This is a quick and basic question. I looked online (Wikipedia articles, Wolfram, etc..., and poked inside of Hestenes and Snygg's books, but couldn't easily pull out an answer). I'm going to define this problem as simply as possible for two 2-dimensional vectors, and in turn I would like a simple (computational) answer.

Let $*$ denote the geometric product, $\cdot$ denote the dot product, and $\wedge$ denote the wedge product. Have $\hat x = e_1 $ and $\hat y = \hat e_2$. That is, they are part of a standard orthonormal basis. The geometric product is sometimes given as

$$ \hat x * \hat y = \hat x \cdot \hat y + \hat x \wedge \hat y $$

I'd prefer this "ground up approach", where we don't treat the geometric product as the primitive element and in particular define the dot product explicitly as $\hat x \cdot \hat x = \hat y \cdot \hat y = 1$ and $\hat x \cdot \hat y = 0$.

So far so good, but I'm confused on how Hestenes and others would define our dot product explicitly (computationally) for the dot product of two exterior products, namely

$$ \left( \hat x \wedge \hat y \right) \cdot \left( \hat x \wedge \hat y\right)$$

Initially I thought that the dot of two wedges would look similar to the determinant of a two dimensional matrix, that is an alternating sum like

$$ \left( \hat x \wedge \hat y \right) \cdot \left( \hat x \wedge \hat y\right) = \left( \hat x \cdot \hat x \right) \left( \hat y \cdot \hat y \right) - \left( \hat x \cdot \hat y\right) \left( \hat y \cdot \hat x\right) = (1) (1) - (0)(0) = 1$$

However, I believe this definition would violate associativity, no? It's defined fairly early due to the supposed associativity that

$$ \hat x * \hat y * \hat x * \hat y = - \hat y * \left( \hat x * \hat x \right)* \hat y = - \hat y * \hat y = -1$$

However from my (I believe erroneous) definition of the inner product, I would have that

$$ (\hat x * \hat y) * (\hat x * \hat y) = (\hat x \wedge \hat y) * (\hat x \wedge \hat y) = (\hat x \wedge \hat y) \cdot (\hat x \wedge \hat y) = 1 $$

Which is $\neq 1$.

I have two subquestions, the second which is a slightly more general restatement of the first:

(I) Using my simple example, how would you properly compute $(\hat x \wedge \hat y) \cdot (\hat x \wedge \hat y)$?

(II) How would you compute $(\hat u \wedge \hat v) \cdot (\hat u \wedge \hat v)$ for slightly more general vectors in $\mathbb R^2$ of the form $\hat u = u_1 \, \hat e_1 + u_2 \, \hat e_2$ and $\hat v = v_1 \, \hat e_1 + v_2 \, \hat e_2$?

From the rest of my problem statement, it should be evident that I want a straightforward computational example for both. Thank you in advance for the help.

  • $\begingroup$ Your definition of the geometric product only applies to the product of vectors so you can't use that formula for your example. The usual definition of inner product on the exterior product is just as you imagine though, and this works for general vectors and even for higher degree products. $\endgroup$
    – Callum
    Commented Apr 12 at 15:36
  • $\begingroup$ @Callum Is there a computational example you can give me for what $\left( \hat x \wedge \hat y \right) * \left( \hat x \wedge \hat y \right)$ looks like if it's not a sum of dot and wedge? If you or someone else gives me that as an answer, I'll mark it as the answer, since that's ultimately what I'm after. $\endgroup$
    – Nate
    Commented Apr 12 at 16:26

1 Answer 1


I am not aware of a simple formula for the product of higher graded pieces of the Clifford algebra (a.k.a. geometric algebra). The formula $xy = x\cdot y + x\wedge y$ only works if $x,y$ are vectors. But we can use the basic properties to deduce what must be happening. Note the Clifford product is defined as the tensor product on a vector space $V$ "quotiented out" by the relation $v \otimes v = v\cdot v$ or equivalently $v\otimes w + w\otimes v = 2 v\cdot w$ where $v,w\in V$.

So we can take your basis $x = e_1,y = e_2$ and deduce all the relations.

To calculate $x(x\wedge y)$ you could expand this into $$x \otimes (x \wedge y) = \frac{1}{2}(x\otimes x \otimes y - x \otimes y \otimes x).$$

Then quotient rules come into play as $x\otimes x = 1$ and $x \otimes y \otimes x = (x\cdot y)x - y \otimes x \otimes x = (x\cdot y)x - y = - y$. So $x(xy) = \frac{1}{2}(y + y) = y$.

Note we could also have written it as $x(xy)$ more directly as $x\otimes x \otimes y = (x \cdot x)y = y$ but I wanted to explicitly use $v \wedge w = \frac{1}{2}(v\otimes w - w \otimes v)$ so that you can see what might be different when $v\wedge w \neq vw$.

You could also deduce this even more concisely by associativity: $x(xy) = (xx)y = y$. Note associativity follows directly from the associativity of the tensor product so we don't need to have a formula to prove this.

Now we can also find $y(xy)$ in similar fashion as $-x$. Then for the final product $(x\wedge y)(x\wedge y) = (xy)(xy)$ you can compute this again using associativity as $x(y(xy)) = x(-x) = -1$ or using the quotient relations:

$$ x\otimes y\otimes x\otimes y = 2(x,y)x\otimes y - y\otimes x \otimes x \otimes y = -(x\cdot x)y\otimes y = -y\otimes y = -1.$$

For more general vectors you could expand out as I did in my first example or note simply:

$$(v\wedge w)^2 = (vw - v\cdot w)^2 = (vw)^2 - 2(v\cdot w)vw + (v\cdot w)^2$$

and $(vw)^2 = vwvw = -vwwv = -(w \cdot w)vv = -(v\cdot v)(w\cdot w)$.

So we get

$$ (v\wedge w)^2 = - 2(v\cdot w)vw + (v\cdot w)^2 - (v\cdot v)(w\cdot w)$$

Note that constant term does work out to be minus the inner product of $v \wedge w$ with itself.

One last thing is that I could probably have dropped the tensor product symbols everywhere and this would all still make sense but I think it helps separate out what's going on at first look.

  • $\begingroup$ Thanks! Incidentally, does all this "machinery" just guarantee that I can view the geometric product associatively, and that as someone more interested in the application then the theory, I can now just always use the geometric product "piece by piece, vector by vector" and assume associativity and drive the car, knowing this "machinery" in the background makes it work? $\endgroup$
    – Nate
    Commented Apr 12 at 22:03
  • $\begingroup$ @Nate The definition I've used guarantees associativity automatically. And since this is an algebra (and so the product is bilinear) we can always break this down into pieces and add them back up at the end. Then you can either work in a basis and just find the multiplication table for it and work that way or use the quotient relations directly to simplify things. $\endgroup$
    – Callum
    Commented Apr 13 at 10:51

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