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.