TITLE: Simple bivectors in four dimensions
QUESTION [1 upvotes]: I am trying to characterize simple bivectors in four dimensions, i.e. elements $B \in \bigwedge^2 \mathbb{R}^4$ such that $B = a \wedge b$ for two vectors $a, b \in \mathbb{R}^4$. In the book Clifford Algebras and spinors by Lounesto Pertti, I found the following:
 I can see why the square of any simple bivector is real since we have the identity $(a \wedge b)^2 = -|a \wedge b |^2$. However, I cannot prove the second statement, i.e. If the square of a bivector is real, then it is simple.
Writing $e_{ij} = e_i \wedge e_j$ and choosing $\{e_{14}, e_{24}, e_{34}, e_{23}, e_{31}, e_{12}\}$
as a basis of $\bigwedge^2\mathbb{R}^4$ (I have specific reasons to choose this slightly atypical basis), I find by direct computation that
$B^2 = -|B|^2 + 2(B_{12}B_{34} + B_{14} B_{23} + B_{31}B_{24})e_{1234}$, where $e_{1234} = e_1 e_2 e_3 e_4$ denotes the pseudo scalar in the Clifford algebra of $\mathbb{R}^4$. Yet, I don't manage to conclude from that.
As a more general approach, I thought of using the relationship between simple rotations of $\mathbb{R}^4$ and simple bivectors. In fact, the simple bivectors form a double cover of the simple rotations, so the geometry of the simple bivectors should be something like the choice of a plane in $\mathbb{R}^4$ and the choice of an angle $\theta \in [- \pi, +\pi]$, i.e.
$$
\text{simple bivectors } \simeq Gr(2, 4) \times [- \pi, +\pi].
$$
Is the latter more or less correct? And how can this help me to characterize more precisely simple bivectors in $\mathbb{R}^4$?

REPLY [0 votes]: $
\newcommand\form[1]{\langle#1\rangle}
$It is much, much easier than the other answers are making it out to be. Note that $B^2$ must be an even element; so
$$
  B^2 = \form{B^2}_0 + \form{B^2}_2 + \form{B^2}_4
      = B\cdot B + B\times B + B\wedge B
      = B\cdot B + B\wedge B.
$$
where $\form{\cdot}_r$ is the grade $r$ projection and $X\times Y = \tfrac12(XY - YX)$ is the commutator product. (It is true for any $k$-vector $X$ that $\form{BX}_k = B\times X$.
If $B$ is simple, then clearly we're left with $B^2 = B\cdot B$, a scalar.
If $B^2$ is scalar, then $B\wedge B = 0$. We may write $B$ as a sum of simple bivectors
$$
  B = B_1 + B_2
$$
Hence
$$
  0 = B\wedge B = 2B_1\wedge B_2
$$
so the planes represented by $B_1$ and $B_2$ must intersect in a line represented by a vector $b$; so $B$ lives in a 3D subspace and $B$ is factorable. More concretely, there are $b_1, b_2$ such that $B_1 = b\wedge b_1$ and $B_2 = b\wedge b_2$, so
$$
  B = b\wedge b_1 + b\wedge b_2 = b\wedge(b_1 + b_2).
$$
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