数理解析研究所講究録 1378 巻 2004 年 121-137

121
Why and how geometric algebra should be taught at highschool. Experiences and proposals.
Ramon Gonzalez Calvet
I.E.S. Pere Calders
U.A.B.,
campus08193 Cerdanyola del Valles
Spain
Research as a source of new pedagogic methods in mathematics

There is a close dependence between pedagogic methods and lines of research.
In fact, there exists a continuous interchange between new advances in science and new
methods of teaching it. The present lines of research will become the new subjects toteach. I may give some examples: differential calculus, algebra, arithmetic, etc.
First example: differential calculus
Newton and Leibniz discovered differential (and integral) calculus, which was
also developed by other researchers such as L’Hdpital, Bernoulli and Euler. In the XVII
and XVIII centuries, differential calculus was an active field of research. In the XIX
century it becomes an academic subject at university level. In the XX century Newton’s
fluxions (derivatives) and Leibniz’s integrals and differentials are taught to 16-17-yearold pupils at high schools. Aside of the main relation between the derivative and theprimitive, I enumerate other theorems taught at these levels:

Chain rule for derivatives of composite ffinctions.
Barrow’s rule for the definite integral.
Rolle’s theorem.
Mean value theorem.
theorem.
L’H\^opital ’s rule to calculate indeterminate limits.
$\mathrm{C}\mathrm{a}\mathrm{u}\mathrm{c}\mathrm{h}\mathrm{y}’\mathrm{s}$Second example: linear algebra
In the field of linear algebra, let us recall that matrices were discovered by
Arthur Cayley1 although Leibniz already worked with deteminants2. Nowadays,
foundations of linear algebra are taught at high schools:

Operations with matrices
Determinants.
Solving the systems of linear equations with determinants.
Gauss’ method to solve systems and invert matrices.
ROuch6-Fr0benius theorem.
memoir on the theory of matrices (1858).
See E. J. Aiton, Leibniz, a biography, Ad $m$ Hilger Ltd. (Redcliffe Way, 1985), pp. 125-127. Leibniz, 1684.
outlined a definitive formulation of the rules of determinants in a manuscript dated in January1$\Lambda$

2$22\mathrm{n}\mathrm{d}$122Third example: Ancients’ geometry
We also have other examples. Archimedes discovered that the volumes of thecone, sphere and cylinder having the same diameter and altitude are in the ratio 1:2:3.
This aphorism was engraved in his tomb. Geometry was an active field of researchduring the Hellenistic period. Now we are teaching the formulas of the volumes to 1214-year-0ld pupils but also other geometric theorems such as:

Theorems about angles.
Theorem of the isosceles triangle.
Formulas to calculate the area of planar figures.
Formulas to calculate volumes.

Fourth example: arithmetic
The field of arithmetic has also been incorporated to the curricula of highschools. Among others we teach (12-year-0ld pupils):

Fundamental theorem of arithmetic (decomposition of a number in prime

factors).

Maximum common divisor and minimum common multiple.
Divisibility criteria.

Fractions3.
Decimal$\mathrm{f}\mathrm{f}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}^{4}r$

In algebra, nowadays we use the algebraic notation mainly developed by Vtete.
As a last example, I teach in the subject of computer science the binary numbers,
which were firstly worked by Leibniz5.
So we must expect that any new field of mathematics, now object of research, ina little time will become an academic subject, also at high school. We may remember
Ernst Haeckel’s aphorism: “ontogeny recapitulates phytogeny”. Although this statement
is nowadays considered a falsehood, we may adapt its sense and say:

The field of geometric algebra will also follow this path. Those new areas now
under the scope of research will be soon learned by our pupils also at high school. In
fact the main problem of geometric algebra is the blockade it suffers mainly atuniversity (at least in Spain, although we have the feeling that this is a general situation

all over the world). So our contributions to geometric algebra will soon become an
academic subject if we are able to overcome this blockade.
However every coin has two sides and this is also applicable to the interaction
between pedagogy and research. My experience teaching mathematics (16 years) has
3

Already used by the Egyptians. See B. L. van der Waerden, Geometry and Algebra in Ancient, 1983).
4
Simon Stevin introduced the decimal positional fiactions in De Thiende {The Dime) in 1585.
5G. W. Leibniz, Explication de l’arithmetique binaire, qui se sert des seuls caracteres 0et 1, avec desremarques sur son utility et sur ce qu’elle donne le sens des anciennes figures chinoises de fohy ,Memoir de 1’Acad. des Sciences (1703), Mathematische Schriften, vol. , p. 223, Georg Olms Verlag
Civilizations, p. 165. Springer Verlag (Berlin, Heidelb$\mathrm{r}\mathrm{g}$$\langle\langle$$\rangle\rangle$$\mathrm{I}\mathrm{I}$(Hildesheim, 1971).

