= Solution
Use the <Levi-Civita connection> $\nabla$ of the <Riemannian metric> $g$ and fix the convention
$$
R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,\qquad R_4(X,Y,Z,W)=g(R(X,Y)Z,W).
$$
The <Leibniz rule> shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of $\Lambda^2T^*M\otimes\operatorname{End}(TM)$, the <Riemann curvature tensor>, and $R_4$ is a covariant <tensor field>. This convention gives positive <sectional curvature> on the round sphere.
Its algebraic symmetries are
$$
\begin{aligned}
R_4(X,Y,Z,W)&=-R_4(Y,X,Z,W),\\
R_4(X,Y,Z,W)&=-R_4(X,Y,W,Z),\\
R(X,Y)Z+R(Y,Z)X+R(Z,X)Y&=0,\\
R_4(X,Y,Z,W)&=R_4(Z,W,X,Y).
\end{aligned}
$$
The first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the <Levi-Civita connection> with the <Riemannian metric>. The left side is zero and the differentiated terms combine to
$$
g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0.
$$
For the <first Bianchi identity>, use <normal coordinates> at a point $p$, so the <Christoffel symbols> vanish there. Torsion-freeness gives $\Gamma^\ell_{ij}=\Gamma^\ell_{ji}$, and
$$
R(\partial_i,\partial_j)\partial_k\big|_p
=(\partial_i\Gamma^\ell_{jk}-\partial_j\Gamma^\ell_{ik})\partial_\ell\big|_p.
$$
The cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.
To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The <first Bianchi identity> and the two antisymmetries give
$$
A=R_{bcda}+R_{cadb}
=2B-R_{dbca}-R_{adcb}.
$$
The <first Bianchi identity> applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.
There is also the differential <second Bianchi identity>,
$$
(\nabla_XR)(Y,Z)+(\nabla_YR)(Z,X)+(\nabla_ZR)(X,Y)=0.
$$
For its proof take commuting coordinate fields, normal at $p$. The <Jacobi identity> for operator commutators says $[\nabla_X,[\nabla_Y,\nabla_Z]]+\text{cyclic}=0$. On these commuting fields, $[\nabla_Y,\nabla_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.
For a two-dimensional plane $\sigma=\operatorname{span}\{X,Y\}\subset T_pM$, define the <sectional curvature> by
$$
\boxed{K(\sigma)=\frac{g(R(X,Y)Y,X)}{|X|^2|Y|^2-g(X,Y)^2}.}
$$
The denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an <orthonormal basis> of $\sigma$, it is simply $g(R(X,Y)Y,X)$.
The <Ricci curvature> is the trace
$$
\operatorname{Ric}(X,Y)=\operatorname{tr}\{Z\mapsto R(Z,X)Y\}
=\sum_{i=1}^n g(R(e_i,X)Y,e_i),
$$
where $(e_i)$ is any <orthonormal basis> of $T_pM$. This is independent of the basis because it is a trace. Applying the <first Bianchi identity> to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\operatorname{Ric}(X,Y)-\operatorname{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence <Ricci curvature> is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an <orthonormal basis>,
$$
\boxed{\operatorname{Ric}(v,v)=\sum_{i=2}^n K(\operatorname{span}\{v,e_i\}).}
$$
For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X|^2$. This determines the entire <Ricci curvature> by the <polarization identity>:
$$
\operatorname{Ric}(X,Y)=\tfrac14\bigl(\operatorname{Ric}(X+Y,X+Y)-\operatorname{Ric}(X-Y,X-Y)\bigr).
$$
Finally, the <scalar curvature> is the metric trace of <Ricci curvature>. Taking its diagonal entries in an <orthonormal basis> and pairing the equal contributions from $(i,j)$ and $(j,i)$ yields
$$
\boxed{\operatorname{Scal}=\sum_i\operatorname{Ric}(e_i,e_i)=2\sum_{1\leq i<j\leq n}K(\operatorname{span}\{e_i,e_j\}).}
$$
In particular, constant <sectional curvature> $K_0$ gives $\operatorname{Ric}=(n-1)K_0g$ and $\operatorname{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.
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