Can a vector have a wedge product with a scalar? And a geometric product?

Question

One of the necessary requirements of a vector space is to be endowed with the multiplication by a scalar. I am now interested in other operations, the wedge and the geometric product in geometric algebras.

Can a vector V and a scalar s have a wedge product between each other (V ∧ s)?

If yes: can they have a geometric product between each other as well?

If yes, since the geometric product would be defined (V·s)+(V ∧ s), how exactly are we supposed to handle the inner product component (V·s) given that you can't have an inner product between a vector and a scalar?

Answer 1

If you’ve read any reasonable treatment of geometric algebra, they explain that the algebra contains scalars, vectors, and higher ordered elements, their sums and their products.

The only place where $ab=a\cdot b+a\wedge b$ is relevant is where $a,b$ are vectors. It is not something that holds for every $a,b$ in the algebra. I think people usually think of it as a derivative identity, not a defining one.

Answer 2

The formula $$ ab = a\cdot b + a\wedge b $$ is true when $a$ and $b$ are vectors, but is usually very wrong in any other case.

The wedge product of a vector and a scalar is the same as the scalar product.

In the Euclidean case, there is a natural inner product on the entire algebra given by $$ (X, Y) \mapsto \langle\widetilde XY\rangle $$ where $X, Y$ are arbitrary multivectors, $\widetilde X$ is the reverse of $X$, and $\langle\cdot\rangle$ is the scalar part. Applying this inner product to a scalar and a vector yields 0; in fact, it will yield zero whenever $X$ and $Y$ have different grades.

Two good references giving an introduction to geometric algebra are Geometric Algebra for Physicists by Chris Doran and Anthony Lasenby, and Clifford Algebras and Spinors by Pertti Lounesto.