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Anonymous Sun 18 Oct 01:58:16 2020 No.12243766 View Reply Original Report

Quoted By: >>12245492 >>12252304

What happens when octonions are studied in practice? I know quaternions can be used to project a 4d object into 3 dimensions, what would octonions project into? What is the group structure of octonions like when performing morphisms? I assume these questions have been studied extensively and I'm just beginning to study abstract algebra so I'm probably not even asking detailed questions.

Anonymous

Anonymous Sun 18 Oct 2020 12:26:17 No.12245204 Report

Quoted By:

>performing morphisms

Maybe your mom should have performed oral sex on you dad instead?

Anyway, just googled this:

http://homepages.wmich.edu/~drichter/octonions.htm

Should be a decent introduction.

Anonymous

Anonymous Sun 18 Oct 2020 13:28:55 No.12245351 Report

Quoted By:

https://www.youtube.com/watch?v=3BZyds_KFWM&list=PLNxhIPHaOTRZMO1VjJcs7_3dgyJ2qU1yZ

Anonymous

Anonymous Sun 18 Oct 2020 14:22:30 No.12245492 Report

Quoted By: >>12245513 >>12245581

>>12243766
In what class do I learn about quaternions?

Anonymous

Anonymous Sun 18 Oct 2020 14:29:37 No.12245513 Report

Quoted By:

>>12245492
John Beaz has written many blogs about them

Anonymous

Anonymous Sun 18 Oct 2020 14:54:21 No.12245581 Report

Quoted By: >>12245584 >>12245646 >>12250327

>>12245492
the only modern practical use i know of is using quaternions to represent rotations. hamilton reformulated a lot of physics in terms of quaternions, but vectors won out

Anonymous

Anonymous Sun 18 Oct 2020 14:55:35 No.12245584 Report

Quoted By: >>12245646 >>12245718

>>12245581
PS: i'll post a rotation w/ quaternions how-to if interested

Anonymous

Anonymous Sun 18 Oct 2020 15:12:45 No.12245646 Report

Quoted By: >>12245850

>>12245584
>>12245581
Yes I am interested, but I'm a math major so I'm more interested in the math then in any practical use

Anonymous

Anonymous Sun 18 Oct 2020 15:34:28 No.12245718 Report

Quoted By:

>>12245584
Yes but someone is gonna have to translate it into layman's terms

Anonymous

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Anonymous Sun 18 Oct 2020 16:37:00 No.12245850 Report

Quoted By: >>12245871 >>12245884 >>12250209

>>12245646
Sunday Afternoon w/ Quaternions [PART 1]

the fundamental quaternion identity is
$ i^2 = j^2 = k^2 = i j k = -1 $
in general, a quaternion can be written as
$ \tilde{\mathbf{q}} = 1 a + i b + j c + k d $
or in terms of "scalar" and "vector" parts
$ \begin{align} \tilde{\mathbf{q}} &= q + \mathbf{q} \\ q &= 1 a \\ \mathbf{q} &= i b + j c + k d \end{align} $

multiply two quaternions by distribution (maintain the order of the factors!) and application of the fundamental identity. using the 3d vector operations of dot and cross product to organize the result, one finds (problem for the reader, prove this)
$ \tilde{\mathbf{p}} \tilde{\mathbf{q}} = p q - \mathbf{p} \cdot \mathbf{q} + p \mathbf{q} + q \mathbf{p} + \mathbf{p} \times \mathbf{q} $

quaternion conjugation is defined similar to complex conjugation, but unlike complex conjugation can be written in terms of multiplication and addition (prove this)
$ \tilde{\mathbf{q}}^* = 1 a - i b -j c - k d = -\frac{1}{2}(\tilde{\mathbf{q}} + i \tilde{\mathbf{q}} i + j \tilde{\mathbf{q}} j + k \tilde{\mathbf{q}} k) $

quaternion norm/modulus is defined similar to complex norm/modulus, and results in a similar formula (prove this)
$ \| \tilde{\mathbf{q}} \| = \sqrt{\tilde{\mathbf{q}}^* \tilde{\mathbf{q}}} = a^2 + b^2 + c^2 + d^2 $
a quaternion with unit norm/modulus is called a versor
$ \| \check{\mathbf{q}} \| = 1 $

quaternion division is defined like complex division
$ \tilde{\mathbf{q}}^{-1} = \frac{\tilde{\mathbf{q}}^*}{\| \tilde{\mathbf{q}} \|^2} $

finally, there is a quaternion analog to Euler's identity for vector versor $\hat{\mathbf{n}}$
$ e^{\hat{\mathbf{n}} \theta} = \cos \theta + \hat{\mathbf{n}} \sin \theta$
that itself is a versor (prove this)

all the tools needed to formulate quaternion rotation and show that's really what it is are now on the assembled.

