0
votes
1answer
87 views

Bivector as a sum of exterior products of basis vectors.

Prove that for any 2-vector $\alpha=\Lambda^2(V)$ there is a basis $\{e_1,\ldots, e_k\}$ of $V$ such that $\alpha= e_1\wedge e_2 + e_3\wedge e_4 + \ldots +e_{k-1}\wedge e_k$, where $\wedge$ denotes ...
2
votes
3answers
174 views

index free proof of dot product of two wedge products

I am learning geometric algebra, and meet an identy of (edited according to Andrey's comments below) $$ (a\wedge b)\cdot(c\wedge d) = (a \cdot d)(b\cdot c) - (a \cdot c)(b \cdot d)$$ as in wiki ...
4
votes
1answer
207 views

On Chevalley's linear identification of the Clifford algebra $C(\mathbf p)$ with the exterior algebra $\wedge \mathbf p$

In reading Sternberg's notes on Clifford algebras and spin representations (page 148) I encountered the following: "...Consider the linear map $$C(\mathbf p)\rightarrow \wedge \mathbf p, x\mapsto ...
2
votes
1answer
116 views

Associativity of Moyal-like products

The Moyal product of two smooth functions $f,g$ on $\mathbb R^{2n}$ can be defined as $$ f\star g = \exp\left(-\omega^{ij} \frac{\partial}{\partial y^i} \frac{\partial}{\partial z^j}\right) f(y)g(z) ...
3
votes
2answers
265 views

How do I evaluate the Clifford product in dimensions greater than 3?

The Clifford product of a pair of vectors $a,b$ is an associative operation defined by $$ ab = a \cdot b + a \wedge b.$$ In sufficiently low dimensions I am used to being able to define the Clifford ...