Riemann Curvature and Riemannian Submanifolds

Expanding the first slot by [F1]–[F3] gives $R(fX,Y)Z=f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-Y(f){\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_{[X,Y]}Z+Y(f){\mathrm{\nabla }}_{X}Z\mathrm{.}$ The two derivative-of-$f$ terms cancel, leaving $fR(X,Y)Z$. The same calculation in the second slot gives $R(X,fY)Z=X(f){\mathrm{\nabla }}_{Y}Z+f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_{[X,Y]}Z-X(f){\mathrm{\nabla }}_{Y}Z,$ and hence the second displayed identity.

For the third slot, two uses of the section Leibniz rule yield $R(X,Y)(fZ)=fR(X,Y)Z+(X(Yf)-Y(Xf)-[X,Y]f)Z\mathrm{.}$ The coefficient in parentheses is zero by the defining action of the Lie bracket on functions. Hence the third identity holds. Additivity and real homogeneity already follow from the connection and bracket laws, so these three identities prove separate $C^{\mathrm{\infty }}(M)$-linearity.

If $X^{\mathrm{\prime }}$ is another extension of $u$, take a coordinate frame $E_1,\dots ,E_n$ near $p$ and write $X-X^{\mathrm{\prime }}={\sum }_a{f}^a{E}_a$. Since every $f^a(p)=0$, [F1] gives $(R(X-X^{\mathrm{\prime }},Y)Z{)}_p={\sum }_a{f}^a(p)(R(E_a,Y)Z{)}_p=0$. Repeating this argument in the second and third slots proves independence of all three extensions. The same identities, evaluated at $p$, prove real trilinearity of $(u,v,w)\mapsto R_p(u,v)w$.

On a coordinate neighborhood with frame $E_i$ and dual coframe ${\epsilon }^l$, put $R^l{}_{ijk}={\epsilon }^l(R(E_i,E_j)E_k)$. Each coefficient is smooth because the defining curvature expression applies the connection and Lie bracket to smooth fields. The identity $R_p(u,v)w=R^l{}_{ijk}(p)u^i{v}^j{w}^k{E}_l{\mathrm{\mid }}_p$, obtained from [F1], therefore makes the vector-valued representative smooth. Pairing its output with a covector gives the smooth fibrewise multilinear map ${\mathcal{R}}_p$ above, hence a smooth section of $T_3^{1}M$ by [F2].

Interchanging $X$ and $Y$ in [F1] gives $R(Y,X)Z={\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_{[Y,X]}Z$.

The Lie bracket is a commutator of derivations, so $[Y,X]=-[X,Y]$; the connection is real-linear in its differentiating field. Hence the expression in step 1.1 is $-({\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_{[X,Y]}Z)=-R(X,Y)Z$.

Applying the connection Leibniz rule to [F2] twice gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\nabla }}_{{\mathrm{\partial }}_j}{\mathrm{\partial }}_k=({\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}+{\mathrm{\Gamma }}^m{}_{jk}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{im}){\mathrm{\partial }}_{\mathrm{\ell }}$ and, after interchanging $i,j$, ${\mathrm{\nabla }}_{{\mathrm{\partial }}_j}{\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_k=({\mathrm{\partial }}_j{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ik}+{\mathrm{\Gamma }}^m{}_{ik}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jm}){\mathrm{\partial }}_{\mathrm{\ell }}$.

By [F3], the bracket term in [F1] is zero. Subtracting the two expansions from step 1.1 therefore makes the coefficient of ${\mathrm{\partial }}_{\mathrm{\ell }}$ exactly ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}-{\mathrm{\partial }}_j{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ik}+{\mathrm{\Gamma }}^m{}_{jk}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{im}-{\mathrm{\Gamma }}^m{}_{ik}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jm}$, as claimed.

Expanding $R(fX,Y)s$ with [F1]–[F3] produces $fR(X,Y)s-Y(f){\mathrm{\nabla }}_{X}s+Y(f){\mathrm{\nabla }}_{X}s=fR(X,Y)s$; expanding $R(X,fY)s$ produces $fR(X,Y)s+X(f){\mathrm{\nabla }}_{Y}s-X(f){\mathrm{\nabla }}_{Y}s=fR(X,Y)s$. Interchanging $X,Y$ in [F1] and using $[Y,X]=-[X,Y]$ gives $R(Y,X)s=-R(X,Y)s$.

Applying the section Leibniz rule twice gives $R(X,Y)(fs)=fR(X,Y)s+(X(Yf)-Y(Xf)-[X,Y]f)s=fR(X,Y)s$. Thus evaluation at a point depends only on $X_p,Y_p,s_p$, and the result is alternating in the tangent entries.

