# Tag Info

10

The previously mentioned "Geometric Algebra for Computer Science" is a good introduction that concentrates on the algebraic (not calculus related part) of GA. It does have material on GA's application to computer graphics, but the bulk of the text is just on the geometry behind GA. Another possible starting point is "Linear and Geometric Algebra" by Alan ...

8

The classic reference is David Hestenes' New Foundations for Classical Mechanics which is by one of the early developers of geometric algebra. You may find it easier to learn geometric algebra from Geometric Algebra for Physicists by Doran and Lasenby though (I certainly did). The link is to a sample version of chapter 1. A reference that I've never looked ...

8

The Clifford (or "geometric") algebra is an algebra "generated" by a vector space with a bilinear form with some special properties. There are several pockets of physicists and mathematicians experimenting and refining how to use these algebras to do some geometry. If you like, you can think of the real 3-D space as your vector space $V$ with its regular ...

7

If $e_1, e_2, \ldots, e_n$ are an orthogonal basis in $V$, then the relations $$e_i e_j = - e_j e_i$$ hold for $i \ne j$ in the corresponding Clifford algebra. From this, we can derive the relation $$e_A e_B = e_B e_A (-1)^{|A| \cdot |B| - |A \cap B|}$$ where $A$ and $B$ are subsets of $\{1,2,\ldots,n\}$ and $e_A = \prod_{i \in A} e_i$, the product taken in ...

6

The book is not intended for high school students. According to the preface it is intended for "the introductory linear algebra course", a sophomore college course. The preface also recommends a calculus course for "mathematical maturity". Definition 5.9 defines the geometric product of two vectors. The first paragraph of Section 6.1 gives reasons for ...

6

It's a basic fact (here's a proof in the second proposition on page 157) that the tensor product of two central simple algebras is another central simple algebra. A proof should be available wherever central simple algebras are discussed. Another location in Jacobson's Basic Algebra II on page 218-219. Another location in Rowen's Ring theory Theorem ...

6

Take inner product with $\boldsymbol b$ on both sides and solve for $\boldsymbol x \cdot \boldsymbol b$. Replace $\boldsymbol x \cdot \boldsymbol b$ in the original equation by the solution just found. Solve for $\boldsymbol x$.

6

Let's consider a Euclidean 3d space. The clifford algebra on this space consists of multivectors, which have 8 linearly independent components. These components can be broken down as follows: 1 scalar component 3 vector components 3 bivector components, which correspond to the 3 linearly independent planes in a 3d space 1 trivector or pseudoscalar ...

5

There are different meanings of the words Geometric Algebra. One is represented by Artin's book on the reconstruction of algebraic structures (fields, rings) from the geometries that they coordinatize. The other is the use of Clifford algebras, quaternions and related ideas as a formalism for geometry and physics. This is popularized by Hestenes and is ...

4

The short answer No, a general multivector is not graded. Only a portion of the elements of the algebra are assigned grades. Yes blades are among the elements assigned grades. A blade represents a subspace of $V$, and the grade of that blade is the dimension of the subspace the blade represents. Everything I'm about to say is with regards to an $n$ ...

4

The argument I know, and which satisfies me is: Show first that $C_{k+2}\cong C_k'\otimes C_2$ and $C_{k+2}'\cong C_k\otimes C_2'$ by exhibiting isomorphisms. Notice that this implies that $C_4\cong C_2\otimes C_2'$. Using the first point twice, $C_{k+4}\cong C_k\otimes C_2\otimes C_2'$ and, by the second point, this is $C_k\otimes C_4$. Using this last ...

4

Clifford algebras arise in several mathematical contexts (e.g., spin geometry, abstract algebra, algebraic topology etc.). If you're just interested in the algebraic theory, then the prerequisites would probably be a solid background in abstract algebra. For example, I think linear algebra and ring theory are prerequisites but in practice, one should ...

4

At this question you can learn why the Jacobson radical of a Clifford algebra of a finite dimension space with a nondegenerate form is $\{0\}$. Finite dimensional algebras with Jacobson radical zero are semisimple algebras.

4

You could express it as a function. Let your operator be $\underline T$. It could be described by \begin{align*}\underline T(e_1) &= e_1 \\ \underline T(e_2) &= e_1 + e_2 - e_3 \\ \underline T(e_3) &= -e_1 -e_2 + e_3\end{align*} You could instead use dot products to combine this into a single expression. Let $a$ be an arbitrary vector, ...

4

Here's an excerpt from Lasenby, Lasenby and Doran, 1996, A Unified Mathematical Language for Physics and Engineering in the 21st Century: The next crucial stage of the story occurs in 1878 with the work of the English mathematician, William Kingdon Clifford (Clifford 1878). Clifford was one of the few mathematicians who had read and understood ...

