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- Why parity have vectors and pseudovectors? why not only vectors?

**Summary:**Why parity have vectors and pseudovectors? why not only vectors?

I am reading Griffiths "Introduction to elementary particle physics" Ed.1.

The book obviously is an undergraduate introduction.Thus, not much detail is presented, but I cannot get my head around pseudovectors (pseudoscalars as well).

Parity in the book is just an inversion, that is to say, a mirror reflection followed by a ##180^{\text{o}}## rotation. It is easy to show that the parity group is just ##\{P,I\}##, Where ##P## is the inversion matrix (or ##-I##). For a vector ##\text{v}\in R^{3}##:

$$P\text{v}=-\text{v}$$

This is how Griffiths explains it. He follows this by saying that for a cross product ##\text{w}=\text{u}\times\text{v}##, parity does not give ##-\text{w}##, because ##P## is distributive over cross products:

$$P(\text{w})=P(\text{u}\times\text{v})=P(\text{u})\times P(\text{v})=(-\text{u})\times (-\text{v})=\text{u}\times\text{v}=\text{w}$$

It looks to me that he is forcing ##P## to be a linear transformation over cross products. But ##\text{w}## is a vector and the solution must be ##-\text{w}##, one way or another. Unless, ##\text{w}## is not a vector (a pseudovector)**. But, in this sense, what he presented is not a cross product to begin with!

Can someone help me understand why parity have pseudo's?

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*Around three years ago, I read a book about geomtric algebra (didn't complete it), it talked about bivectors and pseudovectors. Could this be the same thing? Because the explanation Griffiths presented is very incomplete in my opinion.*