On a neighborhood with tangent frame $E_i$ and bundle frame $e_a$, each $R(E_i,E_j)e_a$ is a smooth section because [F1] combines connection derivatives and a Lie bracket of smooth inputs. Its smooth frame coefficients are alternating in $i,j$ by step 1.1 and define a smooth section of ${\bigwedge }^2{T}^{\ast }M\otimes \mathrm{End}(E)$ by [F4]–[F6]. Step 2.1 shows that this section acts on arbitrary $X,Y,s$ as the original curvature.

Since coordinate fields commute, expand the defining curvature commutator on $e_j$ with [F2]–[F3]. The coefficient of $e_i$ is ${\mathrm{\partial }}_a({\omega }^i{}_j({\mathrm{\partial }}_b))-{\mathrm{\partial }}_b({\omega }^i{}_j({\mathrm{\partial }}_a))+{\sum }_k({\omega }^i{}_k({\mathrm{\partial }}_a){\omega }^k{}_j({\mathrm{\partial }}_b)-{\omega }^i{}_k({\mathrm{\partial }}_b){\omega }^k{}_j({\mathrm{\partial }}_a))$.

By [F4], the first two terms in step 1.1 are $d{\omega }^i{}_j({\mathrm{\partial }}_a,{\mathrm{\partial }}_b)$; by [F5], the sum is ${\sum }_k({\omega }^i{}_k\wedge {\omega }^k{}_j)({\mathrm{\partial }}_a,{\mathrm{\partial }}_b)$ in precisely the stated matrix order. Both sides are two-forms by [F1], so equality on every coordinate-frame pair proves ${\mathrm{\Omega }}^i{}_j=d{\omega }^i{}_j+{\sum }_k{\omega }^i{}_k\wedge {\omega }^k{}_j$.

From [F2], the local matrix of the End$(E)$ connection is the commutator action ${\mathrm{\nabla }}_X^{\mathrm{End}}A=X(A)+\omega (X)A-A\omega (X)$. Alternating this formula as in the definition of $d^{\mathrm{\nabla }}$, with [F3] ensuring the alternating degrees are preserved, gives for an End$(E)$-valued $k$-form $A$ the local identity $d^{\mathrm{\nabla }}A=dA+\omega \wedge A-(-1{)}^{k}A\wedge \omega$. In particular, $d^{\mathrm{\nabla }}\mathrm{\Omega }=d\mathrm{\Omega }+\omega \wedge \mathrm{\Omega }-\mathrm{\Omega }\wedge \omega$.

Substitute [F1] into step 1.1 and apply the graded Leibniz rule: $d^{\mathrm{\nabla }}\mathrm{\Omega }=d^2\omega +d\omega \wedge \omega -\omega \wedge d\omega +\omega \wedge d\omega +\omega \wedge \omega \wedge \omega -d\omega \wedge \omega -\omega \wedge \omega \wedge \omega =0$ by [F5] and associativity of matrix/wedge multiplication. Since this holds in every local frame, it is the intrinsic identity $d^{\mathrm{\nabla }}\mathrm{\Omega }=0$.

Shrink a chart and bundle trivialization about $p$ so that its coordinate image is a convex ball. Coordinatewise linear interpolation gives a fixed-endpoint homotopy $H(s,t)$ from ${\gamma }_0$ to ${\gamma }_1$; after a common finite subdivision it is smooth on each parameter rectangle. Let $V(s,t)$ be the [F2] parallel section along $t\mapsto H(s,t)$ with $V(s,0)=v$. In the fixed trivialization this is a linear ODE with smooth parameter $s$, so [F4] and uniqueness make $V$ smooth on each rectangle and continuous across the subdivision lines.

Put $W={\mathrm{\nabla }}_{{\mathrm{\partial }}_s}V$ along $H$. Expanding the two covariant derivatives in the fixed frame, using ${\mathrm{\nabla }}_{{\mathrm{\partial }}_t}V=0$ and $[{\mathrm{\partial }}_s,{\mathrm{\partial }}_t]=0$, gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_t}W=-R^{\mathrm{\nabla }}({\mathrm{\partial }}_{s}H,{\mathrm{\partial }}_{t}H)V=0$ by flatness. Because $H(s,0)=x$ and $V(s,0)=v$ are constant in $s$, $W(s,0)=0$; uniqueness in [F2] therefore gives $W=0$ on every rectangle and, successively, across all subdivision lines.

At $t=1$ the base point $H(s,1)=y$ is fixed, so $W(s,1)=0$ says that $s\mapsto V(s,1)\in E_y$ is constant. Hence $P_{{\gamma }_0}v=V(0,1)=V(1,1)=P_{{\gamma }_1}v$ by [F3]. This holds for every $v$, proving equality of the transport maps.

