[Math] How is Hessian tensor on Riemannian manifold related to the Hessian matrix from calculus

differential-geometryriemannian-geometry

From calculus, given a smooth function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$, the Hessian matrix is the $n\times n$ matrix of second partials:

$${\rm Hess}(f)= \bigg( \frac{\partial^{2}f}{\partial x_{i} \partial x_{j}}\bigg)$$

On a Riemannian manifold, one way to define the Hessian tensor of a smooth function $f:M \rightarrow \mathbb{R}$ is by

$${\rm Hess}(f)(X,Y)=X(Yf)-df(\nabla_{X}Y)$$

where $X$ and $Y$ are smooth vector fields on $M$.

I would like to know what relationship the Hessian on a Riemannian manifold has with the Hessian matrix of a function on $\mathbb{R}^{n}$.

Best Answer

The relation between the Riemannian Hessian and the Hessian in Euclidean space is very simple - they are the same. More precisely, the Euclidean Hessian is a particular case of the more general Riemannian Hessian.

Let $(x_1,\ldots,x_n)$ be local coordinates on a neighborhood in a Riemannian manifold $M$, and let $\Gamma_{ij}^k$ denote the Christoffel symbols of the Levi-Civita connection with respect to these coordinates. Let us massage your definition for the Hessian; we consider the vector fields $$X=X^i\frac{\partial}{\partial x_i},\;Y=Y^i\frac{\partial}{\partial x_i}.$$ Then $$\begin{align}X(Y(f))-df(\nabla_XY)&=X^i\frac{\partial}{\partial x_i}\left(Y^j\frac{\partial f}{\partial x_j}\right)-df\left(\nabla_{X^i\frac{\partial}{\partial x_i}}Y^j\frac{\partial}{\partial x_j}\right)\\&=X^i\left(\frac{\partial Y^j}{\partial x_i}\frac{\partial f}{\partial x_j}+Y^j\frac{\partial^2f}{\partial x_i\partial x_j}\right)-df\left(X^i\left(\frac{\partial Y^j}{\partial x_i}\frac{\partial}{\partial x_j}+Y^j\Gamma_{ij}^k\frac{\partial}{\partial x_k}\right)\right)\\&=X^iY^j\frac{\partial^2f}{\partial x_i\partial x_j}-X^iY^j\Gamma_{ij}^k\frac{\partial f}{\partial x_k}.\end{align}$$Now, the Christoffel symbols of the Levi-Civita connection with respect to the usual coordinates on $\mathbb{R}^n$ are all zero. Hence, the Euclidean Hessian matrix of the function $f$ is just the matrix whose $(ij)$ entry is $$\mathrm{Hess}(f)_{ij}=\mathrm{Hess}(f)\left(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\right),$$where the right hand side is the Riemannian Hessian.

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