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Hermann Grassmann

Hermann Günther Grassmann

A course in two strands: the life of a Stettin schoolteacher who invented linear algebra and the exterior product decades before anyone could read him, and the mathematics of vectors, subspaces and oriented volumes that grew from his Ausdehnungslehre.

Born 15 April 1809, Stettin Died 26 September 1877, Stettin Ausdehnungslehre 1844 and 1862 Level advanced undergraduate → graduate
Grassmann’s products as oriented magnitudes. The shaded base is a∧b, an oriented area; the box is a∧b∧c, an oriented volume. Reversing the circulation gives b∧a = −a∧b.
Lesson 00

How to use this course

Hermann Grassmann never held a university post. He taught in secondary schools in Stettin for most of his life, and his main mathematical book was ignored for a generation. Yet it contains vector spaces, linear independence, dimension, the exterior product, the regressive product and the coordinates of subspaces now named after him. Late in life he turned to Sanskrit and found lasting fame among linguists, while mathematicians were only beginning to read him.

Prerequisites for Part II
  • A first course in linear algebra: vector spaces, bases, determinants.
  • Vector calculus in \(\mathbb R^3\): dot and cross products, line integrals.
  • For Lesson 06, some projective geometry is helpful.
  • For Lesson 10, the language of multivariable change of variables.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. Grassmann wrote products in square brackets, \([ab]\) for the outer product and \([a|b]\) for the inner product. We use modern notation, \(a\wedge b\) and \(a\cdot b\), and say when his notation differs.

Reference

Timeline

1809Born 15 April in Stettin, Pomerania (now Szczecin, Poland), third of twelve children of Justus Günter Grassmann, teacher at the Stettin Gymnasium.
1827Leaves the Gymnasium second in his class; studies theology, classical languages, philosophy and literature at Berlin.
1830Returns to Stettin and begins teaching himself mathematics.
1831December: teaching examinations in Berlin; qualified only for the lower classes.
1832Assistant teacher at the Stettin Gymnasium.
1834–35Mathematics teacher at the Gewerbeschule in Berlin, succeeding Jakob Steiner; then the new Otto Schule in Stettin.
1839Second theological examination.
1840Qualifies for the highest school classes with an essay of about 200 pages on the theory of the tides, written with his new vector methods.
1844Die lineale Ausdehnungslehre, ein neuer Zweig der Mathematik.
1845Neue Theorie der Elektrodynamik.
18461 July: wins the Jablonowski Society prize on Leibniz’s geometric characteristic; published 1847.
1847Oberlehrer at the Friedrich Wilhelm Schule. A negative report by Ernst Kummer ends his hope of a university post.
1848–49Publishes a political weekly with his brother Robert.
184912 April: marries Therese Knappe. They have eleven children, seven of whom reach adulthood.
1852Succeeds his late father at the Stettin Gymnasium, with the title of professor.
1853Theory of colour mixing: Grassmann’s laws.
1861Lehrbuch der Arithmetik, with recursive definitions of addition and multiplication.
1862Die Ausdehnungslehre: vollständig und in strenger Form bearbeitet.
1863Paper on aspirates in Sanskrit and Greek: Grassmann’s law.
1867Hermann Hankel’s Theorie der complexen Zahlensysteme brings his work to wider notice.
1873Wörterbuch zum Rig-Veda.
1876Honorary doctorate from Tübingen. First volume of his Rig Veda translation (second 1877).
1877Dies 26 September in Stettin, aged 68.
1878William Kingdon Clifford’s geometric algebra, built on Grassmann’s.
1888Giuseppe Peano gives the first axiomatic definition of a vector space, in a book on Grassmann’s calculus.
1894–1911Collected works edited by Friedrich Engel.
Part I · Formation
Lesson 01

Stettin and Berlin, 1809–1830

Hermann Günther Grassmann was born in Stettin, a Baltic port in the Prussian province of Pomerania, on 15 April 1809. He was the third of twelve children. His father, Justus Günter Grassmann, was an ordained minister who taught mathematics and physics at the Stettin Gymnasium. He wrote school textbooks and did research on crystallography. His mother, Johanne Luise Friederike Medenwald, was a minister’s daughter.

His mother taught him first. He then attended a private school and the Gymnasium where his father taught. In his early years there he was not a strong pupil. He improved steadily and, at eighteen, passed the final examinations second in his school.

Berlin

In 1827 he went to the University of Berlin with his eldest brother to study theology. He also took courses in classical languages, philosophy and literature. No mathematics or physics course of his is recorded. The philological training mattered later. The habit of building a subject from first principles, which he applied to geometry, he applied forty years afterwards to Sanskrit.

Grassmann left Berlin in the autumn of 1830 and went home to Stettin. He was twenty-one, a theologian by training, and he had decided to teach himself mathematics.

Lesson 02

The teacher and the tides, 1830–1840

To teach in a Prussian Gymnasium Grassmann needed a state examination. He sat it in Berlin in December 1831 and passed only at the lower level; the examiners told him to try again for the higher classes. In spring 1832 he became an assistant teacher at the Stettin Gymnasium. In 1834 he passed the first theological examination of the Lutheran church council in Stettin.

That autumn he moved to Berlin to teach mathematics at the Gewerbeschule, a trade school, in the post just vacated by the geometer Jakob Steiner. He stayed about a year, then returned to Stettin to teach at the newly opened Otto Schule. There he taught mathematics, physics, German, Latin and religion. In 1839 he passed his second theological examination. He was qualified as a clergyman and had not yet qualified fully as a mathematics teacher.

The essay on the tides

In 1840 he sat the Berlin examination again. Candidates wrote an essay on a set topic, and Grassmann’s was the theory of the tides. He took the classical treatments of Pierre-Simon Laplace and Joseph-Louis Lagrange and rewrote them in a geometric calculus of his own. The essay of about 200 pages added, subtracted and differentiated directed quantities, and treated functions of them. It was in effect the first vector analysis. The examiners passed him for the highest classes in mathematics, physics, chemistry and mineralogy. They did not see what the essay contained. It was first printed in the collected works edited by Friedrich Engel (1894–1911).

