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Riemannian Metrics Length Distance and Volume — Examples

1 · Prerequisites

2 · Summary

The worked metrics include Euclidean space, the round sphere, products, conformal planes, upper half-space, and a torus defined by periodic charts. Each construction checks positivity and displays its coefficients. Further calculations give circle distance, cross-component infinity, polar volume and divergence, and the full Hodge-star table in three dimensions. The constant-map counterexample also records the empty-source convention.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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The euclidean metric and its musical maps

Example

On Euclidean Rn, v=ividxi, α=iαii, and gradf=i(if)i.

Facts & Assumptions

Given: g=idxidxi, v=ivii, and α=iαidxi.

[F1]

Coordinate criterion for a riemannian metric: A tensor g=i,jgijdxidxj is Riemannian exactly when its coordinate matrix G=(gij) has smooth entries and is symmetric positive definite. Under J=x/y it transforms by Gy=JTGxJ.

[F2]

The musical maps are smooth inverse bundle isomorphisms: :TMTM and :TMTM are smooth inverse bundle isomorphisms.

[F3]

The gradient is characterized by inner products: The gradient is the unique smooth vector field Y satisfying g(Y,X)=Xf for every smooth vector field X.

[F4]

Musical isomorphisms: The musical maps are defined by v=g(v,) and by g(α,w)=α(w) for every tangent vector w.

Verification

technique · direct
1.1

The matrix of g is In, which is smooth, symmetric, and has vTInv=i(vi)2>0 for v0. For every basis vector j, F4 gives v(j)=g(v,j)=vj, hence v=ividxi. If α=iwii, then F4 gives wj=g(α,j)=α(j)=αj, hence α=iαii. Substitution in either order returns the original coefficients, as also required by F2.

F1F2F4given
2.1

Since df=i(if)dxi, the inner-product characterization gives gradf=(df)=i(if)i. In particular f(x)=12i(xi)2 has df=ixidxi and gradf=ixii.

F3step 1.1

Source locator

Lee, Example 13.1, p.328; musical isomorphisms and gradient, p.342. The quadratic-function instance is calculated above.

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The round metric on the sphere as an induced metric

Example

The Euclidean inclusion of Sn induces its round metric. In spherical coordinates on S2, g=dθ2+sin2θdφ2.

Facts & Assumptions

Given: Sn={pRn+1:p=1} and its usual smooth structure; i is inclusion.

[F1]

Pullback of a riemannian metric is riemannian exactly for immersions: Fh is Riemannian if and only if F is an immersion. In general it is positive semidefinite, with radical kerdFp at p.

[F2]

Coordinate criterion for a riemannian metric: A tensor g=i,jgijdxidxj is Riemannian exactly when its coordinate matrix G=(gij) has smooth entries and is symmetric positive definite. Under J=x/y it transforms by Gy=JTGxJ.

Verification

technique · direct
1.1

A tangent vector v to Sn satisfies pv=0 by differentiating p2=1. The inclusion differential sends this vector to the identical Euclidean vector, so is injective. Hence igE is Riemannian, and its value is v,w=vw.

F1given
2.1

For X(θ,φ)=(sinθcosφ,sinθsinφ,cosθ), one has Xθ=(cosθcosφ,cosθsinφ,sinθ) and Xφ=(sinθsinφ,sinθcosφ,0). Their dot products are 1, 0, and sin2θ, respectively. Thus the coordinate matrix is diag(1,sin2θ) on 0<θ<π and an angular interval of length less than 2π.

F2step 1.1
3.1

At either pole the spherical parametrization is not a chart, since Xφ=0. The intrinsic quadratic form remains vv>0 on every nonzero tangent vector by step 1.1; the vanishing coordinate coefficient at a pole therefore does not signify tensor degeneracy.

step 1.1step 2.1

Source locator

Lee, Proposition 13.9, p.331, and Example 13.16, p.333, round metric. The spherical-coordinate dot products are displayed above.

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The product riemannian metric

Example

The product metric on (M,g)×(N,h) is πMg+πNh, with block matrix diag(G,H).

Facts & Assumptions

Given: Finite-dimensional smooth Riemannian manifolds (M,g),(N,h), with their product smooth structure.

