The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors

(Point-normal reduction.) Fix $p$ and choose a point-normal chart at $p$ as in [L3]. By [L3], $R_{abcd}(p)=\frac{1}{2}(g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})(p)$. Use this second-derivative formula only in the chosen point-normal chart. By [L2] the lowered curvature is a tensor; each skew-symmetry, pair-exchange and cyclic-sum identity below says that a fixed sum of slot permutations of that tensor vanishes. If its components vanish in the chosen basis, multilinearity makes that tensor zero at $p$, hence its components vanish in every chart. The same applies to $\mathrm{Ric}-{\mathrm{Ric}}^{\mathsf{T}}$ and $G-G^{\mathsf{T}}$, since metric contraction gives tensors by [L2]. Thus it suffices to verify these particular identities in the point-normal chart at each $p$; no tensor transformation law for metric second derivatives is asserted.

(Skew-symmetries.) Interchanging $a$ and $b$ in the display of step 1.1 negates it, because each of its four terms is a second partial derivative with one of $a,b$ in a fixed slot: $R_{bacd}(p)=\frac{1}{2}(g_{bd,ac}+g_{ac,bd}-g_{ad,bc}-g_{bc,ad})(p)=-R_{abcd}(p)$. Interchanging $c$ and $d$ likewise negates it. This proves (a), the two operator forms following by expanding $g(R(X,Y)Z,W)=R_{abcd}{W}^a{X}^c{Y}^d{Z}^b$ in the same components.

(Pair exchange.) Substituting $(a,b,c,d)\mapsto (c,d,a,b)$ into the display of step 1.1 turns its four terms into $g_{cb,da}$, $g_{da,cb}$, $-g_{db,ca}$ and $-g_{ca,db}$; by the symmetry of $g$ and [L4] these are $g_{bc,ad}$, $g_{ad,bc}$, $-g_{bd,ac}$ and $-g_{ac,bd}$, that is, the four terms of the original display in permuted order. Hence $R_{cdab}(p)=R_{abcd}(p)$, which proves (b).

(First Bianchi identity.) By step 1.1, $2(R_{abcd}+R_{acdb}+R_{adbc})(p)$ equals the sum of the twelve terms obtained by substituting the four-term formula into the three summands. The terms pair off with opposite signs: $g_{ad,bc}$ cancels $g_{ad,cb}$, $g_{bc,ad}$ cancels $g_{cb,ad}$, $-g_{bd,ac}$ cancels $+g_{bd,ac}$, $-g_{ac,bd}$ cancels $+g_{ac,bd}$, $g_{ab,cd}$ cancels $-g_{ab,cd}$, and $g_{cd,ab}$ cancels $-g_{cd,ab}$, each cancellation using [L4]. Hence $R_{abcd}+R_{acdb}+R_{adbc}=0$ at $p$, and since $p$ was arbitrary this proves (c).

(Ricci symmetry.) By [L1] and step 1.1, ${\mathrm{Ric}}_{bd}(p)=\frac{1}{2}{g}^{ac}(p)(g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})(p)$ and, writing out the same expression with $b$ and $d$ interchanged, ${\mathrm{Ric}}_{db}(p)=\frac{1}{2}{g}^{ac}(p)(g_{ab,dc}+g_{dc,ab}-g_{db,ac}-g_{ac,db})(p)$. Relabelling the summed pair $a,c$ as $c,a$ in the second expression and using $g^{ac}=g^{ca}$ [L5], its first term becomes $g^{ac}{g}_{cb,da}=g^{ac}{g}_{bc,ad}$ by [L4], which is the second term of ${\mathrm{Ric}}_{bd}$; its second term becomes $g^{ac}{g}_{da,cb}=g^{ac}{g}_{ad,bc}$, the first term of ${\mathrm{Ric}}_{bd}$; its third term is $g^{ac}{g}_{bd,ac}$, the third term of ${\mathrm{Ric}}_{bd}$; and its fourth term is $g^{ac}{g}_{ac,bd}$, the fourth term. Hence ${\mathrm{Ric}}_{db}(p)={\mathrm{Ric}}_{bd}(p)$.

(Einstein symmetry and conclusion.) By [L1], $G_{ab}(p)={\mathrm{Ric}}_{ab}(p)-\frac{1}{2}S(p)g_{ab}(p)$ and $G_{ba}(p)={\mathrm{Ric}}_{ba}(p)-\frac{1}{2}S(p)g_{ba}(p)$; step 2.4 and the symmetry of $g$ [L4] give $G_{ab}(p)=G_{ba}(p)$. Since $p$ was arbitrary, (d) holds, and with steps 2.1, 2.2 and 2.3 this completes the proof.