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Eugenio Beltrami

Eugenio Beltrami

A course in two strands: the life of a railway clerk from Cremona who became one of the founders of Italian differential geometry, and the mathematics with which he gave Lobachevsky’s geometry a concrete home and proved it as consistent as Euclid’s.

Born 16 November 1835, Cremona Died 18 February 1900, Rome Saggio models of hyperbolic geometry, 1868 Level advanced undergraduate → graduate
Two of Beltrami’s 1868 models. Left: the disk, where hyperbolic lines are chords; through P pass two limiting parallels to l (solid) and infinitely many ultraparallels (dashed). Right: the pseudosphere, a surface of constant curvature −1/k² that carries a piece of the same geometry.
Lesson 00

How to use this course

Eugenio Beltrami is remembered above all for the papers of 1868 that ended the debate over non-Euclidean geometry. Nikolai Lobachevsky and János Bolyai had developed the geometry in which the parallel postulate fails, but nobody had shown that it could not contain a hidden contradiction. Beltrami found it already present inside Euclidean geometry, on surfaces of constant negative curvature and in a disk whose lines are straight chords. Any contradiction in the new geometry would therefore be a contradiction in the old. His name also survives on an operator, an equation, an identity and a class of vector fields, and he gave the first proof of the singular value decomposition.

Prerequisites for Part II
  • Multivariable calculus and linear algebra.
  • Curves and surfaces in \(\mathbb R^3\): first fundamental form, Gaussian curvature, geodesics.
  • Hyperbolic functions \(\sinh,\cosh,\tanh\) and their inverses.
  • The courses on Nikolai Lobachevsky and Giovanni Girolamo Saccheri give the synthetic background for Lessons 06–08.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. We follow the conventions of the Lobachevsky and Saccheri courses: the hyperbolic plane has curvature \(-1/k^2\), its upper half-plane model has metric \(ds^2=k^2(dx^2+dy^2)/y^2\), and in the hyperboloid model with \(k=1\) we use \(\langle x,y\rangle=-x_0y_0+x_1y_1+x_2y_2\) and \(\cosh d(x,y)=-\langle x,y\rangle\).

Reference

Timeline

1835Born 16 November in Cremona, Lombardy, then part of the Austrian Empire, into a family of artists.
1853–56Studies mathematics at the University of Pavia under Francesco Brioschi. Leaves without a degree after expulsion from the Collegio Ghislieri for his Risorgimento sympathies, and for lack of money.
1856–62Secretary to an engineer of the Lombardy–Venetia railway, in Verona and then Milan. Resumes mathematics on his own.
1862First papers, on the differential geometry of curves. Appointed to teach algebra and analytic geometry at Bologna.
1864Chair of geodesy at Pisa. Friendship with Enrico Betti; meets Bernhard Riemann.
1864–65Ricerche di analisi applicata alla geometria: differential parameters.
1865Surfaces mapped onto the plane with geodesics going to straight lines are exactly those of constant curvature.
1866Chair of rational mechanics at Bologna.
1868Saggio di interpretazione della geometria non-euclidea and Teoria fondamentale degli spazii di curvatura costante.
1871Felix Klein places Beltrami’s disk in projective geometry and names the geometry hyperbolic.
1873Chair of rational mechanics at the new University of Rome. Sulle funzioni bilineari: the singular value decomposition.
1876Chair of mathematical physics at Pavia.
1880–82Elasticity in spaces of constant curvature.
1889Draws attention to Saccheri’s Euclides vindicatus. Paper on Maxwell’s electromagnetic theory.
1891Returns to Rome.
1892Honorary member of the London Mathematical Society.
1898President of the Accademia dei Lincei, succeeding Brioschi.
1899Senator of the Kingdom of Italy.
1900Dies 18 February in Rome, aged 64.
1902–20Opere matematiche, four volumes, Milan.
Part I · Formation
Lesson 01

Cremona and Pavia, 1835–1856

Eugenio Beltrami was born in Cremona, in Austrian-ruled Lombardy, on 16 November 1835. His family were artists. His grandfather engraved cameos and precious stones, his father painted miniatures, and his mother, Elisa Barozzi, came from Venice. Music stayed important to him all his life, and he wrote a paper on the mathematical theory of the diatonic scale.

In 1853 he entered the University of Pavia to study mathematics, as a student of the Collegio Ghislieri. His teacher was Francesco Brioschi, one of the young mathematicians who would rebuild Italian mathematics after unification, and a lifelong supporter of Beltrami. Lombardy was restless under Austrian rule, and Beltrami sympathised with the Risorgimento, the movement for Italian independence and unity. In 1856 he was expelled from the Collegio for his political views. Short of money, he left the university without a degree.

Lesson 02

The railway years, 1856–1862

Beltrami found work as secretary to an engineer of the Lombardy–Venetia railway company, first in Verona and then in Milan. For about six years mathematics was a private pursuit. Meanwhile the political map changed around him. After the war of 1859 Lombardy passed from Austria to the Kingdom of Sardinia, and in 1861 the Kingdom of Italy was proclaimed. The new state needed professors for its universities.

In Milan Beltrami returned to serious study. His first papers, on the differential geometry of curves, appeared in 1862. That same year, still without a degree, he was appointed to teach complementary algebra and analytic geometry at the University of Bologna, without the usual competition. He was twenty-six.

