12 2. THE TODA HIERARCHY
as an illustration a few of the polynomials Fg and Gg+i,
FQ
= 1 = Fo,
Gi = bz = Gu
Fi = c i  6  h z = ciF0 + Fi,
(2 40) ° 2 = C l ( ~ 6 " Z ) + ( a _ ) 2 *2 + *2z2= ciGi + G2,
^2 = c2 4 ci(6 + z)fa 2 + (a  ) 2 4 62  6z 4 z2 = c2F0 4 ci A 4 F2,
G3 = c2(b  z) 4 ^((a") 2  a2 4 62  z2) + a26+  (a~)2b~

2(a~)26

63

2a2z

z3
= c2Gx 4 cxG2 4 G3,
etc.
REMARK
2.3. 5mce
6T/
(2.11), (2.34), a enters quadratically in Fg and Gg+i,
the Toda hierarchy (2.13) (respectively (2.38), (2.39),) zs invariant under the sub
stitution
(2.41) a(t)  • a€(t) = {e(n)a(n, t)}„eZ, e(n) G {41,  1 } , n e Z.
This result should be compared with (the last part of) Lemma 3.1 and Lemma ]±.l.
Finally, we specialize to the stationary Toda hierarchy characterized by a =
b = 0 in (2.14) (respectively (2.13)), or more precisely, by commuting difference
expressions
(2.42) [P
2 s + 2
,L]= 0
of order 2g 4 2 and 2, respectively. Equations (2.37) and (2.38) then yield
(2.43) (b + z)(Gg+l  G;+1) = 2(a")
2
F  
2a2F+,
(2.44) G++1+Gg+1 =
2(b++z)Fg+.
Because of (2.42) one computes
[
P
29+2
K
er(L
Z
) ]
= [i2aF9S+
+
G9+l)
l
K
er(Lz)l
={2aFg[G++1 + Gg+1 + 2(6+ +
z)Fg+]S+
(  j +^
+1
4a^F;}
Ker(L
_
z )
=
{Gl+l _ 4 a 2

F
P 
F
9
+
} 
K
e r ( Z , 
2
)
= :
^25+2
A simple calculation, using (2.43) and (2.44) then proves that #29+2 is a lattice
constant and hence a polynomial of degree 2g + 2 with respect to z:
(b + z)(R2g+2  R^g+2)
(2.46) = («»+ z){(Gg+1 + G;+1)(Gg+1  G~+l)  4Fg[a2Fg+  (a") 2 F]}
(2^3)
[G9+1 + G;+1 + 2(6 +
z)Fg}2[a2F+
 (
a
)
2
F" ]
( 2
i
4 )
0.
Thus one infers
23+1
(2.47) R2g+2{Z) = [ J (Z  Em), {Fm}0 m2p+1 C C
m=0