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shown me that the process of teaching is a constant source of new ideas and inspiration
for those that are working in research. My lessons on geometry, analysis and linear
algebra at high school have supplied me with many new ideas to advance in the researchof geometric algebra. I may affirm that the Treatise of plane geometry throughgeometric algebra would not exist if I had not been a teacher of mathematics because of
the lack of the source of inspiration.
The title of the conference is:“How and why geometric algebra should be taught at high school. Experiences
and proposals.”
because we must accelerate the collapse of the barriers preventing the geometric algebra
ffom being reckoned an important academic subject, the high school being a first andmain ffont. At this level, the mind of our pupils is very flexible and receptive. Any seedof geometric algebra sown at this age will give very successffil ffuits whenever there
exists continuity of the teaching at university level. The main question is not thedifficulty of geometric algebra for our pupils, since any topic may be adapted to thecorresponding level. The true problem is the inertia of the departments (of the teachers
themselves), which tend to impart the same subjects that have been always taught.
However do not think that these subjects are a good sample of all the mathematics. Not
at all, but usually they are a partial and slanted view of them. So the renovation, the uptO-date reintroduction of geometry through geometric algebra will be a hard task andlikely we shall have to battle against our own colleagues.
Now I give some examples about how we may teach parts of geometric algebra
to our students. I am mainly considering 16-17-year-0ld pupils, but also you may
lengthen this range to the first courses of university. The transition ffom high school tocollege should be more gradual than the current one. In our school we intend that pupils
do not suffer a crack in mathematics during this transition and I believe that in somedegree we are achieving this purpose. However, if we introduce concepts of geometricalgebra at high school and they do not have continuity at university, what is the yield of
this effort?

Barycentric coordinates

One of the most interesting
approaches to point geometry is the useof barycentric coordinates. Every point
on a line may be written as linear
combination of two given points of thisline (figure 1):$0<\mathrm{k}<1$$\mathrm{k}<0$$\mathrm{P}$$1<\mathrm{k}$$\mathrm{Q}$$\frac{\mathrm{P}}{\mathrm{k}=0}$$\overline{\mathrm{k}=1}$$R=(1-k)P+kQ$

Figure 1
with coefficients whose addition is theunity. These coefficients are thebarycentric coordinates. In the same way, the linear combination of two lines is the
pencil of lines passing through the intersection of both lines. Every line of the pencil(figure 2) has the following general equation:

124$(1-p)(n_{1}x+n_{2}y+c)+p(n_{1}’x+n_{2}’y+c’)=0$

In the space, we may write
any plane belonging to an axial
pencil as linear combination of twoplanes of this pencil. Consider the
problem of finding the plane that
contains a line and an outer point(figure 3). If the line is given as
intersection of two planes, the best
way to solve this problem is to use
the axial pencil of this line. Example:$\mathrm{p}=1$Find the plane containing the line
$2x+3y$ $=4$$\{$

$x+y-z=2$

and the point $(3, 4,$ $-2)$ .

Take the axial pencil for this line. Every plane ofthis pencil has the equation (using barycentric
coordinates):$k(2x+3y-4)+(1-k)(x+y-z-2)=0$
but it must contain the point:$k(2\cdot 3+3\cdot 4-4)+(1-k)(3+4+2-2)=0$$\Rightarrow$$k=-1$and so the searched plane is: $y+2z=0$
In fact, a system of Cartesian coordinates hides the barycentric coordinates ofpoints (figure 4):$R$$=(x, y)=O+xOP+yOQ=(1-x -y)$ $O+xP+yQ$

That is, every point can be written as
linear combination of three non-aligned points$\{O, P,Q\}$ with barycentric coordinates always
summing the unity.
An immediate application of barycentric
coordinates is the calculus of the oriented areaof a triangle $ABC$:
$R$

$S$

$S_{ABC}=$ ; $AB\wedge BC$$= \frac{1}{2}|_{1-x_{\mathrm{C}}-y_{C}}^{1-x_{l4}-y_{A}}1-x_{B}-y_{B}$$x_{\mathrm{C}}x_{B}x_{\Lambda}$$y_{\mathrm{C}}y_{B}y_{4}|$$e_{12}$

125
which vanishes if the three points are aligned.
In the same way, the oriented volume of the tetrahedron having vertexes$D$and may be written using barycentric coordinates:$\nabla=\frac{1}{6}AB\wedge BC\wedge CD=\frac{1}{6}|_{1-x_{D}-y_{D}-z_{D}}^{1-x-y_{A}A}1-x_{C}-y_{C}-z_{C}1-x_{B}-y_{B}-z_{B}A-Z$$x_{D}x_{C}x_{B}x\iota 4$$A$, $B$, $C$$y_{D}y_{C}y_{B}y_{A}$These coordinates are very useffil in projective geometry. A projectivity is:simply defined as a linear transformation of the barycentric$\mathrm{c}\mathrm{o}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}^{6}$$h_{00}$$h_{01}$$h_{02}$$=k\{$ $h_{10}$$h_{11}$$h_{12}$$h_{20}$$h_{21}$$h_{22}$$1-x-y$$\det h_{ij}\neq 0$

$x$

$\{$

$y$

where is a variable number allowing that the transformed barycentric coordinates havea sum equal to the unity.
$k$

Dual coordinates in the dual plane
The concept of pencil of lines leads to the definition of the dual plane, whosebarycentric coordinates are those of the pencils of lines. Given a point base $\{O, P,Q\}$,. For Cartesian coordinates, theandthe dual base is formed by the lines
dual base is $\{-x-y \dagger 1=0, x=0, y=0\}$ . Every line on the plane is expressed as alinear combination of these three lines by means of barycentric coordinates. Let us
calculate the dual Cartesian coordinates of the line $2x+3y+4=0.$ We must solve theidentity:$\overline{OP}$2$x$$+3y+4\equiv a’$$\overline{PQ}$$\overline{QO}$$(-x-y+1)+b’x+c’ y$$lx,y$$x(2+a’-b’)+y(3+a’-c’)+4-a’\equiv 0$

whose solution is:$a’=4$$b’=6$$c’=7$

Dividing by the sum of coefficients we obtain:$\frac{2x+3y+4}{17}\equiv\frac{4}{17}(-x-y+1)+\frac{6}{17}x+\frac{7}{17}y$whence the dual coordinates of this line are obtained as $[b, c]=$ [6/17, 7/17]. Let us see
their meaning. The linear combination of both coordinate axes is a line of the pencil oflines passing through the origin:
6

Treatiseplane geometry through geometric algebra, p. 104.