Lesson 03

Bologna and Pisa, 1862–1873

In 1864 Beltrami moved to the chair of geodesy at Pisa. There he became a close friend of Enrico Betti, the leading analyst in Italy, and met Bernhard Riemann, who spent long periods in Pisa in these years for his health. Riemann’s ideas on curved spaces were not yet in print, but they were in the air around Betti. Beltrami’s own work turned to surfaces. Following Carl Friedrich Gauss and Gabriel Lamé, he studied invariant expressions on a surface (Lesson 04) and asked which surfaces can be mapped to the plane so that geodesics become straight lines (Lesson 05).

In 1866 he returned to Bologna as professor of rational mechanics, a post he held until 1873. In Bologna he wrote the two papers of 1868 for which he is best known. The Saggio di interpretazione della geometria non-euclidea (Essay on the interpretation of non-Euclidean geometry) showed that Lobachevsky’s plane geometry is the geometry of a surface of constant negative curvature. He said he had tried to find a real “substrate” for the doctrine. His colleague Luigi Cremona worried that the argument might be circular, and Beltrami held the paper back. Riemann’s habilitation lecture of 1854, published by Richard Dedekind in 1868, convinced him that his methods were sound. It also showed him how to go beyond two dimensions, which he did in the Teoria fondamentale degli spazii di curvatura costante (Fundamental theory of spaces of constant curvature) the same year. Jules Hoüel soon translated both papers into French.

DiscussionGauss, Lobachevsky and Bolyai were all convinced that non-Euclidean geometry was consistent. Why was a model needed, and what exactly does a model prove? Compare with the construction of the complex numbers as points of the plane.

Part II · The mathematics
Lesson 04

Differential parameters and the Laplace–Beltrami operator

For the general reader. On a curved surface there are no preferred coordinates, so a formula that depends on the choice of coordinates says nothing about the surface itself. Beltrami looked for expressions that do not change when coordinates change. The most important of them generalises the Laplacian, the operator behind heat flow, waves and electric potential, to any curved surface. Today it is called the Laplace–Beltrami operator.

Goals
  • Define Beltrami’s first and second differential parameters.
  • Prove the invariance of the second parameter through Green’s formula.
  • Compute it in the hyperbolic half-plane and meet the Beltrami equation.

Let a surface carry coordinates \(u^1,u^2\) and metric \(ds^2=g_{ij}\,du^i\,du^j\), with inverse matrix \(g^{ij}\) and \(g=\det(g_{ij})\). Lamé had introduced differential parameters in Euclidean space. Beltrami’s Ricerche di analisi applicata alla geometria (1864–65) defined them on any surface.

Definition and theorem (Beltrami)

For smooth functions \(f,h\) put

\[ \Delta_1 f=g^{ij}\,\partial_if\,\partial_jf,\qquad \Delta_2 f=\frac{1}{\sqrt g}\,\partial_i\big(\sqrt g\,g^{ij}\,\partial_jf\big). \]

Both are invariants: their values at a point do not depend on the coordinates. On a compact region \(D\) with boundary curve \(\partial D\), outward unit normal \(n\) and area element \(dA=\sqrt g\,du^1du^2\),

\[ \int_D h\,\Delta_2 f\,dA=-\int_D g^{ij}\,\partial_ih\,\partial_jf\,dA+\oint_{\partial D}h\,\frac{\partial f}{\partial n}\,ds. \]

Proof sketch. \(\Delta_1f=|\nabla f|^2\), the squared length of the gradient, is invariant because \(g^{ij}\) transforms inversely to \(g_{ij}\). For \(\Delta_2\), apply the divergence theorem in the coordinate plane to the vector field \(h\sqrt g\,g^{ij}\partial_jf\). This gives the formula above, in which every term except \(\int h\,\Delta_2 f\,dA\) is visibly invariant. Since \(h\) is arbitrary, \(\Delta_2f\) is invariant too. This is Beltrami’s generalisation of Green’s theorem.

In Euclidean coordinates \(\Delta_2\) is the ordinary Laplacian. On a surface it is now written \(\Delta_g\) or \(\Delta_{LB}\) and works in any dimension with the same formula. Its eigenvalues measure the vibrations of a curved drum, and in the hyperbolic plane they connect to number theory through Atle Selberg’s trace formula.

Isothermal coordinates and the Beltrami equation

Coordinates are isothermal if \(ds^2=\lambda\,(du^2+dv^2)\). Then \(\Delta_2=\lambda^{-1}(\partial_u^2+\partial_v^2)\), so harmonic functions are the same as in the plane. Gauss had shown that such coordinates exist for analytic metrics. Write a general metric with \(z=u+iv\) as \(ds^2=\lambda\,|dz+\mu\,d\bar z|^2\) with \(|\mu|<1\). A function \(w=f(z)\) gives isothermal coordinates exactly when \(dw\) is a multiple of \(dz+\mu\,d\bar z\), that is, when

\[ \frac{\partial f}{\partial\bar z}=\mu\,\frac{\partial f}{\partial z}. \]

This is the Beltrami equation, which arose in his work on complex variables on surfaces. In the twentieth century it became the basic equation of quasiconformal mappings.

Exercise 4.1In the half-plane with \(ds^2=k^2(dx^2+dy^2)/y^2\), compute \(\Delta_2\). Show that \(y^s\) is an eigenfunction and find its eigenvalue.

Show solution
Here \(g_{ij}=(k^2/y^2)\delta_{ij}\), \(\sqrt g=k^2/y^2\), \(g^{ij}=(y^2/k^2)\delta_{ij}\), so \(\sqrt g\,g^{ij}=\delta_{ij}\) and \(\Delta_2=\frac{y^2}{k^2}(\partial_x^2+\partial_y^2)\). Then \(\Delta_2y^s=\frac{y^2}{k^2}s(s-1)y^{s-2}=\frac{s(s-1)}{k^2}y^s\).

