Timeline for Build a topological polytope with a specified CW-structure
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2014 at 17:40 | vote | accept | Prasit | ||
| Aug 19, 2014 at 17:40 | |||||
| Aug 19, 2014 at 15:20 | comment | added | Hugh Thomas | I have edited my answer to clarify that I believe $2(n-2)\choose (n-2)$ is the number of zero-cells in $YT_n$. (My indexing also disagreed with yours, and I fixed that too.) | |
| Aug 19, 2014 at 15:12 | history | edited | Hugh Thomas | CC BY-SA 3.0 |
clarifying what I was counting, fixing indexing
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| Aug 19, 2014 at 14:51 | comment | added | Prasit | I do not know the exact number of elements for $YT_n$. But I can calculate the first few by hand. Note that if $n=2$ then there is exactly one tree with two leaves. If $n=3$ then there are $3$ of them (two binary tree and one corolla with 3 leaves). Child(ren) refers to the vertices. In the above case the tree itself determines the youngest child(ren). If $n=4$, $Q^{4-1}$ is precisely $P^3$. So number of cells is precisely number of cells in $P^3$ which is 13. the map from $P^n \to Q^n$ is homeomorphism for $n\leq 3$. But for $n>3$ the map from $P^n \to Q^n$ should be quotient maps. | |
| Aug 19, 2014 at 14:10 | history | answered | Hugh Thomas | CC BY-SA 3.0 |