128

or

$\frac{6}{13}x+\frac{7}{13}y=0$

6 $x+7y=0$

This line intersects the third base line $-x-y$ $+1=0$ at the point $(7,$ $-6)$ , whose pencil oflines is described by:$a(-x-y+1)+(1-a)( \frac{6}{13}x+\frac{7}{13}y)=0$

Then 2 $x+3y+4=0$ is the line of this pencil determined by $a=$ 4/17.
Three lines are concurrent if the determinant of their barycentric dual
coordinates vanishes. On the other hand, the line at the infinity has dual coordinates[1/3, 1/3], that is, it is the centroid of the dual base.
Also the equation of any conic has a very simple form using barycentic
coordinates :$(1-x-y x y)\mathrm{S}$$(\begin{array}{ll}\mathrm{l}- x-y x y\end{array}\}=0$

where is a symmetric matrix. I have shown7 that the matrix of the tangential conic (theconic plotted in the dual plane by the dual points corresponding to the tangents to thepoint conic) is equal to the inverse of the matrix of the point conic, its equation being:$\mathrm{S}^{\cdot}$$[1-a-b$$a$$b]\mathrm{S}^{-1}$$\{\begin{array}{ll}\mathrm{l}- a-b a b\end{array}\}=0$

where $[a, b]$ are Cartesian coordinates in the dual plane.

Scalar and exterior product in the plane
The metric $geomet\eta$ explained to our pupils is a skew geometry, that is, acensured geometry: only those problems involving scalar product are considered.
However the scalar and exterior products arise in a very symmetric form in geometry.
We may and must teach to our pupils both products:$v$$\cdot$$w$ $=v_{X}w_{X}+v_{y}w_{y}$$v\wedge w=(v_{x}w_{y}-v_{y}w_{X})e_{12}$Then the geometric product may be introduced with the help of complex
numbers :$vw=v\cdot w+v\wedge w$

7

Treatiseplane geometry through geometric algebra, p. 133.

127
A light version may be to use a real, instead of imaginary, exterior product.
Anyway it allows pupils to calculate areas of triangjes without having their altitudes.

Plane trigonometry and the hyperbolic plane
Many years ago I had prepared this table where the identities of the circular andhyperbolic ffinctions are compared. But then I could not imagine in which extent thegeometric algebra develops the analogy between circular and hyperbolic trigonometry.
$6$$\mathrm{R}$$\mathrm{r}\epsilon$$\mathrm{R}$$\mathrm{e}\downarrow j\wedge \mathrm{a}$$\mathrm{d}\mathrm{P}\epsilon \mathrm{R}8b\mathrm{L}$$\alpha$$\ovalbox{\tt\small REJECT}\wedge\kappa^{\iota}\approx+\alpha \mathrm{Y}^{l}-$A1$\epsilon 4^{\iota_{\mathrm{r}}}\Rightarrow*$$4\sim$8$+_{\mathrm{A}1}$$*n-a\backslash x$$\mathrm{e}\mathrm{A}\mathrm{x}$$\mathrm{J}\mathrm{k}$$\mathrm{r}\nu_{1}$r1$n\mathrm{n}\mathrm{z}$$\mathrm{t}_{4}\mathrm{r}**$$u\backslash \mathrm{z}*\mathfrak{n}\cdot \mathrm{k}$$\Omega^{\iota_{3}}$$\sim*\iota_{l}$$‘\prime \mathrm{r}.)\alpha*$$\star 4\mathrm{z}$$\mathrm{k}\mathrm{t}*$$\pi$$\theta\wedge$)$A_{\mathrm{t}1*}\backslash \grave,$4$g$$\mathrm{s}\mathrm{v}\iota\tau\tau$$\prime r[]$$\mathrm{A}\wedge\iota$$u\backslash *$$\approx$$.\{\tau_{t}$Afk$\wedge \mathrm{S}\backslash$$\omega_{\mathrm{I}}\iota u\ddagger*$$\mathrm{J}\sim \mathrm{h}^{\mathrm{t}}$$\mapsto^{\{}-*\mathrm{B}^{\backslash }+$$\mathrm{b}_{\delta^{\mathrm{c}^{\mathcal{R}\mathrm{F}}?}}$,}

4\sim $b.\epsilon I,

Qx$r\mathrm{p}$$4\mathrm{k}_{\lambda*}$

$A$

+k

$\mathrm{f}\kappa\iota_{\}}$\vee kx$\vee \mathrm{A}_{\lambda}\mathrm{A}_{\wedge+}$$\mathrm{s}\ltimes*$$\mathrm{w}$$+^{4\mathrm{k}_{\lambda}4\mathrm{k}}\mathrm{r}$$\epsilon \mathrm{r}$$\mathrm{D}\mathrm{o}\mathrm{b}$2$2*\ltimes \mathrm{x}$$2\mathrm{r}$$\epsilon \mathrm{h}$$1\mathrm{O}\mathrm{L}$$t’|.\sim\llcorner$$\iota\tau’ 2*$.

