Math 2.0

Differential geometry Surfaces in Euclidean space Example

Pseudosphere

Example

For t>0t > 0 and θ∈ℝ\theta \in \R, let x(t,θ)=(sech⁡tcos⁡θ,  sech⁡tsin⁡θ,  t−tanh⁡t).x(t, \theta) = \left(\operatorname{sech} t \cos\theta, \; \operatorname{sech} t \sin\theta, \; t - \tanh t\right). This is the surface of revolution of the tractrix t↦(sech⁡t, t−tanh⁡t)t \mapsto (\operatorname{sech} t, \, t - \tanh t) about the zz-axis, and its image is called the pseudosphere. Its Gaussian curvature is −1-1 at every point.

Check. Write r(t)=sech⁡tr(t) = \operatorname{sech} t and z(t)=t−tanh⁡tz(t) = t - \tanh t. Then r′(t)=−sech⁡ttanh⁡tr'(t) = -\operatorname{sech} t \tanh t and z′(t)=tanh⁡2tz'(t) = \tanh^2 t, so r′2+z′2=tanh⁡2t (sech⁡2t+tanh⁡2t)=tanh⁡2tr'^2 + z'^2 = \tanh^2 t \,(\operatorname{sech}^2 t + \tanh^2 t) = \tanh^2 t. For t>0t > 0 the speed is tanh⁡t>0\tanh t > 0, and the arc length from t=0t = 0 is s=ln⁡cosh⁡ts = \ln\cosh t. So in terms of arc length, r=1/cosh⁡t=e−sr = 1/\cosh t = e^{-s}. For a surface of revolution with profile parametrized by arc length, K=−r′′(s)/r(s)K = -r''(s)/r(s), and here K=−e−s/e−s=−1K = -e^{-s}/e^{-s} = -1.

Remarks

The surface is not complete: as t→0t \to 0 it ends in a sharp circular edge of radius 11, where the parametrization stops being regular. This agrees with Hilbert's theorem, which says that no complete surface of constant curvature −1-1 sits smoothly and isometrically in ℝ3\R^3. Pieces of the pseudosphere are locally isometric to pieces of the hyperbolic plane, so they were used as early models of hyperbolic geometry.

Further reading

  1. Web https://en.wikipedia.org/wiki/Pseudosphere
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MSC53A05, 53A35
IllustratesGaussian curvature
Changed2 November 2026

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