Page 364 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY
P. 364
Unit 29: Geodesics
Proposition 7. Let be a geodesic, and let be a variation of through geodesics. Then the Notes
generator Y = of is a Jacobi field.
.
Proof. As before, denote X and Y We first prove the following identity:
Y , X X K Y g(X,Y)X .
Indeed, in the proof of Lemma 7, it was seen that the left-hand side above is a tensor, i.e., is linear
over functions, and hence depends only on the values of the vector fields X and Y at one point.
Fix that point. If X and Y are linearly dependent, then both sides of the equation above are zero.
Otherwise, X and Y are linearly independent, and it suffices to check the inner product of the
identity against X and Y. Taking inner product with X, both sides are zero, and equation (11)
implies that the inner products with Y are equal. Since X X 0, we get:
0 X X Y X Y , X X X Y K Y g(X,Y)X .
Y
X
X
Thus, Y is a Jacobi field.
We see that Jacobi fields are infinitesimal generators of variations through geodesics. If there is
a non-trivial fixed endpoint variation of through geodesics, then the endpoints of are conjugate
along . Unfortunately, the converse is not true but nevertheless, a non-zero Jacobi field which
vanishes at the endpoints can be perceived as a non-trivial infinitesimal fixed-endpoint variation
of through geodesics. This makes the next proposition all the more important.
Proposition 8. Let be a geodesic, and let Y be a Jacobi field. Then, for any vector field Z along
, we have:
I(Y,Z) g Y,Z a b . ...(16)
In particular, if either Y or Z vanishes at the endpoints, then I(Y,Z) = 0.
Proof. Multiplying the Jacobi equation by Z and integrating, we obtain:
b
0 a g Y,Z K g(Y,Z) g ,Y g ,Z dt
b d
Y,Z g Z K g(Y,Z) g ,Y g ,Z dt
a dt g Y,
Thus, a Jacobi field which vanishes at the endpoints lies in the null space of the index form I
acting on vector fields which vanish at the endpoints.
Theorem 2. Let :[a,b] (U,g) be a geodesic parametrized by arc length, and suppose that
there is a point (c) with a < c < b which is conjugate to (a). Then there is a vector field Z along
such that I(Z) < 0. Consequently, is not locally-length minimizing.
Proof. Define:
Y a t c
V
0 c t b
and let W be a vector field supported in a small neighborhood of c which satisfies
Y(c) 0. We denote the index form of on [a, c] by I , and the index form on [c, b] by
W(c) 1
I . Since V is piecewise smooth, we have, in view of (16):
2
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