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Charles Hermite

Charles Hermite, de l’Académie des sciences

A course in two strands: the life of the student who was nearly shut out of the École Polytechnique and became the teacher of a generation of French analysts, and the work that joined number theory, algebra and elliptic functions, from Hermitian forms to the transcendence of \(e\).

Born 24 December 1822, Dieuze, Lorraine Died 14 January 1901, Paris Transcendence of e 1873 Level advanced undergraduate → graduate
Hermite functions \(\psi_n(x)\propto H_n(x)\,e^{-x^2/2}\), \(n=0,\dots,3\), drawn at heights \(n+\tfrac12\) inside the potential \(x^2/2\): the stationary states of the quantum harmonic oscillator.
Lesson 00

How to use this course

Charles Hermite was the central figure of French mathematics in the second half of the nineteenth century. His results are scattered across the curriculum: Hermitian matrices in linear algebra and quantum mechanics, Hermite polynomials in probability and physics, Hermite interpolation in numerical analysis, Hermite’s constant in the geometry of numbers, and the first proof that a classical constant is transcendental. Behind them is one habit of mind. Hermite moved freely between arithmetic and analysis, and used continuous variables and analytic functions to prove facts about integers.

Prerequisites for Part II
  • Linear algebra, including eigenvalues and quadratic forms.
  • Complex analysis: Liouville’s theorem, residues, periodic functions.
  • Elementary number theory and some Galois theory for Lesson 07.
  • Real analysis (integrals, estimates) for Lessons 08 and 09.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern language, proof sketches and exercises with folded solutions. The order is roughly that of his work: quadratic forms (1847–1855), elliptic and modular functions (1843–1858), transcendence and approximation (1873), and orthogonal polynomials and interpolation (1864–1878).

Reference

Timeline

1822Born 24 December in Dieuze, Lorraine, sixth of seven children of Ferdinand Hermite and Madeleine Lallemand.
c. 1829The family moves to Nancy.
1840–41Collège Louis-le-Grand, Paris, taught by Louis Richard, who had earlier taught Évariste Galois.
1842First papers, in the Nouvelles Annales de Mathématiques, including one on the insolubility of the quintic by radicals. Admitted to the École Polytechnique in 68th place.
1843Leaves the Polytechnique after one year because of conditions imposed over his lame foot. Writes to Carl Jacobi on the division of abelian functions.
1847Baccalauréat and licence.
1848Répétiteur and admissions examiner at the École Polytechnique.
1850Letters to Jacobi on the arithmetic of quadratic forms published in Crelle’s Journal: Hermite’s constant.
1854–55Hermitian forms and their real characteristic roots. Reciprocity law in invariant theory.
1856Falls ill with smallpox; Augustin-Louis Cauchy’s support leads him to Catholicism. Elected to the Académie des sciences on 14 July, succeeding Jacques Binet.
1858Solves the general quintic with elliptic modular functions.
1862Maître de conférences at the École Normale Supérieure, a post created through Louis Pasteur.
1864Paper on a new series expansion of functions: the Hermite polynomials.
1869Succeeds Jean-Marie Duhamel as professor of analysis at the École Polytechnique and at the Faculté des sciences (Sorbonne).
1873Sur la fonction exponentielle: \(e\) is transcendental. Elected to the Royal Society of London.
1876Resigns his chair at the Polytechnique; keeps the Sorbonne.
1882Ferdinand von Lindemann extends Hermite’s method to prove \(\pi\) transcendental.
1890President of the Académie des sciences.
1892Seventieth-birthday jubilee; promoted grand officer of the Légion d’honneur.
1897Retires; Émile Picard succeeds him.
1901Dies 14 January in Paris. Buried in the Montparnasse cemetery.
Part I · Formation
Lesson 01

Dieuze, Nancy, Paris, 1822–1842

Charles Hermite was born on 24 December 1822 in Dieuze, in Lorraine, the sixth of seven children of Ferdinand Hermite and Madeleine Lallemand. His father had trained as an engineer and worked at a salt mine near the town before going into the drapery trade; he also had an artist’s interests. When Charles was about seven the family moved to Nancy. He was born with a deformity of the right foot that made walking difficult all his life.

He went to school at the Collège de Nancy and then in Paris, first at the Collège Henri IV and from 1840 at Louis-le-Grand. There his mathematics teacher was Louis Richard, who had taught Évariste Galois about a decade earlier and recognized the same kind of talent. Hermite did not care for the routine of examination preparation. He read Leonhard Euler, Joseph-Louis Lagrange and Carl Friedrich Gauss instead.

Before entering any grande école he had published two papers in the Nouvelles Annales de Mathématiques. One was a simplified argument that the general equation of the fifth degree cannot be solved by radicals, the result of Niels Henrik Abel and Paolo Ruffini. Sixteen years later he would show how the quintic can be solved by other means (Lesson 07).

Lesson 02

The Polytechnique and the letters to Jacobi, 1842–1847

Tutored by Eugène Catalan, Hermite passed the entrance examination of the École Polytechnique in 1842, ranked 68th. After his first year the school refused to let him continue because of his disability. Friends had the decision reversed, but only under conditions he found unacceptable, and he left without graduating.

He spent the next years working on his own. In January 1843, aged twenty, he wrote to Carl Gustav Jacob Jacobi in Königsberg. Abel had shown how to divide the argument of an elliptic function, expressing \(f(u/n)\) through algebraic equations in \(f(u)\). Hermite extended this to the abelian functions of hyperelliptic integrals that Jacobi was then developing. Jacobi replied with encouragement. A further series of letters, on the arithmetic of quadratic forms, followed and appeared in Crelle’s Journal in 1850.

