Sectional Curvature, Ricci Curvature, and Space Forms
Let \((M,g)\) be a smooth Riemannian manifold. Write \(\nabla\) for its Levi–Civita connection and \(T_pM\) for the tangent space at \(p\). The coordinate conventions are those in tangent spaces and derivations.
The curvature tensor contains more information than is easy to read from its components. Sectional curvature extracts a number from each tangent two-plane, while Ricci curvature averages sectional contributions through a fixed direction. We first examine the zero-curvature case, where a parallel coordinate frame is available under the torsion-free hypothesis.
Definition 1. An affine connection \(\nabla\) on \(TM\) is flat if for all \(p \in M\), there exists a neighbourhood \(U = U_p\) with local coordinates such that all coordinate vector fields \(\pd{}{{x^i}}\) are parallel, i.e. \(\nabla \pd{}{{x^i}} = 0\) for all \(i\).
This is the coordinate-flat convention. Many texts call a connection flat when its curvature alone vanishes; that weaker condition gives local parallel frames, which need not be coordinate frames if torsion is present. Theorem 2 includes torsion precisely because this definition asks for parallel coordinate fields.
Theorem 2. An affine connection \(\nabla\) on \(TM\) is flat if and only if its curvature and torsion vanish identically.
Proof. If all coordinate fields are parallel, all Christoffel symbols vanish on the chart. Coordinate fields commute, so the torsion vanishes, and the curvature formula contains only derivatives and products of zero coefficients. Thus both tensors vanish.
Conversely suppose curvature and torsion vanish. In local coordinates let \(\Gamma_i\) be the matrix with entries \((\Gamma_i)^k{}_j=\Gamma^k_{ij}\). A matrix of parallel-frame columns must solve
This is a system of partial differential equations, not a collection of independent ODEs. To justify solvability, on the product of the coordinate domain with \(\operatorname{GL}(n,\mathbb R)\) define horizontal fields \(H_i\) whose base component is \(\partial_i\) and whose matrix component is \(-\Gamma_iB\). Direct differentiation gives
by zero curvature. Their base components are independent, so they span an involutive rank-\(n\) distribution. Frobenius gives an integral submanifold through \((p,I)\); projection to the coordinate domain has invertible differential there. On a smaller neighbourhood the integral submanifold is therefore a graph \(B(x)\), and tangency to the \(H_i\)'s gives the required system. Its columns define a parallel frame \(E_1,\ldots,E_n\).
Let \(\theta^1,\ldots,\theta^n\) be the dual coframe. It is parallel because differentiating \(\theta^a(E_b)=\delta^a_b\) gives zero. Torsion-freeness implies each \(\theta^a\) is closed:
On a star-shaped coordinate ball a closed form \(\theta=\mu_i(x)dx^i\) has the potential \(z(x)=\int_0^1\mu_i(tx)x^i\,dt\). Differentiating and using \(\partial_j\mu_i=\partial_i\mu_j\) gives \(\partial_jz=\int_0^1\frac d{dt}(t\mu_j(tx))dt=\mu_j(x)\). Apply this to every coframe member. The resulting \(z^a\)'s have independent differentials and hence form local coordinates by the inverse function theorem. Since \(dz^a(E_b)=\delta^a_b\), their coordinate fields are exactly the parallel frame. ◻
We now turn from flatness to scalar measurements of curvature. Normalizing by the squared area of the parallelogram spanned by two tangent vectors makes sectional curvature a quantity associated with the plane itself.
Definition 3. With \(R(X,Y)Z=\nabla_Y\nabla_XZ-\nabla_X\nabla_YZ+\nabla_{[X,Y]}Z\), the sectional curvature of the plane spanned by independent vectors \(X,Y\in T_pM\) is
The denominator is the squared area. Replacing \((X,Y)\) by another basis \((aX+bY,cX+dY)\) multiplies both numerator and denominator by \((ad-bc)^2\), by skew-symmetry in the two curvature pairs. Hence the quotient depends only on the plane. The order \(X,Y\) in the final two slots is needed for positive curvature of a round sphere under this curvature convention.
