Joins and Products of polyhedra · DL Ferrario's Test Web Page

# Joins and Products of polyhedra

## Join $K*L$ of two simplicial complexes

Let $K = (X_ K, \Phi_ K)$ and $L = (X_ L, \Phi_ L)$ be two simplicial complexes ($X_ K$ denotes the set of vertices of $K$ and $\Phi_ K$ the set of simplices). Then the join $K*L$ is the simplicial complex defined as follows. Its set of vertices $X=X_ K \cup X_ L$ is the union of the sets of vertices of $K$ and of $L$, and a non-empty subset of $X$ is a simplex of $K*L$ if and only if it is of type $\sigma \cup \tau$, where $\sigma \in \Phi_ K \cup \{ \emptyset \}$ and $\tau \in \Phi_ L \cup \{\emptyset \}$.

Now, assume that $K$ and $L$ are embedded simplicially in $\RR^k$ and $\RR^l$ respectively, so that $|K|\subset \RR^k$ and $|L|\subset \RR^l$ as in the standard geometric realization functor. The vertices of $K$ are points of a basis of $\RR^k$ and the vertices of $L$ are points of a basis of $\RR^l$. For each $n$-simplex $\sigma$ in $|K|$, denote with $\pi_ \sigma \subset \RR^k$ the affine subspace (of dimension $n$) generated by $\sigma$. For every simplex of $K$, $0\not\in \pi_ \sigma \subset \RR^k$ ; the same for $K$.

Now, consider the following euclidean simplicial complex $J$: its vertices are the pairs $(x,0)$ and $(0,y)$ $\in \RR^k\times \RR^l \cong \RR^{k+l}$, with $x$ vertex of $|K|\subset \RR^k$ and $y$ vertex of $|L|\subset \RR^l$. Simplices, the corresponding simplices of the join $K*L$, i.e. convex closures of unions $\sigma\cup \tau$ as above.

(1)
If $P_ 0, \ldots, P_ n \in \RR^k$ are $n+1$ (affinely) independent points in $\RR^k$, and $Q_ 0,\ldots, Q_ m$ are $m+1$ (affinely) independent points in $\RR^l$, then $(P_ 0,0)$, $\ldots$, $(P_ n,0)$, $(0,Q_ 0)$, $\ldots$ , $(0,Q_ m)$ are $n+m+2$ (affinely) independent points in $\RR^{k+l}$.
(2)
The euclidean simplicial complex $J$ is homeomorphic to the geometric realization of the join $|K*L|$.

Proof: A point $P$ in the interior of a simplex of $J$ is the affine combination $$P = t_ 0 (P_ 0,0) + t_ n (P_ n, 0) + s_ 0(0,Q_ 0) + \ldots + s_ m(0, Q_ m)$$ with $t_ 0 + \ldots + t_ n + s_ 0 +\ldots + s_ m = 1$, and $s_ j > 0$, $t_ j > 0$, where $\{P_ 0, \ldots, P_ n\}$ is an $n$-simplex of $K$ (or empty) and $\{ Q_ 0, \ldots, Q_ m \}$ is an $m$-simplex of $L$ (or empty).

Let $p = t_ 0 + \ldots + t_ n$ and $q = s_ 0 + \ldots + s_ m$. Since $p>0$, and $q>0$ (all coefficients are strictly positive), $$x= \dfrac{1}{p} \left( t_ 0 P_ 0 + \ldots + t_ n P_ n \right)$$ is in the interior of the $n$-simplex
$[P_ 0, \ldots, P_ n]$, and $$y= \dfrac{1}{q} \left( s_ 0 Q_ 0 + \ldots + s_ m Q_ m \right)$$ is in the interior of the $m$-simplex $[Q_ 0, \ldots, Q_ m]$, Hence $$P = p x + qy,$$ with $p>0$, $q>0$ and $p+q=1$ for two (uniquely defined) points $x$ and $y$ in the interiors of two simplices of $K$ and $L$ respectively. Hence $J$ is an euclidean (geometric) simplicial complex, isomorphic as simplicial complex to $K*L$. /qed/

Remark: it follows from $P=px+qy$ that $J$ is the union of all unit segments joining points of $|K|$ to point of $|L|$ (from here the name "join").