Exercise 4.2In the same half-plane, compute \(\Delta_1\) of \(f=k\ln y\). What does the result say about the level curves \(y=\text{const}\)?

Show solution
\(\partial_yf=k/y\), so \(\Delta_1f=\frac{y^2}{k^2}\cdot\frac{k^2}{y^2}=1\). A function with \(|\nabla f|=1\) measures distance, so the level curves are parallel: the curves \(y=c\) are equidistant in the sense that \(y=c\) and \(y=ce^{t}\) are everywhere a distance \(kt\) apart along the orthogonal lines \(x=\text{const}\). They are the horocycles of the Lobachevsky course.

Exercise 4.3For the flat metric \(ds^2=du^2+4\,dv^2\), find \(\mu\) and \(\lambda\) with \(ds^2=\lambda|dz+\mu\,d\bar z|^2\), and solve the Beltrami equation.

Show solution
Try \(\mu=-\tfrac13\): \(dz-\tfrac13d\bar z=\tfrac23du+\tfrac43i\,dv\), with \(|\cdot|^2=\tfrac49(du^2+4dv^2)\). So \(\lambda=\tfrac94\). The function \(f=z-\tfrac13\bar z\) has \(f_{\bar z}=-\tfrac13=\mu f_z\), and \(w=f=\tfrac23u+\tfrac43iv\) satisfies \(|dw|^2=\tfrac49ds^2\): an isothermal coordinate.
Lesson 05

Geodesic maps and constant curvature

For the general reader. A map of the globe cannot keep every distance right. It can, however, send every great circle to a straight line: project from the centre of the earth onto a flat sheet. Beltrami asked which curved surfaces allow a map with this property. His answer, in 1865, was the surfaces of constant curvature, and only those. The negatively curved case pointed straight at Lobachevsky’s geometry.

Goals
  • State Beltrami’s theorem on geodesic maps.
  • Write the geodesic equation for a curve \(v=v(u)\) and read off when lines are geodesics.
  • Construct the maps for the sphere and for curvature \(-1/k^2\).
Theorem (Beltrami, 1865)

A surface admits, near each point, local coordinates \((u,v)\) in which every geodesic is a straight line \(au+bv=c\) if and only if its Gaussian curvature is constant.

The geodesic equation for a graph

With Christoffel symbols \(\Gamma^k_{ij}\) of the metric, a curve \(v=v(u)\) is a geodesic (after reparametrisation) exactly when

\[ v''=\Gamma^1_{22}\,v'^3+\big(2\Gamma^1_{12}-\Gamma^2_{22}\big)v'^2+\big(\Gamma^1_{11}-2\Gamma^2_{12}\big)v'-\Gamma^2_{11}. \]

All non-vertical straight lines (\(v''=0\) for every slope and intercept) are geodesics exactly when the four coefficients vanish identically. Vertical lines need \(\Gamma^1_{22}=0\), already among the conditions. Beltrami showed that these four first-order conditions on the metric, together with their integrability conditions, force the Gaussian curvature to be constant. The calculation is long; a modern route goes through the projective curvature of the connection.

The converse: the models

For the sphere of radius \(R\), central projection onto a tangent plane (the gnomonic projection) sends great circles to lines. In coordinates on that plane the metric is

\[ ds^2=R^2\,\frac{(R^2+v^2)\,du^2-2uv\,du\,dv+(R^2+u^2)\,dv^2}{(R^2+u^2+v^2)^2}. \]

Replacing \(R^2\) by \(-k^2\) throughout, and changing the overall sign, gives Beltrami’s metric of curvature \(-1/k^2\) on the disk \(u^2+v^2<k^2\):

\[ ds^2=k^2\,\frac{(k^2-v^2)\,du^2+2uv\,du\,dv+(k^2-u^2)\,dv^2}{(k^2-u^2-v^2)^2}. \]

Its geodesics are the chords of the disk. Lesson 07 studies this model in detail.

Exercise 5.1Project the sphere \(x^2+y^2+z^2=R^2\) from its centre onto the plane \(z=R\), by \((x,y,z)\mapsto(Rx/z,\ Ry/z)\). Show that great circles go to straight lines. Which part of the sphere is mapped?

Show solution
A great circle lies in a plane \(ax+by+cz=0\). Dividing by \(z\) and multiplying by \(R\) gives \(aU+bV+cR=0\) for the image \((U,V)\), a line. Only the open hemisphere \(z>0\) is mapped; antipodal points have the same image, and the equator goes to infinity. The map is local, as the theorem says.

Exercise 5.2A cylinder and a cone have curvature 0. Explain how to map them so that geodesics become lines. Why does the map fail globally?

Show solution
Cut along a ruling and unroll the surface into the plane. Unrolling is an isometry, so geodesics go to straight lines. Globally, a geodesic on a cylinder can be a helix that wraps around, and on the unrolled sheet it becomes a family of parallel segments, one for each turn. The map is a local isometry, not a global one.

Exercise 5.3Check the geodesic equation on the Euclidean plane in polar coordinates \((u,v)=(r,\theta)\), where \(\Gamma^1_{22}=-r\) and \(\Gamma^2_{12}=1/r\), and all other symbols vanish. Are straight lines in the \((r,\theta)\) plane geodesics?

