n − 2, n − 2∗ , . . . , 1∗ )(1, . . . , n, n∗ , . . . , 1∗ ) .
Using Maya Diagrams, the Jacobi Identity is written as follows,
n−1
−
+
n
n−1∗
n∗
n−1
×
×
×
n
n−1∗
n∗
=0.
Later on it will be explained how if one expresses the solutions to the KP and Toda Lattice equations as
grammian determinants, their bilinear equations become Jacobi Identities.
8
5.3
Perfect Square Formula
Consider the determinant of the n × n antisymmetric matrix,
An = det[aj,k ]1≤j,k≤n .
Using the Jacobi Identity, we can prove two facts,
• An is equal to zero if n is odd.
• An is equal to the perfect square of a polynomial (pfaffian).
According to basic properties of determinants, we know that
det(An ) = (ATn ) ,
det(An ) = det(−An ) = (−1)n det(A) .
It is then clear that if n is odd, then the determinant of A is zero.
Now let D be the determinant of a 2m×2m antisymmetric matrix. Because the determinant of an odd-order
antisymmetric matrix is zero, we have that
D
2m
2m − 1
=D
=0.
2m
2m − 1
Also, by the properties of determinants we see that,
D
2m − 1
2m
= −D
.
2m
2m − 1
Since D is of even-order, the Jacobi Identity from the previous section is of the form,
2m − 1
2m
2m − 1
2m
2m − 1 2m
D
D
−D
D
=D
D.
2m − 1
2m
2m
2m − 1
2m − 1 2m
When simplified, the identity above becomes the following recursive formula,
D
2
2m − 1
2m − 1 2m
=D
D.
2m
2m − 1 2m
We see that for m = 1, Dm=1 is a perfect square,
0
−a1,2
a1,2 = a21,2 .
0 9
(9)
Now we check the case for m = 2 using (9). First note that,
3
Dm=2
3
4
= Dm=1 = a21,2 .
4
Hence
2
3
Dm=2
= a21,2 Dm=2 .
4
By simplifying, we see that Dm=2 is also a perfect square. The same can be done for m = 3, 4, . . . . We conclude
that for any arbitrary m, D is a perfect square.
Note
Consider the n × n identity matrix E and the n × n antisymmetric matrices A and B,
A = (ai,j )1≤i,j≤n ,
B = (bi,j )1≤i,j≤n .
We would like to show that the determinant of E + AB is a perfect square such that,
det(E + AB) = (1, 2, . . . , n, 1∗ , 2∗ , . . . , n∗ )2 .
The pfaffian entries are given by,
(i, j) = aij = −aji ,
∗
(i , j ∗ ) = bij = −bji ,
(i, j ∗ ) = (j ∗ , i) = δij .
Here, δi,j is the Kronecker symbol and it is given as,
(
δij =
1
0
i=j,
otherwise .
From (10), let 1∗ → n + 1 , 2∗ → n + 2 , . . . , n∗ → 2n such that,
(1, 2, . . . , n, 1∗ , 2∗ , . . . , n∗ )2 = (1, 2, . . . , 2n)2 .
is the pfaffian for the determinant of a 2n × 2n antisymmetric matrix.
10
(10)
0
a2,1
..
.
a
(1, 2, . . . , 2n)2 = n,1
an+1,1
an+2,1
.
..
a2n,1
a1,2
0
..
.
an,2
an+1,2
an+2,2
..
.
a2n,2
...
...
..
.
a1,n
a2,n
..
.
...
0
. . . an+1,n
. . . an+2,n
..
.
...
a2n,n
(1, 1)
(1, 2)
...
(2, 1)
(2,
2)
.
..
..
..
.
.
(n, 1)
(n,
2)
...
= (n
+
1,
1)
(n
+
1,
2)
...
(n + 2, 1) (n + 2, 2) . . .
..
..
.
.
(2n, 1)
(2n, 2)
...
(1, 1)
(2, 1)
..
.
(n, 1)
= ∗
(1∗ , 1)
(2 , 1)
.
..
(n∗ , 1)
0
−a1,2
..
.
−a
= 1,n
−1
0
.
..
0
A
= −E
a1,n+1
a2,n+1
..
.
a1,n+2
a2,n+2
..
.
an,n+1
0
an+2,n+1
..
.
an,n+2
an+1,n+2
0
..
.
a2n,n+1
a2n,n+2
(1, n)
(2, n)
..
.
...
an,2n . . . an+1,2n . . . an+2,2n .. ..
.
. . . . a2n,2n ...
...
(1, n + 1)
(2, n + 1)
..
.
...
...
(1, n)
(2, n)
..
.
(1, 1∗ )
(2, 1∗ )
..
.
(1, 2∗ )
(2, 2∗ )
..
.
...
...
(n, 2)
(1∗ , 2)
(2∗ , 2)
..
.
...
...
...
(n, n)
(1∗ , n)
(2∗ , n)
..
.
(n, 1∗ )
(1∗ , 1∗ )
(2∗ , 1∗ )
..
.
(n, 2∗ )
(1∗ , 2∗ )
(2∗ , 2∗ )
..
