# Losing My Orientation

12 May 2017While trying to grok $SO(3) \cong \mathbb{RP}^3$ and understand various parametrizations of $SO(3)$, I wandered a bit (okay maybe more that litte:) ) and started to think about projective space itself. It is well know that real projective spaces alternate between being orientable and non-orientable, as dimension increases. Specifically odd dimensional projective spaces are orientable but even ones are not. For example, $\mathbb{RP}^1$ and $\mathbb{RP}^3$ are orientable but $\mathbb{RP}^2$ is not.

Jeffery Week’s *Shape of Space* gives a Flatland inspired explanation of orientations on the projective plane and projective 3-space. It’s a fantastic book and a wonderful example of storytelling as a teaching method. In this note, I want to invoke Week’s style and describe orientations on real projective spaces of any dimension.

Lets start with a low dimensional example, $\mathbb{RP}^2$. There are a few ways to describe the real projective plane: affine plane with points at infinity, as a sphere with antipodes identified, or a disc with boundary antipodes identified. We can’t embed $\mathbb{RP}^2$ in 3-dimensional space so we’re forced to work with patches. So, the disc model will serve our purposes best.

Suppose Pac-Man lives on the real projective plane. He’s hungry and sees food down the road. The boundary of the disc is identified, albeit in a reverse oriented way (note the arrows), so Pac-Man sees no obstruction.

**What does Pac-Man look like when he crosses the disc boundary?**

Antipodal identification means that as Pac-Man crosses the boundary of the disc, he is flipped about his midline. As he emerges from the left side he appears flipped with his *left side* in view. Of course in 2-dimensional space there is no sidedness, but we can excise the road with Pac-Man ( after all it is just a Möbius Band! ) and embed it in 3-dimensional space to help us visualize Pac-Man’s journey. Another way to think about it without appealing to a 3-dimensional embedding is to imagine Pac-Man passing through the boundary one vertical strip at a time. Each time a strip hits the boundary it is up-down reversed and then sent to the opposite side. I imagine him emerging from the left side like a image being printed by a dot matrix printer. Each passage of the print head depositing a up-down reversed strip of Pac-Man.

**The question remains: Is Pac-Man differently oriented?** Imagine an arrow curving the clockwise direction on Pac-Man’s face prior to his passage through the boundary of the disc. That curving arrow is in the counterclockwise direction when he appears from the left; which, to me is a succinct demonstration that Pac-Man’s orientation is reversed.

…So then what is orientation? Well, we’ve seen that Pac-Man is reverse oriented because he’s flipped, once. That could be a definition: one flip. What about two or three, etc.? The boundary identification induces a motion on Pac-Man that leaves his form and shape unchanged. These motions are described by orthogonal groups for *continuous* objects, like Pac-Man. Though for a more concrete explanation imagine a polygon in 2-dimensions and we only care about how it is rearranged by the topology of the space. These symmetries are described exactly by Dihedral groups; which are composed of flips (reflections) and rotations. It should be clear that rotations are orientation preserving – a clockwise motion rotated remains clockwise. A flip followed by another flip is always a rotation, and rotations form a group. So, we have an answer: **an even # of flips preserves orientation, while an odd # of flips reverses orientation.**

Lets apply this definition to determine the orientation of $\mathbb{RP}^n$ where $n$ is $3$ or more. We can model $\mathbb{RP}^n$ as a $n$-ball with the boundary antipodally identified. Imagine a boxy Pac-Man floating over a road in $B^n$. Again he sees no boundary because of the identification, nonetheless as he passes through he will be altered somehow.

Lets consider $n-1$ dimensional slices of Pac-Man, since the boundary of the ball is a $n-1$-sphere and we can think of Pac-Man passing through the boundary slice by slice. We can imagine the boundary, when viewed heuristically from a $n-1$-dimensional slice, drawn below, as the inside of a sphere. The horizontal axis representing $n-2$ directions and the vertical axis one direction.

Each slice is flipped $1 + # { flips in n-2 directions}$ times; a recursive relation–for simplicity let $n$ replace $n-1$; $n$ matches the dimension of the real projective space of interest.

\[\begin{gather*} f(n) = 1 + f(n-1) \quad \text{ where } \quad f(1) = 0 \Longrightarrow \\ f(n) = n-1 \text{ for } n > 0 \end{gather*}\]Orientation is the parity of $f$, so $n-1 \mod 2$ represents orientation. So 3D Pac-Man’s orientation in $\mathbb{RP}^3$ is preserved! And in general

\[\mathbb{RP}^n \text{ is } \left\{ \begin{array}{lr} \text{orientable} & \text{ if } n \mod 2 \equiv 1 \\ \text{non-orientable} & \text{ if } n \mod 2 \equiv 0 \\ \end{array} \right.\]