The young man was now known in Berlin and Königsberg before he had a French degree. He frequented the home of the mathematician Joseph Bertrand, and later married Bertrand’s sister Louise. In 1847 he finally took the baccalauréat and licence.

Lesson 03

Examiner and academician, 1848–1862

In 1848 Hermite was appointed répétiteur, a tutor, and admissions examiner at the École Polytechnique, the school that had pushed him out. He also taught for a time at the Collège de France as a substitute. These were modest posts, and for more than a decade they were all he held. The research was not modest. Between 1847 and 1858 he published his arithmetic theory of quadratic forms, the Hermitian forms, his reciprocity law in invariant theory, and the solution of the quintic.

In 1849 he submitted a memoir on doubly periodic functions that applied Cauchy’s residue calculus. Cauchy and Charles Sturm reported on it favourably, but it was not published at the time, apparently because of a priority dispute with Joseph Liouville, who was developing the theory of elliptic functions from his own theorem on bounded entire functions.

In 1856 Hermite fell seriously ill with smallpox. Cauchy, a devout Catholic, supported him through the illness, and Hermite, until then indifferent to religion, became a convinced Catholic. On 14 July 1856 the Académie des sciences elected him to the seat left by Jacques Binet. He was 33. In 1862 Louis Pasteur, then director of scientific studies at the École Normale Supérieure, had a post of maître de conférences created for him there, and from 1863 he was also a permanent examiner at the Polytechnique.

Part II · The mathematics
Lesson 04

Quadratic forms and Hermite’s constant

For the general reader. How close can an integer point come to the origin when distance is measured by a quadratic formula? Gauss had answered for two variables. In his letters to Jacobi, Hermite answered for any number of variables, and used the answer to prove finiteness theorems in number theory. This is the beginning of what Hermann Minkowski later called the geometry of numbers.

Goals
  • Prove Hermite’s bound for the minimum of a positive definite quadratic form.
  • Define Hermite’s constant and know its first values.
  • See how the bound gives finiteness of class numbers.

Let \(f(x)=x^{T}Ax\) be a positive definite quadratic form in \(n\) variables with real symmetric matrix \(A\), and let \(\det f=\det A\). Its arithmetic minimum is \(m(f)=\min\{f(x): x\in\mathbb Z^n\setminus\{0\}\}\).

Theorem (Hermite, 1850)

For every positive definite quadratic form \(f\) in \(n\) variables,

\[ m(f)\le\left(\tfrac43\right)^{(n-1)/2}(\det f)^{1/n}. \]

Proof sketch. Induction on \(n\); for \(n=1\) it is an equality. Choose \(v\in\mathbb Z^n\) with \(f(v)=m(f)\). Then \(v\) is primitive, so it extends to a basis of \(\mathbb Z^n\), and after an integral change of variables \(f(x)=m\,(x_1+\ell(x'))^2+g(x')\) with \(x'=(x_2,\dots,x_n)\), \(\ell\) linear and \(g\) a positive form in \(n-1\) variables with \(\det g=\det f/m\). By induction there is \(y'\ne0\) with \(g(y')\le(4/3)^{(n-2)/2}(\det g)^{1/(n-1)}\). Choose \(y_1\in\mathbb Z\) with \(|y_1+\ell(y')|\le\frac12\). Then

\[ m\le f(y)\le\tfrac{m}{4}+g(y'),\quad\text{so}\quad \tfrac34m\le\left(\tfrac43\right)^{(n-2)/2}\left(\tfrac{\det f}{m}\right)^{1/(n-1)}. \]

Raising to the power \(n-1\) and rearranging gives \(m^n\le(4/3)^{n(n-1)/2}\det f\), which is the theorem.

Hermite’s constant

The best possible constant \(\gamma_n=\sup_f m(f)/(\det f)^{1/n}\) is called Hermite’s constant. Hermite’s theorem says \(\gamma_n\le(4/3)^{(n-1)/2}\). Minkowski’s convex body theorem of 1889 gave a bound linear in \(n\), much better for large \(n\). The exact value is known only for \(n\le8\) and \(n=24\): \(\gamma_2^2=4/3\), \(\gamma_3^3=2\), \(\gamma_4^4=4\), up to \(\gamma_8=2\) (the \(E_8\) lattice) and \(\gamma_{24}=4\) (the Leech lattice, proved by Cohn and Kumar in 2009). Finding \(\gamma_n\) is the lattice sphere-packing problem.

Finiteness

The same reduction shows that every positive form with integer coefficients is equivalent under \(\mathrm{GL}_n(\mathbb Z)\) to one whose coefficients are bounded in terms of the determinant. Hence there are finitely many classes of integral forms of given determinant. Hermite applied such arguments to algebraic number fields and proved that only finitely many number fields have a given discriminant.

Exercise 4.1Show that the bound is sharp for \(n=2\) using \(f=x^2+xy+y^2\).

Show solution
The matrix is \(\begin{pmatrix}1&\tfrac12\\ \tfrac12&1\end{pmatrix}\) with determinant \(\tfrac34\). The bound gives \(m\le(4/3)^{1/2}(3/4)^{1/2}=1\). Since \(f\) takes positive integer values on \(\mathbb Z^2\setminus\{0\}\) and \(f(1,0)=1\), \(m=1\). Equality holds: this is the hexagonal lattice, the densest in the plane.

Exercise 4.2Use Hermite’s bound for \(n=2\) to prove that every prime \(p\equiv1\pmod 4\) is a sum of two squares.

