Ames W.F., Roger C. Nonlinear equations in the applied sciences. Volume 185
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230
Benno Fuchssteiner
(ii):
We again use
a
repeated argument over the order
of
covariance. Consider an arbi-
trary covariant
a
and arbitrary vector fields
K,G,
then:
(d
.
da)
8
A'
G
=
(L~da8
G
-
d(da
8
I;)
G
=
Lli(da
G)
-
da
8
L/<G
-
d(da
8
K)
G
=
L/;(da
C)
-
do
LIiG
-
Lc(da
li')
G
+
d(da
0
Ii'
0
G)
=
LI<LGa
-
L~d(a
8
G)
+
LIG,l,p
+
d(a
0
[Ii,G])
-
LcLh-a
+
Lcd(a
8
lip)
+
d(L~a
8
G)
-
d(d(a
0
K)
0
G)
=
d(-Lii(aoG)-ao[G,Ii]+
Lc(aoli)+(L~;a)8G-
d(a0K)oG)
=
d(Lc(a
8
I<)
-
d(a
8
K)
0
G)
=
d2(a
0
I<
8
G)
Again, the beginning
of
our
induction argument is given by the fact that
0
applied
to
zero-forms gives zero.
Definition
4.3:
(i)
A
tensor
T
is said to be
invariant with respect to the vector field
Ii
if
LKT
=
0.
(ii) Observe that condition
(4.5)
for
y
being closed can
be
written in terms
of
the ezterior
derivative as
dy
=
0.
Therefore we define
a
covariant tensor
a
to
be
closed
ifda
=
0.
Remark
4.4:
(i) Gradients are closed because
of
d
.
d
=
0.
If
C
is
the vector field Lie algebra on some
manifold
M
and
3
are the scalarfields, then locally the converse
is
also true (Poincari
Lemma
([25)].
(ii) Observe that
(4.12)
implies that any closed covariant tensora is invariant with respect
to
li
if
and only
ifa
0
Ii
is again closed.
(iii) Let
J
be
some invertibie +-linear operator J
:
L
+
C*.
If
LcJ
=
0
for
all
those
G
with
closed
JG
then J itself mtist be closed.
Proof
of
(ii) and
(iii):
(ii): Direct consequence of
(4.12).
(iii): Because
L*
is
generated by its closed elements and since
J
is invertible we have that
C
is the 3-linear hull of those
G
in
L
such that
JG
is
closed.
For
f
E
.F
we obtain
L(j,,)
J
-
d(
Jfli)
=
f
(Li, J
-
d(
JIi)).
Therefore
LK J
-
d( Jli)
vanishes for
all
li
if it
vanishes for the subset
of
those
G
such that
JG
is closed.
So
the assumption on
J
shows
that the right side
of
(4.12)
vanishes. Hence we have
d(J)
0
Ii
=
0
for all
li.
H
Hamiltonian
Structure
and Integrability
231
Now, let
0
:
L*
-
L
be some antisymmetric T-linear map. We define
a
bracket
in
C*
in the following way:
{YI,TZ}Q
:=
4o7,)72
-
L(Q~)YI
t
d
<
YI,@~Z
>
(4.14)
for
y1,y~
E
L*.
Using (4.12) we can rewrite (4.14) as
Before presenting the basic result
for
these brackets we gather some useful identities.
{YIJZ~Q
:=
L(O~~)TZ
-
(dYI)m(@Yz).
(4.15)
SO
when
y1
is
closed
then by application
of
(4.12) this bracket reduces to
IYI,YZ}O
:=
L(Q~,)~z for closedrl
E
C.
.
(4.16)
Furthermore we easily find with (4.9) how this bracket acts on multiplication with elements
of
3
{%.f?Z}Q
:=
f{'Yl,'YZ}Q
t
(L(Q7,)f)
72
.
