> 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?

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Let $X_d$ be a $d$-manifold which is *NOT* a triangulable manifold but only a topological manifold. 


Question:
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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)?)

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Other warm-up info:
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- 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.