Geometric Algebra Pitfalls

After studying more than I expected to during the holidays, I believe I have finally got my wits together and can now give some clear hints to anyone else who wishes to travel this path.  I highly suggest "Linear and Geometric Algebra" by Alan Macdonald.  Read the previews and details online before considering to purchase this book.  I like it, doesn't mean you will too.

First, and most importantly, write out your multi vectors explicitly.  For example, vector (2, 3, 4) should be written as 2e₁ + 3e₂ + 4e₃.  Recall the multiplication of the e's (or basis as that is what they form) is not associative.

Consider (1e₁ + 2e₂)(3e₁ + 4e₂).

Intuitively, from the good ol' days, we would do: (1e₁3e₁ + 1e14e2 + 2e₂3e₁ + 2e₂4e₂).

Regrouping the scalers together: (3e₁e₁ + 4e₁e₂ + 6e₂e₁ + 8e₂e₂)

Since e₁e₁ = 1: (3 + 4e₁e₂ + 6e₂e1 + 8)

Switching the place of two adjacent e's flips the sign, ie. e₁e₂ = -e₂e₁: (3 + 4e₁e₂ - 6e₁e₂ + 8)

Final grouping: (11 - 2e1e2)

Notice that, save for the special rules surrounding the es, it is very much as was learned before from the days of multiplying two expressions consisting of x and y, such as (2x+y)(4z+8w).

If a problem in geometric algebra seems difficult, my first reaction has been to ensure that everything uses e explicitly.

Second, when reading articles and books, always be sure to know which dimension the given results apply in.  I have yet to find an article that does not give examples or definitions at a lower dimension before jumping to the dimensionless forms thanks to the desire to give out something practical without having to face a wall of mathematics.

Third, the inner (also known as the dot) and outer products are filters on the geometric product.  If the result does not match the product, it is zero, and we move on.

Again, let's return to (1e1 + 2e2)(3e1 + 4e2).
The result with the geometric product was: (11 - 2e1e2).
Let us consider the dot product of (1e1 + 2e2) and (3e1 + 4e2).  The grade of 1e1 is 1 (to obtain the grade, count the number of unique e's after the scalar).  When multiplying two elements together, the dot product says the grade of the result must equal the grade of the second element minus the grade of the first.  So when we take the inner product 1e1 with 3e1 where the grade(3e1) - grade(1e1) = 1 - 1 = 0, therefore the result must be a scalar, which it is (3).

The outer product filter is the sum of the grades.

The filter applies only to the two elements being multiplied together at a time, not to the whole multivectors.  After the product is done, the results are added as usual.

Finally, just keep on finding more sources.  I found it easier to work with the topic once I had a dozen articles each tackling the topic slightly differently.  If my mental model of how the geometric algebra was accurate, then each paper I would read would not conflict with the mental model.

Edits:

1. Since my initial hints were leading people astray.  They exemplified my lack of understanding, which should be resolved.
2. Reorganization for clarity.