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A local maximum of a smooth function $f:M\rightarrow \mathbb{R}$ is a critical point of $f$.

Attempt: Let $p\in M$ be a local maximum of $f$. Let $X_p\in T_pM$. Let $c:(-\epsilon,\epsilon)\rightarrow M$ be curve in $M$ such that $c(0)=X_p$ and $c'(0)=X_p$. Note, $0\in c^{-1}(U)$. Moreover, for any $q\in c^{-1}(U)$, $(f\circ c)(0)=f(p)\geq f(c(q))$ . Therefore $0$ is a local maximum of $f\circ c$ and therefore, $(f\circ c)'(0)=0$. (This is the standard, high school derivative)

Now, let $f_{*,p}: T_pM\rightarrow T_{F(p)}M$ be the differential of $f$ at $p$. Observe that

$f_{*,p}(X_p)=f_{*,p}(c'(0))=(f_p\circ c_0)_*(\frac{d}{dt}|_0)$ by the standard chain rule.

We can easily show that $(f_p\circ c_0)_*(\frac{d}{dt}|_0)=f_{*,p}(X_p)=0$ (linear map).

How can I conclude that the partial derivatives are 0 given a chart $(U,\phi)$ about $p$?

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    $\begingroup$ You already showed $p$ is a critical point (because $f_{*,p} = 0$, i.e it is not a surjective linear transformation $T_pM\to T_{f(p)}\Bbb{R}$). Anyway, to show $\dfrac{\partial f}{\partial x^i}(p) = 0$, consider the curve $c(t) = \phi^{-1}(\phi(p) + t e_i)$, where $e_i \in \Bbb{R}^n$ is $0$ everywhere, $1$ in $i^{th}$ slot (and also recall that $\dfrac{\partial f}{\partial x^i}(p)$ really means $\partial_i(f\circ \phi^{-1})_{\phi(p)}$, in words, you take the $i^{th}$ partial derivative of the chart representative function $f\circ \phi^{-1}$ and evaluate at $\phi(p)$). $\endgroup$ – peek-a-boo Jul 14 at 7:53
  • $\begingroup$ right, if the codomain of a linear map is one dimensional then the linear map is either surjective or 0. Thank you though, I really wanted to see how to do it with partial derivatives. $\endgroup$ – orientablesurface Jul 14 at 7:56
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    $\begingroup$ right, the "partial derivative" $\dfrac{\partial}{\partial x^i}(p) \in T_pM$ is precisely the tangent vector to the curve $c(t) = \phi^{-1}(\phi(p) + te_i)$ at $t=0$, i.e it is $c_{*,0}(1)$ (the push-forward of the unit tangent vector $1 \in T_0\Bbb{R} \cong \Bbb{R}$), and your calculation shows that the derivative along any curve is $0$. $\endgroup$ – peek-a-boo Jul 14 at 8:03

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