## Cartesian product $K\times L$ of two simplicial complexes

Consider now $|K| \subset \RR^k$ and $|L|\subset \RR^l$ euclidean simplicial complexes. Then, the product $|K|\times |L| \subset \RR^k \times \RR^l = \RR^{k+l}$ can be easily defined. But, what are the simplices in $K\times L$? Asssume as usual that there is a total (or, partial but total on simplices) order on the sets of vertices $X_ K$, and the same for $X_ L$.

Let $X=X_ {K\times L}$ be the set pairs $(x,y)$, with $x\in X_ K$ and $y\in X_ L$, i.e. $$X = X_ K \times X_ L .$$ Let $p_ 1 \from X \to X_ K$ and $p_ 2\from X \to X_ L$ the projections on the first and second factor, respectively.

Now, consider the following order relation in $X = X_ X \times X_ L$: $(x,y) \leq (x',y')$ iff $x \leq x' \& y \leq y'$ (it is the product order, which is not a total order).

Or, consider the lexycographical order, which is a total order, defined as $(x,y) \leq (x',y')$ iff $x<x'$ or $x=x'$ and $y\leq y'$.

With this, define an $n$-simplex $\sigma \subset X$ as a subset $$\sigma = \{ (x_ 0,y_ 0), (x_ 1,y_ 1), \ldots, (x_ n,y_ n) \} \subset X_ K \times X_ L$$ such that:

1. $(x_ 0,y_ 0) < (x_ 1, y_ 1) < \ldots (x_ n, y_ n)$ is an ascending chain;
2. $p_ 1(\sigma) \subset X_ K$ is a simplex of $K$ (of dimension $\leq n$), and $x_ 0\leq x_ 1 \ldots \leq x_ n$ (some inequalities might be equalities);
3. $p_ 2(\sigma)\subset X_ L$ is a simplex of $L$ (of dimension $\leq n$), and $y_ 0 \leq y_ 1 \ldots \leq y_ n$ (sime inequalities might be equalities).

