Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 115 / 2

An immersed submanifold of $M$ is a manifold $N$ with an injective immersion $\iota :N\to M$. It is an embedded submanifold when $\iota$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding.

Because the inclusion $\iota :N\hookrightarrow M$ has constant rank $n$, the constant rank theorem gives coordinates $y^1,\dots ,y^n$ on $N$ and $x^1,\dots ,x^m$ on $M$ in whichSince $\iota$ is an embedding, the ambient chart can be shrunk so that it meets no other local sheet of $N$. It then satisfiesand $(x^1,\dots ,x^n)$ restricts to the required chart on $N$. This is a slice chart for an embedded submanifold.

Suppose the immersed subset is not embedded. Using the assumed embedded neighborhoods, there are $p\in N$, a relatively small coordinate neighborhood $U\ni p$, and points $p_j\in N\setminus U$ with $p_j\to p$ in $M$. Choose a bump function $g$ on $N$, supported in $U$, with $g(p)=1$. Then $g(p_j)=0$.

If $g=f{|}_N$ for some smooth $f$ on $M$, continuity gives both $f(p_j)\to f(p)=1$ and $f(p_j)=0$, a contradiction. Thus the extension hypothesis forces the subspace and manifold topologies to agree locally, and the immersion is an embedding. This proves the smooth extension criterion for an immersed submanifold.

If $X$ is tangent to $N$ and $f{|}_N=0$, then the restriction of $f$ to every curve in $N$ is zero, so $(Xf){|}_N=0$.

Conversely, use a slice chart for an embedded submanifold. The functions $x^{n+1},\dots ,x^m$ vanish on $N$. Writing $X=X^i{\partial }_i$, the hypothesis givesThus $X$ has no normal component and is tangent to $N$. Equivalently, $X$ preserves the vanishing ideal of an embedded submanifold.

The Lie bracket of vector fields is the commutator of the corresponding derivations:For $X=X^i{\partial }_i$ and $Y=Y^i{\partial }_i$,

Let $I(N)$ be the vanishing ideal of an embedded submanifold. Tangency says $X(I(N)),Y(I(N))\subseteq I(N)$. Hence, for $h\in I(N)$,so the criterion from part (d) makes $[X,Y]$ tangent to $N$.

In adapted coordinates, the tangential coefficients of the displayed bracket use only the restrictions of the tangential coefficients of $X,Y$ and their derivatives along $N$; all normal coefficients vanish there. Consequentlyso the restriction depends only on $X{|}_N$ and $Y{|}_N$. This is tangency under the Lie bracket.