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Recall that the Lusternik–Schnirelmann category (or LS-category) of a space is the integer $n$ such that there is an open cover by $n+1$ open sets which have nullhomotopic inclusions, and no such cover by fewer open sets. This is quite hard to compute, but for instance, spheres clearly have LS-category equal to 1.

However, sometimes one is interested not just in open covers by (relatively) contractible open sets, but by covers where all finite intersections of such opens are also (relatively) contractible. This is especially true when dealing with higher stacks, if one is interested in efficient cofibrant replacements.

Thus, one might want to define something intermediate, like level-$n$ LS-category of a space, which is one more than the minimum number of open sets one needs to cover said space, such that intersections of at most $n$ opens is (relatively) contractible. Then a finite good open cover gives an upper bound on the level-$n$ LS-category for all $n$. There is also the analogue for good open covers, namely where all finite intersections are (relatively) contractible.

For instance, for spheres one can take a homeomorphism with the boundary of a topological simplex, then take the open cover by puffing up a face of the simplex by some small amount (this is the analogue of covering a circle by three arcs). Thus the level-$n$ LS-category of the $k$-sphere is bounded above by $k+1$.

Has something like this been considered in the literature before?

My interest stems from considering this problem for Lie groups, in particular matrix Lie groups $O(n,\mathbb{K})$, where for certain cases we have explicit covers of the minimum size. Even something like an upper bound would be good.

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  • $\begingroup$ Unrelated question: Does every topological manifold have a good open cover? $\endgroup$ – John Pardon Aug 14 '13 at 2:14
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    $\begingroup$ Just in case you don't know this paper about L-S category of simply connected compact simple Lie groups: arxiv.org/abs/0901.2157 $\endgroup$ – Vít Tuček Aug 14 '13 at 23:27
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In some cases the nerve theorem might help (in the strong case without "relatively"; I also assume that some intersections in the question might be empty instead of contractible, hoping that this the intended definition of a good cover). Apropriate version can be found, for example in [A. Hatcher. Algebraic Topology. Cambridge University Press, Cambridge, 2001., Chapter 4.G]. In case of good covers where all intersections are contractible this yields that $X$ is homotopy equivalent to the nerve of the cover. Thus the number of sets in the cover is bounded from below by the number of vertices of the smallest simplicial complex homotopy equivalent to $X$. In case that the contractibility is assumed only for intersections up to $n$th level, then maybe some more refine versions of the Nerve theorem such as [G. Kalai and R. Meshulam. A topological colorful Helly theorem. Adv. Math., 191(2):305–311, 2005., Theorem 3.2] can be used. (In particular, the nerves of good covers have been studied intensively in conections with Helly-type theorems.)

For example, the nerve theorem is already sufficient to get the exact bounds for spheres as described in the statement of the question via the following approach. One can get that $n+1$ sets are sufficient to cover the $(n-1)$-sphere as explained in the statement of the question. However, if we put the assumption on contractibility of intersections only up to level $n$, we get that $n+1$ sets are sufficient to cover arbitrary $k$-sphere with $k \geq n-1$ by considering $(k-n+1)$ suspensions of the example with $k=n-1$ (one has to be little careful to keep the sets open). Altogether we get that $Cat^n(S^k) \leq \min(k+1,n)$, where $Cat^n(X)$ is the minimum j such that there is a cover of $X$ by open sets $U_0,…,U_j$ which are contractible, borrowing the notation analogous to the notation of the answer of Mark Grant.

Using the Nerve theorem, one can deduce that $Cat^n(S^k) \geq \min(k + 1,n)$: Indeed, for contradiction, let us assume that $Cat^n(S^k) = j$ where $j \leq k$ and $j < n$. Let $\mathcal{U} = \{U_0, \dots, U_j\}$ be the corresponding cover of $S^k$. Since $j < n$ we get that all intersections are contractible. Using the Nerve theorem, $S^k$ is homotopy equivalent to the nerve of $\mathcal{U}$. However, the simplicial complex with the smallest number of vertices homotopy equivalent to $S^k$ is the boundary of $k+1$-simplex, which has $k+2$ vertices, contradicting $j \leq k$.

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    $\begingroup$ Correct on the assumption about the intersections: they should be contractible iff nonempty. $\endgroup$ – David Roberts Aug 15 '13 at 2:01
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Update: Apologies for bumping an old question, but I thought it worth mentioning that such a version of LS-category for good covers has now been studied by several authors (under the name of covering type). The basic reference is

Karoubi, Max; Weibel, Charles A., On the covering type of a space, Enseign. Math. (2) 62, No. 3-4, 457-474 (2016). ZBL1378.55002.

