# Do Betti numbers beyond the first have a “number of cuts” interpretation?

I have heard stated the following

Theorem. If $\Sigma$ is a (orientable) surface, then $\mathrm b_1(\Sigma)$ counts the maximum number of "circular cuts" (embedded circles $C_1,\ldots,C_m$) that you can make on $\Sigma$ without disconnecting it (i.e. with $\Sigma\smallsetminus (C_1\cup\ldots\cup C_m)$ still being connected).

Is there a similar interpretation also for higher Betti numbers $\mathrm b_k(M)$ of manifolds in general?

(I've been asked this by a non-mathematician, but I think it fits MO. If the topologists here think it's too trivial or well known, please move it to MSE).

Edit1: It wouldn't be bad to know if the above "theorem" is actually a theorem, and a reference to a rigorous proof.

Edit2: In the above, by a "generalization" I mean a statement of the form: $b=b_k(M)$ counts the maximum number of embedded orientable submanifolds $N_1,\ldots,N_b$ such that $M\setminus (N_1\cup\ldots\cup N_b)$ has clear topological property $\boldsymbol{\mathrm{P}}$. Maybe property $\boldsymbol{\mathrm{P}}$ could be about some $(k-1)$-connectedness condition? For the $b_1$ case, this would involve $0$-connectedness, i.e. just connectedness, and this is the case for $\dim M=2$ if the theorem I quoted above is true.

• For future commenters/answerers: there was an answer (now disappeared) citing the paper On the Cut Number of a 3-manifold by S.L.Harvey arxiv.org/abs/math/0112193 which may (or may not) be useful for an answer. – Qfwfq May 3 '17 at 20:25
• A second answer (by a different user) just disappeared. See the edit to the OP for some clarification. – Qfwfq May 3 '17 at 20:29
• It seems to me that the basis of this theorem is the fact that two circles in $H_1(\Sigma)$ are homologous iff they cobound a 2-manifold $M$ with boundary inside $\Sigma$, so cutting $\Sigma$ along the two circles splits it into $M$ and $\Sigma - M$. So I would expect the closest version for higher-dim manifolds to hold for $H_{d-1}$. Of course, we have Poincare duality, but if we're thinking in terms of homology, this seems like a codimensional phenomenon. – Kevin Casto May 3 '17 at 22:57

Here is one way that your observation about the first Betti number generalizes to higher-dimensional manifolds.

Suppose that $M$ a compact, orientable smooth manifold and $W$ is a regularly embedded, closed, orientable submanifold of codimension 1. Then if the number of path components of $W$ is greater than the first Betti number $b_1(M)$, then $M \setminus W$ has more path components than $M$.

Here's the proof. Consider the long exact sequence $$\dots \to H_1(M \setminus W) \to H_1(M) \to H_1(M, M \setminus W) \to H_0(M \setminus W) \to H_0(M) \to H_0(M, M \setminus W) \to 0.$$ Since $W$ is regularly embedded, it has a neighborhood diffeomorphic to the (1-dimensional) normal bundle $\nu$ of the embedding. By the excision axiom, there is an isomorphism $$H_k(M, M \setminus W) \cong H_k(\nu, \nu \setminus W).$$ (Here we view $W$ as the same as the zero-section of $\nu$.) Since $W$ and $M$ are both orientable, the normal bundle is orientable. Since the normal bundle is orientable of dimension 1, there is a Thom isomorphism $$H_k(\nu, \nu \setminus W) \cong H_{k-1}(W).$$ Therefore, part of our exact sequence becomes $$\dots \to H_1(M) \to H_0(W) \to H_0(M \setminus W) \to H_0(M) \to 0.$$ The rank of $H_0(W)$ is the number of path components of $W$, so if $W$ has more path components than $b_1(M)$ then the map $H_1(M) \to H_0(W)$ cannot possibly be surjective. If it is not surjective, then the image of $H_0(W)$ in $H_0(M \setminus W)$ is nontrivial. Since this image is the kernel of the next map, the map $H_0(M \setminus W) \to H_0(M)$ is not injective. But this map can only fail to be injective if there are more path components in $M \setminus W$ than there are in $M$.

• The difference between the number of path components of $W$ and $b_1(M)$ actually gives us a lower bound on the number of new path components in $M \setminus W$, and if $b_1(M) = 0$ we get an exact count.

