Riemann Tensor on a Sphere

The unit sphere is curved and R = gijRij should be constant. So lets compute R𝜌σμν, Rij and R. Below r is the angle from the north pole.

ds2 = dr2 + (sin r)22. (r, θ) = (x1, x2).
gij =
10
0(sin r)2
.
g22,1 = 2cos(r)sin(r) and other gab,c’s are 0.
Γa bc = ½(∂cgab + ∂bgca − ∂agbc)
Γ1 22 = −cos(r)sin(r) and Γ2 12 = Γ2 21 = cos(r)sin(r); others are 0.
Γabc = gax Γx bc
Γ122 = −cos(r)sin(r) and Γ221 = Γ212 = cot(r); other Γabc = 0.
R𝜌σμν = ∂μΓ𝜌νσ − ∂νΓ𝜌μσ + Γ𝜌μλΓλνσ − Γ𝜌νλΓλμσ
We enumerate the non-zero terms above:
1Γ122 = sin(r)2−cos(r)2 (+=> R1212; −=> R1221)
1Γ212 = −1 − cot(r)2 (+=> R2211; −=> R2211)
1Γ221 = −1 − cot(r)2 (+=> R2112; −=> R2121)
Γ𝜌μλΓλνσ:
Γ111Γ111 + Γ112Γ211 = 0
Γ111Γ112 + Γ112Γ212 = 0
Γ111Γ121 + Γ112Γ221 = 0
Γ111Γ122 + Γ112Γ222 = 0

Γ121Γ111 + Γ122Γ211 = 0
Γ121Γ112 + Γ122Γ212 = −cos(r)2 (+=> R1221; −=> R1212)
Γ121Γ121 + Γ122Γ221 = −cos(r)2 (+=> R1122; −=> R1122)
Γ121Γ122 + Γ122Γ222 = 0

Γ211Γ111 + Γ212Γ211 = 0
Γ211Γ112 + Γ212Γ212 = cot(r)2 (+=> R2211; −=> R2211)
Γ211Γ121 + Γ212Γ221 = cot(r)2 (+=> R2112; −=> R2121)
Γ211Γ122 + Γ212Γ222 = 0

Γ221Γ111 + Γ222Γ211 = 0
Γ221Γ112 + Γ222Γ212 = 0
Γ221Γ121 + Γ222Γ221 = 0
Γ221Γ122 + Γ222Γ222 = −cos(r)2 (+=> R2222; −=> R2222)

In summary:
R1212 = sin(r)2 − cos(r)2 + cos(r)2 = sin(r)2
R1221 = cos(r)2 − sin(r)2 −cos(r)2 = − sin(r)2
R2112 = −1 − cot(r)2 + cot(r)2 = −1
R2121 = 1 + cot(r)2 − cot(r)2 = 1

A plausible symmetry. Also R1212 = R2121 = sin(r)2 = − R2112 = − R1221.
Rij = Rkikj =
10
0sin(r)2
.
R = gijRij = 2.