# Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?

I'm reading a paper about Hamilton's discovery of quaternions and it explains why he failed in his 'theory of triplets' where he tried to make a vector with $$3$$-dimensions, as an analogy to the complex field, where we can see a number as a $$2$$-dimensional vector. In this paper, he explains why it is impossible to create a field with $$3$$ components, that is an extension of the complex field (in other words, it respects addition, and multiplication in the same way...).

Here it is

As you can see, it goes through all the possibilities and proves that it is impossible. The paper, however, does not explain why $$j^2=-1$$. It could be anything! Why $$-1$$?** The article itself is pretty intuitive, but this aspect kills me. Later, in the article, it says that we should instead consider a 4th component called $$k$$, such that $$k^2=-1$$ (also, $$i$$ and $$j$$ too). Here is the paper.

EDIT: This paper by Rupert Shuttleworth turned out to be extremely helpful (mirrored here on archive.org)

• Certainly if $j^2$ is real we might as well take $j^2=-1$. If $j^2 \geq 0$, then $j^2=a^2$ for some real $a$, and so $a+j$ is a zero-divisor. On the other hand, if $j^2<0$, then $j^2=-a^2$ and so $(j/a)^2=-1$, and there's no harm in taking $j/a$ to be our basis element instead of $j$. I don't know that there's a simple argument for why we ought to be able to choose a basis element $j$ such that $j^2$ is real, but I think that entire passage is dicey anyway, and for similar reasons (the "without loss of generality" seems logically highly suspect to me). Aug 28, 2014 at 6:40
• Why the abelian-groups tag? Am I missing something here? Aug 28, 2014 at 8:17
• Well if you try to generalize complex numbers it seems a fine idea to suppose that there is another pure imaginary number (name it $j$) that will be perpendicularly to the Argand plane and which will verify (like $i$) $j^2=-1$. Hamilton supposed this and showed that a third pure imaginary $k$ was needed in this case. This doesn't mean that you couldn't suppose that $j^2=1$ for example : in this case you'll get a $k$ with $k^2=1$ if I remember well and this will be equivalent to the quaternions ; see quaternion algebra. Aug 28, 2014 at 8:29
• I wonder if there are a couple of typos in the portion maa.org/sites/default/files/pdf/upload_library/46/HOMSIGMAA/… which is faithfully reproduced above. Where it says of one proposed $i$ and $j$, "they are not linearly dependent", was this supposed to say "they are not linearly independent" (and therefore not a basis)? And later, we have the phrase "incompatible with $\mathbb H$ being a field, whereas the structure that the several previous paragraphs were trying to build was named $\mathbb D$. Aug 28, 2014 at 13:19
• @user0112358: If $j^{-1}$ is in $D\backslash \mathbb{C}$, then that only forces the $j$-component of $j^{-1}$ to be nonzero: i.e., $j^{-1}=a+bi+cj$ with $c \ne 0$. It doesn't force the other two components to be zero (and in fact, if such an algebra existed, I think there would be lots of choices of a third basis element $j$ for which they were nonzero). You could try to argue that it's possible to choose a $j$ for which $j^{-1}=cj$, but I think in order to do this you need to understand the product $ij$, which probably leads you in the direction of Travis Willse's $j^2$-free answer... May 17, 2020 at 18:40

It seems to me that we can achieve a contradiction more quickly without assuming this anyway: Since $(1, i, j)$ is a basis for the field, $ij = a + bi + cj$ for some unique $a, b, c \in \mathbb{R}$. Then, on the one hand $i^2 j = -j$, and on the other it is

$i(ij) = i(a + bi + cj) = -b + ai + c(ij) = -b + ai + c(a + bi + cj) = (-b + ac) + (a + bc)i + c^2 j .$

Comparing coefficients of the $j$ term gives that $c^2 = -1$, which is not true for any $c \in \mathbb{R}$.

Anyway, the question is certainly not an idle one: Indeed, it leads to something interesting when carrying out this sort of analysis for four-dimensional algebras over $\mathbb{R}$, which famously yields the quaternion algebra $\mathbb{H}$ (NB that since $\mathbb{H}$ is not commutative, it is not a field but a division algebra).

