Geometric algebras are Clifford algebras over the real numbers. They are applied in geometry and theoretical physics.

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Coordinate free Geometric Algebra vs. Linear Algebra

I think I know what coordinate free means. But I never found in ANY text a good explanation of it or something like: This is the problem solved with coordinates and this is the problem solved without ...
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53 views

Different answers to geometric product problem

I started looking into Clifford Algebras, but I think I am conceptually missing some points. Here is how I understand the geometric product now (leaving out coordinate system independence and the ...
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80 views

What is a multivector?

I know how to visually interpret single parts of a multivector. But what do they look like as a whole? Making an analogy with complex numbers doesn't work.
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Why is the pseudoscalar called pseudoscalar in Geometric Algebra

It makes sense to call it a pseudoscalar in odd dimensions, because it commutes with all other objects. But in even dimensions it anticommutes, why is it still called pseudoscalar? Further I don't ...
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50 views

Anisotropic scaling in geometric/Clifford algebra

Take the geometric algebra over $\Bbb R^n$. Suppose we have a blade multivector in this algebra. Now we want to anisotropically scale this multivector. Is there a general closed-form expression for ...
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71 views

Most effective way to calculate the products of Geometric Algebra

I'm somewhat confused about the different products in GA. It appears to me that the most general and most effective way is just to calculate everything with the geometric product, and then apply the ...
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Geometric product of two bivectors

First the definition of the inner and outer product in terms of grade projection: $A_{r} \cdot B_{s} = \langle A_{r}B_{s} \rangle_{|r-s|}$ $A_{r} \wedge B_{s} = \langle A_{r}B_{s} \rangle_{r+s}$ So ...
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Transformation of a vector via geometric algebra [closed]

$\begin{pmatrix} 1 & 3 & 2 \\ 0 & 2 & 5 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} $ Let's take this ...
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53 views

Book(s) on Algebras (Quaternions)?

Well, lately I've been looking for a book on quaternions but I've realized that quaternions are a particular case of the named Algebras(I think Geometric Algebra). Since here, I've found all kind of ...
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53 views

Decomposition into simple bivectors

According to Wikipedia, any element of $\wedge^2\Bbb R^n$ should be decomposable into $n/2$ simple bivectors for $n$ even or $(n-1)/2$ for $n$ odd. How do I count that? How do I check that ...
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67 views

Geometric product of a bivector and a vector anticommute

I want to prove that $\left(e_1\wedge e_2\right)e_1 = -e_1 \left(e_1\wedge e_2\right)$. I did this in two ways, but I didn't understand why the second way is wrong. First way: $\left(e_1\wedge ...
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61 views

Rotations via Rotors

I am reading this PDF and have a question regarding an example which is given page $9$ with regard to using rotors to perform rotation. I made screenshots for reference: As the text suggest we ...
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73 views

Gradient in geometric calculus

In geometric calculus I see that we can unify the three fundamental derivatives from vector calculus; the gradient, the curl, and the diverge; into one operator. However, $\nabla$ is defined for any ...
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63 views

Visualizing the geometric product?

The exterior product between blades has a relatively clear geometric interpretation: it gives the result of "extending" one factor along the other, with the direction pointing along the first factor ...
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1answer
45 views

Exponents with bivectors

According to the Wikipedia page on bivectors: ...if $B$ is a bivector, then the rotor $R$ is $e^{B/2}$ and rotations are generated [by] $v'=RvR^{-1}$. But how do you take an exponent between a ...
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Basis-Independent Definition of Gradient of Multivector Fields

The definition of the gradient operator on multivector fields in geometric calculus seems to be $$\nabla = \sum_i e_i\partial_i$$ where $\{e_i\}$ is an orthonormal basis. That's useful, but it's ...
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44 views

Why are “innermorphisms” not useful?

I commonly studied type of linear function in geometric algebra is the outermorphism. For reference, here's Wikipedia's definition: Let $f$ be an $\Bbb R$-linear map from $V$ to $W$. The ...
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120 views

Is there a nice meaning to the geometric triple product?

Using geometric algebra, I can easily find the geometric tripleproduct of three vectors $a,b,c \in \mathbb{R}^3$ to be $$abc = a \left(b \cdot c \right) - b \left( c \cdot a \right) + c \left( a ...
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How are these definitions of the inertia tensor the same?

