2. TH E TODA HIERARCHY
Explicitly, one obtains from (2.11),
/i = - 6 + ci,
pi = -2a
2
- c
2
,
/
2
=
a2
+ (a")
2
+
62
- ci6 + c2,
(2.15) g2 = 2a2(b + 6+) - 2cxa2 - c3,
f3 = _(
a
-)
2
(fe-
+
26) -
a2(6+
+ 26) -
63
+ ci(a2 + (a") 2 + 62) - c26 + c3,
etc.
and hence from (2.13),
(2.16)
m
, ( a-a{b-b+) \
n
(2.17)
T L l ( a
'
6 ) = =
U-2a2(6
+ +
6)
+
2(a-)^ + 6-)J
+ C l
V-2[(a-)
2

2
]J
= 0
'
_T ,
7N
/ d-a[b3-(b+)3+2(a-)26-2(a+)2b++a2(6-b+)-(a+)2b++-(a-)26-]
TL2(a, 6) =
V
'
^_2(a-)2[b2+66-+(6-)2
+
(a-)2
+
(a—)2]+2o2[62+66++(b+)2+o2
+
(a+)
V } \-2a2(b++b)+2(a-)2(b+b-)J \-2[(a-)2-a2}J
etc.
represent the first few equations of the Toda hierarchy. Here Q denote summation
constants which naturally arise by solving the resulting difference equations for
gg+i,i in (2.11). Throughout this exposition we will chose these constants ct to be
real-valued. The corresponding homogeneous equations obtained by taking
all summation constants equal to zero, Q = 0, £ G N, are then denoted by
(2.19)
TL,(a,6):=TL,(a,6)|Toda
eg
and similarly we denote by A?+2 := P2g+2(ce = 0), fj := fj(c£ = 0), Qj := gj(ce =
0) the corresponding homogeneous quantities. One verifies
9
(2.20) P2g+2 = ^2
C9-™P2m+2,
C Q = 1.
ra=0
Next we relate the homogeneous quantities fj, #j to certain matrix elements of
L(ty.
LEMMA 2.1. The homogeneous coefficients {fj(t)}ojg+i and {gj(t)}ojg
satisfy
(2.21) fj(n,t) = (6njL(ty6n), 0 j g + 1, n e Z,
(2.22) gj(n,t) =
-2a(n)(6n+uL{t)j6n),
0jg,neZ,
where 6n = {5n,m}mez.
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