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**5**

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111 views

### Does there always exist a sequence of handle moves between handle decompositions that does not increase index? (+ ref. request)

Reference request: Firstly, I'm looking for a proof of the following well-known result about handle decompositions:
($\ast$) Given two handle decompositions of a smooth $n$-manifold $M$, there ...

**2**

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**0**answers

128 views

### Is complex surface in CP(3) a two handlebody?

Consider a complex surface given by homogeneous equation in $\mathbb{C}P^3$. Without loss of generality, take
\begin{equation}
S = \{[x:y:z:w] \in \mathbb{C}P^3~ |~ x^d + y^d + z^d + w^d = 0\}
...

**12**

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**2**answers

557 views

### Does every compact manifold exhibit an almost global chart

Let $M$ be a compact connected manifold.
Is there a chart $\Psi:U \to \mathbb{R}^n$ such that the closure of $U$ is $M$?
This is true for $S^n, T^n, K$, all compact surfaces, etc.
If it is not true in ...

**3**

votes

**2**answers

191 views

### Handlebody decomposition of a 3-manifold adapted to a link

Given a compact connected 3-manifold $M$ with non-empty boundary, and a link $L \subset M$, is there a handlebody decomposition of $M = H^0 \cup (\cup_i H^1_i) \cup \{\text{2-handles}\}$ such that:
...

**5**

votes

**1**answer

308 views

### What are Kirby diagrams of candidate exotic 4-manifolds?

It is an open problem whether there exist smooth manifolds homeomorphic, but not diffeomorphic to the standard $S^4$. The same is true for the 4-torus and several other manifolds. Handle ...

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146 views

### Uniqueness of the Smooth Structure on a Handle Attachment [closed]

I posted this question on math stack exchange and didn't receive an answer. If it is too elementary for this forum I will be happy to delete it.
Let $M^m$ be a smooth manifold with boundary. We may ...

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**2**answers

563 views

### Self-indexing Morse functions on non-compact manifolds

Hi,
given a compact manifold M we can always alter a given Morse function f to a self-indexing one (i.e., one where every critical point c has $f(c) = \operatorname{index}(c)$) - a proof of this may ...