14
$\begingroup$

Let $X$ and $Y$ be two path-connected $n$-dimensional manifolds. Then one can construct their connected sum $X\# Y$ which can be visualised as removing a disc from $X$, removing a disc from $Y$, and gluing the boundaries together using a cylinder (a copy of $S^{n-1}\times[0,1]$).

There is a natural $(n-1)$-sphere in $X\# Y$ due to the cylinder. If $X$ or $Y$ is a sphere, then this $(n-1)$-sphere will bound. Is this the only case in which it bounds?

If $X\# Y$ is a non-trivial connected sum (i.e. neither $X$ nor $Y$ is a sphere), is the map $S^{n-1} \to X\# Y$ essential (i.e. not nullhomotopic)?

Oscar Randal-Williams' answer to this question shows that the inclusions of the sphere into $X\setminus D^{\circ}$ and $Y\setminus D^{\circ}$ are essential.

If the answer is yes, then $\pi_{n-1}(X\# Y) \neq 0$. I don't even know if this is true.

If $X \# Y$ is a non-trivial connected sum of $n$-dimensional manifolds, is $\pi_{n-1}(X\# Y)$ necessarily non-zero?

I believe the answer to both of these questions is yes, but I have not been able to completely convince myself.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ At the level of generality you've asked your question, the answer is no. Let $X$ and $Y$ be inverse homotopy spheres, then $X \# Y$ is an $n$-sphere even though the summands are not. You can do this in any dimension where there are non-trivial homotopy spheres. $\endgroup$
    – Ryan Budney
    Commented Oct 4, 2017 at 5:13
  • 2
    $\begingroup$ @RyanBudney: the question does not seem to be about smooth manifolds. $\endgroup$
    – Oscar Randal-Williams
    Commented Oct 4, 2017 at 13:19

1 Answer 1

Reset to default
15
$\begingroup$

The answer to this is no, in a stronger sense than Ryan's comment above. There are two sources, Kahn, Peter J. Codimension-one imbedded spheres (Invent. Math. 10 1970 44–56), and an appendix Null-homotopic, codimension-one embedded spheres, to a paper of Terng-Thorbergsson (Taut Immersions into Complete Riemannian Manifolds in Tight and taut submanifolds. MSRI Publications, 32, 1997) that I wrote, not knowing Kahn's result. The main theorem of the appendix states:

Suppose that $\imath:S^{n-1} \to M^n$ is a smooth embedding which is null-homotopic. Then either $S$ bounds a homotopy ball on one side, or {1.} M is a rational homology sphere, and therefore X and Y are as well. {2.} The fundamental groups of both X and Y are finite, and at least one of them is trivial.

In the appendix, there are constructions of connected sums of rational homology spheres where the summing sphere is null-homotopic that show that this result is essentially best possible.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Do you know of an example where the second possibility occurs? $\endgroup$
    – Michael Albanese
    Commented Oct 4, 2017 at 18:46
  • $\begingroup$ All the examples I know are in the papers I cited; it's been a long time so I don't remember the constructions. The appendix that I wrote is on my web page people.brandeis.edu/%7eruberman/drpapers/taut.final.gz. I think Kahn's paper also gives concrete examples. I'm actually confused a bit by your question; I think that what I intended was that either there's a homotopy ball on one side, or (1 and 2 hold). $\endgroup$
    – Danny Ruberman
    Commented Oct 4, 2017 at 19:48
  • $\begingroup$ My bad. I didn't realise that the alternative to bounding a homotopy ball on one side is 1 and 2. $\endgroup$
    – Michael Albanese
    Commented Oct 4, 2017 at 21:56

You must log in to answer this question.

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