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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

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 ...


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 ...


5

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 ...


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

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

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

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 $n$ ...


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

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

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 $$\begin{equation} a \cdot \partial F = a(x) \cdot \partial_x ...


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

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

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'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 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 ...


2

Denote by $C\ell(V,\beta)$ the Clifford algebra of $V$ with bilinear form $\beta$. Also define the set $\ker(\beta):=\{v\in V\mid \forall x\in V,\beta(v,x)=0\}$. It turns out that the ideal generated by $\ker(\beta)$ is exactly the Jacobson radical $C\ell(V,\beta)$, which I guess I'll denote as $rad(C\ell(V,\beta))$. Now the bilinear form $\beta$ induces 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

Let $I=e_1\dots e_n$ be a pseudoscalar of $cl(V,Q)$. Since, $we_i=-e_i w$ for all $i$, we get $wI=(-1)^nIw$. On the other hand, in general, we have $Iw=a^{n-1}(w)I$, where $a$ is the grading involution of $cl(V,Q)$. By assumption, $w$ is an odd element, hence $a(w)=-w$. Therefore, comparing the two equation, we get $2wI=0$. Suppose that $char(F)\neq2$. It is ...


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 ...


2

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 ...


2

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 ...


2

The Brauer group of $\mathbb R$ is $Br(\mathbb R)\cong\mathbb Z/2\mathbb Z$; one way to see this is to note that there are exactly two (iso)classes of central simple f.d. real algebras, in view of Frobenius' theorem which classifies them, so the there is no choice for the group structure. It follows that the tensor square of every central simple real ...


2

Let $\{e_1, \dots, e_{n+2}\}$ be an orthonormal basis for $\mathbb{R}^{n+2}$ (with respect to the standard inner product), $\{e^\prime_1, \dots, e^\prime_n\}$ the standard generators for $C^\prime_n$ and $\{e^{\prime\prime}_1, e^{\prime\prime}_2\}$ the standard generators for $C_2$. Now we define a map $$f: \mathbb{R}^{n+2} \longrightarrow C^\prime_n \otimes ...


2

Let's take the exmaple of $G_{1,1,0}$ as you did and let $e_1, e_2$ be an orthonormal basis for $V$ with the first squaring to 1 and the second squaring to -1. Let's look at $C$ and $D$ corresponding to $e_1$: $$C=\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\\\end{bmatrix} ...


2

Since both sides are multilinear (linear in each variable when the rest are fixed), it is enough to check it for combinations of a basis. There are not too many distinct cases: e.g. when $a=b$ (and say, $=e_i$), then we get $0$ on both sides. More cases to check: $\ (e_i\land e_j)\cdot (e_i\land e_j)$, $\ (e_i\land e_j)\cdot (e_i\land e_k)$, $\ (e_i\land ...


2

Sure. You can argue it pretty well just from symmetry. As was said by Berci, it has to be linear in each term, and it has to be antisymmetric under exchange of $ a $ and $ b $ or $c$ and $d$. I'm not sure if you can show that the above is the only such relation. It certainly make me feel more comfortable with the identity at least. But, regardless, you can ...



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