Page 360 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 29: Geodesics

The second identity now follows by taking inner product with W. Notes
Let

R ijkl  R  , ,   k , l  ,
i
j
z
i
i
i
i
be the components of the Riemann tensor. The previous proposition shows that if X  x  i ,Y  y  i ,Z   i ,W  w  i ,
z
i
,
i
i
i
X  x  i ,Y  y  i ,Z   i ,W  w  then
i
j
i
k
l
R(W, Z, X, Y)  w z x y R ijkl ,
that is, the value of R(W, Z, X, Y) at a point u depends only on the values of W, Z, X, and Y at u.
Proposition 3. The components R of the Riemann curvature tensor of any metric g satisfy the
ijkl
following identities:
R ijkl   R ijlk   R jikl  R klij ...(1)
R ijkl  R iljk  R iklj  0. ...(2)
Proof. We first prove (2). Since  k , l  0, it suffices to prove
,



 k , l     k      l ,   0. ...(3)


j 
k
j
j
l
m
m
Note that the symmetry    imply that      .
lj
j l
j
l
jl
Thus, we can write:

,
           j .
j
j
k
k
l
k
l
l
Permuting the indices cyclically, and adding, we get (3.19). The first identity in (3.23) is obvious
from Definition 5. We now prove the identity:
R = R .
jikl
ijkl
We observe that:
     
 g  k  j , i    k  g   j ,  l j , k i 
i
l
l
,
  k  g l ij   g  j ,  i    l  g    i   g  j ,   i 
l
k
j
k
l
   l g   k  g  j ,  i    l  g  j ,  i   g  j ,   i  .
l
k
l
k
k
ji
It is easy to see that the first term, and the next two taken together, are symmetric in k and l.
Thus, interchanging k and l, and subtracting, we get:

,
R ijkl    g  k , l  ,  i   g  j  ,  l ,  i   g   l  ,    R jikl .
   
j
k
i
j
k
The last identity in (1) now follows from the first two and (2). We prove that B = R – R = 0.
klij
ijkl
ijkl
Note that B satisfies (1) as well as B = –B . Now, in view of the identities already established,
ijkl
ijkl
klij
we see that:
R = –R – R = –R – R = R + R – R = B + R ,
ljik
likj
iklj
iklj
lkji
iklj
iljk
ljik
klij
ijkl
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