Definition 4. For an orthonormal basis \(e_1,\ldots,e_n\) of \(T_pM\), define
This is a bilinear form on tangent vectors, not a function of a plane. It is independent of the orthonormal basis because it is a trace; under an orthogonal basis change, summing the two basis-change coefficients gives the Kronecker delta. Pair symmetry of curvature gives symmetry in \(X,Y\). For a unit vector \(X=e_n\), it equals \(\sum_{i<n}\kappa(X,e_i)\).
Clearly, if we know the sectional curvature of all planes containing \(p \in M\), then we can recover the Riemann curvature tensor. The calculation uses polarization: expand the curvature expression on sums of vectors, then use the curvature symmetries to recover the mixed terms.
Lemma 5 (Curvature from sectional curvatures). Set \(C(X,Y,Z,W)=\langle R(X,Y)Z,W\rangle\) and \(Q(X,Y)=C(X,Y,X,Y)\). The sectional curvatures determine \(Q\), since \(Q(X,Y)=\kappa(X,Y)|X\wedge Y|^2\) for independent vectors and \(Q=0\) for dependent vectors. Define
Then the full curvature tensor is recovered by
Proof. Bilinearity and pair symmetry give
Polarizing the remaining repeated slot gives
To see exactly how the cyclic identity enters, put \(a=C(X,Y,Z,W)\), \(b=C(X,Z,Y,W)\), and \(c=C(X,W,Y,Z)\). The first Bianchi identity and pair symmetry give \(a+c-b=0\). Formula (5) and skew-symmetry in the last pair give
Their difference is \(2a+b-c=3a\), proving (3). Thus no mixed curvature entry remains undetermined. In particular, if all sectional curvatures at \(p\) equal \(f(p)\), compare with the algebraic tensor \(R_0(X,Y)Z=f(p)(\langle X,Z\rangle Y-\langle Y,Z\rangle X)\): it has the same \(Q\), and polarization proves \(R=R_0\) at that point. ◻
Definition 6. Let \(M\) be a differentiable manifold. \(M\) is called a space form, or a space of constant sectional curvature, if \(\kappa (X,Y) = K = \text{constant}\) for all linearly independent tangent vectors \(X,Y \in T_p M\), where \(p \in M\).
We will see some consequences of constant sectional curvature in the sequel. Note that some texts simply refer to these as “spaces of constant curvature”. For now, we note a significant consequence of isotropy.
Theorem 7 (Schur). Let \(M\) be a connected Riemannian manifold of dimension \(n\geq3\). If every tangent two-plane at \(p\) has the same sectional curvature \(f(p)\), then \(f\) is constant on \(M\). Without connectedness, it is constant on each component.
Proof. By Lemma 5,
The function \(f\) is smooth: locally evaluate the smooth curvature tensor on a smooth orthonormal pair. Metric compatibility then gives
At any point choose three mutually orthonormal vectors \(U,X,Y\), possible because \(n\geq3\). In the second Bianchi identity
the three terms reduce to \((Uf)Y\), zero, and \(-(Yf)U\). Taking the inner product with \(Y\) gives \(Uf=0\). For any chosen unit \(U\), two perpendicular unit vectors can be selected, so \(df=0\) at every point. Connectedness makes \(f\) globally constant. In dimension two, there is only one tangent two-plane at each point and its curvature may vary, which explains the dimension hypothesis. ◻
Theorem 8 (Schur's lemma for Einstein metrics). Let \(M\) be a connected Riemannian manifold of dimension \(n\geq3\). If \(\operatorname{Ric}=f g\) for a smooth function \(f\), then \(f\) is constant. Equivalently, if \(\operatorname{Ric}(X,X)=f(p)\) for every unit \(X\in T_pM\), polarization gives the same hypothesis. Such a metric is Einstein.
