I am reading some geometric algebra notes. They all started from some axioms. But I am still confused on the motivation to add inner product and wedge product together by defining
$$ ab = a\cdot b + a \wedge b$$ Yes, it can be done like complex numbers, but what will we lose if we deal with inner product and wedge product separately? What are some examples to show the advantage of geometric product vs other methods?
[Math] What’s the motivation to add inner product and wedge product together in geometric product
abstract-algebraclifford-algebrasdifferential-geometrygeometric-algebrassoft-question
Related Solutions
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.
Some authors define the geometric product in terms of the dot and wedge product, which are introduced separately. I think that accentuates an apples vs oranges view. Suppose instead you expand a geometric product in terms of coordinates, with $ \mathbf{a} = \sum_{i = 1}^N a_i \mathbf{e}_i, \mathbf{b} = \sum_{i = 1}^N b_i \mathbf{e}_i $, so that the product is $$\mathbf{a} \mathbf{b}= \sum_{i, j = 1}^N a_i b_j \mathbf{e}_i \mathbf{e}_j= \sum_{i = 1}^N a_i b_i \mathbf{e}_i \mathbf{e}_i+ \sum_{1 \le i \ne j \le N}^N a_i b_j \mathbf{e}_i \mathbf{e}_j.$$ An axiomatic presentation of geometric algebra defines the square of a vector as $ \mathbf{x}^2 = \left\lVert {\mathbf{x}} \right\rVert^2 $ (the contraction axiom.). An immediate consequence of this axiom is that $ \mathbf{e}_i \mathbf{e}_i = 1$. Another consequence of the axiom is that any two orthogonal vectors, such as $ \mathbf{e}_i, \mathbf{e}_j $ for $ i \ne j $ anticommute. That is, for $ i \ne j $ $$\mathbf{e}_i \mathbf{e}_j = - \mathbf{e}_j \mathbf{e}_i.$$ Utilizing these consequences of the contraction axiom, we see that the geometric product splits into two irreducible portions $$\mathbf{a} \mathbf{b}=\sum_{i = 1}^N a_i b_i+ \sum_{1 \le i < j \le N}^N (a_i b_j - b_i a_j) \mathbf{e}_i \mathbf{e}_j.$$ The first sum (the symmetric sum) is a scalar, which we recognize as the dot product $ \mathbf{a} \cdot \mathbf{b}$, and the second (the antisymmetric sum) is something else. We call this a bivector, or identify it as the wedge product $\mathbf{a} \wedge \mathbf{b}$.
In this sense, the dot and wedge product sum representation of a geometric product, are just groupings of terms of a larger integrated product.
Another way of reconciling the fact that we appear able to add two unlike entities, is to recast the geometric product in polar form. To do so, consider a decomposition of a geometric product in terms of constituent unit vectors $$\mathbf{a} \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert \left( { \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} + \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right),$$ and assume that we are interested in the non-trivial case where $ \mathbf{a} $ and $ \mathbf{b} $ are not colinear (where the product reduces to just $ \mathbf{a} \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert $). It can be shown that the square of a wedge product is always non-positive, so it is reasonable to define the length of a wedge product like so $$\left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert = \sqrt{-(\hat{\mathbf{a}} \wedge \hat{\mathbf{b}})^2}.$$
We can use this to massage the dot plus wedge unit vector sum above into $$\mathbf{a} \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert \left( { \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} +\frac{\hat{\mathbf{a}} \wedge \hat{\mathbf{b}} }{\left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert}\left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert} \right).$$
The sum has two scalar factors of interest, the dot product $ \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} $ and the length of the wedge product $ \left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert $. Viewed geometrically, these are the respective projections onto two perpendicular axes, as crudely sketched in the figure
That is, we can make the identifications $$\hat{\mathbf{a}} \cdot \hat{\mathbf{b}} = \cos\theta$$ $$\left\lVert { \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right\rVert = \sin\theta.$$
(Aside: Admittedly, I've pulled this sine/wedge identification out of a black hat, but it follows logically from study of projection and rejection in geometric algebra. The black hat magic trick may at least be verified by computing the length of the "rejection" component of the vector $\hat{\mathbf{a}}$, that is, $\hat{\mathbf{a}} - \hat{\mathbf{b}} \left( {\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}} \right)$, which has squared length $ 1 - \left( {\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}} \right)^2$. Expanding $ -\left( { \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right)^2 = -\left( { \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right) \cdot \left( { \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right) = -\hat{\mathbf{a}} \cdot \left( { \hat{\mathbf{b}} \cdot \left( { \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right) } \right) $ produces the same result.)
Inserting the trigonometric identification of these two scalars into the expansion of the geometric product, we now have $$\mathbf{a} \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert \left( { \cos\theta +\frac{\hat{\mathbf{a}} \wedge \hat{\mathbf{b}} }{\left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert}\sin\theta} \right).$$ This has a complex structure that can be called out explicitly by making the identification $$\mathbf{i} \equiv\frac{\hat{\mathbf{a}} \wedge \hat{\mathbf{b}} }{\left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert},$$ where by our definition of the length of a wedge product $ \mathbf{i}^2 = -1 $. With such an identification, we see that the multivector factor of a geometric product has a complex exponential structure $$\begin{aligned}\mathbf{a} \mathbf{b}= \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert \left( { \cos\theta + \mathbf{i} \sin\theta } \right)= \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert e^{\mathbf{i} \theta }.\end{aligned}$$
In this view of the geometric product, while we initially added two apparently dissimilar objects, this was really no less foreign than adding real and imaginary portions of a complex number, and we see that the geometric product can be viewed as a scaled rotation operator operating in the plane spanned by the two vectors.
In 3D, the wedge and the cross products are related by what is called a duality relationship, relating a bivector that can be interpreted as an oriented plane, and the normal to that plane. Algebraically, this relationship is $$\mathbf{a} \wedge \mathbf{b} = I (\mathbf{a} \times \mathbf{b}),$$ where $ I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 $ is a unit trivector (often called the 3D pseudoscalar), which also satisfies $ I^2 = -1 $. With the usual normal notation for the cross product $ \mathbf{a} \times \mathbf{b} = \hat{\mathbf{n}} \left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert \sin\theta $ we see our unit bivector $\mathbf{i}$, is related to the cross product normal-direction by $\mathbf{i} = I \hat{\mathbf{n}} $. A rough characterization of this is that $ \mathbf{i} $ is a unit (oriented) plane that is spanned by $ \mathbf{a}, \mathbf{b} $ normal to $ \hat{\mathbf{n}}$.
The intuition of that the geometric product and the Lagrange identity are related is on the mark. There is a wedge product generalization of the Lagrange identity in geometric algebra. The 3D form stated in the question follows from the duality relationship of the wedge and cross products.
Best Answer
Here's an excerpt from Lasenby, Lasenby and Doran, 1996, A Unified Mathematical Language for Physics and Engineering in the 21st Century: