1
$\begingroup$

If there is a function $f : M \to \mathbb R$ then the critical point is given as a point where

$d f = 0$

$df$ being 1-form (btw am I right here?). Is there a coordinate independent formulation of a criteria to determine if this point is a local maximum or minimum (or a saddle point)?

  • 0
    There are coordinate free definitions of second derivatives. Positive/negative definiteness of this object is a coordinate free criterion.2012-07-18

2 Answers 2

2

The generalization of the Hessian matrix to functions on smooth manifolds is

$H(X,Y) = X (Yf) - df (\nabla_X Y)$

where $X$ and $Y$ are vector fields, i.e. the Hessian is a bilinear form. The definition for positive/negative definiteness for bilinear forms is the usual one. $H$ is positive definite if

$H(X,X)>0$

for all vector fields $X$, and similarly for negative definite. As usual, a critical point is a local maximum if $H$ is negative definite, a local minimum if $H$ is positive definite, and a saddle otherwise.

2

Let $p$ be a critical point for a smooth function $f:M\to \mathbb{R}.$
Let $(x_\,\ldots,x_n)$ be an arbitrary smooth coordinate chart around $p$ on $M.$
From multivariate calculus we know that a sufficient condition for $p$ to be a local maximum (resp. minimum) of $f$ is the positiveness (resp. negativeness) of the Hessian $H(f,p)$ of $f$ at $p$ which is the bilinear map on $T_pM$ defined locally by $H(f,p)=\left.\frac{\partial^2f}{\partial x_i\partial x_j}dx^i\otimes dx^j\right|_p,$ here the Einstein convention on summation is working.

However, as Thomas commented, the Hessian of a function at a critical point has a coordinate-free espression.
Infact, $H(f,p): T_pM\times T_pM\to\mathbb{R}$ is characterized by $H(f,p)(X(p),Y(p))=(\left.\mathcal{L}_X(\mathcal{L}_Y f))\right|_p$ for any smooth vector fields $X$ and $Y$ on $M$ around $p.$


Note that without a Riemannian metric on $M$ you cannot invariantly define the Hessian of a function at a non-critical point.

  • 0
    and we know the signature of a quadratic form is co-ordinate independent by Sylvester's law of inertia.2012-07-18