.
...
...
...
(n∗ , 2) . . .
(n∗ , n)
(n∗ , 1∗ )
(n∗ , 2∗ ) . . .
−a2,n
0
−1
..
.
0
. . . a1,n
. . . a2,n
..
..
.
.
...
0
...
0
...
0
..
..
.
.
1
0
..
.
0
1
..
.
...
...
..
.
0
0
−b1,2
..
.
0
b1,2
0
..
.
...
...
...
..
.
−1
−b1,n
−b2,n
...
...
(1, n + 2)
(2, n + 2)
..
.
(n, n)
(n, n + 1)
(n, n + 2)
(n + 1, n) (n + 1, n + 1) (n + 1, n + 2)
(n + 2, n) (n + 2, n + 1) (n + 2, n + 2)
..
..
..
.
.
.
(2n, n)
(2n, n + 1)
(2n, n + 2)
(1, 2)
(2, 2)
..
.
a1,2
0
..
.
a1,2n
a2,2n
..
.
E
E AB
+
E
=
= B
−B
(1, n∗ ) (2, n∗ ) ..
.
∗ (n, n ) (1∗ , n∗ ) (2∗ , n∗ ) ..
.
(n∗ , n∗ )
1 b1,n b2,n .. . 0 0
0
..
.
A = E + AB .
E
Therefore we see that,
det(E + AB) = (1, 2, . . . , n, 1∗ , 2∗ , . . . , n∗ )2 .
11
...
...
...
...
...
...
(n, 2n) (n + 1, 2n)
(n + 2, 2n)
..
.
(2n, 2n) (1, 2n)
(2, 2n)
..
.
5.4
Bordered Determinants
Consider the following,
A = (ai,j )1≤i,j≤n ,
D = |A| = det(A) ,
∆i,j = (−1)i+j D
i
,
j
where ∆i,j is the cofactor of D with respect to ai,j . Then the following expression for Bordered Determinants
holds
a1,1
a2,1
..
.
an,1
y1
a1,2
a2,2
..
.
a1,3
a2,3
..
.
...
...
a1,n
a2,n
..
.
an,2
y2
an,3
y3
...
...
an,n
yn
x1 x2 n
.. = |A|z − X ∆ x y .
i,j i j
. i,j=1
xn z
Expanding the left-hand side with respect to the (n + 1)th column,
n
X
i+n+1 |A|z −
(−1)
i=1
a1,1
a2,1
..
.
a1,2
a2,2
..
.
...
...
a1,n
a2,n
..
.
ai−1,1
i
ai+1,1
..
.
ai−1,2
...
ai−1,n
ai+1,2
..
.
...
ai+1,n
..
.
an,1
y1
an,2
y2
...
...
an,n
yn
xi .
(11)
Now consider the expansion of the determinant above with respect to the nth row
a1,1
..
.
i
.
..
an,1
y1
a1,2
..
.
...
..
.
an,2
y2
a1,n
..
.
..
.
...
...
an,n
yn
a1,1
..
X
.
n
j+n =
(−1)
i
j=1
.
..
an,1
12
a1,2
..
.
...
j
...
..
.
an,2
a1,n
..
.
..
.
...
...
an,n
yj .
Substituting the result above into expression (11) gives
a1,1
..
.
n
X
i+j+1 |A|z +
(−1)
i
.
i,j=1
..
an,1
n
X
= |A|z +
a1,2
..
.
...
j
...
..
.
an,2
a1,n
..
.
..
.
...
...
an,n
xi yj
∆i,j xi yj .
i,j=1
By simply moving the columns we may also rewrite the expression as follows,
z
x1
x2
..
.
xn
y1
a1,1
a2,1
..
.
y2
a1,2
a2,2
..
.
y3
a1,3
a2,3
..
.
...
...
...
an,1
an,2
an,3
...
yn a1,n n
X
a2,n = |A|z −
∆i,j xi yj .
.. i,j=1
. an,n The equation above will prove to be fundamental when computing the derivative of a determinant expression.
Next we consider a special case. Let z = 1 and employ Gaussian elimination to simplify. Hence
1
x1
x2
..
.
xn
y1
a1,1
a2,1
..
.
y2
a1,2
a2,2
..
.
y3
a1,3
a2,3
..
.
...
...
...
an,1
an,2
an,3
...
a1,1 − x1 y1
a2,1 − x2 y1
=
..
.
an,1 − xn y1
yn 1
a1,n 0
a2,n =
.. 0
. ..
.
an,n 0
a1,2 − x1 y2
a2,2 − x2 y2
..
.
...
...
an,2 − xn y2
...
y1
a1,1 − x1 y1
a2,1 − x2 y1
..
.
y2
a1,2 − x1 y2
a2,2 − x2 y2
..
.
...
...
...
an,1 − xn y1
an,2 − xn y2
...
yn
a1,n − x1 yn a2,n − x2 yn ..
.
an,n − xn yn a1,n − x1 yn n
X
a2,n − x2 yn ∆i,j xi yj .
= |A| −
..
