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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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geometric product between nonorthogonal basis and its reciprocal basis

This is related to the question of wedge product between nonorthogonal basis and its reciprocal basis in geometric algebra. If {$e_i$} is a set of basis that are not necessarily orthogonal, and {$e^i$} is the corresponding reciprocal basis, we…
ahala
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Help me evaluate this geometric product

We are in the geometric algebra generated by the vector space $R^{1,1}$. Consider three vectors, $e_+, e_-, e$ where $e_+^2=1$, $ e_-^2=-1$, $e_+.e_-=0$ and $e=e_+ +e_-$. It is straightforward to verify that $e^2=0$, making $e$ a null vector. I wish…
njt
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Adjoint of a Linear Function Acting on Bivectors

I've been self studying Doran and Lasenby's Geometric Algebra for Physicists, and I'm getting stuck on a small derivation in section 4.4.3 about the adjoint of linear functions. It's not big enough to keep me from proceeding through the rest of the…
Itserpol
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If $A = P(A) = \langle A\rangle_r$, why is $\dot \partial (\dot x \cdot A) = A\cdot \partial x$ (Geometric algebra)

Let $x$ be a vector of an $n$-dimensional subspace $\mathcal A_n$ with $P$ the projection operator onto this subspace. Then, let $A= P(A) = \langle A\rangle_r$ be an $r$-vector. Then, I know that $\dot \partial (\dot x \cdot A) = A\cdot \partial x$…
Rodrigo
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Show that $\partial x^2 = 2P(x)$ and $\partial \wedge x = 0$ (Geometric algebra)

Show that $\partial x^2 = 2P(x)$ and $\partial \wedge x = 0$ (Geometric algebra)I’m trying to show equations (2-1.32) and (2-1.33) on page 51 of 20 of Hestenes and Sobczyk’s “Clifford Algebra to Geometric Calculus”. $\partial |x|^2 = \partial x^2 =…
Rodrigo
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Why is the inner product of two simple vectors simple? (Geometric algebra)

I’m trying to understand the reason for the assertion on page 20 of Hestenes and Sobczyk’s “Clifford Algebra to Geometric Calculus” that If $B$ is a simple $s$-vector, then $B\cdot A$ [where $A$ is a simple $n$-vector] is simple. According to page…
Rodrigo
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How to get scalar product from geometric product in Geometric Algebra?

I read that geometric product $ab$ can be decomposed into 2 parts (symmetric and antisymmetric). But I can't understand why symmetric part is a scalar product. I mean following symmetric part: $\frac{1}{2} (ab+ba) = \frac{1}{2}((a+b)^2-a^2-b^2)$ Why…
Mike_bb
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Is there a single term that generalizes the names of the individual vector and scalar types in a multivector?

In three dimensions, a multivector consists of a scalar, a vector, a bivector and a tri-vector. Is there a term that generalizes these names? For example, in an $n$-dimensional space, can I use the terminology $k$-vector (where $0\leq k\leq n$) to…
HelloGoodbye
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Missing steps+intuition for geometric algebra manipulations

I'm working my way through Doran and Lasenby's Geometric Algebra for Physicists, but I've run into some trouble in chapter three with a couple of derivations where I can't follow the steps. The first one is equation (3.126): $$A\cdot(x\wedge(x\cdot…
G. Paily
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Rotor Identity $ \frac{1+ba}{|a+b|} = e^{-B\theta /2} $

To prove:the identity given above where $ a, b $ are vectors, $ B $ is the unit bivector in the $ a\wedge b $ plane and $\theta $ is the angle between $ a$ and $ b$. (From "Geometric Algebra for Physicists" by Doran and Lasenby). Expanding the…
user997712
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Multivectors which commute with every vector in $ \mathbb G^n $

Let $ \mathcal C $ be the set of multivectors which commute with every vector in $ \mathbb G^n $. Show that: 1) When n is even, then $ \mathcal C $ is the set of all scalars $ a $. 2) When n is odd, then $ \mathcal C $ is the set of all scalars plus…
user997712
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Question about Leo Dorst book on Geometric algebra for Computer Science

In section 7.4.3 Dorst brings a result that in Euclidean and Minkowski spaces, a bivector $B$ can be written as a sum of commuting 2-blades, and therefore $e^B=e^{\mathbf B_1}\cdots e^{\mathbf B_k}$. Each of $e^{\mathbf B_i}$ is a rotor, so it is a…
alexanderyaacov
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Deriving rotation of 3d vector by 3d rotor

I am attempting to implement a geometric algebra based Rotor in a math code library I am writing. In order to do so, I'm attempting to derive the final result of both pre-multiplying a 'rotor' with a vector and then post-multiplying the inverse of…
Termhn
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Converting $p(\theta,t)=e^{a\wedge c\theta}(a+be^{a\wedge bt})e^{-a\wedge c\theta}$ from geometric to vector algebra

Let $a,b,c\in\mathbb{R}^3$ with $a\wedge b\wedge c\ne 0$ and $a^2>b^2$. What is the form of this equation $p(\theta,t)=e^{a\wedge c\theta}(a+be^{a\wedge bt})e^{-a\wedge c\theta}$ in standard vector coordinates, where $\theta,t\in\mathbb{R}$. I am…
Karambwan
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Given $\textbf{a}\perp\textbf{b}$, $\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\mid \textbf{a}\mid^{2} \textbf{b}$ (geometric algebra)

I want to prove, given $\textbf{a}\perp\textbf{b}$, $$\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\mid \textbf{a}\mid^{2} \textbf{b}$$ I realize this is just a matter of…
roshoka
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