(4.17)
Let
0
:
C'
-,
C
be
.F-linear and antisymmetric, then the following
are
Theorem
4.5:
equivalent:
(
i)
{
,
}Q
defines
a
Lie algebra among fhe closed elements
of
C'
ii)
~{YI,
yz}~
=
[Odyl
,
Odyz] for
all
closed yl,
yz
E
C*.
(iii)
{
,
}Q
defines
a
Lie algebm in
C'
(
iv)
WYI,
YZ}Q
=
[%I,
OTZ]
for
all
-11,
YZ
E
L*
(
v) LQ,(~)
=
0
for
all
closed
y
E
C*.
(
vi)
For
arbitrary
y
E
C*
uJe have
that
0
is
invariant with respect to
07
if
and only
if
dy
*(O-yl)
(072)
=
0
for
all
y1,yz
E
C'.
Proof
In the following we omit, for simplicity, the sub-0 in the brackets
{
}@.
(i)
that
(ii): Observe t1ia.t by use
of
(4.14) then the antisymmetry
of
0
implies for closed
y's
{yl,rZ}
=
L(Qd71)?Z
=
-L(Qd72)71.
(4.18)
So,
{
,
}
is antisymmetric anyway, and only the Jacobi identity has to be proved in order
to show that this is
a
Lie algebra. Using (4.18) we find for the triple bracket
{y3,{y13yZ}}
=
-LQ{,,,n}73
=
L(E)73)L(@71)y2
=
(@~3)~(@n)y1
'
232
Benno
Fuchssteiner
Now, using for the cyclic sum suitable representations obtained from this formula we find
with (4.2)
{?3{%?yZ})
t
=
L((3d{71.n})Y3
-k
L(Qd7~)L(Od7~)%
-
L(0dn)L(Qd71)73
=
{L(e{-,l,n))
-
[L(Q~-,~)(
L(odn)l}h
=
{L(Q{m,x.})
-
L[Qd-r~,0dw1)r3
(4.19)
Since
73
was
arbitrary, the Jacobi identity for the triple product can only hold if
L(Qd{7~.n))
-
L[0d7~.0n1
=
(4.20)
which
is
equivalent to (ii) since
Ii
+
LI~
was assumed to be injective. On the other hand,
whenever (ii) (and (4.20) as
a
consequence) holds then the cyclic sum obviously must be
equal to zero, which implies the Jacobi identity.
The implications
(iv)
-
(ii) and (iii)
-
(i) are obvious.
(ii)
-
(iv) and (i)
+
(iii) are either done by direct computation
or
by using remark 4.1
together with (4.17).
To
see this observe that (i) gives
a
Lie algebra
(Lf,,
{
}),
where
L:,
are
the closed elements
in
L*
and that (4.17) gives
a
Lie algebra homomorphism
7
+
LI,
=
Le7
from
L:,
into the derivations
of
3.
Now
taking the unique extension (described in remark
4.1) to the 3-module generated by
Lf,
(which by definition is equal to
L*),
one obtains the
Lie algebra defined by (4.14). The fact that the homomorphism
0
extends from
L:,
to
L*
is due to the property that
0L;
=
LQ?.
(v)
e
(ii):
By
(4.16) the condition
0L(~?)yl
=
L(e7)(0yl)
SOT
all
closed
y,y~
E
L*
is equivalent to (ii). And by the product rule this
is
equivalent to (v).
(4.21)
(vi)
e
(iv): Using (4.15) and the antisymmetry of
0
we find for arbitrary
-y,-yl,-yz
E
P
<
7l,(LQ70)7Z
>
+dYl
*(@72)*(@7l)
=<
?'19[@7,@721-
@{y,yZ)
>
(4.22).
Equating the right side of (4.22) to zero is equivalent to (iv) and equating its left side to
zero is equivalent to (v).
