Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I ran into this problem when studying Morse theory. My professor referred to the torus and riemann surface of genus 2 as an example. And after some manipulating on the long exact sequence of cohomology groups he came to the conclusion that attaching an n-cell(passing a critical point of index n) would either kill a class of m-1 or give birth to a class of m. Denoting $\mathcal{M}^{\pm}=f^{-1}(-\infty, p\pm\epsilon)$, and the index of critical point p is m, here is the long exact sequence of cohomology group:

$0\rightarrow\cdots\rightarrow H^{\ast-1}(\mathcal{M}^-)\rightarrow H^{\ast}(\mathcal{M}^+,\mathcal{M}^-)\rightarrow H^{\ast}(\mathcal{M}^+)\rightarrow H^{\ast}(\mathcal{M}^-)\rightarrow H^{\ast+1}(\mathcal{M}^+,\mathcal{M}^-)\rightarrow\cdots\rightarrow0$

My first question is, is it true that the map $H^{\ast}(\mathcal{M}^+,\mathcal{M}^-)\rightarrow H^{\ast}(\mathcal{M}^+)$ is either injective or the 0 map? Why or why not? And also why $H^{\ast}(\mathcal{M}^+,\mathcal{M}^-)$ is $\mathbb{Q}$ for $\ast=m$ and 0 otherwise?

Secondly, if this is true, we will arrive at the desired results. Either

$0\rightarrow H^{m-1}(\mathcal{M}^+)\rightarrow H^{m-1}(\mathcal{M}^-)\rightarrow0\rightarrow \mathbb{Q}\rightarrow H^{m}(\mathcal{M}^+)\rightarrow H^{m}(\mathcal{M}^-)\rightarrow0$


$0\rightarrow H^{m-1}(\mathcal{M}^+)\rightarrow H^{m-1}(\mathcal{M}^-)\rightarrow\mathbb{Q}\rightarrow0$

Thus the "either kill or give birth" statement is obtained. But what I cannot understand is how to view this inuitively, say, how can you tell whether attaching a cell will kill a class or give a new class from a geometric point of view? My professor tried to explain this using torus but I still fail to see it. I just didn't know how he managed to "see" it. I would really appreciate it if someone could kindly tell me how to see it. Thanks.

share|improve this question
$H^*$ being $\mathbb Q$ implies that the map is either injective or $0$, since maps from $\mathbb Q$ to $\mathbb Q$-vector spaces are injective. Doesn't $H^*$ being $\mathbb Q$ follow from the fact that $\mathcal M^{+}$ minus $\mathcal M^{-}$ contains just a single critical point, in Morse theory? Or a single cell, in CW complex cohomology? What cohomology theory are you using here? –  Will Sawin Oct 6 '12 at 3:47
Formally, it kills a class if it's the image under the coboundary map of that class, and it creates a new class if it's not the image under the coboundary map of any class. So a picture of whether it kills or attaches a class is a picture of the coboundary map. So in CW theory, whether it kills a class depends on whether the attaching map is homologically trivial or not. In Morse theory, it depends on the flow lines coming out of that critical point and whether they cancel each other. –  Will Sawin Oct 6 '12 at 3:51
add comment

3 Answers

up vote 3 down vote accepted

First of all, by Morse theory you know that $\mathcal{M}^+$ is obtained by attaching an $m$-cell to $\mathcal{M}^-$, i.e. $\mathcal{M}^+=\mathcal{M}^-\cup_{\phi}e$, where $e=D^m$ is an $m$-cell and $\phi:\partial(e)=S^{m-1}\to \mathcal{M}^-$ is some attaching map. Therefore you can compute $H^{\*}(\mathcal{M}^+,\mathcal{M}^-)$ by excision (you excise $\mathcal{M}^-\setminus\phi(S^{m-1})$), and get that $H^{\*}(\mathcal{M}^+,\mathcal{M}^-)\simeq H^{\*}(D^m,S^{m-1})$ that has exactly one nonzero cohomology in degree $m$.

As for the second question, I think it's more intuitive looking at homology (it's the same since we are looking at coefficients in a field). There you have the sequence

$0\to H_m(\mathcal{M}^-)\to H_m(\mathcal{M}^+)\to H_m(\mathcal{M}^+,\mathcal{M}^-)\stackrel{\partial}{\to} H_{m-1}(\mathcal{M}^-)\to M_{m-1}(\mathcal{M}^+)\to 0$.

Moreover, the generator of the unique nonzero class of $H_m(\mathcal{M}^+,\mathcal{M}^-)$ can be thought of the fundamental class $[e]$ of the cell you attached, and $\partial[e]=\phi_*[\partial e]$. In other words, the boundary operator sends $[e]$ essentially to the class of its topological boundary. If $\partial[e]=0$, it means that $\phi_{\*}[\partial e]$ is a boundary in $\mathcal{M}^-$, i.e. there exists an $m$-cycle $e_-$ entirely contained in $\mathcal{M}^-$ such that $\partial[e]=\partial[e_-]$. In particular $\partial[e-e_-]=0$ and you can say that $e$ can be completed to a cycle, contained in $\mathcal{M}^-$ away from the critical point}.