Show solution
The coefficients are \(\Gamma^1_{22}=-r\ne0\) and \(\Gamma^1_{11}-2\Gamma^2_{12}=-2/r\ne0\). So the equation \(\theta''=-r\theta'^3-\tfrac2r\theta'\) is not \(\theta''=0\), and lines in the \((r,\theta)\) plane are not geodesics (for example \(\theta=r\), a straight line in the \((r,\theta)\) plane, is a spiral in the Euclidean plane). The theorem is about the existence of some good coordinates, not about every coordinate system.
Lesson 06

The pseudosphere

For the general reader. Pull a weight across a table by a string of fixed length, starting with the weight beside a straight edge. The path it traces is a tractrix. Spin that curve about the edge and you get a trumpet-shaped surface, the pseudosphere. Beltrami showed that its intrinsic geometry is exactly a piece of Lobachevsky’s plane. Geodesic triangles drawn on it have angle sums less than two right angles, with the defect proportional to their area.

Goals
  • Show that the pseudosphere has constant curvature \(-1/k^2\).
  • Identify it isometrically with a region of the upper half-plane.
  • See why no complete surface in space carries the whole hyperbolic plane.

Ferdinand Minding had studied surfaces of constant negative curvature in 1839 and 1840, and found that their trigonometry is obtained from spherical trigonometry by replacing the radius with an imaginary one. Delfino Codazzi continued the work. Nobody linked these surfaces to the parallel postulate until Beltrami.

Proposition

The pseudosphere of radius \(k\),

\[ (t,\phi)\mapsto\big(k\,\mathrm{sech}\,t\cos\phi,\ k\,\mathrm{sech}\,t\sin\phi,\ k(t-\tanh t)\big),\qquad t>0, \]

has Gaussian curvature \(-1/k^2\). With \(x=\phi\) and \(y=\cosh t\), its metric becomes

\[ ds^2=\frac{k^2\,(dx^2+dy^2)}{y^2},\qquad y>1, \]

so it is isometric to the region \(y>1\) of the hyperbolic upper half-plane, with \(x\) taken modulo \(2\pi\).

Derivation. For a surface of revolution with profile \((r(t),z(t))\) the metric is \((r'^2+z'^2)\,dt^2+r^2\,d\phi^2\). Here \(r=k\,\mathrm{sech}\,t\), \(r'=-k\,\mathrm{sech}\,t\tanh t\), and \(z'=k(1-\mathrm{sech}^2t)=k\tanh^2t\). So \(r'^2+z'^2=k^2\tanh^2t\,(\mathrm{sech}^2t+\tanh^2t)=k^2\tanh^2t\). With \(y=\cosh t\), \(dy=\sinh t\,dt\) and \(\tanh^2t\,dt^2=dy^2/y^2\), while \(r^2=k^2/y^2\). The half-plane metric has curvature \(-1/k^2\) (Exercise 6.3), and the profile is a tractrix: the tangent segment from the curve to the axis has constant length \(k\).

The meridians \(x=\text{const}\) are geodesics, all parallel in Lobachevsky’s sense towards the open end of the trumpet. The circles \(y=\text{const}\) are horocycles. The rim \(t=0\) is a singular edge where the surface cannot be continued smoothly.

Theorem (Hilbert, 1901)

There is no complete regular surface immersed in \(\mathbb R^3\) whose Gaussian curvature is a negative constant.

So every surface model of Lobachevsky’s plane in ordinary space shows only part of it. This is why Beltrami’s disk (Lesson 07), which carries the whole plane with an abstract metric, was the decisive step.

Exercise 6.1Find the total area of the pseudosphere and compare it with a sphere of radius \(k\).

Show solution
\(\displaystyle\int_0^{2\pi}\!\!\int_1^\infty\frac{k^2}{y^2}\,dy\,dx=2\pi k^2\), half the area \(4\pi k^2\) of the sphere. The solid it bounds also has half the volume of the ball, \(\tfrac23\pi k^3\).

Exercise 6.2The parallels \(y=1\) and \(y=e^{s}\) of the pseudosphere are horocycles. Compute their lengths and the distance between them along a meridian, and compare with the concentric horocycle lemma of the Lobachevsky course.

Show solution
Along \(y=c\), \(ds=k\,dx/c\), so the lengths are \(2\pi k\) and \(2\pi ke^{-s}\). Along a meridian \(ds=k\,dy/y\), so the distance is \(ks\). With \(x=ks\) the ratio of lengths is \(e^{-x/k}\), exactly Lobachevsky’s lemma \(s'=s\,e^{-x/k}\).

Exercise 6.3For a metric \(ds^2=\lambda(dx^2+dy^2)\), the Gaussian curvature is \(K=-\frac{1}{2\lambda}(\partial_x^2+\partial_y^2)\ln\lambda\). Verify \(K=-1/k^2\) for the half-plane.

Show solution
\(\ln\lambda=2\ln k-2\ln y\), so \((\partial_x^2+\partial_y^2)\ln\lambda=2/y^2\). Then \(K=-\frac{y^2}{2k^2}\cdot\frac{2}{y^2}=-\frac1{k^2}\).
Lesson 07

The disk model and the consistency of Lobachevsky’s plane

For the general reader. Beltrami’s decisive picture is a disk. Its “lines” are ordinary straight chords. Through a point outside a chord, infinitely many other chords miss it, so the parallel postulate fails, yet every other axiom of Euclid holds once distance is measured in the right way. A contradiction in Lobachevsky’s geometry would show up as a contradiction about chords in a Euclidean disk. Non-Euclidean geometry is therefore exactly as safe as Euclid’s.

Goals
  • State Beltrami’s 1868 result on the disk.
  • Compute distances by the cross-ratio formula.
  • Recover the angle of parallelism in the model.
Theorem (Beltrami, 1868)

The open disk \(u^2+v^2<k^2\) with the metric of Lesson 05 is a complete surface of constant curvature \(-1/k^2\). Its geodesics are exactly the open chords. With these geodesics as lines it satisfies all the axioms of plane geometry except the parallel postulate, and it is a model of Lobachevsky’s plane: every theorem of Lobachevsky’s geometry is true in it.

