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- Thread starter cpsinkule
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so am I correct when thinking that the "sum" of multiple simplicies is nothing more than the union of them?

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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?

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I am not sure I can answer your last question without a bit more context.

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so basically it is just linear combinations of the l-simplicies?

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Yes. With integer coefficients.

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lavinia

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The simplices need to be oriented. Then you are correct.so basically it is just linear combinations of the l-simplicies?

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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.Conceptually, I know that it takes a n-simplex to its boundary n-1 simplex, ,but quantitatively I am confused by the actual formula.

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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 n**a** + m**b** is the same as the symbol, m**b** + n**a**

and (n+m)**a** is the same as n**a** + m**a. **One also says that 0**a** is the same as 0**b** 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.

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

and (n+m)

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 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.

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thanks for all the great replies, they've helped a lot

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