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Questions tagged [geometric-algebras]

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

Geometric algebras are Clifford algebras over the real numbers. They can be used as a tool to study vector algebra, and they can be applied to problems in geometry and theoretical physics.

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Reference: Is a simple k-vector an equivalence class?

My understanding of a simple k-vector is that it is the wedge product of k vectors. Also, two simple k-vectors are the same, when their magnitude, attitude and orientation match. Now my question is, could I just define a simple k-vector in this…
dontknow3
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What exactly is the difference between right and left contraction in Geometric algebra?

We motivated the original definition of $A\rfloor B$ in (3.6) in terms of geometrical subspaces as “A taken out of B”. This is clearly asymmetrical in A and B, and we could also have used the same geometrical intuition to define an operation $ B…
tryst with freedom
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Vector in the line of intersection of two planes

In the context of Geometric Algebra, in $ \mathbb{R}^3 $: Let $A$ and $B$ be bivectors (representing planes). Show that $ (\langle AB \rangle_2)^* $ is a vector in the line of intersection of the two planes, where the $ ^* $ represents the dual. My…
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Show that $ B \mathbf I = (-1)^{k(n-1)}\mathbf I B $

where $B$ is a $k$-vector in $n$ dimensional space and $ \mathbf I $ is the pseudoscalar.
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Show that $ \mathbf u^2 \mathbf v^2 = (\mathbf u \cdot \mathbf v)^2 - (\mathbf u \wedge \mathbf v)^2 $

where $ \mathbf u $ and $ \mathbf v $ are vectors. From Linear and Geometric Algebra by Alan Macdonald.
user997712
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Motivation behind inner and exterior product of a vector with a k-grade blade in geometric algebra

The inner product of a vector with a k-grade blade is given as: $$ a \cdot A_k = \frac12 \left( aA_k-(-1)^kA_k a\right)$$ And the exterior product as: $$ a \wedge A_k= \frac12 \left( a A_k+ (-1)^kA_k a \right)$$ What are the motivation for the…
tryst with freedom
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What is the difference between an r-blade and an r-vector (geometric algebra)

I'm following this intro to geometric algebra and I'm a bit confused about what the difference between these two objects is. The author states (p. 12) Let r > 1; then an r-blade or simple r-vector is a product of r orthogonal (thus anticommuting)…
dylnan
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Derivation of rotation formula in $ \mathbb R^3 $

In geometric algebra, for deriving the rotations formula in $ \mathbb R^3 $ I see the following steps ($ \mathbf i $ is the plane of rotation, $ \mathbf u $ is the vector to be rotated, with components $ \mathbf u_{\parallel} $ in the plane of…
user997712
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Geometric Algebra : Intersection of Two Planes in R4

I'm working in a four dimensional Euclidean space. I have two planes, each represented by a bivector. How do I find the intersection of these two planes? This will be either 0, a vector, or a bivector. I could do this if I could find the smallest…
user11865
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Inverse of a Versor

Let $A=a_1a_2\ldots a_N$ be a versor, where $a_i$ is a vector for all $i$. Let $A^\dagger$ denote the reversion of $A$. Let $a_ia_j$ denote the geometric product of vectors $a_i$ and $a_j$. According to the book on Geometric Algebra for Computer…
NicNic8
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Limits of overdot notation in geometric algebra

In geometric calculus the over dot notation is used to denote the proper way to do the vector differentiation of a multivector product - $$ \nabla (AB) = (\nabla A)B + (\dot{\nabla}A)\dot{B} $$ The question is does make any sense to have more than…
user2761381
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Conditions for a C*-algebra to be Abelian

Please help me with this problem. I don't know how to get into proof. Let $A$ be a $C^*$-Algebra. Show that if the condition $0 \leq x \leq y,~~ x,y\in A,$ implies $x^2 \leq y^2,$ then $A$ is abelian.
Yaghoub Ahmadizar
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Finding Dimension of quotient ring

Let $I=(X, Y) \subset k[X,Y]$. show that $\mbox{dim}_k(k[X,Y] / I^n) = 1+2+...+n=n(n+1) /2 $ Here k is algebracally closed field. And $(X,Y) $ is ideal in polynomial ring $k[X,Y]$ generated by $ X, Y$. Actually I can't figure out how to find a…
Peterson Author
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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 this can give an inverse vector. I was wondering…
Tom Ryan
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Are subspaces also subalgebras in Geometric Algebra?

Is a subspace of GA(n) closed under the geometric product? Say we let a k-blade represent a subspace of GA(n), where k < n. Does that also represent a subalgebra of GA(n)? I can see that we won't get any higher dimensional elements. But I'm not…
user137731
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