Now, both choices of order (the lexycographical or the product order) on $X_ K \times X_ L$ yield the same set of simplices. In fact, if $(x_ 0,y_ 0) < (x_ 1, y_ 1) < \ldots (x_ n, y_ n)$ is an ascending chain in the lexycographical order, then 2. and 3. imply that it is also an ascending chain in the product order. On the other hand, if $(x_ 0,y_ 0) < (x_ 1, y_ 1) < \ldots (x_ n, y_ n)$ is an ascending chain in the product order, for each pair of consecutive terms $(x,y) < (x',y')$ either $x=x'$ (and therefore $y<y'$, and hence (x,y) precedes $(x',y')$ in the lexycographical order), or $x<x'$ (and hence $(x,y)$ precedes (x',y')$in the lexycographical order). In Ferrario-Piccinini1, page 99, in the proof it is omitted to mention the simplices$p_ 1$and$p_ 2$need to be ordered, as above. Otherwise, more chains would appear ordered as ascending with the lexycographical order, than those ascending with the product order. Example: Let$\Delta^1 \times \Delta^1$be the product of the standard$1$-simplex$\Delta^1$(with vertices$A,B$) and the standard$1$-simplex$\Delta^1$(with vertices$0,1). Vertices are pairs $$A=(A,0), B=(B,0), A'=(A,1), B'=(B,1).$$ Ascending chains are the following two: \begin{aligned} A < B < B' \\ A < A' < B' \end{aligned} Example: Let\Delta^2 \times \Delta^1$be the product of the standard$2$-simplex$\Delta^2$(with vertices$A,B,C$and the standard$1$-simplex$\Delta^1$(with vertices$0,1). Vertices are pairs $$A=(A,0), B=(B,0), C=(C,0), A'=(A,1), B'=(B,1), C'(C,1).$$ Ascending chains are the following three: \begin{aligned} A < B < C < C' \\ A < B < B' < C' \\ A < A' < B' < C' \end{aligned} Example: As above, the ascending chains of\Delta^3 \times \Delta^1$, with vertices$A,B,C,D,A',B',C',D'are the following: \begin{aligned} A < B < C < D < D' \\ A < B < C < C' < D' \\ A < B < B' < C' < D' \\ A < A' < B' < C' < D' \end{aligned} Consider any geometric realizations|K|\subset \RR^k$and$|L|\subset \RR^l$. (3) The cartesian product$|K|\times |L| \subset \RR^k \times \RR^l$is an euclidean simplicial complex, with as simplices those of$K\times L$. Proof: It suffices to show that axiom S2, page 44 of Ferrario-Piccinini, holds. More precisely, that if$s_ 1$and$s_ 2$are two simplices with non-disjoint interiors$\mathring{s}_ 1 \cap \mathring{s}_ 2 \neq \emptyset\impliess_ 1 = s_ 2$, and that all simplices have vertices which are (affinely) independent in$\RR^k \times \RR^l$. To this end, take$(p,q) \in |K|\times |L|$. There are two unique simplices$\sigma = [P_ 0,\ldots, P_ n]$in$K$and$\tau=[Q_ 0 ,\ldots, Q_ m]$in$L$such that$p \in \mathring{\sigma}$and$q\in \mathring{\tau}$, namely$p = x_ 0 P_ 0+ \ldots + x_ n P_ n$with$x_ j >0$and$x_ 0 + \ldots + x_ n = 1$, and$q = y_ 0 Q_ 0+ \ldots + y_ m Q_ m$with$y_ i >0$and$y_ 0 + \ldots + y_ m = 1$. Now, consider the