Think of the usual example of the torus, with the morse function being the heigth $z$. Let $p_1$ be the second critical point (the first critical point of index one) with critical value $z_1$, let $\mathcal{M}^-=z^{-1}([0,z_1-\epsilon])$ and $\mathcal{M}^+=z^{-1}([0,z_1+\epsilon])$. Then $\mathcal{M}^+$ is obtained by attaching a $1$-cell $e$, that is a little segment passing through $p_1$ and "going down". Its boundary $\partial[e]$ is just two points in $\mathcal{M}^-$. You can check that $\partial:H_{1}(\mathcal{M}^+,\mathcal{M}^-)\to H_0(\mathcal{M}^-)$ is zero, and in fact $\partial[e]$ is the boundary of an arc $e_-$ entirely contained in $\mathcal{M^-}$ (for example the arc through the minimum point). it's quite clear that we just "completed" $[e]$ to a cycle $[e]-[e_-]$ (that you can see as a closed nontrivial closed curve in the torus) that is "almost entirely contained in $\mathcal{M}^-$ away from the critical point.

If $\partial\neq 0$, this means that we are killing some cohomology, that we created at some point before. Take for example the sphere $S^2$, embedded in $\mathbb{R}^3$ in the shape of a "$\bigcap$", and again with $z$ as a Morse function (just wiggle it a bit so that the two minima have different values). Then from bottom-up, you first see two index $0$ points (the two minima), that give you two 0-cells. Then, as you go to the third critical point (the first critical point of index 1), you notice that $\partial \neq 0$ and so the contribution of that critical point is to kill something we created before. In some sense then, this morse function is not "efficient", in the sense that creates too many critical points, that will be killed later anyways. An "efficient" Morse function only creates critical points when they really contribute to the homology of the space. "Efficient" Morse functions are called perfect.

Here is the formal definition of perfect morse function: as you said, every critical point of index $m$, might give rise to at most one new generator of $H_{\*}(\mathcal{M}_+)$. So if $c_m$ denotes the number of critical points of index $m$, we have that $c_m\geq \dim H_{\_*}(M)$, where $M$ is our manifold. The Morse function is called perfect if the inequality above is an equality, for every $m$.

share|improve this answer
Thank you Marco! I found your answers most inspiring and detailed. I think I have got what I want. Thank you very much! –  xuxuzhu Oct 7 '12 at 17:54
add comment

Since the OP apparently already knows the formal argument using the exact sequence, excision, etc., and is just looking for visualization, it may be helpful to give the following visual but partial explanation; it's essentially contained already in Marco Radeschi's answer, and I'm just extracting the visual part. Like Marco, I find it easier to visualize homology rather than cohomology, so I'll talk about the homology situation. You're adjoining an $m$-cell by attaching its boundary to some cycle $C$ in $M_-$. After you do that, this cycle certainly certainly represents 0 in the homology of $M_+$, i.e., it's a boundary, namely the boundary of your newly added cell. So you could say you've killed $C$, except for one detail: $C$ might already have been 0 in the homology of $M_-$, before you added your $m$-cell. In this case, $C$ would be the boundary of some chain in $M_-$, and that chain, combined with your new $m$-cell, would form an $m$-cycle in $M_+$. That gives the new element in the $m$-dimensional homology.

Of course, this visual picture lacks a lot of rigor. For example, how do we know that nothing else happens to the homology beyond killing an $m-1$-dimensional class or creating an $m$-dimensional one? To make the argument rigorous, I think you really need to go beyond visualization and work through the formalities (exact sequence, etc.).

share|improve this answer
When I last took algebraic topology, my professor (Shmuel Weinberger) referred to this phenomenon with the mnemonic "when you kill the dead you create a ghost". Apparently it was memorable! –  Ryan Reich Oct 6 '12 at 14:01
Thank you for the help. This is exactly what I can understand, but I still what some more ituitive ways to "visualize" through an examine of some easy examples such as the torus or something. Thank you all the same for such a kind help! –  xuxuzhu Oct 7 '12 at 17:56
add comment

$\newcommand{\pa}{\partial}$ $\newcommand{\bQ}{\mathbb{Q}}$ Since $M^+$ is homotopic to a space obtained obtained from $M^-$ by attaching an $n$-cell we deduce from the excision theorem

$$ H^\bullet(M^+, M^-)\cong H^\bullet(D^n,\pa D^n). $$

The latter group is nontrivial only in dimension $n$ where it is isomorphic with the coefficient ring $\mathbb{Q}$. Now observe that anylinear map from $\bQ$ to any $\bQ$-vector space is either injective, or trivial.

Have a look at these notes for a Morse theory class I taught a few years ago. Maybe they could help.

share|improve this answer
Thank you for your answering of my first question. I have already been reading your book and find it quite helpful! –  xuxuzhu Oct 7 '12 at 17:53
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.