Gauss equation

For a Euclidean embedded submanifold,It expresses intrinsic curvature in terms of the second fundamental form.

For an embedded submanifold of a Riemannian manifold, use $R(v,w)y={\nabla }_v{\nabla }_{w}y-{\nabla }_w{\nabla }_{v}y-{\nabla }_{[v,w]}y$. The Gauss formula and tangential derivative of a normal field giveSubtract the expression with $v,w$ interchanged and the bracket derivative, whose normal part pairs to zero. This proves the displayed curvature identity. For flat Euclidean ambient space, its ambient curvature term vanishes and one obtains the ordinary Gauss equation. Declaring the four-slot convention $R(x,y,v,w)=⟨R(v,w)y,x⟩$ avoids sign ambiguities when permuting arguments.

Let the unit normal satisfy $n^a{n}_a=\epsilon =\pm 1$, let $h_{ab}=g_{ab}-\epsilon n_a{n}_b$, and set $K_{ab}=h_a{}^c{h}_b{}^d{\nabla }_c{n}_d$. The normal component of an ambient derivative of tangent fields is $-\epsilon K(X,Y)n$. Substitute that decomposition into the curvature commutator and project tangentially to obtain the displayed equation. Its scalar contraction is ${}^{\Sigma }R=R-2\epsilon R_{ab}{n}^a{n}^b+\epsilon (K^2-K_{ab}{K}^{ab})$. Both metric signature and curvature-slot convention must be specified when assigning a sign.

Here $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, the negative of the convention in the ordinary Gauss equation in a curved ambient manifold. Insert the Gauss formula and Weingarten formula into this definition and project tangentially to obtain the displayed identity. Positive sectional curvature uses $⟨R(X,Y)X,Y⟩/|X\wedge Y{|}^2$ with this convention. Thus a Euclidean surface still has Gaussian curvature $K=\det A_{\xi }$; reversing the operator sign without changing the slot order would give the wrong sign.

A surface locally swept out by ambient geodesics has $II(T,T)=0$ in its unit ruling direction $T$. For a perpendicular unit tangent $E$, the Gauss equation in a curved ambient manifold therefore gives the displayed identity. Its Gaussian curvature is at most the ambient sectional curvature in its tangent plane, with equality precisely when $II(T,E)=0$. An independent variation argument applies the Jacobi field equation to the length of a perpendicular connecting field; its ambient derivative has the extra normal component $II(T,E)$. The Euclidean surface $(t,s,ts)$ is ruled by straight lines and has Gaussian curvature $-(1+t^2+s^2{)}^{-2}<0$.

For a two-plane $\sigma =\mathrm{span}(X,Y)$ in a Riemannian tangent space,

Ricci curvature is the trace of the Riemann curvature tensor in its first and third arguments. For a unit tangent vector $X$ and an orthonormal basis $X,e_2,\dots ,e_n$,

For a complete connected $n$-dimensional Riemannian manifold with nonnegative Ricci curvature, the displayed ratio is nonincreasing, tends to one at zero, and is at most one. In particular $\mathrm{Vol}B(p,r)\le {\omega }_n{r}^n$. In geodesic polar coordinates, the radial Jacobi field determinant is at most the Euclidean density $r^{n-1}$ until the cut time. The trace of the radial Riccati equation for distance spheres, with Cauchy-Schwarz inequality, proves this comparison. Equality for every radius forces the Euclidean polar metric and absence of a cut locus.

For a complete connected $n$-dimensional Riemannian manifold with $\mathrm{Ric}\ge (n-1)kg$, $k>0$, put $D=\pi /\sqrt{k}$ andHere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere, and $v_k(r)$ is the Riemannian volume of a radius-$r$ ball in the round curvature-$k$ sphere. The Bishop-Gromov inequality asserts that $V_p(r)/v_k(r)$ is nonincreasing for $0<r<D$, tends to one at zero, and is at most one. In geodesic polar coordinates, the radial volume density before the cut time is at most $s_k(t{)}^{n-1}$. The bound uses trace Ricci curvature; with normalized Ricci curvature it reads $\overline{\mathrm{Ric}}\ge kg$.