• If we toss out the word "orientable" everywhere but replace homology with mod-2 homology everywhere, we get a similar bound using the mod-2 Betti number. This sees that we can remove the nonorientable submanifold $\Bbb{RP}^2 \subset \Bbb{RP}^3$ and still have a connected space, even though this can't happen for orientable submanifolds of $\Bbb{RP}^3$.

• As given, the argument is only in one direction. It gives no guarantees that we can find an oriented submanifold $W$ with $b_1(M)$ connected components so that $M \setminus W$ is still connected.

• If we replace codimension 1 in this argument with codimension $r$, we instead get an exact sequence $$\dots \to H_r(M) \to H_0(W) \to H_{r-1}(M \setminus W) \to H_{r-1}(M) \to 0.$$ By the same argument, this allows us to say: if $b_r(M)$ is less than the number of connected components of $W$, the map $H_{r-1}(M \setminus W) \to H_{r-1}(M)$ cannot be injective. This is not quite as punchy as the statement for path components, but the de Rham theorem tells us that it does still indicate something about the existence of new differential forms on $M \setminus W$.

• I don't think, in the original problem, that you wanted the submanifolds to be disjoint. If instead we have connected, closed, regularly embedded, orientable submanifolds $W_1,\dots,W_d$ of codimension $k$ which intersect transversely, one can show by a more complicated inductive argument with the (relative) Mayer-Vietoris sequence that $H_k(M, M \setminus \cup W_i)$ has rank at least $d$, and find that the map $H_{d-1}(M \setminus \cup W_i) \to H_{d-1}(M)$ has kernel of rank at least $d - b_k(M)$.

• Cool! Note that by Hurewicz, this gives the following answer to "Edit2": if $M$ is $(r-1)$-connected, then $b_r(M)$ counts the number of disjoint codim-$r$ manifolds you can remove before $\pi_{r-1}(M \setminus W)$ is nontrivial. – Kevin Casto May 4 '17 at 3:06
• Great! I haven't read through all the details yet, but it feels like it's the most complete answer one can hope to give to a question of this kind (at least in this degree of generality) – Qfwfq May 4 '17 at 15:20
• @Qfwfq You're correct, I wasn't careful about ordering terms there. – Tyler Lawson May 4 '17 at 19:05
• @KevinCasto: maybe it's nitpicking, but in the light of the third bullet point (i.e. we don't know about "existence") I think it'd be more correct to say that, if $M$ is $(r-1)$-connected, $b_r(M)+1$ is the minimal number of (disjoint codim-$r$) cuts that you can perform while having nontrivial $b_{r-1}$ (or $\pi_{r-1}$). I.e. in principle we don't know if for every (or even for any) such $W$ with $\leq b_r(M)$ components $b_{r-1}$ (resp. $\pi_{r-1}$) of the complement will be zero. – Qfwfq May 4 '17 at 20:43
• Shouldn't the last sentence of the last bullet point read: the map $H_{k-1}(M\setminus\cup W_i)\to H_{k-1}(M)$ has kernel of rank at least $d-b_k(M)$ (i.e. with a $k-1$ instead of the $d-1$)? – Qfwfq May 4 '17 at 21:06

Edit. I misunderstood the question, so what follows may not be relevant, but it may be useful anyway.

The cut number $c(M)$ of a manifold $M$ is usually defined as the maximum number of disjoint two-sided codimension-1 submanifolds whose complement is connected. By collapsing such a configuration to a graph, one sees that $\pi_1(M)$ surjects onto the free group with $c(M)$ generators. This implies that $H_1(M)$ surjects onto the abelian group of rank $c(M)$, and therefore $$b_1(M) \geq c(M).$$ However, it happens that these two numbers are often distinct. See for instance this paper of Harvey.

(Collapsing to a graph means that you pick disjoint tubular neighbourhoods $S\times [-1,1]$ of the connected two-sided submanifold $S\subset M$, then you collapse the complement of all these tubular neighbourhoods to a point $p$, and also each factor $S\times [-1,1]$ to an arc $[-1,1]$ with endpoints in $p$.)

• What does it mean to "collapse the configuration to a graph" (as far as I understand from the first lines of the introduction, Harvey counts the number of connected components, that is, the number of disjoint codim one submanifolds)? Is there any difference if we don't require our codimension one submanifolds to be disjoint? – Qfwfq May 3 '17 at 13:37
• Also, does $b_1(M)=c(M)$ indeed hold for $M$ a (orientable) surface? Rigorous references for this? (If true, I imagine it's a very classical/ well known fact) – Qfwfq May 3 '17 at 13:42