Post #1487037
2026-04-07 18:40 UTC
Replies (12)
-
@boarders@mathstodon.xyz 2026-04-07 18:48
@hallasurvivor very fun observation! I also can't quite figure out what goes wrong precisely. I think you can do some version of euclidean division on the quaternion polynomial ring, but that doesn't seem to imply it is a UFD
-
@trebor@types.pl 2026-04-07 18:49
@hallasurvivor How do you even define polynomials over quaternions? If we define it as a formal sum of aq^n where q is a variable, then evaluation is no longer a homomorphism. If we allow terms such as aqbqc, then factorization of a polynomial with root q0 into g(q) (q - q0) completely breaks, because division breaks (because like terms can't cancel, for example aqb - cqd doesn't simplify for generic quaternions a,b,c,d).
-
@jdw@mathstodon.xyz 2026-04-07 20:16
@hallasurvivor You are probably used to the fact that many matrices can have the same characteristic polynomial, i.e. the fact that a monic polynomial can have more roots than its degree in a non-commutative ring. This cannot happen in commutative rings (the (skew) field property doesn't play a role here). As @trebor remarks, this is related to the fact that evaluation of polynomials is not a homomorphism, a phenomenon that sometimes tricks people into thinking that the Cayley Hamilton theorem should be a triviality which follows from substituting X = A formally in the equation chi(X) = det(X - A).
-
@stylus@social.afront.org 2026-04-07 20:33
@hallasurvivor The real-analytic proof given on wikipedia shows that a higher degree polynomial with real coefficients 'can always be divided by some quadratic polynomial with real coefficients'. Together with the complex conjugate theorem, it proves the result over the complex numbers. However, for quaternions it's simply not true by identity: There are infinitely many square roots of -1, given by combination of the i/j/k basis vectors on a unit sphere instead of just 2. Rather than ±i, it's the set of i,j,k such that ||i+j²+k²|| = 1 I see that this doesn't exactly respond to your question about "where is the use of commutativity". To pursue that thread I guess I'd look down at the quadratic formula. When it gets to the step where you have to take the square root of a negative number, you just have to use the rule for quaternions, not complex numbers: you have to pick pars of numbers that are like specific kinds of conjugates, of which ±i, ±j and ±k would be the prototypical examples. You can say that 1j and 1k are both roots of x²+1 in some sense but unlike when you say -i and +i are roots of the same equation it is NOT the case that jk+1=0 (or kj+1=0) So, I think in that sense it's also true that the conjugate rule holds, you still have to use conjugate pairs of roots that multiply together to satisfy the original quadratic equation. When it comes to polynomials with quaternion coefficients, sorry, my head explodes.
-
@at@mathstodon.xyz 2026-04-07 21:48
@hallasurvivor I vaguely remembered something about all unit vectors being roots of -1 in the quaternions, and I only just remembered a fun way to see it, which is that _every_ quaterion q is a root of a quadratic polynomial over 𝐑, namely χ=X²-tr(q)X+det(q), where tr(q)=2Re(q)=2a is twice the real part of q=a+ib+jc+kd and det(q)=a²+b²+c²+d² is the square of the absolute value of q (χ is the characteristic polynomial à la https://arxiv.org/abs/math/0203276), so taking a=0 gives you an S² worth of roots. 𝐑 is the center of 𝐇, so the map 𝐑[X]→𝐇 that evaluates in q is a ring homomorphism, and its kernel (the ideal of polynomials with q as a root) is generated by χ for q∉𝐑, which is a (roundabout) way of seeing that that only the q∉𝐑 with Re(q)=0 and |q|=1 make the cut. Incidentally χ factors over the (commutative) subring 𝐑[q]≅𝐂 generated by 𝐑 and q as χ=(X-q)(X-q*), since q*=2Re(q)-q is in there and we have tr(q)=q+q* and det(q)=qq*=q*q.