1
Let us have
a
closer
look
at condition (v): If
0
is
invertible, then the condition therein
imposed on
y
is equivalent to
dy
=
0,
i.e.
y
must be closed. Hence, for
J
:=
0-I
condi-
tion (v) means that
J
is li-invariant
(Ii
E
L)
if and only if
JK
is closed. Looking now
at 4.4 (ii) and (iii) we see that for an invertible
0
one of the equivalent conditions of the
theorem above is fulfilled
if
and only if
J
=
0-I
is
closed. In the finite dimensional theory
Hamiltonian Structure
and
Integrability
233
the antisymmetric closed invertible
J
are called symplectic.
So,
loosely speaking,
0
has
the algebraic behaviour
of
the inverse
of
a
symplectic operator. Therefore, if one
of
the
conditions of Theorem
4.5
is fulfilled,
0
is said to be implectic,
a
name which stands
for
inverse-symplectic. Sometimes instead
of
implectic, the name
Poisson
tensor
is
chosen.
For
reasons which will become obvious in the next section the
{
,
}e
are called the Poisson
brackets with respect to
0.
We decided not to insist on the invertibility condition because of the infinite dimen-
sional nature
of
our manifolds,
so
we have to extend the notion
symplectic
for this more
general situation: An operator
J
:
C
-
L*
is called symplectic if it is antisymmetric and
closed, and if in addition its kernel
ker(J)
=
{G
E
ClJG
=
0)
is
a
Lie ideal in
C.
Being
a
Lie ideal means
of
course that
[Ii,G]
E
ker(J)
for all
G
E
ker(J)
and
K
E
L.
This addi-
tional ideal-condition is automatically fulfilled
if
J
is invertible because then
ker(J)
=
0.
In
a
analogy to implectic operators one can use symplectic operators
J
to construct in
JC
suitable Poisson brackets: Take
y1
,yz
E
JC
and choose
Gl,G2
such that
y,
=
JG,,
i
=
1,2.
Then we define
{Yl,Y2}(J)
:=
k,(Yz)-
k,(Yl)+
<
YI,G
>
'
(4.23)
Rewriting that with
(4.12)
and using
d(J)
=
0
we obtain
{Yi,yz}(J)=
Lc,(Gz)-d(JGi)*Gz
=
JLc,Gz+d(J)~Gi~Gz=J[Gi,Gz].
Since
ker(J)
is an ideal with respect to
[
,
]
the bracket
{
,
}(J)
does not depend on the
choice of
GI
,Gz. Furthermore, because
J
is one-to-one from the equivalence classes modulo
ker(J)
to
JC,
the bracket
{
,
}(J)
must be
a
Lie algebra such that
J
is
a
homomorphism
from
{
,
}(J)
into
[
,
1
We may summarize this section:
An implectic opemtor makes out oft' a Lie algebm
module
(C*,{
,
}e,.F),
with corresponding 7-derivations L;
,y
L*
such that
0
is
a
homomorphism from this Lie algebra module into
(L,[
,
],3).
Hew homomorphism means
that for
all
71~
72
E
C'
we have
0{Yl,YZ}O
=
[OYl,0YZ]
(4.24)
(4.25)
In general, it is highly desirable to construct
for
given tensors suitable Lie algebra elements
which leave these tensors invariant. Indeed, as we will see, the search for symmetries,
conservation laws, inva.riant spectral problems a.nd the like may be subsumed under this
general theme. In view
of
that problem it is obvious that symplectic and implectic tensors
must play
a
fundamental role because for them
the
closed elements in
.P
imediatly give rise
to such invariances.
234
Benno
Fuchssteiner
5
Hamiltonian and Bi-Hamiltonian Fields
In
this section we always assume that
0
is an implectic operator.
For
some
of
the results
presented here the reader is referred to the fundamental papers
([lo,]
[ll,]
[12,] [14,] [15)].
Comparing the different
Poisson
brackets defined in Sections
3
and
4,
one discovers
that they are defined
on
rather different spaces. However, these two notions are easily
connected if,
as
before
in
the concrete situation,
a
bracket
among the elements of
3
is
defined
as
follows:
One easily sees that
V
maps the 3-brackets into the C'-brackets:
V{fi,fZ)0
=
{VfI,Vf210
.
