We are in the geometric algebra generated by the vector space $R^{1,1}$. Consider three vectors, $e_+, e_-, e$ where $e_+^2=1$, $ e_-^2=-1$, $e_+.e_-=0$ and $e=e_+ +e_-$. It is straightforward to verify that $e^2=0$, making $e$ a null vector.
I wish to evaluate $(e_+e)^2$.
Begin with the geometric product $e_+e=e_+.e + e_+ \wedge e$.
We can evaluate $e_+.e$ using the rules of the inner product: $e_+.e=\frac{1}{2}[(e_++e)^2-e_+^2-e^2]=\frac{1}{2}[(2e_++e_-)^2-1]=0$.
Thus $e_+e= e_+ \wedge e$ and $ee_+= e \wedge e_+$, implying $e_+e=-ee_+$ since for any two vectors $a\wedge b = -b \wedge a$. This means $(e_+e)^2=(e_+e)(e_+e)=-e_+eee_+=0$.
However if we note that $e_+ \wedge e=e_+ \wedge (e_++e_-)=e_+ \wedge e_-$ (since the outer product of any vector with itself is zero), then $e_+ e= e_+ \wedge e_- = e_+e_-$ which means $(e_+e)^2=(e_+e_-)^2=(e_+e_-)(e_+e_-)=-e_+e_-e_-e_+=1$.
Clearly, I am making an error somewhere. Will some kind soul not help me find it?
I think you made a mistake evaluating the dot product $e_+ \cdot e$:
$$e_+ \cdot e = \frac{1}{2} [(2 e_+ + e_-)^2 -1] = \frac{1}{2} [4 e_+^2 + 4 e_+ \cdot e_- + e_-^2 - 1] = \frac{1}{2} [4 +0 -1-1] = 1$$
This should be easy to verify by just using the geometric product's associativity rules:
$$e_+ e = e_+ (e_+ + e_-) =e_+^2 + e_+ e_- = 1 + e_+ e_-$$
The square is then
$$(1 + e_+e_-)^2 = 1 + 2e_+ e_-+(e_+e_-)^2 = 2 (1 + e_+ e_-)$$
If you divide through by 4, you get that $(e_+ e/2)^2 = e_+ e/2$; it's an idempotent. That's why this quantity is interesting to study.