A Riemannian manifold $(M, g)$ has harmonic Weyl curvature iff its Schouten tensor is Codazzi, and if there exists $f: M \to \mathbb{R}$ such that $W(\bullet, \bullet, \bullet, \nabla f) = 0$, one says $(M, g, f)$ has zero radial Weyl curvature. Some papers show that under certain conditions in addition to zero radial curvature the manifold decomposes as a warped product (see, for instance, this paper). I'm interested in the reciprocal: if $M = B \times_h N$ is a warped product (which does not reduce to a Riemannian product) with harmonic Weyl curvature (which, in particular, implies that $N$ is Einstein), is it true that there exists a non constant $f: M \to \mathbb{R}$ (or non constant $f: B \to \mathbb{R}$) such that $W(\bullet, \bullet, \bullet, \nabla f) = 0$? I think it might be useful to suppose $B = I \times_{\tilde{h}} \tilde{N}$ itself is a warped product over a one-dimensional base and $h$ depends only on $I$.
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$\begingroup$ Have you computed the case where $B$ is one dimensional? Harmonic Weyl is equivalent to the Cotton tensor vanishing. The Ricci curvature of a warped product with one dimensional $B$ and Einstein $N$ can be computed very explicitly. So can the Riemann and Weyl curvatures. Whether what you are looking for holds should be fairly obvious. $\endgroup$– Willie WongCommented Dec 6, 2023 at 16:17
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$\begingroup$ I thought about that but I haven't had the time yet to do these computations, I'm temporarily obligated to other stuff at the moment. But I do suspect if $B$ is one-dimensional, then $W(\bullet, \bullet, \bullet, \nabla h) = 0$. $\endgroup$– Matheus AndradeCommented Dec 6, 2023 at 17:27
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