10 2. TH E TODA HIERARCHY

PROOF .

We abbreviate

(2.23) fj(n) = («n,L$n), g3{n) = -2a{n){8n+1,U 8n).

Then

(2.24) /

i + 1

= (L6n,IS6n) = -\{9,+9j) ~ bf3

and similarly,

(2.25) gj+l = -b9j - 2 a

2

/ / + h =

-b+gj

- 2a

2

/, + h+,

where

(2.26) hj{n) = -2a(n)o(n - l)(fin+i,L^«n_i).

Eliminating h3 in (2.25) results in

(2.27) gj+1 - gj+1 = -2[a

2

/+ - (a")

2

/"] - b[9j - gj}.

By inspection, (2.24) and (2.27) are equivalent to (2.11). In order to determine

which solution of (2.11) has been found (i.e., determine the summation constants

ci,.. . , cg) we temporarily assign the weight one to a(n) and 6(n), n G Z. Then j3

and pj+i have weight j and hence

(2.28) c0 = 1, Cj = 0, 1 j $

completing the proof. •

Now we are in the position to reveal the connections with the usual approach

to the Toda equations. It suffices to consider the homogeneous case.

LEMMA 2.2. The homogeneous Lax operator P2g+2 satisfies

(2.29) P2g+2(t) = [L(ty+1}+ - [L{t)»+1].

(cf. the notation in (2.6)).

PROOF.

We use induction on g. g = 0 is trivial. Suppose (2.29) holds for

g = 0,... ,g0 - 1. By (2.10) we have

(2.30) P2go+2(t) = P290(t)L(t) + [ggo(t) + 2a(t)fgo(t)S+] - fgo(t)L(t) + fgo+1(t).

In order to prove (2.29) one considers (5m, P2gQ+2{t)8n) and makes the case distinc-

tions m n - l , m = n - l , m = n,m = n + l , m n + l . Explicitly, one verifies,

for instance, in the case ra = n,

(^m

5

^2

5 o

+2^n )

= (8n, P2go(a6n-i + a~8n+i - b6n)) + ggo(n) + b(n)fgo(n) 4- f9o+i{n)

= (5n, [(£*)+ - (L*°)_](a«n_i 4- a~6

n + 1

- bSn)) + pPo(n) + 6(n)/Po(n) + fgo+i(n)

= (6n, (L*°)+ a-«

n +

i) - (fin, (L5o)_ afin_!) + &0(n) + b(n)f9o(n) + / ,

0 +

i W

=

a(n)(fin,L50fin+i)

- a(n - l)(5n,L*°5n_i) + &0(n) + b(n)fgo(n) + /

9 o +

i(n)

= 2 ^ o W + 2 ^ n ~ ^ + b n^9o(n) + 4)+i( n ) = 0

(2.31)