$\sim*_{b}$$\mathrm{x}$A$\downarrow\backslash \sim$$\mathrm{b}^{\mathrm{x}\mathrm{h}\vee}$–L lnhk$1_{\nu_{\iota}}\iota_{\lambda}$$<\mathrm{L}*$$*$$e_{\wedge}\mathrm{v}\iota \mathrm{X}$$\mathrm{x}$$u\backslash$$A_{\wedge}$$*\mathrm{n}\mathrm{b}$$\mathrm{w}$$\mathrm{A}^{1}\sim.\iota\epsilon.\iota_{1}$$\mathrm{L}\mathrm{X}$$\mathrm{k}l_{\mathrm{X}\sim}\mathrm{Z}kl$$*$$\mathrm{b}_{\grave{C}}f\mathrm{x}$$\frac{\mathrm{z}}{\triangleleft}\mathrm{h}_{\frac{\mathrm{z}}{f}}\mathrm{h}_{e}$$\mathrm{e}\neq\ltimes’ \mathrm{x}$\kappa$$+$\mathrm{e}\mathrm{L}\mathrm{t}$

$*$

.

1TA

$\mathrm{H}$$\mathrm{t}\mathrm{t}\mathrm{i}\theta\llcorner T\zeta$$\mathrm{t}$$\nu r\mathrm{f}^{rr}\sim‘\iota\kappa\tau$$\sim \mathrm{t}$$n\mathrm{r}\backslash \mathrm{z}IC$$\wedge^{4}\mathrm{X}\mathrm{B}$$|$A$\pi^{\overline{A*u\tau}\mathrm{x}}$$,\backslash \wedge$..

$\mathrm{t}$$\underline{*\iota}$$\mathrm{t}$$*\overline{A\mathit{1}1\mathrm{r}\backslash }\lambda$$\mathrm{u}\backslash$$.\hat{\}^{l}’\approx$$\mathrm{J}\mathrm{x}r*\mathrm{M}\mathrm{t}$$-\iota$

A\sim -$\Lambda.\sim$-$\{\mathrm{k}$ゞ$k,$d\backslash \sim-\epsilon[searrow]\backslash$$d_{\wedge}\kappa\taubackslash \backslash$$\backslash d*xA\cdot*t$$\tilde{\tau}-$$*\wedge \mathrm{i}($$\backslash$$l$”$4\vee$$e_{\mathrm{J}\mathrm{A}}$$\downarrow\ _{\vee}^{\mathrm{Y}}$*$L$${}_{1}P\lambda*$$\wedge \mathrm{A}\kappa$$\mathrm{f}\mathrm{f}\mathrm{i}^{\mathrm{I}}$$t$; fQtJ$\iota_{\grave{n}\aleph}\mathrm{r}_{\mathrm{I}}\neq\vee\succ$$\eta\{\triangleleft_{\mathrm{B}}$$h\iota^{1}r_{\backslash }’\mathrm{L}$$\prime \mathrm{r}\wedge\dot{\zeta}$

1$\ell r\not\supset \mathrm{E}4\sim$$\wedge*‘ \mathrm{r}’ \mathrm{x}$$qq_{\mathrm{h}}l_{d}$

ffi

},

$\epsilon \mathrm{h}\iota$&q$\sim$$\alpha$$‘ W_{L}$$\{\Lambda \mathrm{e})$$\vee^{\mathrm{t}}$$\dot{\prime}$,*$\mathrm{x}_{\mathrm{L}}\vee^{\mathrm{t}}$$\mathrm{A}_{\mathrm{X}}$$\tau_{-ln\backslash ^{\mathrm{Y}44}’ \mathrm{v}[]\backslash }\sim\sim \mathrm{r}$$m-”*\mathrm{s}^{\mathrm{t}}\gamma_{-\prime.\backslash }\mathrm{r}4.4,\mathrm{v}[]\backslash \tilde{\tau}\sim \mathrm{r}$\approx$\mathrm{n}$$\iota$’ $5$$\mathrm{A}_{\tau}^{-\prime}$.

A.

$\mathrm{t}\check{r}\mathrm{c}\dot{\mathrm{A}}$$\tau\sim$dA$+\mathrm{h}$$\mathrm{A}\backslash _{r}$$\lambda--\simeq\#\dot{\mathrm{A}}^{\mathrm{g}}.\succ-$$\mathrm{x}_{\mathrm{L}}-\simeq\#\dot{\mathrm{A}}^{\mathrm{g}}.\succ\sim_{\mathrm{t}}$$\mathrm{L}$$\mathrm{t}$$\mathrm{a}-|$$\sigma \mathrm{k}\backslash$$A_{\tau \mathrm{Y}}$$A_{\tau\mathrm{Y}*}$$\hat{\prime}\wp$$\hat{\prime}\wp \mathrm{z}\sim\triangleright\iota \mathfrak{n}*$$\mathrm{z}\sim\triangleright\tau_{-}\iota \mathfrak{n}$$y\backslash \mathrm{h}\ddagger_{*\mathrm{b}_{\xi^{\mathrm{h}}\mathrm{f}}}\iota\kappa$$\sigma^{\mathrm{t}}\mathrm{c}$$\mathrm{L}$$\backslash$$\mathrm{L}$$\mathrm{t}$$\mathrm{Y}$$*\backslash _{\eta=}\mathrm{s}$$\mathrm{x}4\mathrm{t}$,$u$$\mathrm{A}_{*}$$\mathrm{x}\mathrm{z}4^{(}\backslash$,2$\iota \mathrm{k}$$\mathrm{z}\vee\tau_{\mathrm{t}}6\mathrm{y}_{-t}^{\mathrm{g}-\vee\succ}$$e\backslash _{4}$$JLZ\_{l}$’$\mathrm{A}\sim \mathrm{A}\backslash \}$$\mathrm{I}|$

gure

128
In the same way as complex numbers are those naturally associated with circulartrigonometry, hyperbolic numbers are those naturally associated with hyperbolic

trigonometry :$z=a+be_{12}$$z=a+be_{1}$$e_{12}^{2}=-1$$e_{1}^{2}=1$

So we have the analogous of Euler’s and De Moivre’s identities:$\exp(xe_{12})=\cos x+e_{12}\sin x$$(\cos x+e_{12}\sin x)^{n}=\cos$ n\kappa \dagger $e_{12}\sin$ n\kappa$\exp(xe_{1})=\cosh x$ \dagger$e_{1}$

siffi$\mathrm{x}$$(\cosh x+e_{1}\sinh x)^{n}=\cosh nx+e_{1}$ $\sinh$nx

A Euclidean angle is defined as quotient of the arc length divided by the radiusof the circle (figure 6):$a= \frac{s}{r}=\frac{2A}{r^{2}}$and it is proportional to the area of the circularsector.

