It is known that the existence of a function with hessian proportional to the metric implies that the metric is a warped product metric. Is the reciprocal true as well? I.e, if $(B \times N, g = g_B + h^2 g_N)$ is a warped product, do there always exist functions $\varphi, \psi$ such that $\nabla^2 \varphi = \psi g$?
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Yes. (With the caveat that if there is a function such that $\nabla^2\varphi = \psi g$, then necessarily it is a warped product over a one-dimensional base, so your question should really require $\dim B=1$.)
Consider the warped product $(B^1 \times N, g := dt^2 + h^2(t) g_N)$. Let $f=f(t)$. A direct computation shows that $$ \nabla^2f = f^{\prime\prime}(t) dt^2 + (hh^\prime f^\prime)(t) g_N . $$ It follows that if $(\ln f^\prime)^\prime = (\ln h)^\prime$, then $\nabla^2f = f^{\prime\prime} g$. The ODE is easy to solve.
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$\begingroup$ Thanks! But why must the base be one-dimensional? I can't see the reason. $\endgroup$ Commented Dec 1, 2023 at 1:52
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$\begingroup$ @MatheusAndrade in the accepted answer to the question you linked to, there is a reference. But here's a quick discussion. Suppose $\nabla^2\varphi = \psi g$, let $X = \nabla\varphi$ the gradient vector field, multiplying both sides by $X$ you find $\nabla_X X = \psi X$ and hence $X$ is geodesic. Let $Y,Z$ be such that $Y(\varphi) = Z(\varphi) = 0$, then $\langle \nabla_Y X,Z\rangle = \psi \langle Y,Z\rangle$ tells us that the level sets of $\varphi$ are totally umbilic. So unless $\psi = 0$ you cannot expect the base to have higher dimensions. $\endgroup$ Commented Dec 1, 2023 at 2:42
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$\begingroup$ And when $\psi = 0$ the example of $\mathbb{R}^n$ shows that it can also be a warped product over a higher dimensional base, but it is still definitely a warped product over a one dimensional base. $\endgroup$ Commented Dec 1, 2023 at 2:44
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1$\begingroup$ The above is not a full proof; it is an argument providing a moral justification. The key is that the base manifold embeds as a totally geodesic submanifold of the warped product. So even if the level sets of $\varphi$ factor, the fact that these leaves have extrinsic curvature indicates that you cannot group part of it together into the base. $\endgroup$ Commented Dec 1, 2023 at 3:00
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1$\begingroup$ @MatheusAndrade I posted a full proof of the warped product decomposition at the other question. Hopefully that helps you. $\endgroup$ Commented Dec 1, 2023 at 4:12