Notes:Delta complex/Formal attempt

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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|1N - 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:
    1. Sn(K) - the set of n-simplices of K
    2. The "gluing maps", Gf, which can be enumerated as follows:
      • Let m, nN0 be given and be such that mn
        • Then for each fI(m,n) there exists a Gf:Sn(K)Sm(K) such that:
          1. If f=Id#(n+1) then Gf=IdSn(K), and
          2. If fI(m,n) and gI(n,j) then Ggf=GfGg

That's it!

Problems

  1. 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 kN0) 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.
  2. I need to show that we have a Hatcher-Δ-complex if and only if we have one of these.

Gluing process

  • Let m,nN be given such that mn.
    • Let fI(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+1Rn+1 which is a linear map defined by its action on a basis as Lf(ei):=ef(i) where eiRwhatever 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:
          • Lf:Rm+1Lf(Rm+1) is a linear isomorphism
        • As Rm+1 is finite dimensional we see that Lf is a continuous map, so forth. As would be Lf itself of course.
        • Notice that Lf|Δm:ΔmSome m-face of Δn
        • 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:σthe m-simplex to which the m-face of σ given by f corresponds to

(see paper notes. Will write this again later)


Notes

  1. Jump up This basically means:
    • x,y#(m+1)[x<yf(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 eiRn then eiRm with mn of course. We identify Rm with a subspace of Rn where nm spanned by the first m basis vectors. It's not that big of a leap, so shouldn't require any more discussion