Questions on Quaternion Algebra (introductory stuff)

Question

I am a relatively new Mathematics student who understands about complex numbers and how they work. I am currently trying to create a 3D computer graphics engine and I heard that quaternion algebra may be useful in various operations. As a result I am trying to learn it.

I understand that quaternions are like the extension of the complex number system to the 4th dimension (i.e. there are not only reals and imaginaries (2D), but there are also hyperimaginary (j) and hyperhyperimaginary (k) numbers.

I understand the fundamental rule of quaternions (i.e. that $i^2=j^2=k^2=ijk=-1$ and that the quaternion algebra is noncommutative). I have 2 questions:

a) Why is quaternion algebra noncommutative?

b) Is there anything particularly 'special' about j and k?

I am self-taught so try to keep answers as simple as possible.

Answer 1

It's not really meant to be a complete answer, but you ask why the quaternion algebra is non-commutative. Well, if it is commutative, it would not be very useful for 3D graphics. The reason we use quaternions is because rotations in 3D can be mapped to them and thus give a very efficient way of computing rotations. More efficient than with say matrices.

Now, take a book and hold it in your hands, flat in front of you. Perform a 90° turn of the book towards yourself, then a 90° turn to the left. Remember the orientation of the book w.r.t. yourself. Go back to your initial orientation and perform the two operations in the reverse order. Notice something? That's right, the end result is different. Rotations are non-commutative. Hence quaternions better be if they're going to be useful for rotational algebra.

Answer 2

It's an interesting endeavour in group theory to qualify which properties go to which number systems. It can be shown that as the number system gets more complex, it becomes more powerful, but it loses functionality.

For example, the complex numbers are not ordered. The quaternions are not commutative. If you want to really boggle your mind, the octonions are noncommutative and nonassociative.

Quaternions are not commutative because the product of two imaginary units are order specific. It has to be this way for the system to work.

j and k are the same as i. They are each an imaginary unit in their own axis.

Answer 3

a) Why is quaternion algebra noncommutative?

The defining relations already give you that $ij=k\neq -k=ji$, so that is sufficient to show that they don't commute.

One good way to place the quaternions is that they are linked to $3$-d rotations in a way similar to the link between complex numbers and $2$-d Euclidean rotations. While the $2$-d rotation group is Abelian, the $3$-d rotation group is not.

b) Is there anything particularly 'special' about j and k?

Among the most salient features of $i,j$ and $k$ are these:

• They are all length 1 according to the norm in $\Bbb H$
• They anticommute (that is, they pairwise satisfy $ab=-ba$.)
• Those two properties encode the cross product into the quaternion product.

This is enough to make them a prototypical orthogonal basis. In fact, that is the main use of the quaternions: the subspaces generated by $i$, $j$ and $k$ respectively are made to correspond with the $x,y$ and $z$ axes respectively, and each nonzero quaternion induces a $3$-d rotation on those axes.

Now, that isn't to say that $i,j$ and $k$ are somehow unique with respect to this. Actually, you can take any nonzero quaternion $x$ and take $xix^{-1},xjx^{-1},xkx^{-1}$, and this triple has exactly the same properties that I have bulleted above. They will still generate the quaternion algebra. All that has been done is that the original $i-j-k$ triad has been rotated with a $3$-d rotation, but otherwise the geometric configuration of the three has not been substantially changed. The transformation $q\mapsto xqx^{-1}$ is just a ring automorphism of $\Bbb H$ under which the 'ordinary' generating elements get mapped to some stranger looking elements.

Anyhow, the point of choosing our generators $i,j,k$ this way is to make the most out of what the product does. If you visit the wiki link above, you will find how this choice encodes not only the cross product but also the dot product into quaternion multiplication.