Show solution
Choose \(a\) with \(a^2\equiv-1\pmod p\) (possible because \(p\equiv1\bmod4\)). The lattice \(L=\{(x,y)\in\mathbb Z^2: y\equiv ax\bmod p\}\) has index \(p\), so the form \(x^2+y^2\) restricted to a basis of \(L\) has determinant \(p^2\). Hermite’s bound gives a non-zero \((x,y)\in L\) with \(x^2+y^2\le(4/3)^{1/2}p<2p\). For points of \(L\), \(x^2+y^2\equiv x^2(1+a^2)\equiv0\pmod p\). So \(x^2+y^2=p\).
Lesson 05

Hermitian forms and reciprocity

For the general reader. A symmetric real matrix has real eigenvalues. Hermite saw that the right analogue for complex numbers is not a symmetric matrix but one equal to its own conjugate transpose. These Hermitian matrices are now the observables of quantum mechanics. In the same years he found a surprising symmetry in the invariant theory of Arthur Cayley and James Joseph Sylvester.

Goals
  • Define Hermitian forms and prove their eigenvalues are real.
  • State Hermite’s reciprocity law and verify it by counting partitions.

A Hermitian form in \(n\) complex variables is \(h(z)=\sum_{i,j}a_{ij}z_i\bar z_j\) with \(a_{ji}=\bar a_{ij}\), that is \(h(z)=z^{*}Az\) with \(A^{*}=A\). Its values are real. Hermite introduced such forms in 1854–55, in arithmetic work on forms with complex integer coefficients, and extended his reduction theory of Lesson 04 to them.

Theorem (Hermite, 1855)

If \(A\) is Hermitian, every root of \(\det(\lambda I-A)=0\) is real. Moreover \(A=UDU^{*}\) with \(U\) unitary and \(D\) real diagonal.

Proof. If \(Av=\lambda v\) with \(v\ne0\), then \(\lambda\,v^{*}v=v^{*}Av=(v^{*}Av)^{*}=\bar\lambda\,v^{*}v\), and \(v^{*}v>0\), so \(\lambda=\bar\lambda\). If \(v\) is an eigenvector, the orthogonal complement \(v^\perp\) is invariant under \(A\), since \(w\perp v\) implies \(\langle Aw,v\rangle=\langle w,Av\rangle=\bar\lambda\langle w,v\rangle=0\). Induction on dimension gives an orthonormal eigenbasis. Cauchy had proved the real symmetric case in 1829; Hermite’s remark extended it.

The reciprocity law

Recall from the Cayley course that the invariants of degree \(d\) of a binary form of order \(n\) are counted by \(p(nd/2;d,n)-p(nd/2-1;d,n)\), where \(p(w;d,n)\) counts partitions of \(w\) into at most \(d\) parts, each at most \(n\).

Theorem (Hermite reciprocity, 1854)

The number of linearly independent invariants of degree \(d\) of the binary form of order \(n\) equals the number of invariants of degree \(n\) of the binary form of order \(d\). In modern terms, \(\mathrm{Sym}^d(\mathrm{Sym}^n\mathbb C^2)\cong\mathrm{Sym}^n(\mathrm{Sym}^d\mathbb C^2)\) as representations of \(\mathrm{SL}_2(\mathbb C)\).

Proof via partitions. A partition into at most \(d\) parts each at most \(n\) is a Young diagram inside a \(d\times n\) box. Transposing the diagram gives a partition into at most \(n\) parts each at most \(d\), of the same size. So \(p(w;d,n)=p(w;n,d)\), and the counting formula is symmetric in \(d\) and \(n\). The same argument matches the weight multiplicities of the two representations, which determine them.

Exercise 5.1Find the eigenvalues of \(A=\begin{pmatrix}2&1-i\\1+i&3\end{pmatrix}\) and check they are real.

Show solution
\(\operatorname{tr}A=5\), \(\det A=6-(1-i)(1+i)=6-2=4\). So \(\lambda^2-5\lambda+4=0\), \(\lambda=1,4\).

Exercise 5.2The binary quartic has exactly one invariant of degree 3 (Cayley’s \(J\)). Use reciprocity to deduce how many invariants of degree 4 the binary cubic has, and identify it.

Show solution
Reciprocity with \(d=3,\ n=4\) gives one invariant of degree 4 for the cubic. It is the discriminant of the cubic, which has degree 4 in the coefficients.

Exercise 5.3Show that a Hermitian form \(h\) is positive definite if and only if all eigenvalues of \(A\) are positive, and that \(h(z)=|z_1|^2+|z_2|^2+|z_1+z_2|^2\) is positive definite.

Show solution
With \(A=UDU^{*}\) and \(w=U^{*}z\), \(h(z)=\sum_i\lambda_i|w_i|^2\), positive for all \(z\ne0\) exactly when all \(\lambda_i>0\). For the example, \(h\) is a sum of squared moduli and vanishes only when \(z_1=z_2=0\). Its matrix \(\begin{pmatrix}2&1\\1&2\end{pmatrix}\) has eigenvalues 1 and 3.
Lesson 06

Elliptic functions and theta

For the general reader. The sine function repeats itself every \(2\pi\). Elliptic functions repeat in two independent complex directions, so that their values are determined by a parallelogram. Hermite’s first research, and much of his teaching, concerned these functions and the theta series from which Jacobi built them.

Goals
  • Prove the basic theorems on elliptic functions with residues.
  • Understand the transformation of theta under \(\tau\mapsto\tau+1\) and \(\tau\mapsto-1/\tau\).

