Vector division with quaternions?

[Johnny5]

While trying to learn about rotation matrices I came across this:

 

Quaternion Tutorial

 

If you go to page 15, you will see this:

 

Definition of a Quaternion

 

A quaternion is the geometrical quotient of two vectors.

 

Let A denote a vector, and let B denote a vector.

 

Q = quaternion = A/B

 

Tom Mattson said that vector division isn't defined. Apparently it is.

 

[Tom Mattson]

Tom Mattson said that vector division isn't defined.

 

I said a lot of things in that thread' date=' and I can't possibly hope to recall all of it without re-reading it, but the gist of what I was saying is that there is no definition of vector division [i']as a multiplicative inverse[/i], with reference to the already-defined multiplicative operations with vectors. Sure you can give meaning to the symbols S=A/B, and you can even call it a quotient if you want to. But if we want to interpret division by B as multiplication by B^-1, then what is B^-1? And what is the multiplicative operation?

 

Apparently it is.

 

Is it? I didn't look at the whole thing, but where exactly does he do anything more than simply write S=A/B? You did that much when talking about Newton's 2nd law (m=F/a).

[matt grime]

SO, if you "divide" two vectors, presumably in R^3, you get something that is in a 4-dimensional real space, or a 2-d complex space, or the quarternions. Hmm. And you think that's a reasonable notion of division of two vectors, which Tom, as most poeple would, took to mean as some operation on R^3 that 'undoes' some multiplicative operation. Of course, since we can embed R^3 into the quarternions in any number of ways we can take inverses there - again the operation will not be closed with respect to the embedding.

[Johnny5]

SO, if you "divide" two vectors, presumably in R^3, you get something that is in a 4-dimensional real space, or a 2-d complex space, or the quarternions. Hmm. And you think that's a reasonable notion of division of two vectors, which Tom, as most poeple would, took to mean as some operation on R^3 that 'undoes' some multiplicative operation. Of course, since we can embed R^3 into the quarternions in any number of ways we can take inverses there - again the operation will not be closed with respect to the embedding.

 

Well I don't know that much about quaternions yet, but if i find a good article then i will.

 

I do know that a quaternion is composed of four elements:

 

[s,i,j,k]

 

I know that S represents an amount of rotation, an angle.

 

And that i,j,k represent unit vectors, an orthogonal basis.

Lets see what else do i know...Oh, I know the thing Hamilton supposedly scribbled down, on that bridge in Ireland...

 

i^2=j^2=k^2=ijk=-1

 

so clearly, i^2 is an imaginary unit vector, as with j, and k.

 

So the 'space' as it were is complex, rather than strictly real.

 

In fact, the three dimensions are actually in Imaginary 3D space, yet to visualize that one is going to think 3 dimensionally anyways, so I'm not sure about all this yet.

 

As to your question Matt, I don't know enough about them yet, to know what set they are closed on.

 

The fourth element of a quaternion is an angle, which doesnt constitute an extra spatial dimension, it constitutes a rotation.

 

I'm only just beginning to even wonder what Hamilton was trying to do, to say much about quaternions. Or Gibbs, see I'm not even sure who did what, why, and when.

[matt grime]

Right, let's clear up a few things straight away.

 

1. S is not a rotation, but it can be *interpreted* that way if you so choose I imagine. They are *just* a 4-dimensional real division algebra. Saying the first is an angle is like saying the real numbers are lengths. If it helps to visualize them in a certain way feel free to do so but don't confuse the visuallization with them as abstract objects. The unit quartrnions aer iso to the 4-sphere, and thus SU(2), and various things like the hopf fibration can be described using them.

 

2. The quarternions (denoted H for hamilton) *are* real 4-dimensional vector space, they are also a 2-dimensional complex vector space. I could have used the word algebra too instead of vector space.

 

3. The elements i,j,k *are not* a priori orthogonal, since there is no such thing as a canonincal inner product, though there is an obvious one with respect to which we can delcare them orthogonal.

 

 

4. Hamilton was attempting to define a 3-d space over the reals that was a field, liek C is a 2-d real vector space that is also a field. He realized that in fact it needed to be 4-d, and that it couldn't be commutative, hence only a division algebra. we also have the 8-d octonions.

[Johnny5]

Right' date=' let's clear up a few things straight away.

