Geometric algebra

Related: Clifford algebra.

An Intuitive Introduction

For a more in-depth explanation, you can refer to "What is Geometric Algebra?" in the Calibre Library. In a nutshell, the exterior algebra can be constructed for any vector space V, using V itself rather than its dual V. This results in a graded algebra:

Λ(V)=Λ0(V)Λ1(V)Λ2(V)Λn(V)

The elements of this algebra can be interpreted as scalars, vectors, bivectors, trivectors, etc. A bivector uv can be understood as the plane spanned by u,v along with the area of the parallelogram and the orientation from u to v. Similar interpretations can be made for a k-vector.

In the algebra Λ(V), we have operations like addition, scalar multiplication, inner product for vectors, and wedge product for any k-vectors. All of these can be generalized by a new operation known as the geometric product. For vectors, it takes the following form:

uv=uv+uv

Geometric Algebra in 2D and Complex Numbers

When we set V=R2, it can be shown that G2=Λ(R2) naturally contains complex numbers. Given a basis {e1,e2}, we have:

e1e1=1,e2e2=1,e1e2=e2e1,(e1e2)2=1,

Therefore, we interpret e1e2 as the imaginary unit I. Furthermore, it can be shown that:

Note: Although eIθ acts on Λ1(R2) to produce a rotation of θ, which seems natural, this behavior is not generalizable to all dimensions.

Complex number has a doubled "personality": they represent vectors, on the one hand, and operations on vectors, on the other. From the Geometric Algebra viewpoint, everything is more coherent, since in G2 we have:

Geometric Algebra in 3D and Quaternions

When we take V=R3, G3=Λ(R3) contains quaternions H in a natural way. With a fixed basis {e1,e2,e3}, it turns out that e2e3,e1e3,e1e3 behave like the imaginary units i,j,k in H. Therefore:

Axiomatic approach

We can invert the entire construction process to define a geometric algebra as follows:
Definition:
A geometric algebra is a set G equipped with two composition laws—addition and multiplication (referred to as the geometric product)—that meet the following criteria:

  1. G is a (potentially non-commutative) graded ring with a unit element. The neutral elements for addition and multiplication are denoted by 0 and 1, respectively.

  2. The grade-0 elements, G0, constitute a field with characteristic 0 that includes the neutral elements 0 and 1. The elements of G0 commute under multiplication with any element in G, meaning they are contained in the center of G. Hence, G can be viewed as an associative algebra over G0.

    Elements of G0 are called scalars or 0-vectors.

  3. G includes a subset G1 that is closed under addition, and its elements are termed 1-vectors.

    The square of a 1-vector is a scalar, and if a0,

    a2=|a|20
  4. For aG1 and AkGk, we can define the inner product as

    aAk=12(aAk(1)kAka)

    and the outer product as

    aAk=12(aAk+(1)kAka)

    Consequently,

    aAk=aAk+aAk

    We assume that:

    • aAkGk1
    • aAkGk+1
  5. There exists a certain nN such that for every AnGn,

aAn=0

and

Gj

for j=0,1,,n.

Some remarks:

Geometric interpretations

Two key insights (see this video):

  1. "The geometric product contracts parallel parts and join perpendicular parts of shapes".
  2. "The geometric product of two vectors a and b represents a linear transformation bringing a to something proportional to b".

More specific interpretations

Moreover,wecanfind$e1$and$e2$withnorm1andorthogonal,and$e1$parallelto$c$suchthatthereexists$λR$:

b\wedge c=\lambda e_1\wedge e_2.

Infact,givenanyothervector$x$insidetheplane${b,c}$,youcansupportthebivectorover$x$.Thatis,thereare$e1$and$e2$ofnorm1,suchthat$e1$isparallelto$x$and$e2$isorthogonalto$x$and

b\wedge c=\lambda e_1 ' \wedge e_2 '.

