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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!

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    $\begingroup$ Please define first what you call the Hessian matrix. $\endgroup$
    – abx
    Commented Feb 8, 2023 at 10:40
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    $\begingroup$ 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$. $\endgroup$ Commented Feb 8, 2023 at 11:28
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    $\begingroup$ 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. $\endgroup$ Commented Feb 8, 2023 at 14:31
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    $\begingroup$ @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 $\mathbb{R}^n$. $\endgroup$ Commented Feb 8, 2023 at 18:10

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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.

Now, the existence of a homothety vector for a complete Riemannian metric implies that the metric is flat.

Finally, in the flat coordinates the equation $\nabla_i \nabla_j\phi = \delta_{ij}$ can be immediately solved

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