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I am learning geometric algebra from David Hestene's New foundations to classical mechanics, in it the geometric product of two vector is defined the following way:

$$ ab = a \cdot b + a \wedge b$$

And, the product of a vector with an $r$ blade is given as:

$$ aA_r = a \cdot A_r + a \wedge A_r$$

Where $a$ is the vector and $A_r$ is the $r$ blade. Now, what would be the product rule for an $s$ blade with an $r$ blade?

$$ A_r \wedge A_s=???$$

How would the answer change with the ambient dimensions changing?

tryst with freedom
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1 Answers1

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The best way to view all of these identities is consequences of grade selection. We say that scalars have grade 0, vectors grade 1, bivectors are grade 2, and so forth. Examples, assuming that $ \left\{ { \mathbf{e}_1, \mathbf{e}_2, \cdots } \right\} $ is an orthonormal Euclidean basis, then $$-3, \pi, 7, 9, \cdots$$ are all scalars with grade 0, $$ \mathbf{e}_1, \mathbf{e}_2, 3 \mathbf{e}_3 + \pi \mathbf{e}_7, \cdots$$ are all vectors with grade 1, $$ \mathbf{e}_1 \mathbf{e}_2, \mathbf{e}_2 \mathbf{e}_3, 3 \mathbf{e}_3 \mathbf{e}_7 + \pi \mathbf{e}_7 \mathbf{e}_1, \cdots$$ are all bivectors and have grade 2. We write $$ {\left\langle{{A}}\right\rangle}_{{k}}$$ as the selection of all grade $ k $ elements from the multivector $ A $. For example, if $$ A = -3 + \pi+ 7+ 9 + \mathbf{e}_1+ \mathbf{e}_2+ 3 \mathbf{e}_3 + \pi \mathbf{e}_7 + \mathbf{e}_1 \mathbf{e}_2+ \mathbf{e}_2 \mathbf{e}_3+ 3 \mathbf{e}_3 \mathbf{e}_7 + \pi \mathbf{e}_7 \mathbf{e}_1,$$ then $$ {\left\langle{{A}}\right\rangle}_{{0}} = \pi+ 13,$$ $$ {\left\langle{{A}}\right\rangle}_{{1}} = \mathbf{e}_1+ \mathbf{e}_2+ 3 \mathbf{e}_3 + \pi \mathbf{e}_7 ,$$ $$ {\left\langle{{A}}\right\rangle}_{{2}} = \mathbf{e}_1 \mathbf{e}_2+ \mathbf{e}_2 \mathbf{e}_3+ 3 \mathbf{e}_3 \mathbf{e}_7 + \pi \mathbf{e}_7 \mathbf{e}_1,$$ and $ {\left\langle{{A}}\right\rangle}_{{k}} $ for all $ k > 2 $ is zero (in this case.)

Now, we can consider the geometric product of two vectors $ a = \sum_i x_i \mathbf{e}_i $ and $ b = \sum_i y_i \mathbf{e}_i $ $$\begin{aligned} a b &= \sum_{i,j} x_i \mathbf{e}_i y_j \mathbf{e}_j \\ &= \sum_{i = j} x_i y_i \mathbf{e}_i^2 + \sum_{i \ne j} x_i y_j \mathbf{e}_i \mathbf{e}_j \\ &= \sum_{i} x_i y_i + \sum_{i \ne j} x_i y_j \mathbf{e}_i \mathbf{e}_j \\ &= \sum_{i} x_i y_i + \sum_{i < j} \left( { x_i y_j - x_j y_i } \right) \mathbf{e}_i \mathbf{e}_j \\ \end{aligned}$$ Observe that the first sum is a scalar, and the second sum has only grade two terms. This means that for any product of two vectors we must have $$ a b = {\left\langle{{a b}}\right\rangle}_{{0}} + {\left\langle{{a b}}\right\rangle}_{{2}}.$$ Clearly we wish to identify the scalar component as the dot product, and we define the bivector component to be the wedge product: $$ a \cdot b = {\left\langle{{ a b }}\right\rangle}_{{0}},$$ $$ a \wedge b = {\left\langle{{ a b }}\right\rangle}_{{2}}.$$

You should be able to convince yourself that a product of a vector and r-blade $ A_r $ (an entity with only grade r components) must have only grades $ r-1 $ and $ r + 1 $, so $$ a A_r = {\left\langle{{a A_r}}\right\rangle}_{{r-1}} + {\left\langle{{a A_r}}\right\rangle}_{{r+1}},$$ so we define $$ a \cdot A_r = {\left\langle{{ a A_r }}\right\rangle}_{{r-1}},$$ $$ a \wedge A_r = {\left\langle{{ a A_r }}\right\rangle}_{{r+1}},$$ so that $$ a A_r = a \cdot A_r + a \wedge A_r.$$

Still more generally, answering your final question, one can define the dot and wedge products of two blades $ A_r $, $ B_s $ with grades $ r, s $ respectively, as $$ A_r \cdot B_s = {\left\langle{{ A_r B_s }}\right\rangle}_{{\left\lvert {r-s} \right\rvert}},$$ $$ A_r \wedge B_s = {\left\langle{{ A_r B_s }}\right\rangle}_{{r+s}}.$$ Note that there is some variability in conventions and notations for dot product like operators, but these are definitions that are consistent with those used in New Foundations.

Peeter Joot
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  • So, the geometric product is most fundamental because it just involves the algebra bash. But, you cheated a bit. You used basis, the book hasn't introduced it (to my knowledge) – tryst with freedom Sep 04 '21 at 05:30
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    Just because geometric algebra allows for coordinate free logic in some places, you need not be afraid to use a basis nor coordinates when appropriate. Both are fundamental aspects of vector spaces, and every geometric product space is built upon a vector space augmented with a dot product. – Peeter Joot Sep 04 '21 at 13:21
  • Perhaps not directly relate: But does GA assume an orthonormal basis? @Peter Joot? – tryst with freedom Sep 04 '21 at 15:33
  • Just as in vector algebra, one can use a non-orthonormal basis if desired. That will often be the case if you are using a curvilinear coordinate system, where the basis direction and magnitude varies point to point. However, you can always take a non-orthonormal basis, and run Gram-Schmidt to find an orthonormal one (but care is required if the vector space for the algebra can have null vectors). – Peeter Joot Sep 04 '21 at 19:20
  • also, if you are using studying relativistic, projective, or conformal geometric algebras, your basis vectors are liable to have either +,-, or 0 squares. However, the identification of the scalar part of a vector product will still be the dot product, even for such non-Euclidean spaces. – Peeter Joot Sep 04 '21 at 19:23