In the preface to his 1844 book, Grassmann traced the idea back to a simple observation about signed segments on a line: if \(A,B,C\) are points, then \(AB+BC=AC\) holds whatever their order, provided segments carry a direction. Extending this rule from a line to the plane and to space, and then asking what a product of segments should be, led him to the whole theory.

Lesson 03

The Ausdehnungslehre of 1844

Grassmann began writing in spring 1842 and finished in autumn 1843. The book appeared in Leipzig in 1844 as Die lineale Ausdehnungslehre, ein neuer Zweig der Mathematik (Linear extension theory, a new branch of mathematics). Its long subtitle promised applications to statics, mechanics, magnetism and crystallography. The first part develops a general theory of “extensive magnitudes” before any coordinates appear, in a philosophical style that made it very hard to read.

Reception

Almost nobody read it. Grassmann sent a copy to August Ferdinand Möbius in Leipzig, who was working on related ideas in his barycentric calculus. Möbius admitted he could not get through it and declined to review it. He did suggest that Grassmann enter a prize competition of the Jablonowski Society in Leipzig, which asked for a development of the geometric calculus that Gottfried Wilhelm Leibniz had sketched in a letter to Christiaan Huygens. Grassmann’s essay, Geometrische Analyse, was the only entry and won the prize on 1 July 1846. It was published in 1847 with a commentary by Möbius, who as a judge criticised its abstraction.

In May 1847 Grassmann asked the Prussian Ministry of Education to consider him for a university chair. The ministry asked Ernst Kummer for an opinion. Kummer’s report on the prize essay found good material “expressed in a deficient form”, and the request failed. Grassmann remained a schoolteacher, and in 1847 he received the title Oberlehrer at the Friedrich Wilhelm Schule in Stettin. Years later his publisher told him that about 600 unsold copies of the 1844 book had been pulped as waste paper.

DiscussionGrassmann’s book was rejected less for its mathematics than for its style. Find one other example of a major idea delayed by its presentation, and compare the two cases.

Part II · The mathematics
Lesson 04

Extensive magnitudes and dimension

For the general reader. Arrows in the plane can be added tip to tail and stretched by numbers. Grassmann saw that these two operations alone support a complete theory, with no need for a picture or a fixed number of dimensions. His “extensive magnitudes” are what we now call vectors, and his systems of them are vector spaces of any finite dimension.

Goals
  • Read Grassmann’s definitions in modern language.
  • Prove that any \(n+1\) vectors in an \(n\)-dimensional system are dependent.
  • Prove Grassmann’s dimension formula for two subspaces.

Units and dependence

Grassmann starts from a set of “units” \(e_1,\dots,e_n\) and forms all magnitudes \(a=\alpha_1e_1+\cdots+\alpha_ne_n\) with real coefficients. They are added and multiplied by numbers coefficient-wise. A set of magnitudes is dependent if one of them can be derived numerically from the others, that is, if some non-trivial linear combination vanishes. A system of magnitudes closed under these operations he calls a system of the \(m\)-th level (Stufe) when it has \(m\) independent generators. In 1888 Giuseppe Peano turned this into the axioms of a vector space.

Theorem (Grassmann, 1844 and 1862)

Let \(V\) be spanned by \(n\) magnitudes. Then any \(n+1\) magnitudes in \(V\) are dependent. Consequently every two bases of \(V\) have the same number of elements, the dimension \(\dim V\), and every independent set extends to a basis.

Proof sketch. Induct on \(n\). Write \(n+1\) vectors \(v_0,\dots,v_n\) in terms of the spanning set \(e_1,\dots,e_n\). If no \(v_i\) involves \(e_n\), apply induction to the span of \(e_1,\dots,e_{n-1}\). Otherwise say \(v_0\) involves \(e_n\); subtracting suitable multiples of \(v_0\) from \(v_1,\dots,v_n\) gives \(n\) vectors in the span of \(e_1,\dots,e_{n-1}\), which are dependent by induction, and the dependence lifts to one among the \(v_i\).

Theorem (Grassmann’s dimension formula)

If \(U\) and \(W\) are subspaces of a finite-dimensional vector space, then

\[ \dim(U+W)+\dim(U\cap W)=\dim U+\dim W. \]

Proof. Choose a basis \(c_1,\dots,c_k\) of \(U\cap W\). Extend it to a basis \(c_1,\dots,c_k,u_1,\dots,u_p\) of \(U\) and to a basis \(c_1,\dots,c_k,w_1,\dots,w_q\) of \(W\). The \(k+p+q\) vectors together span \(U+W\). They are independent: if \(\sum\gamma_ic_i+\sum\alpha_ju_j+\sum\beta_lw_l=0\), then \(\sum\beta_lw_l\) lies in \(U\cap W\), so it is a combination of the \(c_i\), forcing all \(\beta_l=0\), and then all other coefficients vanish. Hence \(\dim(U+W)=k+p+q=(k+p)+(k+q)-k\).

Grassmann used the formula as a statement of projective geometry: two planes in three-dimensional space meet in at least a line, because \(2+2-3=1\) in the vector space whose lines are the points.

Exercise 4.1In \(\mathbb R^3\) let \(U=\operatorname{span}(e_1+e_2,\ e_3)\) and \(W=\operatorname{span}(e_1,\ e_2+e_3)\). Find \(\dim(U+W)\) and a basis of \(U\cap W\).