In the same way a hyperbolic angle isdefined as quotient of the arc length of theequilateral hyperbola $x^{2}-y^{2}=r^{2}$ divided by its
radius in the pseudO-Euclidean plane (figure 7):$r$$\psi$$= \frac{s}{r}=\frac{2A}{r^{2}}$Of course, we cannot represent hyperbolicarguments with circle arcs as made in

Euclidean triangles. So a new sketch of
hyperbolic arguments by means of arcs ofhyperbola is needed. In figure 8 we see atypical hyperbolic triangle suitably drawn
to show that the addition of the three; and figure 9 shows theangles isrelation between arguments in different$-\pi$ $e_{12}$qua 計 an 区

Figure 6

128
Also the laws of sines, cosines and tangents for a generic Euclidean triangle(figure 10):$|4$$==\underline{|b|}\underline{|c|}$

sina

sin7

$\sin\beta$$a^{2}=b^{2}+c^{2}$$-2$$|/\mathrm{y}||c|$$\mathrm{c}o\mathrm{s}a$$\frac{|a|+|b|}{|a|-|b|}=\frac{\tan\frac{a+\beta}{2}}{\tan\frac{a-\beta}{2}}$Figure 10
have their analogous laws for hyperbolic triangles:$\underline{|a|}\underline{|b|}\underline{|c|}==$$\sinh a$$\sinh\beta$

siffi

$r$

$a^{2}=b^{2}+c^{2}-2$ $|b||cl|\cosh a$$\frac{|a|+|b|}{|a|-|b|}=\frac{\tanh\frac}{\tanh\frac{a+\beta a-\beta 2}{2}}$Some examples and tests showed me that they work with the suitable orientationof the hyperbolic arguments displayed in figure 9.
We also explain to our pupils the power of a point with respect to a circle, whichis the value obtained when the coordinates of the point are substituted in the equation of
the circle.
$PRPR’=PSPS’=PO^{2}-OS^{2}=(x-x_{0})^{2}+$ $(y-)$$)^{2}-r^{2}$
$0$

130

In the hyperbolic plane, the power of a point with respect to a hyperbola withconstant radius is also constant. The proof is as follows: the yellow and blue hyperbolic
triangles in figure 12 are similar because the hyperbolic argument (in the hyperbolic
plane) is constant and equal to a half of the arc length $R’S$ of the hyperbola. Opposite
similarity may be written using geometric product:$\psi$’$PR(PS’)^{-1}=(PR’)^{-1}$PSwhich implies:$PR’PR=PSPS’=PO^{2}-OS^{2}=(x-x_{0})^{2}-(y-y_{0})^{2}$-$r^{2}$

Rank of a matrix, exterior product and systems of equations
We explain to our pupils that the rank of a matrix is the number of linearly
independent rows or columns. In the method that uses determinants to find the rank, wetake an element of the matrix and we add rows and columns always taking that matrix
having non-null deteminant:$\mathrm{M}=(\begin{array}{llll}\mathrm{l} \mathrm{l} 2 33 \mathrm{l} 3 2-3 \mathrm{l} 0 5\end{array}\}$$\iota$“Je puis demonstrer que ce que.

$\mathrm{v}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{e}\mathrm{n}\mathrm{v}\mathrm{o}\mathrm{y}6\mathrm{u}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{a}\mathrm{y}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{f}_{1}\mathrm{g}\mathrm{u}\mathrm{r}\mathrm{e}$$\mathrm{j}’ \mathrm{a}\mathrm{y}\mathrm{a}\mathrm{v}\mathrm{a}\mathrm{n}\mathrm{c}6$suit de ma caracteristique lineare ou geometrique dont jepure ml’Aierlgemebernet pauitis$\mathrm{n}\mathrm{e}\mathrm{f}\mathrm{j}\mathrm{e}$$\mathrm{c}\mathrm{e}\mathrm{q}\mathrm{C}\mathrm{a}\mathrm{r}$(.ul des lignes
c’est a dire que ce$\mathrm{j}\mathrm{a}\mathrm{m}\mathrm{a}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{q}\mathrm{u}\mathrm{e}S\mathrm{e}\mathrm{x}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{r}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{a}+y\mathrm{r}\mathrm{c}\mathrm{e}$$\mathrm{a}\mathrm{e}\mathrm{q}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{c}$$\mathrm{t}\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{a}\mathrm{m}\mathrm{r}\mathrm{e}\mathrm{o}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{l}’ \mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{u}$cercle, il faut expliquer par la figure ce que c’est que ce et ,droites,sontdont Tune est perpendiculaire A Tautre et Tune commence par le centre, Pautre par la circonference de la
figure).” (letter to Christian Huygens without date but likely written by Leibniz in December 1680) Der
Briefivechsel von Gottfried Wilhelm Leibniz mit Mathematikern vol.VII, p. 580. Herausgegeben von C. I.
Gerhardt, George Olms Verlagsbuchhandlung (Hildesheim’ 1962).
$x$

