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Need some help understanding boundary operator on simplicies

  1. Feb 2, 2015 #1
    I am currently reading up on some algebraic topology\differential geometry and have reached the section on de Rham theory. This is my first encounter with such notions and I am a little confused by what is meant when one applies a boundary operator to a simplex. Conceptually, I know that it takes a n-simplex to its boundary n-1 simplex, ,but quantitatively I am confused by the actual formula. How does one add simplicies? Do you sum their respective coordinate values, or do the list of points ie. v0,v1,v2,v3 constitute an algebraic "object" which is to be thought of as a whole and not it's constituent points? I apologize if I am not making much sense here .
     
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  3. Feb 2, 2015 #2

    Orodruin

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    The sum of a number of n-simplices is called an n-chain. You can view this as a collection of oriented simplices. It does not have many particular properties other than the fact that you can use algebraic rules such as subtracting, adding, and multiply by integers. One important thing with regards to orientation is how you can switch the simplex vertices around and that a minus sign appears when you change the orientation of the simplex.
     
  4. Feb 2, 2015 #3
    so am I correct when thinking that the "sum" of multiple simplicies is nothing more than the union of them?
     
  5. Feb 2, 2015 #4

    Orodruin

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    No, at least not completely. In a chain, you can have the same simplex appearing more than once (and if you integrate over that chain you integrate over that simplex twice) or appear with a negative coefficient. Therefore the usual union and intersections do not really describe the arithmetics you can do with a chain. For example, if you have a simplex (or chain) ##s##, then ##s+s = 2s##, while ##s \cup s = s##.
     
  6. Feb 2, 2015 #5
    thanks for the reply, also the book I am reading doesn't specify the structure of the C(k) group of chains. What is the action of this group? all it says is that it is generated by an n-simplex and that any element of this group is called an n-chain. I guess I just have no idea what the group action is? is it addition of simplicies?
     
    Last edited: Feb 2, 2015
  7. Feb 2, 2015 #6

    Orodruin

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    More accurately, it is the addition of chains. The set of simplices simply form a set of generators.

    I am not sure I can answer your last question without a bit more context.
     
  8. Feb 2, 2015 #7
    well, the book says this: "we now assign an orientation to every l-simplex |σi| of the simplical complex K and write it as <σi>. We denote by Cl(K) the free abelian group generated by <σi>. Any element of this group is called an l-chain of K." my question is, how does σi generate this group? is this group infinite or finite? the index i here is not specified to be finite or infinite either.
     
  9. Feb 2, 2015 #8

    Orodruin

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    Are you familiar with what a free abelian group is? If not it might help to start there as you will get a lot of the definitions and properties from it. If you do not have a textbook that describes this, you can always try http://en.wikipedia.org/wiki/Free_abelian_group
     
  10. Feb 2, 2015 #9
    so basically it is just linear combinations of the l-simplicies?
     
  11. Feb 2, 2015 #10

    Orodruin

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    Yes. With integer coefficients.
     
  12. Feb 2, 2015 #11

    lavinia

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    The simplices need to be oriented. Then you are correct.
     
  13. Feb 2, 2015 #12
    Maybe you realize this already, but it's a union of n-1-simplices, not one of them. Like for a tetrahedron (3-simplex), you have the four faces, each one being a 2-simplex. The alternating sign in the formula is also a bit perplexing. One way to think of it is in terms of lower-dimensional examples, thinking of the orientation as a counter-clockwise/clockwise rule, but you still need a bit of a leap of faith that it keeps working in higher dimensions (hard to talk about without pictures). One thing is that it will make the boundary of a boundary 0, as it must be, because everything cancels out in pairs due to the alternating signs, so it sort of works, but there's something unsatisfying about it. I haven't thought through the details of this stuff lately, but I think I did manage to make it crystal clear to myself at one point. It's unlikely I will have time to sort it out again in the next couple days, since I have a job interview to prepare for, but if I do, maybe I'll take the opportunity to post again and explain it.
     
  14. Feb 3, 2015 #13

    lavinia

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    An orientation of a simplex may be thought of as an ordering of its vertices and is usually symbolized by a k-tuple, ##< σ_{j_{1}}, ... σ_{j_{k}}>##.

    For instance for a 2-simplex, ##< σ_{0},σ_{1},σ_{2}>## and ##< σ_{1},σ_{2},σ_{0}>## are possible orientations.

    Just as with orientation of space, two k-tuples are said to determine the same orientation of they differ by an even permutation of the vertices. and the opposite orientation if they differ by an odd permutation. For instance, the two orientations of the two simplex above are the same.

    Oriented simplices are just symbols and as such one can think of the formal free abelian group generated by them. This group is completely formal.

    More generally whenever one has a collection of symbols, one can think of the free abelian group that they generate. This group is just all formal finite integer linear combinations of the symbols with the requirement that the symbols must commute and that multiplication by integers is associative.

    So if a and b are symbols then one says that na + mb is the same as the symbol, mb + na
    and (n+m)a is the same as na + ma. One also says that 0a is the same as 0b and denote this as 0.

    With simplices one treats opposite orientations as the negatives of each other in the chain group.
    For instance,

    ## <σ_{0},σ_{1},σ_{2}> = - <σ_{1},σ_{0},σ_{2}>## and so their sum is equal to zero.


    The boundary operator

    The boundary of an oriented k-simplex is defined to be a signed sum of the oriented simplices that are obtained by removing one vertex at a time. These are oriented faces of ##<σ>##.

    I have seen two ways of defining the boundary.

    One way is to take the alternating sum of the oriented k-1 faces by removing one vertex at a time sequentially..
    For instance,

    ##∂ <σ_{0},σ_{1},σ_{2}> = <σ_{1},σ_{2}> - <σ_{0},σ_{2}> + <σ_{0},σ_{1}>##

    The other way is to preassign orientations to each simplex and then define an "incidence number" that lines the boundary simplices up in the right way.

    I am not sure which method your book uses.
     
    Last edited: Feb 3, 2015
  15. Feb 3, 2015 #14
    thanks for all the great replies, they've helped a lot
     
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