4

One big advantage is in the conception of some geometric transformations. So-called "rejections" are a good example. For instance, given a vector $v$, the part of a vector $a$ that is orthogonal to $v$ is $a - (v \cdot a) v^{-1}$. This defines a linear map. When you get to finding the rejection on a blade $V$, you might, in ordinary vector algebra, have ...

4

This uses a typical abuse of notation to talk about $x$ as a point on the manifold as well as a one-parameter function $x(\tau)$ that generates a curve on the manifold. Your edition seems to be a bit different from mine. Mine says The derivative of $F$ in the direction $a$ is defined by $$a \cdot \partial F = a(x) \cdot \partial_x ... 3 It's not easy to give a proof without knowing exactly what your definitions of the Clifford and exterior algebras are, so the following argument is still a little handwaving. But let's say that we somehow have defined what the exterior algebra of a vector space V is; we know that its elements are multivectors, and the rule which generates everything is ... 3 Let me address this more on the side of how linear algebra is presented in some GA material. In traditional linear algebra, you use a lot of matrices and column/row vectors because this gives you an easy way to compute the action of a linear map or operator on a vector. What I want to emphasize is that this is a representation. It's a way of talking about ... 3 To briefly synthesize the comments by Willie Wong and Eric O. Korman above, there are no (natural number) solutions to the original equation you posted unless you meant \{\sigma_i,\sigma_j\}=2I\delta_{ij}. Eric's comment applies if we are talking about integer matrices, and it's true that once you find an integer matrix solution (a,b), then ... 3 When first starting out, I found some of Alan Macdonald's introductory material to "geometric algebra" very useful for developing intuition. To make a gross overgeneralization, geometric algebras are basically the low dimensional Clifford algebras over \Bbb R that are most relevant to 2-d and 3-d geometry, and even some 4-d relativistic geometry. I think ... 3 part 1) The differential forms approach is indeed very powerful, what Hestenes points out in his "From Clifford Algebra to Geometric Calculus" is that to give a complete treatment of differential geometry of manifolds you need various structures. In the book you will find an alternative. The starting point (as was pointed out above) is the notion of a ... 3 I'm sorry for not reacting for a long time. I have to admit my question was a bit vague, but I have found the solution, I was looking for, now. There is a complete isomorphism-invariant \theta(\Phi)\in\mathbb{Z}_2 of the irreducible, complex, finite dimensional representations \Phi of the Clifford-Algebra Cl(V) of an odd-dimensional, complex, ... 3 I do not believe there is without passing through some alternate representation (quaternion, matrix, ...). This is one of the known disadvantages of axis-angle compared to the others, while an advantage is the triviality of inversion (simply negate the angle or the axis). 3 In general, infinitely many. Take for example \Bbb H, which is a Clifford algebra for a suitable choice of metric on \Bbb R^2. Pick any unit length quaternion u with real part 0. Then (u\cos(\pi/2)-\sin(\pi/2))^2=1. There are as many choices for u as there are points on the unit sphere in 3-space. Each of these will describe an involution ... 3 I think the problem here is the tail wagging the dog: you've defined the spin 1/2 representation of a rotation and then found that this is indeed also the rotor that performs rotations in a bilinear fashion. I think it's better to look at it the other way around: one can construct a rotor-based transformation that happens to be a rotation, and then prove ... 3 I like Porteous's Clifford Algebras and the Classical Groups for a purely mathematical perspective. Pertti Lounesto's Clifford Algebras and Spinors is also really good and talks about applications to physics. 3 Besides the books by Hestenes, Hestenes and Sobczyk, Dorst, Doran and Lasenby, Porteous, Lounesto, and Baylis, you should find a rather accessible paper by Eric Chisolm on ArXiv.org. You will find its abstract at the following URL. http://arxiv.org/abs/1205.5935 I believe that paper meets your criteria for containing the theorems and proofs, as well as a ... 2 The Lorentz group, SO(1,3), is non-compact, thus their representations are not unitary (in general). Therefore, if you have a spinor, \psi\in \mathcal{S}, transforming as \psi\mapsto S\psi, it follows that the construction$$ \psi^\dagger \psi \mapsto \psi^\dagger S^\dagger S \psi \neq\psi^\dagger \psi, since $S^\dagger\neq S^{-1}$. This tells us ...

2

Start with the geometric product of vectors. This uses the usual properties. For vectors $a, b, c$, $aa$ is a scalar, $a(bc) = (ab)c$, and $a(b+c) = ab + ac$. For a scalar $\alpha$, $(\alpha a)b = \alpha(ab)$. A $k$-blade is a geometric product of $k$ anticommuting vectors. Typically this is written in terms of wedge products, which is why this can ...

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