$y$

131
11
31
-3 1

$3-$
$1\neq 0arrow|\begin{array}{ll}\mathrm{l} \mathrm{l}3 \mathrm{l}\end{array}|$$=-2\neq 0arrow$$-331$$111$$\tau 1-$

$2|20$$2|\begin{array}{lllll}=0arrow 3 1 2 ’ 1 \sigma\end{array}|l-\mathrm{Y}=0$,$\Rightarrow$rank$\mathrm{M}=2$

If all the determinants obtained adding one row and one column to a non-null ndimensional determinant are zero then all the other $n+1$ dimensional determinants are
null. This ensures that the rank of the matrix is equal to .
In the alternative method using exterior product we have:$n$($e_{1}$\dagger $e_{2}+$2e3$+$ $3e_{4}$) ($3e_{1}+e_{2}+$ 3e3$xs$$+2e_{4}$)$=$$=-2e_{1}\wedge e_{2}-3e_{1}\wedge e_{3}-7e_{1}\wedge e_{4}+e_{2}\wedge e_{3}-e_{2}\wedge e_{4}-5e_{3}\wedge e_{4}$The coefficients of the exterior product are the corresponding minors of the matrix.
Now we take the exterior product with the third vector:$(e_{1}+e_{2}+2e_{3}+3e_{4})$ $\wedge(3e_{1}+e_{2}+3e_{3}+2e_{4})\wedge(-3e_{1}+e_{2}+5e_{4})=0$

So we conclude that the rank of the matrix is 2.
Note that the determinants of a given order , which are the components of theexterior product of vectors are not linearly independent although they are orthogonal
ffom the point of view of the Pythagorean theorem. I solely wish to manifest that thefollowing words are not synonyms in geometric algebra: linearly independent
orthogonal perpendicular. Further explanations can be provided if required.
Another application of the exterior product is the resolution of systems ofequations.
$\mathrm{M}$

$n$

$n$

$\neq$$\neq$$a_{11}x_{1}.\cdot.+$$+a_{1n}x_{n}.\cdot.=b_{1}$$a_{n1}x_{1}+$$+a_{nn}x_{n}=b_{n}$$\{$which may be written as a vectorial equality:$x_{1}v_{1}+\cdots+x_{n}v_{n}=b$

where

$1$

$v_{i}=$$a_{ni}a_{1i}.\cdot.]$

and $b=$$1$$b_{n}b_{1}.\cdot.]$. In order to solve it, we take the exterior product withall the other vectors:$x_{i}v_{1}$$\wedge$$v_{2}$$\wedge\cdot.$.

$\wedge$$v_{n}=v_{\mathfrak{l}}\wedge\cdot$

whence it follows Cramer’s rule:..

$\wedge$$v_{j-1}\wedge b$$\wedge$$\wedge\cdot.$$v_{i+\mathrm{t}}$.

$\wedge$$v_{n}$

132$x_{i}= \frac{v_{1}\wedge\cdots\wedge \mathcal{V}_{j-1}\wedge b\wedge v_{i*1}\wedge\cdots\wedge \mathcal{V}_{\hslash}}{\mathcal{V}_{1}\wedge \mathcal{V}_{2}\wedge\cdots \mathcal{V}_{n}}=\frac{\det(_{\mathcal{V}_{1}\cdots \mathcal{V}_{j-1}}bv_{i+1}\cdots v_{n})}{\det(_{\mathcal{V}_{1}\mathcal{V}_{2}\cdots \mathcal{V}_{n}})}\wedge$

Change of coordinates

One set of coordinates may be easily changed to another set of coordinates withthe exterior product. For example, let us see the change ffom Cartesian to sphericalcoordinates :$x$$=r\sin\theta\cos\varphi$$z=r\cos\theta$$y=r\sin\theta\sin\varphi$$dx=\sin\theta\cos\varphi dr+r\cos\theta\cos\varphi d\theta-r\sin\theta\sin\varphi d\varphi$$dy=\sin\theta\sin\varphi dr+r\cos\theta\sin\varphi d\theta+r\sin\theta\cos\varphi d\varphi$-rsin$\thetad\theta$
d\theta$$dz=\cos\theta$$dz=\cos\theta $dr-r\mathrm{s}i\mathrm{n}\theta
dr$When taking the exterior product of the three differentials we obtain the volume

element:$dV=\$$\wedge\phi$$\wedge dz=r^{2}\sin\theta
dz=r^{2}$ $\mathrm{s}i\mathrm{n}fl$$\wedge
dr\wedge
d\theta\wedge$dr\wedge
d\theta\wedged\varphi$
d\varphi$

But we can apply it to any geometric element such as the sufface
surface element:=rsin2$\theta dr\wedged\varphi+r^{2}\sin\theta\cos\theta
d\theta\wedge d\varphi$
d\varphi$$dx\wedge\phi=r\sin^{2}\theta dr\wedged\varphi+r^{2}\sin\theta\cos\theta
d\theta\wedge$dx\wedge\phi$Osintp$dr\wedge
d\theta+r^{2}$$dy\wedge
dr\wedge
d\theta+r^{2}$$dy\wedge dz=-r\sin\varphi
dz=$$dz\wedge\$ $=r\cos\varphi dr\wedge d\theta+r^{2}$