[F1]

Coordinate criterion for a riemannian metric: A tensor g=i,jgijdxidxj is Riemannian exactly when its coordinate matrix G=(gij) has smooth entries and is symmetric positive definite. Under J=x/y it transforms by Gy=JTGxJ.

[F2]

Products of smooth manifolds have a canonical product smooth structure: Let (M,S) and (N,T) be smooth manifolds of dimensions m and n. Then M×N with the product topology is a topological (m+n)-manifold. If A and B are smooth atlases with [A]=S and [B]=T, then the set of product charts A×B:={(V×W, φ×ψ):(V,φ)A, (W,ψ)B} is a smooth atlas on M×N, and the maximal atlas it generates is independent of the presenting atlases: it depends only on S and T. This maximal atlas is the product smooth structure of M×N.

Verification

technique · direct
1.1

In product coordinates a tangent vector is (v,w), and the projection differentials send it to v and w. Thus the sum of pullbacks evaluates on two vectors as g(v,v)+h(w,w), and has the stated block diagonal matrix. The product charts are smooth, and each coefficient is a smooth coefficient of g or h composed with a projection.

F2given
2.1

For (v,w)(0,0) at least one vector is nonzero. The sum g(v,v)+h(w,w) is therefore strictly positive, since each summand is nonnegative and the corresponding nonzero summand is positive. Symmetry holds term by term, so the coordinate criterion proves this is Riemannian. For the concrete product of two Euclidean lines the matrix is diag(1,1) and the squared norm of (3,4) is 9+16=25.

F1step 1.1

Source locator

Lee, Example 13.2 and equation (13.1), p.329, product metrics.

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A conformal metric on the plane

Example

For g=e2u(dx2+dy2) on R2, gradgf=e2u(fxx+fyy) and μg=e2udxdy.

Facts & Assumptions

Given: u,fC(R2).

[F1]

Conformal equivalence of riemannian metrics: Two Riemannian metrics are conformally equivalent if g~=e2ug for a smooth real function u on M. The positive smooth factor preserves the metric condition of def-riemannian-metric-and-riemannian-manifold. Equivalently g~=fg for smooth f>0, since u=12logf. Reflexivity uses u=0, reversal uses u, and composing rescalings adds their functions.

[F2]

Riemannian gradient: For a smooth real function f, its Riemannian gradient is gradgf=(df). prop-exterior-derivative-of-a-function-is-its-differential identifies df(X)=Xf. The smooth bundle isomorphism in thm-the-musical-maps-are-smooth-inverse-bundle-isomorphisms therefore makes the gradient a smooth vector field. In coordinates (gradgf)i=jgijjf. Constants, and all functions in dimension zero, have zero gradient.

[F3]

Riemannian volume density: The Riemannian volume density is μg=detGxdx1dxn in coordinates. The matrix is that of prop-coordinate-criterion-for-a-riemannian-metric, so its determinant is positive and smooth. The density frames and their absolute-Jacobian law are def-density-bundle-and-smooth-density. In dimension zero take the empty determinant to be one, giving weight one at every point, independently of orientation. The compatibility of these local formulas is proved in lem-the-riemannian-volume-density-is-coordinate-independent.

Verification

technique · direct
1.1

The factor e2u is smooth and strictly positive. The matrix is G=e2uI2, its inverse is e2uI2, and vTGv=e2u(vx2+vy2)>0 for v0, so this is the stated conformal metric.

F1given
2.1

Put Y=e2u(fxx+fyy). For every V=ax+by, g(Y,V)=fxa+fyb=df(V), so Y is the gradient. Since detG=e4u, its positive square root is e2u, giving the asserted density.

F2F3step 1.1
3.1

For the explicit instance u(x,y)=x and f(x,y)=y, these formulas yield g=e2x(dx2+dy2), gradgy=e2xy, and μg=e2xdxdy. At (0,0) the gradient is y and the density coefficient is 1.

step 2.1

Source locator

Lee, p.328, coordinate metric criterion; p.342, gradient characterization; Proposition 15.31, p.390, coordinate volume coefficient. The conformal instance and its determinant are derived above.

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The hyperbolic upper half space metric

Example

For n1, on {(x1,,xn1,y):y>0} the metric g=y2(i<ndxi2+dy2) has density yndx1dxn1dy.