I'm looking for some help in understanding the inertia tensor (not the physics, just the math). I'm trying to figure out how to convert between the wedge product and tensor product definitions. ...
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Homotopic properties of the Spin group from geometric algebra

There are two possible ways to define the $\mathrm{Spin}(n)$ group of Euclidean $n$-space from $\mathrm{Pin}(n)$. First is that $\mathrm{Spin}(n)$ is the identity component of $\mathrm{Pin}(n)$. ...
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62 views

Clifford algebra and Spin group of 4-dimensional Euclidean space

I’m seeking for a straightforward construction of well-known $\mathrm{Spin}(4) = \mathrm{Spin}(3)\times\mathrm{Spin}(3)$ isomorphism using geometric algebra-based definition of “Spin”, without ...
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2answers
60 views

Applying geometric calculus to a matrix expression

Let $f(\mathbf x) = \mathbf x^T A \mathbf x + \mathbf b^T\mathbf x + c$, where $A$ is a square matrix, $\mathbf x, \mathbf b$ are column matrices and $c$ is a constant. Then the gradient of this ...
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138 views

About the definition of norm in Clifford algebra

I have seen two definitions for the norm in the Clifford algebra $\mathrm{Cℓ}_{p, q, r}$. According to one of them $\Vert x\Vert = ⟨x. x^\dagger⟩_0$, where the dagger stands for the reversal of the ...
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Struggling with connection between Clifford Algebra (/GA) and their matrix generators

As I thought I understood things, the Gamma matricies behave as the 4 orthogonal unit vectors of the Clifford algebra $\mathcal{Cl}_{1,3}(\mathbb C)$, (also the Pauli matricies are for the 3 of ...
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63 views

Squaring a vector using geometric algebra

I'm doing research involving clifford algebra and I'm having difficulty understanding this one axiom: $a^2 = g(a,a)$. It states that this is the square of a vector and dividing the original vector by ...
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(Geometric algebra) Acceleration of a particle with constant speed as a bivector-vector inner product

I've been working on (self-studying) Geometric Algebra for Physicists which, sadly, has no solutions manual. This is not a problem in general, but I feel like one of my solutions for a question asked ...
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154 views

Geometric algebra: Rotors

I've been (slowly) working my way through a book on geometric algebra and have found one part particularly confusing. I can understand the equation $e_1e_2=\exp(e_1e_2 \pi/2)$ Where the substiution ...
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101 views

From Orthogonal vectors to Useful Bivector

If we have set of orthogonal vectors (X) can we form a set of orthogonal bivectors from that set? I am trying to find if there is a way to get 'more information' from an orthogonal matrix by some ...
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253 views

Geometric Algebra/ Calculus for Physics

I don't know if this would be a better question for physics.SE, but I'll try here first: There is at least one good book on classical mechanics using the geometric algebra/ calculus (GA): New ...
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71 views

Why is the grade of the wedge product of two arbitrary blades the sum of the two blades' grades independently?

I'm reading Geometric Algebra For Computer Science, An Object Oriented Approach to Geometry and it says that this is true of any two arbitrary blades. $\ grade( \textbf{ A} \wedge \textbf{B})=grade( ...
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Comparison of Hilbert space tensor product and wedge product

For Hilbert Spaces: $$(|0\rangle + |1\rangle)\otimes (|0\rangle + |1\rangle) = |00\rangle + |01\rangle + |10\rangle + |11\rangle.$$ where all results are column vectors \begin{eqnarray*} 0 ...
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Norm on a Geometric Algebra

In the literature, for example "New Foundations for Classical Mechanics" by David Hestenes, the author introduces a function on the Geometric Algebra $$||M||^2=\langle M M^\dagger \rangle_0,$$ where ...
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Geometric algebra and quantum field theory

How does the reformulation of QFT with GA look like? I read that GA can be applied to almost every kind of physics, but QFT is rarely mentioned. Is there a lot of research going on in this direction ...
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196 views

I would like to see a De Morgan's Law example from Geometric Algebra

thanks for your time and effort. I really appreciate your help. In the documents "www.cs.bham.ac.uk/~fauserb/pdf/theses/Habilschrift.pdf" page 25 by B. Fauser and "Grade Free Product Formulae from ...
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Product between multivectors

