advanced28 min studyLesson 451 in Mathematics

Riemannian metrics and length functionals

Turn smoothly varying tangent-space inner products into invariant curve lengths, energy, and an intrinsic distance.

On this page

What you’ll understand

  • Interpret a positive-definite metric tensor independently of coordinates.
  • Distinguish length, energy, and distance and prove their basic inequality.
  • Compute lengths in a constant anisotropic metric and explain parameter dependence.

A map's coordinate grid is not a ruler. The same displacement can have different coordinate components in different charts, while its physical or geometric length must remain unchanged. A Riemannian metric supplies that missing ruler on each tangent space and varies smoothly from point to point. Integrating its local measurements along curves creates an intrinsic geometry without requiring the manifold to sit inside a surrounding Euclidean space.

A metric is a field of positive inner products

Let M be a smooth finite-dimensional manifold. A Riemannian metric assigns a positive-definite symmetric bilinear form gpg_p to each tangent space, with smooth components in every chart. For a vector with coordinate column v, squared norm is vTGvv^{\mathsf T}Gv, where G is the metric matrix at that point. Off-diagonal entries describe nonorthogonal coordinate directions, not a failure of positive definiteness. Unlike a Lorentzian metric, every nonzero tangent vector has strictly positive squared norm.

Under a coordinate change, vector components and the metric matrix transform together, leaving their pairing unchanged. Thus an identity matrix is a statement about a chosen frame, not a coordinate-independent definition of flatness. One can choose an orthonormal frame at a point for every Riemannian metric, including a curved one. An embedded surface inherits a metric by restricting the ambient inner product to tangent vectors, but intrinsic constructions below need only g and the smooth manifold structure.

gp(v,w)=∑i,jgij(p)viwj,∥v∥g=gp(v,v)g_p(v,w)=\sum_{i,j}g_{ij}(p)v^iw^j,\qquad \|v\|_g=\sqrt{g_p(v,v)}
The smooth matrix (gᵢⱼ) is symmetric positive definite in every coordinate chart; vector and metric transformations preserve this scalar pairing.↔ Scroll wide equations horizontally.

Length ignores speed, while energy remembers it

For a piecewise smooth curve on a finite parameter interval, length integrates its metric speed. An orientation-preserving regular reparameterization leaves that length unchanged because the speed change cancels the parameter differential. Energy instead integrates half the squared speed and depends on how the same geometric path is traversed. Fixing the parameter duration makes energy a useful variational quantity, but changing that duration changes its numerical value even when the path and length stay the same.

The Cauchy–Schwarz inequality bounds the square of the integrated speed by duration times the integrated squared speed. Equality holds precisely when speed is constant almost everywhere. Thus among parameterizations of a fixed regular path on a fixed duration, constant speed minimizes energy. This statement concerns parameterization of one path; finding a shortest path among different curves is another optimization problem. It also does not imply a globally shortest curve exists between every pair without further geometric assumptions.

L(γ)=∫0T∥γ˙∥g dt,E(γ)=12∫0T∥γ˙∥g2 dt,L(γ)2≤2TE(γ)L(\gamma)=\int_0^T\|\dot\gamma\|_g\,dt,\qquad E(\gamma)=\frac12\int_0^T\|\dot\gamma\|_g^2\,dt,\qquad L(\gamma)^2\le2T E(\gamma)
The duration T is positive. Equality in the last inequality characterizes constant speed almost everywhere.↔ Scroll wide equations horizontally.

Distance is an infimum over admissible curves

On a connected Riemannian manifold, define the distance between two points as the infimum of lengths of piecewise smooth connecting curves. Connected smooth manifolds are path connected, so this infimum is finite. Positive definiteness and local comparison with Euclidean coordinates establish positive distance for distinct points, while reversing and concatenating curves give symmetry and the triangle inequality. The resulting metric induces the original manifold topology, even though its numerical distances depend on g.

An infimum need not be attained on an incomplete manifold: a sequence of routes can approach a missing point and approach a limiting length without realizing it. Later completeness theorems supply conditions guaranteeing minimizing geodesics. Constant rescaling illustrates the distinction between coordinates and geometry: replacing g by c²g for positive c multiplies all lengths and distances by c and energies by c². Merely relabeling coordinates changes no intrinsic measurement when the tensor is transformed correctly.

A plane with direction-dependent costs

On the plane with constant metric g=4 dx2+dy2g=4\,dx^2+dy^2, find the distance from (0,0) to (1,2), the unit-duration straight path's energy, and the length of a horizontal-then-vertical route.

  1. The linear map (x,y)↦(2x,y) converts this metric to the ordinary Euclidean metric. The endpoint displacement becomes (2,2), so its Euclidean straight-line length is √8=2√2 and no other connecting route is shorter.

  2. The straight parameterization γ(t)=(t,2t), for zero through one, has squared metric speed 4+4=8. Its energy is half of eight, equal to four; its squared length eight agrees with twice that energy.

  3. The horizontal leg has metric length two, and the vertical leg also has length two, for a total of four. This longer broken route reaches the same endpoints but is not distance minimizing.

The result

The distance is 2√2, straight-path energy is four for duration one, and the coordinate-axis detour has length four.

Explore the model

Interactive lab

Interactive lab · matrix

A constant pullback metric

Pause and Reset, then set diagonal scale a = 2 and shear b = 0. For A=[[a,b],[0,1]], predict the pullback metric is AᵀA=diag(4,1) and a vector (1,1) has length √5. Compare the transformed square and area scale two. Reset and set both a = 1 and b = 1: the same vector still maps to (2,1), but the metric matrix and area scale differ. The instrument shows constant linear maps, not a position-dependent curved metric.

A constant pullback metricThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOne vector's measured length does not determine a whole metric; the pullback inner product is AᵀA, whereas its area factor is |det A|.

Check your understanding

What changes when the same regular geometric path is traversed at a nonconstant speed over a fixed duration?

Try it yourself

For g=9 dx2+4 dy2g=9\,dx^2+4\,dy^2, compute the length and energy of the constant-speed straight path from (0,0) to (2,−1) traversed in time T=2. Verify the length-energy equality.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The velocity is the endpoint displacement divided by two. Compute its squared metric norm before integrating.

Compare with the solution

The velocity is (1,−1/2), with squared metric norm 9+4(1/4)=109+4(1/4)=10. Thus length is 2102\sqrt{10} and energy is (1/2)(2)(10)=10(1/2)(2)(10)=10. Squared length is forty, equal to 2TE=2(2)(10)=402TE=2(2)(10)=40. Equivalently, transform the displacement by (x,y)↦(3x,2y) to obtain (6,−2), of norm √40. The energy value uses duration two; the same path traversed in duration one would have the same length but twice the energy.

Take it with you

  • A metric tensor measures tangent vectors invariantly when coordinates transform consistently.
  • Length is geometric; energy also depends on speed and duration.
  • Intrinsic distance is a curve-length infimum whose attainment requires additional hypotheses.

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