Proof. The contracted second Bianchi identity is
Here is the contraction with our signs. At a point use a normal orthonormal frame and write \(C_{abcd}=\langle R(e_a,e_b)e_c,e_d\rangle\), so \(\operatorname{Ric}_{bd}=\sum_a C_{abad}\). Contracting the second Bianchi identity in its derivative index and the third curvature slot gives
Now set \(m=k\) and sum over \(k\). The first term is \(\sum_i\nabla_i\operatorname{Ric}_{ji}\), the second is \(-\nabla_j\operatorname{Scal}\), and the third is another copy of the first. This proves the contracted identity.
Under \(\operatorname{Ric}=fg\), metric compatibility gives \(\operatorname{div}\operatorname{Ric}=df\) and \(\operatorname{Scal}=nf\). Hence \((n-2)df=0\). Since \(n\geq3\), \(df=0\), and connectedness finishes the proof. A trace equation for Ricci cannot be “uncontracted” to specify the entire Riemann tensor; Einstein metrics need not have constant sectional curvature. ◻
Lemma 9 (Ricci as an average). If \(n\geq2\), \(|X|=1\), and \(A\) is the uniform average of sectional curvatures of planes containing \(X\), then \(\operatorname{Ric}(X,X)=(n-1)A\).
Proof. Complete \(X=e_n\) to an orthonormal basis and let \(u=\sum_{j=1}^{n-1}\alpha_je_j\), with \(\alpha\) uniformly distributed on \(S^{n-2}\). Then \(X,u\) is an orthonormal pair. The vectors \(u\) and \(-u\) represent the same plane and give the same curvature, so averaging over this sphere gives the desired plane average.
Reflection of one coordinate makes \(\mathbb E\alpha_j\alpha_k=0\) for \(j\ne k\). Permutation invariance makes all diagonal moments equal, and \(\sum_j\alpha_j^2=1\) gives \(\mathbb E\alpha_j^2=1/(n-1)\). Therefore
The omitted \(j=n\) term is zero, so the sum is \(\operatorname{Ric}(X,X)\). For \(n=2\), the sphere is \(S^0=\{-1,1\}\) with equal weights and the same calculation applies. No beta-integral normalization or curvature model is needed. ◻
1. Space Forms
Spaces of constant sectional curvature provide explicit models for comparison geometry. The next theorem identifies their complete simply connected models. Passing to the universal cover separates the local curvature condition from the possible global identifications.
Theorem 10 (Simply connected space-form models). Let \(M\) be a connected complete Riemannian manifold without boundary of dimension \(n\geq2\), with constant sectional curvature \(K\). Its universal cover with the lifted metric is isometric to the complete simply connected model: Euclidean \(\mathbb R^n\) for \(K=0\), the sphere of radius \(K^{-1/2}\) for \(K>0\), or hyperbolic space with metric scaled to curvature \(K\) for \(K<0\). In particular the normalized cases \(K=-1,0,1\) give \(\mathbb H^n,\mathbb R^n,\mathbb S^n\), respectively. In dimension one the complete simply connected model is the real line; sectional curvature is not defined on any tangent two-plane there.
We need a uniqueness fact for local isometries: their value and differential at one point determine them on a connected domain.
Lemma 11. Let \(M\) and \(N\) be connected, differentiable manifolds. Let \(f_i : M \to N\) be local isometries for \(i = 1,2\). If there exists a \(p \in M\) such that \(f_1 (p) = f_2 (p)\), and \((\dif f_1)_p = (\dif f_2)_p\), then \(f_1 = f_2\) on \(M\).
Proof. A local isometry preserves the Levi–Civita connection, by uniqueness of the metric torsion-free connection, and therefore maps geodesics with their affine parameters to geodesics. Consequently, for sufficiently small \(v\in T_pM\),
Equal values and differentials at \(p\) therefore imply equality on a normal neighbourhood of \(p\), and then equality of their differentials on that neighbourhood.
