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Working through "Geometric Algebra for Computer Science: An Object Oriented Approach" on my own. The problem comes from an example on page $130$. This should be an easy calculation, why I'm all the more frustrated.

Calculating it using the general formula (see image) I get the wrong answer.

enter image description here

I do understand how to get $2$ by just multiplying the lengths of the vectors on either side of the wedge product, since the vectors are orthogonal, i.e., (left side) dot (right side) $= 0$. But want to understand where my mistake is in above calculation.

Thanks for your help!

TShiong
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Mark
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  • Welcome to MSE. Please post your question with all the supporting details so that we do not have to click on a link please. As specified in the MSE Guidelines [tour]. – Mike Aug 29 '24 at 22:46
  • This is still unclear, too much shortcuts in your worksheet. – Mike Aug 29 '24 at 23:11
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    They're not orthogonal. The dot product is $(e_1+e_2)\cdot(e_2+e_3)=1$. The norm is $\sqrt3$. – mr_e_man Aug 30 '24 at 00:03
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    For any multivector $$A=a+b_1\mathbf e_1+b_2\mathbf e_2+b_3\mathbf e_3+c_1\mathbf e_2\mathbf e_3+c_2\mathbf e_3\mathbf e_1+c_3\mathbf e_1\mathbf e_2+d,\mathbf e_1\mathbf e_2\mathbf e_3$$ where $a,b_n,c_n,d$ are scalars, the norm is easily calculated as $$\lVert A\rVert=\sqrt{a^2+b_1^2+b_2^2+b_3^2+c_1^2+c_2^2+c_3^2+d^2}.$$ I'm not sure if you're supposed to learn this earlier or later in the book. -- See also my answer to https://math.stackexchange.com/questions/3198338/inner-product-structure-on-geometric-algebra – mr_e_man Aug 30 '24 at 00:30
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    The close votes are completely unwarranted. The question is as clear as it could possibly be. – Nicholas Todoroff Aug 30 '24 at 00:32
  • Great, thanks @mr_e_man for confirming and pointing out the mistake on orthogonality. Much appreciated! – Mark Aug 30 '24 at 04:31
  • @NicholasTodoroff: I didn't down-vote, but it's possible such down-votes are based on the mathematics being included as an image as opposed to being typeset. I don't agree with that, but it does happen. – Brian Tung Aug 30 '24 at 07:34

1 Answers1

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As mr_e_man says, the vectors are not orthogonal, $$ (e_1 + e_2)\cdot(e_2 + e_3) = e_1\cdot e_2 + e_1\cdot e_3 + e_2\cdot e_2 + e_2\cdot e_3 = 0 + 0 + 1 + 0 \ne 0 $$ and $\sqrt3$ is correct.

This doesn't effect your final answer, but you're missing a minus sign. The inner product with a vector is an antiderivation, you have to grade-reverse the factor it "passes over": $$ v\cdot(A\wedge B) = (v\cdot A)\wedge B + \color{red}{\hat A}\wedge(v\cdot B). $$ So you should have $$ e_2\cdot(e_3\wedge e_2) = \color{red}-e_3. $$ Somehow you magically bring the minus sign back in the end though.

A much easier way is to just use $v = e_1 + e_2$ and $w = e_2 + e_3$ directly: $$ A\cdot\widetilde A = (v\wedge w)\cdot(w\wedge v) = v\cdot[w^2v - (w\cdot v)w] = w^2v^2 - (w\cdot v)^2 = (2)(2) - (1)(1) = 3. $$

Nicholas Todoroff
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  • Excellent, good to know my calculation was correct. The orthogonality and missing the negative (when scalar is on right of wedge) were silly mistakes. Thank you for showing me a better way to calculate, treating vectors as fundamental things rather than going straight to components. I need to get in the habit of think like that. Much appreciated! – Mark Aug 30 '24 at 04:30
  • @Mark the negative has nothing to do with "where the scalar is"; perhaps you just mean this as a sort of mnemonic, but I would avoid that sort of thinking. This is purely a result of how the dot product and wedge product interact. The wedge product is commutative on scalars ($A\wedge 1 = 1\wedge A = A$ for all $A$) and if say $A$ is a bivector then we have no minus sign in the following: $$e_1\cdot(A\wedge e_1) = (e_1\cdot A)\wedge e_1 \color{red}+ A.$$ – Nicholas Todoroff Aug 30 '24 at 05:21
  • Oh, somehow I had it, A wedge k = - k wedge A, due to the anticommutavity of the wedge product. Thank you! – Mark Aug 30 '24 at 16:01
  • @Mark It is only guaranteed to be anticommutative between two vectors. It then follows that if $A$ is a $k$-vector and $B$ is an $l$-vector that $$A\wedge B = (-1)^{kl}B\wedge A.$$ You prove this by assuming $A, B$ are blades (which you can do because $\wedge$ is bilinear) and then anticommuting all the vector factors past each other as needed. – Nicholas Todoroff Aug 30 '24 at 16:32