Does a cone in $\mathbb{R}^3$ admit a warped geometry? Let
$$C = \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 \leq r^2 z^2 \text{ and } z > 0 \}$$
be a certain convex cone in $\mathbb{R}^3$, where $r > 0$.
Let $D$ the the closed unit disk in $\mathbb{R}^2$. Is it possible to find a Riemannian metric $g$ on $D \times (0, \infty)$ of the form $g = f(t)^2 dx^2 + dt^2$ ($dx^2$ is the flat metric on $D$ and $dt^2$ is the usual metric on $\mathbb{R}$) such that $C$ is isometric to $(D \times (0, \infty), g)$?
 A: Here is an answer to the title "does the cone admit a warped product metric" and the answer is yes:
One natural wraped product on $C$ will be induced by a spherical section. In $\mathbb{R}^3$, the polar decomposition gives a natural metric on $\mathbb{R}^3 \setminus \{0\}$ which is
$$
\mathrm{d}r^2 + r^2 \overset{\circ}{g}
$$
where $\overset{\circ}{g}$ is the round metric of the unit sphere. It gives an isometry between $(0,\infty)\times \mathbb{S}^2$ and $\mathbb{R}^3\setminus\{0\}$.
As $C$ is a natural riemannian submanifold of $\mathbb{R}^3 \setminus \{0\}$, in polar coordinates, it is isometric to $(0,\infty) \times ({C\cap\mathbb{S}^2})$. Note that $C\cap \mathbb{S}^2$ is a "round" disk. Thus, one can find a diffeomorphism $f:D \to C\cap \mathbb{S}^2$ and pull-back the round metric on $D$.
One then has an isometry $(0,\infty) \times D \to C$, in which metric is the warped metric
$$
\mathrm{d}r^2 + r^2 f^*\left(\overset{\circ}{g}|_{C\cap \mathbb{S}^2} \right)
$$
So one has a wraped metric on $(0,\infty)\times D$. But indeed, the metric on $D$ here is not flat (it has sectionnal curvature $1$).
Now, here is a remark, more than an answer, to the question "can we give a product warped metric that involves the flat metric on the cone":
Note that if $\mathrm{d}r^2 + f(r)g_D$ is a warped product of the cone involving the flat metric, in these coordinates $(r,p) \in (0,\infty)\times D$, the level surfaces $\{r_0\}\times D$ are orthogonal to the geodesics $(0,\infty)\times\{p_0\}$. Then, the $"r"$ coordinate will not be the distance to the origin in $\mathbb{R}^3$. Otherwise, these radial geodesics in $\mathbb{R}^3$ will be orthogonal to a same flat disk, which is not true. This shows that there exists in the cone a flat disk orthogonal to a familly geodesics in $\mathbb{R}^3$, and the cone would be foliated by a family of geodesic rays all orthogonal to the same flat disk. This gives me the intuition that the answer will be "no", but I do not have a formal proof for this right now, and I may be wrong.
