# k-dimensional Regular Submanifold of k-dimensional manifold

The statement is: Prove any $$k$$-dimensional regular submanifold of a $$k$$-dimensional manifold is an open subset. The hint is show that any submersion is an open map.
A regular sub-manifold $$S$$ of $$M$$ is such that $$\forall p \in S$$, there is a chart $$(U ,\phi)$$, $$\phi : U \rightarrow V$$ such that $$\phi(U \cap S) = (\mathbb{R}^{k} \times \{ 0 \}^{n-k}) \cap V$$. A submersion of manifolds $$f : M \rightarrow N$$ is a smooth map with $$D_{p}(f) : T_{p} M \rightarrow T_{f(p)}N$$ surjective $$\forall p \in S$$.

I don't really have much intuition for either of these concepts. Assuming the hint is true. How do I show that the regular submanifold must be an open set. I am thinking that both the manifolds having dimension $$k$$ implies some submersion between them, and perhaps this would show that the submanifold is an open set, but I am rather confused.

## 1 Answer

The inclusion map is a submersion, since the derivative is an isomorphism at each point, hence a surjection. (Presumably you've proved in class or as an exercise that submersions are always open maps. In this case, you could apply the inverse function theorem directly.)

• So $f : S \rightarrow M$ is the inclusion map. I just want to know in more detail why $f$ is a submersion. What exactly does it mean for the derivative to be an isomorphism? Does it mean that for any vector $w$ in the tangent space of $f(p)$, there is a vector in the tangent space of $p$ such that $D_{p}(f)(v) = w$? Is $f$ an isomorphism because $D_{p}(f) : T_{p}(M) \rightarrow T_{f(p)}(M) = T_{p}(M)$ is the identity map? Also, where do we use the assumption that $S$ is $k$-dimension and regular? Is the inclusion map not a surjection without these assumptions, and if so why? – user100101212 Feb 14 '20 at 0:06
• The derivative of the inclusion map $M\hookrightarrow N$ is in general an injection $T_pM\to T_pN$. Since the manifolds have the same dimension, an injective linear map becomes an isomorphism. – Ted Shifrin Feb 14 '20 at 0:14
• So are we appealing to the fact that an injective linear map between vector spaces of equal finite dimension is an isomorphism? What is the dimension of $T_{p}(S)$, where $p \in S$. Is it $k-1$? Are we saying for fixed $p \in S$, $T_{p}(S)$ is isomorphic to $T_{p}(M)$ because the derivative is an injective linear map between vector spaces of equal dimension? This then would yield that the inclusion map is a submersion. Lastly, where do we use the assumption that $S$ is a regular submanifold (do we need it)? – user100101212 Feb 14 '20 at 1:08
• There can't be a non-regular submanifold when they have the same dimension, but the definition allows you to check immediately that the inclusion map is a local diffeomorphism. – Ted Shifrin Feb 14 '20 at 1:55