# How is Hessian tensor on Riemannian manifold related to the Hessian matrix from calculus?

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 $\mathbb{R}^{n}$.

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.

• But then the first term of the Hessian, $X(Yf)$, would be enough to qualify it as an "invariant hessian", since this alone yields the elements of the Hessian matrix. Commented Aug 23, 2019 at 17:26
• @gemini I was thinking the same thing initially, but that is incorrect, because the statement made is "the Christoffel symbols... with respect to the usual coordinates on $\Bbb{R}^n$ are all zero". But note that the $\Gamma$'s can be zero in one coordinate system but non-zero in another (since they are NOT the component functions of a tensor field). So, we really need the whole expression. Another way of writing this definition of is that $\text{Hess}(f) = \nabla (\nabla f)$, where this is an equality of $(0,2)$ tensor fields (and $\nabla$ is say the Levi-civita connection of a metric) Commented Jan 3, 2020 at 20:03

That Hessian matrix is useful for analyzing the behaviour of critical points of $$f$$, where $$df_p = 0$$. More precisely, let $$M$$ be a smooth manifold and $$\nabla$$ be a linear connection in $$TM$$. If $$f\colon M \to \Bbb R$$ is a smooth map, $$p \in M$$ is a critical point of $$f$$, and $$(x^i)_{i=1}^n$$ is a system of coordinates around $$p$$, we have the local expression $${\rm Hess}\,f_p(\partial_i\big|_p,\partial_j\big|_p) = \partial_i\big|_p(\partial_j f) - \require{cancel} \cancelto{0}{df_p(\nabla_{\partial_i}\partial_j)} = \frac{\partial^2f}{\partial x^i\partial x^j}(p),$$and so $${\rm Hess}\,f_p = \sum_{i,j=1}^n \frac{\partial^2f}{\partial x^i\partial x^j}(p) \,dx^i\big|_p\otimes dx^j\big|_p$$

At a generic point $$p$$, the bilinear map $${\rm Hess}\,f_p$$ depends on the choice of connection (because of the $$\nabla$$ in the second term). But if $$p$$ is a critical point, then the map $${\rm Hess}\, f_p$$ becomes independent of the choice of connection. If $$p$$ is not critical, one usually picks $$\nabla$$ to be the Levi-Civita connection of some Riemannian metric in $$M$$. But a priori, you can use any connection (which can lead to some awkward things, such as the Hessian not being symmetric if $$\nabla$$ has torsion).

• 1) I'm saying that if ${\rm Hess}^\nabla(f)(X,Y) \doteq X(Y(f)) - {\rm d}f(\nabla_XY)$, $\nabla'$ is another connection, and $p$ is a critical point of $f$, then ${\rm Hess}^\nabla(f)_p = {\rm Hess}^{\nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric). Commented Jan 31, 2019 at 4:38
• Sure, no problem. Commented Jan 31, 2019 at 4:41
• Thanks, I suggested an edit which is now in the edit queue. Commented Jan 31, 2019 at 4:49
• Thank you, your edit was very good, I have already approved it :-) Commented Jan 31, 2019 at 4:52