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I would like to know how to solve general polynomial equations over $\mathbb{H}$.

For example, how do I solve something like $x^2 = 20$, $x^2 + 2x = -30$, etc.?

I'm able to go as far as saying something like $x = \{h \in \mathbb{H} \mid h^2 = 20\}$, but I can't determine concrete quaternionic values.

Raiyan Chowdhury
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    Well, the two real numbers $\pm\sqrt{20}$ still exist in $\Bbb H$. So that's two solutions to the first equation. A similar thing can be said for the second one. I'm not entirely certain whether quadratic equations have at most two solutions, though. – Arthur Sep 21 '20 at 23:15
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    @Arthur: they have infinitely many solutions in general. Any polynomial with real coefficients has the property that its solutions in $\mathbb{H}$ can be conjugated by any nonzero element of $\mathbb{H}$. For example there are uncountably many solutions to $x^2 = -1$. – Qiaochu Yuan Sep 21 '20 at 23:32
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    @Raiyan: $\mathbb{C}$ embeds as a subfield of $\mathbb{H}$ (in uncountably many different ways); these are the maximal subfields, and every solution to a quadratic equation with real coefficients in $\mathbb{H}$ lies in one of these subfields. It's less clear what to do with polynomials with quaternionic coefficients that don't commute, for example $x^2 + ix + j$. – Qiaochu Yuan Sep 21 '20 at 23:33
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    A polynomial with real coefficients will have a set of solutions over $\mathbb{C}$: $$\lambda_1,\cdots, \lambda_n, a_1+ib_1,\cdots, a_m+ib_m$$ with the $\lambda_i,a_i,b_i\in \mathbb{R}$. The solutions over $\mathbb{H}$ will be precisely: $$\lambda_1,\cdots, \lambda_n, {a_1+ub_1,\cdots, a_m+ub_m|,,uu^=1,u+u^=0}.$$ – tkf Sep 22 '20 at 00:21

2 Answers2

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As mentioned in various comments, solving general expressions involving an indeterminate $x$ and quaternions and addition and multiplication is hard. I avoid the word polynomial as it is a little ambiguous in this context. Note for example that $xi-ix+2k=0$ has $x=j$ as a solution, whilst $ix-ix+2k=0$ has no solutions.

However all your examples in the question are polynomials with real coefficients, and solving these is easy, again as stated in various comments. In case this was your actual question I will expand the comments a little:

Given any polynomial $P$ in $x$ with real coefficients, a general quaternion solution will have the form $q=a+b_1i+b_2j+b_3k$ with $a,b_1,b_2,b_3\in\mathbb{R}$. If $q$ not real let $$b=\sqrt{b_1^2+b_2^2+b_3^2},\qquad u=\frac1b (b_1i+b_2j+b_3k).$$

Thus $q=a+bu$. Here $u^2=-1$ (when you square $u$ all the cross terms cancel and you are left with $-1$). Now $u$ cannot be a root of any polynomial (in $y$) over the reals that is not divisible by $1+y^2$, as $1,u$ are linearly independent over $\mathbb{R}$. Thus $a+ib$ is also a root of $P$.

We conclude that if $q$ is a root of $P$ then either $q\in \mathbb{R}$ and is a real root of $P$, or $q=a+bu$ for a complex root $a+ib$ of $P$ and $u$ a unit vector in the (3 dimensional) plane spanned by $i,j,k$.

Thus if the roots of $P$ over $\mathbb{C}$ are: $$\lambda_1,\cdots\lambda_n, a_1+ib_1,\cdots,a_m+ib_m,$$ for $\lambda_r,a_r,b_r\in \mathbb{R}$, then the roots of $P$ over $\mathbb{H}$ are precisely: $$\lambda_1,\cdots\lambda_n, \{a_1+ub_1,\cdots,a_m+ub_m|u \in S^2\}, $$ where $S^2=\{u_1i+u_2j+u_3k|\,\, u_1,u_2,u_3\in \mathbb{R}, u_1^2+u_2^2+u_3^2=1\}$.

tkf
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  • So essentially I'd first need to determine the solutions over $\mathbb{C}$? – Raiyan Chowdhury Sep 23 '20 at 17:55
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    Yes, but instead of the complex (non-real) roots occuring in conjugate pairs, they occur in "conjugate spheres". – tkf Sep 23 '20 at 18:14
  • This is a nice summary IMHO. – Jyrki Lahtonen Oct 30 '20 at 03:23
  • @JyrkiLahtonen It had everything I needed, but I was recently wondering if anything more is known in the community for general quaternionic equations, i.e. not just for real coefficients. – Raiyan Chowdhury Oct 30 '20 at 03:43
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    @RaiyanChowdhury You may find something interesting here or here. Ignore my answer, it simply reiterates the first paragraph of tkf's nice answer. But buried in those threads there are references to results on the existence of quaternion solutions to a certain type of polynomial equations. – Jyrki Lahtonen Oct 30 '20 at 07:12
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With quaternions you have to consider some preliminary things:

  • General quaternionic polynomials could potentially look something like this: $$1 - qiq^2jqkq$$ due to non commutativity. For simplicity, quaternionic polynomials are divided in 4 classes: left polynomials, of the form: $$\phi(q) = \sum_{m=0}^n a_mq^m$$ right polynomials: $$\phi(q) = \sum_{m=0}^n q^ma_m$$ General linear polynomials, of the form: $$\phi(q) = a_1qa_2q\dots q a_n + \dots$$ (where the ... signify that we can sum other monomials of that form) and finally general quaternionic polynomials, whose general form might be quite difficult to write.

For left/right and general linear quaternionic polynomials the fundamental theorem of algebra is valid (check page 90 and page 156), however, it is not true in general that polynomial equations of degree n have at most n solutions. One easy counter-example of this is the equation

$$q^2 + 1 = 0$$

which has a continuum of solutions (i.e the solution set is infinite and not countable). However, despite this fact, we can still build a relationship between the degree of the equation and the solutions; this relationship, however, won't be between the degree and the cardinality but rather between the degree and the topological structure of the solution set. For more info about this I recommend checking this resource at page 92.

hsnbrn12
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