 

1. S is not a rotation, but it can be *interpreted* that way if you so choose I imagine. They are *just* a 4-dimensional real division algebra. Saying the first is an angle is like saying the real numbers are lengths. If it helps to visualize them in a certain way feel free to do so but don't confuse the visuallization with them as abstract objects. The unit quartrnions aer iso to the 4-sphere, and thus SU(2), and various things like the hopf fibration can be described using them.

 

2. The quarternions (denoted H for hamilton) *are* real 4-dimensional vector space, they are also a 2-dimensional complex vector space. I could have used the word algebra too instead of vector space.

 

3. The elements i,j,k *are not* a priori orthogonal, since there is no such thing as a canonincal inner product, though there is an obvious one with respect to which we can delcare them orthogonal.

 

 

4. Hamilton was attempting to define a 3-d space over the reals that was a field, liek C is a 2-d real vector space that is also a field. He realized that in fact it needed to be 4-d, and that it couldn't be commutative, hence only a division algebra. we also have the 8-d octonions.[/quote']

 

Ok thanks for the answers, its a lot to figure out all at once, but you are helping.

 

Thanks again.

 

4 dimensional real division algebra

 

4 element real division algebra sounds better.

 

Unit quaternions are isomorphic to the 4-sphere, hence to SU(2).

 

SU(2) group is the set of unitary 2x2 hermitian matrices

The terminology just keeps getting more bizarre by the minute

 

Non-commutative instantons on the 4-sphere from quantum groups

 

So what the hell is an instanton?

 

 

(I am thinking do these people know what they are talking about)

[matt grime]

"4 element divison algebra" would indicate that it had 4 elements. There are an infinite number of quarternions (as there must be in any non-zero real vector space), so I really advise you to stop inventing notation that makes no sense.

[Johnny5]

"4 element divison algebra" would indicate that it had 4 elements. There are an infinite number of quarternions (as there must be in any non-zero real vector space), so I really advise you to stop inventing notation that makes no sense.

 

You should be telling that to the quaterniation folk, not me.

 

On a serious note, can you perform vector division with quaternions?

 

You have two 3D vectors in three dimensional Euclidean space, and you take their quotient, and you get a 4D quaternion in a ? space?

[matt grime]

Eh? What on earth? You wanted to call it a 4 element real vector space, no one else.

 

It's a divisoin algebra, Johnny.....

 

Only you seem to want to divide elements of R^3, so why don't you decide what you mean.

[Johnny5]

It's a divisoin algebra' date=' Johnny.....

[/quote']

 

 

You make it sound trivial.

 

I've just read through the quaternion manual, and i saw them referring to

 

tensors, versors, vectors, and scalars.

 

The terminology seems worked out by one person.

 

Is the usage of the term 'tensor', in this manual, that of the tensor calculus?

 

Tensor operator applies tension to a vector.

Versor operator applies version to a vector.

 

It's in the manual.

[matt grime]

It is trivial Johnny since in a division algebra one can divide by definition. The quartnions possess a multiplicative norm.

[Johnny5]

It is trivial Johnny since in a division algebra one can divide by definition. The quartnions possess a multiplicative norm.

 

Matt let me ask you straight out, do you understand the things in that manual?

 

Because here is what I'm getting from you:

 

1. 'it' is a division algebra.

2. Figure out what your division is going to be of.

[matt grime]

I have no idea, nor do I care, what is in this manual that you have and that I do not, and you are being obfuscatory to say the least when talking about this mythical manual.

 

The Quarternions are famously a divisoin algebra, they are the 'it' I refer to, and I do understand them very well, as will most mathematicians: the unit integral quarternions and D_8 are the smallest non-isomorphic groups with the same character table for instance.

 

For someone who is using "vector" without bothering to state which vector space he is referring to that is a very strange post you just made.

 

I know how to divide quarternions, but you are just asking how to "divide vectors" what vectors; you are being very unclear as to what vectors you mean to divide. Do you mean the vectors in H, the quarternions, or not? I suspect you mean, from you original post, that you want to divide two vectors in R^3 in some sense, though you haven't been clear on this.

 

It is you that needs to start explaining yourself clearly.

[Johnny5]

you are being very unclear as to what vectors you mean to divide.

 

It's not deliberate i assure you.

 

Do you mean the vectors in H' date=' the quarternions, or not? I suspect you mean, from you original post, that you want to divide two vectors in R^3 in some sense, though you haven't been clear on this.

 

It is you that needs to start explaining yourself clearly.[/quote']

 

 

I mean whats in the manual, there's a link in the first post to it.

 

They speak about

 

Versors

 

And I quote:

 

A versor is a quotient of two non-parallel vectors of equal length.

