Reconstructing a manifold from critical points I am teaching theoretical calculus this semester, and on the last discussion section we were discussing critical points of functions. I explained the idea of Morse theory, and a student of mine asked me a question that I couldn't answer. I don't know a lot about the Morse theory, so the question might actually be easy. I would really appreciate if you can help me, or at least give me a reference.
Suppose you are given an ordered set of signatures (i.e. number of $+$ and $-$ of the hessian) $\{(a_1,b_1),\dots,(a_r,b_r)\}$, that is supposed to be a set of critical points of some Morse function on a would be a $k$-manifold. The question is the following:

When there exists a manifold with a Morse function having a given set of signatures of critical points?

It is easy to see that the set of signatures must have signatures of the form $(k,0)$ and $(0,k)$, since any function on a compact manifold must have minimum and maximum.
Also, we can't start with, say, $(k-1,1)$, since you must start with the point of minimum, which must be of signature $(k,0)$.
Also, it is not true that we can always construct a manifold with given ordered set of signatures. For example, take the set $\{ (2,0) , (1,1), (0,2) \}$. Following the algorithm, first we attach a 0-cell, then we attach a 1-cell. Topologically it will be equivalent to a letter U made out of a tube (cylinder). But then you need two "caps" to make it into a closed compact thing, but we have only one critical point left. 
I have no idea what are the conditions when we can actually construct a required manifold.
I've heard (but I am not sure if it is true) that if we have passed a sertain number of critical points in out reconstructing algorithm (maybe more than $r/2+1$), then there is unique way to finish the procedure to get a closed compact manifold. If this is correct, is it still true that we can always get a manifold having any set of the first $r/2+1$ signatures? 
Thank you very much!
 A: Here's a lemma that partially answers the question for odd-dimensional manifolds. It's certainly not the case that $n_0,\ldots, n_k$ uniquely determines $n_{k+1},\ldots, n_{2k+1}$ though, since you can always make an equivalent handle decomposition by introducing a canceling pair of critical points. The even case is trickier since there exists a middle dimension.
Lemma: Let $n_0,\ldots, n_k$ be a sequence of nonnegative integers with $n_0>1$. Then there is a $2k+1$-manifold with $n_i$ critical points of index $i$ for $i\leq k$.
Proof:
We build a manifold with boundary by attaching handles. We start with $n_0$ 0-handles. Then we attach $n_1$ $1$-handles in some way. In general, we want to be attaching the $i$-handles along regions $\partial(D^{i})\times D^{n-i}$, and in order to proceed, we need to know such regions exist in the boundary. Since $i<2k+1$, it is easy to find an embedded $S^{i-1}$ inside the $2k$-dimensional boundary. Indeed you can find it inside some open set homeomorphic to a $(2k)$-ball. Then a regular neighborhood is isomorphic to $S^{i-1}\times D^{n-i}$. Okay, so in this way we've built up a manifold with boundary, $M$, that has all the correct critical points up to index $k$. Now take the double of this manifold $D(M)=M\cup_{\partial M}M$. Critical points of index $i$ turn into critical points of index $2k+1-i$ when you turn $M$ upside down, so $D(M)$ has critical points of index $n_0,\ldots, n_k,n_k,n_{k-1},\ldots n_0$. $\Box$ 
