0-chain Boundary
Can anyone explain how adding two vertices in a connected graph to create a $0$-Chain is the boundary of some 1-dimensional chain? I know that the definition of boundary is the collection of $n+1$ faces, and each face is an $n-1$ simplice, but can someone explain the connection?
topological-data-analysis
asked Nov 27 at 0:26
LexyFidds
226
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Can anyone explain how adding two vertices in a connected graph to create a $0$-Chain is the boundary of some 1-dimensional chain? I know that the definition of boundary is the collection of $n+1$ faces, and each face is an $n-1$ simplice, but can someone explain the connection?
topological-data-analysis
asked Nov 27 at 0:26
LexyFidds
226
add a comment |
Can anyone explain how adding two vertices in a connected graph to create a $0$-Chain is the boundary of some 1-dimensional chain? I know that the definition of boundary is the collection of $n+1$ faces, and each face is an $n-1$ simplice, but can someone explain the connection?
topological-data-analysis
asked Nov 27 at 0:26
LexyFidds
226
Can anyone explain how adding two vertices in a connected graph to create a $0$-Chain is the boundary of some 1-dimensional chain? I know that the definition of boundary is the collection of $n+1$ faces, and each face is an $n-1$ simplice, but can someone explain the connection?
topological-data-analysis
topological-data-analysis
asked Nov 27 at 0:26
LexyFidds
226
asked Nov 27 at 0:26
LexyFidds
226
asked Nov 27 at 0:26
LexyFidds
226
asked Nov 27 at 0:26
LexyFidds
226
asked Nov 27 at 0:26
LexyFidds
226
226
add a comment |
add a comment |
1 Answer
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Yes. Suppose the two vertices are $u$ and $w$. Since the graph is connected, then there is a path
$$u=v_0,v_1,v_2,ldots,v_{k-2},v_{k-1},v_k=w$$
of vertices in the graph, connecting vertex $u$ to $w$, such that an edge exists in the graph between $v_i$ and $v_{i+1}$ for all $i$. Hence,
$$[v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k]$$
is a 1-chain. Its boundary, with $mathbb{Z}/2mathbb{Z}$ coefficients, is
begin{align}
&partial([v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k])\
=&partial[v_0,v_1]+partial[v_1,v_2]+ldots+partial[v_{k-2},v_{k-1}]+partial[v_{k-1},v_k]\
=&(v_0+v_1)+(v_1+v_2)+ldots+(v_{k-2}+v_{k-1})+(v_{k-1}+v_k)\
=&v_0+2v_1+2v_1+ldots+2v_{k-2}+2v_{k-1}+v_k\
=&v_0+v_k
end{align}
since we are using $mathbb{Z}/2mathbb{Z}$ coefficients. So we have found a 1-chain whose boundary is $v_0+v_k=u+w$.
answered Nov 29 at 14:06
Henry Adams
1164
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Yes. Suppose the two vertices are $u$ and $w$. Since the graph is connected, then there is a path
$$u=v_0,v_1,v_2,ldots,v_{k-2},v_{k-1},v_k=w$$
of vertices in the graph, connecting vertex $u$ to $w$, such that an edge exists in the graph between $v_i$ and $v_{i+1}$ for all $i$. Hence,
$$[v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k]$$
is a 1-chain. Its boundary, with $mathbb{Z}/2mathbb{Z}$ coefficients, is
begin{align}
&partial([v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k])\
=&partial[v_0,v_1]+partial[v_1,v_2]+ldots+partial[v_{k-2},v_{k-1}]+partial[v_{k-1},v_k]\
=&(v_0+v_1)+(v_1+v_2)+ldots+(v_{k-2}+v_{k-1})+(v_{k-1}+v_k)\
=&v_0+2v_1+2v_1+ldots+2v_{k-2}+2v_{k-1}+v_k\
=&v_0+v_k
end{align}
since we are using $mathbb{Z}/2mathbb{Z}$ coefficients. So we have found a 1-chain whose boundary is $v_0+v_k=u+w$.
