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In scalar and vector form, a quaternion can be represented as $a=(q_0,{\bf{q}})$. The definition of quaternion multiplication is: $ab=(q_0,{\bf{q}})(p_0,{\bf{p}})=(p_0q_0-{\bf{q}}\cdot{\bf{p}},q_0{\bf{p}}+p_0{\bf{q}}+{\bf{q}}\times{\bf{p}})$$^{[1]}$.

Now I encountered a problem when I want to prove associativity of quaternion multiplication under this form. Specifically, a non-associative cross production is included in the multiplication definition which seems to be unable to reconcile with the fact that quaternion multiplication is associative. Following is my tentative proof.

Let $a=(q_0,{\bf{q}}), b=(p_0,{\bf{p}})$ and $c=(r_0,{\bf{r}})$.

Then

$(ab)c=(p_0q_0-{\bf{q}}\cdot{\bf{p}},q_0{\bf{p}}+p_0{\bf{q}}+{\bf{q}}\times{\bf{p}})(r_0,{\bf{r}})$ $=(p_0q_0r_0-r_0{\bf{q}}\cdot{\bf{p}}-q_0{\bf{p}}\cdot{\bf{r}}-p_0{\bf{q}}\cdot{\bf{r}}-({\bf{q}}\times{\bf{p}})\cdot{\bf{r}},$ $p_0q_0{\bf{r}}-{\bf{q}}\cdot{\bf{p}}\cdot{\bf{r}}+q_0r_0{\bf{p}}+p_0r_0{\bf{q}}+r_0{\bf{q}}\times{\bf{p}}+q_0{\bf{p}}\times{\bf{r}}+p_0{\bf{q}}\times{\bf{r}}+({\bf{q}}\times{\bf{p}})\times{\bf{r}})$,

and

$a(bc)=(q_0,{\bf{q}})(p_0r_0-{\bf{p}}\cdot{\bf{r}},p_0{\bf{r}}+r_0{\bf{p}}+{\bf{p}}\times{\bf{r}})$ $=(p_0q_0r_0-q_0{\bf{p}}\cdot{\bf{r}}-p_0{\bf{q}}\cdot{\bf{r}}-r_0{\bf{q}}\cdot{\bf{p}}-{\bf{q}}\cdot({\bf{p}}\times{\bf{r}}),$ $q_0p_0{\bf{r}}+q_0r_0{\bf{p}}+q_0{\bf{p}}\times{\bf{r}}+p_0r_0{\bf{q}}-{\bf{q}}\cdot{\bf{p}}\cdot{\bf{r}}+p_0{\bf{q}}\times{\bf{r}}+r_0{\bf{q}}\times{\bf{p}}+{\bf{q}}\times({\bf{p}}\times{\bf{r}}))$.

All terms in $(ab)c$ and $a(bc)$ are the same (note specifically that $({\bf{q}}\times{\bf{p}})\cdot{\bf{r}}={\bf{q}}\cdot({\bf{p}}\times{\bf{r}})$ which is established by interchange of lines in the determinant), except the last cross product terms: $({\bf{q}}\times{\bf{p}})\times{\bf{r}}$ in $(ab)c$ and ${\bf{q}}\times({\bf{p}}\times{\bf{r}})$ in $a(bc)$. We know that cross product does not satisfy associativity, so the above two cross products are not equal in general, i.e., $({\bf{q}}\times{\bf{p}})\times{\bf{r}}\ne{\bf{q}}\times({\bf{p}}\times{\bf{r}})$, which will lead to inequality of the two vector components of $(ab)c$ and $a(bc)$. Then associativity will not hold for quaternion multiplication. I believe associativity is not true for cross product but true for quaternion multiplication, so where did I made mistakes? Please stick to the scalar and vector form of quaternion representation. For the sake of the tedious latex formulas I input, please help me with this issue. Thanks a lot.

[1]Animating Rotation with Quaternion Curves, Ken Shoemake, SIGGRAPH 1985.

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  • $\begingroup$ I'm not sure that this is the problem, but notice that you've written ${\bf{q}}\cdot{\bf{p}}\cdot{\bf{r}}$, which is ill-defined. Try being more careful with that, and see if it cancels anything else in the expression. $\endgroup$ – Chris Culter Jan 26 '16 at 23:09
  • $\begingroup$ dot product is associative, so parenthesis does not matter, that's why I can write it in this way which is well-defined. But cross product is not; parenthesis does matter. $\endgroup$ – user5280911 Jan 26 '16 at 23:12
  • $\begingroup$ simplify a little with $p_0=q_0=r_0=1$ and use that $p.(r\times q) = -r.(q\times p)$ $\endgroup$ – reuns Jan 26 '16 at 23:15
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    $\begingroup$ The dot product is not associative. First of all, it doesn't make sense to write a repeated dot product, since the dot product of two vectors is not a vector. But even if you interpret it as a scalar product, the value of $\hat x\cdot\hat x\cdot\hat y$ depends on the order of evaluation. $\endgroup$ – Chris Culter Jan 26 '16 at 23:20
  • $\begingroup$ @Chris Culter: You are right. I've been thinking dot product satisfies associativity. This is where I made the mistake.Thank you for pointing it out $\endgroup$ – user5280911 Jan 26 '16 at 23:37
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$(q \cdot p)r$, not $q \cdot p \cdot r$. $q \cdot p$ is a scalar, so this is just scalar multiplication. Fix those, then use $q \times (p \times r) = p (q \cdot r) - r (q \cdot p)$. This is called the BAC CAB rule.

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  • $\begingroup$ Thank you for the answer. PS: where did you learn the "BAC CAB rule"? My humble linear algebra textbook doesn't even mention a word about it. Sigh~ $\endgroup$ – user5280911 Jan 26 '16 at 23:40
  • $\begingroup$ We use it all the time when studying classical electrodynamics. $\endgroup$ – Aurey Jan 26 '16 at 23:42

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