Let \(\Lambda=\mathbb Z\omega_1+\mathbb Z\omega_2\) be a lattice in \(\mathbb C\). An elliptic function is a meromorphic \(f\) with \(f(z+\omega)=f(z)\) for all \(\omega\in\Lambda\).

Theorem (Liouville; residue proofs as in Hermite)
  1. A holomorphic elliptic function is constant.
  2. The sum of the residues of \(f\) in a period parallelogram is 0.
  3. A non-constant elliptic function takes every value the same number of times, counted with multiplicity, in a period parallelogram; this number, the order, is at least 2.

Proof. (1) A holomorphic elliptic function is bounded, since it is continuous on a compact parallelogram, so it is constant by Liouville’s theorem. (2) Integrate \(f\) around the boundary of a parallelogram with no poles on it. Opposite sides cancel by periodicity, so the integral, and hence the sum of residues, is 0. (3) Apply (2) to \(f'/(f-c)\), which is elliptic: the number of zeros of \(f-c\) minus the number of poles is 0. If the order were 1, \(f\) would have a single simple pole with non-zero residue, contradicting (2).

Hermite’s 1848 work showed that every elliptic function is a quotient of entire functions that are periodic in one direction and quasi-periodic in the other; theta functions are the basic examples. He later gave the decomposition of an elliptic function into simple elements, analogous to partial fractions, that bears his name.

Theta functions

For \(\operatorname{Im}\tau>0\) let

\[ \vartheta(z,\tau)=\sum_{n\in\mathbb Z}e^{\pi i n^2\tau+2\pi i n z}. \]

It satisfies \(\vartheta(z+1,\tau)=\vartheta(z,\tau)\) and \(\vartheta(z+\tau,\tau)=e^{-\pi i\tau-2\pi i z}\vartheta(z,\tau)\). Under the modular group it transforms by

\[ \vartheta(0,\tau+2)=\vartheta(0,\tau),\qquad \vartheta\!\left(0,-\tfrac1\tau\right)=\sqrt{-i\tau}\;\vartheta(0,\tau). \]

In 1858 Hermite determined the eighth roots of unity that appear in the transformation of the four Jacobi theta functions under a general element of \(\mathrm{SL}_2(\mathbb Z)\). This was the technical heart of his solution of the quintic.

Exercise 6.1Show directly from the series that \(\vartheta(z+\tau,\tau)=e^{-\pi i\tau-2\pi iz}\vartheta(z,\tau)\).

Show solution
\(\vartheta(z+\tau,\tau)=\sum_n e^{\pi i(n^2+2n)\tau+2\pi inz}\). Write \(n^2+2n=(n+1)^2-1\) and \(2\pi inz=2\pi i(n+1)z-2\pi iz\). Then the sum is \(e^{-\pi i\tau-2\pi iz}\sum_n e^{\pi i(n+1)^2\tau+2\pi i(n+1)z}=e^{-\pi i\tau-2\pi iz}\vartheta(z,\tau)\).

Exercise 6.2Show that \(\vartheta(z,\tau)\) has exactly one zero in the parallelogram with vertices \(0,1,1+\tau,\tau\) (suitably translated), and that it lies at \(z=\tfrac12+\tfrac\tau2\).

Show solution
Integrate \(\vartheta'/\vartheta\) around the boundary. The sides \([0,1]\) and \([\tau,1+\tau]\) differ by the quasi-periodicity: \(\frac{\vartheta'}{\vartheta}(z+\tau)=\frac{\vartheta'}{\vartheta}(z)-2\pi i\), giving a contribution of \(2\pi i\); the other pair cancels. So there is one zero. At \(z_0=\tfrac12+\tfrac\tau2\), pair the terms \(n\) and \(-1-n\): the exponents are \(\pi in^2\tau+\pi in+\pi in\tau\) and \(\pi i(n+1)^2\tau-\pi i(n+1)-\pi i(n+1)\tau\), both equal to \(\pi i(n^2+n)\tau+\pi i n\) up to the factor \(e^{-\pi i}=-1\) in the second. The terms cancel in pairs.
Lesson 07

Solving the quintic

For the general reader. Abel and Galois showed that the general equation of degree five cannot be solved using only arithmetic and root extraction. That leaves open whether some other, richer function can do the job, in the way that the cosine solves the cubic. In 1858 Hermite showed that elliptic modular functions solve the quintic. Leopold Kronecker and Francesco Brioschi reached the same goal by related routes the same year.

Goals
  • Reduce the general quintic to Bring–Jerrard form.
  • Understand the analogy with the trigonometric solution of the cubic.
  • See the structure of Hermite’s solution via the modular equation of degree 6.

The model: cubics and cosines

The cubic \(t^3-3t-2a=0\) with \(|a|\le1\) is solved by \(t=2\cos\big(\tfrac13\arccos a\big)\). The function \(\cos\) has the property that \(\cos3\theta\) is a polynomial in \(\cos\theta\); dividing the argument by 3 solves a cubic. Hermite’s solution replaces trigonometric functions by modular functions and division of the argument by transformation of order 5.

Reduction

Theorem (Bring 1786; Jerrard 1834)

By a Tschirnhaus transformation \(y=\sum_{k=0}^4c_kx^k\) whose coefficients are found by solving equations of degree at most 3, every quintic can be brought to the form \(y^5-y-a=0\) (or to \(y^5=0\) or \(y^5-a=0\) in degenerate cases).