Show solution
The four vectors span \(\mathbb R^3\): \(e_1\) is there, \(e_2=(e_1+e_2)-e_1\), and \(e_3\) is there. So \(\dim(U+W)=3\) and \(\dim(U\cap W)=2+2-3=1\). Solve \(a(e_1+e_2)+be_3=ce_1+d(e_2+e_3)\): comparing coefficients gives \(a=c\), \(a=d\), \(b=d\). So \(U\cap W\) is spanned by \(e_1+e_2+e_3\).

Exercise 4.2Show that two distinct 2-dimensional subspaces of \(\mathbb R^3\) always meet in a line, while two 2-dimensional subspaces of \(\mathbb R^4\) can meet only in \(0\). Translate both statements into projective geometry.

Show solution
In \(\mathbb R^3\), distinct planes have \(\dim(U+W)=3\), so \(\dim(U\cap W)=1\). In \(\mathbb R^4\), take \(U=\operatorname{span}(e_1,e_2)\) and \(W=\operatorname{span}(e_3,e_4)\): \(U+W=\mathbb R^4\) and \(\dim(U\cap W)=0\). Projectively: two distinct lines in the projective plane meet in a point; two lines in projective 3-space can be skew.
Lesson 05

The exterior product

For the general reader. Two arrows span a parallelogram, and three span a box. Grassmann treated these figures as products of the arrows, with a size and an orientation. Swapping two factors reverses the orientation and so changes the sign. A product vanishes exactly when the arrows are squeezed into a lower dimension. Determinants, areas and volumes all follow from this one rule.

Goals
  • Define the exterior algebra and compute in it.
  • Prove that a product of vectors vanishes exactly when they are dependent.
  • Recover the determinant as the coefficient of a top-degree product.

The rules

Grassmann called it the outer (äußeres) product and wrote it \([ab]\). Its defining properties are bilinearity, associativity and the rule \(a\wedge a=0\) for every vector \(a\). Expanding \((a+b)\wedge(a+b)=0\) gives anticommutativity, \(a\wedge b=-\,b\wedge a\).

Theorem (exterior algebra)

Let \(V\) have basis \(e_1,\dots,e_n\). There is an associative algebra \(\Lambda V=\bigoplus_{k=0}^n\Lambda^kV\), generated by \(V\), whose product \(\wedge\) is bilinear and satisfies \(v\wedge v=0\) for \(v\in V\). The products \(e_{i_1}\wedge\cdots\wedge e_{i_k}\) with \(i_1<\cdots<i_k\) form a basis of \(\Lambda^kV\), so \(\dim\Lambda^kV=\binom nk\) and \(\dim\Lambda V=2^n\). Moreover:

  1. \(v_1\wedge\cdots\wedge v_k=0\) if and only if \(v_1,\dots,v_k\) are linearly dependent;
  2. if \(v_j=\sum_i a_{ij}e_i\) for \(j=1,\dots,n\), then \(v_1\wedge\cdots\wedge v_n=\det(a_{ij})\,e_1\wedge\cdots\wedge e_n\).

Proof sketch. For (2), expand by multilinearity. A term \(e_{\sigma(1)}\wedge\cdots\wedge e_{\sigma(n)}\) survives only if \(\sigma\) is a permutation, and reordering it to \(e_1\wedge\cdots\wedge e_n\) costs the sign of \(\sigma\). The coefficient is \(\sum_\sigma\operatorname{sgn}(\sigma)\prod_j a_{\sigma(j)j}\), the determinant. For (1), if the \(v_j\) are dependent, write one as a combination of the others and expand; every term repeats a factor. If they are independent, extend them to a basis; by (2) applied in that basis, the full product is non-zero, so the partial product \(v_1\wedge\cdots\wedge v_k\) is non-zero too.

The geometric meaning: in the plane, \(a\wedge b=(a_1b_2-a_2b_1)\,e_1\wedge e_2\), and the coefficient is the signed area of the parallelogram on \(a\) and \(b\). In space, \(a\wedge b\wedge c\) is the signed volume of the parallelepiped. Grassmann also multiplied points, not only vectors. The product of two points represents the segment joining them, bound to its line, and the product of three points a bounded piece of a plane. This is the projective version of the same algebra, one dimension up.

Exercise 5.1Compute \((2e_1+e_2)\wedge(e_1+3e_2)\) and compare with a determinant.

Show solution
\(2e_1\wedge e_1+6e_1\wedge e_2+e_2\wedge e_1+3e_2\wedge e_2=6e_1\wedge e_2-e_1\wedge e_2=5\,e_1\wedge e_2\). Indeed \(\det\begin{pmatrix}2&1\\1&3\end{pmatrix}=5\).

Exercise 5.2Let \(a=(1,2,0)\), \(b=(0,1,1)\), \(c=(1,0,1)\) in \(\mathbb R^3\). Compute \(a\wedge b\) and \(a\wedge b\wedge c\). Are \(a,b,c\) independent?

Show solution
\(a\wedge b=(1\cdot1-2\cdot0)\,e_{12}+(1\cdot1-0\cdot0)\,e_{13}+(2\cdot1-0\cdot1)\,e_{23}=e_{12}+e_{13}+2e_{23}\), writing \(e_{ij}=e_i\wedge e_j\). Then \(a\wedge b\wedge c=(e_{12}+e_{13}+2e_{23})\wedge(e_1+e_3)\). Only \(e_{12}\wedge e_3=e_{123}\) and \(2e_{23}\wedge e_1=2e_{123}\) survive, giving \(3e_{123}\). The coefficient is non-zero, so the vectors are independent.

Exercise 5.3Show that if \(\omega\in\Lambda^kV\) and \(\eta\in\Lambda^lV\), then \(\eta\wedge\omega=(-1)^{kl}\,\omega\wedge\eta\). Deduce that a 2-vector commutes with everything.

