316 7. THE REDUCED DISTANCE
Hence (7.62) and Y (0) = 0 imply
1 -2 (6f£) (1) -yl;; \7yY,X) + VTRc (Y, Y)I;
= foT VT ( (% 7 Re+ 2 ~ Re) (Y, Y) + ~\7~,YR) dT
+ foT VT(\R(Y,X)Y,X)-1Rc(Y)l^2 ) dT
+ foT VT (-2 (\7y Re) (Y, X) + 2 (\7 x Re) (Y, Y)) dT
+ foT VTl\7xY +Rc(Y)J^2 dT,where we used \7 x Y = \7y X. Let H (X, Y) denote the matrix Harnack
expression
H (X, Y) ~ -2 (%
7Re) (Y, Y) - \7~,YR + 2 JRc (Y)J^2 -t Re (Y, Y)
(7.63) - 2 (R (Y, X) Y, X) - 4 (\7 x Re) (Y, Y) + 4 (\7y Re) (Y, X).
By substituting the definition of H (X, Y) in the above formula, we obtain
(Sf£) (r) - 2yf;; (\7y Y, X) (r) + 2yi;;Rc (Y, Y) (r)
(7.64) = -foT VTH (X, Y) dT + foT 2VT J\7xY +Re (Y)J^2 dT.
An even nicer form isLEMMA 7.37 (£-Second variation - version 2). Let r E (0, T) and let
I: [O, r]---+ M be an £-geodesic. IfY (T) ~ gs Is (T), for a smooth variation
Is of 1, satisfies Y (0) = 0, then
(6f£) (r) - 2yf;; \7yY,X) (r) + 2yi;;Rc (Y, Y) (r) = IY ~1
2(7.65) - foT VTH (X, Y) dT + foT 2VT 1\7 x y +Re (Y) - 2~ Y,
2
dT.PROOF. Sincel\7xY +Rc(Y)- 2 ~Yl
2= IV' x Y + Re (Y) 12 -! (\\7 x Y, Y) + \Re (Y) , Y)) + ~ IY 12
T 4T= J\7xY +Re (Y)J^2 - 2_~ JYl^2 + ~ JYl^2 ,
2T dT 4T