A What Is the Role of 'n' and the Virgule in the Wedge Product Formalism?

[Bill_1]

TL;DR

Formalism of wedge product

Hi guys!
I have some difficulty to correctly understand the following formalism.

Specifically the using of virgule and the "n" in aink expression.

Does the virgule have some rapport with permutation or combination?
And what represent the n in the expression?

Here you can find the book: https://www2.tulane.edu/~ftbirtel/wedge product.pdf

Thanks for your help!

 

[jedishrfu]

are you referring to the wedge operator ^ as a virgule?

It's not. The wedge operator is a real operator used in differential geometry to represent the exterior product between two differential forms.

It's an anti-commutative operator, i.e., a^b = - b^a, where a and b are differential forms of the same type.

https://en.wikipedia.org/wiki/Exterior_algebra

 

[fresh_42]

Your formula is simple the distributive law, or the bilinear property of the wedge product:
$$
(\vec{a}+\vec{b})\wedge (\vec{c}+\vec{d}) = \vec{a}\wedge (\vec{c}+\vec{d})+\vec{b}\wedge (\vec{c}+\vec{d})=\vec{a}\wedge \vec{c}+\vec{a}\wedge \vec{d}+\vec{b}\wedge \vec{c}+\vec{b}\wedge \vec{d}
$$
and the anti-kommutativity $a\wedge b=-b\wedge a$ that allows us to sort the vectors at the cost of one minus sign per permutation. $n$ is the dimension of the vector spaces where the $v_j$ and basisvectors $e_k$ were taken from. And of course
$$
\vec{v}\wedge \alpha\vec{w}=\alpha\vec{v}\wedge \vec{w}\, , \,(\alpha \in \mathbb{R})
$$

Note that $\vec{v}\wedge \vec{v}=0.$ The wedge product over linearly dependent vectors is zero.

 

[WWGD]

I think the " virgule", which means " comma" in French, represents indices and subindices in your pic. The wedge operator is part of Geometric Algebra, where you glue objects together ( parallelepipeds), to build larger objects. Why don't you look it up and ask followups?