Show solution
By bilinearity it suffices to check products of basis vectors. Moving each of the \(l\) factors of \(\eta\) past the \(k\) factors of \(\omega\) takes \(kl\) transpositions, each contributing \(-1\). If \(k=2\) then \(kl\) is even for every \(l\).
Lesson 06

Subspaces as products: the Grassmannian

For the general reader. A plane through the origin can be described by any two arrows lying in it, and there are endlessly many choices. Their exterior product, though, is the same up to a number. Grassmann used this to give every line, plane or higher subspace a set of coordinates. The space of all subspaces of a given dimension, now called a Grassmannian, is one of the central objects of geometry and topology.

Goals
  • Attach coordinates to a subspace and prove they determine it.
  • Find the quadratic relations that a product of vectors must satisfy.
  • Use the regressive product to intersect subspaces.
Theorem (Grassmann coordinates)

Let \(W\subseteq V\) be a \(k\)-dimensional subspace with basis \(w_1,\dots,w_k\). The \(k\)-vector \(\omega_W=w_1\wedge\cdots\wedge w_k\) changes only by the factor \(\det g\) when the basis is changed by \(g\in\mathrm{GL}_k\), and \(W=\{v\in V: v\wedge\omega_W=0\}\). Hence \(W\mapsto[\omega_W]\) embeds the set \(G(k,V)\) of \(k\)-dimensional subspaces into the projective space \(\mathbb P(\Lambda^kV)\).

The first claim is Lesson 05 (2) applied inside \(W\). The second is Lesson 05 (1): \(v\wedge w_1\wedge\cdots\wedge w_k=0\) exactly when \(v\) is dependent on the \(w_i\). In coordinates, the components of \(\omega_W\) are the \(k\times k\) minors of the \(n\times k\) matrix with columns \(w_i\). For lines in projective 3-space (\(k=2\), \(n=4\)) they are the six coordinates that Julius Plücker introduced in the 1860s, which is why they are usually called Plücker coordinates. Grassmann had them in 1844.

Which \(k\)-vectors come from subspaces?

A \(k\)-vector is decomposable if it is a product of \(k\) vectors. Not every one is.

Proposition (the Plücker relation for lines)

A 2-vector \(\omega\) is decomposable if and only if \(\omega\wedge\omega=0\). When \(\dim V=4\) and \(\omega=\sum_{i<j}p_{ij}\,e_i\wedge e_j\), this is the single equation

\[ p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=0. \]

So \(G(2,4)\) is a smooth quadric hypersurface in \(\mathbb P^5\), of dimension \(4\). In general \(G(k,n)\) has dimension \(k(n-k)\).

If \(\omega=a\wedge b\) then \(\omega\wedge\omega=a\wedge b\wedge a\wedge b=0\). Conversely, a 2-vector can always be brought by a change of basis to the normal form \(e_1\wedge e_2+e_3\wedge e_4+\cdots+e_{2r-1}\wedge e_{2r}\), whose square is non-zero when \(r\ge2\). Expanding \(\omega\wedge\omega\) gives \(2(p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23})\,e_1\wedge e_2\wedge e_3\wedge e_4\). The dimension count \(k(n-k)\): a subspace is the column space of an \(n\times k\) matrix which, after row reduction, has a \(k\times k\) identity block and \(k(n-k)\) free entries.

The regressive product

Grassmann’s second product goes the other way. The outer product of a \(p\)-dimensional and a \(q\)-dimensional subspace that meet only in 0 represents their sum. His regressive product represents their intersection when they together span \(V\). In modern terms, fix an orientation and the Hodge star of Lesson 07, and set \(A\vee B=\star^{-1}(\star A\wedge\star B)\). If \(U+W=V\), then \(\omega_U\vee\omega_W\) is a non-zero multiple of \(\omega_{U\cap W}\), of degree \(\dim U+\dim W-n\). This is Grassmann’s dimension formula of Lesson 04 written as an algebraic identity.

Exercise 6.1Find the Plücker coordinates of the line in \(\mathbb P^3\) through the points \([1:1:0:0]\) and \([0:1:1:2]\), and check the relation.

Show solution
With \(u=(1,1,0,0)\), \(v=(0,1,1,2)\) and \(p_{ij}=u_iv_j-u_jv_i\): \(p_{12}=1\), \(p_{13}=1\), \(p_{14}=2\), \(p_{23}=1\), \(p_{24}=2\), \(p_{34}=0\). Then \(p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=0-2+2=0\).

Exercise 6.2Show that \(\omega=e_1\wedge e_2+e_3\wedge e_4\) is not the product of two vectors.

Show solution
\(\omega\wedge\omega=e_{12}\wedge e_{34}+e_{34}\wedge e_{12}=2e_{1234}\ne0\). Equivalently \(p_{12}=p_{34}=1\) and all others vanish, so \(p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=1\ne0\).

Exercise 6.3In \(\mathbb R^3\) with the standard orientation, use \(A\vee B=\star^{-1}(\star A\wedge\star B)\) to intersect the planes \(U=\operatorname{span}(e_1,e_2)\) and \(W=\operatorname{span}(e_2,e_3)\). (Use \(\star e_{12}=e_3\), \(\star e_{23}=e_1\), \(\star e_2=e_3\wedge e_1\).)

Show solution
\(\star e_{12}\wedge\star e_{23}=e_3\wedge e_1\), and \(\star^{-1}(e_3\wedge e_1)=e_2\). So \(e_{12}\vee e_{23}=e_2\): the planes meet in the \(e_2\)-axis, of dimension \(2+2-3=1\).
Lesson 07

Complement and inner product (1862)

For the general reader. After the failure of 1844, Grassmann rewrote the whole theory in 1862 in the dry style of a textbook: definitions, theorems, proofs. The new version starts from coordinates and adds a measure of length and angle. He introduced the complement of a magnitude, a kind of perpendicular partner, and used it to define an inner product of areas and volumes as well as of arrows.

