Prove that the tesseract graph is non-planar

1.6k Views Asked by At

The tesseract graph may be defined in various ways. I'm thinking of it as a subset lattice of the set $\{a,b,c,d\}$, where two subsets are adjacent if they differ in size by one, and one contains the other.

The tesseract is listed as non-planar on Wolfram MathWorld, but there is no proof given. I'm trying to prove it using Kuratowski's theorem, but I'm stuck. I feel certain I'm not going to find a copy of $K_5$ in the tesseract, but it seems there might be a copy of $K_{3,3}$ hiding in there. Does anyone see it?

Or would using Euler's formula be a better approach?

1

There are 1 best solutions below

0
On BEST ANSWER

Tesseract1

Looking at the tesseract above, contract a single edge in the red cycle (so it's a $3$-cycle), contract both orange cycles to a single point, and contract all the green edges to a single edge. This will result in $K_5$, so by Wagner's Theorem, the tesseract is non-planar.

Alternatively, if you really want to use Kuratowski's Theorem, here's a subgraph homeomorphic to $K_{3,3}$.

Tesseract2