Welcome to MathOverflow

Question

2

I know that, in a manifold of dimension $\geq$ 5,there can exist polyhedra P and Q that are homeomorphic but not piecewise-linear homeomorphic. Can this happen if P and Q are compact subsets of $R^{n}$ and the homeomorphism maps $R^{n}$ to itself?

edited Mar 23 2012 at 3:35

asked Mar 23 2012 at 1:32

Ernest Davis
85●3

3

Suppose that $P, Q$ are $k$-dimensional polyhedra in ${\mathbb R}^n$ which are homeomorphic but not PL homeomorphic. If $2k+2\le n$ then the homeomorphism $f: P\to Q$ extends to a homeomorphism ${\mathbb R}^n\to {\mathbb R}^n$ by H.Gluck's theorem 1.3 from his "Embeddings in the trivial range" paper (Annals of Math., 1965). Gluck's result deals with embeddings to general manifolds, so the case when the target is ${\mathbb R}^n$ may have been known earlier. – Misha Mar 23 2012 at 4:42

Misha · Accepted Answer · 2012-03-23 13:23:04Z UTC

The answer is: Yes, this can happen. Suppose that $P, Q$ are $k$-dimensional polyhedra in ${\mathbb R}^n$ which are homeomorphic but not PL homeomorphic. If $2k+2\le n$ (and you can always increase the dimension of the ambient Euclidean space) then the homeomorphism $f:P\to Q$ extends to a homeomorphism ${\mathbb R}^n\to {\mathbb R}^n$ by theorem 1.3 from H.Gluck's "Embeddings in the trivial range" (Annals of Math., 1965). Gluck's result deals with embeddings to general manifolds, so the case when the target is ${\mathbb R}^n$ may have been known earlier.

If H is a homeomorphism from $R^{n}$ to itself, and P and H(P) are compact polyhedra, is H(P) piecewise-linear homeomorphic to P?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.