Notes:Delta complex/Formal attempt
From Maths
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 [ilmath]\{a,b,c\} [/ilmath] for a triangle and such.
Notations:
- Let [ilmath]\#(n):\eq\{1,\ldots,n\}\subset\mathbb{N} [/ilmath] - I did want to use [ilmath]C(n)[/ilmath] for "count" or "consecutive" but given the context that'd be a poor choice!
- Consider [ilmath]\#(n)[/ilmath] as a poset in its own right (in fact a total order is in play) with the "usual" ordering on [ilmath]\mathbb{N} [/ilmath] it inherits. This is a standard substructure construction.
- Let [ilmath]K[/ilmath] be our Delta complex, let us sidestep defining exactly what this is now, as a tuple of sets.
- Let [ilmath]S_n(K)[/ilmath] be the set of [ilmath]n[/ilmath]-simplices of [ilmath]K[/ilmath]
- Let [ilmath]I(m,n)[/ilmath] be defined to be equal the collection of all injective monotonic functions of the form [ilmath]f:\#(m+1)\rightarrow\#(n+1)[/ilmath]^{[Note 1]}
- The [ilmath]+1[/ilmath] comes from the definition: [ilmath]\text{Dim}(\sigma):\eq\vert\sigma\vert - 1\in\mathbb{N} [/ilmath] - we take care with the case [ilmath]\sigma\eq\emptyset[/ilmath] as I'm developing a framework including this and come up with 2 "null objects" that do not alter the theory, for now [ilmath]\text{Dim}(\emptyset)\eq -1[/ilmath] will do. It wont matter.
- [ilmath]\Delta^m[/ilmath] be the standard [ilmath]m[/ilmath]-simplex in [ilmath]\mathbb{R}^{m+1} [/ilmath]
- [ilmath]G(n,m)[/ilmath] - this is our goal, it's a collection of a bunch of maps of the form [ilmath]G:S_n(K)\rightarrow S_m(K)[/ilmath] {{Caveat|Notice the flip of [ilmath]n[/ilmath] and [ilmath]m[/ilmath]) with certain properties.
- Our goal is to find a bijection, say [ilmath]F:I(m,n)\rightarrow G(n,m)[/ilmath]
First stab
Definition:
- The "gluing data" of a [ilmath]\Delta[/ilmath]-complex corresponds to two parts:
- [ilmath]S_n(K)[/ilmath] - the set of [ilmath]n[/ilmath]-simplices of [ilmath]K[/ilmath]
- The "gluing maps", [ilmath]G_f[/ilmath], which can be enumerated as follows:
- Let [ilmath]m,\ n\in\mathbb{N}_0[/ilmath] be given and be such that [ilmath]m\le n[/ilmath]
- Then for each [ilmath]f\in I(m,n)[/ilmath] there exists a [ilmath]G_f:S_n(K)\rightarrow S_m(K)[/ilmath] such that:
- If [ilmath]f\eq \text{Id}_{\#(n+1)} [/ilmath] then [ilmath]G_f\eq\text{Id}_{S_n(K)} [/ilmath], and
- If [ilmath]f\in I(m,n)[/ilmath] and [ilmath]g\in I(n,j)[/ilmath] then [ilmath]G_{g\circ f}\eq G_f\circ G_g[/ilmath]
- Then for each [ilmath]f\in I(m,n)[/ilmath] there exists a [ilmath]G_f:S_n(K)\rightarrow S_m(K)[/ilmath] such that:
- Let [ilmath]m,\ n\in\mathbb{N}_0[/ilmath] be given and be such that [ilmath]m\le n[/ilmath]
That's it!
Problems
- I need to form a statement (and then prove it) which shows that we need only consider [ilmath]m\eq k[/ilmath] and [ilmath]n\eq k+1[/ilmath] cases (for [ilmath]k\in\mathbb{N}_0[/ilmath]) we don't need all of them, that statement 2 of the [ilmath]G_f[/ilmath] function definition ensures the result is consistent. It's pretty obvious but I'm not sure how to phrase it.
- I need to show that we have a Hatcher-[ilmath]\Delta[/ilmath]-complex if and only if we have one of these.
Gluing process
- Let [ilmath]m,n\in\mathbb{N} [/ilmath] be given such that [ilmath]m\le n[/ilmath].