Goals
  • Define Grassmann’s complement, today’s Hodge star.
  • Prove the Gram determinant formula for the inner product of \(k\)-vectors.
  • Derive the cross product and the Lagrange identity.

The complement

Take orthonormal units \(e_1,\dots,e_n\). For an increasing multi-index \(I\) with complement \(I^c\), Grassmann defines the complement \(|e_I=\pm e_{I^c}\), with the sign chosen so that \(e_I\wedge|e_I=e_1\wedge\cdots\wedge e_n\), and extends linearly. Today this is the Hodge star, \(\star\omega=|\omega\). The inner product of two \(k\)-vectors is the coefficient of \(e_1\wedge\cdots\wedge e_n\) in \(\omega\wedge|\eta\), which Grassmann wrote \([\omega|\eta]\).

Theorem (inner product of \(k\)-vectors)

For vectors \(a_1,\dots,a_k,b_1,\dots,b_k\) in Euclidean \(\mathbb R^n\),

\[ \langle a_1\wedge\cdots\wedge a_k,\ b_1\wedge\cdots\wedge b_k\rangle=\det\big(a_i\cdot b_j\big)_{i,j=1}^{k}. \]

In particular \(|a_1\wedge\cdots\wedge a_k|^2\) is the Gram determinant, the squared \(k\)-dimensional volume of the parallelotope on the \(a_i\).

Proof sketch. The basis \(e_I\) is orthonormal for this inner product. The coefficient of \(e_I\) in \(a_1\wedge\cdots\wedge a_k\) is the minor \(\det(A_I)\) of the \(n\times k\) matrix \(A\) of the \(a_i\). So the inner product equals \(\sum_I\det(A_I)\det(B_I)\), and the Cauchy–Binet formula identifies this sum with \(\det(A^{\mathsf T}B)=\det(a_i\cdot b_j)\).

For \(k=2\) in \(\mathbb R^3\) this gives the Lagrange identity \(|a\wedge b|^2=|a|^2|b|^2-(a\cdot b)^2\). The complement turns the 2-vector \(a\wedge b\) into a vector, the cross product: \(a\times b=|(a\wedge b)\). This identification works only in three dimensions, which is one reason Grassmann’s algebra is the more general tool.

The 1862 book also contains a treatment of Pfaff’s problem on differential expressions \(\sum X_i\,dx_i\), an early application of his algebra to analysis that points toward the calculus of differential forms (Lesson 10).

Exercise 7.1Find the area of the triangle in \(\mathbb R^3\) with vertices \(0\), \(a=(1,2,2)\) and \(b=(2,0,1)\) in two ways: by the Gram determinant and from the components of \(a\wedge b\).

Show solution
\(|a|^2=9\), \(|b|^2=5\), \(a\cdot b=4\), so \(|a\wedge b|^2=45-16=29\). Components: \(e_{12}\): \(1\cdot0-2\cdot2=-4\); \(e_{13}\): \(1\cdot1-2\cdot2=-3\); \(e_{23}\): \(2\cdot1-2\cdot0=2\); squares sum to \(16+9+4=29\). The area is \(\tfrac12\sqrt{29}\).

Exercise 7.2In \(\mathbb R^3\), find the complements \(|e_{12}\), \(|e_{13}\), \(|e_{23}\) and show that \(|(a\wedge b)=a\times b\).

Show solution
We need \(e_I\wedge|e_I=e_{123}\). So \(|e_{12}=e_3\), \(|e_{23}=e_1\), and \(|e_{13}=-e_2\) because \(e_{13}\wedge e_2=-e_{123}\). With \(a\wedge b=c_{12}e_{12}+c_{13}e_{13}+c_{23}e_{23}\), \(|(a\wedge b)=(c_{23},-c_{13},c_{12})=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)=a\times b\).
Lesson 08

Electrodynamics and colour

For the general reader. Grassmann used his vector methods on two problems of physics. In 1845 he proposed a law for the force between small pieces of electric current, simpler than the one André-Marie Ampère had given. In 1853 he showed that the mixing of coloured lights obeys the rules of a three-dimensional vector space. Both results are still taught, the first as the form of the magnetic force in every textbook, the second as Grassmann’s laws of colorimetry.

Goals
  • Compare Grassmann’s and Ampère’s force laws for current elements.
  • Prove they agree for closed circuits.
  • State Grassmann’s laws of colour mixing as linear algebra.

Current elements (1845)

Let current \(I_1\) flow in an element \(d\mathbf l_1\) at \(\mathbf x_1\) and \(I_2\) in \(d\mathbf l_2\) at \(\mathbf x_2\). Put \(\mathbf r=\mathbf x_2-\mathbf x_1\), \(r=|\mathbf r|\), \(\hat{\mathbf r}=\mathbf r/r\). In modern SI form, Ampère’s law and Grassmann’s law for the force on element 2 are

\[ \begin{aligned} d\mathbf F^{\mathrm{A}}_{2}&=-\frac{\mu_0I_1I_2}{4\pi r^2}\,\hat{\mathbf r}\,\big[2\,d\mathbf l_1\cdot d\mathbf l_2-3(\hat{\mathbf r}\cdot d\mathbf l_1)(\hat{\mathbf r}\cdot d\mathbf l_2)\big],\\ d\mathbf F^{\mathrm{G}}_{2}&=\frac{\mu_0I_1I_2}{4\pi r^2}\,d\mathbf l_2\times\big(d\mathbf l_1\times\hat{\mathbf r}\big). \end{aligned} \]

Grassmann’s law is what one gets by combining the Biot–Savart field of element 1 with the force \(I\,d\mathbf l\times\mathbf B\) on element 2. It does not obey Newton’s third law element by element. Ampère’s does.