 

But you are right i mean two arbitrary vectors in R³.

 

 

Suppose that you take the quotient of two non-parallel vectors successfully. Because they were not pointing in the same direction, there is a remainder, which will have some direction, hence the remainder is a vector. Furthermore, since the vectors you took the quotient of had equal magnitudes, the quotient, which is a vector, should be a unit vector?

[Johnny5]

Ok, I managed to find an 1899 reprint of Hamilton's lectures on quaternions.

 

On page 60 he says:

 

"But, there was still another view of the whole subject, sketched long afterwards in another communication to the R.I. Academy, on which it is unecessary to say more than a few words in this place, because it is, in substance, the view adopted in the following Lectures, and developed with some fullness in them: namely, that view according to which a QUATERNION is considered as the QUOTIENT of two directed lines in tridimensional space."

 

So the ratio of two vectors is a quaternion.

In the case where the magnitudes of the two vectors are equal, the ratio is called a versor, which is another name for a unit vector.

 

Then on the very next page, he begins do describe what he calls, "The method of Argand".

[matt grime]

And what is your point? Note it states "two directed lines", not necessarily vectors in the sense we mean, though we'd have to guess what the old terminology meant. Of course as Hamilton invented vectors after inventing quarternions (yes, this surprised me too, but is apparently true) we cannot state he invented quarternions in order to "divide vectors".

[Johnny5]

This is my first post using the new server.

 

[math]

\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}

[/math]

 

Great you have latex back.

 

Also, i think the new colors are great, they give the forum an appealing look, easy to read.

 

Ok so as for quaternions.

 

It doesn't make sense that Hamilton would invent quaternions before inventing vectors. In fact, it doesnt make sense to say that he invented vectors at all, since he didn't, Newton was using them, perhaps Newton didn't call them vectors, but certainly they were being used prior to Hamilton's birth.

 

So it does indeed make sense that Hamilton developed quaternions in order to divide one vector by another. In fact, it makes perfect sense. Anything else, doesn't make sense.

[Johnny5]

Let Q1 denote quaternion (w1,x1,y1,z1).

Let Q2 denote quaternion (w2,x2,y2,z2).

 

Let the product of Q1 and Q2 follow the normal rules of multiplication of one polynomial by another, so that we have:

 

(Q1)(Q2)= (w1,x1,y1,z1)(w2,x2,y2,z2)

(Q1)(Q2)= (w1+x1i+y1j+z1k)(w2+x2i+y2j+z2k)

 

Consider the case where w1=w2=0.

 

(Q1)(Q2)= (0+x1i+y1j+z1k)(0+x2i+y2j+z2k)

 

Following the rules of multiplication of one polynomial by another we have:

 

x1ix2i+x1iy2j+x1iz2k+

y1jx2i+y1jy2j+y1jz2k+

z1kx2i+z1ky2j+z1kz2k

 

 

Now, here is the formula for the vector cross product:

 

V1 X V2 = <a,b,c> X <d,e,f> =

 

i(bf-ce)-j(af-cd)+k(ae-bd)

 

Here is Q1Q2 again:

 

x1ix2i+x1iy2j+x1iz2k+

y1jx2i+y1jy2j+y1jz2k+

z1kx2i+z1ky2j+z1kz2k

 

Which is equivalent to:

 

x1ix2i+y1jy2j+z1kz2k +

 

x1iy2j+x1iz2k+

y1jx2i+y1jz2k+

z1kx2i+z1ky2j

[Johnny5]

Let Q1 denote quaternion (w1' date='x1,y1,z1).

Let Q2 denote quaternion (w2,x2,y2,z2).

 

Let the product of Q1 and Q2 follow the normal rules of multiplication of one polynomial by another, so that we have:

 

(Q1)(Q2)= (w1,x1,y1,z1)(w2,x2,y2,z2)

(Q1)(Q2)= (w1+x1i+y1j+z1k)(w2+x2i+y2j+z2k)

 

Consider the case where w1=w2=0.