Further bounds and explicit calculations were given by Dejan Govc, Wacław Marzantowicz, Petar Pavešić in their preprint https://arxiv.org/abs/1710.03333. It's worth noting that the covering type of some surfaces, such as the Klein bottle, is still unknown.


Here's a trivial observation. If you have a categorical open cover $U_0,\ldots , U_k$ of $X$ (such that each inclusion $U_i\hookrightarrow X$ is null-homotopic), then the inclusion of any $n$-fold intersection of sets in the cover factors as $$ U_{i_1}\cap\cdots\cap U_{i_n}\hookrightarrow U_{i_1}\hookrightarrow X $$ and is therefore null-homotopic. It follows that $\mathrm{cat}^n(X)\le \mathrm{cat}(X)$ for all $n$, where $\mathrm{cat}^n$ is your level-$n$ category.

Well, this is true if we read all the parenthetical "relatively"s in your question. If you want the intersections to be contractible (rather than just contractible in $X$), as in a good open cover, then you get a different notion, which should be more related to the strong category, or geometric category. Maybe you know already, the strong category of $X$, denoted $\mathrm{Cat}(X)$, is the minimum $k$ such that there is a cover of $X$ by open sets $U_0,\ldots ,U_k$ which are contractible.

By the way, if your notion has been studied prior to 2003, it should be in the book on LS-category by Cornea--Lupton--Oprea--Tanré. I don't recall seeing it there, however.

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  • $\begingroup$ Ah, well, that was stupid of me. However, the stronger notion is the one that is actually used for higher stacks etc. Thanks for pointing out strong category for me. $\endgroup$ – David Roberts Aug 14 '13 at 7:47
  • $\begingroup$ Hmm, it turns out that $cat(X) \leq Cat(X) \leq cat(X)+1$. Good to know. $\endgroup$ – David Roberts Aug 14 '13 at 7:55
  • $\begingroup$ Yes. It should be noted though that in a cover by contractible opens, the intersections need not be contractible (cf the sphere). So your definition seems to be new and worthy of study. $\endgroup$ – Mark Grant Aug 14 '13 at 8:06
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This is mostly a comment that would never fit (I apologize for taking an answer space for that). My point is that although the category of a space might be hard to compute, it is very important to emphasize that it is easy to estimate (at least when your space is a smooth manifold, which is mostly what I care about anyways). This is quite useful in critical point theory to estimate the number of critical points you get for functions on such domains.

Let me be more precise. Denote by $cat(M)$ the Lusternik-Schnirelmann category of $M$ (for me it is the number of contractible sets in $M$ that you need -- so our definitions differ by 1). Just a quick justification here: I like this definition better because every smooth function $f\colon M\to\mathbb R$ has at least $cat(M)$ critical points (e.g., every function on the sphere has at least 2 critical points), so it seems more convenient to me. Denote by $cl(M)$ its cup-length, i.e., the maximum number of graded elements in $H^*(M,R)$ of positive degree that whose cup product is a non-zero result, where $R$ is some ring with unit.

Then, although I couldn't find it on Wikipedia, it is well-known that:

Prop: If $M$ is a manifold, then $cat(M)\geq cl(M)+1$.

Proof: Let $n=cat(M)$. WLOG, $n<\infty$. If $A_i$ are open sets that cover $M$ such that the morphism $H^*(M)\to H^*(A_i)$ is trivial, then from the exact sequence $$\cdots\to H^*(M,A_i)\to H^*(M)\to H^*(A_i)\to\cdots$$ we have that if $\xi_i\in H^*(M)$ is a cohomology class of positive degree, then $\xi_i$ has a representative in $H^*(M,A_i)$. Thus, for $n$ classes $\xi_1,\dots,\xi_n$ of positive degree, $\xi_1\cup\dots\cup\xi_n$ has a representative in $H^*(M,\cup A_i)=\{0\}$.

For example, it follows immediately that:

  • $cat(S^2)\geq 2$;
  • $cat(T^2)\geq 3$;
  • $cat(\mathbb R P^n)\geq n+1$;

and it is actually not hard to prove that equality holds in all above cases. So, using the above, one can compute explicitly the category of "nice" spaces that a typical geometer would be interested in...

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  • $\begingroup$ Thanks, Renato. This is the general flavour of how one starts to compute $cat(M)$ for $M$ a manifold - calculate the cup length, and also find an atlas; this gives the lower and upper bounds on $M$. An analogue of the above proposition for my $n$-fold LS-category would be nice. $\endgroup$ – David Roberts Aug 14 '13 at 2:44

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