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 {a,b,c} for a triangle and such.
Notations:
- Let #(n):={1,…,n}⊂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 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 Sn(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)→#(n+1)[Note 1]
- The +1 comes from the definition: Dim(σ):=|σ|−1∈N - we take care with the case σ=∅ as I'm developing a framework including this and come up with 2 "null objects" that do not alter the theory, for now Dim(∅)=−1 will do. It wont matter.
- Δm be the standard m-simplex in Rm+1
- G(n,m) - this is our goal, it's a collection of a bunch of maps of the form G:Sn(K)→Sm(K) {{Caveat|Notice the flip of n and m) with certain properties.
- Our goal is to find a bijection, say F:I(m,n)→G(n,m)
First stab
Definition:
- The "gluing data" of a Δ-complex corresponds to two parts:
- Sn(K) - the set of n-simplices of K
- The "gluing maps", Gf, which can be enumerated as follows:
- Let m, n∈N0 be given and be such that m≤n
- Then for each f∈I(m,n) there exists a Gf:Sn(K)→Sm(K) such that:
- If f=Id#(n+1) then Gf=IdSn(K), and
- If f∈I(m,n) and g∈I(n,j) then Gg∘f=Gf∘Gg
- Then for each f∈I(m,n) there exists a Gf:Sn(K)→Sm(K) such that:
- Let m, n∈N0 be given and be such that m≤n
That's it!
Problems
- I need to form a statement (and then prove it) which shows that we need only consider m=k and n=k+1 cases (for k∈N0) we don't need all of them, that statement 2 of the Gf 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-Δ-complex if and only if we have one of these.
Gluing process
- Let m,n∈N be given such that m≤n.
- Let f∈I(m,n) be given, so f:#(m+1)→#(n+1) is an injection and is monotonic - as per the definition of I(m,n).
- We associate f with Lf:Rm+1→Rn+1 which is a linear map defined by its action on a basis as Lf(ei):=ef(i) where ei∈Rwhatever is a tuple that has 0 in every entry except the ith which has 1; as usual.[Note 2]
- It is fairly easy to see that Ker(Mf)={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:Rm+1→Lf(Rm+1) is a linear isomorphism
- As Rm+1 is finite dimensional we see that L′f is a continuous map, so forth. As would be Lf itself of course.
- Notice that L′f|Δm:Δm→Some m-face of Δn
- 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 Gf:Sn(K)→Sm(K) by Gf:σ↦the m-simplex to which the m-face of σ given by f corresponds to
- It is fairly easy to see that Ker(Mf)={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:
- We associate f with Lf:Rm+1→Rn+1 which is a linear map defined by its action on a basis as Lf(ei):=ef(i) where ei∈Rwhatever is a tuple that has 0 in every entry except the ith which has 1; as usual.[Note 2]
- Let f∈I(m,n) be given, so f:#(m+1)→#(n+1) is an injection and is monotonic - as per the definition of I(m,n).
(see paper notes. Will write this again later)
Notes
- Jump up ↑ This basically means:
- ∀x,y∈#(m+1)[x<y⟹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!
- Caveat:Not proved yet
- ∀x,y∈#(m+1)[x<y⟹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)
- Jump up ↑ There's some abuse of notation going on here, as if ei∈Rn then ei∉Rm with m≠n of course. We identify Rm with a subspace of Rn where n≥m spanned by the first m basis vectors. It's not that big of a leap, so shouldn't require any more discussion