Why does $i$ have infinitely many conjugates in $\mathbb{H}$?

Question

Browsing this question: Why are the solutions of polynomial equations so unconstrained over the quaternions?, the pdf linked in the comments says that the infinitely many conjugates of $i$ in $\mathbb{H}$ are roots to $x^2+1$.

I get that they're roots, but how do we know that the conjugacy class of $i$ is in fact infinite?

Calculate $$(\cos\alpha +\sin\alpha j)^{-1}i(\cos\alpha +\sin\alpha j).$$ — Jyrki Lahtonen, Jul 11 '12 at 17:01

score 1 · Answer 1 · answered Jul 11 '12 at 17:25

1

This should be a comment but it came up too long.

I think there's a confusion here: the quaternion $\,i\,$ has one unique quaternionian conjugate in $\,\Bbb H\,$, as we can deduce from this definition

The sense in which the term "conjugate" seems to be used in the OP is that the minimal polynomial of $\,i\,$ over $\,\Bbb H\,$ has infinite other roots, all of which are "conjugate" to (i.e., roots of the same minimal polynomial of) $\,i\,$

So the question doesn't seem to be connected to conjugacy classes a la group theory but with roots of minimal polynomials.

answered Jul 11 '12 at 17:25

DonAntonio

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Hmm. Are you thinking about conjugates in a group of 8 elements? I was thinking about conjugates in the infinite group of invertible quaternions. Edit: Oh, you were referring to the antiautomorphism of $\mathbb{H}$. Sorry. Still, in the group theoretic sense $i$ has infinitely many conjugates. – Jyrki Lahtonen Jul 14 '12 at 19:23
As far as I know $,\Bbb H,$ usually denotes the infinite division ring of quaternions, and $,Q_8,$ the group of quaternions of order $,8,$...and no: I wasn't referring to group theoretic meaning but to the quaternionian meaning as in the linked definition. – DonAntonio Jul 15 '12 at 00:03
Quite. I realized what you had meant, but decided to leave the comment anyway. Talking about a conjugacy class does suggest that the OP is interested in the group theoretic version of "conjugate". – Jyrki Lahtonen Jul 15 '12 at 06:02

score 0 · Answer 2 · answered Jul 11 '12 at 17:45

$$(a+bi+cj+dk)^2+1=0$$ $$a^2+1 + b^2i^2+c^2j^2+d^2k^2+2abi+2acj+2adk+(bcij+cbji)+(cdjk+dckj)+(bdik+dbki)=0$$ The last three brackets are all $0$. The $i,j,k$ components must all be zero as well, which we can do by setting $a=0$. We're left with $$1-b^2-c^2-d^2=0$$ which can be satisfied for infinitely many $b,c,d$.

DonAntonio seems to have covered the issue about quaternion conjugates quite nicely, if that's more what you were asking.

Why does $i$ have infinitely many conjugates in $\mathbb{H}$?

2 Answers2