The metric is an abstract one; Beltrami did not need a surface in space. Rescaling by \((u,v)=k\,x\) moves the model to the unit disk \(|x|<1\), where the metric reads \(ds^2=k^2\big[(1-|x|^2)|dx|^2+(x\cdot dx)^2\big]/(1-|x|^2)^2\). We use the unit disk from now on. A direct way to see everything at once is the hyperboloid of the Saccheri course: central projection from the origin, \((x_0,x_1,x_2)\mapsto(x_1/x_0,\ x_2/x_0)\), carries the hyperboloid onto the unit disk. Geodesics of the hyperboloid are its sections by planes through the origin, and these project to chords.

Distance formula

Let the chord through \(P\) and \(Q\) meet the boundary circle at \(A\) (beyond \(P\)) and \(B\) (beyond \(Q\)). Then, with Euclidean lengths on the right,

\[ d(P,Q)=\frac{k}{2}\,\ln\frac{|AQ|\,|BP|}{|AP|\,|BQ|}. \]

The quantity inside the logarithm is a cross-ratio. Arthur Cayley had defined projective metrics through cross-ratios in 1859. In 1871 Felix Klein recognised Beltrami’s disk as the case where the absolute conic is a real circle, and the model is now often called the Beltrami–Klein or Klein model. Angles in the model are not Euclidean angles, except at the centre. Two chords are perpendicular when one passes through the pole of the other with respect to the circle.

Exercise 7.1Use the distance formula to show that the point at Euclidean radius \(r\) of the unit disk is at hyperbolic distance \(k\,\operatorname{artanh} r\) from the centre. Check this against the hyperboloid when \(k=1\).

Show solution
Take \(P=0\), \(Q=(r,0)\), \(A=(-1,0)\), \(B=(1,0)\): \(|AQ|=1+r\), \(|BP|=1\), \(|AP|=1\), \(|BQ|=1-r\). So \(d=\frac k2\ln\frac{1+r}{1-r}=k\operatorname{artanh}r\). On the hyperboloid (\(k=1\)), the point at distance \(d\) from \((1,0,0)\) is \((\cosh d,\sinh d,0)\), which projects to \(r=\tanh d\).

Exercise 7.2In the unit disk, let \(l\) be the chord \(u=t\) with \(0<t<1\), and \(P\) the centre. Show that the perpendicular from \(P\) to \(l\) runs along the \(u\)-axis, find the two limiting parallels to \(l\) through \(P\), and recover Lobachevsky’s formula \(\cos\Pi(x)=\tanh(x/k)\).

Show solution
The pole of the chord \(u=t\) is \((1/t,0)\), on the \(u\)-axis, so the \(u\)-axis is perpendicular to \(l\) at \(Q=(t,0)\). By Exercise 7.1, \(x=d(P,Q)=k\operatorname{artanh}t\), so \(t=\tanh(x/k)\). The limiting parallels are the chords from \(P\) to the endpoints \((t,\pm\sqrt{1-t^2})\) of \(l\). At the centre the model’s angles are Euclidean, so the angle between the \(u\)-axis and a parallel satisfies \(\cos\Pi=t=\tanh(x/k)\).

Exercise 7.3Explain why the disk model proves that the parallel postulate cannot be derived from Euclid’s other axioms.

Show solution
Suppose a derivation existed. It would use only the other axioms, which all hold in the disk model, so the parallel postulate would hold in the model. But through the centre pass infinitely many chords missing the chord \(u=\tfrac12\) (any chord through 0 at an angle to the \(u\)-axis of more than \(\pi/3\)). This is a contradiction, provided Euclidean geometry itself is consistent, since the model is built inside it.
Lesson 08

Spaces of constant curvature in n dimensions

For the general reader. The Saggio handled only plane geometry. After reading Riemann’s lecture on the foundations of geometry, Beltrami extended his models to space of any number of dimensions. In the same paper he described several pictures of the same geometry, including the half-space and the ball in which angles are drawn correctly. These later carried Henri Poincaré’s name.

Goals
  • Write Riemann’s metric of constant curvature and Beltrami’s models of it.
  • Pass between the projective (Klein) ball and the conformal (Poincaré) ball.
  • Compute distances in the conformal models.

Riemann had written the metric of constant curvature \(\alpha\) in \(n\) dimensions as \(ds^2=|dx|^2/\big(1+\tfrac{\alpha}{4}|x|^2\big)^2\). Beltrami took \(\alpha=-1/k^2\) and studied the resulting spaces systematically.

Theorem (Beltrami, 1868, in modern form)

The following Riemannian manifolds are isometric to one another, and each is a complete, simply connected space of constant curvature \(-1/k^2\):

  1. the projective ball \(|x|<1\) with \(ds^2=k^2\,\dfrac{(1-|x|^2)\,|dx|^2+(x\cdot dx)^2}{(1-|x|^2)^2}\), whose geodesics are chords;
  2. the conformal ball \(|x|<1\) with \(ds^2=\dfrac{4k^2\,|dx|^2}{(1-|x|^2)^2}\), whose geodesics are arcs of circles meeting the boundary at right angles;
  3. the half-space \(x_n>0\) with \(ds^2=\dfrac{k^2\,|dx|^2}{x_n^2}\);
  4. the upper sheet of the hyperboloid \(-x_0^2+x_1^2+\cdots+x_n^2=-k^2\) with the metric induced from \(-dx_0^2+dx_1^2+\cdots+dx_n^2\).

