Page 362 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY
P. 362
Unit 29: Geodesics
Corollary 1. The Riemann curvature tensor of any metric g on a surface is given by: Notes
R ijkl K g g g g jk . ...(11)
ik
il
i l
Proof. Denote the right-hand side of (11) by S , and note that it satisfies (1). Thus, the same
ijkl
comment which follows Proposition 8 applies and the only non-zero components of S are
ijkl
determined by S 1212 :
S 1212 = S 2112 = S 2121 = S 1221 .
In view of (11), we have R 1212 = S 1212 , thus it follows that R = S ijkl
ijkl
In particular, we conclude that:
R(Z,W,X,Y) K g(W,X)g(Z,Y) g(W,Y)g(Z,X) . ...(12)
29.3 The Second Variation of Arc length
In this section, we study the additional condition E (0) 0 necessary for a minimum. This leads
to the notion of Jacobi fields and conjugate points.
Proposition 5. Let :[a,b] U be a geodesic parametrized by arc length on the Riemannian
surface (U, g), and let be a fixed-endpoint variation of with generator Y. Then, we have:
b 2 2 2
E (0) Y K Y g ,Y dt, ...(13)
a
where K is the Gauss curvature of g.
Before we prove this proposition, we offer a second proof of the first variation formula:
E (0) a b g ,Y dt, ...(14)
which is more in spirit with our derivation of the second variation formula. First note that if
is a fixed-endpoint variation of with generator = Y, and with X, then [X, Y] = 0. Here Y
denotes the vector field along rather than just along . Indeed, since X = d (d/dt) and
Y = d(d/ds), it follows, that for any smooth function f on U, we have
d d
,
[X,Y] f dt ds f 0.
,
i
In view of the symmetry i kj this implies:
jk
Y X X Y [X,Y] 0.
We can now calculate:
1 b b
E (s) 2 Y g(X,X)dt a g Y X,X dt a g X ,X dt
Y
b d b b
b
a
a dt g(Y,X)dt a g Y, X X dt g(Y,X) a g Y, X X dt
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