43

This is not really an answer, but rather a meta-answer as to why there exist many conventions in the first place.
The symmetric monoidal category $\mathit{sVect}$ of super-vector spaces has a non-trivial involution $J$.
The symmetric monoidal functor $J:\mathit{sVect}\to \mathit{sVect}$ is the identity at the level of objects and at the level of morphisms.
...

answered Oct 29 '14 at 15:28

André Henriques

37.8k55 gold badges101101 silver badges232232 bronze badges

22

As Igor mentions in the comments, this is really a question about deformations of the multiplication map of the exterior algebra in the space of associative multiplications. Since this is a pretty general story, let me try to sketch how it works in this case.
Let's fix some field $k$ and a $k$-algebra $(A,\mu)$. We want to understand deformations of $(A,\mu)...

20

Here is one conceptual description of the relationship. (I wouldn't call it an explanation; I don't really know why it's true, except that it's because Bott periodicity is true.)
KO-theory is the first Weiss-derivative of the $K$-theory of Clifford algebras.
More precisely: given a real inner product space $V$, we get a category $M(V)=\mathrm{Mod}(Cl(V))$ ...

18

The $p$-completed algebraic $K$-theory of the algebraic closure of $\mathbb{Q}_p$, i.e., $K(\bar{\mathbb{Q}}_p; \mathbb{Z}_p)$, is equivalent to its second loop space, up to an issue about path components. This is due to Suslin. The descent to $\mathbb{Q}_p$ is more subtle than the descent from $\mathbb{C}$ to $\mathbb{R}$, because the absolute Galois group ...

ag.algebraic-geometry at.algebraic-topology p-adic-analysis clifford-algebras rigid-analytic-geometry

16

To my mind, it has to do with p.d.e.'s and to your background, geometry or physics. Let me explain. $\newcommand{\pa}{\partial}$
Fix $\epsilon=\pm 1$. Define the Clifford algebra using the $\epsilon$ rule
$$ c_jc_k+c_kc_j=2\epsilon\delta_{jk} $$
and we form the Dirac operator
$$ D=\sum_k c_k\pa_{x^k} $$
Its square is
$$D^2=\epsilon\sum_k \pa_{...

answered Oct 29 '14 at 13:07

Liviu Nicolaescu

29.7k22 gold badges7070 silver badges131131 bronze badges

14

In addition to the answer of Bertram Arnold, let me point out that there is a very explicit formula for "fermionic" Weyl-Moyal product. Let us assume that your vector space $V$ (or module) is defined over a ring containing the rationals. This is sort of crucial in the following and in prime characteristic the story is slightly different.
Part of the sport ...

11

Here is another proposal. The key will be to enlarge the category of quadratic vector spaces fairly substantially. Here are three hints leading towards the proposal:
Clifford algebras can and should be thought of as $\mathbb{Z}_2$-graded; for example, Clifford algebras over $\mathbb{R}$ give every element of the $\mathbb{Z}_2$-graded Brauer group / Brauer-...

answered Sep 9 '14 at 16:03

Qiaochu Yuan

98.7k2929 gold badges355355 silver badges643643 bronze badges

11

It may be worthwhile to vary the question ever so slightly to read
Question: Is there a conceptual explanation for the tight relationship (or some aspect thereof) between K-theory and superalgebra?
This seems to have a deep answer as follows:
The $\mathbb{Z}_2$-grading in superalgebra/supergeometry is usefully identified as the low degree pieces of $\...

11

Here's a standard explicit formula: Let $\mathbb{O}\simeq\mathbb{R}^8$ denote the algebra of octonions, and for $x\in\mathbb{O}$, let $L_x$ (respectively $R_x)$ denote the linear map from $\mathbb{O}$ to itself generated by left (respectively, right) multiplication by $x$ and let $C:\mathbb{O}\to\mathbb{O}$ be conjugation in the octonions. Now define $\rho(...

answered Jan 19 '15 at 19:05

Robert Bryant

84.1k77 gold badges254254 silver badges358358 bronze badges

11

Let $(M,g)$ be an orientable pseudo-riemannian manifold. Each tangent space $T_xM$ is a pseudo-euclidean space and hence has an associated Clifford algebra $CL(T_xM)$, which is the fibre at $x\in M$ of the Clifford bundle $Cl(TM)$. If the manifold is spin (a topological condition which says that the oriented orthonormal frame bundle lifts to a spin bundle) ...