Facts & Assumptions

Given: The upper half-space with its Euclidean open-subset smooth structure.

[F1]

Coordinate criterion for a riemannian metric: A tensor g=i,jgijdxidxj is Riemannian exactly when its coordinate matrix G=(gij) has smooth entries and is symmetric positive definite. Under J=x/y it transforms by Gy=JTGxJ.

[F2]

Riemannian volume density: The Riemannian volume density is μg=detGxdx1dxn in coordinates. The matrix is that of prop-coordinate-criterion-for-a-riemannian-metric, so its determinant is positive and smooth. The density frames and their absolute-Jacobian law are def-density-bundle-and-smooth-density. In dimension zero take the empty determinant to be one, giving weight one at every point, independently of orientation. The compatibility of these local formulas is proved in lem-the-riemannian-volume-density-is-coordinate-independent.

Verification

technique · direct
1.1

For y>0, y2 is smooth and positive. Hence G=y2In is a smooth symmetric positive-definite matrix, since vTGv=y2ivi2>0 when v0. It defines a Riemannian metric.

F1given
2.1

Its determinant is y2n and the positive square root is yn. The density definition therefore gives the asserted formula. At (0,,0,2) the matrix is 14In and the density coefficient is 2n.

F2step 1.1

Source locator

Lee, p.328, coordinate positive-definiteness criterion; Proposition 15.31, p.390, volume coefficient; pp.430–431, Riemannian density. The upper-half-space coefficients are computed above; no curvature or completeness statement is asserted.

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The flat torus metric from periodic euclidean coordinates

Example

The periodic Euclidean coordinates on Rn/Zn=(R/Z)n define a metric locally equal to idxi2, called the flat torus metric.

Facts & Assumptions

Given: The integer translation action on Rn and quotient map q.

[F1]

Coordinate criterion for a riemannian metric: A tensor g=i,jgijdxidxj is Riemannian exactly when its coordinate matrix G=(gij) has smooth entries and is symmetric positive definite. Under J=x/y it transforms by Gy=JTGxJ.

Verification

technique · direct
1.1

The map q is open since q1q(U)=mZn(U+m) is open when U is. Its restriction to a box of side lengths less than 1 is injective and hence a homeomorphism onto its open image. On overlap components the inverse charts differ by a constant integer translation. These smooth maps have derivative In.

given
2.1

For two distinct orbits represented by x,y, their displacement vectors xym never vanish. Only finitely many mZn have xym1, because every coordinate of such an integer vector is bounded. Taking the minimum of 1 and these finitely many positive lengths gives δ>0. Images of radius-δ/3 balls around x,y are disjoint, proving Hausdorffness. The images of rational boxes form a countable basis. Thus the charts in step 1.1 define a smooth manifold.

step 1.1
3.1

Since all transition derivatives are In, the local tensors idxi2 agree on overlaps. They glue to a smooth tensor with positive-definite identity matrix in every quotient chart. The coordinate criterion proves it is Riemannian. For example in dimension two the local vector 21+32 has squared norm 4+9=13, independent of the integer translate chosen for the chart. Local equality to the Euclidean metric is the flatness meant here.

F1step 1.1step 2.1

Source locator

Lee, p.332, definition of flatness as local Euclidean isometry and Theorem 13.14(b). The particular torus quotient atlas and metric descent are proved above.

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Length and distance on the circle

Example

On the unit circle with induced metric, d(eia,eib)=minkZba+2πk. Antipodes have two distinct minimizing semicircles.

Facts & Assumptions

Given: Real angles a,b, and the circle parametrization eit=(cost,sint).

[F1]

Riemannian speed and length: The Riemannian speed on a C1 piece is γ˙(t)g=gγ(t)(γ˙(t),γ˙(t)). Its length is Lg(γ)=jtj1tjγ˙(t)gdt. The curve convention is def-piecewise-c-one-curve-on-a-manifold and the norm is def-pointwise-norm-and-angle-from-a-riemannian-metric. Each integrand is continuous on its closed piece with the one-sided endpoint derivative, hence Riemann integrable and nonnegative. Values chosen at the finitely many corners do not change its integral. For a singleton interval the empty sum is zero; a constant curve also has zero length. Partition independence is established next.

