Questions tagged [clifford-algebras]

Clifford algebras are associative algebras constructed from quadratic forms on vector spaces. They can be viewed as generalizations of the real numbers, complex numbers, and quaternions. These algebras have applications in geometry and theoretical physics.

Given an $F$ vector space $V$ with a quadratic form $Q:V\rightarrow F$, the Clifford algebra $C\ell(V,Q)$ is a quotient of the tensor algebra of $V$ by the ideal generated by elements of the form $v\otimes v - Q(v)\cdot 1$ for $v\in V$.

The algebras are eponymous for William Kingdon Clifford's work with them at the end of his life, although he himself called them "geometric algebras." As the name suggests, these algebras have applications in geometry.

Examples:

  1. When $Q=0$, the algebra $C\ell(V,Q)$ is the exterior algebra of $V$.

  2. When $F=\Bbb R$ and $V=\Bbb R^2$ and $Q$ has the signature $(1,-1)$, the algebra $C\ell(V,Q)\cong \Bbb C$ as $\Bbb R$ algebras.

  3. When $F=\Bbb R$ and $V=\Bbb R^4$ and $Q$ has the signature $(1,-1,-1,-1)$, the algebra $C\ell(V,Q)\cong \Bbb H$ as $\Bbb R$ algebras.

Clifford algebras over $\Bbb R$ and $\Bbb C$ have been used in recent decades for theoretical physics. Advocates of so-called geometric algebra interpret elements of the algebra geometrically as vectors and subspaces of $V$. This interpretation allows convenient execution of tasks like projection, reflection and rotation to be done with the algebra product.

Given an ordered basis $\{b_i\mid i\in 1\dots n \}$ for finite dimensional $V$, a standard basis for $C\ell(V,Q)$ is given by forming all possible products of basis elements such that the indices appearing are strictly ascending from left to right. Thus $b_1\otimes b_3\otimes b_4$ is a basis element, but $b_1\otimes b_4\otimes b_3$ is not. The empty product of basis elements is taken to be the identity of the algebra. If $V$ is $n$ dimensional, then $C\ell(V,Q)$ will be $2^n$ dimensional.

With the standard basis, the Clifford algebra has a natural grading. The grade-$k$ subspace is the subspace generated by all elements of the standard basis which are products of exactly $k$ of the $b_i$. Grade $k$ elements are called k-vectors or k-blades, and are interpreted by geometers as $k$-volume elements.

553 questions
8
votes
1 answer

Clifford Algebra Multiplication Intuition

I discovered Clifford Algebra recently and I am deeply impressed with its explanatory power and geometrical intuitiveness. I've been playing with the GAViewer Home Page: http://www.science.uva.nl/research/ias/ga/viewer/content_viewer.html Download…
slehar
  • 298
4
votes
1 answer

Injection of vector space in clifford algebra

Let $K$ be a field (of characteristic $\neq 2$ if that matters) and $V$ a finite dimensional vector space over $K$, $dim(V)=n$. Let $g$ be a symmetric bilinear form on $V$. By $Cl(V,g)$ I denote the clifford algebra associated to $V$ and…
zarder
  • 73
4
votes
1 answer

Can vectors act as Identity under Clifford Multiplication?

Let $\newcommand{\CCl}{\mathbb{C}l}\CCl_{p,q}$ be the complex Clifford-Algebra associated to the Minkowski space $\mathbb{R}^{p,q}$ of signature $(p,q)$, where we consider $\mathbb{R}^{p,q}$ as a linear subspace of $\CCl_{p,q}$ in the usual way.…
4
votes
0 answers

Definition of the twisting space of a Clifford module

Let $V$ be an even-dimensional euclidean vector space, $S$ the spinor module and $E$ a module of the Clifford algebra $C(V)$. Then $$W=\mathrm{Hom}_{C(V)}(S,E)$$ is called the twisting space of $E$. Now here is the issue: In the book Heat Kernels…
Filippo
  • 3,536
3
votes
1 answer

Why Clifford algebras do not have basis vectors squaring to i?

