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I have been listening to many videos and reading but I am very confused. Firstly, I read that quaternions exist in $\mathbb{R}^4$ which would seem to exclude imaginary numbers completely (I would have guessed). But secondly, if a quaternion is sort of like a complex number, why do we need $j$ and $k,$ why can't we just use $i$ throughout ($a+ib+jc+kd$) is different somehow than ($a+ib+ic+id$) and if so, in what way does $i$ differ from $j$ and $k$?

Also, if quaternions have something to do with rotation is 3D space, why is a quaternion in a hypersphere? If we want to do something with rotations in on the plane, is 3D space involved?

md2perpe
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releseabe
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  • The quaternions are different from the complex numbers, and $j$, $k$ are not equal to $i$ (though considered individually, $i$, $j$, and $k$ play similar roles). They are a generalization of the complex numbers (complex numbers are a subset). You could just use $i$ throughout your formula, but then you would be working in the complex numbers and not the quaternions. – xxxxxxxxx Nov 15 '20 at 08:06
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    I think the question would benefit from having more focus. All the parts have been answered already on our site, but not necessarily in the same place. See also this for some of the parts not handled in the first link. – Jyrki Lahtonen Nov 15 '20 at 09:13
  • Anyway, a quick explanation to the last question: To determine a rotation in 3D you need to specify the direction of an axis in 3D and the angle of rotation. That's four parameters right there. We can drop one by scaling the axes to a unit vector, but that's where 4D comes from (we can similarly adjust the scale and only use unit quaternions). – Jyrki Lahtonen Nov 15 '20 at 09:17
  • @JyrkiLahtonen. It can be difficult to focus on one thing when you are confused. – md2perpe Nov 15 '20 at 11:23
  • @md2perpe Understood. However, the goal of the site is to collect answers to questions in a searchable way rather than cure all possible combinations of confusions. There is a reason we have needs more focus as a closing reason. It is up to whoever wants to answer to help with the search. – Jyrki Lahtonen Nov 15 '20 at 11:31
  • In other words, I am not pointing fingers at the asker :-) – Jyrki Lahtonen Nov 15 '20 at 11:36
  • @JyrkiLahtonen. Kyllä. – md2perpe Nov 15 '20 at 11:55

3 Answers3

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Quaternions build a number system similar, but distinct from the complex numbers. Every quaternion may be written in the form $a + bi + cj + dk$ and $i,k,j$ being pairwise distinct make sure that this representation is in fact unique. If $i=j=k$ then we would have $$a + bi + cj + dk = a + (b+c+d)i + 0j +0k$$ violating uniqueness. You can think of it as adding three independent imaginary axes to the real line $\Bbb R$, ie making it a 4-dimensional vector space. But the real clue is that, just like the complex numbers have $i^2=-1$, quaternions come with algebraic relations (I think it was along the lines of $i^2 = j^2 = ijk = -1$), which allow you to have a sort of multiplication on that vector space. It turns out this multiplication is very handy to describe composing rotations.

Finally let’s answer the question, why quaternions require to be in $\Bbb R^4$ rather than $\Bbb R^3$. Quaternions mimick specifying an axis of rotation $r\in \Bbb R^3$ as well as an angle $\alpha \in \Bbb R$, which adds up to be a quantity in $\Bbb R^4$. Of cause you can use Euler Angles, which require one dimension less, but those lack most of the cool properties of quaternions...

Jonas Linssen
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Firstly, I read that quaternions exist in $\mathbb{R}^4$ which would seem to exclude imaginary numbers completely (I would have guessed).

The text didn't mean that quaternions are real, but probably that they are 4-dimensional over the reals: $t+xi+yj+zk$ has 4 real parameters $t,x,y,z$.

But secondly, if a quaternion is sort of like a complex number, why do we need $j$ and $k,$ why can't we just use $i$ throughout ($a+ib+jc+kd$) is different somehow than ($a+ib+ic+id$) and if so, in what way does $i$ differ from $j$ and $k$?

It is sort of like a complex number, but it has two extra imaginary dimensions. It is an extended type of complex number. Note that $ijk=-1$ while $iii=-i.$ We therefore can not just replace $j$ and $k$ with $i$.

Also, if quaternions have something to do with rotation is 3D space, why is a quaternion in a hypersphere?

A rotation an angle $\theta$ around a direction $n=(n_x,n_y,n_z),$ where $|n|:=\sqrt{n_x^2+n_y^2+n_z^2}=1,$ can be described using a quaternion $r=\cos\theta+n_x \sin\theta\,i+n_y\sin\theta\,j+n_z\sin\theta\,k$. The magnitude of this is $$ |r|=\sqrt{\cos^2\theta+n_x^2 \sin^2\theta+n_y^2\sin^2\theta+n_z^2\sin^2\theta} = 1. $$ This means that $r$ as a 4-tuple $(\cos\theta, n_x\sin\theta, n_y\sin\theta, n_z\sin\theta)$ lies on the hypersphere $S^3 \subset \mathbb{R}^4.$

If we want to do something with rotations in on the plane, is 3D space involved?

For rotations in a plane, we can do with the ordinary complex numbers. Let $(x,y)\in\mathbb{R}^2$ be a point in the plane and set $z=x+iy.$ To rotate the point the angle $\theta$ around origin, just multiply $z$ with $e^{i\theta} = \cos\theta + \sin\theta\,i$ and take the real and imaginary parts of the result.

md2perpe
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I think your question reflects some basic misunderstandings about what mathematical objects are. In particular the misunderstanding that there is such a thing as "what a mathematical object is".

