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Mapping of the standard k-simplex in R^n to X

  1. Feb 3, 2015 #1
    My book denotes by σ:Δk→X for some suitable topological space X a standard k-simplex of X. It then describes the free abelian group generated by such σ's as the group of k-chains on X. It is not clear to me what is meant by a chain for a map σ. I understand a chain in Rn to be sums of integer linear combination of simplicies, but I cannot wrap my head around what is meant by chains of mappings of Δk. Is each σ mapping Δk to different neighborhoods on X? Is the group action for chains function composition of each σ? Currently, I am understanding a chain to be a mapping of Δk multiple times into X such that they connect at their boundaries (or perhaps overlap?), but I feel like this is not the correct view. Any insight would be greatly appreciated.

    EDIT: I incorrectly put R^n in the title, it should be R^k to match the dimension of the simplex
  2. jcsd
  3. Feb 3, 2015 #2


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    As I already wrote the k chains are a purely symbolic group. There is no group action.
    I am not sure why you think there is.

    if you take simplices to be maps of the standard simplex in Euclidean space then just form the formal free abelian group on all of these maps.

    In this case, the boundary of the simplex is the restriction of the mapping to the boundary of the standard simplex in R^k.

    Before I thought you were talking about a simplicial complex. Sorry for the confusion.
    But the story here is essentially the same.

    A map of the standard simplex in Euclidean space can be thought of as the generator of a purely formal free abelian group. One jus associates a symbol for this map and defines a group of symbols as I described in you previous post. When you consider all possible maps - in de Rham theory they are smooth maps - then one can formally talk about the free abelian group that they generate. This is an infinite dimensional free abelian group.

    Orientation comes from the standard orientation of the standard k-simplex in Euclidean space.
    The boundary of the standard simplex is the formal alternating sum of its k-1 faces taken in sequential order just as I described in the other post. The boundary of the mapping is the alternating sum of its restriction to each k-1 face.

    So for the standard 2 simplex σ = <0,(1,0),(0,1)>

    ∂σ = <(1,0),(0,1)> - <0,(0,1)> + <0,(1,0)>

    If f is a smooth map of σ into a manifold, then ∂f = f(∂σ) = f(<(1,0),(0,1)>)- f(<0,(0,1)>) + f(<0,(1,0)>)
    Last edited: Feb 4, 2015
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