I’ve had two recent thoughts about the fourth dimension.
The first relates to Euler’s Formula, which says that the difference between the sum of the vertices and faces of a convex polyhedron and its edges is always 2 (that is, \(v + f – e = 2\)). The Number Devil presents this slightly differently: The sum of the vertices and enclosed spaces of a connected graph and its segments is always 1 (that is, \(v + f – e = 1\)). Can we generalize this?
For one dimension, that would suggest that \(v + f – e = 0\). How many enclosed spaces will that imply? For a connected graph, there will always be one more point than there are segments, so \(e + 1 + f – e = 1 + f = 0\), meaning there are -1 enclosed spaces. This is impossible, but its impossibility reflects the fact that one dimension can’t have enclosed spaces in the first place. If we include the segment that connects the endpoints, we have as many segments as vertices (making the case of two points degenerate, since we’d be counting the same segment twice), and we have zero enclosed spaces. Either way, though, we illustrate that it’s impossible to have a “face” in one dimension. Regardless, let’s adjust the formula to remove the f and count only the minimal segments: \(v = e + 1\).
For four dimensions, that would suggest that \(v + f – e = 3\). Let’s test this with the a pentachoron and a tesseract. A pentachoron has five vertices, each of which is connected to each other ((\5 \times 4 \div 2\)) for a total of ten edges. Every set of three points represents a face, hence \(5 \times 4 \times 3 \div 6 = 10\) faces. So \(5 + 10 – 10 = 5\). This is not what we predicted.
A tesseract has sixteen vertices. Unlike the pentachoron, not all of these are connected: There are eight interconnected cubes, each with twelve edges and six faces. Each edge is shared by three cubes ((\12 \times 8 \div 3 = 32\)), while each face is shared by two cubes ((\6 \times 8 \div 2 = 24\)). This gives us \(16 + 24 – 32 = 8\). As with the pentachoron, this is equal to the number of cells.
A check of a table of values for other polychora shows the same pattern: The fourth-dimension equivalent of Euler’s Formula is \(v + f = e + c\), where c represents the number of cells. For 5-dimensional shapes, the equivalent appears to be \(v + f + F = c + e + 2\), where F represents facets.
To summarize, this leaves us with, for each dimension (considering convex shapes for three or more dimensions):
- v = e + 1
- v + f = e + 1
- v + f = e + 2
- v + f = e + c
- v + f + F = e + c + 2
Analyses for polytopes in higher dimensions, but this should be sufficient to show that there’s a pattern, albeit a complex one.
An independent thought: Twice this week, I’ve seen a demonstration of a gravity well, meant to represent the pull of the sun according to Einstein’s theory of relativity. A gravity well demonstration consists of a heavy weight in a stretched elastic material. The weight distorts the fabric, causing other things to roll in a curved rather than straight manner. I thought about how the Martian made things disappear in Stranger in a Strange Land, and how gravity wells would be better modelled if we could create a fourth dimensional model: A three dimensional fabric would be stretched by an object, pulling the fabric of space towards it. I doubt there’s anything novel about this approach, but it helps me understand the basic concept.