embeddings$\iota\from \sigma \approx \Delta^n \to \RR^n$and$\iota' \from \tau \approx \Delta^m \to \RR^m$, defined as $$X_ \alpha = \sum_ {j=0}^\alpha x_ j \implies x_ j = X_ j - X_ {j-1},$$ $$Y_ \beta = \sum_ {i=0}^\beta y_ i \implies y_ i = Y_ i - Y_ {i-1}$$ with$\alpha = 0 \ldots n-1$and$\beta=0\ldots m-1$. Note that$\iota \sigma $is the set of all elements$(X_ 0,\ldots, X_ {n-1})$of$\RR^n$such that $$0 \leq X_ 0 \leq X_ 1 \leq \ldots \leq X_ n = 1,$$ and the same for$\iota' \tau $as$0\leq Y_ 0 \leq \ldots \leq Y_ m = 1$. By definition, moreover, in the interior of$\sigma$and$\tau$the inequality $$0 < X_ 0 < X_ 1 < \ldots < X_ {n-1} < 1$$ $$0< Y_ 0 < Y_ 1 \ldots < Y_ {m-1} < 1$$ hold. The images of the vertices$\iota P_ j$have coordinates$(0,\ldots, 0, 1, \ldots, 1)$, made with$j$zeroes and$n-j$ones. The same,$\iota' Q_ i$has as coordinates a list of$i$zeroes and$m-i$ones. Consider the union of all the values of the$X$and$Y$coordinates, together with$0$and$1$, i.e. $$S = \{ 0, X_ 0, X_ 1, \ldots, X_ {n-1}, Y_ 0, Y_ 1, \ldots, Y_ {m-1}, 1 \} \subset [0,1] \subset \RR.$$ This is a set of$\# S$distinct numbers, with$\# S \leq n+m+2$(since it is not a priori required for them to be distinct), and hence$S = \{ 0, \xi_ 0, \xi_ 1, \ldots, \xi_ {s} = 1 \}$with $$0 < \xi_ 0 < \xi_ 1 < \ldots < \xi_ {s-1} < 1=\xi_ {s}.$$ Now, for each index$r=0,\ldots, s$, consider the sets $$S_ X(r) = \{ \alpha : X_ \alpha < \xi_ r \}$$ (the set of all indices$\alpha=0\ldots n-1$such that$X_ \alpha < \xi_ r$) and $$S_ Y(r) = \{ \beta : Y_ \beta < \xi_ r \}$$ (the set of all indices$\beta =0\ldots m-1$such that$Y_ \beta < \xi_ r$). What happens is that if$r=0$, then for all$\alpha$and$\beta$one has$\xi_ 0\leq X_ \alpha$and$\xi_ 0\leq Y_ \beta$, and hence $$S_ X(0) = S_ Y(0) = \emptyset.$$ On the other hand, if$r=s$, then$\xi_ s = 1$and $$S_ X(s) = \{0,\ldots, n-1 \}$$ and $$S_ Y(s) = \{0,\ldots, m-1 \}.$$ Ranging over$r=0,\ldots, s$, they yield ascending chains of subsets $$\emptyset =S_ X(0) \subset S_ X(1) \subset \ldots \subset S_ X(s) = \{0,\ldots, n-1 \}$$ $$\emptyset =S_ Y(0) \subset S_ Y(1) \subset \ldots \subset S_ Y(s) = \{0,\ldots, m-1 \}$$ where the next subset either coincides with the preceding or has one additional element. Finally, let$j_ r$denote the number of elements of$S_ X(r)$, and let$i_ r$denote the number of elements of$S_ Y(r)$. As above, for$r=0$one has$j_ 0 = i_ 0 = 0$, and for$r=s$one has$j_ {s}= n$,$i_ {s}=m$. Furthermore, $$(j_ r, i_ r) < (j_ {r+1}, i_ {r+1} )$$ in the product order, since by definition $$j_ 0 = 0 \leq j_ 1 \leq \ldots \leq j_ {s} = n$$ $$i_ 0 = 0 \leq i_ 1 \leq \ldots \leq i_ {s} = m$$ and it cannot be that both$j_ r = j_ {r+1}$and$i_ r = i_ {r+1}$hold. The sets equalities$\{j_ 0, j_ 1, \ldots, j_ s\} = \{0,1,\ldots, n-1\}$,$\{i_ 0,i_ 1, \ldots, i_ s \} = \{0,1,\ldots, m-1 \}$hold, since when$j_ r \neq j_ {r+1}$, then$j_ r+1 = j_ {r+1}$, and the same for$i_ r$and$i_ {r+1}$. Furthermore, for each$\alpha = 0 , \ldots, n$let$J_ \alpha$be defined as $$J_ \alpha = \max \{ r = 0 \ldots s : j_ r = \alpha \}$$ and analogously for each$\beta =0, \ldots, m$let$I_ \beta$be defined as $$I_ \beta = \max \{ r=0\ldots s : i_ r = \beta \}.