.
i,j=1
an,n − xn yn Now we impose the following constrains,
• The matrix entries are antisymmetric such that aj,i = −ai,j .
• xi = yi .
Thus we see that, ∆j,i = −∆i,j . Then the following holds for the second term in (12),
n
X
∆i,j xi yj =
i,j=1
n
X
∆i,j yi yj = −
i,j=1
n
X
i,j=1
∆j,i yj yi = −
n
X
∆i,j yi yj = 0 .
i,j=1
Therefore,
a1,1 − y1 y1
a2,1 − y2 y1
..
.
an,1 − yn y1
a1,2 − y1 y2
a2,2 − y2 y2
..
.
...
...
an,2 − yn y2
...
a1,n − y1 yn a2,n − y2 yn = |A| = (1, 2, . . . , 2n)2 .
..
.
an,n − yn yn 13
(12)
From this we see that adding yi yj to each entry ai,j does not change the value of the pfaffian corresponding to
the determinant of A.
6
6.1
Pfaffian Formulae
Pfaffian Identities
We first consider the following basic identity,
M
X
(−1)j (b0 , b1 , . . . , bbj , . . . , bM )(bj , c0 , c1 , . . . , cN )
j=0
N
X
=
(−1)k (b0 , b1 , . . . , bM , ck )(c0 , c1 , . . . , cbk , . . . , cN ) .
(13)
j=0
We prove this by first expanding (bj , c0 , c1 , . . . , cN ) with respect to bj
l.h.s. =
M
N
X
X
(−1)j (b0 , b1 , . . . , bbj , . . . , bM )
(−1)k (bj , ck )(c0 , . . . , cbk , . . . , cN )
j=0
=
k=0
M X
N
X
(−1)j+k (b0 , . . . , bbj , . . . , bM )(bj , ck )(c0 , . . . , cbk , . . . , cN ) .
j=0 k=0
Now expand (b0 , b1 , . . . , bM , ck ) with respect to ck ,
r.h.s. =
N
X
(−1)k
(−1)j (b0 , . . . , bbj , . . . , bM )(bj , ck )(c0 , . . . , b
ck , . . . , cN )
j=0
k=0
=
M
X
M
N X
X
(−1)k+j (b0 , . . . , bbj , . . . , bM )(bj , ck )(c0 , . . . , b
ck , . . . , cN ) .
k=0 j=0
Since the l.h.s. and r.h.s. are equal, identity (13) holds. Next, in order to derive another identity, we first
impose the following constrains for identity (13).
M = 2n, N = 2m + 2n − 2 ,
b0 = a1 , b1 = 1, b2 = 2, . . . , bM = b2n = 2n ,
c0 = a2 , c1 = a3 , c2 = a4 , . . . , c2m−2 = a2m ,
c2m−1 = 1, c2m , . . . , cN = c2m+2n−2 = 2n .
Then identity (13) may be rewritten as,
2n
X
(−1)j (a1 , 1, 2, . . . , b
j, . . . , 2n)(j, a2 , a3 , . . . , a2m , 1, 2, . . . , 2n)
j=a1 ,1,...
=
2m+2n−2
X
(−1)k (a1 , 1, 2, . . . , 2n, ak+2 )(a2 , a3 , . . . , a
bk+2 , . . . , a2m 1, 2, . . . , 2n) .
k=0
14
For the l.h.s. we have,
(1, 2, . . . , 2n)(a1 , a2 , . . . , a2m , 1, 2, . . . , 2n) +
2n
X
(−1)j (a1 , 1, 2, . . . , b
j, . . . , 2n)(j, a2 , a3 , . . . , a2m , 1, 2, . . . , 2n) .
j=1
However, notice that the pfaffian (j, a2 , a3 , . . . , a2m , 1, 2, . . . , 2n) goes to zero for all j’s.
For the r.h.s. we see that,
=
2m−2
X
(−1)k (a1 , 1, 2, . . . , 2n, ak+2 )(a2 , a3 , . . . , a
bk+2 , . . . , a2m 1, 2, . . . , 2n)
k=0
=
2m
X
(−1)k (a1 , 1, 2, . . . , 2n, ak )(a2 , a3 , . . . , a
bk , . . . , a2m 1, 2, . . . , 2n) .
k=2
Setting the l.h.s. and r.h.s. equal to each other we obtain the following identity,
(1, 2, . . . , 2n)(a1 , a2 , . . . , a2m , 1, 2, . . . , 2n)
=
2m
X
(−1)k (a1 , 1, 2, . . . , 2n, ak )(a2 , a3 , . . . , a
bk , . . . , a2m 1, 2, . . . , 2n) .
k=2
Additionally, we impose yet another set of constrains on identity (13).
M = 2n − 2, N = m + 2n − 1 ,
b0 = 1, b1 = 2, b2 = 3, . . . , bM = b2n−2 = 2n − 1 ,
c0 = a1 , c1 = a2 , c2 = a3 , . . . , cm−1 = am ,
cm = 1, cm+1 , . . . , cN = cm+2n−1 = 2n .