(5.2)
This suggests that these 3-brackets also form
a
Lie algebra. Indeed this
is
the case, the
proof is literally almost the same
as
the proof
(i)
e
(ii)
in the last theorem7.
Definition
5.1:
(i) Elements
li
E
C
which are
of
the form
li
=
0y
with closed
y
and implectic
0
are
called
hamiltonian
(with respect
to
0).
(ii) Given some symplectic
J,
then elements
li
E
C
such that
JlC
is closed are called
inverse-hamiltonians
(with respect to
J).
(iii) Let
0
be
implectic and
J
be closed, and assume that
J
is not the inverse
of
0.
Then
some
li
is called a
bi-hamiltonian field
(with respect to
0
and
J)
if
JIC
is
closed
and
if
there is
some
closed
y
such that
li
=
07.
The power of bi-hamiltonian fields is seen from:
Observation
5.2:
Consider
a
bi-hamiltonian field
6'
=
07,
Jli
=
71,
y
and
71
being
closed
as described above, and define
lin+l
=
(0J)"li.
invariant with respect to
li.
Then all the
lin
and
7,
:=
Jli,
are
Proof
'An interesting question
is
whether
or
not the Jacob; identity
for
the 3-brackets is
a
sufficient condition lo
guarantee that
8
is
implectic. In
most
situations this is indeed the case.
One only needs some kind
of
Poincd
Lemma
to
show that.
'Observe that
lor
invertible
0
or
J
the
notions hamiltonian and inverse-hamiltonian
do
coincide. This is the
readon why in the hite dimensional theory the notion inverse-hamiltonian does not appear. since there sympleclic
lorms
are usually assumed to be non-degenerate.
In
the infinite dimensional situation however, the assumption
of
nondegeneracy is not advised because invertibility
of
operators usually involves topo~ogid considerations and depends
very
much
on the spaces under consideration.
Hamiltonian Structure and Integrability
235
Using the bi-hamiltonian nature of
li
we see from Remark
4.4
(ii) and from Theorem
4.5
(v) that
0
and
J
are invariant with respect to
li.
Hence, by the product rule,
all
K,
have
to be invariant with respect to
li.
Let us now illustrate the notions and techniques of the last chapter in the light of the
concrete situation which was considered in Sections
2
and
3.
Standard situation
:
Let
M
be
a
Cm-manifold, we consider Cw-tensor fields on
M.
In particular let
L
be the
vector fields,
3
be the scalar fields and let
L*
be those Cm-maps from
M
into the cotangent-
bundle which are given by assigning to
u
E
A4
a
continuous linear functional on the tangent
space
T,M
at
u.
In order to carry out computations with Lie derivatives we should show what these
look
like in charts.
As
we already know, for
a
scalar field
f
the Lie derivative is the usual
gradient
LA^
=<
Vf,A
>=
/‘[A]
LAY
=
y‘[A]
+
A’+?
(5.3)
and the application
of
L.4
to a co-vector field
y
is found to be
(5.4)
where
A‘+
denotes the transpose
of
the operator
A’
with respect to the duality between
tangent and cotangent space.
Furthermore, if
0
:
T’Af
-+
TM
and
J
:
TM
+
T‘M
are two-times contravariant
and two-times covariant, respectively, then their Lie derivatives are
LAO
=
@’[A]
-
OA”
-
A’O
LAJ
=
J’[A]
+
A‘+J
+
JA‘
LA@
=
@’[A]
-
A’@
+
@A’
.
(5.5)
(5.6)
Finally the Lie derivative for an operator
@
:
TM
-
TM
is equal to
(5.7)
Fix some
li
E
L,
then
f
E
.F
and
G
E
C
are invariant with respect to
li
if and only
if
f
is
a
conserved quantity and
G
a
symmetry group generator, respectively. Using the
Lie derivatives, we see that
0
is iniplectic
if
and only if condition
3
(ii) of Theorem
3.2
is
fulfilled. In the same way we obtain
as
equivalent condition for
J
being closed that
<
J(u)”G~]G~,G~
>
+
<
J(u)’[G~]G~,GI
>
+
<
J(u)’(G~]G~,Gz
>=
0
(5.8)
must hold for all
GlrG2,G3
E
L.