Hermite’s solution

Let \(k\) be the modulus of an elliptic function with \(k^2=\lambda(\tau)\), and set \(u=k^{1/4}\), a modular function of \(\tau\). Jacobi’s modular equation of order 5 relates \(u=u(\tau)\) and \(v=u(5\tau)\):

\[ u^6-v^6+5u^2v^2(u^2-v^2)+4uv(1-u^4v^4)=0. \]

For fixed \(u\) this has six roots \(v_0,\dots,v_5\), corresponding to the six transformations of order 5. In modern terms the monodromy group of this sextic is \(\mathrm{PSL}_2(\mathbb F_5)\cong A_5\). Galois had stated, in his last letter of 1832, that the modular equation of degree 6 can be lowered to degree 5. Hermite wrote down explicit products of differences of the \(v_j\), five in number, that are permuted among themselves by the monodromy, and showed that they satisfy a quintic of Bring–Jerrard form \(y^5-y-a=0\) after rescaling, with \(a\) an explicit algebraic function of the modulus \(k\). Given \(a\), one solves an algebraic equation for \(k\), then finds \(\tau\) by elliptic integrals, and the roots are values of modular functions at \(\tau\) and its transforms.

Felix Klein’s Lectures on the Icosahedron (1884) recast the whole construction through the symmetries of the icosahedron.

Exercise 7.1Solve \(t^3-3t-1=0\) trigonometrically.

Show solution
Here \(2a=1\), \(a=\tfrac12=\cos(\pi/3)\). From \(4\cos^3\theta-3\cos\theta=\cos3\theta\) we get that \(t=2\cos\theta\) satisfies \(t^3-3t=2\cos3\theta\). So \(t=2\cos\big(\tfrac{\pi/3+2\pi k}{3}\big)\), \(k=0,1,2\): \(t=2\cos\tfrac\pi9,\ 2\cos\tfrac{7\pi}9,\ 2\cos\tfrac{13\pi}9\).

Exercise 7.2Show that \(A_5\) has no subgroup of index 2, 3 or 4, and explain why this means the sextic modular equation has no resolvent of degree below 5, while \(\mathrm{PSL}_2(\mathbb F_5)\) does act on 5 points.

Show solution
A subgroup of index \(m\) gives a homomorphism \(A_5\to S_m\), which is injective because \(A_5\) is simple; for \(m\le4\), \(|S_m|\le24<60\), a contradiction. A resolvent of degree \(m\) corresponds to a subgroup of index \(m\), so the minimum is 5. Subgroups of index 5 exist: \(A_5\) contains \(A_4\) of order 12. Through \(\mathrm{PSL}_2(\mathbb F_5)\cong A_5\), this gives the action on five points that Galois noticed and Hermite made explicit.
Lesson 08

The transcendence of e

For the general reader. A number is algebraic if it solves a polynomial equation with integer coefficients, and transcendental otherwise. Joseph Liouville constructed the first transcendental numbers in 1844, but they were made for the purpose. In 1873 Hermite proved that \(e\), the base of natural logarithms, is transcendental. Nine years later Lindemann used the same method to prove that \(\pi\) is transcendental, which settled the ancient problem of squaring the circle.

Goals
  • Prove that \(e\) is transcendental.
  • Understand Hermite’s simultaneous rational approximations.
  • Prove that \(\pi\) is irrational by a Hermite-style integral.
Theorem (Hermite, 1873)

The number \(e\) is transcendental: there are no integers \(a_0,\dots,a_n\), not all zero, with \(a_0+a_1e+\cdots+a_ne^n=0\).

Hermite’s idea

Hermite found polynomials \(P_0,\dots,P_n\) such that \(P_0(x)e^{kx}-P_k(x)\) is very small near \(x=0\) for every \(k\), and evaluated at \(x=1\). This gives rational numbers \(P_k(1)/P_0(1)\) that approximate \(e,e^2,\dots,e^n\) simultaneously, and too well for a linear relation to survive. These are now called Hermite–Padé approximants, after Hermite’s student Henri Padé. The streamlined proof below follows David Hilbert (1893) and Adolf Hurwitz.

Proof

For a polynomial \(f\) of degree \(m\) put \(F=f+f'+\cdots+f^{(m)}\). Integration by parts gives

\[ e^{k}F(0)-F(k)=e^{k}\int_0^k e^{-x}f(x)\,dx. \]

Suppose \(\sum_{k=0}^n a_ke^k=0\) with \(a_0\ne0\) (divide by a power of \(e\) if needed). Multiply the identity by \(a_k\) and sum:

\[ -\sum_{k=0}^n a_kF(k)=\sum_{k=0}^n a_ke^{k}\int_0^k e^{-x}f(x)\,dx. \]

Take a large prime \(p>\max(n,|a_0|)\) and \(f(x)=\frac{1}{(p-1)!}x^{p-1}(x-1)^p\cdots(x-n)^p\). Every derivative \(f^{(j)}(k)\) for \(k=1,\dots,n\) is an integer divisible by \(p\), and so is \(f^{(j)}(0)\) for \(j\ne p-1\), while \(f^{(p-1)}(0)=(\pm n!)^p\), which is not divisible by \(p\). So the left side is an integer congruent to \(-a_0(\pm n!)^p\not\equiv0\pmod p\), hence of absolute value at least 1. On \([0,n]\), \(|f(x)|\le n^{(n+1)p}/(p-1)!\), so the right side is at most \(C\cdot c^p/(p-1)!\) for constants \(C,c\), which tends to 0. For large \(p\) this is a contradiction.

