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Hot answers tagged clifford-algebras

13

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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 $\boldsymbol x$.

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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 ...

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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 ...

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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.

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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 ...

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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 ...

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As much as I love Clifford / geometric algebra, I don't think it is of use here. CA allows you to deal in a basis-free way with multidimensional subspaces as algebraic objects, making various complicated derivations more elegant and transparent. In the case of neural nets, however, we really are dealing with a plain vector of parameters. If $\theta \in ... 5 The formula$A \cdot (B \times C) = \textrm{Det}(A,B,C)$shows this the cross product can be thought of as the transpose of the linear map$\textrm{Det}(\cdot,B,C)$. Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that$A \times B=\star(A^\flat \wedge B^\flat)^\sharp$. This is the connection between the ... 5 Not every element of a Clifford algebra has the form$v$for$v$a vector. In general an element of a Clifford algebra is a sum of products of such things, and most of these aren't invertible. Explicitly, let$\text{Cliff}(p, q)$be the Clifford algebra over$\mathbb{R}$generated by$p$generators$e_1, \dots e_p$satisfying$e_i^2 = -1$and$q$... 4 The argument I know, and which satisfies me is: Show first that$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$. Using this last ... 4 Besides the books by Hestenes, Hestenes and Sobczyk, Dorst, Doran and Lasenby, Porteous, Lounesto, and Baylis, you should find a rather accessible paper by Eric Chisolm on ArXiv.org. You will find its abstract at the following URL. http://arxiv.org/abs/1205.5935 I believe that paper meets your criteria for containing the theorems and proofs, as well as a ... 4 part 1) The differential forms approach is indeed very powerful, what Hestenes points out in his "From Clifford Algebra to Geometric Calculus" is that to give a complete treatment of differential geometry of manifolds you need various structures. In the book you will find an alternative. The starting point (as was pointed out above) is the notion of a ... 4 The short answer No, a general multivector is not graded. Only a portion of the elements of the algebra are assigned grades. Yes blades are among the elements assigned grades. A blade represents a subspace of$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$... 4 At this question you can learn why the Jacobson radical of a Clifford algebra of a finite dimension space with a nondegenerate form is$\{0\}$. Finite dimensional algebras with Jacobson radical zero are semisimple algebras. 4 Clifford algebras arise in several mathematical contexts (e.g., spin geometry, abstract algebra, algebraic topology etc.). If you're just interested in the algebraic theory, then the prerequisites would probably be a solid background in abstract algebra. For example, I think linear algebra and ring theory are prerequisites but in practice, one should ... 4 You could express it as a function. Let your operator be$\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. Leta$be an arbitrary vector, ... 3 I'm sorry for not reacting for a long time. I have to admit my question was a bit vague, but I have found the solution, I was looking for, now. There is a complete isomorphism-invariant$\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 $$\begin{eqnarray*} 2{\bf a}{\bf b} &=& (e_2+e_{12})(1+e_1) \\ &=& e_2+e_{12}+e_2 e_1+e_{12}e_1 \\ &=& e_2 + e_{12} +(-e_{12}) + (-e_2) \\ &=& 0 \end{eqnarray*}$$ 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 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 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 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 When you say$c = ab$is a bivector, you're taking$a \perp b$implicitly. The geometric product will in general have both terms:$ab = a \cdot b + a \wedge b\$, remember? A scalar and a bivector. It is, admittedly, pretty hard to visualize what that is geometrically. Adding vectors to vectors or bivectors to bivectors is sensible; adding vectors and ...

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