What's the motivation to add inner product and wedge product together in geometric product

Question

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?

Answer 1

Here's an excerpt from Lasenby, Lasenby and Doran, 1996, A Unified Mathematical Language for Physics and Engineering in the 21st Century:

The next crucial stage of the story occurs in 1878 with the work of the English mathematician, William Kingdon Clifford (Clifford 1878). Clifford was one of the few mathematicians who had read and understood Grassmann's work, and in an attempt to unite the algebras of Hamilton and Grassmann into a single structure, he introduced his own geometric algebra. In this algebra we have a single geometric product formed by uniting the inner and outer products—this is associative like Grassmann's product but also invertible, like products in Hamilton's algebra. In Clifford's geometric algebra an equation of the type $\mathbf{ab}=C$ has the solution $\mathbf{b}=\mathbf{a}^{-1}C$, where $\mathbf{a}^{-1}$ exists and is known as the inverse of a. Neither the inner or outer product possess this invertibility on their own. Much of the power of geometric algebra lies in this property of invertibility.

Clifford's algebra combined all the advantages of quaternions with those of vector geometry, [...]

Answer 2

In addition to invertibility, as mentioned by Joe, geometric operations can be expressed in simple, co-ordinate free expressions using the geometric product.
For instance, Rotation: $$ R_{i\theta}(A) = e^{-i\theta/2}Ae^{i\theta/2} $$ rotates the blade $ A $ by an angle $ \theta $ in the plane of the bivector $ i $.
Reflection: $$ F_B(A) = (-1)^{j(k+1)}BAB^{-1} $$ reflects the $j$-blade $A$ in the $k$-blade $B$.
Projection: $$ P_B(A) = (A\cdot B)B^{-1}$$ projects the blade $ A $ onto the blade $ B $

Answer 3

It is, perhaps, misleading to even call this addition. It is no more (and no less) addition than it is addition to add $5 e_1$ and $3 e_2$. You might say, "Of course we can add those. They're members of the same vector space; you just add corresponding components."

Well, we can do the same thing with multivectors. You just have $2^n$ components corresponding to $2^n$ basis blades. In this sense, the addition operations we're doing are actually quite pedestrian. The problem with viewing it as a $2^n$ dimensioned vector space is that you no longer have the clear geometric interpretation of elements, which is why this picture is often avoided. Still, you could easily say that all the geometric product is doing is giving us a meaningful multiplication operation between these vector space elements.

You ask about "motivation" for adding two disparate things together. I don't know if that's the right word. I'm no authority on the history, but I think you need to turn the picture on its head. It's much easier to start with the axioms of the geometric product and explore the consequences and how those consequences are useful.

The geometric product allows us quite a bit of compactness of notation. For example, the following integrals come up often in discussions of the fundamental theorem of calculus:

$$\oint G(r-r') \, dS' \, A(r') = \int_V \dot G(r-r') \, dV' \cdot \dot \nabla' A(r') + \int_V G(r-r') \, dV' \cdot \nabla' A(r')$$

when $A$ is, for example, a vector field with nonzero curl, there's actually quite a lot going on in the LHS than you might think. Without the geometric product's ability to combine dot and wedge products, we would have to do something like

$$\langle G (dS') A \rangle_2 = (G \cdot dS') \wedge A + (G \wedge dS') \cdot A$$

And if the vector field has nonzero divergence also, then we also have the expressions

$$\langle G (ds') A \rangle_0 = (G \cdot dS') \cdot A$$

on the left. Without the implicit ability to add multivectors of different grades, we would have to write two separate integrals to capture the full description of the theorem.

This is also apparent when writing certain differential equations. For example, Maxwell's equations in vacuum can be simplified to

$$\nabla F = J$$

for a vector field $J$ and bivector field $F$, which tells us immediately that $\nabla \wedge F = 0$ as a consequence.

Will you be fundamentally unable to do tensor algebra and mathematics without the ability to add multivectors? Well, no. You can always separate equations in GA out into their component grades, and this is exactly what ends up happening when you do stuff in index notation or in differential forms. Still, the ability to describe several equations at once, with each grade describing its own independent equation, is just as powerful as the ability to break down a vector equation into each of its components' equations.