Singular $n$-simplex, unknown notation, homology
I would like to know what is denoted in the singular simplices paragraph by $e_0,...,e_n$ here:
$$[p_0,p_1,...,p_n]=[sigma(e_0),...,sigma(e_n)]$$ ?
How this matches with simplicial identities $d_i$ and $s_i$ ?
continuity simplex
edited Nov 22 '18 at 10:42
Saad
19.7k92252
asked Nov 21 '18 at 12:50
user122424
1,0881616
add a comment |
I would like to know what is denoted in the singular simplices paragraph by $e_0,...,e_n$ here:
$$[p_0,p_1,...,p_n]=[sigma(e_0),...,sigma(e_n)]$$ ?
How this matches with simplicial identities $d_i$ and $s_i$ ?
continuity simplex
edited Nov 22 '18 at 10:42
Saad
19.7k92252
asked Nov 21 '18 at 12:50
user122424
1,0881616
add a comment |
I would like to know what is denoted in the singular simplices paragraph by $e_0,...,e_n$ here:
$$[p_0,p_1,...,p_n]=[sigma(e_0),...,sigma(e_n)]$$ ?
How this matches with simplicial identities $d_i$ and $s_i$ ?
continuity simplex
edited Nov 22 '18 at 10:42
Saad
19.7k92252
asked Nov 21 '18 at 12:50
user122424
1,0881616
I would like to know what is denoted in the singular simplices paragraph by $e_0,...,e_n$ here:
$$[p_0,p_1,...,p_n]=[sigma(e_0),...,sigma(e_n)]$$ ?
How this matches with simplicial identities $d_i$ and $s_i$ ?
continuity simplex
continuity simplex
edited Nov 22 '18 at 10:42
Saad
19.7k92252
asked Nov 21 '18 at 12:50
user122424
1,0881616
edited Nov 22 '18 at 10:42
Saad
19.7k92252
asked Nov 21 '18 at 12:50
user122424
1,0881616
edited Nov 22 '18 at 10:42
Saad
19.7k92252
edited Nov 22 '18 at 10:42
Saad
19.7k92252
edited Nov 22 '18 at 10:42
Saad
19.7k92252
19.7k92252
asked Nov 21 '18 at 12:50
user122424
1,0881616
asked Nov 21 '18 at 12:50
user122424
1,0881616
asked Nov 21 '18 at 12:50
user122424
1,0881616
1,0881616
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$Delta^n$ is the standard $n$-simplex in $mathbb{R}^{n+1}$, i.e. the convex hull of the $n+1$ elements $e_0,dots,e_n$ forming the standard basis of $mathbb{R}^{n+1}$ ($e_k = (delta_{k0},dots,delta_{kn})$ with $delta_{kk} = 1$ and $delta_{kj} = 0$ for $k ne j$).
If $sigma : Delta^n to X$ is a singular $n$-simplex, then the $p_i = sigma(e_i)$ are (not necessarily distint) points of $X$. Hence the collection $[p_0,dots,p_n]$ is not a complete description of $sigma$, but only a symbolic notation. Defining
$$partial sigma = sum_{k=0}^n (-1)^k [p_0,dots,p_{k-1},p_{k+1},dots,p_n]$$
is therefore also only symbolic. In fact, $[p_0,dots,p_{k-1},p_{k+1},dots,p_n]$ denotes the $(n-1)$-simplex $partial_k sigma : Delta^{n-1} to X$ (the $k$-face of $sigma$) defined by
$$partial_k sigma = sigma circ s_k ,$$
where $s_k : Delta^{n-1} to Delta^n$ is the restriction of the linear map determined by $s_k(e_j) = e_j$ for $j < k$, $s_k(e_j) = e_{j+1}$ for $j ge k$. This map embeds $Delta^{n-1}$ as the $k$-th face of $Delta^n$, i.e. the convex hull of $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
add a comment |
The point $e_i$ is just
$$
e_i=begin{pmatrix}
0\
vdots\
0\
1\
0\
vdots\
0
end{pmatrix}inmathbb R^{n+1}
$$
with the entry $1$ in the $i$-th row. These points are exactly the vertices of the $n$-simplex $Delta_n$.
