Is Associativity Key in Simplifying Multivector Products in Geometric Algebra?

[JonnyMaddox]

Hi, I just want to see if I understood this. Since the geometric product is associative and so on we can write for two multivectors A and B given by

A= \alpha_{0}+\alpha_{1}e_{1}+\alpha_{2}e_{2}+\alpha_{3}e_{1}\wedge e_{2}
B= \beta_{0}+ \beta_{1}e_{1}+\beta_{2}e_{2}+\beta_{3}e_{1}\wedge e_{2}

the geometric product multiplication as

AB=M=\mu_{0}+\mu_{1}e_{1}+\mu_{2}e_{2}+\mu_{3}e_{1}e_{2}

Where for example \mu_{0}=\alpha_{0}\beta_{0}+\alpha_{1}\beta_{1}+\alpha_{2}\beta_{2}-\alpha_{3}\beta_{3} and so on.

Now let's take an example with beautiful vectors with numbers like a = (e_{1}+2e_{2}), a_{1}=(2e_{1}+3e_{3}), a_{2}=(2e_{1}+0e_{2})

So aa_{1}= 8- e_1\wedge e_{2} Now what if I multiply this with a_{2}?

(8-e_{1}\wedge e_{2})(2e_{1})? Is it just 16e_{1}-(e_{1}\wedge e_{2})(2e_{1}) ? My logic behind this is that one can symbolically (GP) multiply multivectors and then opens up his list with like 300 different products or something and then evaluates each of the products? Is this right?

 

[stevendaryl]

Hi, I just want to see if I understood this. Since the geometric product is associative and so on we can write for two multivectors A and B given by

A= \alpha_{0}+\alpha_{1}e_{1}+\alpha_{2}e_{2}+\alpha_{3}e_{1}\wedge e_{2}
B= \beta_{0}+ \beta_{1}e_{1}+\beta_{2}e_{2}+\beta_{3}e_{1}\wedge e_{2}

the geometric product multiplication as

AB=M=\mu_{0}+\mu_{1}e_{1}+\mu_{2}e_{2}+\mu_{3}e_{1}e_{2}

Where for example \mu_{0}=\alpha_{0}\beta_{0}+\alpha_{1}\beta_{1}+\alpha_{2}\beta_{2}-\alpha_{3}\beta_{3} and so on.

Now let's take an example with beautiful vectors with numbers like a = (e_{1}+2e_{2}), a_{1}=(2e_{1}+3e_{3}), a_{2}=(2e_{1}+0e_{2})

So aa_{1}= 8- e_1\wedge e_{2} Now what if I multiply this with a_{2}?

(8-e_{1}\wedge e_{2})(2e_{1})? Is it just 16e_{1}-(e_{1}\wedge e_{2})(2e_{1}) ? My logic behind this is that one can symbolically (GP) multiply multivectors and then opens up his list with like 300 different products or something and then evaluates each of the products? Is this right?

To me, if you're dealing with geometric algebra, it's better to leave out the \wedge. For vectors, A \wedge B = \frac{1}{2}(AB - BA). If A and B are orthogonal, then A \wedge B = AB

So aa_{1}= 8- e_1 e_{2}.
Then when we multiply by 2 e_1 you just get:

8 e_1 - e_1 e_2 e_1

You can simplify using e_2 e_1 = -e_1 e_2 to get:
8 e_1 + e_1 e_1 e_2

Then you can use the fact that e_1 e_1 = 1 to get:
8 e_1 + e_2

(Note: this is assuming that your basis vectors are orthonormal:
e_i e_j + e_j e_i = 2 \delta_{ij})

 

[JonnyMaddox]

Hi stevendaryl,
I think I got it. If we can reformulate everything as GP then we can always use this simple relations between the orthogonal basis vectors, hm that was more then obvious...ok thank you.

When you multiply 8-e_{1}e_{2} by 2e_{1} where did you leave the factor of 2 in the result? I think it should be a rotation by 90 degrees anti-clockwise and dilation by factor 2?

Thx for your reply !

 

[stevendaryl]

Hi stevendaryl,
I think I got it. If we can reformulate everything as GP then we can always use this simple relations between the orthogonal basis vectors, hm that was more then obvious...ok thank you.

When you multiply 8-e_{1}e_{2} by 2e_{1} where did you leave the factor of 2 in the result? I think it should be a rotation by 90 degrees anti-clockwise and dilation by factor 2?

Thx for your reply !

That was a typo, sorry.