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I tried to solve another exercise and I would be grateful if someone could tell me if my answer is right. This is the exercise:

Characterize the pairs $p,q \in \mathbb H$ such that $pq = -qp$.

I can see that the unit quaternion paris $(i,k), (i,j), (j,k)$ anticommute.

Since the real part of two purely imaginary quaternions is the dot product and the imaginary part the vector product it follows that purely imaginary quaternions anticommute.

If $a+p, b+q$ are two quaternions with real parts $a,b$ then $(a+p)(b+q) = ab + aq + bp + pq$. The only way for this to anticommute is if $ab=0$. Therefore I conclude that all pairs of purely imaginary quaternions anticommute.

Is this correct?

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  • $\begingroup$ One small problem: you don't know that $x_iy_i=0$ for $i=2,3,4$ - you've shown that that must be so in the pure-imaginary case, but that doesn't mean that you can use the result you derived for the purely-imaginary case in the case where $x$ and $y$ aren't purely imaginary. $\endgroup$ – Steven Stadnicki Nov 14 '14 at 4:40
  • $\begingroup$ What's more, even in the purely imaginary case you haven't shown that each product $x_iy_i=0$; instead, all you have is that their sum - the dot product - is zero. $\endgroup$ – Steven Stadnicki Nov 14 '14 at 4:42
  • $\begingroup$ What I wrote above doesn't work at all thanks for pointing it out. I see now that pairs $(i,j), (i,k) $ and $(j,k)$ anti commute. But what other pairs? $\endgroup$ – learner Nov 14 '14 at 4:50
  • $\begingroup$ @StevenStadnicki I tried to fix it. I edited my question. $\endgroup$ – learner Nov 14 '14 at 5:05
  • $\begingroup$ Reduce to the case of the purely imaginary quaternions $Q$, then use the action of $Q$ on itself by conjugation (i.e., $q \to z^{-1}qz$ for some $z$, not conjugation as in complex conjugate) to reduce to the case $p = i$. After that, you can just do a quick explicit calculation. $\endgroup$ – anomaly Nov 15 '14 at 0:47
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Claim: $q_1$ anticommutes with $q_2$ if and only if either is zero, or else both have zero real-part and their imaginary parts are orthogonal.

Representing our quaternions using a vector for the imaginary part, then the product of $q_1 = (r_1, \vec{v_1})$ and $q_2 = (r_2, \vec{v_2})$ is $(r_1r_2 - \vec{v_1}\cdot\vec{v_2}, \vec{v_1}\times\vec{v_2}+r_1\vec{v_2}+r_2\vec{v_1})$.

First, if either $\vec{v_1}$ or $\vec{v_2}$ are zero then it quickly follows that the corresponding $r_1$ or $r_2$ is also zero (or else, the other quaternion is also zero). So we can restrict ourselves to the case where both $\vec{v_1}$ and $\vec{v_2}$ are non-zero.

Considering real and imaginary parts of $q_1q_2 + q_2q_1$ (which must be zero if the quaternions anticommute):

The real part: $2r_1r_2 - 2\vec{v_1}\cdot\vec{v_2} = 0$ which implies $r_1r_2 = \vec{v_1}\cdot\vec{v_2}$.

The imaginary part: $\vec{v_1}\times\vec{v_2} + \vec{v_2}\times\vec{v_1} + 2r_1\vec{v_2} + 2r_2\vec{v_1} = 2r_1\vec{v_2} + 2r_2\vec{v_1} = 0$ (using the anticommutativity of vector cross product).

Dotting the last equation with $\vec{v_1}$ and substituting gives us:

$r_1\vec{v_1}\cdot\vec{v_2} + r_2\vec{v_1}\cdot\vec{v_1} = 0$

$r_1^2r_2 + r_2\|\vec{v_1}\|^2 = r_2(r_1^2 + \|\vec{v_1}\|^2) = 0$

Since we've assumed $\vec{v_1}$ is non-zero, this can only be satisfied if $r_2=0$. By symmetry we get $r_1=0$ and our conclusion that both real parts must be zero.

Finally, we can use the equation $r_1r_2 = \vec{v_1}\cdot\vec{v_2}$ to deduce that the two imaginary parts must be orthogonal.

That the condition is sufficient (that is, if both real parts are zero and the imaginary parts are orthogonal then the quaternions anticommute) can be observed from the definition of mulitplication.

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  • $\begingroup$ Thank you for your help. Using your answer I wrote my own proof. Can you please check it? I will accept your answer and then give you the bounty. I will now post my proof in an answer to this thread. Thanks in advance for proof-reading. $\endgroup$ – learner Nov 27 '14 at 2:41
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We write the product as $$ pq = (r_p r_q - v_p \cdot v_q, r_p v_q + r_q v_p + v_p \times v_q)$$

Considering the imaginary part we see that since the crossproduct is antisymmetric, $\operatorname{Im}{pq}=-\operatorname{Im}{qp}$ if and only if $r_pv_q + r_q v_p = 0$.

This can happen in exactly 3 ways: if either $r_p = r_q = 0$, if $v_p = v_q = 0$ or if $v_p, v_q$ are linearly dependent.

If $r_p = r_q = 0$, $p,q$ are purely imaginary. Since the real part of the product of two purely imaginary quaternions is given by the dot product and the dot product is symmetric the real part commutes if and only if $v_p \cdot v_q = 0$. This happens if and only if the angle between the two vectors is $90$ degrees. Hence $v_p, v_q$ must be orthogonal if $r_p=r_q =0$.