Under the hypotheses of Bishop volume comparison with a positive-curvature model, the Bonnet-Myers theorem gives diameter at most $D$. If total Riemannian volume equals $v_k(D)$, every comparison ball ratio is one. In a small normal ball the pointwise radial density bound therefore becomes equality. Writing $h={\partial }_r\log J$, the traced radial Riccati equation for distance spheres and Cauchy-Schwarz inequality force the radial operator to equal $(s_k^{\prime }/s_k)I$. The angular metric is consequently $s_k(r{)}^2{g}_{S^{n-1}}$, so the metric is locally round. The classification of complete positive constant-curvature manifolds gives a quotient $S_k^n/\Gamma$, whose volume is $v_k(D)/|\Gamma |$. Equality forces $\Gamma$ trivial.

Assume the trace Ricci curvature satisfies $\mathrm{Ric}\ge (n-1)kg$, with $k>0$. If points $p,q$ have Riemannian distance $D=\pi /\sqrt{k}$, the balls $B(p,r)$ and $B(q,D-r)$ are disjoint. Bishop volume comparison with a positive-curvature model bounds their volumes below by $\mathrm{Vol}(M)v_k(r)/v_k(D)$ and $\mathrm{Vol}(M)v_k(D-r)/v_k(D)$. Since $v_k(r)+v_k(D-r)=v_k(D)$, both inequalities must be equalities. Taking $r\downarrow 0$ gives $\mathrm{Vol}M=v_k(D)$, and spherical rigidity of maximal total volume yields the round sphere.

The Bishop-Gromov inequality forces every volume ratio to equal one if the asymptotic ratio is one. Equality in the radial Jacobian comparison forces the radial operator $A(X)={\nabla }_X{\partial }_r$ of each distance sphere to be $r^{-1}I$ (it is the negative of the outward shape operator in the convention $S=-\nabla \nu$), by the traced radial Riccati equation for distance spheres and equality in Cauchy-Schwarz inequality. Any finite cut time would remove a positive cone of radial volume at a later radius, contradicting equality. Thus the Riemannian exponential map is global and one-to-one, and the polar metric is $d{r}^2+r^2{g}_{S^{n-1}}$, the Euclidean metric.

For a Riemannian manifold of dimension $n\ge 2$, the normalized directional Ricci curvature averages, rather than sums, the $n-1$ sectional curvatures through a unit vector. Thus the Bonnet-Myers theorem gives $\mathrm{diam}M\le \pi /\sqrt{k}$ from $\overline{\mathrm{Ric}}\ge kg$. With trace Ricci curvature the corresponding hypothesis is $\mathrm{Ric}\ge (n-1)kg$. The distinction changes positive constants but not the signs of Ricci curvature.

At a point of a Riemannian manifold, Ricci curvature is isotropic when $\mathrm{Ric}(v,v)/g(v,v)$ is independent of nonzero tangent direction. Polarization makes this equivalent to the displayed tensor identity. In dimension three, sectional curvatures from Ricci curvature in dimension three gives every sectional curvature at that point the value $c_p/2$. In higher dimensions isotropic Ricci curvature does not determine every two-plane curvature.

An Einstein manifold has Ricci curvature equal to a constant multiple of its metric. In the Riemannian setting, the round unit sphere has $\Lambda =n-1$. This condition is weaker than constant sectional curvature in higher dimensions. The definition can also be applied to pseudo-Riemannian metrics, but positivity is assumed in the completeness theorems discussed here.

A Riemannian or pseudo-Riemannian metric is Einstein when its Ricci tensor is a constant scalar multiple of the metric. Constant-sectional-curvature metrics satisfy this condition with $\Lambda =(d-1)k$. The Killing-form metric on a real semisimple Lie group is another example, with $\Lambda =-1/4$ for $g=B$ in the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$.

The product Riemannian metric of two round unit spheres has Ricci curvature equal to its metric, since the factor Ricci tensors equal their metrics and the product connection splits. Its directional Ricci curvatures are all one. A two-plane tangent to a single sphere has sectional curvature one, while a mixed plane has sectional curvature zero. This four-dimensional Einstein manifold therefore need not have constant sectional curvature.

For an orthonormal triple and distinct $i,j,k$, tracing sectional curvature gives ${\mathrm{Ric}}_{ii}=K_{ij}+K_{ik}$. Solving the three equations gives the displayed formula. Equivalently $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2$. This pointwise assertion does not require the triple to diagonalize the Ricci curvature.

Scalar curvature is the complete trace of the Riemann curvature tensor:An $n$-manifold of constant sectional curvature $K$ has $\mathrm{Scal}=n(n-1)K$.