-
@francisbrb@mathstodon.xyz 2026-04-07 21:54
@hallasurvivor This is a nice observation. Here is how I think about Puzzle 1: The unit quaternions are isomorphic to \( \mathbb{R}^3 \), and their product is given in terms of the dot product and cross product. Writing a general quaternion as \( (a, \vec{v}) \), with \( a \in \mathbb{R}, \vec{v} \in \mathbb{R}^3 \), the product can be written as (0,𝑣⃗)(0,𝑢⃗)=(−𝑣⃗⋅𝑢⃗,𝑣⃗×𝑢⃗). So taking \( \vec{v} \in S^2 \) to be any unit imaginary quaternion, we find that its square is \( -1 \).
-
@antoinechambertloir@mathstodon.xyz 2026-04-07 22:02
@hallasurvivor Puzzle 1. Since ij=-ji etc., one has (a i + b j + ck)^2 = -(a^2 + b^2 + c^2). Puzzle 2. This is tricky and probably explained not precisely enough in the literature. The point is that while you can decide to evaluate a quaternionic polynomial f = sum q_n T^n at a quaternion a, setting f(a) = sum q_n a^n, this map f -> f(a) is not a ring morphism. So when you manage to divide a polynomial f which vanishes at a by the polynomial T-a, which you can do, you get f = g · (T-a), but you can't deduce that f(b)=g(b) · (b-a). In particular, there is no relation between the roots of f and those of g and the classic induction breaks down.
-
@francisbrb@mathstodon.xyz 2026-04-07 22:22
@hallasurvivor This is related to the fact that \( Q[x] \) does not have unique factorization. When dealing with polynomial rings over commutative fields, we can prove unique factorization as a consequence of the Euclidean algorithm. And this doesn't work for skew-fields. The key point seems to be that you need to multiply things like \( ax^m bx^n = ab x^{m+n} \). This requires you to commute the indeterminate \( x \) with \( b \). Perhaps there's something more conceptual...
-
@fibresquare@mathstodon.xyz 2026-04-08 18:15
@hallasurvivor This argument is a bit over the top, but it's the first thing I thought of. Let G be the unit norm quaternions. Then -1 generates an order 2 subgroup which is the center of G. We have that i^2 = -1, and hence for any g in G, g i g^(-1) is also a square root of -1. We have a Lie group action of G on itself by conjugation, and so the orbit of i consists of square roots. But this orbit is a subspace homeomorphic to G/Stab(i). Stab(i) is a closed subgroup of G and hence an embedded Lie subgroup. Thus, since G is connected, if it is not G itself, G/Stab(i) is uncountable. But j is not in Stab(i). You can also show that G/Stab(i) is homeomorphic to S^2 with a bit more work. Essentially, the adjoint representation Ad: G -> GL(g) will show that Stab(i) is the inverse image of the elements of Ad(G) = SO(3) which fix the subspace spanned by i (I'm identifying g with the imaginary quaternions), so rotations with some fixed axis of rotation. But this is the stabiliser for the natural action of SO(3) on S^2, hence G/Stab(i) is homeomorphic to S^2.
-
@RefurioAnachro@mathstodon.xyz 2026-04-09 13:35
The following is in analogy to i²=-1 from complex numbers. I know that imaginary quaternions can be written using any orthonormal basis. Which means they are spherically symmetric. In fact, the symmetries cover the unit sphere twice! So we can take any solution and rotate it continuously. Since I still felt surprised I wanted to get my hand on some of them: Take any point (a,b) on the unit cirle. It means a²+b²=1. Expanding (ai+bj)² we get a²i²+abij+abji+b²k². Now, also by definition, i²=j²=-1 and ij=-ji. So we can simplify to get -a²-b², which is what we wanted: Using a subspace of just two imaginary dimensions we get a circle of solutions! I think we can simply use general imaginary unit quaternions and do the algebra. Appealing to symmetry seems odd at this point, because symmetry alone already solves the problem. @hallasurvivor
-
@counting_is_hard@mathstodon.xyz 2026-04-09 15:32
@hallasurvivor I do not like this fact. Let me just sweep it under the carpet where no-one can find it
-
@energyscholar@mathstodon.xyz 2026-04-09 19:15
@hallasurvivor The solution set is S² — the unit sphere in the pure imaginary quaternions. Every root is conjugate to i via the action of SU(2) on Im(H). A polynomial whose roots form an orbit, not a finite set. That same S² is the Bloch sphere. Non-commutativity turns algebra into geometry.