Anonymous

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Anonymous Sun 18 Oct 2020 16:44:15 No.12245871 Report

Quoted By: >>12246715

>>12245850
wow

Anonymous

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Anonymous Sun 18 Oct 2020 16:47:55 No.12245884 Report

Quoted By: >>12245887 >>12250209

>>12245850

Sunday Afternoon w/ Quaternions [PART 2]

suppose you have a spatial vector $\mathbf{r} \in \mathbb{R}^3$ that you want to rotate around unit vector $\hat{\mathbf{n}}$ by angle $\theta$ to get the transformed vector $\mathbf{r}'$ .

represent a spatial vector with a quaternion with scalar part equal to 0 and vector part equal to the spatial vector's rectangular coordinates
$ \mathbf{r} = \hat{\mathbf{x}} x + \hat{\mathbf{y}} y + \hat{\mathbf{z}} z \sim \mathbf{i} x + \mathbf{j} y + \mathbf{k} z $

with this representation, define the versor $\check{\mathbf{q}}$ as follows
$ \check{\mathbf{q}} = e^{\hat{\mathbf{n}} \frac{\theta}{2}} = \cos \frac{\theta}{2} + \hat{\mathbf{n}} \sin \frac{\theta}{2} $

the following quaternion equation is then equivalent to the desired transformation
$ \mathbf{r}' = \check{\mathbf{q}} \mathbf{r} \check{\mathbf{q}}^{-1} $

Anonymous

Anonymous Sun 18 Oct 2020 16:49:07 No.12245887 Report

Quoted By: >>12245901 >>12250209

>>12245884
proof:
$ \begin{align} \check{\mathbf{q}} \mathbf{r} \check{\mathbf{q}}^{-1} &= (\cos \frac{\theta}{2} + \hat{\mathbf{n}} \sin \frac{\theta}{2}) \mathbf{r} (\cos \frac{\theta}{2} - \hat{\mathbf{n}} \sin \frac{\theta}{2}) \\ &= \mathbf{r} \cos^2 \frac{\theta}{2} - \mathbf{r} \hat{\mathbf{n}} \cos \frac{\theta}{2} \sin \frac{\theta}{2} + \hat{\mathbf{n}} \mathbf{r} \cos \frac{\theta}{2} \sin \frac{\theta}{2} - \hat{\mathbf{n}} \mathbf{r} \hat{\mathbf{n}} \sin^2 \frac{\theta}{2} \\ &= \mathbf{r} \cos^2 \frac{\theta}{2} + (\hat{\mathbf{n}} \mathbf{r} - \mathbf{r} \hat{\mathbf{n}}) \cos \frac{\theta}{2} \sin \frac{\theta}{2} - (\mathbf{r} -2 \hat{\mathbf{n}} (\hat{\mathbf{n}} \cdot \mathbf{r})) \sin^2 \frac{\theta}{2} \\ &= \mathbf{r} \cos^2 \frac{\theta}{2} + 2(\hat{\mathbf{n}} \times \mathbf{r}) \cos \frac{\theta}{2} \sin \frac{\theta}{2} - \mathbf{r} \sin^2 \frac{\theta}{2} + 2 \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r}) \sin^2 \frac{\theta}{2} \\ &= \mathbf{r} (\cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}) + 2 (\hat{\mathbf{n}} \times \mathbf{r}) (\cos \frac{\theta}{2} \sin\frac{\theta}{2}) + \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r}) (2 \sin^2 \frac{\theta}{2}) \\ &= \mathbf{r} \cos \theta + (\hat{\mathbf{n}} \times \mathbf{r}) \sin \theta + \hat{\mathbf{n}} (\hat{\mathbf{n}} \cdot \mathbf{r})(1 - \cos \theta) \\ &= \mathbf{r} \cos \theta + (\hat{\mathbf{n}} \times \mathbf{r}) \sin \theta + \hat{\mathbf{n}} (\hat{\mathbf{n}} \cdot \mathbf{r}) - \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r}) \cos \theta \\ &= (\mathbf{r} - \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r})) \cos \theta + (\hat{\mathbf{n}} \times \mathbf{r}) \sin \theta + \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r}) \\ \check{\mathbf{q}} \mathbf{r} \check{\mathbf{q}}^{-1} &= \mathbf{r}_\perp \cos \theta + (\hat{\mathbf{n}} \times \mathbf{r}) \sin \theta + \mathbf{r}_\parallel \end{align} $
the last equation is the rodrigues rotation equation