Hermite himself doubted that the method would extend to \(\pi\). Lindemann proved in 1882 that \(e^\alpha\) is transcendental for every non-zero algebraic \(\alpha\); since \(e^{i\pi}=-1\), \(i\pi\) and hence \(\pi\) is transcendental. Karl Weierstrass gave the general Lindemann–Weierstrass theorem in 1885.

Exercise 8.1Prove that \(e\) is irrational using \(e=\sum 1/k!\).

Show solution
If \(e=a/b\), then \(N=b!\big(e-\sum_{k=0}^b\frac1{k!}\big)\) is a positive integer. But \(N=\sum_{k>b}\frac{b!}{k!}<\frac1{b+1}+\frac1{(b+1)^2}+\cdots=\frac1b\le1\), a contradiction.

Exercise 8.2(Niven, 1947, after Hermite.) Suppose \(\pi=a/b\). Let \(f(x)=\frac{x^n(a-bx)^n}{n!}\). Given that \(f\) and all its derivatives take integer values at \(0\) and \(\pi\), show that \(I_n=\int_0^\pi f(x)\sin x\,dx\) is a positive integer, and derive a contradiction.

Show solution
Let \(F=f-f''+f^{(4)}-\cdots\). Then \((F'\sin x-F\cos x)'=(F''+F)\sin x=f\sin x\), so \(I_n=F(\pi)+F(0)\), an integer. It is positive because the integrand is positive on \((0,\pi)\). But \(0<f(x)\le\frac{(\pi a)^n}{n!}\) on \([0,\pi]\), so \(I_n\le\pi\frac{(\pi a)^n}{n!}\to0\). For large \(n\), \(0<I_n<1\), impossible. (The integrality claim holds because \(f(\pi-x)=f(x)\) and \(f\) has integer coefficients divided by \(n!\) with lowest degree \(n\).)
Lesson 09

Hermite polynomials and interpolation

For the general reader. The bell-shaped Gaussian curve has a family of companion polynomials that let any reasonable function be expanded in Gaussian “harmonics”. Pierre-Simon Laplace and Pafnuty Chebyshev met these polynomials first; Hermite’s 1864 paper studied them systematically, including in several variables. Seventy years later they appeared as the wave functions of the quantum harmonic oscillator, drawn at the top of this page.

Goals
  • Derive the generating function, recurrence, differential equation and orthogonality of Hermite polynomials.
  • Solve the quantum harmonic oscillator.
  • State and use Hermite interpolation.

Definition and properties

The (physicists’) Hermite polynomials are

\[ H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2},\qquad H_0=1,\ H_1=2x,\ H_2=4x^2-2,\ H_3=8x^3-12x. \]
Theorem
  1. Generating function: \(e^{2xt-t^2}=\sum_{n\ge0}H_n(x)\frac{t^n}{n!}\).
  2. Recurrence: \(H_{n+1}=2xH_n-2nH_{n-1}\), and \(H_n'=2nH_{n-1}\).
  3. Differential equation: \(H_n''-2xH_n'+2nH_n=0\).
  4. Orthogonality: \(\int_{-\infty}^{\infty}H_m(x)H_n(x)e^{-x^2}dx=\sqrt\pi\,2^nn!\,\delta_{mn}\).

Proofs. (1) By Taylor’s theorem, \(e^{-(x-t)^2}=\sum_n\frac{t^n}{n!}\frac{d^n}{dt^n}e^{-(x-t)^2}\big|_{t=0}=\sum_n\frac{t^n}{n!}(-1)^n\frac{d^n}{dx^n}e^{-x^2}\); multiply by \(e^{x^2}\). (2) Differentiate the generating function \(G=e^{2xt-t^2}\) in \(t\) and in \(x\): \(\partial_tG=(2x-2t)G\) and \(\partial_xG=2tG\), and compare coefficients. (3) Combine the two relations in (2). (4) For \(m<n\), integrate by parts \(n\) times: \(\int H_m(-1)^n\frac{d^n}{dx^n}e^{-x^2}dx=\int H_m^{(n)}e^{-x^2}dx=0\). For \(m=n\), \(H_n^{(n)}=2^nn!\) and \(\int e^{-x^2}=\sqrt\pi\).

The quantum oscillator

The Schrödinger equation \(-\tfrac12\psi''+\tfrac12x^2\psi=E\psi\) (units with \(\hbar=m=\omega=1\)) has square-integrable solutions exactly when \(E=n+\tfrac12\), and then \(\psi_n(x)=c_nH_n(x)e^{-x^2/2}\). Substituting \(\psi=He^{-x^2/2}\) turns the equation into \(H''-2xH'+(2E-1)H=0\), Hermite’s equation with \(2n=2E-1\).

Hermite interpolation (1878)

Theorem (Hermite)

Given distinct nodes \(x_1,\dots,x_k\), multiplicities \(m_1,\dots,m_k\ge1\) with \(N=\sum m_i\), and prescribed values \(y_i^{(j)}\) for \(0\le j<m_i\), there is a unique polynomial \(P\) of degree less than \(N\) with \(P^{(j)}(x_i)=y_i^{(j)}\). If \(f\in C^N\) and \(P\) interpolates \(f\) in this sense, then for each \(x\) there is \(\xi\) in the span of the nodes and \(x\) with

\[ f(x)-P(x)=\frac{f^{(N)}(\xi)}{N!}\prod_{i=1}^k(x-x_i)^{m_i}. \]

Uniqueness: the difference of two interpolants has \(N\) zeros counted with multiplicity and degree below \(N\). Existence then follows because the linear map from polynomials of degree below \(N\) to the \(N\) data is injective between spaces of the same dimension. Hermite derived it, with the error term, from a contour integral generalizing Lagrange’s formula.