Namely, two mathematical objects can be exactly the same even if they are completely different. Look at some examples:

  • (a) Remainders when dividing by $12$, with addition modulo $12$ (e.g. $6+7=1$); (b) Rotations around origin by a multiple of $30^\circ$ with respect to composition of rotations; (c) Multiplication of nonzero remainders modulo $13$. Those three are different objects, yet they obey the same arithmetic rules - all three have a structure of a group, all three have $12$ elements, and in all three there is one element ("generator") which produces all elements when you use the operation repeatedly. (In (c) you can take the remainder $2$ as such element, as $2^0,2^1,\ldots 2^{11}$ are all different $\pmod{13}$.)
  • (a) Complex numbers and (b) Pairs of real numbers: $\mathbb R\times\mathbb R=\{(a,b)\mid a,b\in\mathbb R\}$, with addition given as $(a,b)+(c,d)=(a+c,b+d)$ and multiplication given by $(a,b)(c,d)=(ac-bd, ad+bc)$. Obviously, in the second case $(a,0)$ "behaves like" the real complex number $a$ and $(0,b)$ "behaves like" a pure imaginary number $bi$. (In fact, $(0,1)$ "behaves like" $i$.)

What is a moral of this? It is this: we just don't want to distinguish two mathematical objects if they are exactly the same, provided that we forget the nature of the objects they are made of. In algebra, we formalize this notion of being the same "up to the nature of the constituents" using the definition of a isomorphism. An isomorphism is a bijection between the underlying sets for two algebraic structures, which maps results of operations into results of operations. If there is an isomorphism between two algebraic structures, we call those two structures "isomorphic". This means that you can replace the elements of one with the elements of the other and all the calculation rules remain the same - in other words, they are "the same up to the nature of the constituents". Thus, in mathematics, we don't particularly care which one example of an algebraic structure (out of many mutually isomorphic ones) we get to analyze.

For example, if you make a bijection $f:\mathbb R\times\mathbb R\to\mathbb C$, given by $f(a,b)=a+bi$ - this is one isomorphism between $\mathbb R\times \mathbb R$ (with addition and multiplication as given above) and $\mathbb C$ (with complex number addition and multiplication). In my first example, identify the remainder $x\pmod{12}$ with a rotation by $x\cdot 30^\circ$ and with the remainder of $2^x\pmod{13}$.

Now back to quaternions. There is no point in musing about what they really are. You just need one possible definition of quaternions and the operations between them. Anything that is isomorphic to that example can equally be called "quaternions", and anything that is not isomorphic to that example is, well, not the quaternions.

For one possible definition, take $\mathbb R^4$ (set of quadruplets of real numbers) and define operations:

$$(a,b,c,d)+(p,q,r,s):=(a+pb+q,c+r,d+s)$$ $$(a,b,c,d)(p,q,r,s):=(ap-bq-cr-ds,aq+bp+cr-ds,ar-bs+cp+dq,as+br-cq+dp)$$

What you get are quaternions. You may want to call the elements $(0,1,0,0),(0,0,1,0),(0,0,0,1)$ the names $i,j,k$, respectively, and you may want to call the element $(a,0,0,0)$ just $a$ (for $a\in \mathbb R$) - silently identifying $\mathbb R\times\{0\}\times\{0\}\times\{0\}$ with $\mathbb R$ via isomorphism $a\mapsto (a,0,0,0)$, in which case you can prove $(a,b,c,d)=a+bi+cj+dk$ and go from there. Of course, you can identify some rotations of $3D$ space with quaternions - again via an isomorphism of the set of rotations (with respect to composition) to a particular subset of quaternions (with respect to multiplication).

Isomorphic structures have all the properties identical, which now lets you prove that quaternions are not isomorphic to complex numbers. (Using tools of linear algebra: quaternions are of dimension $4$ over reals, while complex numbers are of dimension $2$.) Also, in quaternions "of the kind explained above" (i.e. over $\mathbb R^4$) we have $i=(0,1,0,0)\ne(0,0,1,0)=j$) so in no isomorphic structure you can ever have $i=j$ because isomorphisms are bijections - i.e. they must be "one to one".

My bigger point is: do not spend time thinking what quaternions really are. Study their properties. The same properties will be the properties of any particular instance of quaternions - i.e. of any one of the many mutually isomorphic structures of quaternions. A part of a training of a mathematician is to be able to silently and seamlessly switch from one structure to another isomorphic structure, we do it all the time, and you should be able to do so as well.

  • Note there is also an isomorphism of $\mathbb C$ onto a subset $\mathbb R\times\mathbb R\times{0}\times{0}$ given by $a+bi\mapsto (a,b,0,0)$, identfying complex $i$ with quaternion $i$... but, curiously enough, there is another isomorphism mapping $\mathbb C$ onto $\mathbb R\times{0}\times\mathbb R\times{0}$ mapping $a+bi\mapsto(a,0,b,0)$ effectively identifying complex $i$ with quaternion $j$ ... and ther is even another one identifying complex $i$ with quaternion $k$. Thus, the complex numbers can be seen as a subset of quaternions, sort of, in three different ways at once. –  Nov 15 '20 at 11:05
  • i think knowing what kind of problems, often physical problems, the originators of a kind of math were trying to solve is useful. it is my understanding that in the 19th century and before, math tended not to exist in isolation. maybe i should read more about why Hamilton was doing this. – releseabe Nov 15 '20 at 11:18
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    Actually, if ${\bf u}=u_1i+u_2j+u_3k$ is any unit vector in $\Bbb{R}^3$, then $({\bf u})^2=-1$ as a quaternion. And we have infinitely many ways of mapping a copy of $\Bbb{C}$ into $\Bbb{H}$ by a homomorphism of rings: $a+bi\mapsto a+b{\bf u}$. – Jyrki Lahtonen Nov 15 '20 at 11:34