I can't find good information on the internet.
In my mathematical physics class the definition of a vector was given as:
That object with magnitude and direction which doesn't change under inversions. Pseudovectors do change.
Scalars are that magnitude that don't change with inversions. Pseudoscalars do change.
Inversions were loosely defined as inverting all components of a vector.
The examples were given as:
Say $\vec{A}=\vec{B}\times\vec{C}$; under the inversion it becomes $\vec{-B}\times\vec{-C}$ which is again, $\vec{A}$ which then has to be a pseudovector.
Say $A=\vec{a}\cdot\vec{b}\times\vec{c}$ then under inversion $\vec{-a}\cdot\vec{-b}\times\vec{-c}$, which is $-A$ and has to be a pseudoscalar.
I don't get it though. Why are inversions done on the individual vectors and not to the whole product?
What really is an inversion?
Where could I read more about this?
Thanks.
In the context of abstract mathematics, these are definitions of how to transform objects. More specifically, the rule is: if you invert your coordinate axis (taking $ \hat{x} \rightarrow - \hat{x} $ etc.), then the components of a vector negate, the components of a pseudovector stay the same.
These definitions are motivated from objects in physics. It is a law/fact/assumption of physics whether a certain physical quantity is a pseudo vector/scalar or genuine vector/scalar. The laws of physics relating different quantities are then preserved under coordinate inversion - meaning precisely that if you transform each guy appearing in the equation by his transformation rule, the equation should stay the same. This is why in those examples, the inversions are done on individual vectors.
For example, it is a fact that position and velocity are vectors and angular momentum is a pseudovector. It had to be this way, because the quantities are related by $ \vec{L} = \vec{r} \times m \vec{v} $ . If $ \vec{L} $ was defined to be a vector then upon coordinate inversion $ \vec{L} = \vec{r} \times m\vec{v} \rightarrow - \vec{L} = \vec{r} \times m \vec{v} $, which changes the equation. With $ \vec{L} $ defined as a pseudovector, there is not this problem.