### Massoud Saidi

Shared publicly -**point and grok teaching**:

*Part 2: Multiplication in geometric algebra*. While i still think my previous post on

*Clifford algebra*is a nice point of first contact, i had to greatly shorten my notes. In this post i'll try and convey how these spaces feel, and show you some computations.

Part 1:

*"Into Clifford algebra"*, if you haven't, look here:

https://plus.google.com/u/0/115434895453136495635/posts/1CtiDXou6Rq

**Legend**: We have

*cliffs*

**a**,

**b**where

**a**has

*components*named

**xᵢ**, and

**b**'s are named

**yᵢ**.

A 3d

*cliff*has 8 components. A product

**a·b**expands to 8·8 = 64 subfactors, mucho tedious! But we can look at grade-n

*subspaces*(n-blades), and try to understand multiplying

*blades*. Cl(3,0) has 4

*blades*:

**a₀ = x₀**

**a₁ = x₁e₁ + x₂e₂ + x₃e₃**

**a₂ = x₁₂e₁₂ + x₂₃e₂₃ + x₁₃e₁₃**

**a₃ = x₁₂₃e₁₂₃**

And multiplying bladewise is straightforward:

**a·b = (a₀ + a₁ + a₂ + a₃)·(b₀ + b₁ + b₂ + b₃)**

The result's just a big sum with every pairing of

**aᵢbⱼ**. So it really is enough to understand multiplying blades. The product of anything with a 0-blade is simple stretching by

**x₀**:

**a₀·b₁ = x₀·(y₁e₁ + y₂e₂ + y₃e₃)**

Multiplying 1-blades

**a₁·b₁**is equivalent to simultaneously calculating

*dot*- and

*wedge product*of the respective

*vectors*, returning the

*dot product*as

*scalar*, and the

*wedge product*as 2-blade:

**c = a₁·b₁**consists of:

**c₀ = a₁∙b₁**

**c₂ = a₁∧b₁**

Nice as that may be, i didn't know to wedge multiply before and the above would have told me nothing. And it's also kind of a special case, so let's look at an

*inhomogeneous*product

**a₁·b₂**instead. For a quick impression we can just expand one side, leaving the other as is:

**a₁·b₂ = x₁e₁·b₂ + x₂e₂·b₂ + x₃e₃·b₂**

The following will lead to a single parametrized summation expression, which is nicely short, and obscure. We're tracking here what happens to the inputs of a multiplication, but there's an easier way to compute a

*geometric product*i'll show you further down.

Where was i? Here, the three summands seem similar, and we'll try to look at a single one first:

**x₁e₁·b₂ = x₁e₁·(y₁₂e₁₂ + y₂₃e₂₃ + y₃₁e₃₁)**

**= x₁y₁₂e₁e₁₂ + x₁y₂₃e₁e₂₃ + x₁y₃₁e₁e₃₁**

**= x₁y₁₂e₂ + x₁y₂₃e₁₂₃ – x₁y₃₁e₃**

By the way, in the last step i've been using these identities:

**e₁e₁ = 1**

**e₁e₂₃ = e₁₂₃**

**e₁e₃₁ = e₁e₃e₁ = –e₁e₁e₃ = –e₃**

You see, those

**xᵢ**and

**yⱼ**again appear in all possible pairings

**xᵢyⱼ**. Here's a short and crisp summation form:

**a·b = ∑ᵢ∑ⱼ xᵢ·yⱼ·eᵢ·eⱼ**

**eᵢ·eⱼ**is the multiplication table i invited you to calculate yourself in the first post. I've seen people getting clever to encode it, but instead of computing where it all goes, we could also work in reverse...

**a better way to multiply**

So you

*do*want to see the componentwise results. The scalar part is again extra-simple:

**z₀ = (x₀y₀ + x₁y₁ ... + x₁₂y₁₂ + ... + x₁₂₃y₁₂₃ )·1**

The next one

**z₁**is such that we must sum those that have a single 1 in their index, and zero or two of the others. I first listed all the

**xᵢ**, and then made up the

**yⱼ**to get single remaining indices of 1. Here's the end result:

**x₀y₁**

**x₁y₀**

**–x₂y₁₂**

**x₃y₃₁**

**x₁₂y₂**

**–x₂₃y₁₂₃**

**–x₃₁y₃**

**x₁₂₃y₂₃**

I added minus signs whenever i'd need to swap indices an odd number of times to get doubles to cancel. Take for example

**x₂₃**: We need to add a 1 and remove 2 and 3, so the other has to be

**y₁₂₃**. To get the signe we only need to look at the indices:

23123

23213

22313

22133

Three swaps in between, thats an odd number,

**x₂₃y₁₂₃**gets a minus sign. Let me write down

**z₁**a bit more horizontally for you:

**z₁ = (x₀y₁ + x₁y₀ – x₂y₁₂ + x₃y₃₁ + x₁₂y₂ – x₂₃y₁₂₃ – x₃₁y₃ + x₁₂₃y₂₃)·e₁**

Note that for higher grades you further need to bring the indices into the right order, but thats all then. This has been even easier, i'm sure you can now compute the other

**zᵢ**! Have fun!

**references**

*Cayley graph*and multiplication table i found on

**Martin Baker**'s excellent introduction to

*geometric algebra*. Note that the magenta products are

*commutative*, the turqoise ones

*anticommute*[update: fixed, had it the other way around before]:

http://www.euclideanspace.com/maths/algebra/clifford/d3/

The picture of

**David Hestenes**is from this interesting article by

**Emily Hanford**about Hestenes:

*"The Problem with Lecturing"*

http://americanradioworks.publicradio.org/features/tomorrows-college/lectures/problem-with-lecturing.html

#sundayscience or #sundaymathematics rather:

#computing products in #geometric - or #clifford #algebra

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