If $v_p = v_q = 0$ then we are multiplying real numbers. These anticommute if and only if they are zero. Hence in this case, $p=q=0$.

If $v_p,v_q$ are linearly dependent with $$ r_p v_q + r_q v_p = 0$$ Then also $$ r_p v_q \cdot v_p + r_q \|v_p\|^2 = 0$$

The real part $r_p r_q - v_p \cdot v_q $ vanishes if and only if $r_p r_q = v_p \cdot v_q$ which can happen only if

$$ r_p^2 r_q + r_q \|v_p\|^2 = r_q (r_p^2 + \|v_p\|^2)= 0$$

hence $r_q$ must be zero. Doing the same argument with $v_q$ yields $r_p=0$. Hence if $v_p$ and $v_q$ are linearly dependent the product $pq$ anticommutes if both real parts are zero. But in this case we have shown that $v_p$ and $v_q$ must be orthogonal, a contradiction to the linear dependence of $v_p,v_q$. Hence $v_p, v_q$ cannot be linearly dependent.

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    $\begingroup$ There's a missing fourth case when one of $p$ or $q$ is zero, but otherwise the proof looks right to me. $\endgroup$ – Paul Hankin Nov 27 '14 at 3:49
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Therefore I conclude that all pairs of purely imaginary quaternions anticommute.

Well, $i$ doesn't anticommute with itself, so this can't be right. Just stating "the only way this can happen is..." is a huge gap without justification.


Here's a method that avoids getting down into the real coefficients (for the most part.)

Of course, the situation is clear when $p$ or $q$ is zero (everything anticommutes with them), so let's assume $p\neq 0$ and $q\neq 0$. Then it is possible to rewrite the equation as $pqp^{-1}=-q$ or $qpq^{-1}=-p$.

It's easy to show that the conjugation map created by a nonzero quaternion leaves elements of $\Bbb R$ invariant and leaves the subspace $i\Bbb R\oplus j\Bbb R\oplus k\Bbb R$ invariant with this lemma:

Lemma: for a nonzero quaternion $p$, $q$ and $pqp^{-1}$ have exactly the same real part.

Proof-hint: Use the identity $p^{-1}=\frac{\overline{p}}{p\overline{p}}$ where the overline denotes quaternion conjugation, and $\Re(x)=\frac12(x+\overline{x})$ to show that $\Re(pqp^{-1})=\Re(q)$ directly.

For $p$ and $-p$ to have the same real part, that real part must be zero. By symmetry, the same can be said for $q$.

Now we can continue geometrically (as your Lie-algebra tag suggests we should). The most important interpretation of the quaternions in Lie algebra is that they act by conjugation on the subspace of quaternions with real part zero to produce proper rotations of $\Bbb R^3$, where $\Bbb R^3$ is identified with the quaternions with real part $0$.$^\ast$

Concentrating on the transformation $R(-)=p(-)p^{-1}$ for a moment, we see that (geometrically) this produces a proper orthogonal rotation of $\pi$ radians around the axis $p$. As we know, this rotation by $\pi$ in $\Bbb R^3$ has an eigenspace for the value $-1$ of dimension $2$, the plane of rotation normal to $p$, and this is where $q$ must lie. Any $q$ in that plane whatsoever is an eigenvector for $-1$, and hence a possibility for $q$.

In summary, if $p=0$, everything anticommutes with $p$, and if $p\neq 0$ it must have real part $0$ and the other quaternions with real part zero which are orthogonal to $p$ (in $\Bbb R^3$) are exactly the $q$'s which anticommute with $p$.


$^\ast$ There is even a convenient formula for these rotations: $\cos(\theta/2)+u\sin(\theta/2)$ produces a rotation of $\theta$ radians around the axis $u$ where $u$ is a unit-length quaternion with zero real part. In our case, $\cos(\pi/2)+\frac{p}{|p|}\sin(\pi/2)=\frac{p}{|p|}$ is something we've already seen (conjugating by an element and by its normalization produce the same map.)

It should be noted that this isn't the only quaternion that produces this rotation, though.

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Here an answer that uses only elementary algebra. We put $p=x+y\mathbf{i}+z\mathbf{j}+t\mathbf{k}$ and $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ then we compute the products $pq $ and $qp$ remembering that $$ \mathbf{i}\mathbf{j}=\mathbf{k}\quad \mathbf{j}\mathbf{k}=\mathbf{i}\quad \mathbf{k}\mathbf{i}=\mathbf{j} $$ and $$ \mathbf{j}\mathbf{i}=-\mathbf{k}\quad \mathbf{k}\mathbf{j}=-\mathbf{i}\quad \mathbf{i}\mathbf{k}=-\mathbf{j} $$ Putting $pq+qp=0$, and since $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$, are linearly independent, we have the linear system: $$ \begin{cases} bx+ay +0+0 =0\\ cx+0+az+0=0\\ dx+0+0+at=0\\ ax-by-cz-dt=0 \end{cases} $$ that has solutions $$\begin{split} a \ne 0 & \rightarrow x(a^2+b^2+c^2+d^2)=0 \rightarrow x=y=z=t=0\\ a=0 & \rightarrow x=0 \land by+cz+dt=0\\ \end{split} $$ that is the same result as in @Anonymous , since $ by+cz+dt$ is the inner product of the vectors corresponding to the imaginary parts of $p$ and $q$ and it's $=0$ iff this imaginary parts are orthogonals.

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