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Question: If $X_4$ is a non-triangulable topological (TOP) manifold,

  1. can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable manifold?

  2. can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a PL manifold?

  3. can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a smooth DIFF manifold?

Note that we have smooth (DIFF) ⊂ PL ⊂ triangulable ⊂ TOP.

(If $X_4$ spin or non-spin manifold makes a difference for the answer, then we should discuss the differences.) The $I^1$ means a 1-dimensional finite internal.


Let $X_4$ be a $4$-manifold which is NOT a triangulable manifold but only a topological manifold.


Other warm-up info:

  • If $X_4$ is the non-triangulable Freedman's E8 topological manifold, then $X_{4+𝑘}=X_4\times T^𝑘$ is triangulable, but not piecewise linear (PL).

  • Any orientable 5-manifold is triangulable.

This question is a more specific version of the previous one focusing on $d=4$ only.

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    $\begingroup$ I guess that the other question may not be answered, that is why I asked a specific simplified question. Thank you! (also I ask the PL, which we have: smooth $\subset$ PL $\subset$ triangulable. $\endgroup$
    – wonderich
    Commented Mar 18, 2021 at 0:45
  • $\begingroup$ It is more likely that this question can have an immediate answer by an expert. (The other question - I am not sure who can answer.) $\endgroup$
    – wonderich
    Commented Mar 18, 2021 at 0:46
  • $\begingroup$ This question is not simplified. On the contrary, it is one of the difficult cases of the other question. You have already got answers to the other one. $\endgroup$ Commented Mar 18, 2021 at 12:03

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