Beltrami linked the ball models through a hemisphere: vertical projection from a hemisphere onto its equatorial ball gives the projective model, and stereographic projection gives the conformal one. The upshot is a single formula relating the radii of the same point in the two balls.

Proposition

A point at hyperbolic distance \(d\) from the centre has Euclidean radius \(r_K=\tanh(d/k)\) in the projective ball and \(r_P=\tanh(d/2k)\) in the conformal ball. Hence \(r_K=\dfrac{2r_P}{1+r_P^2}\).

The first claim is Exercise 7.1. The second is Exercise 8.1. The relation is the double-angle formula for \(\tanh\). Poincaré met the conformal disk and half-plane in 1882 through automorphic functions, independently of Beltrami. Both groups of models are now standard, and hyperbolic 3-space, in the half-space form, is the setting of William Thurston’s geometrisation of three-manifolds.

Exercise 8.1In the conformal disk \(ds^2=4k^2|dx|^2/(1-|x|^2)^2\), compute the distance from 0 to radius \(r\) along a diameter, and verify \(r_P=\tanh(d/2k)\).

Show solution
\(d=\int_0^r\frac{2k\,d\rho}{1-\rho^2}=2k\operatorname{artanh}r\), so \(r=\tanh(d/2k)\). Then \(\tanh(d/k)=\frac{2\tanh(d/2k)}{1+\tanh^2(d/2k)}=\frac{2r_P}{1+r_P^2}\).

Exercise 8.2In the half-plane \(ds^2=k^2(dx^2+dy^2)/y^2\), find the distance between \((0,1)\) and \((0,e^{s})\), and show that the vertical segment is shorter than any other path between them.

Show solution
Along the segment \(ds=k\,dy/y\), giving length \(ks\). For any path \(\gamma\), \(\int_\gamma k\frac{\sqrt{dx^2+dy^2}}{y}\ge\int_\gamma k\frac{|dy|}{y}\ge k\int_1^{e^s}\frac{dy}{y}=ks\), with equality only when \(dx=0\) and \(y\) is monotone.

Exercise 8.3The Cayley transform \(z\mapsto w=\dfrac{z-i}{z+i}\) maps the upper half-plane onto the unit disk. Show it carries \(k^2|dz|^2/y^2\) to \(4k^2|dw|^2/(1-|w|^2)^2\).

Show solution
\(dw=\frac{2i\,dz}{(z+i)^2}\), so \(|dw|=\frac{2|dz|}{|z+i|^2}\). Also \(1-|w|^2=\frac{|z+i|^2-|z-i|^2}{|z+i|^2}=\frac{4y}{|z+i|^2}\). Then \(\frac{4k^2|dw|^2}{(1-|w|^2)^2}=\frac{4k^2\cdot4|dz|^2/|z+i|^4}{16y^2/|z+i|^4}=\frac{k^2|dz|^2}{y^2}\).
Lesson 09

Bilinear forms and the singular value decomposition

For the general reader. Every linear transformation of space, however it twists and shears, is a rotation, followed by stretching along perpendicular axes, followed by another rotation. The stretching factors are called singular values. Beltrami proved this in 1873 in a paper written for students. Today the result underlies data compression, statistics, image processing and much of numerical computation.

Goals
  • State the singular value decomposition.
  • Follow Beltrami’s proof through the matrix \(A^{\mathsf T}A\).
  • Compute a decomposition by hand.

Beltrami’s paper Sulle funzioni bilineari appeared in the Giornale di Matematiche, the student journal where the Saggio had also been published. He considered a real bilinear form \(f(x,y)=x^{\mathsf T}Ay=\sum a_{ij}x_iy_j\) and asked for orthogonal substitutions \(x=U\xi\), \(y=V\eta\) that reduce it to a sum of products \(\sum\sigma_m\xi_m\eta_m\).

Theorem (Beltrami, 1873; Jordan, 1874)

For every real \(n\times n\) matrix \(A\) there are orthogonal matrices \(U,V\) and real numbers \(\sigma_1\ge\cdots\ge\sigma_n\ge0\) such that

\[ A=U\,\Sigma\,V^{\mathsf T},\qquad \Sigma=\operatorname{diag}(\sigma_1,\dots,\sigma_n). \]

The \(\sigma_i^2\) are the eigenvalues of \(A^{\mathsf T}A\), and the columns of \(V\) are corresponding orthonormal eigenvectors.

Beltrami’s argument. If \(A=U\Sigma V^{\mathsf T}\), then \(A^{\mathsf T}A=V\Sigma^2V^{\mathsf T}\) and \(AA^{\mathsf T}=U\Sigma^2U^{\mathsf T}\). So the \(\sigma_i^2\) must be roots of \(\det(A^{\mathsf T}A-\lambda I)=0\). Beltrami showed these roots are real and positive, because \(A^{\mathsf T}A\) is symmetric and \(x^{\mathsf T}A^{\mathsf T}Ax=|Ax|^2\) is a sum of squares. Taking \(V\) to diagonalise \(A^{\mathsf T}A\) and \(u_i=Av_i/\sigma_i\), the \(u_i\) are orthonormal, since \(u_i\cdot u_j=v_i^{\mathsf T}A^{\mathsf T}Av_j/(\sigma_i\sigma_j)=\delta_{ij}\). Beltrami assumed \(\det A\ne0\) and did not treat repeated roots. Camille Jordan found the decomposition independently in 1874. The general case, including zero and repeated singular values, follows by completing the \(u_i\) to an orthonormal basis.

Exercise 9.1Find the singular value decomposition of \(A=\begin{pmatrix}3&0\\4&5\end{pmatrix}\).