The l.h.s. of identity (13) becomes,
l.h.s. =
2n−2
X
(−1)j (1, 2, . . . , b
j+1 , . . . , 2n − 1)(j + 1, a1 , a2 , . . . , am , 1, 2, . . . , 2n) .
j=0
Here we see that the l.h.s. becomes zero for all values of j.
Now we consider the r.h.s.,
r.h.s. =
m+2n−1
X
(−1)k (1, 2, . . . , 2n − 1, ak+1 )(a1 , a2 , . . . , b
ak+1 , . . . , am , 1, 2, . . . , 2n) .
k=0
Note that for k = m, . . . , m + 2n − 2 the r.h.s. goes to zero. Thus we may rewrite the expression above as
follows,
m−1
X
(−1)k (1, 2, . . . , 2n − 1, ak+1 )(a1 , a2 , . . . , b
ak+1 , . . . , am , 1, 2, . . . , 2n)
k=0
+ (1, 2, . . . , 2n − 1, 2n)(a1 , a2 , . . . , am , 1, 2, . . . , 2n − 1) .
15
Setting the two sides equal to each other and moving the second term to the left we see that,
m−1
X
(−1)k+1 (1, 2, . . . , 2n − 1, ak+1 )(a1 , a2 , . . . , b
ak+1 , . . . , am , 1, 2, . . . , 2n)
k=0
= (1, 2, . . . , 2n)(a1 , a2 , . . . , am , 1, 2, . . . , 2n − 1) .
Next we fix the index k such that k = 1, . . . , m,
m
X
(−1)k (1, 2, . . . , 2n − 1, ak )(a1 , a2 , . . . , b
ak , . . . , am , 1, 2, . . . , 2n)
k=1
=
m
X
(−1)k+1 (ak , 1, 2, . . . , 2n − 1)(a1 , a2 , . . . , b
ak , . . . , am , 1, 2, . . . , 2n) .
k=1
Finally we obtain the following new identity,
(1, 2, . . . , 2n)(a1 , a2 , . . . , am , 1, 2, . . . , 2n − 1)
m
X
=
(−1)k+1 (ak , 1, 2, . . . , 2n − 1)(a1 , a2 , . . . , b
ak , . . . , am , 1, 2, . . . , 2n) .
k=1
6.2
Expansion Formula for (a1 , a2 , 1, 2, . . . , 2n)
Let (a1 , a2 ) = 0. Then the n + 1th order pfaffian (a1 , a2 , 1, 2, . . . , 2n) can be expanded in the following two
ways,
(a1 , a2 ,1, 2, . . . , 2n)
X
=
(−1)j+k−1 (a1 , a2 , j, k)(1, 2, . . . , b
j, . . . , b
k, . . . , 2n) ,
(14)
1≤j<k≤2n
(a1 , a2 ,1, 2, . . . , 2n)
=
2n
X
h
i
(−1)j (a1 , a2 , 1, j)(2, 3, . . . , b
j, . . . , 2n) + (1, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n) .
(15)
j=2
Expansion Formula (14)
By expanding the pfaffian (a1 , a2 , 1, 2, . . . , 2n) by the first term a1 , and then by the second term a2 , we have
that
(a1 , a2 , 1, 2, . . . , 2n) =
2n
X
(−1)j (a1 , j)(a2 , 1, 2, . . . , b
j, . . . , 2n)
j=a2 ,1,...
=
2n
X
(−1)j (a1 , j)(a2 , 1, . . . , b
j, . . . , 2n)
j=1
=
2n
X
(−1)j (a1 , j)
j=1
=
2n X
2n
X
2n
X
(−1)k (a2 , k)(1, . . . , b
j, . . . , b
k, . . . , 2n)
k=1
(−1)j+k (a1 , j)(a2 , k)(1, . . . , b
j, . . . , b
k, . . . , 2n) .
j=1 k=1
16
Notice that in the equation above j 6= k. Otherwise the pfaffian (1, . . . , b
j, . . . , b
k, . . . , 2n) would have an oddnumber of elements. Consider the case where j = 1 and k = 2 such that,
(−1)1+2 (a1 , 1)(a2 , 2)( b
1, b
2, 3, . . . , 2n) .
j k
Similarly, there is also the case where j = 2 and k = 1,
(−1)2+1 (a1 , 2)(a2 , 1)( b
2, b
1, 3, . . . , 2n) .
j k
We would like to factor the two expressions above,
1, b
2, 3, . . . , 2n) .
(−1)1+2 (a1 , 1)(a2 , 2) − (a1 , 2)(a2 , 1) ( b
j k
With this in mind, we only consider the sums where j < k such that,
(a1 , a2 ,1, 2, . . . , 2n)
h
i
X
=
(−1)j+k (a1 , j)(a2 , k) − (a1 , k)(a2 , j) (1, . . . , b
j, . . . , b
k, . . . , 2n) .
1≤j<k≤2n
Next, consider the relation,
(a1 , a2 , j, k) = −(a1 , j)(a2 , k) + (a1 , k)(a2 , j) .
