By Eric Chisolm

This can be an advent to geometric algebra, an alternative choice to conventional vector algebra that expands on it in ways:

1. as well as scalars and vectors, it defines new items representing subspaces of any dimension.

2. It defines a product that is strongly influenced by means of geometry and will be taken among any gadgets. for instance, the made from vectors taken in a definite method represents their universal plane.

This method used to be invented by way of William Clifford and is usually referred to as Clifford algebra. it is really older than the vector algebra that we use this day (due to Gibbs) and contains it as a subset. through the years, a variety of components of Clifford algebra were reinvented independently via many folks who stumbled on they wanted it, frequently no longer knowing that every one these components belonged in a single method. this implies that Clifford had the appropriate proposal, and that geometric algebra, now not the diminished model we use this day, merits to be the traditional "vector algebra." My aim in those notes is to explain geometric algebra from that viewpoint and illustrate its usefulness. The notes are paintings in growth; i will preserve including new themes as I research them myself.

https://arxiv.org/abs/1205.5935

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**Additional info for Geometric Algebra**

**Sample text**

Therefore we proceed as follows. 1. Let b1 = a1 . 2. For each j starting with 1, let Bj = b1 ∧ · · · ∧ bj . 3. Then let bj+1 = aj+1 ∧ Bj Bj−1 . This procedure will work only if each Bj is invertible, which is why it is normally used only in Euclidean spaces. If Ar is a blade, then A⊥ r represents the orthogonal complement of Ar . That means that orthogonal projection into A⊥ should equal orthogonal rejection from Ar . Using Eqs. (181) and (183), this is easy to r show directly. ⊥ −1 −1 a ⌋ A⊥ = (a ∧ Ar )⊥ (A⊥ r (Ar ) r) = a ∧ Ar I −1 IA−1 r = a ∧ Ar A−1 r .

1. (You may wonder why I did this instead of just leaving in the (−1)r . ) Another way to arrive at this formula is to write Ar = a1 a2 · · · ar and reflect a along each of the aj in succession. Once again, an expression in terms of vectors generalizes to subspaces with only minimal change. Reflections in subspaces also preserve inner products; the proof is very similar to Eqs. (238). 2. Reflecting a multivector in a subspace Now that I can reflect vectors, I can reflect subspaces too: the reflection of subspace Bs in subspace Ar is found by taking every vector from Bs , reflecting it in Ar , and seeing what subspace you get.

A) If Ar and Bs are orthogonal, then Ar Bs = Ar ∧ Bs . (b) The converse is true if either (1) r = 1 or s = 1 or (2) Ar or Bs is invertible. Proof. To begin, I note that Ar can be written a1 a2 · · · ar where the ai are orthogonal to each other; similarly, Bs can be expressed b1 b2 · · · bs where the bj are also orthogonal to each other. Now suppose that Ar and Bs are orthogonal; then all of the ai and bj are orthogonal to each other as well, so now I can use the rule that the product of orthogonal vectors equals their outer product to get Ar Bs = a1 a2 · · · ar b1 b2 · · · bs = a1 ∧ a2 ∧ · · · ∧ ar ∧ b 1 ∧ b 2 ∧ · · · ∧ b s = Ar ∧ Bs .