answered Jan 9 '15 at 18:27

José Figueroa-O'Farrill

30.2k44 gold badges8585 silver badges161161 bronze badges

10

$C_{n,0}$ is either a full matrix algebra over $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$, or the direct sum of two such algebras that are isomorphic. The exact description depends on the residue class of $n$ (mod $8$) and can be found in textbooks or on wikipedia http://en.wikipedia.org/wiki/Classification_of_Clifford_algebras
$\mathrm{Mat}_{k \times k}(\...

answered Mar 10 '15 at 6:33

Dave Witte Morris

3,72111 gold badge1313 silver badges1919 bronze badges

10

I would think that these these notes by Akhil Mathew provide the "exact definition" you are asking for:
In response to the follow-up question "which is the first Clifford module used in the physics context":
Two different representations (modules) of the Clifford algebra were studied in early work on the Dirac equation. Paul Dirac himself used a ...

answered Jan 9 '15 at 18:24

Carlo Beenakker

113k1111 gold badges271271 silver badges390390 bronze badges

10

$\def\RR{\mathbb{R}}$This is an answer to flesh out what I wrote in comments. Let $V$ be a real vector space with positive definite inner product. For a unit vector $\vec{u} \in V$, let $s_{\vec{u}}$ be the reflection over $\vec{v}$. Here is something we might hope for:
Vague hope: Can we build a ring $R$ whose additive structure comes from the vector ...

answered Sep 25 '19 at 14:43

David E Speyer

125k1010 gold badges318318 silver badges600600 bronze badges

9

For any finite-dimensional associative unital $\mathbb{R}$-algebra $A$, the set $S$ of noninvertible elements is the zero-set of a polynomial. Namely, let $f:A \to Hom (A,A)$ be the map $a \mapsto ( x \mapsto ax)$ to the linear endomorphisms of $a$. $f$ is an injective algebra homomorphism, since $f(a)=0$ means $ax =0$ for all $x \in A$, in particular $a= a1 ...

answered Dec 13 '14 at 15:47

Johannes Ebert

18.9k22 gold badges6262 silver badges105105 bronze badges

7

A very good place to read about this is the 3. chapter of the book "Heat kernels and Dirac operators" by Nicole, Getzler & Vergne.
Up to the wrong sign in your second definition, 1. and 2. are equivalent, as the symbol just collects the highest order terms. Also you obtain from the basic properties of the symbol map that the bundle which admits a Dirac ...

7

To put the problem to a rest, I add my comment as an answer which is, in a nutshell, $Spin(6,2)\cong Spin(4,\mathbb H)$. The existence of this isomorphism follows from the isomorphism of the Satake diagrams and simple connectedness.
The tricky thing is the definition of a spin group over the quaternions which is explained in much greater generality in ...

6

I am not aware of a single formula, only recursive formulas involving the "shuffle basis" for Cayley-Dickson spaces. The shuffle basis is not the one commonly used by most researchers. For the shuffle basis, $e_0=1$ and $e_{2n}=(e_n,0)$ and $e_{2n+1}=(0,e_n)$. Furthermore, the product $e_pe_q=\pm e_{pq}$ where $pq$ is defined as the 'exclusive or' of the ...

6

Keep in mind that any finite-dimensional representation of a Lie group determines a finite-dimensional representation of its Lie algebra, and for a connected Lie group the induced Lie algebra representation determines the Lie group representation.
However, every finite-dimensional representation of $\operatorname{Lie}(\mathrm{Mp}(2,\mathbb R)) = \mathfrak{...

answered Jan 16 '13 at 5:46

Theo Johnson-Freyd

45.9k88 gold badges9898 silver badges281281 bronze badges

6

I don't think that, strictly speaking, the question by the OP has been properly addressed. Although it is true that real Clifford algebras are isomorphic (through a non-canonical isomorphism of unital associative algebras) to some matrix algebra or direct sum thereof, Clifford algebras are more than just matrix algebras.
Let $(V,g)$ be a real regular ...

6

To answer this question, we have to make a few rather natural assumptions, which I give in a slightly different notation.
Let the spinor bundle $S$ be associated to a Euclidean vector bundle $E$.
Assume that both bundles are locally trivialised over some open coordinate patch $U$ by orthonormal / unitary frames $e_1,\dots, e_n$, $\psi_1,\dots,\psi_N$, such ...

answered Dec 20 '15 at 11:40

Sebastian Goette

6,16422 gold badges2626 silver badges5454 bronze badges

5

You can find all what you want to know in the chapter 1 of the book X.Ma and G. Marinescu:Holomorphic Morse Inequalities and Bergman Kernels.
In Kahler case, all the metric connection is the chern connection.
And for the general case(hermitian manifold), the connection is called Bismut connection (see section 1.2.3). And the Weitzenböck formula is in ...

