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Exercise 7.1

(a) Components of the Riemannian curvature tensor

R ( i , j ) k = i j k j i k [ i , j ] k = 0

R ( i , j ) k = i ( l Γ j , k l l ) j ( l Γ i , k l l )

R ( i , j ) k = l [ i ( Γ j , k l l ) j ( Γ i , k l l ) ]

R ( i , j ) k = l ( [ ( i Γ j , k l j Γ i , k l ) l + m ( Γ j , k l Γ i , l m Γ i , k l Γ j , l m ) m ] ) R i , j , k l l

R i , j , k , l = m g l m R i , j , k m

(b) Simplifications in normal coordinates

In normal coordinates, centered at a point p M :

So the previous expression becomes R i , j , k , l = i Γ j k l j Γ i k l .

The formula page 70 is Γ i , j k = 1 2 l g k l ( i g j l + j g i l l g i , j ) , so we get

m Γ i , j k = 1 2 l m g k l ( i g j l + j g i l l g i , j ) + g k l ( m i g j l + m j g i l m l g i , j )

m Γ i , j k p = 1 2 l 0 ( i g j l + j g i l l g i , j ) + δ k l ( m i g j l + m j g i l m l g i , j )

m Γ i , j k p = 1 2 ( m i g j k + m j g i k m k g i j )

We can rename the indices to get the expected derivative of the Christoffel symbols:

i Γ j , k l p = 1 2 ( i j g k l + i k g j l i l g j k )

j Γ i , k l p = 1 2 ( j i g k l + j k g i l j l g i k )

Finally, substracting this two expressions give:

R i , j , k , l p = i Γ j , k l p j Γ i , k l p = 1 2 ( j l g i k + i k g j l i l g j k j k g i l ) p