Assume $M1,M2\subseteq \mathbb R^n$ are two smooth submanifolds such that $M1\subseteq M2$. How can I prove that $dim(M1)\le dim(M2)$?

What I know: $M \subseteq \mathbb R^n$ is a k-dimensional submanifold if for every $p \in M$ there is an open set $U \subseteq \mathbb R^n$ which is a neighborhood of p, an open set $V \subseteq \mathbb R^k$, and a smooth and regular map $r:V \to \mathbb R^n$ such that $r:V \to r(V)$ is a homeomorphism and $M \cap U=r(V)$.

I know that the dimension of a submanifold is well defined but I don't see how it can help me, because $M1$ and $M2$ are two different submanifolds.


1 Answer 1


I think it's simpler if you work with the definition of submanifolds given in Differential Forms and Applications by Do Carmo:

$\textbf{Definition.}$ Let $M^m$ and $N^n$ be differentiable manifolds. A differentiable map $\phi: M \longrightarrow N$ is an $\textit{immersion}$ if $d \phi_p: T_pM \longrightarrow T_{\phi(p)}N$ is injective for all $p \in M$. If, an addition, $\phi$ is a homeomorphism onto $\phi(M) \subset N$, where $\phi(M)$ has the topology induced by $N$, $\phi$ is an $\textit{embedding}$. If $M \subset N$ and the inclusion map $i: M \longrightarrow N$ is an embedding, we say that $M$ is a $\textit{submanifold}$ of $N$.

I don't know what definition of manifold you are using, but I will use the definition of Do Carmo's book, which is given here.

Now, let be $\{ (U_{\alpha},f_{\alpha}) \}$ a differentiable structure for $M^m$,
$\{ (V_{\alpha},g_{\alpha}) \}$ a differentiable structure for $N^n$ and consider the inclusion map $i: M \longrightarrow N$. Since $M \subset N$ is a submanifold, $di_p: T_pM \longrightarrow T_pN$ is injective for all $p \in M$. If $q = f_{\alpha}^{-1}(p)$, then follows that

$$\dim_{\mathbb{R}} \mathbb{R}^n = \dim_{\mathbb{R}} \text{Ker} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right) + \dim_{\mathbb{R}} \text{Rank} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right)$$

by the Rank-Nullity Theorem. Thus,

$$\dim_{\mathbb{R}} \mathbb{R}^m = \dim_{\mathbb{R}} \text{Rank} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right) \leq \dim_{\mathbb{R}} \mathbb{R}^n,$$

because $d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q$ is injective and $\text{Rank} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right)$ is a subspace of $\mathbb{R}^n$. Since $\dim_{\mathbb{R}} \mathbb{R}^n = \dim_{\mathbb{R}} T_pN$ and $\dim_{\mathbb{R}} \mathbb{R}^m = \dim_{\mathbb{R}} T_pM$ (by the definition of manifold given by Do Carmo), we have that $\dim_{\mathbb{R}} T_pM \leq \dim_{\mathbb{R}} T_pN$.

  • $\begingroup$ Thanks for the answer, I appreciate it. Though I guess I don't have the knowledge to understand your solution, at least for now. Anyway, I managed to solve the problem on my own-it took very long time (and my solution is pretty long) but in the end I succeeded. But still thanks for answering, this can help other users who will read it. $\endgroup$
    – Mark
    Oct 9, 2018 at 14:20

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