5

Update: "The eight Cayley-Dickson doubling products", Adv. Appl. Clifford Algebras 26 (2016) pp 529–551, doi:10.1007/s00006-015-0638-6, arXiv:1707.07318 I now find that 4 of those 8 should also be discarded: each allows zero divisors at the eight dimensional stage. The 4 remaining products are denoted in the paper as $P_0$, $P_3$ (the standard doubling ...

5

Consider the matrix $\sum_i a_iR_i$. One can show that it sends every vector of length 1 to a vector of length $\sqrt{\sum_i a_i^2}$. It follows that if the norm of $a=(a_i)$ is 1, then the matrix $\sum_i a_iR_i$ is orthogonal. In particular, all the matrices $R_i$ are orthogonal.
By multiplication from the left or from the right with an orthogonal matrix, ...

5

The essence of "geometric algebra" (better known as Kahler-Atiyah
algebra) is the classical Chevalley-Riesz isomorphism, which presents
the Clifford algebra of a quadratic space $(V,h)$ as a deformation
quantization of the exterior algebra of $V$. The systematic use of this
presentation allows for the automatic translation of various
computations ...

4

I think that you want to look at the relevant passages in Spin Geometry by Lawson and Michelsohn, particularly Chapter IV, Sections 9 and 10, where they explain in general how the square of a spinor can be written as a sum of differential forms. This gives a general way to see how having a parallel spinor (in any dimension) defines a set of parallel forms ...

answered Mar 5 '15 at 9:52

Robert Bryant

84.1k77 gold badges254254 silver badges358358 bronze badges

4

The answer for both questions 1, 2 is yes, the metric connection in Proposition 9.1.27 is the Levi-Civita induced connection if the background manifold is Kahler. When $X$ is only a Hermitian manifold, the result is not true. One such example is discussed in great detail in Bismut's paper
A local index theorem for non-Kahler manifolds, Math. ...

answered Jul 24 '13 at 13:41

Liviu Nicolaescu

29.7k22 gold badges7070 silver badges131131 bronze badges

4

A little bit of what you want can be found in Chapter 5 of Gracia-Bondia, Varilly, and Figueroa's book Elements of Noncommutative Geometry. They don't say much about subalgebras, I think, but they do prove (Lemma 5.7, page 182) the result that bivectors in the Clifford algebra $Cl(V)$ are closed under taking commutators, and that the adjoint action of ...

4

About 15 years ago I used Karoubi's description of $K$-theory to solve prove some cutting and pasting formula for the index families of elliptic problems. To do so I needed rephrase Karoubi's theory into something more flexible and more computable. Some of the interpretations I found might be relevant to your question. I will briefly describe ...

answered May 21 '13 at 10:04

Liviu Nicolaescu

29.7k22 gold badges7070 silver badges131131 bronze badges

4

One more example: for $d=8$ based on octonions. And this are all possible dimensions. $R_1$ takes vectors of length one to vectors of length one hence it is an isometry. By multiplying from the left by $R_1^{-1}$ we may assume that $R_1=\mathop{\rm Id}$. Then for every $v\in S^{d-1}$, the unit sphere in $\mathbb{R}^d$, the map $v\mapsto(R_2v,R_3v,\ldots,R_dv)...

answered Aug 19 '15 at 12:32

Adam Przeździecki

2,88611 gold badge1515 silver badges2424 bronze badges

4

This is an alternative way to see this question.
By argument given by the current answer
given by Ehud Meir we
may assume that $R_1$ is the identity matrix.
Then the hypotheses imply that for $i=2,\cdots,d$ the maps
$v\rightarrow R_i v$ are $d-1$ independent vector fields on the unit
sphere $S^{d-1}$.
But it is known that the maximal number of independent ...

Only top voted, non community-wiki answers of a minimum length are eligible

#### Related Tags

clifford-algebras × 100spin-geometry × 17

rt.representation-theory × 16

ra.rings-and-algebras × 15

dg.differential-geometry × 13

reference-request × 12

linear-algebra × 10

lie-groups × 10

mp.mathematical-physics × 10

kt.k-theory-and-homology × 9

at.algebraic-topology × 7

ag.algebraic-geometry × 6

quaternions × 6

octonions × 5

riemannian-geometry × 4

ct.category-theory × 3

cv.complex-variables × 3

homotopy-theory × 3

lie-algebras × 3

oa.operator-algebras × 3

noncommutative-algebra × 3

nt.number-theory × 2

gr.group-theory × 2

mg.metric-geometry × 2

matrices × 2