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 $\mathring{s}_ 1 \cap \mathring{s}_ 2 \neq \emptyset$ $\implies$ $s_ 1 = s_ 2$, 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

Let $K$ 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]