# Geometric product

1. Aug 21, 2014

Hey JO,
I'm reading a book on geometric algebra and in the beginning (there was light, jk) a simple calculation is shown:
Geometric product is defined as:
$ab = a \cdot b + a \wedge b$
or
$ba = a \cdot b - a\wedge b$

Now
$(a\wedge b)(a \wedge b)=(ab-a \cdot b)(a\cdot b - ba)$
$=-ab^{2}a-(a \cdot b)^{2}+a \cdot b(ab+ba)$
$=(a \cdot b)^{2}-a^{2}b^{2}$
$=-a^{2}b^{2}sin^{2}(\phi)$

I think this term $a \cdot b(ab+ba)$ has to vanish somehow, but it is $(a\cdot b)^{2}$ and that doesn't make sense :( Any suggestions?

Ok I know the answer, the term is $2(a\cdot b)^{2}$. But thank you for your attention !

Last edited: Aug 21, 2014
2. Aug 22, 2014

### Staff: Mentor

So how did you get from the next to last step to your solution?

3. Aug 22, 2014

### jhae2.718

From the original post I'm going to assume the question is answered, so here are the steps to the final solution for anyone interested (noting that I've only done a bit of work with geometric algebra after seeing a talk by David Hestenes...):

The quantity $\a\b + \b\a$ reduces to $2\a\cdot\b$ as stated, so the last term becomes $(\a\cdot\b)(2\a\cdot\b)$ or $2(\a\cdot\b)^2$. The whole RHS then reduces to $(\a\cdot\b)^2 - \a^2\b^2$.

The contraction property states that $\a\a = \a^2 = |\a|^2$, so this becomes $|\a\cdot\b|^2 - a^2b^2$. The dot operator is just the inner product, so $|\a\cdot\b| = ab\cos\phi$. Using the Pythagorean identity (there may be a more geometric algebra way to do this, but it's late and I am tired), the whole thing then reduces to $-a^2b^2\sin^2\phi$.

4. Aug 23, 2014

### Staff: Mentor

Thanks for the explanation, I was thinking the math was more involved with some condition we weren't told about.

5. Aug 23, 2014

### jhae2.718

No problem, it gave me a reason to dig out my copy of New Foundations for Classical Mechanics and play with that. :D

6. Aug 26, 2014

In the book the unit trivector is defined like this: $(e_{1}e_{2})e_{3}=e_{1}e_{2}e_{3}$
But that would mean $(e_{1}e_{2})e_{3}= (e_{1} \wedge e_{2})\cdot e_{3}+(e_{1} \wedge e_{2} \wedge e_{3})$ But I thougt it is just $e_{1} \wedge e_{2} \wedge e_{3}$? I could somehow imagine in my head that the plane spanned by $e_{1} \wedge e_{2}$ is perpendicular to the line $e_{3}$ but I'm not sure that it works like that. Is that right? Would make sense.
Ok but why is this true $(e_{1}\wedge e_{2})e_{1}=(-e_{2}e_{1})e_{1}=-e_{2}e_{1}e_{1}=-e_{2}$ because $(e_{1}\wedge e_{2})e_{1}=(e_{1} \wedge e_{2})\cdot e_{1}+ e_{1}\wedge e_{2} \wedge e_{1}$ where the last term is zero. I don't know how the term $e_{1} \wedge e_{2}$ interacts as dot product with $e_{1}$. Similar is $(e_{1}\wedge e_{2})(e_{2}\wedge e_{3})=e_{1}e_{3}$ why isn't this just zero? Because $(e_{1}\wedge e_{2})(e_{2} \wedge e_{3})=(e_{1}\wedge e_{2}) \cdot (e_{2} \wedge e_{3})+e_{1}\wedge e_{2} \wedge e_{2} \wedge e_{3}$ any help?