Let \(A\) be the set of points where both their values and differentials agree. It is nonempty by hypothesis and open by the preceding argument applied at each point of \(A\). It is closed: equality of values passes to limits by continuity, and in common coordinate charts around a limiting pair of image points, equality of the derivative matrices also passes to the limit. Connectedness of \(M\) gives \(A=M\). The exponential on the right belongs to the target manifold and is based at the image point, not at \(p\). ◻
Proof of Theorem 10. The universal cover has the pulled-back metric, so its projection is a local isometry and its curvature is still \(K\). It is complete: a geodesic upstairs projects to a geodesic downstairs, which exists for all time by Hopf–Rinow; the path-lifting property of a covering lifts this extension, and local isometry shows the lift continues the original geodesic. Thus the cover is geodesically complete and hence metrically complete. Denote it by \(\widetilde M\).
We first justify the local isometries used in the construction. Put
Along a unit radial geodesic, an angular Jacobi field with initial data \(J(0)=0\), \(D_rJ(0)=w\perp\dot\gamma(0)\) solves \(D_r^2J+KJ=0\), because \(R(\dot\gamma,J)\dot\gamma=KJ\). After parallel transport to the initial tangent space, ODE uniqueness gives \(J(r)=S_K(r)P_rw\). The Gauss lemma makes radial and angular directions orthogonal. Therefore the exponential pullback metric is
This holds wherever the radial exponential is a local diffeomorphism; the displayed differential shows that it is one for \(r>0\) with \(S_K(r)\ne0\), and at zero its derivative is the identity. This computation proves that radial maps between two curvature-\(K\) manifolds with an initial tangent-space isometry are local isometries on the indicated domains.
For \(K\leq0\), let \(\Xi\) be the Euclidean or scaled hyperbolic model, choose \(p\in\Xi\), \(\widetilde p\in\widetilde M\), and a linear isometry \(I:T_p\Xi\to T_{\widetilde p}\widetilde M\). The model exponential is a global diffeomorphism (in hyperbolic polar coordinates its radial lines cover the model uniquely). Completeness defines \(\exp_{\widetilde p}\) on the entire tangent space. Hence
is globally defined and is a local isometry by the pullback-metric calculation.
We use the following covering fact. A local isometry from a connected complete Riemannian manifold to a connected Riemannian manifold is a surjective covering map. To prove it, lift a piecewise smooth path starting at an image point by local inverses. If its maximal lift stops at a finite parameter, the remaining path lengths tend to zero, so the lift is Cauchy; completeness supplies a limiting point and a local inverse extends it. Every point of the connected target is reachable by such a path, proving surjectivity. Take a sufficiently small normal ball about any target point. For each point in its fibre, lift its radial geodesics to obtain an inverse sheet. The sheets are open by the local inverse theorem, disjoint by uniqueness of lifts (reverse the radial path), and cover the preimage by lifting the reverse radial paths. This is exactly an evenly covered neighbourhood. A connected covering of a simply connected space has one sheet, by lifting homotopies of paths. Applying this to \(F\) gives a diffeomorphism; a bijective local isometry is an isometry.
For \(K>0\), rescale to \(K=1\). Let \(\Xi=S^n\) and let \(q\) be antipodal to \(p\). The model exponential gives a diffeomorphism from the tangent ball of radius \(\pi\) onto \(S^n\setminus\{q\}\). The same radial formula therefore gives a local isometry \(F\) on this punctured sphere, because \(\sin r>0\) for \(0<r<\pi\). Choose \(p'\ne p,q\), put \(\widetilde p'=F(p')\), and use \(I'=(dF)_{p'}\) to construct \(F'\) on the sphere minus the antipode \(q'\) of \(p'\). The overlap \(S^n\setminus\{q,q'\}\) is connected for \(n\geq2\). At \(p'\), both maps have the same value and differential, so Lemma 11 makes them agree on the whole overlap. They patch to a local isometry \(G:S^n\to\widetilde M\). The sphere is complete, so the covering fact applies again; simple connectivity of the target makes \(G\) a global isometry. Rescaling the metric gives radius \(K^{-1/2}\) and completes the positive-curvature case. ◻
Reading: Carmo (1992), Riemannian geometry.