WLOG,suppose$x$isofnorm1.Then$x=cos(θ)e1+sin(θ)e2$.Sotake

e_1'=x=cos(\theta)e_1+sin(\theta)e_2

and

e_2 '=-sin(\theta)e_1+cos(\theta)e_2

Itstrivialtocheckthat$e1e2=e1e2$and$e1e2=0$.Avector$x$multipliedbyabivector$ab$.First,bythepreviouspoint,wecanthinkin$ab$like$λe1e2$.Now,wediscern:1.$x$islinearcombinationofof$e1$and$e2$.Then$xab=xλe1e2$isacontractionbyafactor$λ$and90degreesturnof$x$.2.$x$isorthogonaloftheplanegeneratedby$e1$and$e2$.Wehavethat

x a\wedge b= x\wedge a \wedge b

a trivector. 3. So if $x$ is a sum of a parallel component and a orthogonal component, $x$ multiplied by $a\wedge b$ gives rise to a rotation of the projection of $x$ in $\{a,b\}$, and the creation of a trivector $x_{\perp}a\wedge b= x_{\perp}\wedge a \wedge b$. ## Angles and trigonometry In GA, bivectors represent 2-dimensional directions. Any simple bivector can be written as

\theta e_1e_2

with$θR$and${ei}$unitaryandorthogonalvectors.Theunitaryvectorsspecifythedirectionitself,and$θ$isthesize(analogoustothelengthofavector,thatisa1dimensionaldirection).Giventwovectors$a$and$b$(thatwetaketobeunitary,wlog)wesaythatthesimplebivector$θe1e2$is the angle formed by themif

ab=e^{\theta e_1 e_2}

wherewetakeasadefinition

e^A=1+A+A^2 /2+\cdots=\lim_N (1+\frac{A}{N})^N

Realanalysisletusassurethatthisexpressioniswelldefinedwhen$A$isa2blade.Infact,since$e1e2e1e2=1$,wecanwrite

e^{\theta e_1 e_2}=cos(\theta)+e_1 e_2sin(\theta)

where$cos$and$sin$aredefinedbytheirpowerseries.Observethat$b=aeθe1e2$andthen

b=a(cos(\theta)+e_1 e_2sin(\theta))=a\cdot cos(\theta)+ ae_1e_2 \cdot sin(\theta)

andfromhereweconcludethat$a$isintheplanegeneratedby$e1$and$e2$,sinceotherwisewewouldhaveatrivectorintherighthandsideoftheaboveequation.Ofcourse,$b$isalsoinsidetheplane.Thefactthat$ab=eθe1e2$canbeinterpretedasfollows.Wehavethat

b=a e^{\theta e_1 e_2}\cong a(1+\frac{\theta e_1 e_2}{N})^N=a+\frac{\theta}{N}a_{\perp}+\cdots

whosemeaningisthatwearriveto$b$from$a$byapplyingaspecialinfinite process.Weneedsuchaninfiniteprocessbecauseweareproducing a circlewithstraight lines operations.![Pastedimage20240412120246.png](/img/user/imagenes/PastedOntheotherhand,observethat

ab=a\cdot b+a\wedge b=

=\mu +\lambda e_1e_2

andissatisfiedthat$μ=cos(θ)$and$λ=sin(θ)$.Since$baab=1$wecanshowthat

cos^2(\theta)+sin^2(\theta)=1

Inthiscontext,wecandefine$π$asthelowestrealpositivenumber$θR$suchthat

e_1 e_2=e^{\frac{\theta}{2} e_1 e_2}

Now,naturalquestionsarise:could$π$bethesameforeverypairofunitaryorthogonalvectors$e1,e2$?Moreover,istheonlyone?Wecandealwiththeunicityquestionbymeansofrealanalysisbystudyingthepairofequations$$cos(θ/2)=0sin(θ/2)=1

and we will arrive to the usual conclusions.

On the other hand, suppose e1e2 are other unitary and orthogonal vectors. Then

eπ2e1e2=cos(π/2)+e1e2sin(π/2)=e1e2

So π is universal for every 2-direction.

From here we can conclude usual facts like

eπi=1

where we have called i=e1e2 to resemble the famous expression. To see it, take any unitary v spanned by e1,e2. And then, since

v=ve1e2e1e2=veπ2e1e2eπ2e1e2=veπe1e2

and therefore

eπi=1

Observe that we have used that

eAeB=eA+B

but this is true whenever AB=BA.

\bigbreak


IDEA TO FOLLOW: Could be non-simple bivectors the angle between two r-blades? This way, the bivector J leading to the complex structure of a complexificated vector space

J=e1f1+e2f2++enfn=J1+J2++Jn

is the support for angles between two prefered n-blades, in the same way that for complex numbers i=e1e2 is the support for angles.

More on this: there is a formula for the total angle between subspaces

cosθA,B=cosθ1cosθ2cosθr

and this reminds me the product of exponentials when we decompose a sum. At the same time, it reminds me the probabilty of an intersection.