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When I reflect $v$ through the hyperplane orthogonal to $v-w$, where $w = f(v)$ and $f$ is an isometry on $\mathbb{R}^n$ that preserves the origin. I believe I am supposed to get $w$, however whenever I do the calculation I get: $$\text{Ref}_{v-w}(v) = -(v-w)v(v-w)^{-1} = \frac{-(v-w)v(v-w)}{(v-w)^2} = $$ $$= \frac{(w-v)v(v-w)}{(v-w)^2} = \frac{(wv-v^2)(v-w)}{(v-w)^2} = \frac{wv - v^2}{v-w} = -v$$ No matter what I do I can't seem to get away from this conclusion. Can anybody help me understand what I'm doing wrong?

Emilio Minichiello
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    Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error. – KCd Jul 22 '18 at 03:35
  • But what about this result https://math.stackexchange.com/questions/465304/reflection-representation-of-isometry ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong. – Emilio Minichiello Jul 22 '18 at 03:48
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    There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one. – amd Jul 22 '18 at 04:52
  • How do I know which one ive chosen? – Emilio Minichiello Jul 22 '18 at 05:26
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    To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1. – KCd Jul 22 '18 at 05:31
  • This is great, thank you so much. – Emilio Minichiello Jul 22 '18 at 05:40

2 Answers2

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Here's a less convoluted version of your calculation that should make your error more immediately obvious: $$\mathrm{Ref}_{v-w}(v) = −(v−w)v(v−w)^{−1} = -\frac{(v−w)v}{v−w} = -v$$ In short, you've tricked yourself by using fraction notation in a non-commutative algebra. Since geometric algebra is non-commutative, there's a difference between $ab^{-1}$ and $b^{-1}a$ (if that were not the case, you could simply exchange the last two terms in the reflection formula, and then cancel $(v-w)$ with $(v-w)^{-1}$).

Now if you choose to use fraction notation $\frac{a}{b}$ (which is a very bad idea for non-commutative algebra to begin with), you have to decide on a meaning for it. Now in the step $$-(v-w)v(v-w)^{-1} = \frac{-(v-w)v(v-w)}{(v-w)^2}$$ you used the definition $\frac{a}{b} = ab^{-1}$ (it's the only definition under which that equation would hold), but in the step $$\frac{wv - v^2}{v-w} = -v$$ you used the definition $\frac{a}{b} = b^{-1}a$ (again, that's the only definition where that step would be valid). So essentially, in those two steps together, you exchanged the factors in $v(v-w)^{-1}$; all the other stuff only serves to obfuscate this. And thus your result is easily explained, as indeed $$-(v-w)(v-w)^{-1}v = -v,$$ it's just no longer the reflection of $v$ on a hyperplane.

Bottom line: When dealing with non-commutative algebra, don't use fraction notation. It only invites errors such as this one.

celtschk
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  • Ah okay. So my calculation should be $-\frac{(v-w)v(v-w)}{|v-w|^2} = - \frac{(|v|^2v - w |v|^2 - |v|^2w + wvw)}{|v-w|^2} = - \frac{(|v|^2(v-2w) + wvw)}{|v-w|^2}$ right? What can I do now? I should be getting $w$ right? – Emilio Minichiello Jul 22 '18 at 12:25
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I found the answer in Guide to Geometric Algebra in Practice by Dorst and Lasenby. As celtschk explains, my use of fraction notation messed me up. This is the correct derivation.

$$-(v-w)v(v-w)^{-1} = (-(v-w)v)(v-w)^{-1} = (-(v-w)((v+w)-w))(v-w)^{-1}=(-(v^2 - w^2) + w(v-w))(v-w)^{-1} = (w(v-w) - (v^2-w^2))(v-w)^{-1} = w - (v^2 - w^2)(v-w)^{-1} = w$$ Because $v^2 = |v|^2 = |w|^2 = w^2$ since $f(v) = w$ is an isometry.

Emilio Minichiello
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  • This is much longer than it needs to be. From your last line $v^2=w^2$, it follows that $$(w-v)v(v-w)^{-1}=(wv-v^2)(v-w)^{-1}=(wv-w^2)(v-w)^{-1}=w(v-w)(v-w)^{-1}=w.$$ – mr_e_man Dec 10 '18 at 02:41