# Are j and k on different imaginary planes than i?

I'm trying to understand Quaternions. So I understand that a Quaternion is written like $xi+yj+zk+w$. I also understand that $i^2 = j^2 = k^2 = ijk = -1$, and how that can be used to derive equations such as $ij = k$ and $jk = i$.

One things that confuses me is that $i$ is not equal to $j$ which is not equal to $k$. I can say $i^2 = -1$ and $j^2 = -1$ but I can't say $ij = -1$.

Correct me if I'm misunderstanding something, but why do they seem to have the same product when squared yet all three must be multiplied to equal the product of any one of them squared? Are they supposed to be on different imaginary planes?

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Here is meta.math.stackexchange.com/questions/5020/… for displaying functions. – Arbuja Jan 20 at 15:12
In the sense described in Lee Mosher's answer that is true. But there are actually infinitely many complex planes inside the quaternions. If $a,b,c$ are real numbers such that $a^2+b^2+c^2=1$, then the quaternion $u=ai+bj+ck$ satisfies $u^2=-1$. So $1$ and $u$ also span a copy of the complex plane. So there is a 2-dimensional family of complex planes - all sharing the same real axis, but their imaginary axes can point at any direction orthogonal to the real axis. – Jyrki Lahtonen Jan 20 at 17:33
The complex numbers $i$, $j$, and $k$ are different roots of -1, but that by itself implies nothing about geometry, planes, or anything else. IFF you are using them to represent different basis vectors—because they can be represented that way, see answers below—THEN yes. – jvriesem Jan 20 at 17:55

That is correct, $i$, $j$ and $k$ are contained in three separate complex planes contained in the quaternion numbers.

Just as the complex numbers $\mathbb{C}$ can be thought of as a 2-dimensional vector space over $\mathbb{R}$ with basis $1,i$, the quaternions $\mathbb{H}$ are a 4-dimensional vector space over $\mathbb{R}$ with basis $1,i,j,k$.

In particular, the plane spanned by $1,i$, the plane spanned by $1,j$, and the plane spanned by $1,k$ are different subplanes, any two of which intersect each other in the real line spanned by $1$.

So you can indeed think of these three planes as three separate copies of the complex numbers embedded in the quaternion numbers.

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One way to view quaternions is using matrices: $$1=\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}$$ $$i=\begin{bmatrix} 0&1&0&0\\ -1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{bmatrix}$$ $$j=\begin{bmatrix} 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0 \end{bmatrix}$$ $$k=\begin{bmatrix} 0&0&0&1\\ 0&0&-1&0\\ 0&1&0&0\\ -1&0&0&0 \end{bmatrix}$$ We can view this as $4$ orthogonal vectors which span a $4$ dimensional subspace of $\mathbb{R}^{16}$.

So yes, these can be viewed as $4$ orthogonal basis vectors, $3$ of which are orthogonal to the reals, thus "imaginary".

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Downvoter: What is wrong with providing a different view of the quaternions? – robjohn Jan 20 at 17:08
Wasn't the downvoter, but I don't really understand what you're saying about the matrices. How did you turn a complex number into a matrix?! – Mehrdad Jan 20 at 17:55
@Mehrdad One can define a quaternion (not a complex number) as a 4-d matrix, and have all the math check out; so, they are considered equivalent. I believe robjohn said much the same thing in his last sentence. – Cᴏɴᴏʀ O'Bʀɪᴇɴ Jan 20 at 18:43
@Mehrdad: you can also look at this Wikipedia link. Using the matrices above, we get $i^2=j^2=k^2=-1$ and $ij=k$, $jk=i$, and $ki=j$. These relationships, in addition to those naturally satisfied by matrices, make these isomorphic to the quaternions. – robjohn Jan 20 at 18:51

The operation of quaternion multiplication (product) is a generalization of multiplication of complex numbers, which in turn is a generalization of multiplication of real numbers. It is not "just" a multiplication. Let's look at it in detail.

There is of course a precise definition and theorems describing its properties but in this context it's enough to know that quaternions can be compared to orthogonal unit vectors. Their [vector] product is equal to a rotation, thus $i*j=k$. With each rotation you move into a new orthogonal dimension. After three rotations you land again on a real axis but the vector points to $-1$. So to illustrate it we can write: $k^2=k*k=i*j*k=1*i*j*k=-1$

The same logic applied to the complex $i$ which represented a rotation by 90 degrees in two dimensional space and so $1*i*i=-1$ as a result of two rotations i.e. by 180 degrees.

I hope now it is obvious why $i*j$ not equals $-1$? After two rotations in four dimensional space we have not arrived back to a real axis yet!

This is absolutely equivalent to the explanation with the matrix representation but I find it more intuitive.

Here are some further reference for those concepts http://physics.info/vector-multiplication https://en.m.wikipedia.org/wiki/Cayley–Dickson_construction

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