The clifford-algebras tag has no wiki summary.

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### Why do Clifford algebras determine $KO$ (and $K$-)-theory?

In the paper "Clifford modules" by Atiyah-Bott-Shapiro, they construct a family of Clifford algebras $C_k$ over the real numbers, so that $C_k$ is the algebra associated to a negative definite form on ...

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### What is the appropriate setting for Cauchy's Integral Formula?

For a $C^1$ function $f:U\to\mathbb{C}$, where $D\subset U\subseteq \mathbb{C}$ with piecewise $C^1$ boundary $\partial D$, we have the following generalized Cauchy integral formula:
$$
f(\zeta) = ...

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### What is the symmetric monoidal functor from Clifford algebras to invertible K-module spectra?

There ought to be a symmetric monoidal functor from the symmetric monoidal $2$-groupoid whose objects are Morita-invertible real superalgebras (precisely the Clifford algebras), morphisms are ...

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### Examples of Clifford Modules

For a Riemannian manifold $M$, a Clifford module is a bundle over $M$ that is, fiberwise, a representation of the Clifford algebra bundle $Cl(TM)$ and has a connection compatible with this action. ...

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### Element in spin group

I've got the following question: why is it true (if it really is?), that if I have a unitary element $u$ in the (real) Clifford algebra $Cl(V,g)$ which is even and the operator $\varphi(u)$ defined ...

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### Orders of Clifford algebra

Let $C_n$ be the Clifford algebra over $\mathbb{Q}$ associated to negative definite quadratic form $-I_n$ (i.e. $-x_1^2-\dots-x_n^2$). Let $\mathcal{O}$ be a $\mathbb{Z}$-order of $C_n$.
Q1) Is it ...

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### What is the Atiyah-Bott-Shapiro map for a bundle of *complex* quadratic forms?

In order to ask the question in the title more precisely, let me
recall some standard stuff introduced in [1; Atiyah, Bott, Shapiro].
Suppose $X$ is a compact CW complex and $V \to X$ is an oriented ...

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### Discriminants of Clifford algebras

I have a Clifford algebra defined over a field of characteristic not equal to 2. Is there a formula for its discriminant in terms of the corresponding symmetric bilinear form (or in terms of its ...

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### Clifford algebra is graded separable

Let $D$ be an algebra of odd differential operators on a free module $V$, this algebra is isomorphic to the Clifford algebra $Cl(V^* \oplus V)$. Let $m$ denote multiplication map $$m : D\otimes D \to ...

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### fast multipole method and geometric algebra

Hello,
I just learned about fast multipole method(FMM) from this article http://math.nyu.edu/faculty/greengar/shortcourse_fmm.pdf and I really liked the use of complex numbers in 2d. But I didn't ...

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### Norm of the operator acting on spinor bundle

Please forgive me if the question is too elementary, but however I was unable to manage by myself. The question comes from J.Varilly, H.Figueroa and J. Gracia-Bondia book "Elements of noncommutative ...

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### Homomorphism of algebra of “Clifford-valued continuous functions”

I am interested in the general question of
When is a map between algebra of "Clifford-valued continuous functions" homomorphism?
As a starter, I would like to first understand the case for ...

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### Properties of Clifford Algebras

It is a well known fact that Clifford algebras, $Cl(p,q)$, have similar properties depending on $(p-q)\mod 8$.
In most of the places I have found a proof of the theorem, explicit representations of ...

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### How can the Cayley-table for the elements of basis of a Cayley-Dickson algebra be summarized in an algebraic expression?

One would be able to construct a Cayley table that has all $e_i$ elements of the basis of algebra $A$ where $0<i<\dim A$ such that $e_0=1$ and $e_1=i$ and $e_2=j$ and so on. I'm looking for an ...

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132 views

### Need help determining whether a certain map is a $C^\ast$ homomorphism

Hello, I need help determining whether the map I defined between two algebras is a well-defined homomorphism of $C^\ast$-algebras. I ran into this problem while trying to define a "rotation map" ...

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### General criterion for homomorphism between Clifford Algebras

I understand that there is a universal property which tells me that given a Clifford algebra $Cl(V, q)$, for a linear map $f: V \to A$ ($A$ any associative algebra) satisfying $f(v)f(v) = Q(v)$, $f$ ...