Clifford's algebras have basis elements squaring either to $1$, $0$ or $-1$. Why not $i$? Is that already covered by $1$, $0$ or $-1$? If there is an algebra squaring to $i$, what is his name? I have interest on the geometric interpretation of the…
Colim
  • 173
2
votes
0 answers

An expression for $e_1e_2e_3e_4$

In a real Clifford algebra $\mathbb{Cl}(2,2)$ over ${\mathbb R}^4$ with the quadratic form defined on the orthogonal basis $e_1,e_2,e_3,e_4$ by $e_1^2=1, e_2^2=1, e_3^2=-1$, and $e_4^2=-1$, find an even number of vectors $a_i$ ($i=1,2,\dots,2k$)…
Andrey Sokolov
  • 1,536
  • 10
  • 15
2
votes
1 answer

Geometric/Clifford algebra - intuition behind this derivation of a formula that recovers a rotor from a vector basis and its transformation

This short manuscript by Francisco G. Montoya presents and proves a general formula that gives the rotor (up to a sign change) that effects a certain rotation transformation, from a given vector basis and its rotated image. Although I could follow…
jvf
  • 491
2
votes
1 answer

The "geometric product" of a frame and its reciprocal frame

On p.102 of their book "Geometric Algebra for Physicists", Doran and Lasenby start an argument with "Since $e_{i}e^{i} = n$, ...". Here $e_{i}$ is an arbitrary (not necessarily orthonormal) frame in an n-dimensional inner product space of arbitrary…
Dullard
  • 23
  • 5
2
votes
1 answer

Homomorphism of Clifford algebras

Assume that $(X, \mathbb{R}), (Y, \mathbb{R})$ are a vector space over $\mathbb{R}$, together with a linear map $f : X\rightarrow Y$ and pullback quadratic form $(\hat{f}q)(x) = q(f(x))$ for every $x\in X$. Prop: There exists an algebra homomorphism…
Deane
  • 21
2
votes
1 answer

What is the center of a Clifford algebra

I have for short interess at Clifford algebras, and I've read on the german wikipedia that the center, that the set $$\{x\in CL| x\cdot y= y\cdot x \;\forall y\in CL\}$$ was $$\mathbb{R}\cdot 1_{CL}$$ But I couldn't figure out why. Does someone…
2
votes
1 answer

Clarification of example of Clifford Algebra

I just started studying Clifford algebras and I am puzzled by the following example. Let $X$ be a Hilbert space with $\mathrm{dim}\ X = 1$. Let $\{e_1\}$ be the basis for $X$. Then the Clifford algebra $\mathcal{C}(X)$ consists of all elements of…
Konrad Burnik
  • 265
  • 1
  • 9
2
votes
1 answer

Length of a Clifford Poly-vector?

The length of a vector is defined as: $$ ||\mathbf{v}||^2=\mathbf{v}\cdot \mathbf{v} $$ In the case that $\mathbf{v}:=a\hat{x}+b\hat{y}$ is expressed in an orthogonal basis using $\hat{x}$ and $\hat{y}$ as the generators of a Clifford algebra…
Anon21
  • 2,581
2
votes
0 answers

Question about Clifford algebras (reference Matrix Groups: An Introduction to Lie Group Theory, Baker)

Reference: Page 140 of the book in the title. I know this question may be difficult to answer on here, but regarding $\mathbb{R}^{n}$ as a subset of the nth Clifford algebra $Cl_{n}$ for each $u \in \Gamma_{n}$, the function $\rho_{u}:…
Tuo
  • 4,556
2
votes
1 answer

What's the argument of a quaternion?

Suppose I use the Cayley-Dickson construction on complex numbers to obtain the quaternions. I then have the "real" and the "imaginary" parts of the quaternions. The argument of a complex number $x$ is then $\mathrm{atan2}(\mathrm{Im}(x),…
2
votes
1 answer

Vectors in Clifford Algebra

I'm studying Clifford Algebra $\mathcal{Cl}_2$ and got stuck in an exercise: Let $\mathbf{a}=e_2+e_{12},\quad \mathbf{b}=(1/2)(1+e_1).$ Compute $\mathbf{ab}$. The answer is zero, but I can't get to it and I'm having trouble putting $\mathbf{b}$ in…
1
2 3 4