- A smooth manifold $M$ is a topological manifold with compatible smooth atlas (in the following all manifolds are assumed to be $n$-dimensional, smooth, oriented, closed, and without boundary.)
- A critical point $q \in M$ of a smooth function $f: M \rightarrow \mathbb{R}$ is a zero of the differential $d f$.
- The Hessian $H_{f}(q)$ of $f$ at a critical point $q \in M$ is the matrix of second derivatives.
How to show Hessian is independent of co-ordinate system at critical point?
For instance : If $f(x,y) = x^2-y^2$ then $f_x = 2x$ and $f_y = -2y$ which are zero at $(0,0)$ and $f_{xx}=2$ and $f_{yy}=-2$ and $f_{xy}=f_{yx}=0$. Now if we change coordinates to $u = x-y$ and $v=x+y$ so that $f(u,v) = uv$ one has still $f_u = v, f_v = u$ which are zero at $(0,0)$. But now $f_{uu}= f_{vv}=0,f_{uv}=1=f_{vu}$. It is clear Hessian matrix is not same. Also its determinet not same. Although both detereminent are non vanishing. I do not know in which sense Hessian is independent of co-ordinate system at critical point?
Any help or hint will be highly appreciated.