Joins of simplicial sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:44:11Z http://mathoverflow.net/feeds/question/7134 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7134/joins-of-simplicial-sets Joins of simplicial sets Harry Gindi 2009-11-29T13:41:13Z 2009-12-05T18:51:17Z <p>Why doesn't the join operation on the category of simplicial sets commute up to unique isomorphism? I mean, aren't products and coproducts commutative up to isomorphism? That leads me to conclude at first glance that the join is commutative, but it's not. Recall, given two simplicial sets $S$ and $S'$, we define the join to be the simplicial set such that for all finite nonempty totally ordered sets $J$, $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$ Where $\forall (i \in I \land i' \in I') i &lt; i'$, which implies that $I$ and $I'$ are disjoint. </p> <p>Now the thing is, clearly my conclusion is stupid, because we use the fact that it doesn't commute to distinguish between over quasi-categories and under quasi-categories. Where did I go wrong?</p> <p>I hope this is up to the standards of MO, but if it's not, I'll delete the topic. </p> http://mathoverflow.net/questions/7134/joins-of-simplicial-sets/7138#7138 Answer by Tyler Lawson for Joins of simplicial sets Tyler Lawson 2009-11-29T14:03:51Z 2009-11-29T22:11:51Z <p>Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.</p> <p>EDIT: Some elaboration.</p> <p>For a simplicial set $T$, let's write $T_n$ for the "n-simplices", i.e. the value of on the ordered set $\{0,1,...,n\}$; these together with the maps between them determine the functor $T$ completely. (Your formula for the join requires the convention that $T$ takes the empty set to a single point.)</p> <p>Given $S$ and $S'$, let's determine the 0- and 1-simplices of the join.</p> <p>First, $(S \star S')_0$. There are exactly two ways to write $\{0\} = I \cup I'$ in an order preserving way as indexed by the coproduct: either $I'$ is empty and $I$ is everything, or vice versa. Thus $(S \star S')_0 = S_0 \cup S'_0$ accordingly. The zero-simplices of the join are the zero-simplices of the original simplicial set.</p> <p>Next, the 1-simplicies. Similarly $$(S \star S')_1 = S(\{0,1\}) \cup (S(\{0\}) \times S'(\{1\})) \cup S'(\{0,1\})= S_1 \cup (S_0 \times S'_0) \cup S'_1$$ There are then 3 types of 1-simplices: the 1-simplices from S, those from S', and for each choice of a point of S and a point of S' there is a new 1-simplex.</p> <p>The two boundary maps $(S \star S')_1 \to (S \star S')_0$ are induced by the inclusions of $\{0\}$ and $\{1\}$ into $\{0,1\}$ (the "back" and "front" boundaries respectively). In particular, on the new 1-simplices $S_0 \times S'_0$ the back boundary is the projection to $S_0$ and the front boundary is projection to $S'_0$. There is <em>asymmetry</em> here because the only ways we're allowed to decompose $\{0,1\}$ in the coproduct have $I$ (the subset corresponding to $S$) first and $I'$ second. None of the "new" edges start at a vertex of $S'$ and end at a vertex of $S$.</p> http://mathoverflow.net/questions/7134/joins-of-simplicial-sets/7237#7237 Answer by Emily Riehl for Joins of simplicial sets Emily Riehl 2009-11-30T05:00:58Z 2009-11-30T05:00:58Z <p>It might be helpful to work through some simple examples. You probably know that &Delta;<sup>n</sup> &#9733; &Delta;<sup>k</sup> = &Delta;<sup>n+k+1</sup>. This has to do with the ordinal sum: one way of defining joins is as a restriction of the monoidal structure on <i> augmented </i> simplicial sets, which are contravarient functors from the category &Delta;<sub>+</sub> of all finite ordinals (including the empty ordinal) into sets. The category &Delta;<sub>+</sub> has a monoidal structure given by ordinary addition with &empty; as the unit, and this induces the aforementioned monoidal structure on augmented simplicial sets. The thing we call <i>n</i> when we are talking about simplicial sets is really the ordinal <i>n</i>+1, so the formula above holds because </p> <p>(<i>n</i>+1) + (<i>k</i>+1) = (<i>n</i>+<i>k</i>+1)+1.