# Inequality about dimensions of submanifolds

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.

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 $$\dim_{\mathbb{R}} T_pM \leq \dim_{\mathbb{R}} T_pN$$.

• 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.
– Mark
Oct 9, 2018 at 14:20