TITLE: Definitions of Hessian in Riemannian Geometry
QUESTION [55 upvotes]: I am wondering if there is any quick way to see the following two definitions of Hessian coincide with each other without using local coordinates.

$\operatorname{Hess}(f)(X,Y)= \langle \nabla_X \operatorname{grad}f,Y \ \rangle$; and
$\operatorname{Hess}(f)(X,Y)=X (Yf) - (\nabla_XY) f$.

REPLY [3 votes]: On one hand, you can define the Hessian without mentioning the gradient (so you don't have to raise or lower indices):
First, define the covariant derivative of a $1$-form $\theta$ as follows: It should satisfy the "product rule",
$$ X\langle \theta, Y\rangle = \langle \nabla_X\theta, Y\rangle + \langle \theta, \nabla_XY\rangle. $$
Therefore, $\nabla_X\theta$ is the $1$-form satisfying
$$ \langle \nabla_X\theta, Y\rangle = X\langle\theta,Y\rangle - \langle\theta,\nabla_XY\rangle. $$
It's easy to check that this is a $2$-tensor.
The Hessian is simply the covariant derivative of $df$. In particular,
$$ \langle \nabla^2f, X\otimes Y\rangle = \langle \nabla_Xdf, Y\rangle = X\langle df,Y\rangle - \langle df,\nabla_XY\rangle $$
On the other hand, the gradient of $f$ is defined by its property that for any vector $Y$,
$$
\langle df, Y\rangle = g(\nabla f, Y),
$$
where $g$ is the Riemannian metric. Therefore, by the "product rule",
$$
Xg(Z,Y) = g(\nabla_XZ,Y) + g(X,\nabla_XY),
$$
it follows that
$$
\langle\nabla^2f,X\otimes Y\rangle = Xg(\nabla f, Y) - g(\nabla f,\nabla_XY) = g(\nabla_X\nabla f, Y)
$$