2
$\begingroup$

A Householder reflection of a vector $v$ along a direction $n$ is given by the formula \begin{eqnarray} v' = v - 2 \frac{n \cdot v}{|n|^2} n. \end{eqnarray}

If $n$ is unitary then $v'=v-2 (n \cdot v) n$. Clearly for $v=(2,3)$ and $n=(0,1)$ we find that $v'=(2,-3)$ which in fact is the Householder reflection of $(2,3)$ along the vector $(0,1)$.

According to Geometric Algebra (GA) a Housholder reflection is defined as $-nvn^{-1}$. Since $n$ is unitary $n^{-1}=n$ and we can write $v'=-nvn$. However when I compute this formula with the vectors above, I get different results.

Let me explain: \begin{eqnarray} n v = (0,1) \cdot (2,3) + (0,1) \wedge (2,3) = 3 - 2 e_1 \wedge e_2. \end{eqnarray} and so

\begin{eqnarray} v' = -nvn^{-1} = -(3 -2 e_1 \wedge e_2) n = -3n + 2 e_1 \wedge e_2 n. \end{eqnarray}

My concern is that, since $e_1 \wedge e_2$ is equivalent to a 90 degree counterclockwise rotation then we find that the second (last) term here is $(-2,0)$ and so the result is $v'=-3(0,1)+2(-1,0)=(-2,-3)$ and not $(2,-3)$ as expected.

Probably I have a stupid error such a sign somewhere or a deep problem on understanding the geometric algebraic product.

I would appreciate any help on this matter.

Thanks.

$\endgroup$
2
$\begingroup$

The problem is apparently with your interpretation, because the math is working out. $$ -nvn^{-1} = -(3 -2 e_1 \wedge e_2) n = -3n + 2 e_1 \wedge e_2 n=-3e_2+2e_1e_2e_2=2e_1-3e_2$$.

I’m not even sure why you were using the product identity with the wedge. You may as well just compute it directly:

$$ -nvn =-e_2(2e_1+3e_2)e_2=-2e_2e_1e_2-3e_2e_2e_2=2e_1-3e_2 $$

It seems to me that multiplication by $e_1\wedge e_2$ on the left creates a clockwise rotation by 90, since it maps $e_2$ to $ e_1$ and $e_1$ to $-e_2$.

$\endgroup$
  • $\begingroup$ So, is it wrong to say that $e_1 e_2 u$ (I get your point. I can remove the wedge symbol here) is a $\pi/2$ positive rotation on $u$? I thought $e_1 e_2$ behaved as the imaginary complex number $i=\sqrt{-1}$. Thanks. $\endgroup$ – Herman Jaramillo Apr 29 '18 at 21:22
  • $\begingroup$ $| e_1 e_2 |^2 = (e_1 e_2)^*(e_1 e_2)= e_2 e_1 e_1 e_2 = 1$ so $e_1 e_2 = \sqrt{-1}$. Right? $\endgroup$ – Herman Jaramillo Apr 29 '18 at 21:39
  • $\begingroup$ Now the problem reduces to the following: Assuming the associative law is valid: $(e_1 e_2) e_2=-e_1$, $e_1(e_2 e_2)=e_1 1=e_1$. Why is the first interpratation wrong? Thanks. $\endgroup$ – Herman Jaramillo Apr 29 '18 at 21:45
  • $\begingroup$ @HermanJaramillo If you multiply on the right with it, it will send $e_1\to e_2$ and $e_2\to -e_1$. To model complthe x multiplication, you’d have to write numbers with a real part and a complex part using $e_1\wedge e_2$, not as a linear combination of $e_1$ and $e_2$. So I think you are mixing interpretations. $\endgroup$ – rschwieb Apr 29 '18 at 22:16
  • 1
    $\begingroup$ That is correct. Previously I did not understand your notation $\langle, \rangle$ because I was thinking on inner products but you are talking about spanning a space. Now since 1 is a scalar it commutes with $e_1 e_2$, but if instead of 1 you pick a vector $a$ then $\langle a, e_1 e_2 \rangle$ spans a space where only if $e_1 e_2$ is at the right , it resembles a counterclockwise rotation on $a$. On the left would be a clockwise rotation. I am thinking on $\mathbb{R}^2$ $\endgroup$ – Herman Jaramillo May 2 '18 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.