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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 triangulable manifold but only a topological manifold.

Question:

  1. Is this true that $X_d \times T^k$ can always be triangulable? (The $T^k$ is a $k$-torus.)

  2. Is this true that $X_d \times I^k$ can always be triangulable? (The $I^k$ is a finite width interval in $k$ dimensions.)

  3. Is this true that $X_d \times \mathbb{R}^k$ can always be triangulable? (The $R^k$ is a real space in $k$ dimensions.)

  • If not true, could $X_d \times T^k$, $X_d \times I^k$, or $X_d \times \mathbb{R}^k$ sometimes be triangulable? under what criteria? (for example, for a certain dimension $d$? for a certain bound on $k$? or when $X_d$ has a certain structure (like Spin)?)

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.

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    $\begingroup$ As long as $d \geq 7$ the answer is negative as a consequence of Theorems 1 and 2 of Galewski-Stern, Classifications of simplicial triangulations of topological manifolds (Annals 111, 1980, p.1-34). I suggest you read that paper if you haven't already. For reference, the claim about 5-manifolds may be found in Galewski-Stern's A universal 5-manifold with respect to triangulations. $\endgroup$
    – skupers
    Commented Mar 18, 2021 at 0:17

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