Theorem

For two closed circuits \(C_1,C_2\), the total forces given by the two laws coincide:

\[ \mathbf F_2=-\frac{\mu_0I_1I_2}{4\pi}\oint_{C_1}\oint_{C_2}\frac{\hat{\mathbf r}}{r^2}\,(d\mathbf l_1\cdot d\mathbf l_2). \]

Proof sketch for Grassmann’s law. By the triple product rule, \(d\mathbf l_2\times(d\mathbf l_1\times\hat{\mathbf r})=d\mathbf l_1\,(d\mathbf l_2\cdot\hat{\mathbf r})-\hat{\mathbf r}\,(d\mathbf l_1\cdot d\mathbf l_2)\). Along \(C_2\), with \(\mathbf x_1\) fixed, \((\hat{\mathbf r}\cdot d\mathbf l_2)/r^2=-d(1/r)\), an exact differential, so the first term integrates to zero around \(C_2\). The second term is the stated integrand. The same formula follows from Ampère’s law after an integration by parts, and it is antisymmetric under exchanging the circuits, so Newton’s third law holds for whole circuits. Experiments with closed circuits cannot distinguish the two laws.

Colour mixing (1853)

Hermann von Helmholtz had reported in 1852 that he could find only one pair of complementary spectral colours. Grassmann replied in 1853 with a short theory of colour mixing. He argued from a few postulates, including continuity, that every spectral colour must have a complementary one. Helmholtz’s later measurements found the pairs. Grassmann’s postulates are now stated as follows.

Grassmann’s laws (modern form)

Write \(X\sim Y\) when two lights look identical to a standard observer, and \(X+Y\) for their superposition. Then \(\sim\) is an equivalence relation compatible with superposition and with scaling of intensity: if \(X\sim Y\) then \(X+Z\sim Y+Z\) and \(\lambda X\sim\lambda Y\) for \(\lambda>0\). There are three primaries \(R,G,B\) such that every light \(X\) satisfies a unique match \(X+\sum_{c<0}|c|P_c\sim\sum_{c>0}c\,P_c\), written \(X\sim r R+g G+b B\) with real (possibly negative) coefficients.

So colour classes form a convex cone in a three-dimensional real vector space, and the map from light spectra to tristimulus values \((r,g,b)\) is linear. A negative coefficient means that primary had to be added to the test light rather than to the mixture. Different spectra with the same tristimulus values are metamers.

Exercise 8.1Two current elements lie on the same straight line and point in the same direction. Compare the forces given by the two laws.

Show solution
Here \(d\mathbf l_1=dl_1\,\hat{\mathbf r}\) and \(d\mathbf l_2=dl_2\,\hat{\mathbf r}\). Grassmann: \(d\mathbf l_1\times\hat{\mathbf r}=0\), so no force. Ampère: the bracket is \(2\,dl_1dl_2-3\,dl_1dl_2=-dl_1dl_2\), so \(d\mathbf F^{\mathrm A}_2=+\frac{\mu_0I_1I_2}{4\pi r^2}dl_1dl_2\,\hat{\mathbf r}\), a repulsion. Ampère’s longitudinal forces were the subject of experiments for over a century.

Exercise 8.2Two parallel current elements \(d\mathbf l_1=d\mathbf l_2=dl\,\hat{\mathbf z}\) are separated by \(\mathbf r=r\hat{\mathbf x}\). Compute both forces on element 2.

Show solution
Grassmann: \(\hat{\mathbf z}\times\hat{\mathbf x}=\hat{\mathbf y}\) and \(\hat{\mathbf z}\times\hat{\mathbf y}=-\hat{\mathbf x}\), so \(d\mathbf F^{\mathrm G}_2=-\frac{\mu_0I_1I_2dl^2}{4\pi r^2}\hat{\mathbf x}\): attraction for like currents. Ampère: \(\hat{\mathbf r}\cdot d\mathbf l_i=0\), so \(d\mathbf F^{\mathrm A}_2=-\frac{2\mu_0I_1I_2dl^2}{4\pi r^2}\hat{\mathbf x}\): also attraction, twice as strong. The element laws differ; the circuit totals agree.

Exercise 8.3Light \(X\) matches \(2R+G+0.5B\) and light \(Y\) matches \(R+2G-0.5B\). What does the negative coefficient mean in the laboratory, and which combination of primaries matches \(X+Y\)?

Show solution
For \(Y\), the observer had to add \(0.5B\) to \(Y\) itself to match \(R+2G\): \(Y+0.5B\sim R+2G\). By additivity, \(X+Y\sim3R+3G+0B\). The blue contributions cancel, and the mixture is matched by red and green alone.
Lesson 09

Arithmetic by recursion (1861)

For the general reader. Why is \(a+b=b+a\)? Most people take it as obvious. In his 1861 school textbook of arithmetic, Grassmann proved it. He defined addition and multiplication step by step, starting from the operation of adding one, and derived all the familiar laws by induction. The same method reappeared a generation later in the foundations of arithmetic of Richard Dedekind and Giuseppe Peano.

Goals
  • State Grassmann’s recursive definitions.
  • Prove associativity and commutativity of addition by induction.
  • Prove the distributive law.

Grassmann built the integers from a unit \(e\) by repeated addition and subtraction. Restricting to the positive integers, with \(1\) as the unit, his definitions read:

\[ a+(b+1)=(a+b)+1,\qquad a\cdot1=a,\qquad a\cdot(b+1)=a\cdot b+a. \]

Each defines the operation for \(b+1\) once it is known for \(b\). The only tool of proof is induction: a statement true for \(1\), and true for \(b+1\) whenever true for \(b\), is true for every positive integer.

Theorem (Grassmann, 1861)

From these definitions and induction alone, addition and multiplication are associative and commutative, and multiplication distributes over addition.