</p> <p>Of course, this example doesn't illustrate the asymmetry you asked about, but this one will:</p> <p>&#8706;&Delta;<sup>n</sup> &#9733; &Delta;<sup>0</sup> = &Lambda;<sup>n+1</sup>[n+1] while &Delta;<sup>0</sup> &#9733; &#8706;&Delta;<sup>n</sup> = &Lambda;<sup>0</sup>[n+1].</p> <p>To work out the details, you'll need to understand how the face maps of <i>S&#9733;T</i> are defined, as alluded to above. Here's my notation: <i>(S&#9733;T)<sub>n</sub> = S<sub>n</sub></i> &cup; <i>T<sub>n</sub></i> &cup; &#40;&cup;<sub> j+k = n+1 </sub> <i> S<sub>j</sub> &times; T<sub>k</sub></i> &#41;.</p> <p>The <i>i</i>-th boundary map <i>d<sub>i</sub> : (S&#9733;T)<sub>n</sub> &rarr; (S&#9733;T)<sub>n-1</sub></i> is defined on <i>S<sub>n</sub></i> and <i>T<sub>n</sub></i> using the <i>i</i>-th boundary map on <i>S</i> and <i>T</i>. Given &sigma;&isin;<i>S<sub>j</sub></i> and &tau;&isin;<i>T<sub>k</sub></i> , we have:</p> <p>d<sub>i</sub> (&sigma;, &tau;) = (d<sub>i</sub> &sigma;,&tau;) if i &le; j, j &ne; 0. <br> d<sub>i</sub> (&sigma;, &tau;) = (&sigma;,d<sub>i-j-1</sub> &tau;) if i &gt; j, k &ne; 0.</p> <p>If <i>j</i> = 0, <i>d</i><sub>0</sub>(&sigma;, &tau;) = &tau; &isin; <i>T<sub>n-1</sub> &sub; (S&#9733;T)<sub>n-1</sub></i>. If <i>k</i> = 0, <i>d<sub>n</sub></i>(&sigma;, &tau;) = &sigma; &isin;<i>S<sub>n-1</sub> &sub; (S&#9733;T)<sub>n-1</sub></i> .</p> <p>Try this out for <i>n</i> = 1 or 2 first, to get a feel for things. While these sorts of computations can be quite annoying, I find they do really help me develop my intuition. Best of luck!</p> http://mathoverflow.net/questions/7134/joins-of-simplicial-sets/7879#7879 Answer by Peter LeFanu Lumsdaine for Joins of simplicial sets Peter LeFanu Lumsdaine 2009-12-05T18:01:12Z 2009-12-05T18:51:17Z <p>As mentioned in the other answers, the join of simplicial sets is closely related to "ordered disjoint union", or "concatenation", of (totally, partially, pre-) ordered sets. You can use this both to get simple examples of its non-commutativity, and to help reconcile that with the intuition that it <em>should</em> be commutative.</p> <p>Any order $X$ can be seen as the simplicial set whose $n$-simplices are chains $(x_0 \leq \ldots \leq x_n)$ from $X$. That is, there's a full and faithful "nerve" embedding $N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$. Now if $X$ and $Y$ are orders, seen as their nerves, $X \star Y$ is exactly (the nreve of) their ordered disjoint union.</p> <p>So e.g. $1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$ is an easy and intuitive example of the non-commutativity.</p> <p>On the other hand, like you say, looking at the definition, there is an immediate intuition that it <em>should</em> be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets $X$, $Y$,</p> <p>$(X \star Y)^\mathrm{op} \cong Y^\mathrm{op} \star X^\mathrm{op}.$</p> <p>So commuting $\star$ distributes over $\mathrm{op}$: so in a sense, the only asymmetry in $\star$ is an asymmetry of variance. This is nice and intuitive for ordered sets, and easily shown for all simplicial sets.</p>