Why does the square of a bivector give its magnitude squared? I'm trying to learn geometric algebra, but I've gotten stuck on the magnitude of a bivector. In this video, the following derivation is made:
For $u,v \in \mathbb{R}^n, u^2v^2 = uv^2u = uvvu$
$= (u\cdot v + u \wedge v)(u\cdot v - u \wedge v)$
$=(u\cdot v)^2 - (u\wedge v)^2$
$= u^2 v^2 \cos^2(\theta_{uv}) - (u\wedge v)^2$
$\Rightarrow (u\wedge v)^2 = u^2v^2[\cos^2(\theta_{uv}) - 1] = -u^2v^2\sin^2(\theta_{uv})$
The author then concludes that $u\wedge v$ squares to a negative scalar, and as such is neither a scalar or a vector itself, but a new object (bivector). I understand that, but they then claim that the object $u\wedge v$ has the magnitude $|u||v|\sin(\theta_{uv})$. I understand that a bivector represents an oriented area, but I don't see how it follows that $(u\wedge v)^2 = |u\wedge v|^2$?
Wikipedia makes this claim as well here
Any help much appreciated.
Edit:
I understand that the bivector squares to a real number with magnitude $|-u^2 v^2 \sin^2(\theta_{uv})|^2 = (|u||v|\sin(\theta_{uv}))^2$. I don't see how that means that the bivector itself has magnitude $|u\wedge v| = |u||v|\sin(\theta_{uv})$.
 A: I'm under the impression that you're perfectly OK with $(u\wedge v)^2=-u^2v^2\sin^2(\theta_{uv})$.
I don't think anyone's claiming that $(u\wedge v)^2=|u\wedge v|^2$.
Consider that $i^2=-1$, and we would say that $i^2$ has magnitude $1$ but we would not say that $i^2=|-1|^2$.
There are two things: the real number $(u\wedge v)^2$ and its magnitude squared $|u\wedge v|^2$.  It is not hard to compute $|-u^2v^2\sin^2(\theta_{uv})|^2=(|u||v|\sin(\theta_{uv}))^2$, and that quantity inside the square on the right is the familiar quantity that gives the magnitude of the cross product or, if you like, the volume of the spanned parallelpiped.
Are you perhaps thinking that every element of the geometric algebra squares to a positive number that is its magnitude?  It's important to remember that in the usual Euclidean signature geometric algebra, $v^2=|v|^2$ certainly for the vectors in $V$ but no such guarantee is made for the bivectors.  The magnitude defined for a bivector is related to but not the same thing as the original norm on $V$, which it extends.
A: Let's start with your original expression  $u^{2} v^{2}=u v^{2} u=u v v u$.
$$
\begin{aligned}
u^{2} v^{2} &=(u \cdot v+u \wedge v)(v \cdot u+v \wedge u)=(u \cdot v+u \wedge u)(u \cdot v-u \wedge v) \\
u^{2}v^{2}&=(u \cdot v)^{2}-(u \wedge v)^{2} \\
u^{2} v^{2} &=u^{2} v^{2} \cos ^{2}(\theta)-(u \wedge v)^{2}
\end{aligned}
$$
Isolating $(u \wedge v)^{2}$
$$\begin{aligned}
u^{2} v^{2}\left(1-\cos ^{2} \theta\right)&=-(u \wedge v)^{2} \\
u^{2} v^{2} \sin ^{2} \theta&=-(u \wedge v)^{2}
\end{aligned}
$$
we can see that the square of a bivector must be negative, as $u^{2}v^{2}\sin^{2}(\theta)$ is non-negative.
To calculate the magnitude of $u \wedge v$ it helps to remember the modulus of a complex number $|z| = \sqrt{zz^{*}}$.
$$
|u \wedge v|^{2}=-(u \wedge v)^{2}=(u \wedge v)(u \wedge v)^{\dagger}
$$
$$
|u \wedge v|=\sqrt{(u \wedge v)(v \wedge u)}=\sqrt{(u \wedge v) \cdot(v \wedge u)}
$$
$$
\begin{array}{l}
=\sqrt{u \cdot(v \cdot(v \wedge u))} \\
=\sqrt{u \cdot((v \cdot v) u-v(v \cdot u))} \\
=\sqrt{v^{2} u^{2}-(u \cdot v)^{2}}\\
=|u||v|\sin(\theta)
\end{array}
$$