I just want to see if I understood this. Since the geometric product is associative and so on we can write for two multivectors A and B given by $A= ...
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what is vector $(\vec{a}\cdot \vec{b})\vec{c} + (\vec{b}\cdot \vec{c})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$

Suppose we have three non orthogonal vectors in $R^3$ as $\vec{a}, \vec{b}, \vec{c}$. The vector of $(\vec{b}\cdot \vec{c})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ is in the plane spanned by ...
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124 views

Proof that geometric product is associative

Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the ...
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137 views

Geometric Product

I have a problem with the geometric product: In my book the unit trivector is defined like this: $(e_{1}e_{2})e_{3}=e_{1}e_{2}e_{3}$ But that would mean $(e_{1}e_{2})e_{3}= (e_{1} \wedge e_{2})\cdot ...
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derivative formula $\nabla \times (\mathbf{a} \times \mathbf{r}) = \nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = (n-1)\mathbf{a}$

Assume $\mathbf{r}=\mathbf{x}−\mathbf{x}′$ is the position vector in $\mathbb{R}^n$, for constant $\mathbf{a}$, we have $$\nabla \times (\mathbf{a} \times \mathbf{r}) = \nabla \cdot(\mathbf{a} ...
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derivative $\nabla \frac{\mathbf{r}}{r^k}$ in the context of Geometric Calculus

Suppose $\mathbf{r} = \mathbf{x - x'}$ is the position vector in $\mathbf{R^n}$, and $r = |\mathbf{r}| = |\mathbf{x - x'}|$. Do we have $\nabla \frac{\mathbf{r}}{r^k} = \frac{n-k-1}{r^k}$ or $\nabla ...
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258 views

A question on the dual relationship between the regressive product and the exterior product

I am trying to understand the following sentence, which I came across in a book: The underlying beauty of the Ausdehnungslehre is due to this symmetry [the duality between the regressive and ...
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86 views

Can every element of a geometric algebra be represented as the product of vectors?

An element of the n-dimensional geometric algebra $\Bbb G^n$ over the reals is of the form $M = \langle M \rangle + \langle M \rangle_1 + \langle M \rangle_2 + \cdots + \langle M \rangle_n$. My ...
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Projections in Geometric Algebra

Given a geometric algebra $G$. Let $I, A,B$ be an $n,r$ and $s$ blade, respectively. Let $P(*)=(*\rfloor I) I^{-1}$ be the projection operator onto $I$. Suppose that $P(A)=A$. The problem is to ...
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How to define integration over the boundary of a curve?

When learning about Stokes' theorem ($\int_{\partial \Omega} \omega=\int_{\Omega} \mathrm d \omega$), we are told that it is just a generalization of the 2nd Fundamental Theorem of Calculus $(\int_a^b ...
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Proof that every noninvertiable multivector has an idempotent factor.

Prove that every multivector which does not have an inverse has an idempotent for a factor. Define an idempotent as a multivector $A$ with the property that $A^2=A$ and $A \neq 1$. I can ...
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518 views

What good is the Commutator product?

In geometric algebra, the commutator product is defined as $A \times B = \frac 1 2 (AB - BA)$. From linear algebra, I remember that the commutator of matrices is $[A, B] = AB - BA$ and the commutator ...
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Question about domains of Geometric (Clifford) Calculus functions

For those of you familiar with Geometric Calculus, and in particular, the book Vector and Geometric Calculus by Alan MacDonald, maybe you can explain something to me. On pg 16, MacDonald defines the ...
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Proving det(A) property with outermorphism definition

Let A be a square matrix. Prove that exchanging two columns of A changes the sign of $\det(A)$. **Note: I'm pretty sure this is supposed to be "adjacent columns" Source: "Linear and Geometric ...
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How is rejection defined: Question from *Linear and Geometric Algebra*

The rejection $rej_B(a) = (a \wedge B)/B$ is a linear transformation and so has an outermorphism extension. Show that the extension cannot serve as a rejection operator on $G^n$. Hint: Use the ...
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Rank and Nullity of Projection of Multivectors onto k-Blades

Describe the image and kernel of the projection function & verify the rank-nullity theorem: The projection function is defined as $P_B(M) = {(M \cdot B)}\ /\ B$ where M is a multivector in ...