 

(Q1)(Q2)= (0+x1i+y1j+z1k)(0+x2i+y2j+z2k)

 

Following the rules of multiplication of one polynomial by another we have:

 

x1ix2i+x1iy2j+x1iz2k+

y1jx2i+y1jy2j+y1jz2k+

z1kx2i+z1ky2j+z1kz2k

 

 

Now, here is the formula for the vector cross product:

 

V1 X V2 = <a,b,c> X <d,e,f> =

 

i(bf-ce)-j(af-cd)+k(ae-bd)

 

Here is Q1Q2 again:

 

x1ix2i+x1iy2j+x1iz2k+

y1jx2i+y1jy2j+y1jz2k+

z1kx2i+z1ky2j+z1kz2k

 

Which is equivalent to:

 

x1ix2i+y1jy2j+z1kz2k +

 

x1iy2j+x1iz2k+

y1jx2i+y1jz2k+

z1kx2i+z1ky2j

 

I didn't get to finish this yesterday, but I was trying to figure something out. I worked on it last night. Here it is:

 

In a right handed coordinate system we have:

 

i X j = k

i X k = -j

j X k = i

 

and also

 

j X i = -k

k X i = j

k X j = -i

 

Now, look at the expansion of the product of two quaternions again:

 

 

x1ix2i+y1jy2j+z1kz2k +

 

x1iy2j +x1iz2k+

y1jx2i+y1jz2k+

z1kx2i+z1ky2j

 

Let it be the case that:

 

x1iy2j = x1y2ij=x1y2(i X j) = x1x2 k

 

x1iz2k= x1z2ik=x1z2(i X k) = x1z2 (-j) = -x1z2 j

 

y1jx2i = y1x2ji=y1x2(j X i) = y1x2 (-k) = - y1x2 k

 

y1jz2k = y1z2jk = y1z2(j X k) = y1z2 i

 

z1kx2i = z1x2 (ki) = z1x2 (k X i) = z1x2 j

 

z1ky2j = z1y2 (kj) = z1y2 (k X j) = z1y2 (-i) = -z1y2 i

 

From which it follows that the product of two quaternions Q1,Q2 is given by:

 

x1ix2i+y1jy2j+z1kz2k +

 

x1x2 k -x1z2 j - y1x2 k + y1z2 i + z1x2 j -z1y2 i

 

Which leads to this:

 

x1ix2i + y1jy2j + z1kz2k +

 

(y1z2-z1y2)i-(x1z2-z1x2)j+(x1x2 - y1x2)k

 

As you can see the three terms above are for the cross product of two vectors.

 

So to recapitulate.

 

Suppose that we are given two quaternions Q1, and Q2:

 

Q1 = (w1,x1,y1,z1)

Q2 = (w2,x2,y2,z2)

 

For now consider the case where w1=w2=0. The product of Q1 with Q2 leads to this:

 

(Q1)(Q2)= (0+x1i+y1j+z1k)(0+x2i+y2j+z2k)

 

Which leads to:

 

(Q1)(Q2) = [x1ix2i + y1jy2j +z1kz2k'] + V1 X V2

 

Where:

 

V1 X V2 = <x1,y1,z1> X <x2,y2,z2> = (y1z2-z1y2)i-(x1z2-z1x2)j+(x1x2 - y1x2)k

 

Now, since we assumed that quaternion multiplication is associative, it follows that:

 

 

(Q1)(Q2) = [x1x2ii + y1y2jj +z1z2kk] + V1 X V2

 

Now, because i was using cross product to handle

 

ij as i X j

ik as i X k

jk as j X k

 

and so on, it follows that i need to handle

 

ii as i X i

jj as j X j

kk as k X k

 

So lets start with evaluating i X i.

 

We can use the determinant so that we don't make any mistake.

 

[math]

i \times i = \left| \begin{array}{ccc}i & j & k\\1 & 0 & 0\\1 & 0 & 0\end{array}\right|

[/math]

 

Which you can see is equal to zero.

 

But now if we interpret ii as 1, jj as 1, and kk as 1, we have:

 

(Q1)(Q2) = [x1x2ii + y1y2jj +z1z2kk] + V1 X V2 =

[x1x2 + y1y2 +z1z2] + V1 X V2

 

And the first three terms are the scalar product or dot product of two vectors V1,V2. That is:

 

[math] Q_1 Q_2 = V_1 \bullet V_2 + V_1 \times V_2 [/math]

 

But the above formula is only true provided that:

 

ii=1

jj=1

kk=1

 

But that is not how Hamilton interpreted them.

 

In fact, the story is that he carved the following on some bridge in Ireland:

 

ii=-1

jj=-1

kk=-1

 

If we take that as the case then the product of two quaternions would have been given by:

 

(Q1)(Q2) = [-x1x2 - y1y2 -z1z2] + V1 X V2 =

-[x1x2 + y1y2 +z1z2] + V1 X V2

 

From which it follows that:

 

[math] Q_1 Q_2 = -V_1 \bullet V_2 + V_1 \times V_2 [/math]

 

Yet, in using the formula for the determinant, what we should have used was:

 

ii=0

jj=0

kk=0

 

So what is the correct interpretation of quaternion multiplication?