The product Riemannian metric of the round unit two-sphere and a circle has Ricci curvature equal to $g_{S^2}$ on the sphere directions and zero on the circle direction. Its scalar curvature is constantly two, but no single scalar multiple of the whole metric equals that Ricci tensor. In two dimensions the identity $\mathrm{Ric}=\mathrm{Scal}g/2$ does make constant scalar curvature equivalent to an Einstein metric.

This is the positive-definite metric case of a Ricci-flat manifold.

A Ricci-flat Riemannian manifold has identically zero Ricci tensor. Every flat Riemannian manifold is Ricci-flat, while the converse can fail in dimension at least four.

For a connected Riemannian manifold of dimension $n\ge 2$, assume geodesic completeness and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$. The Riemannian distance then has diameter at most $\pi /\sqrt{\kappa }$; the manifold is compact and its fundamental group is finite. The dimension restriction excludes the vacuous one-dimensional Ricci bound.

The graph $z=x^2+y^2$ in Euclidean three-space has the displayed induced Riemannian metric and positive Gaussian curvature. It is smooth at its vertex; the polar-coordinate singularity is only a coordinate artifact. The metric dominates the Euclidean plane metric, so upward stability of Riemannian completeness proves completeness. It has infinite diameter. In dimension two Ricci curvature equals Gaussian curvature times the metric, so this is a counterexample to compactness under merely pointwise positive Ricci curvature. Its curvature tends to zero at infinity.

If a connected complete Riemannian manifold has $\mathrm{Ric}\ge (n-1)\kappa g$ with $\kappa >0$, its universal cover with the lifted metric is complete and satisfies the same bound. The Bonnet-Myers theorem makes that cover compact. A covering fiber is closed and discrete and hence finite; the number of its elements equals the order of the fundamental group. Applying the diameter theorem only to the base would not prove this conclusion: the positive lower bound must also be used on the cover.

This metric pulls back the round spherical metric by $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$. It has unit Gaussian curvature and Ricci curvature equal to its metric, yet the meridian reaches a missing boundary in finite time. Every curve between $(0,0)$ and $(0,L)$ has length at least $|L|/\sqrt{2}$, so its diameter is infinite. It shows that the positive Ricci bound alone does not replace geodesic completeness in the Bonnet-Myers theorem.

This is a Riemannian splitting theorem whose hypotheses include a globally minimizing geodesic line.

A complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line splits isometrically as $N\times \mathbb{R}$.

If a connected closed manifold $M^3$ has a noncontractible universal cover, then $M\times \mathbb{R}$ has no complete metric making it a Ricci-flat Riemannian manifold. The product has two ends and hence contains a line in a Riemannian manifold for any complete metric. The Cheeger-Gromoll splitting theorem would split it isometrically as $N^3\times \mathbb{R}$. Its three-dimensional factor is Ricci-flat and therefore flat by three-dimensional curvature from the Ricci tensor. The complete simply connected cover of the product would be Euclidean four-space. But that cover is $\tilde{M}\times \mathbb{R}$, which retracts onto $\tilde{M}$, contradicting noncontractibility.

A Riemannian manifold is flat when every sectional curvature is zero. A complete simply connected flat $n$-manifold is isometric to Euclidean space ${\mathbb{R}}^n$.

On a compact flat Riemannian manifold, the integrated Bochner identity for Killing vector fields makes every Killing vector field parallel. Parallel vector fields are exactly the fixed vectors of the linear holonomy. Thus the six orientable closed flat three-manifold types have identity-component isometry dimensions three for the torus, one for each cyclic nontrivial holonomy, and zero for the Hantzsche-Wendt manifold.

The theorem classifies compact flat manifolds through their crystallographic covering groups.

For a compact flat Riemannian manifold ${\mathbb{R}}^n/\Gamma$, the translations in $\Gamma$ form a normal full-rank Euclidean lattice of finite index. The quotient is its finite linear holonomy. Affine conjugacy of the torsion-free crystallographic groups classifies these manifolds up to affine diffeomorphism. An abelian $\Gamma$ must consist entirely of translations: commuting with every lattice translation forces each linear part to fix a spanning set of ${\mathbb{R}}^n$.

