The Unapologetic Mathematician

Regular and Critical Points

Let $f:M^m\to N^n$ be a smooth map between manifolds. We say that a point $p\in M$ is a “regular point” if the derivative $f_{*p}$ has rank $n$; otherwise, we say that $p$ is a “critical point”. A point $q\in N$ is called a “regular value” if its preimage $f^{-1}(q)$ contains no critical points.

The first thing to notice is that this is only nontrivial if $m\geq n$. If $m then $f_{*p}$ can have rank at most $m$, and thus every point is critical. Another observation is that is $q\notin f(M)$ then $q$ is automatically regular; if its preimage is empty then it cannot contain any critical points.

Regular values are useful because of the generalization of the first part of the implicit function theorem: if $q$ is a regular value of $f:M\to N$, then $A=f^{-1}(q)\subseteq M$ is a topological manifold of dimension $m-n$. Or, to put it another way, $A$ is a submanifold of “codimension” $n=\dim(N)$. Further, there is a unique differentiable structure for which $A$ is a smooth submanifold of $M$.

Indeed, let $(V,y)$ be a coordinate patch around $q$ with $y(q)=0$. Given $p\in A$, pick a coordinate patch $(U,x)$ of $M$ with $x(p)=0$. Let $\pi_1:\mathbb{R}^m\to\mathbb{R}^n$ be the projection onto the first $n$ components; let $\pi_2:\mathbb{R}^m\to\mathbb{R}^{m-n}$ be the projection onto the last $m-n$ components; an let $\iota_2:\mathbb{R}^{m-n}\to\mathbb{R}^m$ be the inclusion of the subspace whose first $m$ components are $0$.

Now, we can write down the composition $y\circ f\circ x^{-1}$. Since this has (by assumption) maximal rank at $0\in\mathbb{R}^m$, the implicit function theorem tells us that there is a coordinate patch $(W,h)$ in a neighborhood of $0$ such that $y\circ f\circ x^{-1}\circ h=\pi_1\vert_W$. So we can set $\tilde{W}=\pi_2(W)$, which is open in $\mathbb{R}^{m-n}$, and get

$\displaystyle y\circ f\circ x^{-1}\circ h\circ\iota_2\vert_{\tilde{W}}=\pi_1\circ\iota_2\vert_{\tilde{W}}=0$

Setting $z=x^{-1}\circ h\circ\iota_2\vert_{\tilde{W}}$ we conclude that $z(\tilde{W})\subseteq A$, since all these points are sent by $f$ to the preimage $y^{-1}(0)=q$.

Now we claim that $z(\tilde{W})$ is not just any subset of $A$, but in fact $z(\tilde{W})=A\cap x^{-1}(h(W))$. Clearly $z(\tilde{W})$ is contained in this intersection, since

$\displaystyle z(\tilde{W})=x^{-1}(h(\iota_2(\tilde{W})))=x^{-1}(h(W\cap(0\times\mathbb{R}^{m-n})))$

On the other hand, if $\tilde{p}$ is in this intersection, then $\tilde{p}=x^{-1}(h(u))$ for a unique $u\in W$ — unique because $x$ and $h$ are both coordinate maps and thus invertible — and we have

$\displaystyle0=y(f(\tilde{p}))=y(f(x^{-1}(h(u))))=\pi_1(u)$

meaning that the first $n$ components of $u$ must be $0$, and thus $u\in(0\times\tilde{W})$. Thus $\tilde{p}\in z(\tilde{W})$.

Therefore $z$ maps $\tilde{W}\subseteq\mathbb{R}^{m-n}$ homeomorphically onto a neighborhood of $p\in A$ in the subspace topology induced by $M$. But this means that $(z(\tilde{W}),z^{-1})$ acts as a coordinate patch on $A$! Since every point $p\in A$ can be found in some local coordinate patch, $A$ is a topological manifold. For its differentiable structure we’ll just take the one induced by these patches.

Finally, we have to check that the inclusion $\iota:A\to M$ is smooth, so $A$ is a smooth submanifold — that its differentiable structure is compatible with that of $M$. But this is easy, since at any point $p$ we can go through the above process and get all these functions. We check smoothness by using local coordinates $x$ on $M$ and $z^{-1}$ on $A$, concluding that $x\circ\iota\circ(z^{-1})^{-1}=h\circ\iota_2$, which is clearly smooth.

April 21, 2011 - Posted by | Differential Topology, Topology

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