Given this, you can compute that the $d_i$ and $s_i$ satisfy indeed the simplicial identities.
answered Nov 21 '18 at 13:38
Fumera
235
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$Delta^n$ is the standard $n$-simplex in $mathbb{R}^{n+1}$, i.e. the convex hull of the $n+1$ elements $e_0,dots,e_n$ forming the standard basis of $mathbb{R}^{n+1}$ ($e_k = (delta_{k0},dots,delta_{kn})$ with $delta_{kk} = 1$ and $delta_{kj} = 0$ for $k ne j$).
If $sigma : Delta^n to X$ is a singular $n$-simplex, then the $p_i = sigma(e_i)$ are (not necessarily distint) points of $X$. Hence the collection $[p_0,dots,p_n]$ is not a complete description of $sigma$, but only a symbolic notation. Defining
$$partial sigma = sum_{k=0}^n (-1)^k [p_0,dots,p_{k-1},p_{k+1},dots,p_n]$$
is therefore also only symbolic. In fact, $[p_0,dots,p_{k-1},p_{k+1},dots,p_n]$ denotes the $(n-1)$-simplex $partial_k sigma : Delta^{n-1} to X$ (the $k$-face of $sigma$) defined by
$$partial_k sigma = sigma circ s_k ,$$
where $s_k : Delta^{n-1} to Delta^n$ is the restriction of the linear map determined by $s_k(e_j) = e_j$ for $j < k$, $s_k(e_j) = e_{j+1}$ for $j ge k$. This map embeds $Delta^{n-1}$ as the $k$-th face of $Delta^n$, i.e. the convex hull of $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
add a comment |
$Delta^n$ is the standard $n$-simplex in $mathbb{R}^{n+1}$, i.e. the convex hull of the $n+1$ elements $e_0,dots,e_n$ forming the standard basis of $mathbb{R}^{n+1}$ ($e_k = (delta_{k0},dots,delta_{kn})$ with $delta_{kk} = 1$ and $delta_{kj} = 0$ for $k ne j$).
If $sigma : Delta^n to X$ is a singular $n$-simplex, then the $p_i = sigma(e_i)$ are (not necessarily distint) points of $X$. Hence the collection $[p_0,dots,p_n]$ is not a complete description of $sigma$, but only a symbolic notation. Defining
$$partial sigma = sum_{k=0}^n (-1)^k [p_0,dots,p_{k-1},p_{k+1},dots,p_n]$$
is therefore also only symbolic. In fact, $[p_0,dots,p_{k-1},p_{k+1},dots,p_n]$ denotes the $(n-1)$-simplex $partial_k sigma : Delta^{n-1} to X$ (the $k$-face of $sigma$) defined by
$$partial_k sigma = sigma circ s_k ,$$
where $s_k : Delta^{n-1} to Delta^n$ is the restriction of the linear map determined by $s_k(e_j) = e_j$ for $j < k$, $s_k(e_j) = e_{j+1}$ for $j ge k$. This map embeds $Delta^{n-1}$ as the $k$-th face of $Delta^n$, i.e. the convex hull of $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
add a comment |
$Delta^n$ is the standard $n$-simplex in $mathbb{R}^{n+1}$, i.e. the convex hull of the $n+1$ elements $e_0,dots,e_n$ forming the standard basis of $mathbb{R}^{n+1}$ ($e_k = (delta_{k0},dots,delta_{kn})$ with $delta_{kk} = 1$ and $delta_{kj} = 0$ for $k ne j$).