Therefore we obtain expansion formula (14),
(a1 , a2 ,1, 2, . . . , 2n)
X
=
(−1)j+k−1 (a1 , a2 , j, k)(1, 2, . . . , b
j, . . . , b
k, . . . , 2n) .
1≤j<k≤2n
Expansion Formula (15)
Expand pfaffian (a1 , a2 , 1, 2, . . . , 2n) with respect to the index 1 so that we get
(a1 , a2 ,1, 2, . . . , 2n)
=
2n
X
(−1)j (1, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n)
j=a1 ,a2 ,2,...
= (1, a1 )(a2 , 2, 3, . . . , 2n) − (1, a2 )(a1 , 2, 3, . . . , 2n)
+
2n
X
(−1)j (1, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n) .
j=2
17
We expand the r.h.s. even further and seek to simplify the expression,
= (1, a1 )
2n
X
(−1)j (a2 , j)(2, 3, . . . , b
j, . . . , 2n)
j=2
− (1, a1 )
2n
X
(−1)j (a2 , j)(2, 3, . . . , b
j, . . . , 2n)
j=2
+
2n
X
(−1)j (1, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n)
j=2
=
2n
X
(−1)j
h
(1, a1 )(a2 , j) − (1, a2 )(a1 , j) (2, 3, . . . , b
j, . . . , 2n)
j=2
i
+ (1, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n) .
Notice that by the relation,
(a1 , a2 , 1, j) = −(a1 , 1)(a2 , j) + (a1 , j)(a2 , 1)
= (1, a1 )(a2 , j) − (1, a2 )(a1 , j) ,
we obtain expansion formula (5),
(a1 , a2 ,1, 2, . . . , 2n)
=
2n
X
h
i
(−1)j (a1 , a2 , 1, j)(2, 3, . . . , b
j, . . . , 2n) + (1, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n) .
j=2
Later on these formulas will come in useful to understand the derivative of pfaffian expressions.
6.3
The Addition Formula
We introduce a new special pfaffian (1, 2, . . . , 2n)c with special entries (i, j)c given by,
(i, j)c = (i, j) + λ(a, b, i, j) .
Also, let (a, b) = 0. Then we claim that for the special pfaffian above there is an addition formula such that,
(1, 2, . . . , 2n)c = (1, 2, . . . , 2n) + λ(a, b, 1, 2, . . . , 2n) .
where λ is a parameter.
Proof.
We see that for n = 1, equation (16) holds,
(1, 2)c = (1, 2) + λ(a, b, 1, 2) .
18
(16)
Next we assume that equation (16) holds for the n + 1th order special pfaffian,
(2, 3, . . . , b
j, . . . , 2n)c = (2, 3, . . . , b
j, . . . , 2n) + λ(a, b, 2, 3, . . . , b
j, . . . , 2n) .
(17)
By expanding equation (16) with respect to index 1 we have that,
2n
X
(1, 2, . . . , 2n)c =
(−1)j (1, j)c (2, 3, . . . , b
j, . . . , 2n)c .
(18)
j=2
Substituting equation (17) into equation (18) yields the following expression,
(1, 2, . . . ,2n)c =
2n
h
i
X
(−1)j (1, j)c (2, . . . , b
j, . . . , 2n) + λ(a, b, 2, . . . , b
j, . . . , 2n)
j=2
=
2n
X
h
ih
i
(−1)j (1, j) + λ(a, b, 1, j) (2, . . . , b
j, . . . , 2n) + λ(a, b, 2, . . . , b
j, . . . , 2n)
j=2
=
2n
X
h
j, . . . , 2n) + λ (a, b, 1, j)(2, . . . , b
j, . . . , 2n) + (1, j)(a, b, 2, . . . , b
j, . . . , 2n)
(−1)j (1, j)(2, . . . , b
j=2
+ λ2 (a, b, 1, j)(a, b, 2, . . . , b
j, . . . , 2n) .
(19)
In the equation above, the coefficient of λ0 is simply the expansion formula for (1, 2, . . . , 2n) with respect to
the first index,
2n
X
(−1)j (1, j)(2, 3, . . . , b
j, . . . , 2n) = (1, 2, . . . , 2n) .
j=2
For the coefficient of λ, we recall an expansion formula from the previous section,
(a1 , a2 ,1, 2, . . . , 2n)
=
2n
h
i
X
(−1)j (a1 , a2 , 1, 2)(2, 3, . . . , b
j, . . . , 2n) + (i, j)(a1 , a2 , 2, 3, . . . , b
j, . . . , 2n) .
(20)
j=2
Thus we see that the coefficient of λ becomes,
2n
X
h
i
(−1)j (a, b, 1, 2)(2, 3, . . . , b
j, . . . , 2n) + (i, j)(a, b, 2, 3, . . . , b
j, . . . , 2n)
j=2
= (a, b, 1, 2, . . . , 2n) .
In order to simplify the coefficient of λ2 , we first consider the following pfaffian expanded using formula (20),
(a, b, 1, 2, . . . , 2n, 2n + 1, 2n + 2)
2n+2
h
X
=
(−1)j (1, j)(a, b, 2, . . . , b
j, . . . , 2n, 2n + 1, 2n + 2)
j=2
i
+ (a, b, 1, j)(2, 3, . . . , b
j, . . . , 2n, 2n + 1, 2n + 2) .