Show solution
\(A^{\mathsf T}A=\begin{pmatrix}25&20\\20&25\end{pmatrix}\), with eigenvalues 45 and 5 and eigenvectors \(v_1=\tfrac1{\sqrt2}(1,1)\), \(v_2=\tfrac1{\sqrt2}(1,-1)\). So \(\sigma_1=3\sqrt5\), \(\sigma_2=\sqrt5\). Then \(u_1=Av_1/\sigma_1=\tfrac1{\sqrt{10}}(1,3)\) and \(u_2=Av_2/\sigma_2=\tfrac1{\sqrt{10}}(3,-1)\), which are orthonormal. Check: \(\sigma_1\sigma_2=15=|\det A|\).

Exercise 9.2Show that \(\sigma_1=\max_{|x|=1}|Ax|\), the operator norm, and that the unit circle is mapped by \(A\) onto an ellipse with semi-axes \(\sigma_1,\sigma_2\).

Show solution
Write \(x=V\xi\) with \(|\xi|=|x|\). Then \(|Ax|=|U\Sigma\xi|=|\Sigma\xi|=\big(\sum\sigma_i^2\xi_i^2\big)^{1/2}\le\sigma_1|\xi|\), with equality at \(\xi=e_1\). The image of the unit circle under \(\Sigma\) is the ellipse \(\eta_1^2/\sigma_1^2+\eta_2^2/\sigma_2^2=1\), and \(U\) rotates or reflects it.
Lesson 10

Mathematical physics

For the general reader. From the 1870s Beltrami was a professor of mechanics and mathematical physics, and most of his later papers are about physics: fluids, elasticity, heat, potential theory and James Clerk Maxwell’s electromagnetism. Three of his results are still in daily use: an identity that solves many problems of the calculus of variations, a class of fluid flows that swirl along their own direction of motion, and the equations that make a field of stresses physically possible.

Goals
  • Prove and use the Beltrami identity.
  • Recognise Beltrami fields.
  • State the Beltrami–Michell compatibility equations.
Beltrami identity

If \(L=L(y,y')\) does not depend explicitly on \(x\), then every solution of the Euler–Lagrange equation \(\frac{d}{dx}L_{y'}=L_y\) satisfies

\[ L-y'\,\frac{\partial L}{\partial y'}=C \]

for a constant \(C\).

Proof. \(\frac{d}{dx}\big(L-y'L_{y'}\big)=L_yy'+L_{y'}y''-y''L_{y'}-y'\frac{d}{dx}L_{y'}=y'\big(L_y-\frac{d}{dx}L_{y'}\big)=0\). In mechanics, with \(x\) as time, this is conservation of energy, a special case of the theorem of Emmy Noether.

Beltrami fields

A vector field \(\mathbf v\) is a Beltrami field if \(\nabla\times\mathbf v=\lambda\,\mathbf v\) for a scalar \(\lambda\). The name recalls Beltrami’s research on the kinematics of fluids, begun in 1871–74. In a Beltrami flow the vorticity is parallel to the velocity, so the Lamb vector \((\nabla\times\mathbf v)\times\mathbf v\) vanishes. The same fields are the force-free magnetic fields of solar and plasma physics.

Elasticity

A stress field in an elastic body must satisfy the equilibrium equations, and it must also come from an actual displacement. For a homogeneous isotropic body without body forces, with Poisson’s ratio \(\nu\), the second requirement becomes the Beltrami–Michell compatibility equations, later extended by John Henry Michell to include body forces:

\[ (1+\nu)\,\nabla^2\sigma_{ij}+\partial_i\partial_j\sigma_{kk}=0. \]

In 1880–82 Beltrami also asked what becomes of elasticity in a space of constant curvature. He showed that Lamé’s fundamental formulas depend on space being Euclidean. In 1889 he gave a mechanical treatment of Maxwell’s equations, starting from d’Alembert’s principle.

Exercise 10.1Rotating a curve \(y(x)>0\) about the \(x\)-axis gives a surface of area \(2\pi\int y\sqrt{1+y'^2}\,dx\). Use the Beltrami identity to find the minimising curves.

Show solution
\(L=y\sqrt{1+y'^2}\), \(L_{y'}=yy'/\sqrt{1+y'^2}\), so \(L-y'L_{y'}=y/\sqrt{1+y'^2}=c\). Then \(y'=\sqrt{y^2/c^2-1}\), and separating variables gives \(y=c\cosh\frac{x-x_0}{c}\): a catenary, whose surface of revolution is the catenoid.

Exercise 10.2Show that \(\mathbf v=(\sin z,\ \cos z,\ 0)\) is a Beltrami field and find \(\lambda\). Is it divergence-free?

Show solution
\(\nabla\times\mathbf v=(\partial_yv_z-\partial_zv_y,\ \partial_zv_x-\partial_xv_z,\ \partial_xv_y-\partial_yv_x)=(\sin z,\ \cos z,\ 0)=\mathbf v\), so \(\lambda=1\). \(\nabla\cdot\mathbf v=0\), as it must be: taking the divergence of \(\nabla\times\mathbf v=\lambda\mathbf v\) with constant \(\lambda\) gives \(\lambda\nabla\cdot\mathbf v=0\).

Exercise 10.3Take the trace of the Beltrami–Michell equations. What does it say about \(\sigma_{kk}\), and what does it imply for each \(\sigma_{ij}\)?

