# Deriving reduction formula in Geometric Algebra

I am trying to learn Geometric Algebra by going through the book "New Foundations for Classical Mechanics" by David Hestenes.

I was reading the part about reduction formula (shown below) but couldn't get the result the shown in the book. Can someone show me how iterating (1.15) gives the reduction formula?

notation in the book:

• dot (.) is inner product
• circumflex (^) is outer product
• ab mean geometric product of a and b
• inner and outer product have precedence over geometric product unless indicated by parentheses

Thank you

# update 1

I can easily apply (1.15) for the first iteration and get

$$a \cdot (a_1 \wedge a_2 \wedge \cdots \wedge a_r) = (a \cdot a_1) (a_2 \wedge \cdots \wedge a_r) - a_1 \wedge (a \cdot (a_2 \wedge a_3 \cdots \wedge a_r))$$

I can see that I should apply (1.15) to the term

$$- a_1 \wedge (a \cdot (a_2 \wedge a_3 \cdots \wedge a_r))$$

but there is a $$- a_1 \wedge$$ in the term which will get inherited if I apply (1.15) directly, what should I do to get rid of that?

## 1 Answer

Let $$b=a_1$$ and $$C_{r-1}=a_2\wedge a_3\wedge\dots\wedge a_r$$. Then, from (1.15) it follows that $$a\cdot(a_1\wedge a_2\wedge\dots\wedge a_r)=a\cdot a_1 a_2\wedge\dots\wedge a_r$$ $$-a_1\wedge(a\cdot a_2\wedge\dots\wedge a_r).$$ Next you apply the same (1.15) to the second factor in the second piece above with $$b=a_2$$ and $$C_{r-2}=a_3\wedge a_4\wedge\dots\wedge a_r$$ to get $$=a\cdot a_1 a_2\wedge\dots\wedge a_r-a\cdot a_2 a_1\wedge a_3\dots\wedge a_r+$$ $$+a\cdot a_3 a_1\wedge a_2\wedge a_4\wedge\dots\wedge a_r+\textrm{next term}.$$ One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $$C_1$$ which is a vector, and the inner product with $$C_1$$ is the usual dot product of vectors.

Here is how you proceed: $$-a_1\wedge(a\cdot (a_2\wedge\dots\wedge a_r))=$$ $$=-a_1\wedge[(a\cdot a_2a_3\wedge\dots\wedge a_r)-a_2\wedge(a\cdot(a_3\wedge a_4\dots\wedge a_r))]=$$ $$=-(a\cdot a_2)a_1\wedge a_3\wedge\dots\wedge a_r+a_1\wedge a_2\wedge(a\cdot(a_3\wedge a_4\dots\wedge a_r))$$ Observe that the "$$\cdot$$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $$a\cdot a_3$$ which can be pulled out and another term of the form $$a_1\wedge a_2\wedge (a_3\cdot(a_4\wedge a_5\dots\wedge a_r))$$ where $$\cdot$$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.

• I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem. – Eric Feb 3 at 7:16
• @Eric I just improved my answer to include the details of the computations. Please let me know. – GReyes Feb 4 at 0:44
• Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D – Eric Feb 4 at 6:51