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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?

njt
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1 Answers1

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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.

Muphrid
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  • Ah yes, less sleepy eyes see the mistake now.

    But now if we let $a=ee_+/2$ then $a=a^2=a^3=a^n$ and $e^{a}=1 + a + a/2! + a/3! +... = 1+ae^1$ or $e^{a}=1+e^1ee_+/2$. But Hestenes (p.20) has it that $e^{ec/2}=1+ec/2$ for any vector $c$, which means if we let $c=e_+$, then $e^{a}=1+ee_+/2$. I am still off by a factor of $e^1$?

     – njt Feb 18 '14 at 14:28
  • Not 100% sure, but I think $c$ is a vector in the base space, with no $e_+, e_-$ components. This is supposed to apply toward conformal models of translations, so $c$ should be a direction you can translate along. – Muphrid Feb 18 '14 at 15:26
  • After careful re-reading (actually, another paper (p.7) made the whole thing clearer), I believe you are right. The vector $e$ must be perpendicular to all regular vectors in the Euclidean space, so that $e.c=0$ and $ec=-ce$, in which case $(ce)^2=-(ce)(ec)=0$ and $e^{ec}=1+ec$. Thanks for your help! – njt Feb 19 '14 at 01:28