Exercise 9.1Compute \(H_4\) from the recurrence and check the differential equation for \(n=2\).

Show solution
\(H_4=2xH_3-6H_2=16x^4-24x^2-24x^2+12=16x^4-48x^2+12\). For \(n=2\): \(H_2''=8\), \(-2xH_2'=-16x^2\), \(4H_2=16x^2-8\); the sum is 0.

Exercise 9.2Find the cubic \(P\) with \(P(0)=0,\ P'(0)=1,\ P(1)=1,\ P'(1)=0\).

Show solution
Write \(P=x+ax^2+bx^3\), which already satisfies the conditions at 0. Then \(P(1)=1+a+b=1\) and \(P'(1)=1+2a+3b=0\), so \(b=-a\) and \(1-a=0\): \(a=1,\ b=-1\). \(P(x)=x+x^2-x^3\).

Exercise 9.3Show that \(\int_{-\infty}^\infty x^2\,\psi_0(x)^2dx=\tfrac12\) when \(\psi_0=\pi^{-1/4}e^{-x^2/2}\), and interpret it.

Show solution
\(\int x^2e^{-x^2}dx=\tfrac{\sqrt\pi}2\), so the integral is \(\pi^{-1/2}\cdot\tfrac{\sqrt\pi}2=\tfrac12\). The mean potential energy \(\langle x^2/2\rangle\) is \(\tfrac14\), half the ground-state energy \(\tfrac12\), as the virial theorem predicts.
Lesson 10

Integer matrices and the Hermite normal form

For the general reader. Gaussian elimination solves linear equations over the rational numbers, but it divides, and division is not allowed when the unknowns must be whole numbers. In 1851 Hermite showed that every integer matrix can still be brought to a canonical triangular shape using only invertible integer operations. The result is the integer counterpart of row-reduced echelon form, and it is how computers handle lattices today.

Goals
  • State and prove existence and uniqueness of the Hermite normal form.
  • Use it to compare lattices and solve linear Diophantine systems.

A square integer matrix \(U\) is unimodular if \(\det U=\pm1\); equivalently \(U^{-1}\) also has integer entries. Unimodular matrices form the group \(\mathrm{GL}_n(\mathbb Z)\). Multiplying by \(U\) on the left performs invertible integer row operations: swapping rows, negating a row, and adding an integer multiple of one row to another.

Theorem (Hermite, 1851)

Let \(A\) be an \(n\times n\) integer matrix with \(\det A\ne0\). There is a unique matrix \(H=UA\) with \(U\in\mathrm{GL}_n(\mathbb Z)\) such that \(H\) is upper triangular, its diagonal entries are positive, and each entry above the diagonal satisfies \(0\le h_{ij}<h_{jj}\) for \(i<j\).

Existence. Work on the first column. While two of its entries are non-zero, subtract an integer multiple of the row with the smaller entry (in absolute value) from the other; this is the Euclidean algorithm, and it ends with a single non-zero entry, equal to \(\pm\gcd\) of the column. Move that row to the top and make the entry positive. Repeat on the lower-right submatrix. Finally, for each column \(j\), reduce the entries above \(h_{jj}\) modulo \(h_{jj}\) by subtracting multiples of row \(j\).

Uniqueness. If \(H\) and \(H'=VH\) are both normal forms with \(V\) unimodular, then \(V=H'H^{-1}\) is upper triangular with positive diagonal entries whose product is \(|\det V|=1\), so all are 1. Suppose \(V\ne I\) and take the lowest row \(i\) with an off-diagonal entry \(v_{ik}\ne0\), \(k>i\) smallest. Then \(h'_{ik}=h_{ik}+v_{ik}h_{kk}\), and both \(h_{ik}\) and \(h'_{ik}\) lie in \([0,h_{kk})\), which forces \(v_{ik}=0\), a contradiction.

The rows of \(A\) and of \(H\) generate the same lattice in \(\mathbb Z^n\), so the theorem gives a canonical basis of every full-rank sublattice of \(\mathbb Z^n\). Two integer matrices generate the same lattice exactly when they have the same Hermite normal form. The diagonal gives the index: \([\mathbb Z^n:L]=\prod_j h_{jj}=|\det A|\). Henry Smith’s normal form of 1861, which also uses column operations, gives the elementary divisors.

Exercise 10.1Find the Hermite normal form of \(A=\begin{pmatrix}4&6\\6&3\end{pmatrix}\).

Show solution
Rows \((4,6)\) and \((6,3)\). Row 2 minus row 1 gives \((2,-3)\); then row 1 minus 2·(new row 2) gives \((0,12)\). Swap to get \(\begin{pmatrix}2&-3\\0&12\end{pmatrix}\). Reduce the entry above the diagonal modulo 12 by adding row 2: \(-3+12=9\). So \(H=\begin{pmatrix}2&9\\0&12\end{pmatrix}\). Check: \(\det H=24=|\det A|=|12-36|\).

Exercise 10.2Do the vectors \((1,2)\) and \((3,1)\) generate the same lattice as \((1,2)\) and \((2,-1)\)? Decide using the normal form.

