# Local maximum and critical points smooth manifolds surjectivity

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 $$p$$?

• 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 $\phi(p)$). – peek-a-boo Jul 14 at 7:53
• 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. – orientablesurface Jul 14 at 7:56
• 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 $0$. – peek-a-boo Jul 14 at 8:03