Hence Section
4
generalizes the situation considered in
Section
3.
Now, we consider again the dynamical system
=
K(u)
.
236
Benno Fuchssteiner
In analogy to Definition 5.1 we call (1.1)
a
bi-hamiltonian system
if there are closed
J(u),
implectic
O(u)
and
closed
yo,y1
such that
I<
=
070
and
JA'
=
y1
.
(5.9)
If that is the case then we obtain as consequence of 5.2
Observation
5.3:
Define inductively
Iil
=
I<
and
Iin+l
=
OJK,
(5.10)
yn+l
=
J%n
(5.11)
then all the
A',,
and
yn
are
invariant with respect
to
K.
Hence the
I<,,
are symmetry group
genemtors
for
(1.1)
and,
if
the
y,,
are gmdients, then they are gmdients
of
conserved quan-
tities
for
the evolution equation
(1.1).
Example
5.4:
Korteweg-de Vries equation
Consider the situation
as
in
Example 3.5. We observe that the inverse
of
D
(D-lJ)(z)
:=
1
f(O@
00
is
a
well defined operator
D-'
:
S
-
9.
The Korteweg-de Vries equation
U(
=
I<@)
:=
GUU,
+
u,,,
(3.8)
is
a
bi-hamiltonian system since for the implectic
0
=
D3
+
2(Du
+
uD) and the
symplectic
J
=
D-I we have that (5.9) is fulfilled when
yo
and
y1
are taken to be the
gradients of
10
and
11
as given
in
(3.10)
and (3.11), respectively. Hence putting
I<,,+]
:=
(@),,I<,
where
@
=
D2
+
2DuD-I
+
221, (5.12)
then all these
li,
are symmetry group generators. Since
8
is
recursively generating sym-
metry group generators, this usually is called
a
recursion operator
[20(]sometimes also
a
strong symmetry,
for example
in
[21,] [5,][3)].
0
Although Observation 5.3 is very useful for constructing symmetry groups, there still re-
mains
a
major problem is the analysis of the system (1.1). Ultimately, we are interested in
constructing suitable coordinates (action variables) for the flow,
so
we need scalar quantities
which are invariant under the flow. Certainly, action variables give rise to symmetry group
generators, however the converse is not always true, because the corresponding co-vector
fields may not be closed. Therefore the question whether
or
not the fields
y,
are closed is
of great importance. If that happens then, by the Poincarb lemma, one can, at least locally,
construct suitable coordinates. And
if,
as
in the example, the manifold under consideration
is
a vector space then even the construction of global potentials is
a
simple excercise:
Lemma
5.5:
Let the manifold
of
all
u's
be
a
vector space.
Then
a
co-vector field
y(u)
is the gradient
of
a
scalar field
F(
u)
if
and only
if
y
is
closed.
Proof
Hamiltonian Structure and Integrability
237
Since gradients are closed we only have to prove the existence of
F
for closed
y.
So
let
y(u)
be closed. Consider the scalar field
F(U)
=
l1
4
y(Xu),u
>
dA.
(5.13)
Take some arbitrary
v(u)
in the tangent bundle. Then by (5.4) we obtain
{A
<
y’(Xu)[v],u
>
+
<
y(Xu),v
>}dX
=
1
{<
Xy‘(Xu)(u],v
>
+
<
y(Xu),v
>}dX
(5.14)
=
J1
L{<
Xy(Xu),v
>}dX
=
<
y(u),v
>
o
dX
This proves
VF(u)
=
y(u)
since
v(u)
was arbitrary
H
Observe that this proof can be generalized to the situation where the manifold can be
parametrized by
a
star-shaped subset of
a
vector space.
