Show that $ \mathbf u^2 \mathbf v^2 = (\mathbf u \cdot \mathbf v)^2 - (\mathbf u \wedge \mathbf v)^2 $

Question

where $ \mathbf u $ and $ \mathbf v $ are vectors. From Linear and Geometric Algebra by Alan Macdonald.

Answer 1

You can use

$$ \begin{align} (u\cdot v)^2 - (u\wedge v)^2 & = \frac{1}{4} \left( (uv+vu)^2 - (uv-vu)^2\right) \\ & = \frac{1}{4} \left(uvuv + uvvu + vuuv + vuvu - uvuv + uvvu + vuuv - vuvu \right) \\ & = \frac{1}{2} \left( uvvu + vuuv\right) \\ & = \frac{1}{2} \left( u^2v^2+u^2v^2\right) \\ & = u^2v^2 \end{align} $$

The third line to the fourth line is true because

$$uvvu = uv^2u = uuv^2 = u^2v^2$$

where we can re-order terms because $u^2$, $v^2$ are scalars.

Answer 2

In a paper i read $$a\cdot b=\dfrac {ab+ba}{2}=b\cdot a$$

and $$a\wedge b=\dfrac {ab-ba}{2}$$ $$(u\cdot v)^2-(u\wedge v)^2$$ $$\left(\dfrac {uv+vu}{2}\right)^2-\left(\dfrac {uv-vu}{2}\right)^2$$ $$\left(\dfrac {uv+vu}{2}-\dfrac {uv-vu}{2}\right)\left(\dfrac {uv+vu}{2}+\dfrac {uv-vu}{2}\right)$$ $$(vu)(uv)$$ $$(uv)^2$$

Answer 3

This is straightforward, then, using grade projection and associativity of the geometric product. Consider the product $vuuv$:

$$vuuv = u^2 v^2 = \langle vuuv \rangle_0$$

On the other hand, you can group the products like so

$$vuuv = (vu)(uv) = (v \cdot u) (u \cdot v) + (v \wedge u)(u \wedge v) = (u \cdot v)^2 - (u \wedge v)^2$$

Because you know from the first equation that $vuuv$ is a scalar, this is all you need to consider--for instance, there are some bivector terms that I didn't write down, but you can argue that the overall bivector component must be zero, and thus you don't even need to compute them.

Edit: this basic technique of using grade projection and associativity is very useful. You can, for instance, prove the BAC-CAB rule without using index notation this way (but you do have to approach it with some cyclic products, which is kinda lame). It's also very useful when you start doing geometric calculus to prove some basic vector calculus identities that would otherwise be cumbersome.