$$ Hence for each$\alpha = 1, \ldots, n$, $$\{ r : j_ r = \alpha \} = \{ r : J_ {\alpha -1} < r \leq J_ \alpha \}$$ and $$\{ r : j_ r = 0 \} = \{ r : 0\leq r \leq J_ 0 \}$$ (where we set$J_ {-1} = -1$). Also, if$r=J_ {\hat \alpha}$, then$j_ r = \hat \alpha$and$j_ {r+1} = \hat\alpha +1$, and hence by the fact that the number of indices$\alpha$such that$X_ \alpha < \xi_ r$is equal to$\hat\alpha$and the number of indices$\alpha$such that$X_ \alpha < \xi_ {r+1}$is equal to$\hat \alpha + 1$, it follows that$\xi_ {r} = X_ {\hat\alpha}$, or equivalently that $$\xi_ {J_ \alpha} = X_ \alpha$$ Similar equalities hold for$I_ \beta$: $$\xi_ {I_ \beta} = Y _ \beta.$$ This choice determines a unique ascending chain of vertices $$(P_ 0,Q_ 0) = (P_ {j_ 0}, Q_ {i_ 0} ) < (P_ {j_ 1}, Q_ {i_ 1}) < \ldots (P_ {j_ {s}}, Q_ {i_ s}),$$ with$\{P_ {j_ 0}, P_ {j_ 1}, \ldots, P_ {j_ s} \} = \{ P_ 0, P_ 1, \ldots P_ n \}$and$\{Q_ {i_ 0}, Q_ {i_ 1}, \ldots, Q_ {i_ s} \}  = \{ Q_ 0, Q_ 1, \ldots Q_ m \}$, and therefore$ \{ (P_ {j_ 0}, Q_ {i_ 0} ),
(P_ {j_ 1}, Q_ {i_ 1}) ,
\ldots (P_ {j_ {s}}, Q_ {i_ s}) \} $is an$s$-simplex in the cartesian product$K\times L$. Now, actually$(p,q)$is in the interior of such$s$-simplex. To prove it, for$r=1\ldots s$let$\lambda_ r$be defined as$\lambda_ r = \xi_ r - \xi_ {r-1}$, and$\lambda_ 0 = \xi_ 0$. For each$r=0\ldots, s$,$\lambda_ r >0$, and $$\sum_ {r=0}^s \lambda_ r = \xi_ 0 + (\xi_ 1 - \xi_ 0) + \ldots + (\xi_ s - \xi_ {s-1}) = \xi_ s = 1.$$ Or, set$\xi_ {-1} =0$and$\lambda_ 0 = \xi_ 0 - \xi_ {-1}. Now, consider the sum $$\sum_ {J_ {\alpha-1} < r \leq J_ \alpha} \lambda_ r = \sum_ {J_ {\alpha-1} < r \leq J_ \alpha} (\xi_ r - \xi_ {r-1}) = \xi_ {J_ \alpha} - \xi_ {J_ {\alpha-1}} = X_ \alpha - X_ {\alpha -1} = x_ \alpha~,$$ and the sum $$\sum_ {I_ {\beta-1} < r \leq J_ \beta} \lambda_ r = \sum_ {I_ {\beta-1} < r \leq I_ \beta} (\xi_ r - \xi_ {r-1}) = \xi_ {I_ \beta } - \xi_ {I_ {\beta-1}} = Y_ \beta - Y_ {\beta-1} = y_ \beta~.$$ Therefore \begin{aligned} \sum_ {r=0} ^s \lambda_ r (P_ {j_ r}, Q_ {i_ r}) & = \left( \sum_ {r=0} ^s \lambda_ r P_ {j_ r}, \sum_ {r=0} ^s \lambda_ r Q_ {i_ r} \right) \\ & = \left( \sum_ {\alpha=0}^n \left( \sum_ {J_ {\alpha-1} < r \leq J_ \alpha} \lambda_ r \right) P_ \alpha , \sum_ {\beta =0}^m \left( \sum_ {I_ {\beta -1} < r \leq I_ \beta} \lambda_ r \right) Q_ \beta \right) \\ &= \left( \sum_ {j=0}^n x_ j P_ j, \sum_ {i=0}^m y_ j