Hessian matrix vs differential 2-form Could someone clarify the convention that the second derivative of a scalar function $f: \Bbb R^n \rightarrow \Bbb R$ is sometimes defined as a linear operator $D^2f : \Bbb R^n \rightarrow L(\Bbb R^n, L(\Bbb R^n, \Bbb R))$, which is also identified with the Hessian matrix (filled with second order partial derivatives), and sometimes as a differential 2-form?
It seems that both happens in vector calculus. 
Is the definition of derivative just context dependent, and the first example is just a case of "standard derivative" and the second one of the "exterior derivative"?
Or am I really confused about something?
 A: The differential 2-form $df \wedge df$ is always $0$, which is due to that the exterior product (or wedge product) is alternating. The intuition is as follows. One can think of $d x_1 \wedge d x_2$ as the ``oriented area element'' along the directions of $x_1$ and $x_2$. If $x_1 = x_2$, they don't span an area and thus $dx_1 \wedge d x_2$ is zero.
The Hessian matrix of $f$ at $x$ is the second total covariant derivative of $f$ at $x$. The total covariant derivative describes the rate of change along all directions of a local canonical basis. For example, the total covariant derivative of a scalar function at $x$ is a vector in the tangent space at $x$, which gives the gradient of $f$ at $x$. The total covariant derivative of the gradient field is then the Hessian field.
More concretely, if we are in the Euclidean space $\mathbb{R}^n$, the total covariant derivative of $f$ is $\nabla f = (\partial_1 f, \partial_2 f, \cdots, \partial_n f)$, which is the rate of change of $f$ along the directions of the canonical basis. The Hessian is then $ \nabla^2 f = [ \partial_i \partial_j f]_{1 \le  i,j \le n}$, which gathers the rate of change of $ \partial_j f $ along the directions $\partial_i$.