19
Now let a = 2n + 1 and b = 2n + 2 such that,
2n
X
0=
(−1)j (a, b, 1, j)(2, 3, . . . , b
j, . . . , 2n, a, b)
j=2
2n
X
=
(−1)j (a, b, 1, j)(a, b, 2, 3, . . . , b
j, . . . , 2n) .
j=2
Therefore, expression (19) becomes the desired addition formula,
(1, 2, . . . , 2n)c = (1, 2, . . . , 2n) + λ(a, b, 1, 2, . . . , 2n) .
(21)
The addition formula (21) will come up when looking for difference analogues of soliton equations and in
looking for difference formulae for pfaffians.
6.4
Pfaffian Derivatives
For an arbitrary determinant, A ≡ det(ai,j )1≤i,j≤n , the x-derivative is given by
∂
A=
∂x
X
1≤i,j≤n
∂
ai,j ∆i,j .
∂x
(22)
where ∆i,j is the cofactor of A. Now let D be the determinant of an antisymmetric matrix with entries (i, j),
D = det (i, j)
.
1≤i,j≤2n
Then D is the square of a pfaffian such that
D = (1, 2, . . . , 2n)2 .
We derive a formula for the derivative of a pfaffian by substituting D into (22),
∂
D=
∂x
X
1≤i,j≤2n
∂
(i, j)∆i,j .
∂x
For the l.h.s. we have that
∂
∂
D=
(1, 2, . . . , 2n)2
∂x
∂x
∂
= 2 (1, 2, . . . , 2n)
(1, 2, . . . , 2n) .
∂x
20
Since,
∆i,j = Γ(i, j) (1, 2, . . . , 2n) ,
the r.h.s. becomes
X
1≤i,j≤2n
∂
(i, j) Γ(i, j) (1, 2, . . . , 2n) .
∂x
Setting both sides equal to each other yields,
∂
1
(1, 2, . . . , 2n) =
∂x
2
X
1≤i,j≤2n
∂
(i, j) Γ(i, j) .
∂x
(23)
Next, by the properties of the cofactor Γ(i, j), we have that
X
1≤i,j≤2n
∂
(i, j) Γ(i, j) =
∂x
X
1≤i<j≤2n
X
=
1≤i<j≤2n
=2
∂
(i, j) Γ(i, j) +
∂x
X
1≤j<i≤2n
∂
(i, j) Γ(i, j) + (−1)2
∂x
X
1≤i<j≤2n
∂
(j, i) Γ(j, i)
∂x
X
1≤i<j≤2n
∂
(i, j) Γ(i, j)
∂x
∂
(i, j) Γ(i, j) .
∂x
Therefore, from equation (23) we obtain a formula for the derivative of a pfaffian given by
∂
(1, 2, . . . , 2n) =
∂x
X
1≤i<j≤2n
∂
(i, j) Γ(i, j) .
∂x
Next we consider the following N th order determinant
τN = det(ai,j )1≤i,j≤N ,
where
Z
x
ai,j ≡ ci,j +
fi gj dx .
Then the derivative of τN is
∂
τN =
∂x
X
1≤i<j≤N
21
fi gj ∆i,j .
Recall the following expression for Bordered Determinants
z
x1
x2
..
.
xn
y1
a1,1
a2,1
..
.
y2
a1,2
a2,2
..
.
y3
a1,3
a2,3
..
.
...
...
...
an,1
an,2
an,3
...
yn a1,n n
X
a2,n = |A|z −
∆i,j xi yj .
.. i,j=1
. an,n Consider the case where z = 0,
0
x1
x2
..
.
xn
y1
a1,1
a2,1
..
.
y2
a1,2
a2,2
..
.
y3
a1,3
a2,3
..
.
...
...
...
an,1
an,2
an,3
...
yn a1,n n
X
a2,n = −
∆i,j xi yj .
.. i,j=1
. an,n Then τN x can be expressed similarly,
∂
τN
∂x
0
−f1
= −f2
..
.
−fN
g1
a1,1
a2,1
..
.
g2
a1,2
a2,2
..
.
g3
a1,3
a2,3
..
.
...
...
...
aN,1
aN,2
aN,3
...
gN a1,N a2,N .
.. . aN,N (24)
Before computing the second derivative, let us recall another identity of Bordered Determinants,
0
z1
..
.
zn
y1
a1,1
..
.
...
...
an,1
...
0
yn 0
a1,n x
.. − 1
. ..
.
an,n xn
0
z1
= x2
..
.
xn
0
0
b1
..
.
y1
c1
a1,1
..
.
...
...
...
yn
cn
a1,n
..
.
bn
an,1
...
an,n
y1
b1 c 1
a2,1
..
.
...
...
...
an,1
...
0
yn x1
b1 cn a2,n + · · · + ..
.
.. xn−1
. zn
an,n
y1
a1,1
..
