# Hints for an exercise on Morse theory

Exercise: Let $M$ be a $3$-dimensional smooth manifold with boundary $\partial M$ which is a surface of genus $g$. Moreover let $f:M\longrightarrow [0,1]$ be a Morse function with the following properties:

1. $f(\partial M)$ is a regular value.
2. Lets denote with $\mu_i(f)$ the number of critical points of $f$ of index $i$, then $\mu_0(f)=1$ and $\mu_2(f)=\mu_3(f)=0$

Prove that $M$ is connected and determine $\mu_1(f)$.

I tried different approaches to solve it; here you can see my wrong reasonings:

$1)$ Lets reconstruct partially the Morse-Smale complex (over $\mathbb Z/2\mathbb Z$). We have one critical point of index $0$, therefore $C_0(M)\cong\mathbb Z/2\mathbb Z$ and in the same way we can conclude that $C_2(M)=C_3(M)=0$. For compact manifolds we have the following theorem:

Let $M$ be a compact manifold of dimension $n$, then the dimension as vector space of the Morse-Smile homology group $H_n(C^\bullet)$ (or $H_0(C^\bullet)$) is the number of connected components of $M$.

In my case $H_3(C^\bullet)=0$ since $C_3(M)=C_4(M)$ so maybe I'm on the wrong way. Probably I can't use the theorem for two reasons: $M$ has boundary and it is not compact.

$2)$ I tried to use the Morse inequalities but I have two main obstacles, $M$ ha boundary and I don't know the number of critical points of index $1$ (I have to find them).

$3)$ I don't know if I can use the fact that $\chi(M)=0$ if $\text{dim}(M)$ is odd because I've seen this theorem only for compact manifolds without boundary.

$4)$ If $M$ where without boundary I could conclude that it has the homotopy type of a CW-complex with only one $0$-cell. At this point I should prove that this type of $CW$-complexes are connected (it is easy).

I don't know how to use the fact that the homology of $\partial M$ is known, because I don't understand how to relate it with the homology of $M$.

On each connected component, $f$ admits a minimum and a maximum. Let $M_1$ the connected component of $M$ containing the boundary (it is connected, by assumption). If $f(\partial M)$ is a minimum, the maximum of $f$ restricted to this component is achieved at a critical point and $\mu_3(f)\geq 1$. So $f(\partial M)$ is not a minimum and $f$ has a minimum on $M_1$, achieved at a critical point. As $\mu_0(f)=1$, $f$ cannot have other connected components, and $M=M_1$ is connected. Now let $N$ be the double of $M$ glued along the boundary $\chi (N)=2\chi (M)- \chi(\partial M)$. Therefore $0=2\chi (M)- (2-2g)$, $\chi(M)=-g+1= 1-\mu_1+0-0$, $\mu_1=g$
• Does the equation $\chi(M)=\sum_{i=0}^{\text{dim}(M)} (-1)^i\mu_i$ hold also for manifolds with boundary? I've seen it used only for closed manifolds... – Dubious Dec 30 '15 at 12:01
• Certainly. When passing through a critical point of index $i$, you glue a disk (Euler characteristic 1) along a $i-1$ sphere (Euler characteristic $0$ if $i-1$ is odd ($i$ even), $2$ if it is even. So you obtain a space with Euler characteristic +1 if $i$ is even, $-1$ if $i$ is odd. Here we use $\chi(A\cup B)=\chi(A)+\chi(B)-\chi (A\cap B)$ – Thomas Dec 30 '15 at 13:18
• BTW, the argument above also works if $\partial M$ is not connected. If $M_1$ has two connected components in its boundary, a mountain path argument would prove that there exists an critical point of Morse index $2$ : joint the two connected components by a path in $M_1$, and maximize the minimum. At this point the index is $dim M-1=2$. – Thomas Dec 30 '15 at 13:23
• Dear @Thomas look at this question (the comments): math.stackexchange.com/questions/1039302/… It seems that the equation which relates $\chi(M)$ with the critical points doesn't hold in presence of boundary – Dubious Dec 30 '15 at 21:23
• So, I think there is a problem when you write $\chi(M)=1-\mu_1+0+0$. Maybe in this special case works because $f$ is constant on the boundary – Dubious Dec 30 '15 at 21:26