Contents

Idea

The simplicial identities encode the relationships between the face and degeneracy maps in a simplicial object, in particular, in a simplicial set.

Definition

The simplicial identities are the duals to the simplicial relations of coface and codegeneracy maps described at simplex category:

Let $S \in$ sSet with

• face maps $\partial_i : S_n \to S_{n-1}$ obtained by omitting the $i$th vertex;

• degeneracy maps $s_i : S_n \to S_{n+1}$ obtained by repeating the $i$th vertex.

Definition

The simplicial identities satisfied by face and degeneracy maps as above are (whenever these maps are composable as indicated):

1. $\partial_i \circ \partial_j = \partial_{j-1} \circ \partial_i$ if $i \lt j$,

2. $s_i \circ s_j = s_j \circ s_{i-1}$ if $i \gt j$.

3. $\partial_i \circ s_j = \left\{ \array{ s_{j-1} \circ \partial_i & if \; i \lt j \\ id & if \; i = j \; or \; i = j+1 \\ s_j \circ \partial_{i-1} & if\; i \gt j+1 } \right.$

Properties

Relation to nilpotency of differentials

The simplicial identities of def. 1 can be understood as a non-abelian or “unstable” generalization of the identity

$\partial \circ \partial = 0$

satisfied by differentials in chain complexes (in homological algebra).

Write $\mathbb{Z}[S]$ be the simplicial abelian group obtained form $S$ by forming degreewise the free abelian group on the set of $n$-simplices, as discussed at chains on a simplicial set.

Then using these formal linear combinations we can sum up all the $(n+1)$ face maps $\partial_i : S_n \to S_{n-1}$ into a single map:

Definition

The alternating face map differential in degree $n$ of the simplicial set $S$ is the linear map

$\partial : \mathbb{Z}[S_n] \to \mathbb{Z}[S_{n-1}]$

defined on basis elements $\sigma \in S_n$ to be the alternating sum of the simplicial face maps:

(1)$\partial \sigma \coloneqq \sum_{k = 0}^n (-1)^k \partial_k \sigma \,.$

This is the differential of the alternating face map complex of $S$:

Proposition

The simplicial identity def. 1 (1) implies that def. 2 indeed defines a differential in that $\partial \circ \partial = 0$.

Proof

By linearity, it is sufficient to check this on a basis element $\sigma \in S_n$. There we compute as follows:

\begin{aligned} \partial \partial \sigma & = \partial \left( \sum_{j = 0}^n (-1)^j \partial_j \sigma \right) \\ & = \sum_{j=0}^n \sum_{i = 0}^{n-1} (-1)^{i+j} \partial_i \partial_j \sigma \\ & = \sum_{0 \leq i \lt j \leq n} (-1)^{i+j} \partial_i \partial_j \sigma + \sum_{0 \leq j \leq i \lt n} (-1)^{i + j} \partial_i \partial_j \sigma \\ & = \sum_{0 \leq i \lt j \leq n} (-1)^{i+j} \partial_{j-1} \partial_i \sigma + \sum_{0 \leq j \leq i \lt n} (-1)^{i + j} \partial_i \partial_j \sigma \\ & = - \sum_{0 \leq i \leq j \lt n} (-1)^{i+j} \partial_{j} \partial_i \sigma + \sum_{0 \leq j \leq i \lt n} (-1)^{i + j} \partial_i \partial_j \sigma \\ & = 0 \end{aligned} \,.

Here

1. the first equality is (1);

2. the second is (1) together with the linearity of $d$;

3. the third is obtained by decomposing the sum into two summands;

4. the fourth finally uses the simplicial identity def. 1 (1) in the first summand;

5. the fifth relabels the summation index $j$ by $j +1$;

6. the last one observes that the resulting two summands are negatives of each other.

References

For instance definition 1.1 in

Revised on February 23, 2016 05:13:22 by Anonymous Coward (213.21.42.128)