Timeline for Amending flawed "proof" that homology groups are zero
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2023 at 12:54 | comment | added | Sam Hopkins | @HenrikRüping: I just meant that according to the description of the complex in question at the end of Andy Putman's answer, it is clearly flag: a subset of integers being coprime-free is clearly a pairwise condition. (Sorry, I realize I meant "nerve of $X_I$" and not "$X_I$" above.) | |
| Sep 13, 2023 at 12:01 | comment | added | HenrikRüping | It feels like one can then further restrict to only squarefree $n$'s. If we would only allow squarefree $k$'s in this construction, we get an inclusion of a subcomplex. The map sending $k$ to the lcm of its prime factors (I don't know whether that number has a name) defines a retraction and I think both are homotopy equivalences. | |
| Sep 13, 2023 at 12:01 | comment | added | HenrikRüping | @SamHopkins : I dont think the nerve of the cover is flag, it could happen that we find three open sets $U_1,U_2,U_3$ such that any pair has nonempty intersection, but $U_1\cap U_2\cap U_3$ is empty. | |
| Sep 13, 2023 at 3:39 | comment | added | Sam Hopkins | I don’t know if this helps in any way, but since whether a set of vertices determine a face of $X_I$ is defined by a pairwise condition, $X_I$ is what is called a flag simplicial complex (also known as a clique complex). | |
| Sep 13, 2023 at 2:47 | comment | added | Andy Putman | @MarcelK.Goh: When you add a new vertex, the effect is to cone off the link of that vertex. So to prove this inductively, you’ll need to understand the topology of that link. | |
| Sep 13, 2023 at 2:28 | comment | added | Marcel K. Goh | Because there are now far fewer vertices to work with, it almost seems like one can work by induction. There is a simplex consisting of the multiples of every prime, and whenever a new vertex $n$ is added, it becomes a vertex of intersection among all the simplices corresponding to primes that divide it. It feels like this should never cause any empty cycles to appear, but I'm still thinking about how to prove this rigorously. | |
| Sep 13, 2023 at 1:57 | comment | added | Marcel K. Goh | Wow, thanks! I do think this will be easier to work with than the original complex. I think it is okay to keep $1$ in the index set, since by the definition above, the set $\{1\}$ is a maximal coprime-free set (and there are already other isolated vertices for every prime $p$ in $(n/2,n]$, so adding $1$ alongside these primes shouldn't mess things up too much). Besides these primes and $1$, the rest of $\Delta_n$ should be connected. | |
| Sep 13, 2023 at 0:06 | history | answered | Andy Putman | CC BY-SA 4.0 |