answered Nov 29 at 14:06
Henry Adams
1164
add a comment |
Yes. Suppose the two vertices are $u$ and $w$. Since the graph is connected, then there is a path
$$u=v_0,v_1,v_2,ldots,v_{k-2},v_{k-1},v_k=w$$
of vertices in the graph, connecting vertex $u$ to $w$, such that an edge exists in the graph between $v_i$ and $v_{i+1}$ for all $i$. Hence,
$$[v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k]$$
is a 1-chain. Its boundary, with $mathbb{Z}/2mathbb{Z}$ coefficients, is
begin{align}
&partial([v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k])\
=&partial[v_0,v_1]+partial[v_1,v_2]+ldots+partial[v_{k-2},v_{k-1}]+partial[v_{k-1},v_k]\
=&(v_0+v_1)+(v_1+v_2)+ldots+(v_{k-2}+v_{k-1})+(v_{k-1}+v_k)\
=&v_0+2v_1+2v_1+ldots+2v_{k-2}+2v_{k-1}+v_k\
=&v_0+v_k
end{align}
since we are using $mathbb{Z}/2mathbb{Z}$ coefficients. So we have found a 1-chain whose boundary is $v_0+v_k=u+w$.
answered Nov 29 at 14:06
Henry Adams
1164
add a comment |
Yes. Suppose the two vertices are $u$ and $w$. Since the graph is connected, then there is a path
$$u=v_0,v_1,v_2,ldots,v_{k-2},v_{k-1},v_k=w$$
of vertices in the graph, connecting vertex $u$ to $w$, such that an edge exists in the graph between $v_i$ and $v_{i+1}$ for all $i$. Hence,
$$[v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k]$$
is a 1-chain. Its boundary, with $mathbb{Z}/2mathbb{Z}$ coefficients, is
begin{align}
&partial([v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k])\
=&partial[v_0,v_1]+partial[v_1,v_2]+ldots+partial[v_{k-2},v_{k-1}]+partial[v_{k-1},v_k]\
=&(v_0+v_1)+(v_1+v_2)+ldots+(v_{k-2}+v_{k-1})+(v_{k-1}+v_k)\
=&v_0+2v_1+2v_1+ldots+2v_{k-2}+2v_{k-1}+v_k\
=&v_0+v_k
end{align}
since we are using $mathbb{Z}/2mathbb{Z}$ coefficients. So we have found a 1-chain whose boundary is $v_0+v_k=u+w$.
answered Nov 29 at 14:06
Henry Adams
1164
Yes. Suppose the two vertices are $u$ and $w$. Since the graph is connected, then there is a path
$$u=v_0,v_1,v_2,ldots,v_{k-2},v_{k-1},v_k=w$$
of vertices in the graph, connecting vertex $u$ to $w$, such that an edge exists in the graph between $v_i$ and $v_{i+1}$ for all $i$. Hence,
$$[v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k]$$
is a 1-chain. Its boundary, with $mathbb{Z}/2mathbb{Z}$ coefficients, is
begin{align}
&partial([v_0,v_1]+[v_1,v_2]+ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k])\
=&partial[v_0,v_1]+partial[v_1,v_2]+ldots+partial[v_{k-2},v_{k-1}]+partial[v_{k-1},v_k]\
=&(v_0+v_1)+(v_1+v_2)+ldots+(v_{k-2}+v_{k-1})+(v_{k-1}+v_k)\
=&v_0+2v_1+2v_1+ldots+2v_{k-2}+2v_{k-1}+v_k\
=&v_0+v_k
end{align}
since we are using $mathbb{Z}/2mathbb{Z}$ coefficients. So we have found a 1-chain whose boundary is $v_0+v_k=u+w$.
answered Nov 29 at 14:06
Henry Adams
1164
answered Nov 29 at 14:06
Henry Adams
1164
answered Nov 29 at 14:06
Henry Adams
1164
answered Nov 29 at 14:06
Henry Adams
1164
1164
add a comment |
add a comment |
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