Coordinate free definition of hessian

[latentcorpse]

Let M be a manifold and $f: M \rightarrow \mathbb{R}$ be a smooth function such that df=0 at some point $p \in M$. Let $\{ x^\mu \}$ be a coordinate chart defined in a neighbourhood of p. Define

$F_{\mu \nu} = \frac{ \partial f}{ \partial x^\mu \partial x^\nu }$

By considering the transoformation law for components show that $F_{\mu \nu}$ defines a (0,2) tensor, the Hessian of f at p. Construct also a coordinate free definition and demonstrate its tensorial properties.

I don't really know what I'm supposed to do in order to demonstrate that this defines a (0,2) tensor?

 

[BigJDubs]

Could someone help me out? A (0,2) tensor is a tensor of type (0,2), which means that it is a covariant rank-2 tensor field. This can be defined coordinate free as a bilinear map from a vector space V to its dual V*, given by F : V \times V \rightarrow \mathbb{R}such that F(v,w) = F(w,v) and F(cv,w) = cF(v,w) for any v,w \in V and c \in \mathbb{R}. To demonstrate the tensorial properties of the Hessian of f at p, we can start by considering the transformation law for components. Let x'^{\mu'} be a new coordinate chart defined in a neighbourhood of p, then the components of the Hessian can be expressed as F'_{\mu' \nu'} = \frac{\partial f}{\partial x'^{\mu'} \partial x'^{\nu'}}. We can relate these components to the original components by the chain rule as F'_{\mu' \nu'} = \frac{\partial x^\mu}{\partial x'^{\mu'}} \frac{\partial x^\nu}{\partial x'^{\nu'}} F_{\mu \nu}. This shows that the Hessian of f transforms in the same way as a (0,2) tensor, i.e. it is a (0,2) tensor.