2
$\begingroup$

Suppose $n\ge d,$ and $f: \mathbb R^n\to \mathbb R^{n-d}$ is smooth, Then, $Z=f^{-1}(0)$ is a submanifold of dimension $d.$ Fix $a\in Z$ and suppose that $f'(a)$ is surjective. Then, $Ker f'(a)=T_aZ.$

Milnor's proof is a commutative diagram with a one line comment. I have seen this proof before, using paths; i.e., a definition of tangent space which does not coincide with Milnor's. So, I want to do the proof using only the definitions in Milnor's book. My attempt follows.

$g=f|_Z$ is a submersion, so it is equivalent to the canonical submersion $p:\mathbb R^d\to \left \{ 0 \right \}$ and so there is a chart $(U,\phi)$ on $\mathbb R^d$ such that $x\in U\Rightarrow g\circ \phi(x)=p(x)=0.$ Wlog $\phi(0)=a.$ By definition, $T_aZ=Im\ \phi'(0).$

On the other hand, $f$ is a smooth function defined on $\mathbb R^n$ that agrees with $g$ on $Z$ i.e. there is an open $W\in \mathbb R^n$ such that $f:\mathbb R^n\to \mathbb R^{n-d}$ that agrees with $g$ on $W\cap Z.$ Then, by definition, $g'(a)=f'(a)$.

Therefore, $g'(\phi(0))\circ\phi'(0)(x)=f'(\phi(0))\circ\phi'(0)(x)=f'(a)\circ\phi'(0)(x)=0.$

This says now that $T_aZ=Im\ \phi'(0)\subseteq Ker f'(a).$ The other direction follows by a dimensionality argument.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.