Q_ j \right) = (p,q)~. \end{aligned} /qed/ ## Chain homotopy and cylinders LetK$be a simplicial complex,$I\cong \Delta^1$the standard$1$-simplex, and$K\times I$the product of$K$and$I$, i.e. the cylinder on$K$. Now, let$C_ *(K)$and$C_ *(K\times I)$denote the simplicial chain complexes of$K$and$K\times I$respectively. Finally, let$b\from K \to K \times I$and$t\from K \to K \times I$be the simplicial maps defined by $$b(x) = (x,0),\quad t(x) = (x,1)$$ for each$x\in K$. For each vertex$P\in X$, use the notation$P\equiv b(P)$and$P'=t(P)$, so that$P \equiv (P,0)$and$P'=(P,1)$. Let now$s_ n \from C_ n(K) \to C_ {n+1}(K\times I)be the chain complex morphism defined as \begin{aligned} s_ n([P_ 0,\ldots, P_ n]) & = \sum_ {j=0} ^n (-1)^j [P_ 0,\ldots, P_ j, P'_ j, \ldots, P'_ n] \\ &= [P_ 0,P'_ 0, \ldots, P_ n] - [P_ 0,P_ 1, P'_ 1, \ldots, P'_ n] + \ldots + (-1)^n [P_ 0,\ldots, P_ n, P'_ n]~. \end{aligned} Note that \begin{aligned} \partial_ {n+1} [P_ 0,\ldots, P_ j, P'_ j, \ldots, P'_ n] & = \sum_ {i=0}^{j-1} (-1)^i [P_ 0, \ldots, \hat P_ i, \ldots , P_ j, P'_ j, \ldots, P'_ n] \\\ & + (-1)^j [P_ 0, \ldots, \hat P_ j, P'_ j , \ldots, P'_ n] \\ & - (-1)^j [P_ 0, \ldots, P_ j, \hat P' _ j, \ldots, P'_ n] \\ & - \sum_ {i=j-1}^n (-1)^i [P_ 0, \ldots, P_ j, P'_ j, \ldots, \hat P'_ i, \ldots, P'_ n] \end{aligned} and \begin{aligned} s_ {n-1} [P_ 0, \ldots, \hat P_ i, \ldots, P_ n] &= \sum_ {j=0}^ {i-1} (-1)^i [P_ 0, \ldots, P_ j, P'_ j, \ldots, \hat P'_ i, \ldots, P'_ n] \\\ & - \sum_ {j=i+1} ^ n (-1)^j [P_ 0, \ldots, \hat P_ i, \ldots, P_ j, P'_ j, \ldots, P'_ n] \end{aligned} \begin{aligned} \implies \partial_ {n+1} s_ n [P_ 0,\ldots, P_ n] & = \sum_ {j=0} ^n (-1)^j \partial_ {n+1} [P_ 0,\ldots, P_ j, P'_ j, \ldots, P'_ n] \\ & = \sum_ {j=0} ^n \sum_ {i=0}^{j-1} j (-1)^{i+j} [P_ 0, \ldots, \hat P_ i, \ldots , P_ j, P'_ j, \ldots, P'_ n] \\\ & + \sum_ {j=0}^ n (-1)^{j+j} [P_ 0, \ldots, \hat P_ j, P'_ j , \ldots, P'_ n] \\ & - \sum_ {j=0}^ n (-1)^{j+j} [P_ 0, \ldots, P_ j, \hat P' _ j, \ldots, P'_ n] \\ & - \sum_ {j=0} ^n \sum_ {i=j}^n (-1)^{i+j} [P_ 0, \ldots, P_ j, P'_ j, \ldots, \hat P'_ i, \ldots, P'_ n] \\ s_ {n-1} \partial_ n [P_ 0,\ldots, P_ n] &= \sum_ {i=0}^ {n} \sum_ {j=0}^ {i-1} (-1)^{i+j} [P_ 0, \ldots, P_ j, P'_ j, \ldots, \hat P'_ i, \ldots, P'_ n] \\\ & - \sum_ {i=0}^ n \sum_ {j=i+1} ^ n (-1)^{i+j} [P_ 0, \ldots, \hat P_ i, \ldots, P_ j, P'_ j, \ldots, P'_ n] \end{aligned} and \begin{aligned} \implies ( \partial_ {n+1} s_ n + s_ {n-1} \partial_ n ) [P_ 0,\ldots, P_ n] &= \sum_ {j=0}^ n \left( [P_ 0, \ldots, \hat P_ j, P'_ j , \ldots, P'_ n] - \sum_ {j=0}^ n [P_ 0, \ldots, P_ j, \hat P' _ j, \ldots, P'_ n] \right) \\ &= [P'_ 0, \ldots, P'_ n] - [P_ 0,P'_ 1, \ldots, P'_ n] + [P_ 0,P'_ 1, \ldots, P'_ n] - \ldots - [P_ 0, \ldots, P_ n] \\ &= [P_ 0', \ldots, P'_ n] - [P_ 0, \ldots, P_ n] \end{aligned} and therefore $$\partial s + s \partial = t_ * - b_ *.