Imagine a rectangular lattice and the only way to move around is to go from node to node by horizontal and vertical movements. Diagonal movements are not allowed.
Think of a Taxicab on the Manhattan street grid. It certainly can't drive diagonally through a block!
So the node point (m,n) is at a distance m+n from the origin (m and n are integers of course) and the metric on the space is therefore d(m,n)=m+n
We define a circle as usual - a set of point equidistant from a given point. But remember, we now measure distance using our metric. So a circle centered on the origin is a diamond. We define pi as usual as periphery/diameter and we get pi=4 exactly.
In this grid each node is connected to exactly 4 others and pi=4. Is this a coincidence?
Is there a grid where each node is connected to exactly 3 others? Yes, a hexagonal grid has this, and pi=3 exactly.
Is there a grid where each node is connected to exactly 6 others? Yes, a triangular grid has this, and pi=6 exactly.
So the first thing we learn is that the value of pi is dependent on the metric used. In this metric pi=connectivity.
Physics uses the metric d(m,n)=sqrt(m^2+n^2). What does this mean for pi?
The General Case This geometry is not constrained to grids (rectangular, hexagonal, triangular or otherwise). So long as the connectivity is correct the whole thing could be a mesh piled in a giant heap on the floor!
There are just two rules: You can only move between nodes along a connection, and the metric is d(m,n)=m+n.
It's very general. It's topological. It's all about connectivity.
A fun way to imagine this geometry: Think of this space as a fishing net. The knots in the net are the points of space and the rope between the knots are the connections. It does not matter how you handle the net - throw it on the ground in a heap if you wish - the topology is unchanged.
Thought: Could we build physics on such a space and if so what would it look like?
Content written and posted by Ken Abbott firstname.lastname@example.org