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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.

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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

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  • $\begingroup$ i am assuming V is supposed to be the Jacobian, since you mention Vf = 0, and what is U? $\endgroup$ – masszz Nov 5 '13 at 7:58
  • $\begingroup$ Sorry about the notation, $U$ and $V$ are both vectors fields. $\endgroup$ – user99914 Nov 5 '13 at 8:00
  • $\begingroup$ And $Vf = L_V f$. $\endgroup$ – user99914 Nov 5 '13 at 8:01
  • $\begingroup$ got it. Thanks you very much. One more thing, i am asked to find the Hessian in local coordinates, now i just wrote the Hessian matrix (matrix of second partials), is this generally a correct notation? $\endgroup$ – masszz Nov 5 '13 at 8:04
  • $\begingroup$ You are correct only when you are at a critical point. $\endgroup$ – user99914 Nov 5 '13 at 8:08

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