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This page describes aspects of the combinatorics of Cartesian products of simplicial sets, mostly by describing (Prop. ) the non-degenerate simplicices inside a Cartesian product $\Delta[p] \times \Delta[q]$ of basic n-simplices for $n = p,q$. These are enumerate by the $(p,q)$ (un-)shuffles in a manner that for simplicial abelian groups is originally known from the Eilenberg-Zilber map.
(cartesian product of simplicial sets)
For $K, L \,\in\,$ SimplicialSets, their Cartesian product,
is the simplicial set whose $k$th component set is the Cartesian product of Sets of the components of the two factors
and whose face- and degeneracy maps are, similarly, the image under the Cartesian product-functor of the face- and degeneracy maps of the factors:
Since SimplicialSets is a category of presheaves, namely over the simplex category, this is a special case of the general fact that limits of presheaves are computed objectwise.
But it is also immediate to check that (2) with (1) satisfies the defining universal property of the Cartesian product.
(degenerate simplices in product of simplicial sets) Prop. means in particular that a simplex of the form $(s_\alpha(x), s_\beta(y) \in K \times L$ may be non-degenerate even though its two components are each degenerate (see the archetypical Example below).
Indeed, the proposition says that the degenerate simplices in $K \times L$ are precisely those such that their two simplex-components in $K$ and $L$, respectively, are in the image of the same degeneracy map $\alpha = \beta$.
(non-degenerate $(p+q)$-simplices in $\Delta[p] \times \Delta[q]$)
For $p,q \,\in\, \mathbb{N}$, the non-degenerate simplices in the Cartesian product (Prop. )
of standard simplices in sSet correspond, under the Yoneda lemma, to precisely those morphisms of simplicial sets
which satisfy the following equivalent conditions:
as permutations of $(p+q)$ elements they are $(p,q)$-shuffles;
as morphisms of finitely generated categories they take generating morphisms to generating morphisms;
(e.g. Kerodon 2.5.7.2: 00RH)
Such morphisms may hence be represented by paths
on a $(p+1)\times(q+1)$-lattice,
from one corner to its opposite corner,
consisting of $p+q$ unit steps,
each either horizontally or vertically:
From Prop. it is clear (Rem. ) that a simplex $\sigma$ (3) is degenerate precisely if, when regarded as a path as above, it contains a constant step, i.e. one which moves neither horizontally nor vertically. But then – by degree reasons, since we are looking at paths of $p + q$ steps in a lattice of side length $p$ and $q$ – it must be that the path proceeds by $p + q$ unit steps.
(sequence-notation for simplices in a product of simplices)
Written as a pair of $(p+q)$-simplices, one in $\Delta[p]$ and one in $\Delta[q]$, the non-degenerate simplex (3) is a pair of monotone lists of natural numbers
such that one row has a constant step precisely where the other has not.
Here is a slightly alternative way to think of these non-degenerate simplices:
For $X$ some simplicial set $x \in X_p$ some $p$-cell and for $\mu = (\mu_1 \lt \mu_2, \lt \cdots \lt \mu_q)$ a sequence of natural numbers in $\{0, \cdots p+q\}$, write
for the map dual to the sequence
where $\sigma_i$ is the surjective monotone map that repeats the index $i$.
The non-degenerate simplices in the Cartesian product (Prop. )
of simplices in sSet are precisely those of the form
for $(\mu,\nu)$ a $(p,q)$-shuffle and $x, y$ non-degenerate simplices in $\Delta[p]$ and $\Delta[q]$, respectively.
The non-degenerate (n+1)-simplices in the cylinder of the n-simplex
are, according to Prop. and in the notation of Remark , as follows:
and so on.
(non-degenerate simplices in simplicial square)
The complete set of non-degenerate simplices in $\Delta[1] \times \Delta[1]$ is, in specialization of Example , according to Prop. and in the notation of Remark , the following:
See also Friedman 2008, Fig. 20.
The non-degenerate simplices in
are, in specialization of Example , according to Prop. and in the notation of Remark , the following three:
Each such simplex yields a partition of $\{0, \ldots, p+q-1\}$ into two disjoint sets, $\mu_1\lt\ldots \lt\mu_p$, and $\nu_1 \lt \ldots \lt \nu_q$, and vice versa, any such partition yields a simplex. Suppose that we have an array, as above, written
with $0= i_0 \leq i_1 \leq \ldots \leq i_{p+q}= p$, then if $i_k = i_{k+1}$, we put $k$ into the second set, otherwise $k$ is put in the first set. This, of course, leads to an operation that preserves order. For instance, in the above example 2., the $i$-sequence is $(0,1,1,2)$, so there is the single repeat with $k = 1$, and $\nu = \{1\}$.
