# Hessian quadratic form is well defined

Could someone show why for the Hessian to be well defined ($d_{p}^{2}f(v,w) = L_{v}L_{w}f$) we need $p$ to be a critical point.

Let $H_p(U, V) = UV f(p)$. Then $H_p(U, gV) = U(gV f) = (Ug)(Vf) + gUVf = gH_p(U, V)$ as $Vf=0$ at $p$. Thus $H_p$ is $C^\infty$ linear and defines a quadratic form. Note that similar argument shows that $H_p$ is symmetric
• Sorry about the notation, $U$ and $V$ are both vectors fields. – user99914 Nov 5 '13 at 8:00
• And $Vf = L_V f$. – user99914 Nov 5 '13 at 8:01