Hot answers tagged clifford-algebras
7
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 ...
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 ...
4
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 ...
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 ...
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
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
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
First, care is needed! We need the right notion of "op", for behold
$Cl(1) \simeq \mathbb{R}^2 \not\simeq \mathbb{C} \simeq Cl(-1)$, which as algebras are both commutative!
Over $\mathbb{R}$, the Clifford algebras arise naturally in the braided symmetric monoidal category of $\mathbb{Z}/2$-graded vector spaces, or "super-vector spaces" (alternatively...), ...
2
$\def\dd#1{\tfrac{\partial}{\partial #1}}$Observe first that $$\dd{x}\Big(f(x,y)|_{x=y}\Big) = \Bigg(\Big(\dd x+\dd y\Big)f(x,y)\Bigg)\Bigg|_{x=y}$$
Let us write $E(\dd{y}, \dd{z})=\exp(-\omega^{i,j}\dd{y^i}\dd{z^j})$, so that $$
(f\star g)(x)=\Big(E(\dd{x},\dd{y})\cdot\big(f(x)g(y)\big)\Big)\Big|_{x=y}$$ and consequently
\begin{align}
((f\star g)\star ...
2
Here we let $Q(x) = -x^2 - y^2$ and $Q'(x) = x^2 + y^2$, so that
$$C_2 = \mathrm{C}\ell(\mathbb{R}^2, Q)$$
and
$$C^\prime_2 = \mathrm{C}\ell(\mathbb{R}^2, Q').$$
Let $\{e_1, e_2\}$ be an orthornomal basis for $(\mathbb{R}^2,Q)$. Then $C_2$ has generators $\{1, e_1, e_2, e_1 e_2\}$ satisfying the relations
$$e_1^2 = e_2^2 = -1,$$
$$e_1 e_2 = -e_2 e_1.$$
Then ...
2
This is from Cassels, Rational Quadratic Forms, chapter 10, especially pages 177-178. We have a basis $e_1, e_2, e_3$ of a vector space $V$ of dimension 3 over a field. These satisfy
$$ e_i e_i = f(e_i) $$ and
$$ e_i e_j + e_j e_i =0, \; \mbox{when} \; i \neq j, $$ meaning that they are orthogonal. Then a basis of the even Clifford algebra is
$$ 1, \; ...
1
It is enough to consider action of $\epsilon_v \equiv \epsilon(v)$ and $\iota_v \equiv \iota(v)$, $v \in {\bf p}$
on a basis of $\wedge {\bf p}$.
$\epsilon_v(1) \equiv v$;
$\epsilon_v(v_1 \wedge \cdots \wedge v_k) \equiv v \wedge v_1 \wedge \cdots \wedge v_k$.
$\iota_v(1) \equiv 0$;
$\iota_v(v_1 \wedge \cdots \wedge v_k) \equiv
\sum_{j=1}^k (-1)^{j-1} ...
1
By Mazur-Ulam theorem every surjective isometry is an affine map, i.e.
$$
\gamma(x)=x_0+T(x)
$$
for some fixed vector $x_0\in X$ and necessary isometric isomorphism $T\in\mathcal{B}(X)$.
Assume that surjective isometry $\gamma$ is a Clifford isometry, then we have $C\geq 0$ such that for all $x\in X$ we have
$$
\Vert x_0+T(x)-x\Vert=C
$$
I think it's ...
1
First of all we have the fundamental theorem of algebra which says that the the roots of polynomial functions with complex coefficients all lie in the field of complex numbers.
You are intimating that integers are required to solve
$x+a=b$ with $a>b$ are natural numbers.
Fractions are required to solve
$ax=b$ where $a,\,b\in\mathbb{N}$ with $a\neq 0$.
...
1
$\newcommand{\Cl}{\mathrm{C}\ell}$Write
$$\tilde{q} = \langle b_2, \dots, b_n \rangle$$
and let $\{e_1, \dots, e_n\}$ be the standard basis for $K^n$. Define a map
$$f: K^{n-1} = \operatorname{span} \{e_2, \dots, e_n\} \longrightarrow \Cl^0(q)$$
by
$$f(e_i) = e_1 \cdot e_i$$
for $i > 1$ and extending linearly. Given
$$x = \sum_{i = 2}^n x_i e_i \in ...
1
By the structure classification theorem, such Clifford algebras are actually two copies of a matrix ring over $\mathbb{C}$.
Being semismple, all of its modules are direct sums of copies of the simple modules.
The irreducible ones would just be the simple ones. (right?)
(When talking about representations as opposed to modules I always feel like there is a ...
1
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, ...
1
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 ...
1
Too many comments to fit into a comment box:
(1) the initial determination of the isomorphism classes of those Clifford algebras need not be so awful, if one does a two-step induction, then just needing to know how the Hamiltonian quaternions arise, versus the split algebra.
(2) Clifford algebras of non-degenerate quadratic forms on even-dimensional ...
1
There is pretty good information on this at Wikipedia. The recurrence of smaller Clifford algebras in larger ones is a manifestation of Bott periodicity.
1
$\newcommand\rad{\operatorname{rad}}$Let $\beta$ be an arbitrary symmetric form on a vector space $V$, let $\ker\beta$ be its kernel and let $\bar\beta$ be the induced non-degenerate form on $V/\ker\beta$.
Check that there is an surjective algebra map $p:C(\beta)\to C(\bar\beta)$ which is induced by the quotient map $V\to V/\ker\beta$ whose kernel is ...
1
Suppose you begin with an orthonormal basis (orthonormal in the sense that the elements are orthogonal and square to either 1 or -1) for $V$.
Then (if I understand your notation correctly), the standard basis elements square to $\pm1$, so $(\overline{G}_{p,q}[i])^3=\pm[\overline{G}_{p,q}[i]]$ where the sign depends on the grade of the basis element and the ...
1
In the context of Clifford/geometric algebras (which we have been talking about recently), I can give you a crash course on how a special case of a real geometric algebra is built. Keep in mind that what I'm describing is a special case, and there are more possibilities.
Here is a brief introduction to why Clifford algebras for finite dimensional vector ...
1
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 ...
1
I don't want to reprove the existence of Clifford algebras in general here. You will have to go look that up, for a rigorous explanation.
I can give a heuristic explanation, though, on how to think of it.
The quadratic form $Q:V\rightarrow F$ gives rise to a bilinear form $B:V\times V\rightarrow F$ that works like an inner product on $V$. From $Q$ and $V$, ...
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