Associativity of addition. Prove \(a+(b+c)=(a+b)+c\) by induction on \(c\). For \(c=1\) it is the definition. If it holds for \(c\), then

\[ a+\big(b+(c+1)\big)=a+\big((b+c)+1\big)=\big(a+(b+c)\big)+1=\big((a+b)+c\big)+1=(a+b)+(c+1). \]

Each step uses either the definition or the induction hypothesis. This is probably the first appearance of the primitive recursive definitions of the arithmetic operations, nearly thirty years before Dedekind’s Was sind und was sollen die Zahlen? (1888) and Peano’s axioms (1889).

Exercise 9.1Prove that addition is commutative. First show \(1+a=a+1\) by induction on \(a\).

Show solution
Lemma: for \(a=1\) it is trivial; if \(1+a=a+1\) then \(1+(a+1)=(1+a)+1=(a+1)+1\). Main claim, by induction on \(b\): \(a+1=1+a\) is the lemma. If \(a+b=b+a\), then \(a+(b+1)=(a+b)+1=(b+a)+1=b+(a+1)=b+(1+a)=(b+1)+a\), using the definition, the hypothesis, the definition, the lemma and associativity.

Exercise 9.2Prove the left distributive law \(a(b+c)=ab+ac\) by induction on \(c\).

Show solution
For \(c=1\): \(a(b+1)=ab+a=ab+a\cdot1\). If true for \(c\): \(a\big(b+(c+1)\big)=a\big((b+c)+1\big)=a(b+c)+a=(ab+ac)+a=ab+(ac+a)=ab+a(c+1)\), using the definitions, the hypothesis and associativity of addition.
Lesson 10

Clifford, Gibbs and differential forms

For the general reader. Three later developments grew directly from Grassmann’s algebra. William Kingdon Clifford merged it with William Rowan Hamilton’s quaternions into a single “geometric” product. Josiah Willard Gibbs extracted the dot and cross products used in physics today. Élie Cartan made the exterior product the language of integration on curved spaces. In quantum field theory, anticommuting “Grassmann numbers” describe electrons and quarks.

Goals
  • Define Clifford’s product and find the quaternions inside it.
  • See why exterior products appear in change of variables.
  • Meet anticommuting variables.

Clifford’s geometric algebra (1878)

Clifford kept Grassmann’s units but replaced the rule \(e_p\wedge e_p=0\) by \(e_p^2=1\), keeping \(e_pe_q=-e_qe_p\) for \(p\ne q\). For vectors this gives

\[ uv=u\cdot v+u\wedge v, \]

combining Grassmann’s inner and outer products in one associative product.

Proposition

In the Clifford algebra of Euclidean \(\mathbb R^3\), the elements \(i=e_3e_2\), \(j=e_1e_3\), \(k=e_2e_1\) satisfy \(i^2=j^2=k^2=ijk=-1\). The even part \(\operatorname{span}(1,i,j,k)\) is isomorphic to Hamilton’s quaternions.

Differential forms

Under a change of variables \(x=x(u,v)\), \(y=y(u,v)\), the differentials transform linearly: \(dx=x_u\,du+x_v\,dv\), \(dy=y_u\,du+y_v\,dv\). Their exterior product is

\[ dx\wedge dy=(x_uy_v-x_vy_u)\,du\wedge dv, \]

which is exactly the Jacobian rule for double integrals, sign included. Cartan’s exterior differential calculus (1899 onward) treats integrands as elements of the exterior algebra of the cotangent space at each point. Stokes’ theorem then takes the single form \(\int_M d\omega=\int_{\partial M}\omega\).

Grassmann numbers

Variables with \(\theta_i\theta_j=-\theta_j\theta_i\), and hence \(\theta_i^2=0\), generate an exterior algebra. Physicists call them Grassmann variables and use them, with Felix Berezin’s rules of integration, to write path integrals for fermions.

Exercise 10.1Verify the proposition: compute \(i^2\), \(ij\) and \(ijk\).

Show solution
\(i^2=e_3e_2e_3e_2=-e_3e_3e_2e_2=-1\), and likewise \(j^2=k^2=-1\). \(ij=e_3e_2e_1e_3\). Move the last \(e_3\) left past \(e_1\) and \(e_2\) (two sign changes): \(e_3e_3e_2e_1=e_2e_1=k\). Then \(ijk=k^2=-1\).

Exercise 10.2Using \(x=r\cos\phi\), \(y=r\sin\phi\), compute \(dx\wedge dy\).

Show solution
\(dx=\cos\phi\,dr-r\sin\phi\,d\phi\), \(dy=\sin\phi\,dr+r\cos\phi\,d\phi\). Then \(dx\wedge dy=r\cos^2\phi\,dr\wedge d\phi-r\sin^2\phi\,d\phi\wedge dr=r\,dr\wedge d\phi\).

Exercise 10.3In the Clifford algebra of the Euclidean plane, let \(u,v\) be unit vectors at angle \(\theta\) from \(u\) to \(v\). Take \(u=e_1\). Express \(uv\) in terms of \(\theta\) and the bivector \(e_1e_2\), and show that \((e_1e_2)^2=-1\).

Show solution
\(v=\cos\theta\,e_1+\sin\theta\,e_2\), so \(uv=e_1v=\cos\theta+e_1e_2\sin\theta\), while \(vu=\cos\theta-e_1e_2\sin\theta\). Also \((e_1e_2)^2=e_1e_2e_1e_2=-e_1e_1e_2e_2=-1\). The bivector \(e_1e_2\) plays the role of \(i\), and \(uv\) is the complex number \(e^{i\theta}\) in disguise.
Part III · Life and legacy
Lesson 11

Stettin, Sanskrit and the Rig Veda, 1847–1877

The revolution of 1848 drew Grassmann briefly into politics. With his brother Robert he published a weekly paper, the Deutsche Wochenschrift für Staat, Kirche und Volksleben, which argued for German unification under a constitutional monarchy. He grew unhappy with the paper’s direction and withdrew. On 12 April 1849 he married Therese Knappe, a landowner’s daughter. They had eleven children, seven of whom reached adulthood. When his father died in March 1852, Grassmann took over his post at the Stettin Gymnasium, with the title of professor.