.
...
...
an−1,1
b1 cn
...
...
.
an−1,n bn cn yn
a1,n
..
.
Consider the case where bi cj = 0,
0
z1
..
.
zn
y1
a1,1
..
.
...
...
an,1
...
0
yn z
a1,n 1
x
.. = 2
. ..
.
an,n xn
y1
0
a2,1
..
.
...
...
...
an,1
...
0
yn x1
0 a2,n + · · · + ..
.
.. xn−1
. zn
an,n
22
y1
a1,1
..
.
...
...
an−1,1
0
...
...
.
an−1,n 0 yn
a1,n
..
.
(25)
Next we compute the second derivative of τN using basic properties of determinants.
∂2
τN
∂x2
0
∂ −f1
=
.
∂x ..
−fN
0
−f1
= .
..
−fN
0
−f1
= .
..
−fN
0
−f1
= .
..
−fN
g1
a1,1
..
.
...
...
aN,1
...
g1 x
a1,1
..
.
...
...
aN,1
...
g1 x
a1,1
..
.
...
...
aN,1
...
g1 x
a1,1
..
.
...
...
aN,1
...
gN a1,N .. . aN,N 0
gN x −f
a1,N 1 x
.. + −f2
. ..
.
aN,N −fN
g1
a1,1 x
a2,1
..
.
0
gN x −f
a1,N 1 x
.. + −f2
. ..
.
aN,N −fN
0
gN x −f
a1,N 1 x
.. + −f2
. ..
.
aN,N −fN
...
...
...
aN,1
...
g1
f1 g1
a2,1
..
.
...
...
...
aN,1
...
g1
0
a2,1
..
.
...
...
...
aN,1
...
0
gN −f1
a1,N x a2,N + · · · + ..
.
.. −fN −1
. −fN
aN,N x
0
gN −f1
f1 gN a2,N + · · · + ..
.
.. −fN −1
. −fN
aN,N
x
0
gN −f1
0 a2,N + · · · + ..
.
.. −fN −1
. −fN
aN,N
x
g1
a1,1
..
.
...
...
aN −1,1
aN,1 x
...
...
g1
a1,1
..
.
...
...
aN −1,1
fN g1
...
...
g1
a1,1
..
.
...
...
aN −1,1
0
...
...
aN −1,N aN,N gN
a1,N
..
.
x
aN −1,N fN gN gN
a1,N
..
.
.
aN −1,N 0 gN
a1,N
..
.
Now let us rewrite identity (25) using the notation from the expression above
0
f1 x
..
.
fN
x
g1
a1,1
..
.
...
...
aN,1
...
0
gN f
a1,N 1 x
.. = f2
. ..
.
aN,N fN
g1
0
a2,1
..
.
...
...
...
aN,1
...
0
gN f1
0 a2,N + · · · + ..
.
.. fN −1
. fN
aN,N x
g1
a1,1
..
.
...
...
aN −1,1
0
...
...
.
aN −1,N 0 gN
a1,N
..
.
Thus we see that,
∂2
τN
∂x2
0
−f1
= .
..
−fN
g1 x
a1,1
..
.
...
...
aN,1
...
gN x 0
a1,N −f1 x
.. + ..
. .
aN,N −fN
x
g1
a1,1
..
.
...
...
aN,1
...
gN a1,N .. .
. aN,N In a similar way we may compute the third and fourth derivatives. However, we see that these calculations
can be, to some extent, quite tedious. By expressing τN as a pfaffian we can obtain the same results with easier
calculations.
23
First we start by defining the pfaffian entries,
Z
∗
ai,j = (i, j ) = c +
x
fi gj dx ,
∂n
gj ,
∂xn
∂n
(d∗n , i) =
fi ,
∂xn
(i, j) = (i∗ , j ∗ ) = 0 ,
(dn , j ∗ ) =
(d∗m , dn ) = (dn , i) = (d∗m , j ∗ ) = 0 .
where m, n = 1, 2, . . . . Notice that these entries are quite similar to that of a wronskian determinant expressed
as a pfaffian. Then from the section on general determinants, we take that τN may be expressed as follows,
τN = (1, 2, . . . , N, N ∗ , . . . , 2∗ , 1∗ ) .
The first derivative of a pfaffian entry becomes,
∂
(i, j ∗ ) = fi gj = (d0 , d∗0 , i, j ∗ ) .
∂x
We claim that the first derivative of τN is given by,
∂
τN = (d0 , d∗0 , 1, . . . , N, N ∗ , . . . , 1∗ ) .
∂x
Proof.
Recall the expansion formula,
(a1 , a2 , 1, 2, . . . , 2n)
X
=
(−1)j+k−1 (a1 , a2 , j, k)(1, 2, . . . , b
j, . . . , b
k, . . . , 2n) .
