The curvature operator is diagonal in an orthonormal coframe $\{e^a\}$ when every <curvature 2-form> satisfies $\mathcal R^{ab}=K_{ab}e^a\wedge e^b$. Its matrix on the corresponding basis of <2-forms> is then diagonal.

Diagonal curvature operator

Substituting the forms from part (b) into <Cartan's second structure equation> gives the independent mixed-index <curvature 2-forms>
$$
\boxed{\Theta^0{}_1=\frac{2m}{r^3}e^0\wedge e^1,
\qquad
\Theta^0{}_2=-\frac{m}{r^3}e^0\wedge e^2,
\qquad
\Theta^0{}_3=-\frac{m}{r^3}e^0\wedge e^3},
$$
$$
\boxed{\Theta^1{}_2=-\frac{m}{r^3}e^1\wedge e^2,
\qquad
\Theta^1{}_3=-\frac{m}{r^3}e^1\wedge e^3,
\qquad
\Theta^2{}_3=\frac{2m}{r^3}e^2\wedge e^3}.
$$
Every form is proportional to its corresponding basis <2-form>, so the curvature operator is diagonal. The diagonal <Ricci tensor> components are contractions of these sectional curvature coefficients. In each case the coefficients cancel in the pattern $2-1-1=0$, with the Lorentzian sign included when the time direction is contracted. Hence
$$
\boxed{R_{\mu\nu}=0}.
$$
This is consistent with <Schwarzschild spacetime> being a vacuum solution away from $r=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 4 / d / Solution

A 2-form is an alternating covariant tensor field of rank two. Curvature forms and electromagnetic field strengths are standard examples.


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 Incoming links: 2-form