Difference between revisions of "Notes:Delta complex/Formal attempt"

Latest revision as of 14:36, 6 February 2017

Formal attempt

We try and keep everything combinatorial, so keep an abstract simplicial complex in the back of your mind, and a simplex as being like $\{a,b,c\}$ for a triangle and such.

Notations:

• Let $\#(n):\eq\{1,\ldots,n\}\subset\mathbb{N}$ - I did want to use $C(n)$ for "count" or "consecutive" but given the context that'd be a poor choice!
  • Consider $\#(n)$ as a poset in its own right (in fact a total order is in play) with the "usual" ordering on $\mathbb{N}$ it inherits. This is a standard substructure construction.
• Let $K$ be our Delta complex, let us sidestep defining exactly what this is now, as a tuple of sets.
• Let $S_n(K)$ be the set of $n$-simplices of $K$
• Let $I(m,n)$ be defined to be equal the collection of all injective monotonic functions of the form $f:\#(m+1)\rightarrow\#(n+1)$^{[Note 1]}
  • The $+1$ comes from the definition: $\text{Dim}(\sigma):\eq\vert\sigma\vert - 1\in\mathbb{N}$ - we take care with the case $\sigma\eq\emptyset$ as I'm developing a framework including this and come up with 2 "null objects" that do not alter the theory, for now $\text{Dim}(\emptyset)\eq -1$ will do. It wont matter.
• $\Delta^m$ be the standard $m$-simplex in $\mathbb{R}^{m+1}$
• $G(n,m)$ - this is our goal, it's a collection of a bunch of maps of the form $G:S_n(K)\rightarrow S_m(K)$ {{Caveat|Notice the flip of $n$ and $m$) with certain properties.
  • Our goal is to find a bijection, say $F:I(m,n)\rightarrow G(n,m)$

First stab

Definition:

• The "gluing data" of a $\Delta$-complex corresponds to two parts:
  1. $S_n(K)$ - the set of $n$-simplices of $K$
  2. The "gluing maps", $G_f$, which can be enumerated as follows:
     • Let $m,\ n\in\mathbb{N}_0$ be given and be such that $m\le n$
       • Then for each $f\in I(m,n)$ there exists a $G_f:S_n(K)\rightarrow S_m(K)$ such that:
         1. If $f\eq \text{Id}_{\#(n+1)}$ then $G_f\eq\text{Id}_{S_n(K)}$, and
         2. If $f\in I(m,n)$ and $g\in I(n,j)$ then $G_{g\circ f}\eq G_f\circ G_g$

That's it!

Problems

1. I need to form a statement (and then prove it) which shows that we need only consider $m\eq k$ and $n\eq k+1$ cases (for $k\in\mathbb{N}_0$) we don't need all of them, that statement 2 of the $G_f$ function definition ensures the result is consistent. It's pretty obvious but I'm not sure how to phrase it.
2. I need to show that we have a Hatcher-$\Delta$-complex if and only if we have one of these.

Gluing process

• Let $m,n\in\mathbb{N}$ be given such that $m\le n$.
  • Let $f\in I(m,n)$ be given, so $f:\#(m+1)\rightarrow\#(n+1)$ is an injection and is monotonic - as per the definition of $I(m,n)$.
    • We associate $f$ with $L_f:\mathbb{R}^{m+1}\rightarrow\mathbb{R}^{n+1}$ which is a linear map defined by its action on a basis as $L_f(e_i):\eq e_{f(i)}$ where $e_i\in\mathbb{R}^\text{whatever}$ is a tuple that has $0$ in every entry except the $i^\text{th}$ which has $1$; as usual.^{[Note 2]}
      • It is fairly easy to see that $\text{Ker}(M_f)\eq\{0\}$, then by "a linear map is injective if and only if its kernel is trivial" and "the image of a linear map is a vector subspace of the codomain" wee see that:
        • $L_f':\mathbb{R}^{m+1}\rightarrow L_f(\mathbb{R}^{m+1})$ is a linear isomorphism
      • As $\mathbb{R}^{m+1}$ is finite dimensional we see that $L_f'$ is a continuous map, so forth. As would be $L_f$ itself of course.
      • Notice that $L_f'\vert_{\Delta^m}:\Delta^m\rightarrow \text{Some }m\text{-face of }\Delta^n$
        • and that this is a homeomorphism onto its image.
      • This is the idea of our "gluing map" we see we glue some $m$-face of an $n$-simplex to some $m$-simplex that we already have.
        • Define $G_f:S_n(K)\rightarrow S_m(K)$ by $G_f:\sigma\mapsto\text{the }m\text{-simplex to which the }m\text{-face of }\sigma\text{ given by }f\text{ corresponds to}$

(see paper notes. Will write this again later)

Notes

1. Jump up ↑ This basically means:
   • $\forall x,y\in \#(m+1)[x < y\implies f(x)<f(y)]$ - notice the strict ordering used here. This ensures that it is 1-to-1. We can never have equality of $f(x)$ and $f(y)$
     • Caveat:Not proved yet
       TODO: Do the proof!
2. Jump up ↑ There's some abuse of notation going on here, as if $e_i\in\mathbb{R}^n$ then $e_i\notin\mathbb{R}^m$ with $m\neq n$ of course. We identify $\mathbb{R}^m$ with a subspace of $\mathbb{R}^n$ where $n\ge m$ spanned by the first $m$ basis vectors. It's not that big of a leap, so shouldn't require any more discussion