 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/38e/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Define the Ricci tensor by contracting the first and third Riemann indices:

R_{α γ} = R^{μ}_{αμ γ} .

Contract the first Bianchi identity in its upper index and third lower index. The middle term vanishes by antisymmetry in the first pair, while antisymmetry in the final pair turns the last term into

- R_{γ α}

. Thus

R_{α γ} - R_{γ α} = 0,

so the Ricci tensor is symmetric.

For the stated constant-curvature form,

R_{α γ} R = g^{μ β} R_{μα β γ} = (n - 1) K g_{α γ}, = g^{α γ} R_{α γ} = n (n - 1) K .

\nabla^{α} R_{α β} = \frac{1}{2} \nabla_{β} R .

Using

\nabla g = 0

gives

(n - 1) \partial_{β} K = \frac{1}{2} n (n - 1) \partial_{β} K,

\frac{( n - 1 ) ( n - 2 )}{2} \partial_{β} K = 0.

R_{α β} = 3 K g_{α β}, R = 12 K, K is constant .

More generally,

K

is constant for every

n > 2

n = 2

, the contracted Bianchi identity gives no such conclusion, and the Gaussian curvature may vary from point to point.

 New to topics? Read the docs here!