Let $M = \{(x, y, z, w) \in \mathbb{R}^4 \ | \ x^4 + y^4 + z^2 + w^2 = 1\}$ and let $f:M \rightarrow \mathbb{R}$ be given by $f(x, y, z, w) = x^3 - z.$ Then it is clear (I have already proven that is) that $M$ is a manifold. What I am having trouble doing is finding the critical points of $f$. I can find them if $f: \mathbb{R}^4 \rightarrow \mathbb{R}$. Is this helpful?
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You should look for the points of $M$ where the gradient of $f$ is perpendicular to the tangent plane. If $M=\{g=0\}$ this means that the gradient of $f$ is parallel to the gradient of $g$.
Emanuele Paolini
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What is $g$? Not sure what that means. Thanks. – dunkindonuts Jul 08 '13 at 20:53
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Also how do you take the gradient of $f$? $f$ isn't a function from $\mathbb{R}^4$. – dunkindonuts Jul 08 '13 at 20:55
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$g(x,y,z,w)=x^4+y^4+z^2+w^2$ and $f(x,y,z,w)=x^3-z$. – Emanuele Paolini Jul 08 '13 at 21:08
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$f$ is the restriction of a function on $\mathbb{R}^4$. – Neal Jul 09 '13 at 15:59
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In general, if we have a smooth function $f:M\to N$ between manifolds, a point $p\in M$ is called critical if the differential $\mathrm{d} f_p :T_p M\to T_{f(p)} N$ fails to be surjective. In particular, if we have a smooth function $f:X\subseteq \mathbf{R}^n \to Y\subseteq\mathbf{R}^m$, by definition his critical values are that points $p\in X$ for which the jacobian $J_{\tilde f}(p)=0$, where $\tilde f$ is a smooth extension of $f$ to an open euclidean set containing $X$.
Now, in the case of a function $f:M\to \mathbf{R}$, calling $\tilde f$ the extesions the differential acts by euclidean dot product $\mathrm{d}f_p:T_p M\to \mathbf{R}$ as $\mathrm{d}_p (v)=\langle \nabla \tilde f(p),v\rangle$; so differential is not surjective only if $\nabla \tilde f$ vanisches identically.
