SUB-LAPLACIANS WIT H DRIF T 5
Let / G C°°, let xix) e^6,7r^^ be a multiplicative function on G and let
a = (ai,..., an) and b (61,..., 6n). Then
L(xf)=- E *jXXiXjf- E aa(XiX)Xjf- E M**/)*;X
lijq 1M? l5^?i7
+ X] aixXi(f)- ^2
aijfXiXjX+
Y2
aifXiX
= ~X X^ CLijXiXjf-X Yl aiJbiXjU)~X Y2 aiJh3Xif
lz,jfq l 2 n 1*9
+ x X^
a*x*^
~ -fa X^
a^A'
+ fa X^
aibi
127 l ^ * ? i ^ n l i n
= - x X^
aiJxiXjf+x
X!
Ia* ~2
X!
a*A
I
x
^
+
X
E (ai-2(B6)i)X4/-«6,B6)-ft,a»/x.
l i n
Hence
x-1Lx
=
-(^i2
+ - + ^ ) + E
fa'-2
E
a*A]*i
+ E (ai-2(Bb)i)Xi-((b,Bb)-(b,a))
lin
We want to have
a.i-2 (Bb)i = 0, 1 z n
or else that
a = 2Bb.
Since the matrix B is invertible, this is equivalent to
b= l-B^a
2
For this choice of b the constant term becomes
P = - «b, Bb) - (6, a)) = (b, Bb) 0.
Also, if we set
y= E U -
2
E °*A-)*
niq \ 1J™ /
then 7 El) and
X
-
1
L
X
= -(£
1
2
+ ... + £ 2 ) + F + /3
which proves the lemma.
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