He kept working on mathematics. In 1853 Augustin-Louis Cauchy published a method of “algebraic keys” for linear equations whose keys multiply like Grassmann’s units. Grassmann answered in Crelle’s journal in 1855 with a French paper on the different kinds of multiplication, claiming priority over Cauchy and Adhémar Barré de Saint-Venant. The Lehrbuch der Arithmetik appeared in 1861 and the second Ausdehnungslehre in 1862. It too found almost no readers.

The turn to language

After 1862 Grassmann gave most of his research time to comparative philology, which he had studied on his own since about 1849. His 1863 paper on aspirates, in Kuhn’s Zeitschrift für vergleichende Sprachforschung, explained a puzzling pattern in Sanskrit and Greek: a root cannot keep aspirated consonants at both ends, and when two aspirates come close the first loses its aspiration. Greek thríks (hair) beside the genitive trikhós is a classic example. The rule, known as Grassmann’s law, removed a set of apparent exceptions to the sound laws and showed that the Sanskrit forms were not always the oldest.

He then compiled a dictionary to the Rig Veda, the oldest Sanskrit hymn collection. The Wörterbuch zum Rig-Veda appeared in 1873 and was reissued in a third edition as late as 1955. A translation of the Rig Veda followed in two volumes in 1876 and 1877. Linguists honoured him as mathematicians had not. He was elected to the American Oriental Society, and in 1876 the University of Tübingen gave him an honorary doctorate.

Lesson 12

Death and late recognition

Grassmann’s health declined slowly during 1877, and he died of a heart condition in Stettin on 26 September 1877, aged 68. In his last years he saw the first signs that mathematicians were reading him. Hermann Hankel’s Theorie der complexen Zahlensysteme (1867) gave his ideas a central place, and Victor Schlegel, a colleague in Stettin, published an exposition in 1872 and 1875 and a biography in 1878. A new edition of the 1844 book appeared after his death.

Legacy

  • Linear algebra: vector spaces of arbitrary dimension, linear independence, bases, dimension and the dimension formula. Peano’s Calcolo geometrico secondo l’Ausdehnungslehre di H. Grassmann (1888) stated the vector space axioms; Alfred North Whitehead’s Universal Algebra (1898) gave the first systematic English account.
  • Exterior and Clifford algebra: Clifford’s paper of 1878 on applications of Grassmann’s extensive algebra; Cartan’s differential forms; Grassmann variables in quantum field theory.
  • Geometry: Grassmann coordinates and the Grassmannian, now basic in algebraic geometry, topology (classifying spaces of vector bundles) and combinatorics.
  • Vector analysis: Gibbs, who built modern vector analysis in the 1880s, admired Grassmann and pressed for publication of the tides essay. Friedrich Engel edited the collected mathematical and physical works in three volumes, 1894–1911.
  • Linguistics and colour science: Grassmann’s law in Indo-European phonology and Grassmann’s laws in colorimetry, both still under his name.
  • Family: his son Hermann Ernst Grassmann took a doctorate at Halle in 1893 with a thesis applying the extension theory to curves and surfaces, and became a professor at Giessen.

Final projectChoose one theorem from Part II. Write a two-page account for a strong undergraduate that states it in modern form, works one example in full, and explains what Grassmann’s coordinate-free viewpoint adds to the matrix version taught in a first course.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Hermann Günter Grassmann, MacTutor History of Mathematics.
  2. Hermann Grassmann, Wikipedia (with bibliography of his works).
  3. V. Schlegel, Hermann Grassmann: sein Leben und seine Werke, Brockhaus, Leipzig, 1878.
  4. F. Engel, Grassmanns Leben, in Gesammelte mathematische und physikalische Werke, vol. III.2, Teubner, 1911.
  5. H.-J. Petsche, Hermann Graßmann: Biography, Birkhäuser, 2009.
  6. G. Schubring (ed.), Hermann Günther Graßmann (1809–1877): Visionary Mathematician, Scientist and Neohumanist Scholar, Kluwer, 1996.
  7. W. P. Lehmann, A Reader in Nineteenth-Century Historical Indo-European Linguistics, ch. 10: H. Grassmann on aspirates.

Mathematics and physics

  1. H. Grassmann, Die lineale Ausdehnungslehre, ein neuer Zweig der Mathematik, Wigand, Leipzig, 1844.
  2. H. Grassmann, Die Ausdehnungslehre: vollständig und in strenger Form bearbeitet, Berlin, 1862. English translation by L. C. Kannenberg, Extension Theory, American Mathematical Society, 2000.
  3. H. Grassmann, A New Branch of Mathematics: The Ausdehnungslehre of 1844 and Other Works, trans. L. C. Kannenberg, Open Court, 1995.
  4. H. Grassmann, Neue Theorie der Elektrodynamik, Annalen der Physik und Chemie (1845).
  5. H. Grassmann, Zur Theorie der Farbenmischung, Annalen der Physik und Chemie (1853).
  6. H. Grassmann, Lehrbuch der Arithmetik für höhere Lehranstalten, Berlin, 1861.
  7. W. K. Clifford, Applications of Grassmann’s extensive algebra, American Journal of Mathematics 1 (1878).
  8. G. Peano, Calcolo geometrico secondo l’Ausdehnungslehre di H. Grassmann, Turin, 1888.
  9. A. N. Whitehead, A Treatise on Universal Algebra, Cambridge, 1898.
  10. M. J. Crowe, A History of Vector Analysis, University of Notre Dame Press, 1967 (Dover reprint).
  11. D. Fearnley-Sander, Hermann Grassmann and the creation of linear algebra, American Mathematical Monthly 86 (1979).
  12. Grassmann’s laws (colour science) and Grassmann’s law (linguistics), Wikipedia.