6
Compatibility
In this section we introduce the essential structure which will be responsible for the fact
that in most cases of bi-hamiltonian systems the invariant co-vector fields constructed by
use
of
Observation 5.2 are indeed closed. Again we will consider the necessary arguments
at
a
rather general level. This is not done in order to introduce an unreasonable level of
abstraction but rather to get rid
of
unnecessary ballast. Further, it turns out that with a
general formulation of
compatibility
in
Lie
algebras one is more flexible with respect to
applications. In our presentation we follow closely
[9.]
Let,
(C,[
,
1)
be some Lie algebra over
R
or
(c
.
We call
(C,[
,
1)
the
reference
algebra.
Let furthermore
A
be
a
vector space over the same scalars. Now, consider
a
linear
transformation
T:A-C.
from
A
into the reference algebra
C.
We call some (bilinear) product
{
,
}
in
A
(not
necessarily assumed to be
a
Lie product)
a
T-product
if
T
is
a
homomorphism into
(C,
[
,
I),
i.e. if
T{a,b}
=
[Tn,Tb] for
all
a,b
E
A.
(6.1)
To emphasize that some product is
a
T-product we write
[
,
IT
instead of
{
}.
A linear
@
:
C
-t
C
is said to be
hereditary
if
[n,b]a
:=
[@a,b]
+
[a,Qb]
-
Q[a,b]
(6.2)
defines
a
@-product in
L.
Using then
@[a,b]o
=
[@a,Ob],
we see that
@
is hereditary if and
only if
@*[a,b]
+
[@a,@b]
=
@{[@a,b]+
[a,@b]}
for
all
a,b
E
C.
(6.3)
238
Benno Fuchssteiner
Rephrasing
a
notion introduced earlier we call
a
linear map
8
:
L
+
L
invariant
with
respect to
k
E
L
if
8[k,b]
=
[k,Qb]
for
all b
E
L
(6.4)
Theorem
6.1:
Let
8
be
a hereditary ma
which is invariant with res ect to
k.
Then
{8"k(n
E
No}
is an abelian subset
of
(L,
[I).
If
8
is
invertible then &"kln
E
Z
}
is
abelian as well.
For the proof we need:
Lemma
6.2:
Let
8
be
hereditary and let it
be
invariant with respect to
k.
Then
8
is
invariant with respect to
8k
.
If
8
is invertible then
it
is
invariant with respect to
8-'k
as
well. Thus the set
(kJ8
invariant with respect
to
k}
of
all elements which leave
8
invari-
ant is a subalgebra
of
L
which is invariant under the application
of
8
(and
of
8-'
if
8
is
invertible).
Proof
Replace
a
by
k
in
(6.3).
Since
8
is invariant with respect
to
k
the first
and
fourth term
cancel and the equality
reads
[8k,86]
=
8[8k,b]
for
all b
.
(6.5)
This clearly implies that
8
is invariant with respect to
8k.
If
8
is invertible, we replace
a
in
(6.3)
by
8-'k
and apply
8-'
to the remaining two terms.
W
Proof
of
Theorem
6.1:
From the Lemma
6.2
we obtain by induction that
8
is invariant with respect to any
arnk
and
8"k.
Hence using antisymmetry we find
[Ornk.O"k]
=
O"+"[k,k]
=
0
for
aJ1
m,n.
For invertible
a,
in
this argument
8-I
has to replace
8.
W
Remark
6.3:
Let
8
be
hereditary and let al and a2
be
eigenvectors
of
8
,
i.e.
@a,
=
A,a,
,
A,
=scalar,
i
=
1,2.
Then
for
these a, relation
(6.3)
zs
equivalent to
(0
-
XI)(@
-
Az)[al,a2]
=
0.
(6.6)
Hence, when an operator
8
has a spectral resolution, this operator is hereditary
if
and only
zf
all the corresponding spectml projections are algebra homomorphisms.
Now, let
us
return
to
the generd situation
of
maps from
A
into
L.
