- The zero components of the metric connection [Ken, p.195, l.-10].

Proof. Consider G^{m}_{sr }(m, s, r = 0, 1, 2, 3) [64 cases]. There are 13 cases for the non-zero components of the metric connection [Ken, p.195, l.-9-l.-5]. If any two of {m, s, r} are distinct, then G^{m}_{sr}= 0 [24 cases].- m = 0
- s
= 0, 1, 2, 3.

G^{0}_{ss}= 0 [4 cases]. - s = 2, 3.

G^{0}_{s0}= G^{0}_{0s}= 0 [4 cases].

- s
= 0, 1, 2, 3.
- m = 1
- s
= 0, 2, 3.

G^{1}_{s1}= G^{1}_{1s}= 0 [6 cases].

- s
= 0, 2, 3.
- m = 2
### s = 0, 1, 2.

G^{ 2}_{ss}= 0 [3 cases].### s = 0, 3.

G^{2}_{s2}= G^{2}_{2s}= 0 [4 cases].

- m = 0

### s = 0, 1, 2, 3.

G^{ 3}_{ss}= 0 [4 cases].### s = 0.

G^{3}_{s3}= G^{3}_{3s}= 0 [2 cases].

Remark 1. We should not check the cases haphazardly because there are too many cases. Disorganization may easily lead to an incomplete or ineffective count. Only after a systematic and complete count may we discover how easy it is to leave some cases out.

Remark 2. (Taking advantage of the system's special features) Our task is to check that the claim in [Ken, p.197, l.12-l.19] is correct. It is easier than establishing the claim from scratch. We first assume that it is correct and infer that the nonzero elements are included in cases a = g or a = d. Once we obtain this piece of intelligence, we can easily design a strategy to classify the cases effectively. In other words, in the following solution we will classify the cases with this feature and [Ken, p.189, (B.3)] in mind so that our analysis will follow the texture of the Riemann curvature tensor.

Remark 3. We display nontrivial cases in order to show cancellations. A nontrivial case means that at least one of the four terms on the right-hand side of [Ken, p.74, (7.3)] is not zero.

Solution. Consider G

- Consider G
^{ a}_{bad }, where (d ¹ a) [48 cases].

R^{1}_{010}= -mZ/r^{3}; R^{2}_{020}= R^{3}_{030}= mZ/2r^{3}.

R^{0}_{101}= m/r^{3}Z; R^{2}_{121}= R^{3}_{131}= -m/2r^{3}Z.

R^{0}_{202}= R^{1}_{212}= -m/2r; R^{3}_{232}= m/r.

R^{0}_{303}= R^{1}_{313}= -m sin^{2}q /2r; R^{2}_{323}= m sin^{2}q /r.

- b ¹ d.
- b = a [12 cases].
- b
¹ a
[4´2´3
= 24 cases].

R^{3}_{132}= G^{ 3}_{12, 3}- G^{ 3}_{13}_{, 2}+ G^{ 3}_{23}G^{2}_{12}- G^{ 3}_{32}G^{3}_{13}

= 0 - 0 + (cot q)(1/r) - (cot q)(1/r).

R^{3}_{231}= G^{ 3}_{21, 3}- G^{ 3}_{23}_{, 1}+ G^{ 3}_{23}G^{2}_{21}- G^{ 3}_{31}G^{3}_{23}

= 0 - 0 + (cot q)(1/r) - (1/r)(cot q).

- G
^{ a}_{bda}= -G^{ a}_{bad }[Ken, p.189, (B.3)],_{ }where (d ¹ a) [48 cases]. - G
^{ a}_{bgd}= 0, where a ¹ g, d ¹ a and d ¹ g [4´4´3´2 = 96 cases].- a =1.

R^{1}_{332}= -R^{1}_{323}= G^{ 1}_{32, 3}- G^{1}_{32, 2}+ G^{ 1}_{33}G^{3}_{32}- G^{1}_{22}G^{2}_{33}

= 0 - (2r/B) sin q cos q + (-r sin q/B) cot q - (-rB)(-sin q cos q) = 0. -
a =2.

R^{2}_{331}= -R^{2}_{313}= G^{ 2}_{31, 3}- G^{ 2}_{33, 1}+ G^{ 2}_{33}G^{3}_{31}- G^{ 2}_{21}G^{2}_{33}

= 0 - 0 + (-sin q cos q)(1/r) - (1/r)(-sin q cos q) = 0. - a =3.

R^{3}_{312}= -R^{3}_{321}= G^{ 3}_{32, 1}- G^{ 3}_{31, 2}+ G^{ 3}_{31}G^{3}_{32}- G^{3}_{32}G^{ 3}_{31}

=0 - 0 + (1/r) cot q - (cot q)(1/r) = 0.

- a =1.