[F2]

Riemannian distance on a connected manifold: On a connected Riemannian manifold define dg(p,q)=inf{Lg(γ):γ is piecewise C1 from p to q}. Lengths are those of def-riemannian-speed-and-length. For each pair p,q, lem-any-two-points-in-a-connected-smooth-manifold-can-be-joined-by-a-piecewise-c-one-curve supplies a curve, so the set of lengths is nonempty, contains a finite real number and is bounded below by zero. Applying the least-upper-bound property cor-cauchy-reals-lub-complete to the negatives gives a finite nonnegative infimum. On the empty connected manifold this defines the empty distance function; there are no pairs to evaluate. No minimizing curve is part of this definition.

[F3]

Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative: Let a<b. Suppose G:[a,b]R is continuous on [a,b] and differentiable on (a,b). If f:[a,b]R is Riemann integrable and f(x)=G(x)(a<x<b), then abf=G(b)G(a). No derivative of G at either endpoint is assumed, and the two endpoint values assigned to the integrable extension f do not enter the conclusion.

Verification

technique · direct
1.1

For any piecewise C1 circle path, the inverse images of smooth angle arcs form an open cover of its compact parameter interval. A finite subcover has a positive Lebesgue number; subdividing more finely than it, and at the original differentiability breakpoints, puts each piece in one angle arc. Start the first angle at a, and add a multiple of 2π to each successive local angle to match the preceding endpoint. This yields a continuous piecewise C1 lift θ starting at a, ending at b+2πk for some integer k.

given
2.1

Differentiation of (cosθ,sinθ) gives squared speed θ2(sin2θ+cos2θ)=θ2. Consequently L=θθ=ba+2πk, where Newton–Leibniz is applied on each closed smooth piece and the endpoint increments telescope.

F1F3step 1.1
3.1

There is an integer k0 with δ=ba+2πk0[π,π], obtained by rounding (ab)/(2π) to a nearest integer. Every other representative has absolute value at least δ. The path tei(a+tδ) on [0,1] has constant speed δ and attains that lower bound, proving the distance formula.

F2step 2.1
4.1

For ba=π, the representatives δ=π and δ=π both minimize. The paths ei(a+πt) and ei(aπt) have length π and disjoint interior semicircle images. For equal endpoints δ=0, the same construction is a constant path of length zero.

step 3.1

Source locator

Lee, pp. 331 and 337–338, induced metric and distance; the finite angle lift and minimization over integers are proved above.

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A riemannian distance with no cross component finite value

Example

On M=R×{0,1} with metric dx2 on each component, d((x,i),(y,i))=xy and d((x,0),(y,1))=+.

Facts & Assumptions

Given: The two disjoint Euclidean lines with their disjoint-union smooth structure.

[F1]

Extended riemannian distance on a disconnected manifold: The extended Riemannian distance on arbitrary M is the componentwise Riemannian distance when two points are in the same component, and + otherwise. Within each component use thm-riemannian-distance-is-a-metric. Components are open, since small coordinate balls are connected. A continuous curve cannot meet two components because its connected interval image is connected, so the cross-component curve family is empty, with inf=+. This is an extended metric: if two endpoints are in different components, any third point is in a different component from at least one of them, so the triangle inequality has infinite right side. It is a finite metric precisely when there are no distinct components. Empty and singleton manifolds retain their unique distances.

[F2]

Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative: Let a<b. Suppose G:[a,b]R is continuous on [a,b] and differentiable on (a,b). If f:[a,b]R is Riemann integrable and f(x)=G(x)(a<x<b), then abf=G(b)G(a). No derivative of G at either endpoint is assumed, and the two endpoint values assigned to the integrable extension f do not enter the conclusion.

Verification

technique · direct
1.1

The two lines are disjoint open-and-closed components; their usual charts give a Hausdorff second-countable smooth one-manifold and metric coefficient 1>0. A continuous path cannot meet both components, because their inverse images would separate its connected interval. Hence the cross-component admissible family is empty and its length infimum is +. In particular d((0,0),(0,1))=+.

F1given
2.1

Within a component, an admissible path has length γγ=yx, by Newton–Leibniz on its finitely many smooth pieces. The path t(x+t(yx),i) for 0t1 attains this bound. The same-component infimum is therefore yx; for example d((0,0),(3,0))=3.

