42
J. Zhang and G. Khan
Substituting X → L X, Y → LY in Eq. (3.10) yields
∇ L X (L
2 Y ) − ∇ LY (L
2 X ) = LT
∇
L (L X, LY ) + L[L X, LY ],
or, after cancelling an L on both sides
L(∇ L X Y − ∇ LY X ) = T
∇
L (L X, LY ) + [L X, LY ].
Applying L operator on both sides of Eq. (3.10) and adding to the above, we have
L(∇ X (LY ) − ∇ Y (L X) + ∇ L X Y − ∇ LY X )
= L
2 T
∇
L (X, Y ) + L
2
[X, Y ] + T
∇
L (L X, LY ) + [L X, LY ],
or
L([L X, Y ] + T
∇
(L X, Y ) + [X, LY ] + T
∇
(X, LY ))
= L
2 T
∇
L (X, Y ) + T
∇
L (L X, LY ) + L
2
[X, Y ] + [L X, LY ].
Rearranging the terms yields
−L
2
[X, Y ] + L[L X, Y ] + L[X, LY ] − [L X, LY ]
= L
2 T
∇
L (X, Y ) + T
∇
L (L X, LY ) − LT
∇
(L X, Y ) − LT
∇
(X, LY ).
The lefthand side is nothing but N L (X, Y ). So we obtain Eqs. (3.8) and (3.9).
3.2.3 MC1 Versus MC2
A careful examination of expressions (3.8) and (3.9) reveals two different scenarios under which N L vanishes—these scenarios will be referred to as 1st Matching
Condition (MC1) and 2nd Matching Condition (MC2) of (∇, L).
Definition 1 A pair (∇, L) is said to satisfy
(i) 1st Matching Condition, or MC1, if the following holds:
T
∇
(L X, Y ) + T
∇
(X, LY ) = 0;
(3.11)
(ii) 2nd Matching Condition, or MC2, if the following holds:
T
∇
(L X, Y ) = LT
∇
(X, Y ) + (∇ X L)Y − (∇ Y L)X,
(3.12)
or equivalently,
T
∇
(L X, Y ) = L(T
∇
L (X, Y )).
(3.13)