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I'm working on the following exercise

Let $N: \mathbb{H}^*\to\mathbb{R}^*$ be defined by $N(x)=x\overline{x}$. Show that:

a) $[\mathbb{H}^*,\mathbb{H}^*]\subset\mathrm{Ker}(N)$

b)$\forall x\in\mathbb{H}, \exists y\in\mathbb{H}^*$ such that $yx=\overline{x}y$

c) Let $x\in\mathrm{Ker}(N)$, $x\not=-1.$ Prove $\exists y\in\mathbb{H}^*$ such that $$x=[1+\overline{x},y]$$

d) I want to solve this one myself

So I managed to solve a) and b). For part a), let $x,y\in\mathbb{H}^*$, then $[x,y]=xyx^{-1}y^{-1}$, so $$\begin{align}N([x,y])&=N(xyx^{-1}y^{-1})=xyx^{-1}y^{-1}\overline{xyx^{-1}y^{-1}}\\&=xyx^{-1}y^{-1}\overline{x^{-1}y^{-1}}\overline{xy}=N((xy)^{-1})N(xy)=N(xy)^{-1}N(xy)=1,\end{align}$$ so $[x,y]\in\mathrm{Ker}(N)$ and thus $[\mathbb{H}^*,\mathbb{H}^*]\subset\mathrm{Ker}(N)$.

For the second part, I just wrote everything out with the `brute force approach' and managed to show this. Follows from the arithmetic rules for quaternions. The d) part I want to solve myself, but I don't get how I should show part c). Can anyone help me with part c)?

By the way I can deduce part d) from parts a) and c), so I'm quite sure that we need part b) for part c).

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Part (a) can be generalized as follows: If $\phi:G\to A$ is any group homomorphism in which the codomain $A$ is abelian, then $\ker(\phi)$ contains the commutator subgroup $[G,G]$. (In fact, this is considered a "universal property" - the first isomorphism theorem says $G\to A$ factors uniquely into maps $G\to G/[G,G]\to A$ where $G\to G/[G,G]$ is the quotient map.) Proving the general form doesn't involve big expressions with conjugates and the $N$ function.


Part (b) can be generalized too. I am not sure what you meant when you said you brute forced it - but if you wrote out a bunch of equations with the four coordinates then you are working harder than you need to be!

I will use the convention that a quaternion is a formal sum $a+\mathbf{v}$ of a scalar $a$ and a vector $\mathbf{v}$ from the oriented inner product space $\mathbb{R}^3$ (I say oriented so it also has a cross product). Multiplying two quaternions is accomplished using the distributive property and, for vectors (a.k.a. pure imaginary quaternions) the real/imaginary parts of the product are given by $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$.

Some nice things to know:

  • Every quaternion has a polar form $re^{\theta\mathbf{u}}=r\cos\theta+r\sin\theta\,\mathbf{u}$ for some unit vector $\mathbf{u}$ and scalar $r\ge0$. If the quaternion is not real, this representation is unique except for our freedom to swap $(\theta,\mathbf{u})$ with the pair $(-\theta,-\mathbf{u})$ or $(\theta+2\pi,\mathbf{u})$.
  • Unit vectors commute iff they're parallel, and anticommute iff they're perpendicular.
  • Conjugation $e^{\theta\mathbf{u}}\mathbf{v}e^{-\theta\mathbf{u}}$ has the effect of rotating $\mathbf{v}$ around $\mathbf{u}$ by $2\theta$. (Prove this!) Conjugating in general doesn't affect real parts; $\mathrm{Re}(qxq^{-1})=\mathrm{Re}(x)$.

The "phase" of a complex number is well-defined because there is a distinguished choice between $\pm i$, but there is no straightforward way to do that for all antipodal pairs of unit vectors in $\mathbb{R}^3$, so we may as well consider the "phase" $\theta$ of a quaternion to be in the interval $[0,\pi]$. Now see if you can use the above properties to prove this:

Two quaternions in $\mathbb{H}^\times$ are conjugate if and only if they have the same norm and phase.


Part (c) is wrong. To see this, first rewrite

$$ x=(1+\bar{x})y(1+\bar{x})^{-1}y^{-1} $$

$$ xy(1+\bar{x})=(1+\bar{x})y $$

$$ y(1+\bar{x})y^{-1}=\bar{x}(1+\bar{x}) $$

which isn't always possible.

(For instance if $\bar{x}=i$ then $1+\bar{x}$ and $\bar{x}(1+\bar{x})$ have unequal real parts.)

It is not immediately obvious to me if the problem is a typo'd version of a correct one.

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  • $\begingroup$ Many thanks for this answer, really helpful since I didn't know about the polar forms of quaternions. I'll look into this and try if I can prove the identities you gave! I didn't the isomorphism theorems yet and also didn't see group actions yet (next chapter) but I also found a similar proof for part a using $\mathrm{Ker}(N)$ is a normal subgroup of $\mathbb{H}^*$ and a theorem on factor groups in my syllabus. For part c, I think it should be a typo, part d asks to show $[\mathbb{H}^*,\mathbb{H}^*]=\mathrm{Ker}(N)$, which sounds reasonable and might be proven using a similar identity as c. $\endgroup$ – Václav Mordvinov Mar 10 '18 at 20:18
  • $\begingroup$ And indeed I wrote it out using the four coordinates but it was definitely too much work and since it was not really helpful I also skipped a few cases for part b. You're approach is definitely smarter! :) $\endgroup$ – Václav Mordvinov Mar 10 '18 at 20:22
  • $\begingroup$ @VáclavMordvinov There is a nice way to do part (d), but I can't see what it has to do with part (c). Here's a hint for part (d): conjugation is an algebra automorphism, so in particular it interacts nicely with commutators. This is helpful for a "wlog... [something]" approach. $\endgroup$ – anon Mar 10 '18 at 21:30
  • $\begingroup$ Assuming part c is true, we know that for an arbitrary $x\in\mathrm{Ker}(N)$, we have $x\in[\mathbb{H}^*,\mathbb{H}^*]$, thus $\mathrm{Ker}(N)\subset[\mathbb{H}^*,\mathbb{H}^*]$ and we are done using part a. $\endgroup$ – Václav Mordvinov Mar 10 '18 at 21:36
  • $\begingroup$ I'll try you're approach later also! Thanks :) $\endgroup$ – Václav Mordvinov Mar 10 '18 at 21:40

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