$$ EDIT 2018-10-28: Trying to be more explicit/formal. Consider the (symbolic) operator\gamma_ j $from$n$-chains of$K$to$n+1$-chains of$K\times I$, defined as $$\gamma_ j [ P_ 0, \ldots, P_ n ] = [ P_ 0, \ldots, P_ j, P'_ j, \ldots, P'_ n ]$$ and the$i$-th face operator $$F_ i [ P_ 0, \ldots, P_ i, \ldots, P_ n ] = [P_ 0, \ldots, \hat P_ i, \ldots, P_ n ].$$ For$j=0, \ldots, n$and$i=0, \ldots, n+1$, $$F_ i \gamma_ j [ P_ 0, \ldots, P_ n ] = F_ i [ P_ 0, \ldots, P_ j, P'_ j, \ldots, P'_ n ]$$ $$F_ 0 \gamma_ 0 [ P_ 0, \ldots, P_ n ] = [ P'_ 0, \ldots, P'_ n ] ,$$ $$F_ j \gamma_ j [ P_ 0, \ldots, P_ n ] = [ P_ 0, \ldots, P_ {j-1} , P'_ j , \ldots, P'_ n] = F_ {j} \gamma_ {j-1} [ P_ 0, \ldots, P_ n ]$$ $$F_ {n+1} \gamma_ n [ P_ 0, \ldots, P_ n ] = [ P_ 0, \ldots, P_ n ] ,$$ $$i < j \implies F_ i \gamma_ j [ P_ 0, \ldots, P_ n ] = [ P_ 0 , \ldots , \hat P_ i, \ldots, P_ j , P'_ j, \ldots, P'_ n ] = \gamma_ {j-1} F_ {i} [ P_ 0, \ldots, P_ n ];$$ $$i > j+1 \implies F_ i \gamma_ j [P_ 0, \ldots, P_ n] = [ P_ 0 , \ldots , P_ j , P'_ j, \ldots, \hat P'_ {i-1} , \ldots, P'_ n ] = \gamma_ {j} F_ {i-1} [ P_ 0, \ldots, P_ n ].$$ Quindi se$x = [P_ 0, \ldots, P_ n]\$ si ha $$\partial_ {n+1} s_ n x = \sum_ {i=0}^{n+1} (-1)^i \sum_ {j=0}^n (-1)^j F_ i \gamma_ j x = \sum_ {i=0}^{n+1} \sum_ {j=0}^n (-1)^{i+j} F_ i \gamma_ j x$$ $$= \sum_ {j=0}^n \left( \sum_ {i=0}^{j-1} (-1)^{i+j} F_ i \gamma_ j x + (-1)^{2j} F_ {j} \gamma_ j x + (-1)^{2j+1} F_ {j+1} \gamma_ j x + \sum_ {i=j+2}^{n+1} (-1)^{i+j} F_ i \gamma_ j x \right)$$ $$= F_ 0 \gamma_ 0 x - F_ {n+1} \gamma_ n x + \sum_ {j=0}^n \left( \sum_ {i=0}^{j-1} (-1)^{i+j} \gamma_ {j-1} F_ i x + \sum_ {i=j+2}^{n+1} (-1)^{i+j} \gamma_ j F_ {i-1} x \right)$$ $$= F_ 0 \gamma_ 0 x - F_ {n+1} \gamma_ n x + \sum_ {j=1}^{n} \sum_ {i=0}^{j-1} (-1)^{i+j} \gamma_ {j-1} F_ i x + \sum_ {j=0}^{n-1} \sum_ {i=j+2}^{n+1} (-1)^{i+j} \gamma_ j F_ {i-1} x$$ $$= F_ 0 \gamma_ 0 x - F_ {n+1} \gamma_ n x + \sum_ {j=0}^{n-1} \sum_ {i=0}^{j} (-1)^{i+j+1} \gamma_ {j} F_ i x + \sum_ {j=0}^{n-1} \sum_ {i=j+1}^{n} (-1)^{i+j+1} \gamma_ j F_ {i} x$$ $$= F_ 0 \gamma_ 0 x - F_ {n+1} \gamma_ n x - \sum_ {j=0}^{n-1} \sum_ {i=0}^{n} (-1)^{i+j} \gamma_ {j} F_ i x .$$

Quindi \begin{aligned} ( \partial_ {n+1} s_ n + s_ {n-1} \partial_ n ) x &= F_ 0 \gamma_ 0 x - F_ {n+1} \gamma_ n x - \sum_ {j=0}^{n-1} \sum_ {i=0}^{n} (-1)^{i+j} \gamma_ {j} F_ i x + \sum_ {j=0}^{n-1} (-i)^j \sum_ {i=0}^n (-1)^i \gamma_ j F_ i x \\ &= [P_ 0', \ldots, P'_ n] - [P_ 0, \ldots, P_ n] . \end{aligned}

1. Ferrario, Davide L.; Piccinini, Renzo A. Simplicial structures in topology. CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC. Springer, New York, 2011. xvi+243 pp. ISBN: 978-1-4419-7235-4 [return]