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Stream: learning: questions

Topic: homotopy level of a Kan complex

Leopold Schlicht (Nov 05 2021 at 17:37):

How can one define the homotopy level of a Kan complex? Note that this notion is pretty unrelated to the notion of dimension of a Kan complex: take any groupoid with a non-identity morphism, then the nerve of that groupoid will be infinite-dimensional, while its homotopy level should be 1.

Alexander Campbell (Nov 05 2021 at 22:54):

@Leopold Schlicht A Kan complex $X$ "has homotopy level $\leq n$”, or "is an $n$-type”, if for every $k > n+1$, every map of simplicial sets $\partial\Delta^{k} \to X$ extends to a map $\Delta^{k} \to X$, or equivalently, if for every $m > n$, and every base point $x \in X$, the homotopy group $\pi_m(X,x)$ is trivial.

Alexander Campbell (Nov 05 2021 at 22:58):

The second definition needs some tweaking for $n < 0$.

Alexander Campbell (Nov 05 2021 at 23:16):

And of course, a Kan complex "has homotopy level $n$” if it has homotopy level $\leq n$ but not homotopy level $\leq n-1$.

John Baez (Nov 06 2021 at 04:24):

Alexander Campbell said:

The second definition needs some tweaking for $n < 0$.

For example $\pi_0$ is a set, while $\pi_{-1}$ is a mere truth value.

Mike Shulman (Nov 06 2021 at 05:45):

Alexander Campbell said:

And of course, a Kan complex "has homotopy level $n$” if it has homotopy level $\leq n$ but not homotopy level $\leq n-1$.

I don't think that's a very useful definition. I'm more inclined to use "homotopy level $n$" to mean just "is an $n$-type".

Mike Shulman (Nov 06 2021 at 05:46):

John Baez said:

Alexander Campbell said:

The second definition needs some tweaking for $n < 0$.

For example $\pi_0$ is a set, while $\pi_{-1}$ is a mere truth value.

But with that modification it does work, right? Where "trivial" means "isomorphic to the terminal object" of whatever category it lies in?

Alexander Campbell (Nov 06 2021 at 06:30):

@Mike Shulman I agree it's not a useful definition, but I guess to my ear one can be only on one "level" and not on another.

Alexander Campbell (Nov 06 2021 at 06:38):

Mike Shulman said:

John Baez said:

Alexander Campbell said:

The second definition needs some tweaking for $n < 0$.

For example $\pi_0$ is a set, while $\pi_{-1}$ is a mere truth value.

But with that modification it does work, right? Where "trivial" means "isomorphic to the terminal object" of whatever category it lies in?

There are some further subtleties involving basepoints and the empty space.

Mike Shulman (Nov 06 2021 at 14:42):

Alexander Campbell said:

There are some further subtleties involving basepoints and the empty space.

Let's see, so according to the definition as stated, $X$ would be a $(-1)$-type if it is a 0-type and for every $x\in X$ the pointed set $\pi_0(X,x)$ is isomorphic to 1. In other words, it is equivalent to a set, and if that set is inhabited then it is terminal. That seems to me a correct definition of a $(-1)$-type.

You're right that it doesn't work for $n=-2$, though; to get that definition right you can't have the "for all $x\in X$ in there.

Leopold Schlicht (Nov 07 2021 at 18:44):

Thanks! Does somebody know whether the concept of the homotopy level of a Kan complex is discussed in Lurie's Kerodon or Higher Topos Theory? I couldn't find it under the names "homotopy level", "h-level", and "$n$-type".

Reid Barton (Nov 07 2021 at 19:24):

It's definition 2.3.4.15 of the arxiv version of HTT, "$k$-truncated" (doesn't seem to be in Kerodon yet, a bit surprising)

Reid Barton (Nov 07 2021 at 19:25):

also mentioned in the Terminology section before the table of contents

Leopold Schlicht (Nov 13 2021 at 16:05):

Alexander Campbell said:

Leopold Schlicht A Kan complex $X$ "has homotopy level $\leq n$”, or "is an $n$-type”, if for every $k > n+1$, every map of simplicial sets $\partial\Delta^{k} \to X$ extends to a map $\Delta^{k} \to X$, or equivalently, if for every $m > n$, and every base point $x \in X$, the homotopy group $\pi_m(X,x)$ is trivial.

I noticed that the first of these two equivalent conditions makes sense for each simplicial set (not only for Kan complexes). Does one sometimes use the attribute "has homotopy level $\leq n$" for simplicial sets that that don't satisfy the Kan extension property? (If not, why not?)

Reid Barton (Nov 13 2021 at 16:21):

I think it's a reasonable usage but it would no longer be equivalent to a lifting property. Instead, it would mean that a fibrant (Kan) replacement of the simplicial set has homotopy level $\le n$.

Leopold Schlicht (Nov 19 2021 at 19:01):

Reid Barton said:

I think it's a reasonable usage but it would no longer be equivalent to a lifting property. Instead, it would mean that a fibrant (Kan) replacement of the simplicial set has homotopy level $\le n$.

If one wants to define when an $\infty$-category is an $n$-category, is this done in a similar way?

Can one consider the "homotopy category" construction from simplicial sets to 1-categories as a kind of "1-truncation"? How are the other truncations defined?

Leopold Schlicht (Nov 23 2021 at 16:35):

The first question is discussed in 2.3.4 in Higher Topos Theory, but the last question is still open (for me).

Ian Coley (Nov 23 2021 at 16:36):

See Section 2 of the following paper of Raptis https://arxiv.org/pdf/1910.04117.pdf

Leopold Schlicht (Nov 23 2021 at 16:51):

Ah, nice, thanks! Now I recognize that Proposition 2.3.4.12 in Higher Topos Theory already discusses this construction, in particular it exhibits "the homotopy $n$-category of an $\infty$-category" as a left adjoint to the inclusion of $n$-categories into $\infty$-categories.

Leopold Schlicht (Dec 15 2021 at 18:55):

Alexander Campbell said:

Leopold Schlicht A Kan complex $X$ "has homotopy level $\leq n$”, or "is an $n$-type”, if for every $k > n+1$, every map of simplicial sets $\partial\Delta^{k} \to X$ extends to a map $\Delta^{k} \to X$, or equivalently, if for every $m > n$, and every base point $x \in X$, the homotopy group $\pi_m(X,x)$ is trivial.

Fix $n$. If $X$ is any simplicial set with the property that for every $k>n+1$, every map of simplicial sets $\partial \Delta^k\to X$ extends to a map $\Delta^k\to X$, does it follow that $X$ is a Kan complex? I know that in the case $n=-2$ the answer is yes: Kerodon/0076. (By definition, a contractible Kan complex is a simplicial set with the above property for $n=-2$.)

Zhen Lin Low (Dec 15 2021 at 22:12):

No. The nerve of any category has this property for n = 1.

Leopold Schlicht (Dec 16 2021 at 16:00):

Thanks! Is there any $n$ except $1$ for which this is true?

Mike Shulman (Dec 16 2021 at 16:58):

You mean, any $n$ except $-2$?

Mike Shulman (Dec 16 2021 at 16:59):

It should be true for $n=-1$, since a simplicial set with that property is either empty or a contractible Kan complex. And I think it should be true for $n=0$, since a simplicial set with that property is a disjoint union of contractible Kan complexes. But once it fails for $n=1$ it's going to automatically fail for any $n>1$.

Leopold Schlicht (Dec 16 2021 at 17:04):

Ah, thanks!