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This is an old exam problem at my school:

Let $F\colon M\to\mathbb{R}^k$ be a smooth map of smooth manifolds, with coordinate functions $F^1,\dots,F^k$. Let $c\in\mathbb{R}^k$ be a regular value of $F$, and $C=F^{-1}(c)$. If $f\colon M\to\mathbb{R}$ is smooth, prove $f|_C$ has a critical point at $p$ iff there exist $a_i$ such that $$ df_p=\sum a_idF^i_p. $$

In one direction, suppose $df_p=\sum a_idF^i_p$. I know $d(f|_C)_p\colon T_pC\to\mathbb{R}$, so $p$ is a critical point if $d(f|_C)_p$ is identically zero. But $d(F^i|_C)_p\equiv 0$ since $F^i$ is constant on $C$. It follows then that $d(f|_C)_p\equiv 0$.

On the other hand, if $p$ is a critical point of $d(f|_C)_p$, then $df_p$ is zero on $T_pC$. I know $C$ has codimension $k$ in $M$, but I'm not sure how to finally get that $$ df_p=\sum a_idF^i_p $$ for some $a_i$. I think $T_pC$ is a subspace of codimension $k$ in $T_pM$ as well, do the $dF^i_p$ form a basis or something of the complement of $T_pC$ or something? Thanks.

YN Chew
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1 Answers1

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Since $c$ is a regular value, $C=\{x\in M: F(x)=c\}$ is a smooth manifold, and the tangent space is $T_pC=\{u\in T_pM: d_pF(u)=0\}$ (quite naturally the equation of the tangent vectors is the derivative of the equation of the manifold). In other words, $$ T_pC=\ker d_pF=\ker d_pF^1\cap\cdots\cap\ker d_pF^k\subset T_pM, $$ and these $k$ linear forms are independent ($c$ is a regular value), hence the codimension $k$. Now $p$ is a critical point of $f|C$ if and only if $d_p(f|C)=(d_pf)|T_pC\equiv0$. In other words, if and only if the linear form $d_pf$ vanishes on the above intersection. Linear algebra tells us this happens if and only if that form is a linear combination of the others. You have in your argument one half of this equivalence. Note that these $a_i$'s are nothing but Lagrange multipliers of the conditioned critical point.

Jesus RS
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