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12 votes
3 answers
872 views

Can triangulations (or some related combinatorial structure) distinguish smooth structures on $RP^4$?

There are exotic versions of $RP^4$, constructed by Cappell-Shaneson, which are homeomorphic but not diffeomorphic to the standard $RP^4$. One way to distinguish them is via the $\eta$ invariant of $...
Joe's user avatar
  • 545
6 votes
0 answers
209 views

If $X_d$ is a non-triangulable manifold, can $X_d \times T^k$, $X_d \times I^k$, or $X_d \times \mathbb{R}^k$ be a triangulable manifold?

If $X_d$ is a non-triangulable manifold, can $X_d \times T^k$, $X_d \times I^k$, or $X_d \times \mathbb{R}^k$ always be a triangulable manifold? Let $X_d$ be a $d$-manifold which is NOT a ...
wonderich's user avatar
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20 votes
0 answers
540 views

Homeomorphisms of the sphere mapping a geodesic triangulation to another one

Consider the standard Riemannian 2-sphere $S$, equipped with a geodesic triangulation $T$. Let $L(S,T)$ be the space of homeomorphisms of $S$ which map $T$ to a geodesic triangulation. What is the ...
François Laudenbach's user avatar
7 votes
1 answer
260 views

Bordism for oriented triangulable manifolds without smooth differentiable structures

We know the bordism group for oriented smooth differentiable structures such as $\Omega_d^{SO}$ that requires the special orthogonal group structure on the tangent bundle $TM$ of manifold $M$. $$\...
wonderich's user avatar
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4 votes
0 answers
106 views

Retriangulating manifolds via triangulations of low local complexity

Suppose that $M$ is closed, connected PL $n$-manifold. We say that a triangulation of $M$ has local complexity at most $L$ if every zero-cell of $T$ meets at most $L$ $n$-simplices. (An alternative ...
Sam Nead's user avatar
  • 28.2k