0
$\begingroup$

I recently found a video that claimed to give intuition to Quaternions. And to my suprise it nearly did, but I have a few large hang ups on the lack of definitions of certain operations, it also appears that many other resources on the internet suffer the same issues and don't adequately define operators in a way that I could figure this out independently.

First projection on the projection and rejection operators:

While I understand what both do, I do not understand how either can actually be computed given the Wikipedia definitions of the un-generalized geometric product, the contraction product (generalized dot product), and the wedge product (exterior product). These are the same definitions given in the video.

video definition at 36:28

Projection of $v$ onto $u$ is $Geometric Product($$v \bullet u$, $inv(u))$

This implies would need to take the dot product of $u$ and $v$, a scalar, then take the geometric product of that scalar... I don't understand how I'm supposed to do that (it would require taking the dot product of a scalar and taking the wedge product of a scalar and a vector, which makes me think the un-generalized definition is not what I want here?)

video definition at 42:05

Similarly, Rejection of $v$ onto $u$ is $Geometric Product($$v \land u$, $inv(u))$ which if $u$ is unit length, then $v \land u$ $\bullet$ $u$ should be $v$ using the contraction product because $v,u$ ^ $u$ $= 0$, which also doesn't make any sense (clearly the rejection in the example shown is not going to be $v$, the vector being projected itself...)

video definition at 54:13

rotation of $x$ given vector $\theta$ /2 ccwise from $x$, $a$, and $\theta$ /2 ccwise from $a$, $b$ is given by $GeometricProduct(Geometric Product(GeometricProduct(b, inv(a), x), GeometricProduct(a, inv(b)))$

Finally, while it did make sense why we needed to do two reflections to achieve a rotation, and how this absolutely shows quaternion analogies, I did not understand how the function could actually be calculated with out running into bivector scalar and vector scalar geometric products.

I must be misunderstanding something, but I can't figure out how these functions are supposed to be actually evaluated given the current information I have.

$\endgroup$
1
$\begingroup$

I have not seen the video but maybe can help you.

This implies would need to take the dot product of u and v, a scalar, then take the geometric product of that scalar... I don't understand how I'm supposed to do that (it would require taking the dot product of a scalar and taking the wedge product of a scalar and a vector, which makes me think the un-generalized definition is not what I want here?)

No, in the case of a scalar and a vector, the geometric product (gp) reduces to the familiar vector space product-with-scalar, so gp($\alpha$, $u$) = $\alpha u = \alpha (u\cdot e_1) + \alpha (u\cdot e_2) + \alpha (u\cdot e_3)$. So the projection of $v$ on $u$: $(u \cdot v) u^{-1}$ = $\frac{1}{\|u\|^2}(u \cdot v) u$, just the same as in vector space.

Similarly, Rejection of v onto u is GeometricProduct(GeometricProduct(v ∧ u, inv(u))) which if u is unit length, then v ∧ u ∙ u should be v using the contraction product because v,u ^ u = 0, which also doesn't make any sense (clearly the rejection in the example shown is not going to be v, the vector being projected itself...)

Considering $\|u\| = 1$, the rejection $(v \wedge u) u^{-1}$ is equal to $(v \wedge u) u$. The expansion is $(v \wedge u) \cdot u + (v \wedge u) \wedge u$, since $(v \wedge u) \wedge u = 0$ the rejection is $(v \wedge u) \cdot u = -(v \cdot u) \wedge u + v \wedge (u \cdot u)$. The term $(v \cdot u) \wedge u = (v \cdot u) u$ since the wedge product of a scalar and a vector reduces to the familiar vector space product, and since $u \cdot u = 1$, the rejection is $-(v \cdot u) u + v$.

Finally, while it did make sense why we needed to do two reflections to achieve a rotation, and how this absolutely shows quaternion analogies, I did not understand how the function could actually be calculated with out running into bivector scalar and vector scalar geometric products.

These geometric products reduces to the familiar vector space products, they commute and only can change the weight of a k-vector but never its attitude.

$\endgroup$
  • $\begingroup$ What does "familiar vector space product" mean, are you just saying that because you are trying to wedge, and geometric product a scalar and a vector the scalar product is used? Is that a special rule for scalars? This seems to be what was confusing me, math works out and makes sense when you reduce the products to scalar products. $\endgroup$ – opa Dec 29 '17 at 23:31
  • 1
    $\begingroup$ It is by definition. Wedge product of scalar and any multivector is just a scalar product. Similar definition is made for the geometric product of scalar and any multivector. $\endgroup$ – Mauricio Cele Lopez Belon Dec 29 '17 at 23:48
  • $\begingroup$ Ah, ok, that makes sense, all my knowledge was initially based off of that video, and it failed to mention that (as well as several other resources). $\endgroup$ – opa Dec 31 '17 at 0:11
1
$\begingroup$

I don't want to take the time to look at the video. You might find my paper A Survey of Geometric Algebra and Geometric Calculus useful: http://faculty.luther.edu/~macdonal/

$\endgroup$
  • 1
    $\begingroup$ You don't need to watch the entire video, I included links and time stamps to the parts which describe the issues I have trouble with. This is also more relevant for a comment than an answer. $\endgroup$ – opa Dec 21 '17 at 19:29
  • $\begingroup$ so reflection is defined as so ($a$ $\land$ $b$)$inv(a)$, and expanding that by applying the geometric product you get ($a$ $\land$ $b$ $\bullet$ $inv(a)$) + ($a$ $\land$ $b$ $\land$ $inv(a)$, the last part cancels out because expanding $inv(a)$ turns into $a/(a \bullet a)$, thus we will be wedge producting two $a$s resulting in 0. so then we have ($a$ $\land$ $b$ $\bullet$ $inv(a)$), which I have no idea how to compute, is it applying the dot product seperate on both vectors? if so that leaves only $b*a/a*a$, which obviously isn't reflection. $\endgroup$ – opa Dec 22 '17 at 22:53
  • $\begingroup$ Your expansion of the geometric product is not valid. My videos are designed to give an overview of geometric algebra, not to give a working knowledge of the subject. The latter requires reading something designed for that. $\endgroup$ – Alan Macdonald Dec 25 '17 at 19:15

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.