Assume $\mathrm{\nabla }$ is flat, fix $p$, and take a coordinate ball $U$ from [F1], small enough to lie in one bundle trivialization. Let $v_1,\dots ,v_r$ be the basis of $E_p$ induced by that trivialization and define $e_a(q)=P_{p,q}{v}_a$, where [F1] makes the notation independent of the path in $U$. Using the radial coordinate paths, [F4] shows that $e_a$ depends smoothly on $q$. Linear ODE uniqueness makes transport linear, and transport along the reversed path is its inverse; hence the $e_a(q)$ form a basis of $E_q$ and [F3] makes $(e_a)$ a local frame.

For $q\in U$ and a smooth curve $c$ through $q$, concatenate any path from $p$ to $q$ with the segment of $c$. Endpoint independence in [F1] identifies $e_a(c(t))$ with parallel transport of $e_a(q)$ along $c$; [F2] therefore gives ${\mathrm{\nabla }}_{\dot{c}}{e}_a=0$. Every tangent vector is the velocity of such a local curve, so every $e_a$ is parallel. This proves the forward implication.

Conversely, suppose every point has a neighborhood with a parallel frame. On such a neighborhood the defining curvature commutator gives $R^{\mathrm{\nabla }}(X,Y)e_a=0$ because all three covariant derivatives of $e_a$ vanish. At each point the $e_a$ are a basis by [F3], so [F5] gives $R^{\mathrm{\nabla }}(X,Y)=0$ on the whole fibre. These neighborhoods cover $M$, hence the connection is flat.

On any coordinate chart, write $X=X^i{\mathrm{\partial }}_i$ and $Y=Y^i{\mathrm{\partial }}_i$. Applying [F4] to a local smooth function and using the ordinary product rule, the terms with second derivatives cancel and give $[X,Y]=(X^i{\mathrm{\partial }}_i{Y}^j-Y^i{\mathrm{\partial }}_i{X}^j){\mathrm{\partial }}_j\mathrm{.}$ The displayed coefficients are smooth by [F3], so every bracket used below is a smooth vector field.

Expanding the three curvature terms by [F1] and collecting derivatives with the same outer field gives the cyclic sum as $\begin{array}{rl}{\displaystyle }& {\displaystyle {\mathrm{\nabla }}_X({\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_{Z}Y)+{\mathrm{\nabla }}_Y({\mathrm{\nabla }}_{Z}X-{\mathrm{\nabla }}_{X}Z)+{\mathrm{\nabla }}_Z({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X)}\\ {\displaystyle }& {\displaystyle -{\mathrm{\nabla }}_{[X,Y]}Z-{\mathrm{\nabla }}_{[Y,Z]}X-{\mathrm{\nabla }}_{[Z,X]}Y\mathrm{.}}\end{array}$ By [F2] the three parenthesized differences are $[Y,Z]$, $[Z,X]$, and $[X,Y]$.

Acting on an arbitrary local smooth function $f$ and using [F4], expand the twelve resulting third-order compositions in $[X,[Y,Z]]f+[Y,[Z,X]]f+[Z,[X,Y]]f\mathrm{.}$ Each composition occurs once with sign $+$ and once with sign $-$; hence the sum is zero. Since equality of vector fields is local and is detected by their action on smooth functions, the Jacobi identity holds for $X,Y,Z$.

Apply [F2] once more to pair each term ${\mathrm{\nabla }}_X[Y,Z]$ with $-{\mathrm{\nabla }}_{[Y,Z]}X$, and cyclically. The expression from step 1.2 becomes $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]$, which is zero by step 2.1.

Combining [F1] with [F2] gives first-pair skewness, and pairing [F3] with $W$ gives the displayed cyclic identity for $\mathrm{Rm}$.

Metric compatibility [F4] expands the scalar identity $XYg(Z,W)-YXg(Z,W)-[X,Y]g(Z,W)=0$. The mixed terms $g({\mathrm{\nabla }}_{Y}Z,{\mathrm{\nabla }}_{X}W)$ and $g({\mathrm{\nabla }}_{X}Z,{\mathrm{\nabla }}_{Y}W)$ cancel in pairs, leaving $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$. Symmetry of $g$ and [F1] give last-pair skewness.

Write the cyclic identity from step 1.1 for the four ordered triples $(X,Y,Z;W)$, $(Y,Z,W;X)$, $(Z,W,X;Y)$, and $(W,X,Y;Z)$ and add them. Last-pair skewness from step 1.2 cancels the eight terms whose first pair is respectively $(X,Y)$, $(Y,Z)$, $(Z,W)$, or $(W,X)$. The four remaining terms, simplified with both pair skews, give $2\mathrm{Rm}(Y,W,X,Z)-2\mathrm{Rm}(X,Z,Y,W)=0$. Renaming $(X,Z,Y,W)$ as an arbitrary quadruple yields pair interchange.