Show solution
Setting \(i=j\) and summing: \((1+\nu)\nabla^2\sigma_{kk}+\nabla^2\sigma_{kk}=(2+\nu)\nabla^2\sigma_{kk}=0\), so the trace is harmonic. Applying \(\nabla^2\) to the full equation and using \(\nabla^2\partial_i\partial_j\sigma_{kk}=\partial_i\partial_j\nabla^2\sigma_{kk}=0\) shows each stress component is biharmonic: \(\nabla^4\sigma_{ij}=0\).
Part III · Life and legacy
Lesson 11

Rome, Pavia, Rome, 1873–1900

After Rome became the capital of Italy, its university was refounded as a national institution. In 1873 Beltrami took its chair of rational mechanics. Three years later, in 1876, he moved to Pavia, where he had once been a student, as professor of mathematical physics, also teaching higher mechanics. His research now centred on physics: potential theory for ellipsoids and discs, heat conduction, wave propagation and Huygens’ principle, elasticity and electromagnetism. A letter of December 1888 to his younger colleague Ernesto Cesàro sets out his thinking on a mechanical interpretation of Maxwell’s theory.

In 1889 he published a note in the Rendiconti of the Accademia dei Lincei on Euclides ab omni naevo vindicatus, the book of 1733 by the Jesuit Giovanni Girolamo Saccheri. Beltrami showed how much of non-Euclidean geometry Saccheri had derived while trying to refute it. The note rescued Saccheri from obscurity.

In 1891 he returned to Rome for the rest of his career. The London Mathematical Society made him an honorary member in 1892. In 1898 he succeeded his teacher Brioschi as president of the Accademia dei Lincei, and in 1899 he was named a senator of the Kingdom of Italy. He died in Rome on 18 February 1900, aged 64.

Lesson 12

Legacy

His collected Opere matematiche appeared in four volumes in Milan between 1902 and 1920, with a biographical sketch by Cremona in the first. Luigi Bianchi, whose textbooks on differential geometry shaped the Italian school, wrote about him, and the Italian line of Ricci-Curbastro and Levi-Civita grew in a tradition he had helped to found.

  • Non-Euclidean geometry: Beltrami’s models made Lobachevsky’s and Bolyai’s geometry mathematically respectable. Klein (1871) and Poincaré (1882) built on them, and the projective model is called the Beltrami–Klein model. The conformal models, now usually named after Poincaré, appear in his 1868 Teoria fondamentale.
  • Differential geometry and analysis: the Laplace–Beltrami operator, central to spectral geometry and to the theory of automorphic forms; Beltrami’s theorem on geodesic maps, the start of projective differential geometry.
  • Complex analysis: the Beltrami equation, the basic tool of quasiconformal maps and of the Teichmüller theory of Lars Ahlfors and Lipman Bers.
  • Linear algebra: the singular value decomposition, with Jordan.
  • Physics: the Beltrami identity, Beltrami fields in fluid and plasma physics, and the Beltrami–Michell equations of elasticity.
  • History of mathematics: the rediscovery of Saccheri.

Final projectWrite a two-page account for a strong undergraduate that explains how Beltrami’s disk proves the independence of the parallel postulate. State the model, verify two of Euclid’s axioms and the failure of the parallel postulate in it, and work one Lobachevsky formula in full.

Reference

Connections

People linked to Eugenio Beltrami across this collection. A name without a link has no page here yet. See Eugenio Beltrami in the network of all 337 lives →

Teachers and advisors

  • Francesco Brioschi 1853-1856

Students

  • Giovanni Frattini

Influenced by

Influenced

Colleagues and collaborators

Friends

  • Enrico Betti 1864-1866

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Eugenio Beltrami, MacTutor History of Mathematics.
  2. D. J. Struik, Beltrami, Eugenio, Dictionary of Scientific Biography.
  3. Eugenio Beltrami and Eugenio Beltrami (Italian), Wikipedia.
  4. L. Cremona, Eugenio Beltrami, in E. Beltrami, Opere matematiche, vol. I, Hoepli, Milan, 1902.
  5. E. Study, Review of Beltrami’s Opere matematiche, Bull. Amer. Math. Soc. (1909).

Mathematics

  1. E. Beltrami, Ricerche di analisi applicata alla geometria, Giornale di Matematiche 2–3 (1864–65).
  2. E. Beltrami, Risoluzione del problema: riportare i punti di una superficie sopra un piano in modo che le linee geodetiche vengano rappresentate da linee rette, Annali di Matematica 7 (1865).
  3. E. Beltrami, Saggio di interpretazione della geometria non-euclidea, Giornale di Matematiche 6 (1868), 284–312.
  4. E. Beltrami, Teoria fondamentale degli spazii di curvatura costante, Annali di Matematica, ser. II, 2 (1868), 232–255.
  5. J. Stillwell, Sources of Hyperbolic Geometry, AMS/LMS, 1996: English translations of both 1868 papers, with Klein and Poincaré.
  6. E. Beltrami, Sulle funzioni bilineari, Giornale di Matematiche 11 (1873). English translation by D. Boley.
  7. G. W. Stewart, On the early history of the singular value decomposition, SIAM Review 35 (1993).
  8. E. Beltrami, Un precursore italiano di Legendre e di Lobatschewsky, Rendiconti Accad. Lincei (1889).
  9. D. Hilbert, Über Flächen von konstanter Gaußscher Krümmung, Trans. Amer. Math. Soc. 2 (1901).
  10. R. Bonola, Non-Euclidean Geometry, 1906; English translation 1912 (Dover reprint).
  11. J. Gray, Worlds Out of Nothing: A Course in the History of Geometry in the 19th Century, Springer, 2nd ed. 2010.
  12. V. Matveev et al., Beltrami’s theorem via parabolic geometry, arXiv:1711.05980.