Anonymous

Anonymous Sun 18 Oct 2020 16:53:39 No.12245901 Report

Quoted By: >>12250209 >>12250420 >>12252317

>>12245887
one of the terms takes a few lines to work out, so it's included seperately
$ \begin{align} \hat{\mathbf{n}} \mathbf{r} \hat{\mathbf{n}} &= \hat{\mathbf{n}}(\mathbf{r} \times \hat{\mathbf{n}} - \mathbf{r} \cdot \hat{\mathbf{n}}) \\ &= \hat{\mathbf{n}}(\mathbf{r} \times \hat{\mathbf{n}}) - \hat{\mathbf{n}}(\mathbf{r} \cdot \hat{\mathbf{n}}) \\ &= \hat{\mathbf{n}} \times (\mathbf{r} \times \hat{\mathbf{n}}) - \hat{\mathbf{n}} \cdot (\mathbf{r} \times \hat{\mathbf{n}}) - \hat{\mathbf{n}}(\mathbf{r} \cdot \hat{\mathbf{n}}) \\ &= \hat{\mathbf{n}} \times (\mathbf{r} \times \hat{\mathbf{n}}) - \hat{\mathbf{n}} (\mathbf{r} \cdot \hat{\mathbf{n}})\\ &= \mathbf{r} (\hat{\mathbf{n}} \cdot \hat{\mathbf{n}}) - \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r}) - \hat{\mathbf{n}}(\mathbf{r} \cdot \hat{\mathbf{n}}) \\ &= \mathbf{r} - 2 \hat{\mathbf{n}}(\hat{\mathbf{n}} \cdot \mathbf{r}) \end{align} $

I don't pretend to have a deeper understanding of why this works besides "it's equivalent to the Rodrigues formula". the math is mysterious, but really it's just axis-angle rotation in slightly different terms. quaternions are nice because you can interpolate them easily, which is important when modeling or animating smooth rotations. they also don't suffer gimbal lock.

Anonymous

Anonymous Sun 18 Oct 2020 20:55:31 No.12246715 Report

Quoted By:

>>12245871
commit sudoku

Anonymous

Anonymous Mon 19 Oct 2020 07:46:46 No.12248454 Report

Quoted By: >>12255831

hell, forget practical applications, what theoretical applications do octonions have? i know that octionions are the most general set of numbers akin to the complex numbers; that's gotta be important somewhere in math

Anonymous

Anonymous Mon 19 Oct 2020 20:23:13 No.12250209 Report

Quoted By:

>>12245850
>>12245884
>>12245887
>>12245901
Thanks for the quick lesson anon.

Anonymous

Anonymous Mon 19 Oct 2020 20:57:11 No.12250327 Report

Quoted By:

>>12245581
>but vectors won out
If you multiply the pauli matricies by i then you get the quaternions.
This shows they are related to spin in QM.
Maybe hamilton shouldn't have been doubted.

El Arcón

El Arcón Mon 19 Oct 2020 21:19:33 No.12250420 Report

Quoted By: >>12252316

>>12245901
the $tag is an automatic align environment$

vz !!WgTN4IZasIt

vz !!WgTN4IZasIt Tue 20 Oct 2020 11:09:41 No.12252304 Report

Quoted By:

>>12243766
Octonions do not have a group structure (the product isn't associative, (ab)c ? a(bc)), that's one of the reasons they're not used that much. I suppose the unit octonionic sphere is useful to prove things about 7-spheres the same way the quaternionic unit sphere is used to prove things about $S^3$ , just less effectively because groups are cooler. I'm not sure what you mean when you talk about quaternions projecting 4d objects in 3d, though.

El Arcón

El Arcón Tue 20 Oct 2020 11:15:07 No.12252316 Report

Quoted By:

>>12250420
this is wrong

Anonymous

Anonymous Tue 20 Oct 2020 11:15:24 No.12252317 Report

Quoted By: >>12253690

>>12245901
So much of this is simplified once you lift from R^3 into the Geometric Algebra G^3. R^3 is limited by needlessly self-enforced constraint of needing to map various geometric objects (points, vectors, planes) to vectors. G^3 solves this by giving natural geometric and algebraic representation to non-vectorial geometric element.

Anonymous

Anonymous Tue 20 Oct 2020 19:01:46 No.12253690 Report

Quoted By:

>>12252317
i was trying to teach myself about exterior algebra a couple weeks ago. i was mainly motivated to learn it to understand determinates better. i think i almost understood it this time (i've looked into in before), but there are still some things that seemed a little unclear. is algebraic geometry related?

Anonymous

Anonymous Wed 21 Oct 2020 06:41:40 No.12255831 Report

Quoted By:

>>12248454
I'm no expert, but don't the 2^2-nions cap out at octanions? I imagine that would be a case against >8 dimensional systems

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