## What Was Up With That Rotation Trick?June 10th, 2009 Patrick Stein

In my prior post about using Clifford algebras to do plane rotations, I finished with a non-intuitive step at the end. Rather than multiplying on the right by an element representing a rotation of angle $\theta$, I multiplied on the left by an element representing a rotation of angle $\frac{\theta}{2}$ and multiplied on the right by an element representing a rotation of angle $-\frac{\theta}{2}$.

Why did I do this? Well, I mentioned it would be awkward for the two-dimensional case, but that it will be important when we get to three or more dimensions. Well, work for a moment with $\frac{\theta}{2}$ being a quarter rotation (ninety degrees, $\frac{\pi}{2}$ radians). This means our total rotation is going to be a half turn (180 degrees, $\pi$ radians).

For that $\frac{\theta}{2}$, $r = e_1e_2$ and so $\overline{r} = -e_1e_2$. Let’s just look at what it does to our unit vectors $e_1$ and $e_2$ to multiply on the left by $r$ and on the right by $\overline{r}$.

For $e_1$, we get $-e_1e_2e_1e_1e_2 = -e_1e_2e_2 = -e_1$. Similarly, for $e_2$, we get $-e_1e_2e_2e_1e_2 = -e_2$.

So far, we were only working in two dimensions. As such, there wasn’t any $e_3$ to worry about. But, what if there were? What happens to the $z$-coordinate of something if you rotate things parallel to the $xy$-plane? It remains unchanged.

Well, what would happen if we multiplied $e_3$ on the right by $\cos\theta + \sin\theta e_1e_2$? We would end up with $\cos\theta e_3 + \sin\theta e_3e_1e_2 = \cos\theta e_3 + \sin\theta e_1e_2e_3$. We’ve ended up scaling $e_3$ and adding in a trivector $e_1e_2e_3$. We’ve made a mess.

Let’s try it instead with our trick. We’re going to start with $-e_1e_2e_3e_1e_2$. Every time we transpose elements with different subscripts, we flip the sign. Every time we get two elements next to each other with the same subscript, they cancel out. So, switching the $e_3$ with the second $e_1$, we get $e_1e_2e_1e_3e_2$. From there, we can switch the first two elements to get $-e_2e_1e_1e_3e_2$ which is just $-e_2e_3e_2$. We can switch the $e_3$ with the second $e_2$ to get: $e_2e_2e_3$ which is just $e_3$. So, our trick leaves $e_3$ unchanged.

In the above, there is nothing special about the subscript three. It would work for any subscript except one or two. So, the trick allows us to break the rotation up into two parts that still do what we want with $e_1$ and $e_2$ but leave our other directions unchanged (or, maybe it’s easier to think of them as changing them and then changing them right back).