If $sigma : Delta^n to X$ is a singular $n$-simplex, then the $p_i = sigma(e_i)$ are (not necessarily distint) points of $X$. Hence the collection $[p_0,dots,p_n]$ is not a complete description of $sigma$, but only a symbolic notation. Defining
$$partial sigma = sum_{k=0}^n (-1)^k [p_0,dots,p_{k-1},p_{k+1},dots,p_n]$$
is therefore also only symbolic. In fact, $[p_0,dots,p_{k-1},p_{k+1},dots,p_n]$ denotes the $(n-1)$-simplex $partial_k sigma : Delta^{n-1} to X$ (the $k$-face of $sigma$) defined by
$$partial_k sigma = sigma circ s_k ,$$
where $s_k : Delta^{n-1} to Delta^n$ is the restriction of the linear map determined by $s_k(e_j) = e_j$ for $j < k$, $s_k(e_j) = e_{j+1}$ for $j ge k$. This map embeds $Delta^{n-1}$ as the $k$-th face of $Delta^n$, i.e. the convex hull of $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
$Delta^n$ is the standard $n$-simplex in $mathbb{R}^{n+1}$, i.e. the convex hull of the $n+1$ elements $e_0,dots,e_n$ forming the standard basis of $mathbb{R}^{n+1}$ ($e_k = (delta_{k0},dots,delta_{kn})$ with $delta_{kk} = 1$ and $delta_{kj} = 0$ for $k ne j$).
If $sigma : Delta^n to X$ is a singular $n$-simplex, then the $p_i = sigma(e_i)$ are (not necessarily distint) points of $X$. Hence the collection $[p_0,dots,p_n]$ is not a complete description of $sigma$, but only a symbolic notation. Defining
$$partial sigma = sum_{k=0}^n (-1)^k [p_0,dots,p_{k-1},p_{k+1},dots,p_n]$$
is therefore also only symbolic. In fact, $[p_0,dots,p_{k-1},p_{k+1},dots,p_n]$ denotes the $(n-1)$-simplex $partial_k sigma : Delta^{n-1} to X$ (the $k$-face of $sigma$) defined by
$$partial_k sigma = sigma circ s_k ,$$
where $s_k : Delta^{n-1} to Delta^n$ is the restriction of the linear map determined by $s_k(e_j) = e_j$ for $j < k$, $s_k(e_j) = e_{j+1}$ for $j ge k$. This map embeds $Delta^{n-1}$ as the $k$-th face of $Delta^n$, i.e. the convex hull of $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
answered Nov 21 '18 at 14:02
Paul Frost
9,1961631
9,1961631
add a comment |
add a comment |
The point $e_i$ is just
$$
e_i=begin{pmatrix}
0\
vdots\
0\
1\
0\
vdots\
0
end{pmatrix}inmathbb R^{n+1}
$$
with the entry $1$ in the $i$-th row. These points are exactly the vertices of the $n$-simplex $Delta_n$.
Given this, you can compute that the $d_i$ and $s_i$ satisfy indeed the simplicial identities.
answered Nov 21 '18 at 13:38
Fumera
235
add a comment |
The point $e_i$ is just
$$
e_i=begin{pmatrix}
0\
vdots\
0\
1\
0\
vdots\
0
end{pmatrix}inmathbb R^{n+1}
$$
with the entry $1$ in the $i$-th row. These points are exactly the vertices of the $n$-simplex $Delta_n$.
Given this, you can compute that the $d_i$ and $s_i$ satisfy indeed the simplicial identities.
answered Nov 21 '18 at 13:38
Fumera
235
add a comment |
The point $e_i$ is just
$$
e_i=begin{pmatrix}
0\
vdots\
0\
1\
0\
vdots\
0
end{pmatrix}inmathbb R^{n+1}
$$
with the entry $1$ in the $i$-th row. These points are exactly the vertices of the $n$-simplex $Delta_n$.
Given this, you can compute that the $d_i$ and $s_i$ satisfy indeed the simplicial identities.
answered Nov 21 '18 at 13:38
Fumera
235
The point $e_i$ is just
$$
e_i=begin{pmatrix}
0\
vdots\
0\
1\
0\
vdots\
0
end{pmatrix}inmathbb R^{n+1}
$$
with the entry $1$ in the $i$-th row. These points are exactly the vertices of the $n$-simplex $Delta_n$.
Given this, you can compute that the $d_i$ and $s_i$ satisfy indeed the simplicial identities.
answered Nov 21 '18 at 13:38
Fumera
235
answered Nov 21 '18 at 13:38
Fumera
235
answered Nov 21 '18 at 13:38
Fumera
235
answered Nov 21 '18 at 13:38
Fumera
235
235
add a comment |
add a comment |
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