sin2

sin2

tl

$\wedge dr$dr$$\cos\varphi d\varphi$$\theta\cos\varphi
d\varphi\wedge$\theta\cos\varphi d\theta\wedge
d\theta\wedge d\varphi+r\sin\theta\cos\theta\cos\varphi
d\varphi+r\sin\theta\cos$$\theta\sin\varphi d\theta\wedge d\varphi+r\sin\theta\cos\theta\sin\varphi d\varphi\wedge dr$$+\phi$ $\wedge dz+\$ $\wedge\$$dA=\$ $\wedge\phi+\phi$$\wedge\phi$This is a vectorial equation whence one deduces the modulus of the element of area:dA$\wedge\phi|^{2}+|$$\phi$ $\wedge dz|^{2}+|$$\ \wedge\phi$$|^{2}=$$|^{2}+|\phi\wedge
dz|^{2}+|dz\wedge\$$dz\wedge\ |^{2}=$$|dA$ $|2$|2=|$=|\$
$|$

sin2

$\theta|d\theta\wedged\varphi|^{2}+r^{2}$$d\varphi|^{2}+r^{2}$$=r^{2}|dr\wedge
d\theta|^{2}+r^{4}$$=r^{2}|dr\wedge d\theta|^{2}+r^{4}\sin2\theta|d\theta\wedge$

sin2$\theta|d\varphi\wedge dr|^{2}$

So we have the differential of area in spherical coordinates:$dA=rdr\wedge d\theta$ \dagger $r^{2}\sin\theta d\theta\wedge d\varphi+r\sin\theta d\varphi\wedge dr$

Rotations and symmetries$\mathrm{f}\mathrm{o}\mathrm{m}$:The axial symmetry (or reflection) shown in figure 13 may be expressed in the133$PR’=v^{-1}PRv$

where is the direction vector of the axis of symmetry. This is easily proved because
the perpendicular component anticommutes with the vector :
$v$

$v$

$v^{-1}(PR_{[perp]}+PR_{1!})v=v^{-1}v(-PR_{[perp]}+PR_{!\mathrm{I}})=-PR_{[perp]}+PR_{1!}$

If the point belongs to the line, the vector $PR$ and the direction vector
proportional and commute:$R$$PRv=vPR$$\Leftrightarrow$$PRv-v$ $PR=0$$PRv-vPR=0$$\Leftrightarrow$

The last equation is the algebraic equationof a line and shows that vector $PR$ remains
invariant under a reflection in the direction
onlyof the line. That is, the point
belongs to the line when it coincides withthe point reflected in this line. Separating
components we have:$PR\wedge v=0$$\Leftrightarrow$

$v$

are

$PR=v^{-1}$ $PR^{\cdot}v$$PR=v^{-\iota}PRv$

Figure 13$\mathrm{R}’$

$r$

$R$

$\frac{x-x_{P}}{v_{1}}=\frac{y-y_{p}}{v_{2}}$Also, a rotation of angle may be
expressed in a way analogous to axial
symmetries :$a$($v’=v(\cos a+e_{12}\sin a)=$ $\cos$ a/2 $-e_{12}\sin$a/2) ($v$$\cos$

a/2 $+e_{12}\sin a/2$ )The first form is only valid on aplane,
form is general for any
a plane, while the second fom$a/2$
dimension. If we take any complex number with argument /2 we may write:$a$$v’=z^{-1}vz$$z=|z|_{a/2}$

Now we may easily prove whichtransformation is the composition of two
reflections in the directions and :
$v$

$w$

$l’=v^{-1}tv$$t”=w^{-1}t’w=w^{-1}v^{\sim 1}tvw=z^{-1}vz$$t’=w^{-1}t’w=w^{-1}v^{\sim}tvw=z^{-\mathrm{l}}vz1$$z=vw$The product of two vectors is a complex
number and therefore the composition of
two reflections is a rotation over double the angle between both directions (figure 14). If

134
the directions are parallel, the rotation has center at the infinity and becomes atranslation along a distance double of that between the axes of symmetyy.
Of course, the composition of two rotations on the plane is another rotation overan angle equal to the addition of both angles as follows ffom the product of complex
numbers in polar:$\mathrm{f}\mathrm{o}\mathrm{m}$$zl=|z|$ , $|l|,$ $=|z|$ $|t|_{a+\beta}$

In the Euclidean space, each rotation is described in the following way:$v’=$ $(\cos a\mathit{1}2-u\sin a\mathit{1}2)v$($\cos$

a/2 $+u\sin$a/2)

where is a unitary bivector. In general:$u$$v’=q^{-1}vq$

where is any quaternion whose plane is the plane of rotation and whose argument ishalf of the rotation angle. This expression is applied to any element of the algebra ofth -dimensional space such as scalars, vectors, bivectors or volumes. However intechnological applications the use of only bivectors and quaternions is preferred.
The composition of two rotations in any planes is obtained through the product$q$$\mathrm{e}\mathrm{e}$of quaternions:$v’=r^{-1}q^{-1}vqr$

This is the best way to describe rotations and composition of rotations with immediatetechnological interest, such as engineers have discovered long ago.
However there is a difficulty with mirror reflections, which cannot be written inthis way. This has a physical consequence: an asymmetric molecule cannot be
converted into its mirror image, as Louis Pasteur showed in the case of tartaric acid9.
Notable points of a triangle
The conditions of intersection of the medians, bisectors of the sides, the angle
bisectors and the altitudes lead to geometric equations for the notable points of atriangle $PQR$ whose solutions are:$\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}1\mathrm{y}^{1\mathit{0}}$(centroid)$G= \frac{P+Q+R}{3}$$\mathit{0}=-(P^{2}QR$ $+Q^{2}RP+R^{2}$

PQ)$I= \frac{P|QR|+Q|RP|+R|PQ|}{|QR|+|RP|+|PQ|}$’$(2 PQ\wedge QR)^{-1}$(circumcenter)(incenter)

L. Pasteur, Researches on the Molecular Asymmetry of Natural Organic Products (1860), p. 24.
Published by the Alambic Club (Edinburgh, 1905).
10Treatise ofplane geometry through geometric algebra, chap. 8.