1≤j<k≤2n
Using the formula above and the fact that,
∂
(i, j ∗ ) = (d0 , d∗0 , i, j ∗ ) ,
∂x
we have that,
(d0 , d∗0 , 1, . . . , N, N ∗ , . . . , 1∗ ) =
X
(−1)i+j+1 (d0 , d∗0 , i, j) (1, . . . , bi, . . . , b
j, . . . , 1∗ )
1≤j<k≤1∗
=
X
(−1)i+j+1
1≤j<k≤1∗
=
X
1≤j<k≤1∗
∂
(i, j) (1, . . . , bi, . . . , b
j, . . . , 1∗ )
∂x
∂
(i, j)Γ(i, j)
∂x
∂
(1, 2, . . . , N, N ∗ , . . . , 2∗ , 1∗ )
∂x
∂
=
τN .
∂x
=
24
Next we compute the derivative of a pfaffian of the form,
(dm , d∗n , 1, . . . , N, N ∗ , . . . , 1∗ ) .
(26)
Recall the formula for the derivative of a pfaffian,
∂
(1, 2, . . . , 2n) =
∂x
X
1≤i<j≤2n
∂
(−1)i+j+1 (1, . . . , bi, . . . , b
j, . . . , 2n)
(i, j) .
∂x
Thus, for the derivative of pfaffian (26) we have that,
∂
(dm , d∗n , 1, . . . , N, N ∗ , . . . , 1∗ ) =
∂x
X
(−1)i+j+1 (dm , d∗n , 1, . . . , bi, . . . , b
j, . . . , 1∗ )
dm ≤i<j≤1∗
∂
(i, j) .
∂x
We expand the sum above into three sums by fixing i for dm , d∗n and 1, . . . , 1∗ respectively.
X
∂
(−1)j (d∗n , 1, . . . , b
j, . . . , 1∗ )
(dm , j)
∂x
∗
1≤j≤1
i=dm
+
X
∂ ∗
(dn , j)
(−1)j+1 (dm , 1, . . . , b
j, . . . , 1∗ )
∂x
∗
1≤j≤1
i=d∗
n
X
+
1≤i<j≤1∗
∂
(i, j)
(−1)i+j+1 (dm , d∗n , 1, . . . , bi, . . . , b
j, . . . , 1∗ )
∂x
∗
=
1
X
(−1)j (d∗n , 1, . . . , b
j, . . . , 1∗ ) (dm+1 , j)
j=1
∗
+
1
X
(−1)j+1 (dm , 1, . . . , b
j, . . . , 1∗ ) (d∗n+1 , j)
j=1
+
X
(−1)i+j+1 (dm , d∗n , 1, . . . , bi, . . . , b
j, . . . , 1∗ ) (d0 , d∗0 , i, j) .
1≤i<j≤1∗
Now we recall the expansion formulas below,
(1, 2, . . . , 2n) =
2n
X
(−1)j (1, j) (2, 3, . . . , b
j, . . . , 2n)
j=2
(a1 , a2 , 1, 2, . . . , 2n)
X
=
(−1)j+k−1 (a1 , a2 , j, k)(1, 2, . . . , b
j, . . . , b
k, . . . , 2n) .
1≤j<k≤2n
Therefore the derivative formula of pfaffian (26) is given by,
∂
(dm , d∗n , 1, . . . , N, N ∗ , . . . , 1∗ ) = (dm+1 , d∗n , 1, . . . , N, N ∗ , . . . , 1∗ ) + (dm , d∗n+1 , 1, . . . , N, N ∗ , . . . , 1∗ )
∂x
+ (d0 , d∗0 , dm , d∗n , 1, . . . , N, N ∗ , . . . , 1∗ ) .
25
Form now on we use the following abbreviation in a pfaffian,
(. . . , 1, . . . , N, N ∗ , . . . , 1∗ , . . . ) = (. . . , τN , . . . ) .
Now we can quite easily compute the derivatives of τN ,
τN = (1, . . . , N, N ∗ , . . . , 1∗ ) ,
∂
τN
∂x
∂2
τN
∂x2
∂3
τN
∂x3
∂4
τN
∂x4
= (d0 , d∗0 , τN ) ,
= (d1 , d∗0 , τN ) + (d0 , d∗1 , τN ) ,
= (d2 , d∗0 , τN ) + 2(d1 , d∗1 , τN ) + (d0 , d∗2 , τN ) ,
= (d3 , d∗0 , τN ) + (d0 , d∗3 , τN ) + 3(d2 , d∗1 , τN ) + 3(d1 , d∗2 , τN ) + 2(d0 , d∗0 , d1 , d∗1 , τN ) ,
..
.
By simply expressing the pfaffian expressions above as determinants, we see that these results are identical
to those obtained previously. For example, using the pfaffian definitions, the first derivative of τN becomes,
∂
τN
∂x
(d0 , d∗0 )
(1, d∗0 )
∗
= (d0 , d0 , τN ) = .
..
(N, d∗ )
0
(d0 , 1∗ ) . . .
(1, 1∗ ) . . .
..
.
(N, 1∗ )
...
Which coincides with determinant (24).
26
(d0 , N ∗ ) 0
(1, N ∗ ) −f1
= ..
..
.
.
(N, N ∗ ) −fN
g1
a1,1
..
.
...
...
aN,1
...
gN a1,N .. .
. aN,N