
Geometry 
of 
the Melnikov Vector 
107 
defined  by 
(6.1) 
b(7) 
= 
dim 
Wa(z-) 
- 
dim 
Wa(z+). 
Notice from (5.10)  and diagram 
1 
in 
$5 
that the dimension 
m 
of  the Melnikov  vector 
is given  by 
m 
= 
n 
- 
[{n 
- 
dimWa(z-)} 
+ 
dimWa(z+) 
- 
k]. 
Thus 
m, 
k 
and 
b(7) 
satisfy the following relation. 
(6.2) 
m 
= 
k 
+ 
b(7). 
The splitting index 
6(7) 
expresses the 'degeneracy'  of 
7 
in  the following  sense. 
Here we  distinguish the homoclinic and heteroclinic cases. 
(i) Suppose that 
7 
is 
a 
homoclinic orbit.  Then the splitting index 
b(y) 
is always 
zero.  Thus  by  theorem  5.5(ii),  all  of 
k 
column vectors in 
[& 
n 
M&M] 
must  be 
nonzero to guarantee  the transversal intersection.  This situation  is  only  the 
case 
in 
the homoclinic case. 
(ii)  Suppose that 
7 
is 
a 
heteroclinic orbit  in 
R". 
In  this  case  we  have  three 
different situations. 
(ii-1) 
b(7) 
> 
0. 
Then 
m 
= 
k 
+ 
b(7) 
> 
k. 
Thus  theorem  5.5(ii) implies  that 
there is no transversal intersection because the matrix 
[&!&$A41 
is 
of 
size 
m 
x 
k. 
A 
reason  for  this  is  that  dimWa(z+) 
< 
dimWa(z-)  is  equivalent  to  saying  that 
dimWU(z-) 
+ 
dimWa(z+) 
< 
n. 
(ii-2) 
b(7) 
= 
0. 
In this case we  have the same situation 
as 
in the homoclinic case. 
(iii-3) 
6(7) 
< 
0. 
Then 
m 
= 
k+b(~) 
< 
k. 
Thus transversal intersection is possible. 
In this way, we  can classify the possibility and impossibility of  transversal intersection 
by  using the splitting index 
b(7).