135$H=(PP\cdot QR+QQ\cdot RP+RR\cdot PQ)(QR\wedge RP)^{-1}$(orthocenter)$N=(PQRP+QRPQ+RPQR)(4PQ\wedge QR)^{-1}$(center of the nine-

point circle)
Observe that the points lying on the Euler line $(O, H, N)$ have expressions
implying the geometric (Clifford) product (it is not needed for the centroid but wemay also add a factor and a divisor $(PQ\wedge QR)^{-1})$, expressions which cannot be written
using only the scalar product of the sO-called metric geometry. This clearly shows theimportance of the geometric product in geometry and the partial vision and censure withwhich geometry is nowadays taught.
$G$

Some
comments
Lobachevskian geometry

on

the

Also Lobachevsky’s geometry may beeasily explained to our pupils using geometricalgebra. The twO-sheeted hyperboloid withconstant radius in the pseudO-Euclidean spacerealizes Lobachevsky’s geometry:$z^{2}-x^{2}-y^{2}=1$

Starting ffom this point we may deduce
expressions of the change of coordinates forany projection.
Let us see, for instance, thestereographic projection or Poincare’s disk
with more detail. This conformal projection
whose point of view is the pole of one sheet isdisplayed in figures 15 and 16:$\frac{x}{u}=z+1$

Figure 16$\frac{y}{v}=z+1$

where , are the coordinates on the plane ofprojection. The arc length is:
$u$

$v$

$d_{S=} \sqrt{\ ^{2}+\phi^{2}-\ ^{2}}=\frac{2\sqrt{du^{2}+dv^{2}}}{1-u^{2}-v^{2}}$The exterior product allows us to calculate quickly the differential of area:$dA=$$( \ \wedge\phi)^{2}-(\phi\wedge dz)^{2}-(\ \wedge\ )^{2}= \frac{4du\wedge dv}{(1-u^{2}-v^{2})^{2}}$138
Also with the help of exterior product we may deduce the azimuthal equivalentprojection, which preserves areas, by imposing that the modulus of the area differentials
with Cartesian coordinates and in the projection must be equal:$d4^{2}= \frac{1}{z^{2}}$$(\ \wedge\phi)^{2}=(du\wedge dv)^{2}$$u=xf(z)$$v=yf(z)$$u=x\ulcorner$$v=y\sqrt{\frac{2}{z+1}}$

Other projections of the hyperboloid are
the central projection (Beltrami’s disk), whose
point of view is the origin of coordinates, thecylindrical equidistant projection, which uses
Weierstrass’ coordinates (figure 17) analogousof spherical coordinates:$x=\sinh\psi\cos\varphi$$y=\sinh\psi\sin\varphi$$z=\cosh\psi$$ds^{2}=d\psi^{2}+\sinh^{2}\psi d\varphi^{2}$$dA=\sinh\psi d\psi\wedge d\varphi$

the

.fllf

$\cdot$; OF$\mathrm{S}\mathrm{i}’ \mathrm{l}\mathrm{S}\mathrm{f}$$\iota’\iota$

,.SN[

’

$\subset\cdot.\dagger..‘’.\backslash 1*_{J}^{\backslash }\cdot\ddagger.\# 1$$.\iota\cdot\iota\iota t\mathrm{s}\mathrm{c}’ \mathrm{t}\cdot C\grave{*}lJ\mathrm{f}\mathrm{p}.j\mathrm{f}\mathit{1}.411^{\backslash }.\cdot \mathrm{f}\dagger 11\mathrm{C}^{\backslash },\backslash \mathrm{t}\mathrm{t},*\cdot\cdot t\mathrm{t}\}\mathrm{t}\mathrm{t}*$.
”$\hslasharrow\gamma\nu_{\wedge\backslash }^{-_{\mathcal{F}}\hat{\dot{\gamma}}^{\mu_{\hslash}}}*.d^{\sqrt}\vee\acute{.,}\cdot.\nearrow$cylindrical conformal projection(analogous of Mercator’s projection) andthe conic equidistant and conformal
projections11\neg For all of them the exterior
product allows us to calculate the area
differential.
The Treatise of plane geometry
through geometric algebra$\wedge.\cdot\backslash ji^{\tilde{\circ}\iota^{_{*}}}\hslash\grave{\check{i}}^{\grave{\mathrm{Y}}^{l\acute{i}_{\sim}^{\backslash }}}\cdot\cdot’\grave{.}...-,.\cdot\cdot$$\mathrm{r}_{\wedge}^{\wedge}\vee’-\nearrow\backslash ,-\cdot-\backslash _{\backslash }$$\backslash$$\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{A}\mathrm{s}\mathrm{a}$allgebforar,$\mathrm{p}\mathrm{e}\mathrm{d}\mathrm{a}\mathrm{g}\mathrm{o}\mathrm{g}\mathrm{i}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}$$\mathrm{y}\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{h}\mathrm{e}$have available the Treatise of planegeometry through $geome\sigma ic$ algebra.
This book has four parts:
Romon$1_{\grave{j}}^{\cdot}\ell’ \mathrm{n}ri\mathrm{k}.\mathrm{z}\mathrm{f}’\mathrm{a}1,.\mathrm{e}\mathrm{S}$

1. The vector plane and complex
numbers
2. Geometry of the Euclidean plane.
3. PseudO-Euclidean geometry.
11

Treatiseplane geometry through geometric algebra, p. 203.

137
4. Plane projections of three-dimensional spaces.

Thank you very much for listening to me and I will answer your questions as
best as I can.