# Intuition for geometric product being dot + wedge product

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.

• perhaps geometric products are not actually as geometric as the name suggests? – Lord Shark the Unknown Apr 19 at 3:43
• How do you feel about complex numbers? Is a+ib apples and oranges? – rob3c Apr 19 at 22:54
• @Rob3c thats a good point but complex numbers seem less like apples and oranges because its easy to think of “4 + 5i” as similar to the vector $4 \hat{i} + 5 \hat{j}$ or as a rotation. – Kevin Goodman Apr 21 at 13:32
• @KevinGoodman see my answer for how the geometric product can similarly be seen as a rotation operator. – rob3c Apr 22 at 8:15

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.

• In the comments for OP above, robc pointed out that components of complex numbers also have the property of seeming like apples and oranges. So a complex number might also be thought of as a bag like you describe. Is Lagrange’s Identity saying that the inner and outer products are orthogonal components of the vector product, just as the real and imaginary components of a complex number are orthogonal? Is that a reasonable way to think of them? – Kevin Goodman Apr 21 at 16:30
• @KevinGoodman It depends what you mean by "orthogonal". If you just mean "add component-wise", then definitely yes. If you meant something more, I don't think it's very natural to think that way. You'd sort of give up the geometry of geometric algebra to claim that there was a sense in which nonzero scalars are orthogonal to nonzero bivectors (or individual 2-blades/2-wedges). – Mark S. Apr 21 at 19:07
• @MarkS. I meant orthogonal in the same sense that a right triangle has opposite and adjacent sides that are orthogonal because Lagrange's Identity seems to be ~= a restating of Pythagorus applied to the vector product equation. It's in that sense that I think the inner and outer products are orthogonal components of the vector product...? – Kevin Goodman Apr 21 at 21:39
• @KevinGoodman Ah, good point. Yes, length/norm is extended to all sums of "apples and oranges" in a natural way, and we have $|\mathbf a\mathbf b|^2=|\mathbf a\cdot\mathbf b|^2+|\mathbf a\wedge\mathbf b|^2$ when $\mathbf a$ and $\mathbf b$ are vectors, just like the pythagorean theorem. – Mark S. Apr 21 at 21:47
• For what it is worth, $(\mathbf{a}\mathbf{b})^2 = (\mathbf{a} \cdot \mathbf{b})^2 - (\mathbf{a} \wedge \mathbf{b})^2$ with $(\mathbf{a} \wedge \mathbf{b})^2 < 0$ is more basic. (Lagrange's identity). – Alan Macdonald Apr 22 at 4:31

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.

• That’s an interesting angle to come at this from - thanks! Btw, which author did you find that approaches things that way? Incidentally, I think showing ab as the product of the norms times the inner and outer product of the unit vectors is a neat way to quickly see that the vector product produces an area scaled according to the relative orientation of the vectors. Same for when you relate it to Euler’s e^i. A question though: at one point you wrote the norm of the dot product is sin, did you mean wedge product instead? – Kevin Goodman Apr 21 at 16:32
• @KevinGoodman, Yes, I meant the wedge. That's fixed. The axiomatic approach can be found in many sources (one is "Geometric Algebra for Physicists"). The angle interpretation of the product can also be found in many places (I think that Alan MacDonald's "Linear and Geometric Algebra" may have been one of them.) – Peeter Joot Apr 21 at 20:23
• Can you clarify how you recognized that the opposite side of the triangle you drew would be: $\left\lVert {\hat{\mathbf{a}} \wedge \hat{\mathbf{b}}} \right\rVert$? (Unless the hypotenuse was intended to be: $\left\lVert {\mathbf{a}} \right\rVert \left\lVert {\mathbf{b}} \right\rVert$ ?) It's not obvious to be how the previous equation led to that. – Kevin Goodman Apr 21 at 21:57
• I didn't show that, but it's not too hard to do. The rejection component of that vector is $\hat{\mathbf{a}} - \hat{\mathbf{b}} \left( {\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}} \right)$ which has squared length $1 - \left( {\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}} \right)^2$. If you expand $-\left( { \hat{\mathbf{a}} \wedge \hat{\mathbf{b}} } \right)^2$ you'll get the same result. – Peeter Joot Apr 22 at 12:15
• I've edited the answer to at least describe where this magic trick comes from. – Peeter Joot Apr 22 at 12:26

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.

• That’s really cool, but not at all obvious. Lol I’ll definitely take a look at that link! – Kevin Goodman Apr 21 at 16:34

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.