Page 365 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes 2



1
2
1
I(V,W) I (V,W) I (V,W) I (Y,W)      Y(c)  0
It follows that:
I(V + W, V + W) = I(V) + 2I(V,W) +  I(W) = 2I(V,W) +  I(W)
2
2
is negative if  > 0 is small enough. Although V + W is not smooth, there is for any
2 Z 2 C uniformly in  > 0, which differs from

 > 0 a smooth vector field Z , satisfying Y      
V + W only on (c – , c + ). Since the contribution of this interval to both I(V + W, V + W) and

 I Z ,Z tends to zero with , it follows that also  I Z ,Z    0 for  > 0 small enough. Thus,  is



not locally energy-minimizing. Since it is parametrized by arc length, if it was locally length
minimizing, it would by Lemma 6 also be locally energy-minimizing. Thus,  cannot be locally
length-minimizing.
A partial converse is also true: the absence of conjugate points along  guarantees that the index
form is positive definite.
Theorem 3. Let :[a,b]  (U,g) be a geodesic parametrized by arc length, and suppose that no
point  (t), a < t  b, is conjugate to  (a) along . Then the index form I is positive definite.

Proof. Let X    and let Y be a Jacobi field which is perpendicular to X, and vanishes at t = a.
,
Note that the space of such Jacobi fields is 1-dimensional, hence Y is determined up to sign if we

also require that Y(a)  1. Since Y is perpendicular to X, it satisfies the equation:

  X Y KY 0.


X
Furthermore, since Y never vanishes along , the vectors X and Y span T  (t) U for all t  (a, b].
Thus, if Z is any vector field along  which vanishes at the endpoints, then we can write Z = fX +
hY for some functions f and h. Note that f(a) = f(b) = h(b) = 0 and hY(a) = 0. We then have:
I(Z,Z) = I(fX, fX) + 2I(fX, hY) + I(hY, hY).


Since R(X, fX,X, fX) = 0 and  X fX  fX, it follows from (3.31) that:

b b


2 
I(fX,fX)   a  g fX,fX dt   a f dt.
Furthermore,
b

I(fX, hY)   a  g fX, X hY  dt



b

 a 
  g fX,hY |  a b  g  X fX,hY  dt    a b  g fX,hY dt  0.
2
2
Finally, since  X hY   g  X Y, X h Y  h Y , 2  2 it follows from Proposition 3.20 that:
b 2 b 2
2
2 

I(hY,hY)   a h Y dt I(Y,hY)   a h Y dt.

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