Thuses

This was a fun question I thought about once.  My answer is at the end, in case you’d like to try solving the problem yourself.  The question is likely more interesting than my solution.

A well known theorem says that every group occurs as for some topological space .  It’s …

In this blog post, I’ll provide a slick proof of a form of the slice-Bennequin inequality (as outlined by Kronheimer in a mathoverflow answer.) The main ingredient is the adjunction inequality for surfaces embedded in closed 4-manifolds. To obtain the slice-Bennequin inequality (which is a statement about surfaces embedded …