# If there exists a function on a Riemannian manifold such that its Hessian matrix is the identity matrix?

In Euclidean space $$\mathbb{R}^n$$, $$n\geq 2$$, the Hessian matrix of the function $$\frac{|x|^2}{2}$$ is the identity matrix. While on a smooth manifold $$(M^n, g)$$, do there exists a function on $$(M^n, g)$$ such that its Hessian matrix is the identity matrix? Welcome some examples, thanks!

• Please define first what you call the Hessian matrix.
– abx
Commented Feb 8, 2023 at 10:40
• Sorry! Let $\nabla$be the Levi-Civita connection of $(M, g)$. For the differentiable function $v$ on $(M, g)$, define $\nabla v$ as the gradient of $v$. Use $\nabla^ 2 v$ to represent the Hessian matrix of $v$. Locally, it can be expressed as $\nabla_ {i j} v=\nabla_ i\left(\nabla_j v\right)-\Gamma_ {i j}^k \nabla_ k v$. Commented Feb 8, 2023 at 11:28
• It is known that the existence of a (non trivial) function satisfying $\mathrm{Hess V} = V g$ characterize the hyperbolic space (as far as I remember this is proven by X. Wang in "On the uniqueness of the AdS spacetime"). I would not be surprised if the existence of a function satisfying $\mathrm{Hess f} = g$ characterize the Euclidean space. Commented Feb 8, 2023 at 14:31
• @RomainGicquaud: Actually, there are many (incomplete) Riemannian manifolds $(M^n,g)$ that admit a function $f$ that satisfies $\mathrm{Hess}(f) = g$. For example, let $(N^{n-1},h)$ be any Riemannian manifold, let $M= \mathbb{R}^+\times N$, and let $g = \mathrm{d}r^2 + r^2\,h$ be the usual cone metric for $(N,h)$. Then $f = \tfrac12 r^2$ has $\mathrm{Hess}(f) = g$. However, $(M,g)$ won't be complete, and including a point for $r=0$ gives a complete smooth metric only if $(N,g)$ is a unit sphere in $\mathbb{R}^n$. Commented Feb 8, 2023 at 18:10

If a manifold is complete, the existence of the function $$\phi$$ such that $$\nabla_i \nabla_j\phi = g_{ij}$$ implies that the metric is flat and that in a `flat' coordinate system such that the metric is $$(dx^1)^2 +...+ (dx^n)^2$$ the functions is $$\frac{|x|^2}{2} + const$$.
In order to show the flatness, observe that the vector field $$\nabla^i\phi$$ is the homothety vector field.
Indeed, it is known that the Lie derivative of the metric with respect to the vector field $$v^i$$ is given by $$({\mathcal L}g)_{ij}=\frac{1}{2}(\nabla_i v_j + \nabla_j v_i)$$; substituting $$v_i= \nabla_i \phi$$ inside proves the claim.
Finally, in the flat coordinates the equation $$\nabla_i \nabla_j\phi = \delta_{ij}$$ can be immediately solved