Singular $n$-simplex, unknown notation, homology











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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 $s_i$ ?










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    up vote
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    down vote

    favorite












    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 $s_i$ ?










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      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 $s_i$ ?










      share|cite|improve this question















      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






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      edited Nov 22 at 10:42









      user302797

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      asked Nov 21 at 12:50









      user122424

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          $Delta^n$ 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 $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.






          share|cite|improve this answer




























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            0
            down vote













            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.






            share|cite|improve this answer





















              Your Answer





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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

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              up vote
              1
              down vote



              accepted










              $Delta^n$ 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 $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted










                $Delta^n$ 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 $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.






                share|cite|improve this answer























                  up vote
                  1
                  down vote



                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  $Delta^n$ 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 $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.






                  share|cite|improve this answer












                  $Delta^n$ 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 $e_0,dots,e_{k-1},e_{k+1},dots,e_n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 14:02









                  Paul Frost

                  8,3921528




                  8,3921528






















                      up vote
                      0
                      down vote













                      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.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        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.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          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.






                          share|cite|improve this answer












                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 21 at 13:38









                          Fumera

                          235




                          235






























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