Alternate version of this question: is there any formula composed entirely of complex numbers that has a unique quaternion solution (with non-zero $j$ and $k$ components)?

I've struggled to understand quaternions for years now. I very much understand how to "get" from real to imaginary numbers, and then from imaginary to complex numbers. $\sqrt{-1} = i$ is an equation composed of only real numbers that results in an imaginary number. Then $1 + i$ is both an equation that shows how to "get" from real and imaginary numbers to complex numbers, and a representation of the final complex number itself. What I don't understand is the subsequent leap to quaternions.

The classic formula is of course $i^2 = j^2 = k^2 = ijk = -1$. But the existence of $j$ and $k$ do not seem "justified" by this in the same way that $\sqrt{-1} = i$ justifies $i$. The equation $i*i$ equals $j^2$, but it also equals $-1$. There is also $i = jk$, but this is really a way of getting back to complex numbers from quaternions, not the other way around.

I realize quaternions can be thought of as a 4-dimensional analogue to the 2-dimensional complex numbers, and that a complex number is just a quaternion of the form $a + bi + 0*j + 0*k$. But this does not help me understand them. It seems to me that the existence of the 1-dimensional reals and normal mathematical operators means that the 2-dimensional complex numbers have to exist, but there is no such equivalent logical connection between the complex numbers and 4-dimensional quaternions. Do you have to simply accept quaternions axiomatically?

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    $\begingroup$ Complex numbers are algebraically closed, so any polynomial with coefficients in $\mathbb{C}$ will have roots residing in it. Quaternions are also not commutative, so the general theory of field extensions don’t really apply to it. To see why quaternions exist naturally, you might want to look into “division algebras” in which they classified uniquely up to isomorphism $\endgroup$
    – klein4
    Jun 16, 2021 at 14:01
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    $\begingroup$ The original motivation for quaternions was not to solve equations, but to find a good algebraic way to work with rotations in 3 dimensions. $\endgroup$ Jun 16, 2021 at 14:45
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    $\begingroup$ (1) The existence of the quaternions is not an axiom; it's proved from the axioms of set theory, just like the existence of the real numbers and the complex numbers. (2) No individual, non-real quaternion can be uniquely described by an algebraic equation with real coefficients, because the algebra of quaternions is invariant under all rotations of the subspace spanned by $\{i,j,k\}$. (2.5) No individual, non-real complex number can be uniquely described by an algebraic equation with real coefficients, because the algebra of complex numbers is invariant under complex conjugation. $\endgroup$ Jun 16, 2021 at 14:46
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    $\begingroup$ To amplify Andreas' comment (2.5): The equation $x^2 = -1$ has two complex solutions. So, one can't define $i$ as "the unique number that makes $x^2 = -1$ true," since $-i$ also makes $x^2 = -1$ true. You might think, "Well, that's fine, I'll just define $i = \sqrt{-1}$ to be the positive square root of $-1$," but that doesn't make sense either, since the symbol $>$ only makes sense for real numbers (i.e., $\sqrt{-1} > 0$ is completely meaningless). $\endgroup$ Jun 16, 2021 at 16:02
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    $\begingroup$ Once you have the real numbers (constructed, for example, as Dedekind cuts), you usually define complex numbers as ordered pairs of real numbers, with specific operations of addition and multiplication. Define quaternions similarly, as ordered quadruples of real numbers, with specific operations of addition and multiplication. $\endgroup$ Jun 16, 2021 at 19:43

1 Answer 1


Edit This is more a lengthy comment than an actual answer. I realized it doesn’t tackle the actual question, but I don’t want to take it down, since I learned something from the comments it attracted, so maybe someone else does as well.

I think the complex numbers $\Bbb C$ and the quaternions $\Bbb H$ were discovered / are motivated out of very different reasons.

In the case of complex numbers the question was, what happens if one freely adds a solution $i$ to the real polynomial equation $x^2+1=0$. It turns out that this is not some abstract trick, but can be realized in form of the complex numbers. Now it is a theorem that not only every real polynomial decomposes into complex linear factors, but in fact every complex polynomial does as well. This is to say: the complex numbers are algebraically closed.

Now having the complex numbers at hand, you can play a bit with them and notice that complex multiplication describes rotation and scaling of the plane, which makes dealing with such transformations of the plane extremely easy. While developing the mathematics of mechanics it became an important question, whether a similar algebraic description of rotations in the 3dimensional space was possible as well. In this context, Hamilton discovered the quaternions $\Bbb H$. Note that this has nothing to do with polynomial equations anymore.

Now the final question becomes, whether higher dimensional transformations allow a similar algebraic description as those in dimensions two and three do. After some translations one can phrase this as the Hopf Invariant One problem, which was resolved by Adams using tools from algebraic topology: The only calculi of transformations are those given by $\Bbb R, \Bbb C, \Bbb H$ and $\Bbb O$, the last one being the Cayley-Octonions.

  • $\begingroup$ It can also be phrased more simply as Frobenius' Theorem or as Hurwitz' Theorem. $\endgroup$ Jun 16, 2021 at 15:49
  • $\begingroup$ Thanks, I didn’t know about those. I only learned about division algebras in the context of topological K-theory, where they are commonly used as motivation. I like the fact that one can prove the classification without this heavy machinery though :) $\endgroup$ Jun 16, 2021 at 17:29
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    $\begingroup$ The extra work that goes into the algebraic topology proof pays off with a stronger result: The only parallelizable spheres are the ones of dimensions $0,1,3,7$. $\endgroup$ Jun 16, 2021 at 19:55

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