Page 361 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes hence, B = B . Using the symmetries of B , we can rewrite this identity as:
ijkl ljik ijkl
B + B = 0. ...(4)
iklj
ijkl
We now permute the first three indices cyclically:
B + B = 0, ...(5)
kijl
kjli
B + B = 0, ...(6)
jkil
jilk
add (4) to (5) and subtract (6) to get, using the symmetries of B :
ijkl
B + B + B + B – B – B = 2B = 0.
kjli
kjli
iklj
ijkl
ijkl
iklj
iklj
This completes the proof of the proposition.
It follows, that all the non-zero components of the Riemann tensor are determined by R 1212 :
R 1212 = –R 2112 = R 2121 = –R 1221 ,
and all other components are zero. The proposition also implies that for any vectors X, Y, Z, W,
the following identities hold:
R(W, Z, X, Y ) = –R(W, Z, Y, Z) = –R(Z, W, X, Y) = R(X, Y, W, Z), ...(7)
R(W, Z, X, Y) + R(W, Y, Z, X) + R(W, X, Y, Z) = 0. ...(8)
Proposition 4. The components R of the Riemann curvature tensor of any metric g satisfy:
ijkl
g R imkl   ik,l   j il,k       nk . ...(9)
n
mj
n
j
j
j
nl
ik
il
Furthermore, we have:
R
K  1212 , ...(10)
det(g)
where K is the Gauss curvature of g.
Proof. Denote the right-hand side of (9) by S . We have:
j
ikl
 
n
j
     k      j ik ,l    j nl  , j
ik
j
l
i
k
ik
or equivalently:
jm
,
j
n
 j ik ,l     g g     m .
ik
i
k
l
nl
Interchanging k and l and subtracting we get:
jm
S ikl  g jm   g  l , k  ,  m  g R jm milk  g R imkl .
j
i
According (9), we have:
1 1
il
ik
j
ik
K  g S  g g R .
ijkl
2 2
ikj
In view of the comment following Proposition 8, the only non-zero terms in this sum are:
1
11
22
21
12
11
12
22
  R
21
K  g g R  g g R  g g R  g g R   det g  1 ,
2 1212 1221 2112 2121 1212
which implies (10)
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