The popular YouTube channel Numberphile has just released a video featuring yours truly — filming this with Brady Haran was a real pleasure!
A couple of quick mathematical comments: in the video, I compute the Dehn invariant of the cube with side length \(1\) as \(12\otimes \pi/2\) — astute YouTube commenters have already noticed that this is the same as zero! I never really discussed the group structure on \(\mathbb\otimes \mathbb/2\pi\), so I resisted saying this in the video.
For a similar reason, the proof that the Dehn invariant is actually an invariant (under cutting and pasting) sketched in the video is not quite complete. The gap is that I never explain what happens with new edges added (in the middle of what used to be a face) when one makes a cut! The point is that new edges come in pairs with equal length and with dihedral angles which sum to \(\pi\). So the contribution to the Dehn invariant is \[\ell\otimes \theta_1+\ell\otimes \theta_2 =\ell\otimes \pi=0,\] where here I use that \(\ell\otimes \theta=0\) if \(\theta\) is a rational multiple of \(2\pi\).
Anyway, hope you all enjoyed the video!