[Johnny5]

All right here is an article which explains how to add and multiply two quaternions:

 

Addition and multiplication of quaternions

Two such quaternions are added by adding the real parts and the imaginary parts separately:

 

(a + u) + (b + v) = (a + b) + (u + v)

The multiplication of quaternions translates into the following rule:

 

(a + u) (b + v) = (ab − <u, v>) + (av + bu + u × v)

Here, <u, v> denotes the scalar product and u × v the vector product of u and v.

 

Let Q1= (a+u)=quaternion 1

Let Q2 = (b+v)=quaternion 2

 

a is the scalar part of quaternion 1, and u) is the vector part.

b is the scalar part of quaternion 1, and v) is the vector part.

 

So if the above article is correct, then in the case where a=b=0 the answer should be

 

Q1Q1= -(dot product of u with v) + (cross product of u with v)

[reverse]

lol...

 

that's impressive.

 

way above my head.

 

What did he want to use this for in the end anyway...did he have some practical problem to solve?

I'm assuming that you don’t invent a style of calculation unless you need it to do something...like tensor analysis or something (heh)

[matt grime]

It doesn't make sense that Hamilton would invent quaternions before inventing vectors. In fact' date=' it doesnt make sense to say that he invented vectors at all, since he didn't, Newton was using them, perhaps Newton didn't call them vectors, but certainly they were being used prior to Hamilton's birth.

 

So it does indeed make sense that Hamilton developed quaternions in order to divide one vector by another. In fact, it makes perfect sense. Anything else, doesn't make sense.[/quote']

 

You appear to be using "vector" to mean a quantity with length and direction. This is strictly not the sense which are we are talking about: vector algebra or linear algebra if you prefer.

 

We can speculate (some of us in a better informed manner than others it appears) wildly about motivation, but as far as we can tell Hamilton was the first to use the word vector, and along with Gibbs one of the first to develop their abstract theory.

 

 

In modern terminology Hamilton was trying, it is alleged, to extend the construction of C as a vector space over R with an algebra structure into a higher dimensional theor. In particular he is supposed to have attempted to create a set of "numbers" a+bi+cj with similar properties as the complex numbers a+bi. His Eureka moment was the realization that you needed a 4'th object, k, to make it work: no consistent attemp for 3 things is possible within the contraints he wanted:

 

i^2=j^2=-1 is ok, but what do we declare ij to be?

 

if ij = a+bi+cj then multipliying by i on both sides means

 

-j=ai-b+cij=ai-b+a+bi+cj

 

thus c=-1, a-b=0 and a+b=0 or a=b=0 so ij --j, but then i(1+j)=0 so we have zero divisors, which means it can't be a field, thus meaning it isn't that much like C.

 

Introducing k removes this problem, though we lose commutativity so strictly we only have a division algebra not a field.

 

All of these terms would pretty much have been meaningless to Hamilton, I imagine.

[reverse]

Lol.

 

Very good.

 

See I’m interested in this for a practical reason.

You might like this.

 

 

“ Hamilton wrote to his son:

 

Every morning in the early part of the above-cited month [Oct. 1843] on my coming down to breakfast, your brother William Edwin and yourself used to ask me, 'Well, Papa, can you multiply triplets?' Whereto I was always obliged to reply, with a sad shake of the head, 'No, I can only add and subtract them.'

 

We can guess how Hollywood would handle the Brougham Bridge scene in Dublin. Strolling along the Royal Canal with Mrs. H-, he realizes the solution to the problem, jots it down in a notebook. So excited, he took out a knife and carved the answer in the stone of the bridge.

 

Hamilton had found a long sought-after solution, but it was weird, very weird, it was 4D. One of the first things Hamilton did was get rid of the fourth dimension,”

 

 

http://world.std.com/~sweetser/quaternions/intro/history/history.html

[reverse]

and to see the actual thing..

 

http://serge.mehl.free.fr/jpeg/quaternions_jf.jpg

[reverse]

And it’s always nice to see where a concept first occurred ..

Here is the bridge….and Hey…guess what…Octonions also.

 

See below.

 

 

 

http://maths.ucd.ie/photo_gallery/ucd150KJ/ppt/img28.html