# Hex on the real projective plane

Some number of years ago while thinking about the games Y and Hex, I wondered about what such a game would be like on the real projective plane (unaware of Projex).

Setup: the game board is a connected graph embedded on the surface of a sphere, where the graph has antipodal symmetry (so vertices and edges each come in antipodal pairs). A nice example, which is the basis for the Y board, is a subdivided icosahedral graph.

Gameplay: players take turns placing two of their stones on antipodal pairs of empty vertices. The first player to form a path connecting a pair of antipodal points is the winner. Practically speaking, the game can take place on a board like the one illustrated to the left. Stones are placed on interior vertices like usual, but on the boundary stones are placed in antipodal pairs. With this modified setup, the aim of the game is to make a path that connects boundary antipodes.

In other words: given a connected graph embedded in $\mathbb{R}{\mathrm{P}}^{2}$ (with at least one essential cycle), players take turns placing stones on vertices. The first player to make a path that generates ${H}_{1}\left(\mathbb{R}{\mathrm{P}}^{2}\right)$ is the winner. This is similar to to Hex, where each player is aiming to construct a particular generator of ${H}_{1}\left(\left({S}^{1}{\right)}^{2}\right)$.

A nice feature of the game is that one player winning excludes the other player from completing a cycle of their own. A quick way to see this is that a completed cycle is a simple closed curve on the sphere, separating the sphere into two disks. The other player would have to construct a path from one disk to the other, which is impossible. This also has to do with the $\mathbb{Z}/2\mathbb{Z}$ intersection form on $\mathbb{R}{\mathrm{P}}^{2}$.

I got Morgan to play this game at the 2017 Georgia International Topology Conference on a subdivided hemi-icosahedron. One question I had was whether one of the players must win, or whether it was possible for there to be a draw. We thought about it for a few minutes and then forgot about the problem as we went to the next talk.

Yesterday, I brought the game up again, and Ian Agol pointed out that if the graph is a triangulation, the kind of argument for Hex should work because one of the components will have to contain a Mobius band. In this post, I’m going to flesh out what I thought he meant. For now, assume the game board is a triangulation of $\mathbb{R}{\mathrm{P}}^{2}$.

Let’s say some players have played the game with red and blue stones, and for some reason they kept playing until the entire board was been filled. By the exclusion property, it’s not the case that both the players have completed a cycle, so there is a clear winner if there is one at all. Let’s also say we don’t care how many of each color stone there are, since we’re not going to use any sort of parity argument.

There are, up to rotation, only four configurations of stones around each triangle, depending purely on how many of each color stone there are. In the two cases where there are both colors present, there is a well-defined path between midpoints of the triangle’s edges that separate the two colors. By taking all such paths, one gets a collection of loops in $\mathbb{R}{\mathrm{P}}^{2}$ that intersect the edges of the game graph transversely. Each region between the loops contains stones only of a single color, and when crossing a loop that color changes (so the loops define a properly $2$-colored map in $\mathbb{R}{\mathrm{P}}^{2}$).

At this point, we could use the Mayer-Vietoris sequence with a decomposition of the projective plane into two regions whose shared boundary is a $1$-manifold, but I find the following more hands-on normal curve argument interesting.

If there is a loop that bounds a disk, then none of the stones within that disk can possibly contribute to a winning cycle. Furthermore, because of the way the curve is constructed, on just the opposite side from the disk there is a non-winning monochromatic cycle of stones of the opposite color. If we replace all the stones within the disk with the color of the monochromatic cycle, we do not change the winner or whether the game ended in a draw. (The fact that the loop has stones of opposite colors on either side means that a regular neighborhood of it is orientable — and not a Möbius band.)

So, now we may assume that none of the loops bound disks. But, all of the loops have to bound disks! The loop has stones of opposite colors on each side so traces out an orientation-preserving path. Maybe I’ll fill in this detail in later, but this means the path is nullhomotopic, and every nullhomotopic simple closed curve a surface bounds a disk.

Therefore, our simplifying assumption implies the game board contains stones of only one color. This obviously contains a winning cycle, so that player must have been the winner. No draws!

I have another version of this argument on math.stackexchange.com.

## Game board designs

The design for Projex is a subdivided triangle, where I was proposing a subdivided hemi-icosahedron. With the subdivided triangle, it seems like there will be a more intense battle to control the center of the board, where the hemi-icosahedron’s symmetry makes every icosahedral face equally valuable, so early mistakes might not cost the game. I actually don’t know how much this affects gameplay, but maybe I will simulate it someday.