We likewise require $0= j_0 \leq j_1 \leq \ldots \leq j_{p+q}= p$, and put $k$ into the first set if $j_k = j_{k+1}$. For the example, we have the $j$-sequence is $(0,0,1,1)$, so $\mu = \{ 0,2\}$. Of course, from the partition you can get the sequences and conversely. The attentive reader will, of course, have noted that, for example 2., the $\alpha$, we specified was exactly the $\nu$, and the $\beta$ was the $\mu$. This is general with the simplex corresponding to a partition, $(\mu,\nu)$, being given by $(s_{\nu_q}\ldots s_{\nu_1}x_p,s_{\mu_p}\ldots s_{\mu_1}y_q)$.
Each such partition defines a permutation of $\{0,\ldots, p+q-1 \}$. Let us write $\iota_0 : \{ 0, \ldots, q-1\}\to \{0, \ldots, p+q-1\}$ for the order preserving function $\iota_0 (r)= p+r$, whilst $\iota_1 : \{ 0, \ldots, p-1\}\to \{0, \ldots, p+q-1\}$ will denote the inclusion, so $\iota_1(r) = r$. There will be a permutation $\sigma$ of $\{0,\ldots, p+q-1 \}$ such that $\sigma \iota_0(r) = \nu_{r+1}$ and $\sigma\iota_1(r) = \mu_{r+1}$. This means that the permutation looks like
We can thus assign a sign, $sgn(\sigma)$, to each such shuffle, namely the sign of the corresponding permutation.
For our standard examples, we have : 1) $\sigma$ is the identity, 2) $\sigma = \left(\begin{array}{ccc}0&1&2\\0&2&1\end{array}\right)$, i.e. the transposition exchanging 1 and 2, and for 3) $\sigma = \left(\begin{array}{ccc}0&1&2\\1&2&0\end{array}\right)$, a 3-cycle. We thus have, in this case, the signs are +1, -1, and + 1, respectively.
Any pair $(p,q)$ yields a poset relating the various $(p,q)$-shuffles.
Our $(2,1)$-example is really too simple and small to illustrate this well, but the two cases $(3,1)$ and $(2,2)$ do a much better job, but, even so, we first look at the (2,1) example:
(This Hasse diagram has been laid out horizontally to save space. The bottom is to the left. The vertex $(0\lt 1)$ corresponds to the shuffle with $\mu_1 = 0, \mu_1 = 1$, and so on.)
We need here to explain the partial order. We take the ‘$\mu$-signature’ of the shuffle, that is, the ordered set $\mu_1\lt\ldots \lt \mu_p$. (Of course, this determines the $\nu$-signature as that is the complement of $\mu$.)
Given two $(p,q)$-shuffles, represented by $(\mu, \nu)$ and $(\mu\prime,\nu\prime)$, we set
if, for each $i$ in the range $1\leq i\leq p$, $\mu_i \leq \mu_i\prime.$ We refer to this poset as $(Shuff(p,q),\leq)$.
Going to $(Shuff(3,1),\leq)$, the poset for $(3,1)$-shuffles, the fact that $q = 1$ will mean that this poset is again linear:
This is true in general:
If $p = 1$ or $q = 1$, then $(Shuff(p,q),\leq)$ is a linear poset.
If $p = 1$, $\mu = (\mu_1)$ is a singleton, and the poset will be:
For $q = 1$, $\nu = (\nu_1)$, and the poset is
where at each stage one misses out the single $\nu$-value.
In all cases, each position is obtained from the one immediately to its ‘left’ by increasing one value, yet remaining with a shuffle. This is more clearly seen in the (2,2) example, which is no longer linear. First we display the grid in which things are happening.
We can then look at the shuffle poset, noting again that it is not linear:
The left hand shuffle, labelled $(0\lt1)$, corresponds to $\left(\begin{array}{ccccc}0&1&2&2&2\\0&0&0&1&2\end{array}\right)$, so gives the path along the bottom of the square and then up the right hand side.
