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

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?

• What do you mean by "adding them together"? – bradhd Jul 14 '13 at 16:42
• If you can, why not? – Paracosmiste Jul 14 '13 at 18:11
• @Brad, please see my edited question. – ahala Jul 14 '13 at 20:34

## 5 Answers

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, [...]

• Thanks, but the inverse of a vector b seems only using the inner product part. i.e., bb^(_1) = b . b^(_1) since b^b(^-1) = 0. We can define inverse without using the wedge product. I do not see how the wedge product part interacts with the inner product part on the invertibility. – ahala Jul 14 '13 at 21:35
• @ahala The key point is that the geometric product can take $1$ as its multiplicative identity and hence define inverses with respect to $1$ being that identity. Neither the dot nor wedge products have a meaningful multiplicative identity, so you can't say that the inversion only uses one or the other. – Muphrid Jul 14 '13 at 21:41

It isn't so much that the Clifford product is useful as that it's natural.

Instead of defining the Clifford product as $v\cdot w + v\wedge w$, you can define the Clifford algebra as the algebra in which S+S and V+V addition, SS and SV multiplication, and V2 squared norm have their usual meanings from vector calculus, the usual algebraic rules apply (except for commutativity of multiplication), and there are no other constraints. It isn't obvious that this produces anything sensible, but if it does (and indeed it does) then it seems quite natural. Starting from that formulation, the identity $v\cdot w = \tfrac12(\|v+w\|^2 - \|v\|^2 - \|w\|^2)$ gives you $v\cdot w = \tfrac12(vw+wv)$, and you can show that $\tfrac12(vw-wv)$ satisfies all of the axioms of the wedge product, which justifies calling it $v\wedge w$, and adding those together yields $vw = v\cdot w + v\wedge w$.

You could also informally motivate the sum by the fact that $|v\cdot w|^2 = \|v\|^2\|w\|^2 \cos^2 \theta$ and $\|v\wedge w\|^2 = \|v\|^2\|w\|^2 \sin^2 \theta$, so it makes sense to combine them to get a product that satisfies $\|vw\|^2 = \|v\|^2\|w\|^2$ yet also preserves $\theta$ by putting the inner and wedge products in orthogonal components of the result. Straying further into truthiness-based argument, the wedge product is also called the (Grassmann) exterior product, and the (Grassmann) interior product coincides with the inner product for vectors, so you can say that the Clifford product combines the interior and exterior Grassmann products. They sound like they belong together, at least.

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 $B$

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.

Well, actually the perspective should be reversed. Starting from the geometric product $a b$ having an inverse as mentioned $$a a^{-1}=1$$ one can define 2 identities $$a \cdot b= \frac{1}{2} \left(a b + \alpha(a) \, \alpha (b) \right)$$ where $\alpha$ is the reflection automorphism $$a \wedge b= \frac{1}{2} \left(a b - \alpha(a) \, \alpha (b) \right)$$ and this is true for all grades in the algebra.

Then $a \wedge b$ gives an oriented area element, $a \wedge b \wedge c$ gives an oriented volume and so on. The usual linear algebra can also be reformulated using the Geometric algebra.