Intuition for geometric product being dot + wedge product

Question

While I feel quite comfortable with the meaning of the dot and exterior products separately (parallelity and perpendicularity), I struggle to find meaning in the geometric product as the combination of the two given that one’s a scalar and the other is a bivector:

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

I can’t shake the feeling that you can't add apples and oranges and produce something meaningful.

I feel like Lagrange’s Identity is saying something similar for dot and cross products, while at the same time relating them to a circle/pythagoras:

$ \vert a \vert^2 \vert b \vert^2 = \vert a \cdot b \vert^2 + \vert a \times b \vert^2 $

but for some reason it’s just not clicking. I’d love to hear suggestions for how to think about this and what it means.

Answer 1

The most intuitive interpretation of a Geometric Product I have found is from Hestenes who notes that it can be visualized as a directed arc just as a vector can be viewed as a directed line.

For more depth, see page 11 of the following:

Hestenes - Oersted Medal Lecture.

Answer 2

Yes, you are adding apples and oranges. But there is a sense in which one can add apples and oranges: put them together in a bag. The apples and oranges retain their separate identities, but there are "apples + oranges" are in the bag. The situation with the inner and outer products of vectors is analogous: the bag is $ab$ and $a \cdot b$ and $a \wedge b$ are "in" it.

Adapted from my text Linear and Geometric Algebra.

Answer 3

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.

Answer 4

As I alluded to in my original comment, the isomorphism to complex numbers (that other answers also mention) is a good path to thinking about it...assuming complex numbers are ok intuitively, that is! :-) The symmetric dot product part corresponds to the real part of a complex number, and the antisymmetric wedge part corresponds to the imaginary part.

However, i prefer the following intuition for both in terms of how the math works out (without actually doing the math, of course lol):

The geometric product between two vectors produces a geometric operator that can perform a scaled rotation of another vector (or other GA objects via linearity) according to the properties of the relationship it captures between the two vectors: their relative angle and magnitudes. However you label it, the main intuition for visualizing it is that it is an operator with the potential to rotate something, rather than being a rotation itself (or 'directed arc' a la Hestenes, which creates more confusion than clarity IMHO).

To see it easily without cranking through the details, note that the geometric product results in a value with scalar and bivector parts. When multiplying a third vector by the product (now an operator), the operator's scalar part will just create a weighted version of the vector along its same direction, and the operator's bivector part will created a weighted vector in its orthogonal direction, since wedging a vector with a bivector 'cancels' any part in the shared direction. The sum of those two vector 'components' results in the third vector being essentially rotated/scaled, depending on all of the relative magnitudes and angles.

Answer 5

I'll try to explain why there are no apples and oranges, there are only reflections. The apples and oranges appear because in order to actually compute the geometric story I'm about to tell you, we use an orthogonal basis.

Vectors can be thought of as representing (hyper)planes. For example, in 3D space a vector can be used to represent a plane through the origin. Now, a plane $u$ can be reflected in a plane $v$ using

$$ v[u] = -vuv^{-1}, $$

where the minus sign is needed such that when you reflect $v$ in itself, the front and back of the mirror flip ($v[v] = -v$). If we now also perform a reflection in a second plane $w$, we get the rotation

$$ w[v[u]] = (wv)u v^{-1} w^{-1} = (wv) u (wv)^{-1} $$

The composition of two reflections $wv$ is called a bireflection, and could in fact be either a rotation, translation, or a boost. The picture below shows how two intersecting reflections form a rotation, while parallel reflections form a translation.

So the product of two vectors is a bireflection. The apples and oranges only appear because in order to actually compute it, we would have to somehow choose a basis. Staying with the 3D example, we could choose an orthogonal basis $e_1, e_2, e_3$ such that $e_i e_j = \delta_{ij} + e_{ij}$ and represent any plane as $x = \sum_i x^i e_i$.

Now, when we compute the bireflection $wv$ we will get a scalar and bivector part: $$ w v = \sum_{ij}(w^i e_i) (v^j e_j) = \sum_i w^i v^i + \sum_{i \neq j} w^iv^j e_{ij}. $$

So just remember the truth: there are no apples and oranges. For more detail on this approach I would refer to this video, or to the Graded Symmetry Groups paper. Full disclaimer: I'm one of the authors.