# I'm a Numberphile!

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!