From Harvard math qualification exam, 1990.

Let $X$ be a smooth manifold with an open cover $N<\infty$ sets $\{B_{n}\}^{N}_{1}$ which are contractible. Assume that $$\pi_{0}(B_{n}\cap B_{m})\le k, \forall n,m$$ for all $n$ and $m$. Give an upper bound to the first betti number of $X$.

Recall that the first betti number is defined to be the rank of $H_{1}(X)$. The main trouble is how the conditon on $\pi_{0}$ relates to the bound. My thought (perhaps dumb) is to proceed this inductively. But an arbitrally large genus surface can be covered by a large enough open ball that is connected. So in this case $n=2g, k=0, N=1$.

I asked my professor today, and he gave the hint that this is nothing but Cech cohomology, in which the $B_{i}$s resemble the faces in simplicial homology. He suggest since $X$ has such a cover then it is the same as putting a simplicial structure on $X$. So the first betti number must be bounded. But how large is the bound given $N$ and $k$? Imagine a two hole torus covered by a million small open balls with one next to each other, then $k$ should be 2 if the balls are small enough. But this seems to carry over to a 3-hole torus or a torus of any genus. I suspect there is some fundamental misunderstanding on the statement of the problem or concepts behind, so I venture to ask. A closer look at the Cech cohomology article does not help me to come up with a bound.

  • $\begingroup$ Couldn't you try an induction and Mayer Vietoris? I have not thought through it, but it might be worth thinking about... $\endgroup$
    – mland
    Aug 7, 2012 at 9:05
  • 2
    $\begingroup$ I don't think the bound is supposed to be a good one, just a finite one! $\endgroup$
    – Zhen Lin
    Aug 7, 2012 at 9:13
  • 1
    $\begingroup$ The sets in the open over here are intended to be open subsets of the manifold. I think you are thinking of the torus as embedded in Euclidean space, but you should be thinking of the manifold as a topological space in its own right. THe only open covering by one set is the whole torus and that isn't contractible. $\endgroup$
    – Rob Arthan
    Aug 8, 2012 at 5:27
  • 3
    $\begingroup$ On reflection, my estimate was wrong, because I underestimated the number of connected components of $B_i \cap (B_1 \cup \ldots \cup B_{i-1})$ (which in a worst case could be $ik$). With the correction, the Mayer-Vietoris method comes up with the same answer as Zhen Lin (but without the sheaves!). $\endgroup$
    – Rob Arthan
    Aug 8, 2012 at 6:24
  • 1
    $\begingroup$ I see your computation. Assume $B_{i}$ are open subsets of $X$, then we should have $0\rightarrow H_{1}(X)\rightarrow H_{0}(B_{i}\cap (B_{1}\cup...\cup B_{i-1} ))\rightarrow 0$. The rest just follows by standard induction. Since every time we add $B_{i}$ as you noted the worst case is $ik$, $(\sum^{N}_{1}i)k=\frac{N(N-1)}{2}k$. Thanks! $\endgroup$ Aug 8, 2012 at 6:39

1 Answer 1


If $\mathfrak{U}$ is any open cover of $X$, the Čech-to-derived functor spectral sequence assures us of an exact sequence of the form $$0 \longrightarrow H^1(\check{C}^\bullet(\mathfrak{U})) \longrightarrow H^1(X, \mathbb{Z}) \longrightarrow H^0(\check{C}^\bullet(\mathfrak{U}, \mathscr{H}^1(\mathbb{Z}))) \longrightarrow \cdots$$ where $\mathscr{H}^1(\mathbb{Z})$ is the presheaf $U \mapsto H^1(U, \mathbb{Z})$. (Here, $H^1$ refers to sheaf cohomology, but this is the same as singular cohomology when $X$ admits a triangulation.) Since we are assuming the open sets of $\mathfrak{U}$ are contractible, we have $\check{C}^0(\mathfrak{U}, \mathscr{H}^1(\mathbb{Z})) = 0$. So in fact we have an isomorphism $$H^1(\check{C}^\bullet(\mathfrak{U})) \cong H^1(X, \mathbb{Z})$$ and the universal coefficient theorem gives a short exact sequence $$0 \longrightarrow \textrm{Ext}^1(H_0(X, \mathbb{Z}), \mathbb{Z}) \longrightarrow H^1(X, \mathbb{Z}) \longrightarrow \textrm{Hom}(H_1(X, \mathbb{Z}), \mathbb{Z}) \longrightarrow 0$$ but if $X$ is path-connected then $H_0(X, \mathbb{Z})$ is a finitely-generated free abelian group, so $\textrm{Ext}^1(H_0(X, \mathbb{Z}), \mathbb{Z}) = 0$, so there is an isomorphism $$H^1(X, \mathbb{Z}) \cong \textrm{Hom}(H_1(X, \mathbb{Z}), \mathbb{Z})$$ and therefore the ranks of $H^1(X, \mathbb{Z})$ and $H_1(X, \mathbb{Z})$ are equal when finite. Thus, $$b_1(X) = \operatorname{rank} H^1(\check{C}^\bullet(\mathfrak{U}))$$ By simple linear algebra, $\operatorname{rank} H^1(\check{C}^\bullet(\mathfrak{U})) \le \operatorname{rank} \check{C}^1(\mathfrak{U})$, and $$\operatorname{rank} \check{C}^1(\mathfrak{U}) = \sum_{\substack{\{U, V\} \\ U \ne V}} b_0 (U \cap V) \le \frac{1}{2} (N-1) N k$$ by definition of $\check{C}^1(\mathfrak{U})$.

In summary, if $X$ admits a cover by $N$ contractible open subsets whose pairwise intersections have at most $k$ connected components, then: $$b_1(X) \le \frac{1}{2} (N-1) N k$$

  • $\begingroup$ Thanks! I need sometime to understand all you had written, but how could the bound be independent of $k$? $\endgroup$ Aug 7, 2012 at 15:40
  • $\begingroup$ @user32240 Oops, I accidentally assumed $k = 1$. I've fixed my answer. $\endgroup$
    – Zhen Lin
    Aug 7, 2012 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .