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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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6 votes
0 answers
162 views

Can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable, PL or DIFF manifold, if $X_4$ is a non-triangulable manifold? [duplicate]

Question: If $X_4$ is a non-triangulable topological (TOP) manifold, can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable manifold? can $X_4 \times S^1$, $X_4 \...
wonderich's user avatar
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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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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
4 votes
1 answer
2k views

On Thurston's triangulations of sphere

I have two questions from Thurston's paper [1]. In the paper [1], Thurston talks about classifying certain classes of triangulations of the sphere. Here a triangulation of a sphere a Topological ...
hrkrshnn's user avatar
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