4
$\begingroup$

I need to reference the following result. Do you know a good source?

The following conditions on an $n$-dimensional simplicial complex $S$ are equivalent: a) $S$ is an $n$ manifold; b) The link of every vertex of $S$ is homeomorphic to the $(n - 1)$-sphere; c) The link of every $k$-simplex is homeomorphic to the $(n - k - 1)$-sphere.

$\endgroup$
3
  • $\begingroup$ I think this title is a little misleading as the term "simplicial manifold" is usually reserved for a simplicial object in the category of manifolds. $\endgroup$ May 10, 2010 at 11:49
  • $\begingroup$ Changed the title. $\endgroup$ May 10, 2010 at 15:01
  • $\begingroup$ To get a simple correct statement in the category implied by the tags, (a) should refer to S as a combinatorial n-manifold; and (b) and (c) should refer to the links in question as being PL homeomorphic to the appropriate sphere. $\endgroup$ May 10, 2010 at 17:49

2 Answers 2

5
$\begingroup$

The usual term for objects like this is "combinatorial manifolds".

However, the result is not quite true as you have stated. Definitely b is true if and only if c is true, and b implies a. However, a does not imply b. There definitely exist simplicial complexes which do not satisfy b or c but which are topological manifolds. For example, the famous double suspension theorem of Cannon (weaker versions were proved by Edwards) says that if $X$ is a homology $n$-sphere, then the space $Y$ obtained by suspending $X$ twice is homeomorphic to the $(n+2)$-sphere. If neither $X$ nor the suspension of $X$ is an actual sphere (for instance, this holds if $X$ is the Poincare homology sphere), the vertices of $Y$ corresponding to the suspension points will then not satisfy b.

$\endgroup$
2
  • $\begingroup$ Thanks! I saw an analogous result for homology manifolds and generalized homology spheres, and having heard that b) implies a) guessed that the same should hold. $\endgroup$ May 10, 2010 at 15:00
  • 4
    $\begingroup$ One needs to be a little more careful in this mixed category of triangulated topological manifolds. Consideration of the triple suspension of a triangulated homology 3-sphere shows that (b) does not imply (c). $\endgroup$ May 10, 2010 at 17:46
0
$\begingroup$

I'm pretty sure you can find this kind of results in Rourke and Sanderson's "Introduction to Piecewise linear topology".

http://www.amazon.com/Introduction-Piecewise-Linear-Ergebnisse-Mathematik-Grenzgebiete/dp/0387111026

$\endgroup$
1
  • $\begingroup$ It has parts of the claim as exercises on p. 22 and p. 24. $\endgroup$ May 10, 2010 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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