Expanding the alternating definition of $d^{\mathrm{\nabla }}R$ gives three output-derivative terms and the bracket terms $-R([X,Y],Z)+R([X,Z],Y)-R([Y,Z],X)$. Replace every bracket by $[A,B]={\mathrm{\nabla }}_{A}B-{\mathrm{\nabla }}_{B}A$ using [F4], and use skewness of the two-form $R$. The six resulting argument-derivative terms are exactly those subtracted in the tensor covariant derivatives, so $d^{\mathrm{\nabla }}R(X,Y,Z)=({\mathrm{\nabla }}_{X}R)(Y,Z)+({\mathrm{\nabla }}_{Y}R)(Z,X)+({\mathrm{\nabla }}_{Z}R)(X,Y)$.

Specialize [F1] to the tangent bundle to make the left side of step 1.1 zero, proving the first displayed identity. Since the Levi–Civita connection preserves $g$, lowering the output in [F2] commutes with covariant differentiation by [F3]; evaluating the resulting contracted identity on $U,V$ gives the second displayed formula.

Multilinearity and first-pair alternation in [F2] give $\mathrm{Rm}(X^{\mathrm{\prime }},Y^{\mathrm{\prime }},Y^{\mathrm{\prime }},X^{\mathrm{\prime }})=\mathrm{\Delta }\mathrm{Rm}(X,Y,Y^{\mathrm{\prime }},X^{\mathrm{\prime }})$. Applying last-pair alternation to $(Y^{\mathrm{\prime }},X^{\mathrm{\prime }})$ gives a second factor $\mathrm{\Delta }$, so the numerator is ${\mathrm{\Delta }}^2\mathrm{Rm}(X,Y,Y,X)$.

If $G$ is the Gram matrix of $(X,Y)$, the Gram matrix of $(X^{\mathrm{\prime }},Y^{\mathrm{\prime }})$ is $AG{A}^{\mathsf{T}}$ for $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, so its determinant is ${\mathrm{\Delta }}^2\det G$. The nonzero common factor ${\mathrm{\Delta }}^2$ cancels from the quotient in [F1], proving basis independence.

The sectional quotient is intrinsic to a two-plane using only the given pair skews. Indeed, if its ordered basis $(X,Y)$ is changed by a matrix $A\in {\mathrm{G}\mathrm{L}}_2(\mathbb{R})$ with determinant $d$, the alternating-pair numerator $R(X,Y,Y,X)$ becomes $d^{2}R(X,Y,Y,X)$, while the Gram determinant becomes $\det (AG{A}^{\mathsf{T}})=d^2\det G$. Thus the quotient is basis-independent. For independent $X,Y$, equality of the two quotients and positivity of the common Gram denominator give $T(X,Y,Y,X)=0$. For dependent $X,Y$, the same equality follows from first-pair skewness.

Expanding $0=T(X+Y,Z,Z,X+Y)$ and using step 1.1 removes both diagonal terms. Pair interchange followed by the two pair skews identifies the two cross terms, so $2T(X,Z,Z,Y)=0$. Hence $T(X,Z,Z,Y)=0$ for all $X,Y,Z$.

Polarize step 2.1 in $Z$: expanding $0=T(X,Z+W,Z+W,Y)$ leaves $T(X,Z,W,Y)+T(X,W,Z,Y)=0$, so $T$ is also skew in its two middle slots. The Bianchi identity now gives $0=T(X,Y,Z,W)-T(Y,X,Z,W)-T(X,Z,Y,W)=3T(X,Y,Z,W)$ by first-pair and middle-slot skewness. Therefore $T=0$ and $R_1=R_2$.

Define $A_K(X,Y,Z,W)=K(g(Y,Z)g(X,W)-g(X,Z)g(Y,W))$. Directly exchanging arguments shows that $A_K$ is skew in each pair and invariant under pair interchange; its three cyclic terms cancel pairwise, so it has all algebraic curvature symmetries. Moreover $A_K(X,Y,Y,X)=K(g(X,X)g(Y,Y)-g(X,Y{)}^2)$.

If the manifold has constant sectional curvature $K$, step 1.1 shows that $A_K$ and $\mathrm{Rm}$ give the same quotient on every two-plane. By [F2], $\mathrm{Rm}=A_K$. Conversely, if $\mathrm{Rm}=A_K$, division of the last identity in step 1.1 by the positive Gram determinant gives sectional curvature $K$ on every two-plane, which is [F1].

The four-tensor identity says for every $W$ that $g(R(X,Y)Z,W)=g(K(g(Y,Z)X-g(X,Z)Y),W)$. Nondegeneracy of $g$ yields the vector-valued formula, and pairing that formula with $W$ gives the converse equivalence.