Quaternions as an Algebra

I'm lacking some vital understanding about quaternions and algebras in general. If we first define $V=\{a+bi+cj+dk|a,b,c,d\in\mathbb{R}\}$. Then we define scalar multiplication, vector multiplication, addition e.t.c. But as I understand it, aren't $1,i,j,k$ meant to be vectors, meaning that $bi$ for example has to already be defined even for the set to be defined. Very confused! As usual thanks for any replies!

Solution 1:

What you write for $V$ is not a precise definition (strictly speaking). One defines $\mathbb{H} := \mathbb{R}^4$ as a $\mathbb{R}$-vector space, so that quaternions are $4$-tuples $(a,b,c,d)$ of real numbers $a,b,c,d \in \mathbb{R}$. If we define $1 := (1,0,0,0)$, $i := (0,1,0,0)$, $j := (0,0,1,0)$ and $k := (0,0,0,1)$, then by construction we have $(a,b,c,d) = a1 + bi + cj + dk$. After that, one defines the multipliciation in such a way that the relations $i^2=j^2=k^2=-1$, $ij=k$ are satisfied. This is a bit cumbersome, especially when one wants to check that $\mathbb{H}$ is in fact a skew field. A more elegant way is to define $\mathbb{H}$ as the subalgebra $\left\{\begin{pmatrix} z & w \\ -\overline{w} & \overline{z} \end{pmatrix} : z,w \in \mathbb{C}\right\}$ of $M_2(\mathbb{C})$.

Solution 2:

In the quaternion algebra $\;\Bbb H\;,\;\;1,i,j,k\;$ are just abstract elements, and then there are defined operations and relations: $\;i^2=j^2=k^2=-1\;,\;\;ij=k\;$ , etc.

I don't understand what's the problem with seeing these algebra's (or any other's, of course) elements as "vectors": as part of a four dimensional real vector space, all the elements are vectors, and thus you can see $\;bi\;,\;\;b\in\Bbb R\;$ the scalar product of the vector $\;i\;$ with the scalar $\;b\;$ , or as part of a ring, the product of the two ring elements $\;b\,,\,i\;$. Both these products, formally different, are the same since an algebra behaves well in this respect.

In short: $\;\Bbb H\;$ can be seen as a four dimensional real vector space which is also a non-commutative ring (in fact, a division algebra).

Solution 3:

Mull over this: when you define polynomials over $x$ in terms of $x$, do you have trouble accepting $x$ alone as a polynomial?

If it helps, you could just think of $1,i,j,k$ as symbols, with no special properties other than being distinct. Then you define quaternions by

• declaring the underlying set to be formal expresions of the form $a+bi+cj+dk$, whose nicknames are "quaternions."
• declaring all the addition multiplication rules you need to make the implied addition and multiplication mean something official (like $1\cdot x=x$, $ij=-ji$, $ai+bi=(a+b)i$, distributivity, etc.)

When both of these steps are done, the four original symbols are represented in the formal set: e.g. $0\cdot 1+1i+0k+0j$ is the same thing as $i$, and henceforth would be a "quaternion."

Both polynomial rings and quaternions can be viewed as building rings out of generators and relations.

I'm guessing you're familiar with polynomial rings in two variables, like $F[x,y]$. When we build those there is an assumption that $xy=yx$... but do we really need to assume that? No! The notation $F\langle x,y\rangle$ is often used to denote the polynomial ring in "noncommuting indeterminates," and it has more elements in the sense that $xy$ and $yx$ are different things, $xyx$ and $x^2y$ are different, and so on.

(And really, there's nothing stopping us from defining an even more general type of polynomial ring where the indeterinates are nonassociating, and $(xy)z\neq x(yz)$ and so on. But we have to stop digressing for the sake of discussion :) )

But $F\langle x,y \rangle$ and $F[x,y]$ are definitely related: its easy to see that $F[x,y]\cong F\langle x,y \rangle/(xy-yx)$, that is, the polynomial ring is a quotient of the noncommuting polynomial ring. The thing in the ideal we're modding out by is referred to as a relation because it specifies some extra property that we are enforcing in the quotient algebra.

Another example of this would be $\Bbb R[i]/(i^2+1)\cong\Bbb C$: you see, the symbol $i$ (nobody said anything about an imaginary number yet) is forced to have the property $i^2=-1$ in the quotient ring. Thus $i$ in the quotient ring becomes a complex number, along with all the other elements.

By now you're noticing that if this makes sense for complex numbers, then why not quaternions? Indeed, if you take $\Bbb H\cong \Bbb R\langle i,j,k\rangle/(i^2+1,j^2+1,k^2+1,ijk+1)$ since those are all the relations we need to bend the noncommuting polynomial ring into the quaternions. (Those aren't necessarily the best or the only relations that will do the trick, but I guess they are the most familiar.)