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Jul 14, 2012 at 5:51 comment added Andy Putman @Alexander Chernov : No.
Jul 10, 2012 at 4:59 comment added Alexander Chervov The complex is finiite?
Jul 3, 2012 at 16:02 comment added Andy Putman @alvarezpaiva : It's something special about dimension $2$. It's easy to come up with counterexamples in higher dimensions; for instance, you can triangulate $\mathbb{R}^3$ so that it contains $S^2$ as a subcomplex.
Jul 3, 2012 at 5:12 comment added alvarezpaiva This is just in two dimensions or are you giving us the simplest unkown case ?
Jul 3, 2012 at 4:17 history answered Andy Putman CC BY-SA 3.0