Interestingly, one can also try to construct such an algebra taking $i^2 = -1$ but $j^2 = k^2 = 1$ and find there's a coherent and interesting way to define an associative product, giving an algebra $\widetilde{\mathbb{H}}$ sometimes called the split quaternions. This is perhaps nonobviously isomorphic (as an $\mathbb{R}$-algebra) to the ring $M(2, \mathbb{R})$ of $2 \times 2$ real matrices. Unlike $\mathbb{H}$ this is not a division ring (there are nonzero matrices that square to the zero matrix), but like $\mathbb{H}$ it has a nondegenerate quadratic form $Q$ that satisfies $Q(xy) = Q(x) Q(y)$, namely the determinant.

• I have long noticed that one could define an associative (and commutative) product with $i^2=j^2 = -1$ and $k^2 = 1$, but wondered why I never heard about it. I appreciate @Micah 's comment above about getting zero divisors when trying to make a field out of a multiplicative group that contains a second element that squares to a positive real. Would this product I've specified also lead to the split quaternions, or something else? I'm guessing not since, if the split quaternions are isomorphic to the 2 by 2 matrix ring, they're also noncommutative. Nov 15, 2015 at 1:33
• I don't know offhand of a nice $\Bbb R$-algebra that has that property. How do you define $ij$, $jk$, $ki$? Anyway, like you say, $M(2, \Bbb R)$ is noncommutative, and so it is not isomorphic to any commutative ring. Nov 15, 2015 at 22:15
• Like this. I figured I'd try getting Mathematica to symbolically solve for a set of 2x2 matrices that might commute like this, and right away it reduced the equations to several possibilities. I started plugging in values and ... oh, right, I have to tell it none of them equal each other. I told it to run again and it didn't finish, so I don't have high hopes Nov 16, 2015 at 3:02
• Perhaps this ring is $\Bbb C^2$ in disguise? Anyway, this seems like a fine thing to ask as a question here in its own right. Nov 16, 2015 at 9:05
• Yep, that checks out. i $=(i,i)$, j $=(i,-i)$, and k $=(-1,1)$. Nov 16, 2015 at 12:55

I don't know if this is what Hamilton had in mind, but: suppose you've decided that what you want to do is to write down a division algebra $D$ over $\mathbb{R}$ bigger than the complex numbers $\mathbb{C}$. Any element $d \in D$ of $D$ generates a subalgebra $\mathbb{R}[d]$ of $D$ which is a commutative division algebra (in other words, a field) over $\mathbb{R}$ and hence (because $\mathbb{C}$ is algebraically closed) must be isomorphic to either $\mathbb{R}$ or $\mathbb{C}$.

If $d \not \in \mathbb{R}$, then this subalgebra $\mathbb{R}[d]$ must be isomorphic to $\mathbb{C}$. Any such isomorphism realizes $d$ as a complex number $a + bi, b \neq 0$, and hence up to a real change of coordinates we might as well have chosen $d$ to correspond to $i$ under this isomorphism; that is, to satisfy $d^2 = -1$. With a bit more work this argument shows that $D$ is generated by elements that square to $-1$.

• The fact that ${\bf C}$ is algebraically closed is not the most relevant thing, in my opinion: I'd say that the more important fact is that the algebraic degree of the complex numbers over the reals is a prime number (and they are the algebraic closure). This forces the fact that there are no intermediate extensions (and no bigger algebraic ones). Nov 15, 2015 at 23:24
• From reading the text above, it seems to me that he was looking for a field larger than $\mathbb C$ and was proving that one didn't exist. Since the quaternions were the first noncommutative division algebra discovered, it seems to me that they would have helped lead to these now-well-understood properties. Nov 16, 2015 at 3:10
• @Travis: well, that's not clear to me. I think in the old days fields were not required to be commutative (so "field" meant "division ring"), and as you can see Hamilton never uses commutativity. Although he certainly has not exhausted all possibilities for the product $ij$. Very strange. Nov 16, 2015 at 3:14
• Perhaps, but any ring forms a left-algebra over itself though, right? So if they had other division rings, wouldn't they already have other noncommutative algebras? I'd assumed that he avoided commutativity because he was writing this after having figured them out, and by using one less assumption he was able to simultaneously prove that no field over a basis of 3 elements existed, and none existed even if he gave up commutativity, an objection someone would likely raise as he was giving up commutativity while jumping from 2 to 4 elements. Nov 16, 2015 at 3:36
• Check the comments above, under the other Travis' answer. When you choose $j^2=-1, ij=k, ji=k$ and get a commutative set, it's isomorphic to $\mathbb C^2$. Nov 16, 2015 at 12:57