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How can I prove that if $p$ is a critical point of a real-valued fuction $f$, then there is a function $H:T_{(p)}M\to T_{(p)}M$, such that $H$ is bilinear, simetric and:

$\bullet$ $H(X_p, Y_p)=X_p(Yf)=Y_p(Xf)$ for all $X,Y$ vector fields.

$\bullet$ $H(\delta_i|_p,\delta_j|_p)=\frac{\delta^2f}{\delta^2\delta x^i\delta x^j}(p)$ relative to a coordinate system

$\bullet$ $H(v,v)=(\frac{d^2(f\alpha)}{ds^2})(0)$ if $\alpha'(0)=v$

Patricia Suarez Gonzalez
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1 Answers1

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Hint: You should have $H:T_pM\times T_pM\to\mathbb R$. In this form, you can use your first bullet point as a definition by showing that $X_p(Yf)$ depends only on $Y_p$. (Use that $(Yf)(p)=0$ by assumption.) Doing the same for the second expression and showing that the two agree (since $([X,Y]f)(p)=0$), you get the basic properties of $H$. Then verify that the other two bullet points are satisfied automatically.

Andreas Cap
  • 20,577
  • Hi, how can I show that $X_p(yf)$ depends only on $Y_p$? Im trying to prove that if $Y$ and $Z$ are two vector fields such that $Y_p=Z_p$ then $Yf$ and $Zf$ are iqual on a neighborhood of $p$ – Patricia Suarez Gonzalez Mar 23 '17 at 04:22
  • The easiest way to show this is by observing that $Y\mapsto X_p(Yf)$ is linear over smooth functions (i.e. for a smooth function $g$, you get $X_p((gY)f)=g(p)X_p(Yf)$. Imitating the arguments used in the characterization of tensor fields as multilinear operators, you then show that $Y(p)=0$ implies $X_p(Yf)=0$, which implies the result by linearity. (It is not true that for fields $Y$ and $Z$ such that $Y_p=Z_p$ the functions $Yf$ and $Zf$ are equal on a neighborhood of $p$, they only have the same value and derivative in $p$. – Andreas Cap Mar 23 '17 at 07:44