# Automorphisms of the Fano plane

Timer:60:0 hints: colours: layout: portrait/landscape (change for wide enough screens)
FINISHED - WELL DONE!
The timer starts when you start moving one of the 7 discs and stops when you have got them all in the right place (on the correct gray circles), or when it gets to zero, or if you reset.
Hints lets you know if you put a disc in the right place.
You can also just look at the solutions, by clicking here:
solutions: off: on: configurtion 0

#### The Fano plane

The Fano plane is a collection of 7 points and 7 lines. Each line has three points on it, each point is on three lines, every two lines intersect at some point, every two points are on a common line.

#### The automorphisms

An automorphism of the Fano plane is a way of switching round the points in such a way that if points were on a common line before switching, they are still on a common line after switching, etc; all the relationships are the same after switching as before. There are 168 such automorphisms. This is a picture for thinking about what this means, why it is, and playing around with this idea. Playing around with finite projective planes gives us the game "Dobble", also known as "Spot it!".

#### The game

The image on this page is a game. You have 7 cards, these are like cards in Dobble, but you only have 7 cards, and each card only has 3 symbols. These symbols are just dots of different colours, or different patterns. The game is to place each card in the correct place. The correct place is where the dot lies on three lines which have the same colour as the dots on the card. How fast can you do this? There is a timer which starts as soon as you move the first card.
Each time you play the game, an automorphism will be applied to switch round the lines, so the cards will go in different places. You can play the game 168 different times before you get a repeat.
The program has a list of the 168 confgurations which it goes through. If you reload the page it will restart with the first configuration.

#### More maths (undergrad level, linear algebra assumed)

Why are there 168 automorphisms? There are lots of ways to prove this. One method is to use the construction of the projective plane as a projection of three-space onto a surface. For the Fano plane, you can imagine beams of light, from a light source at (0,0,0), pointing in directions (1,1,1), (1,1,0), (1,0,1), (0,1,1), (1,0,0), (0,1,0), (0,0,1), "projected" on to the surface of a sphere. To see where these beams of light hit the sphere, you just scale so they have length 1. You get a configuration of 7 points. Now when you flatten this out, you can get a picture like the triangle with the seven points, shown below. For the Fano plane, we just take the numbers 0 and 1 in our coordinates. For the Fano plane we work modulo 2, so 1+1=0. For other projective planes, we might take a different set of numbers (or use a different construction).
The automorphisms can be interpreted as certain linear maps of three space. These will permute the beams of light, and so permute the points. Linear maps means scaling, rotating, and also reflecting, stretching and shearing maps. But in order to be an automorphism, you have to map the set of these 7 points to itself. Since we're working modulo 2, any 3 by 3 integer matrix, with entries 0 and 1 will map from these points to these points, together with (0,0,0). We don't want to map a point to (0,0,0), since this is not on the sphere. So (rank nullity theorem), we need that the matrix is invertible. One way to check that the matrix is invertible is to check that it maps a basis to another basis. The columns of the matrix correspond to the images of (1,0,0), (0,1,0), (0,0,1). We just want to map these to other sets of three of the seven points, which will form a basis provided they are not colinear. I.e., they form the vertices of a triangle, and don't lie on a straight line (we have to interpret the circle in the figure as a line when we're working mod 2; it's given by x+y+z=0 mod 2).
So, the automorphims correspond to triangles with vertices at three of these points. There are 28 possible triangles. This is because you can choose 3 points to be the vertices, so that's 7 choose 3 = 35. But 7 of these choices will correspond to the 7 straight lines. So, there are 35-7=28 triangles. For each triangle, you can permute its three points in 6 ways (e.g., think of reflections and rotations; or think of the six ways of colouring the vertices of a fixed triangle, with three colours, a different one for each vertex). So we get 28*6 = 168 different automorphisms, and so 168 ways to place our cards on lines, as in this game.

#### Duality and the dual game

A finite projective plane is self dual, in the sense that you'll get the same picture if you swap lines and points, and swap all relations appropriately, e.g., lines A and B meet at point C becomes points A and B are both on line C.
The third colour option corresponds to playing a dual game. Instead of each disc having three colours, corresponding to the three lines the disc lies on, the discs have constant colours, and the lines have three colours, given by the three discs that lie on the line.
Note, the dual game shown is not exactly the dual game of colour option one; there are lots of possible games, as explained below. An exact dual would have a line with red, green, blue, colours, corresponding to the disc with red, green, blue dots, etc. In this example, none of the lines match up with any of the cards. But it's still a dual of some 7 card Dobble game.

#### How many games

Could you use the same colour dots/lines for a different set of cards, and a different Dobble game? Yes. There are 7! ways to arrange these 7 lines, that's 5040 ways. But set of lines can be put in 168 ways for the same game. So there are 30 = 5040/168 different ways to make a set of 7 cards with these colours. But some sets will overlap and have some cards in common. 30*7 = 210, so if we made all the posible games, we'd make 210 cards. But there are only (7 choose 3) = 35 possible cards. If you take any card, e.g., a card with red, blue, green dots, how many different games is it in? If we pick two cards, how many games are they in? How many cards do you have to specify to completely determine a game? Can you find a set of five games, which together contain each of the possible 35 cards, exacly once each? If that is possible, how many ways can you do it?