These are precisely the faithful linear holonomy groups of closed orientable Euclidean three-manifolds. Cyclic cases arise from planar lattice rotations combined with axial screw translations. The noncyclic case is generated by two perpendicular half-turn screws. The list refers to the maximal translation lattice, not to arbitrary finite quotients of a fundamental group.

The closed orientable flat Riemannian manifold with holonomy $C_2\times C_2$ can be presented by Euclidean screws $\alpha (x,y,z)=(x+1/2,-y,-z)$ and $\beta (x,y,z)=(-x,y+1/2,-z+1/2)$. Their squares and the square of their product generate the unit translation lattice. Each nonidentity holonomy coset has a half-integer displacement along its fixed axis, so acts freely. The holonomy has no common fixed vector, giving first Betti number zero and a finite isometry group for every flat metric.

The mapping torus of an order-six rotation of a hexagonal flat torus is a closed orientable flat Riemannian manifold with holonomy $C_6$. The screw rotation has angle $\pi /3$ and axial displacement $1/6$. Its first Betti number is one.

The mapping torus of an order-three rotation of a hexagonal flat torus is a closed orientable flat Riemannian manifold with holonomy $C_3$. A screw rotation through $2\pi /3$ with axial displacement $1/3$, together with the planar lattice, describes its deck group. Its first Betti number is one.

This closed flat Riemannian manifold is the mapping torus of $-I$ on a two-dimensional flat torus. Equivalently, take planar lattice translations and the screw $(v,z)\mapsto (-v,z+1/2)$ in Euclidean three-space. Its square is a unit axial translation, its holonomy is $C_2$, and its first Betti number is one.

The Euclidean transformations $a(x,y,z)=(x+1,y,z)$, $b(x,y,z)=(x,y+1,z)$ and $s(x,y,z)=(-y,x,z+1/4)$ generate a torsion-free cocompact group with $sa{s}^{-1}=b$, $sb{s}^{-1}=a^{-1}$ and $s^4$ equal to unit translation along $z$. Every nontranslation has nonzero axial displacement modulo integers, hence no fixed point. Its maximal translation subgroup is ${\mathbb{Z}}^3$ and its holonomy is $C_4$.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with its quotient Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its volume is the lattice covolume. The dual lattice indexes its complete Fourier series of Laplacian eigenfunctions.

Let ${\Lambda }_1,{\Lambda }_2\subset {\mathbb{R}}^d$ be full-rank Euclidean lattices, and choose $\epsilon$ smaller than every nonzero vector length in either. The shortest vectors of ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$ are exactly the new coordinate vectors and their negatives. Any isometry between the associated flat tori lifts to an affine Euclidean isometry, whose orthogonal part maps the lattices. It must preserve the span of the shortest vectors, hence also its orthogonal complement, and therefore maps ${\Lambda }_1$ to ${\Lambda }_2$. Consequently a nonisometric pair remains nonisometric after adding sufficiently short equal circle factors. The spectrum of a flat torus shows that adding the same factors preserves isospectral manifolds.

With the nonnegative Laplace-Beltrami operator, a frequency $w$ in the dual lattice gives the eigenfunction $e^{2\pi i⟨w,x⟩}$ and eigenvalue $4{\pi }^2|w{|}^2$. Its multiplicity is the number of dual vectors with that norm. Fourier series prove completeness. In dimension two the shortest vector, shortest independent vector and covolume determine a reduced Gram matrix, proving spectral rigidity.

The vector-length multiset of a rank-two Euclidean lattice determines it up to an orthogonal map. Let $a$ be the shortest nonzero length. Removing two copies of each positive integer multiple of $a$ removes one primitive lattice line; the shortest remaining length $b$ is the shortest independent-vector length. Those two vectors form a lattice basis: otherwise a nonzero point in the centred basis parallelogram would be an independent lattice vector of length less than $b$. Lattice-point density determines the covolume $A$. The basis Gram matrix then has diagonal entries $a^2,b^2$ and absolute off-diagonal entry $\sqrt{a^2{b}^2-A^2}$. This proves two-dimensional spectral rigidity for a flat torus through its dual lattice.

On a complete connected Riemannian manifold of nonpositive sectional curvature, the exponential map at every point is a covering map. If the manifold is simply connected, each exponential map is a diffeomorphism from a tangent space onto the manifold.

With outward normal $N(x)=x$, the unit sphere has $A(X,Y)=-⟨X,Y⟩N$. The Gauss equation therefore gives sectional curvature one on every tangent two-plane.