- Let [ilmath]f\in I(m,n)[/ilmath] be given, so [ilmath]f:\#(m+1)\rightarrow\#(n+1)[/ilmath] is an injection and is monotonic - as per the definition of [ilmath]I(m,n)[/ilmath].
- We associate [ilmath]f[/ilmath] with [ilmath]L_f:\mathbb{R}^{m+1}\rightarrow\mathbb{R}^{n+1} [/ilmath] which is a linear map defined by its action on a basis as [ilmath]L_f(e_i):\eq e_{f(i)} [/ilmath] where [ilmath]e_i\in\mathbb{R}^\text{whatever} [/ilmath] is a tuple that has [ilmath]0[/ilmath] in every entry except the [ilmath]i^\text{th} [/ilmath] which has [ilmath]1[/ilmath]; as usual.^{[Note 2]}
- It is fairly easy to see that [ilmath]\text{Ker}(M_f)\eq\{0\} [/ilmath], 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:
- [ilmath]L_f':\mathbb{R}^{m+1}\rightarrow L_f(\mathbb{R}^{m+1})[/ilmath] is a linear isomorphism
- As [ilmath]\mathbb{R}^{m+1} [/ilmath] is finite dimensional we see that [ilmath]L_f'[/ilmath] is a continuous map, so forth. As would be [ilmath]L_f[/ilmath] itself of course.
- Notice that [ilmath]L_f'\vert_{\Delta^m}:\Delta^m\rightarrow \text{Some }m\text{-face of }\Delta^n [/ilmath]
- and that this is a homeomorphism onto its image.
- This is the idea of our "gluing map" we see we glue some [ilmath]m[/ilmath]-face of an [ilmath]n[/ilmath]-simplex to some [ilmath]m[/ilmath]-simplex that we already have.
- Define [ilmath]G_f:S_n(K)\rightarrow S_m(K)[/ilmath] by [ilmath]G_f:\sigma\mapsto\text{the }m\text{-simplex to which the }m\text{-face of }\sigma\text{ given by }f\text{ corresponds to} [/ilmath]
- It is fairly easy to see that [ilmath]\text{Ker}(M_f)\eq\{0\} [/ilmath], 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:
- We associate [ilmath]f[/ilmath] with [ilmath]L_f:\mathbb{R}^{m+1}\rightarrow\mathbb{R}^{n+1} [/ilmath] which is a linear map defined by its action on a basis as [ilmath]L_f(e_i):\eq e_{f(i)} [/ilmath] where [ilmath]e_i\in\mathbb{R}^\text{whatever} [/ilmath] is a tuple that has [ilmath]0[/ilmath] in every entry except the [ilmath]i^\text{th} [/ilmath] which has [ilmath]1[/ilmath]; as usual.^{[Note 2]}
- Let [ilmath]f\in I(m,n)[/ilmath] be given, so [ilmath]f:\#(m+1)\rightarrow\#(n+1)[/ilmath] is an injection and is monotonic - as per the definition of [ilmath]I(m,n)[/ilmath].
(see paper notes. Will write this again later)
Notes
- ↑ This basically means:
- [ilmath]\forall x,y\in \#(m+1)[x < y\implies f(x)<f(y)][/ilmath] - notice the strict ordering used here. This ensures that it is 1-to-1. We can never have equality of [ilmath]f(x)[/ilmath] and [ilmath]f(y)[/ilmath]
- Caveat:Not proved yet TODO: Do the proof!
- Caveat:Not proved yet
- [ilmath]\forall x,y\in \#(m+1)[x < y\implies f(x)<f(y)][/ilmath] - notice the strict ordering used here. This ensures that it is 1-to-1. We can never have equality of [ilmath]f(x)[/ilmath] and [ilmath]f(y)[/ilmath]
- ↑ There's some abuse of notation going on here, as if [ilmath]e_i\in\mathbb{R}^n[/ilmath] then [ilmath]e_i\notin\mathbb{R}^m[/ilmath] with [ilmath]m\neq n[/ilmath] of course. We identify [ilmath]\mathbb{R}^m[/ilmath] with a subspace of [ilmath]\mathbb{R}^n[/ilmath] where [ilmath]n\ge m[/ilmath] spanned by the first [ilmath]m[/ilmath] basis vectors. It's not that big of a leap, so shouldn't require any more discussion