The first change goes to $(0\lt 2)$ and gives a path with 2 steps,
and corresponds to the shuffle, $\left(\begin{array}{ccccc}0&1&1&2&2\\0&0&1&1&2\end{array}\right)$, so at this position, $(0\lt2)$, there is a choice, either increase 0 by 1 to get $(1\lt 2)$ or increase 2 by 1 to get $(0\lt3)$. In the first case, we get the path
going horizontally across the square, and the shuffle, $\left(\begin{array}{ccccc}0&0&1&2&2\\0&1&1&1&2\end{array}\right)$, and in the second case, we get the shuffle: $\left(\begin{array}{ccccc}0&1&1&1&2\\0&0&1&2&2\end{array}\right)$, and the obvious path up the middle of the square:
From $(1 \lt 2)$ or $(0\lt3)$, there is only one way to go, namely to $(1 \lt 3)$ and a 2-step path (left to you), and, finally it is just one step from $(1 \lt 3)$ to $(2 \lt 3)$ and the other extremal path.
In summary, each path corresponds to a simplex of maximal dimension in the product. The poset encodes the simplest relationships between those paths with the links in the Hasse diagram corresponding to simple changes in the path, and adjacency of the two simplices in the product, but note that the poset is usually not linear.
There is a total order on $Shuff(p,q)$ related to the above partial order and which is useful when checking cancellation of terms in non-commutative contexts, as occurs in the Whitehead product formula given by Curtis and attributed to Kan. This is the anti-lexicographic order.
We represent a $(p,q)$-shuffle $(\mu,\nu)$ by an increasing $\mu$-sequence $\mu_1\lt \ldots \lt \mu_p$, (so with complementary $\nu$-sequence $\nu_1\lt \ldots \lt \nu_q$).
We order the $(\mu,\nu)$ using just the $\mu$-sequence as, of course, that determines the $\nu$-sequence completely. Inductively in $p$, we set
(i) if $\mu_p\lt \mu\prime_p$,
or
(ii) if $\mu_p = \mu\prime_p$ and $\mu_p$ is odd,
whilst if $\mu_p = \mu\prime_p$ and $\mu_p$ is even,
where we adopt the notation $(\mu_1,\ldots, \mu_p)$ instead of $(\mu_1\lt\ldots \lt \mu_p)$.
For example, on our (2,2)-shuffles, the total order is
We note the lexicographic order on the sector with $\mu_2 = 2$ is reversed.
For illustrative purposes, we will look at two other examples.
First a generic $(p,1)$ case: the shuffle poset for $(p,1)$ is, more or less,
as, at each position, there is only one transposition that can be applied. The anti-lex total order will correspond exactly to this if $p-1$ is odd, but, if $p-1$ is even, it will reverse the order on the last $p-1$ positions giving
There are various points to note. Firstly that if $p$ is even, the permutation corresponding to $(1, \ldots , p-1)$ is odd. Secondly, the geometric picture is simply a prism, $\Delta[p]\times \Delta[1]$, and we can easily interpret the above order as a filling scheme for the simplices of $(\Delta[p]\times \Delta[1])\times \Delta[1]$, starting with the empty shell of
together with part of the top of the ‘canister’. (In general the total order seems to correspond to some sort of filling scheme although this is not always as clear as here.)
Our second case will be (3,2). Again we will write $(i,j,k)$ instead of $i\lt j\ltk$ in order to save confusion over the various different orders being considered. The shuffle poset, $\mathrm{Shuff}(3,2)$, has Hasse diagram given below:
The subgraph of those with $\mu_3 = 4$, of course, is a copy of our earlier $(2,2)$-example, whilst that with $\mu_3 = 3$, is a copy of $Shuff(3,1)$. Within that we have, on deleting the 3s from the final positions, a copy of $Shuff(2,1)$ and so on. Taking $Shuff(2,1) \times \{3\}$, we can match it with a $Shuff(2,1) \times \{4\}$ within the $\mu_3 = 4$ block, the matching being given by, for instance, $(0,1,3)\leftrightarrow(0,1,4)$, etc. The anti-lex order gives
which is
the ordinal sum, or join, of the anti-lex ordered shuffle sets. This sort of decomposition is quite general.
The shuffle-formula is due (in terms of simplicial abelian groups of chains on a simplicial set, see at Eilenberg-Zilber map) to:
following:
The streamlined perspective of strictly monotonic morphisms $\Delta[p+q] \to \Delta[p] \times \Delta[q]$ is highlighted in
Exposition:
Textbook accounts:
Last revised on August 30, 2021 at 06:04:23. See the history of this page for a list of all contributions to it.