Show solution
For \(\begin{pmatrix}1&2\\3&1\end{pmatrix}\): subtract 3·row 1 from row 2 to get \((0,-5)\), then negate: \(\begin{pmatrix}1&2\\0&5\end{pmatrix}\). For \(\begin{pmatrix}1&2\\2&-1\end{pmatrix}\): subtract 2·row 1 to get \((0,-5)\), then negate: \(\begin{pmatrix}1&2\\0&5\end{pmatrix}\). The normal forms agree, so the lattices are equal; both are \(\{(x,y)\in\mathbb Z^2: y\equiv2x \pmod 5\}\).
Part III · Life and legacy
Lesson 11

The Sorbonne years, 1869–1897

In 1869 Hermite succeeded Jean-Marie Duhamel as professor of analysis at the École Polytechnique and professor at the Faculté des sciences of Paris, the Sorbonne. He gave up the Polytechnique chair in 1876, where Camille Jordan followed him, and kept the Sorbonne chair until 1897. His lectures were celebrated. Henri Poincaré, Émile Picard, Paul Appell and Jacques Hadamard heard him, and his printed Cours d’analyse spread his style. Poincaré described him as the most thorough Platonic realist among the mathematicians he knew, and Hermite wrote to Thomas Stieltjes in 1889 that numbers seemed to him to exist outside us.

His doctoral students included Henri Poincaré, Thomas Stieltjes, Henri Padé, Jules Tannery, Gaston Floquet and Mihailo Petrović. His correspondence was enormous: with Stieltjes, with Gösta Mittag-Leffler (whom he sent to Berlin to study with Weierstrass), with Weierstrass and Kronecker, and with Cayley and Sylvester. Through it he kept French mathematics in close touch with German work in the decades after the war of 1870, when relations between the two countries were bitter.

His later research included linear differential equations with doubly periodic coefficients, notably Lamé’s equation, and continued fractions and approximation. He disliked geometry, and he was famously uneasy with the new set-theoretic analysis; in an 1893 letter to Stieltjes he wrote of turning away in horror from continuous functions without derivatives. He was president of the Académie in 1890. For his seventieth birthday, in December 1892, colleagues and former students held a jubilee in his honour, and he was promoted grand officer of the Légion d’honneur.

DiscussionHermite combined deep Platonist convictions with a dislike of “pathological” functions. Is there a tension between believing that mathematical objects exist independently and refusing to look at some of them? How did the next generation, including his own students, resolve it?

Lesson 12

Death and legacy

Hermite retired in 1897, and his son-in-law Émile Picard succeeded him at the Sorbonne. (His other daughter married the engineer Georges Forestier.) He died in Paris on 14 January 1901, aged 78, and was buried in the Montparnasse cemetery. His Œuvres, edited by Picard, appeared in four volumes from 1905 to 1917, and his correspondence with Stieltjes was published in 1905.

Legacy

  • Number theory: reduction theory and Hermite’s constant began the geometry of numbers, later developed by Minkowski; the finiteness of number fields of bounded discriminant remains a basic theorem.
  • Transcendence: his 1873 method, refined by Lindemann, Weierstrass, Hilbert, Siegel and others, is the ancestor of all modern transcendence theory; Padé approximation grew from it.
  • Algebra and analysis: Hermitian matrices and the spectral theorem for them; the modular solution of the quintic, which led to Klein’s icosahedron lectures; elliptic and theta functions.
  • Applied mathematics and physics: Hermite polynomials in probability (as the orthogonal polynomials of the Gaussian), in quantum mechanics and signal processing; Hermite interpolation and cubic Hermite splines in numerical analysis and computer graphics.
  • Teaching: a generation of French analysts, including Poincaré, Picard, Appell and Stieltjes.
  • Commemoration: the lunar crater Hermite, near the north pole of the Moon, and the Rue Charles Hermite in Paris.

Final projectChoose one of Hermite’s results from Part II. Write a two-page account for a strong undergraduate that states it, works one example in full, and explains how it uses analysis to answer a question about integers or algebraic numbers.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Charles Hermite, MacTutor History of Mathematics.
  2. Charles Hermite, Wikipedia; Charles Hermite, Wikipédia (French).
  3. B. Baillaud and H. Bourget (eds.), Correspondance d’Hermite et de Stieltjes, 2 vols., Gauthier-Villars, 1905.
  4. É. Picard, Charles Hermite, in Œuvres de Charles Hermite, vol. 1, Gauthier-Villars, 1905.

Mathematics

  1. Œuvres de Charles Hermite, ed. É. Picard, 4 vols., Gauthier-Villars, 1905–1917.
  2. C. Hermite, Extraits de lettres à M. Jacobi sur différents objets de la théorie des nombres, J. reine angew. Math. 40 (1850).
  3. C. Hermite, Sur la résolution de l’équation du cinquième degré, Comptes rendus 46 (1858).
  4. C. Hermite, Sur un nouveau développement en série des fonctions, Comptes rendus 58 (1864).
  5. C. Hermite, Sur la fonction exponentielle, Comptes rendus 77 (1873).
  6. C. Hermite, Sur la formule d’interpolation de Lagrange, J. reine angew. Math. 84 (1878).
  7. F. Klein, Vorlesungen über das Ikosaeder, Teubner, 1884.
  8. D. Hilbert, Über die Transcendenz der Zahlen e und π, Math. Ann. 43 (1893).
  9. I. Niven, A simple proof that π is irrational, Bull. Amer. Math. Soc. 53 (1947).
  10. J. H. Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, Springer, 3rd ed. 1999.
  11. H. Cohn and A. Kumar, Optimality and uniqueness of the Leech lattice among lattices, Annals of Math. 170 (2009).
  12. T. J. Stieltjes’ letters, “Marvelous cancellations”: T. J. Stieltjes’ letters concerning the zeta function, arXiv:1410.8592.