Assume we have a Riemannian manifold $(M,g)$, and then define the "cylinder" $M' := M\times\mathbb{R}$. Then $M'$ together with the product metric $g'_p : (v_1,r_1),(v_2,r_2)\mapsto g(v_1,v_2)+r_1\cdot r_2$ for $v_1,v_2\in T_p M$ and $r_1,r_2\in\mathbb{R}$ is again a Riemannian manifold.
Now, let $(e_1,...,e_n,e_r)$ be a $g'$-orthonormal base of $T_{(p,r)}M'$, so that $(e_1,...,e_n)$ is a $g'$-orthonormal basis of $T_pM$, and denote by $\nabla$ the Levi-Civita connection on $M'$.
Does it then hold that $\nabla_{e_i} e_r = 0$ for all $i\in\{1,...,n\}$?
My thought process is that since they are basis vector fields, one should have $[e_i,e_r]=0$ for the commutator, and thus $$\nabla_{e_i} e_r = \nabla_{e_r} e_i + [e_i,e_r] = \nabla_{e_r} e_i = 0,$$ where the last equality follows (intuitively) because the $e_i$ do not change into the radial direction $e_r$.