F1F2step 1.1

Source locator

Lee, pp. 337–338, connected distance and Euclidean calculation; cross-component infinity follows from the declared extended-distance convention.

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Volume density in polar coordinates

Example

In a polar chart of the Euclidean plane, the density is rdrdθ and the positive-oriented volume form is rdrdθ.

Facts & Assumptions

Given: x=rcosθ, y=rsinθ, with r>0 and θ in an open interval of length less than 2π.

[F1]

Riemannian volume density: The Riemannian volume density is μg=detGxdx1dxn in coordinates. The matrix is that of prop-coordinate-criterion-for-a-riemannian-metric, so its determinant is positive and smooth. The density frames and their absolute-Jacobian law are def-density-bundle-and-smooth-density. In dimension zero take the empty determinant to be one, giving weight one at every point, independently of orientation. The compatibility of these local formulas is proved in lem-the-riemannian-volume-density-is-coordinate-independent.

[F2]

The riemannian volume density is coordinate independent: The local Riemannian volume densities glue to a positive smooth density independent of coordinates.

Verification

technique · direct
1.1

Differentiation gives dx=cosθdrrsinθdθ and dy=sinθdr+rcosθdθ. Expanding dx2+dy2, the mixed terms cancel and the diagonal terms sum to dr2+r2dθ2. Thus detG=r2, whose positive square root is r on this domain, and the density is rdrdθ.

F1given
2.1

The wedge expansion gives dxdy=r(cos2θ+sin2θ)drdθ=rdrdθ. The Jacobian is positive, so the chart has the standard orientation and this is its positive volume form. The absolute Jacobian density law agrees with step 1.1. At r=2, both coordinate coefficients are 2. The excluded value r=0 is a failure of polar coordinates, not a zero of the Euclidean density.

F2step 1.1

Source locator

Lee, Example 13.12, p.332, polar metric; Proposition 15.31, p.390, coordinate volume formula; pp.430–431, Riemannian density.

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Hodge star on euclidean three space

Example

In standard oriented Euclidean R3, 1=dxdydz, dx=dydz, dy=dzdx, dz=dxdy, and 2=id in every degree.

Facts & Assumptions

Given: The orthonormal coframe (dx,dy,dz) with volume ν=dxdydz.

[F1]

Hodge star is a smooth bundle isomorphism: The Hodge star exists uniquely and is a smooth bundle isomorphism in every degree 0kn.

[F2]

Hodge star squared sign: On real k-forms, 2=(1)k(nk)id.

Verification

technique · direct
1.1

The complementary-wedge formula gives 1=ν. For the one-forms, dx(dydz)=ν, dy(dzdx)=ν, and dz(dxdy)=ν: the latter two permutations each have two transpositions. These complementary two-forms wedge to zero with either of the other one-form basis vectors because of a repeated factor. Thus they satisfy all pairings in the defining identity and are the displayed stars.

F1given
2.1

Similarly (dxdy)dz=ν, (dxdz)(dy)=ν, and (dydz)dx=ν. Distinct two-form basis vectors have zero pairing and wedge to zero with the listed complementary one-form. Consequently (dxdy)=dz, (dxdz)=dy, (dydz)=dx, and ν=1. For example (2dx+3dy)=2dydz+3dzdx by linearity.

F1step 1.1
3.1

For k=0,1,2,3, the exponent k(3k) is respectively 0,2,2,0, always even. The star-square formula therefore gives 2=id in all these degrees, in agreement with the table.

F2step 1.1step 2.1

Source locator

Lee, Problem 16-18(a–e), pp. 437–438, and Problem 16-19, p. 438, Euclidean Hodge-star computations.

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Divergence in polar coordinates

Example

For the Euclidean metric in polar coordinates, div(Xrr+Xθθ)=r1r(rXr)+θXθ.

Facts & Assumptions

Given: A polar chart with r>0, metric matrix diag(1,r2), and a smooth vector field X.

[F1]

Coordinate formula for riemannian divergence: In coordinates, divgX=(detG)1/2i=1ni((detG)1/2Xi).

Verification

technique · direct
1.1

The positive square root of the metric determinant is r. Substituting into the divergence formula gives divX=r1{r(rXr)+θ(rXθ)}. Since θr=0, the second term is θXθ, proving the formula.