Assume that in
A
we
have
T-
and "-products
[
,
]T
and
[
,
]u,
respectively. These products are said
to
be
compatible
if
{a,b)
:=
[n,bl~
+
[a,b]+
(6.7)
defines
a
(T
+
")-product in
A.
Lemma
6.4:
Let
6
,
]T
and
[
,
10
be
T-
axd *-products, respectively. These products
are compatible
if
an
only
if
T[a,b]~
+
"[a, b]~
=
[Ta, Qb]
+
["a,Tb]
for
all a,b
E
A
(6.8)
Hamiltonian Structure and Integrability 239
Proof:
Observe that
T[a,b]~
=
[Ta,Tb]
and
q[a,b],p
=
[qa,Ob].
So,
(6.8)
is obviously equivalent
to
(T+q)I[a,bI~+[a,bIq}
=
[(T+q)a,(T+q)bl,
(6.9)
which proves the claim.
Observation
6.5:
Let
A
be
a
scalar. Obviously,
[
,
]AT
defined
by
[a,
b
AT
:=
A[a,
b]~
is
a
and
[
,
]A*,
respectively, we see that
(6.8)
remains valid. In other words,
(6.8)
is linear in
q
(as well as in
T
).
Hence, if
[
,
and
[
,
IT2
are compatible
,
and
if
both are compatible
with
[
,
]T
then
[
,
]AT]
+
[
,
1.~~
is
always compatible with
[
,
IT.
Observation
8.6:
Consider the case when the reference algebm is equal to
A,
i.e.
A
=
L
and
put
T
=
I,
q
=
@.
Furthermole, assume that
[
,
]T
is the given product in
(L,[
I),
and
that
[
,
]a
=
{
}
is a second product such that
0
:
(L,{
})
-,
(L,[
1)
is a homomorphism.
Then
(6.8)
holds
if
and only
if
{
}
is the product defined in
(6.3).
Hence,
0
is hereditary
if
and only
if
(C,
{
,
})
and
(L,[
,
1)
are compatible.
In
order to shorten
our
notions
we
call
q
and
T
compatible
if
their
I-
and
T-
products,
[
,
]q
and
[
,
]T
are compatible. By application
of
this notion to the special case
of
hered-
itary operators
@1,@2
we see that and
@2
are compatible
if
and only
@I
t
02
is again
hereditary.
Theorem
6.7:
Consider
maps
T,
'4f
:
A
+
L
and their cormsponding products
[
,
17-
and
[
,
],p.
Assume that
q
is invertible. Then
T
and
Q
are compatible
if
and only
if
@
=
Tq-'
is hereditary.
Proof
Define
a
second product in
L
by
(AT)-product whenever
[
,
IT
is
a
T-product. Now, replactng zn
(6.8)
4
and
[
,
]a
by
A'#
{a,b}
:=
q[I-'a,q-'L]~ for a,b
E
L
.
Then
@
:
(L,{
})
-
(L,[
,
1)
is a homomorphism. Using the definition of
{
,
}
and
(6.1)
(for
@
instead of
T)
we obtain
(It
@)([a,b]
t
{a,b})
=
(T
+
q)q-'([a,b]
+
{a,b))
=
(T
+
'P)(['J-'a,I-'b]q
+
[q-'a,@-'b]~).
For
general
a,
b
E
L
the right side is equal to
[(T
+
q)q-'a,(T
+
q)q-'b]
if and only if
T
and are compatible, but this expression is equal to
[(I
+
0)a,(I
t
@)b
and equal to the
left side if and only
if
I
and
0
are compatible. Hence the compatibility of
T
and
'$
and
that of
I
and
@
are equivalent.
Now
using Observation
6.6
we see that this is equivalent to
0
being hereditary.
H
By similar arguments we find:
Observation
6.8:
I
do commute.
polynomial in
0
is hereditary.
Let
0,
'€'
be conipotible hereditary operators and assume that
0
and
Then
0q
is hereditary. As a consequence,
if
@
be
hereditary, then
any