F1given
2.1

The orthonormal polar frame is er=r, eθ=r1θ. Thus if X=aer+beθ, its coordinate components are Xr=a, Xθ=b/r, and the formula becomes r1r(ra)+r1θb. For X=rr, it gives r1r(r2)=2. For X=θ it gives 0.

step 1.1

Source locator

Lee, p.423, definition of divergence and Exercise 16.31; Example 13.12, p.332, polar metric. The coordinate divergence theorem declared as F1 supplies the local coefficient formula used above.

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A degenerate pullback metric under a constant map

Statement refuted

A constant smooth map always pulls a Riemannian metric back to a Riemannian metric.

Facts & Assumptions

Given: F:MR is constant, M is a smooth manifold of dimension n>0, and the target metric is dy2.

[F1]

Pullback of a riemannian metric as a tensor: For smooth F:MN and a Riemannian metric h on N, its pullback tensor is (Fh)p(v,w)=hF(p)(dFpv,dFpw). This is def-pullback-of-a-covariant-tensor-field for the tensor in def-riemannian-metric-and-riemannian-manifold. It is always symmetric and positive semidefinite; the name does not assert positive definiteness. Smoothness and the precise immersion criterion are established next.

[F2]

Pullback of a riemannian metric is riemannian exactly for immersions: Fh is Riemannian if and only if F is an immersion. In general it is positive semidefinite, with radical kerdFp at p.

[F3]

Smooth manifolds and their smooth charts: A smooth n-manifold is a pair (M,S) in which M is a topological n-manifold (def-topological-manifold-without-boundary) and S is a smooth structure on M: a maximal smooth atlas (thm-each-smooth-atlas-is-contained-in-a-unique-maximal-smooth-atlas). Because thm-each-smooth-atlas-is-contained-in-a-unique-maximal-smooth-atlas sends every smooth atlas to the unique maximal atlas containing it, a smooth manifold is equivalently specified by a topological manifold M together with any one smooth atlas A, the structure being the generated [A]. A chart (U,φ)S is called a smooth chart (or a chart of the smooth structure); its domain is a coordinate domain and its coordinate functions are smooth coordinates on U. When the structure is clear from context, the manifold itself is written M in place of (M,S).

[F4]

Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces: Let nN. For n1, put Rn:=k<nR with its usual topology; for n=0 put R0:={0}, the one-point space. A topological n-manifold without boundary (or briefly an n-manifold) is a topological space M satisfying: 1. M is Hausdorff (def-hausdorff-space); 2. M is second countable (def-second-countable-space); 3. M is locally Euclidean of dimension n: every pM has an open neighbourhood UM homeomorphic to an open subset of Rn (def-homeomorphism-and-open-maps). The empty space satisfies all three conditions vacuously, so M= is an n-manifold for every n; this degenerate instance is kept, and statements about nonempty manifolds name the hypothesis. In dimension zero, condition 3 forces the one-point neighbourhoods of points to be open singletons, so a 0-manifold is exactly a discrete second-countable space with at most countably many points.

Counterexample

technique · direct
1.1

In each coordinate chart, the component of F is constant, so its differential is zero. The pullback formula gives (Fdy2)p(v,w)=dyF(p)2(0,0)=0 for all tangent vectors at every point. Thus the pullback is the zero smooth tensor.

F1given
2.1

If M is nonempty, fix a point p and a chart there. Its first coordinate tangent vector is nonzero because n>0. The zero tensor has quadratic value zero on that vector and is not positive definite, so is not Riemannian. Equivalently dFp=0 is not injective on the positive-dimensional tangent space. In particular M=R and F(x)=0 provide an explicit counterexample: (Fdy2)0(x,x)=0.

F2F3F4step 1.1
3.1

If M is empty, its unique tensor is smooth and the requirement of positive definiteness at every point has no instances, so the pullback is vacuously Riemannian. Together with step 2.1, this shows that for the stated positive dimension it is not Riemannian exactly when M is nonempty. The counterexample uses the nonempty real line, so the empty case does not rescue the universal assertion.

F3F4step 1.1step 2.1

Source locator

Lee, pp. 330–331, pullback metrics and Proposition 13.9; the empty-manifold convention is that of the cited library definitions.

Sources