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Suppose $M$ and $N$ are manifolds of dimension $m$ and $n$ respectively. Let $F : N \rightarrow M$ be a $C^{\infty}$ map. Then $\dim T_{p}N = n $ and $\dim T_{F(p)}M = m$ , where $p \in N$. Now, rank of $F$ at $p$ is defined as the rank of differential $F_{*,p} : T_{p}N \rightarrow T_{F(p)}M$.

Here is my simple problem - I don't get it why do we define rank of $F$ in such a way? What is the need of introducing the concept of Differential here? Can't we simply say that the rank of $F$ is the $\dim F(N)$?

Dark_Knight
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1 Answers1

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The problem is that $F(N)$ is not necessarily a manifold, and hence $\dim F(N)$ is not well-defined in general.

But if $F$ is a smooth embedding, then $F(N)$ is a manifold and $\dim F(N)=\mathrm{rank} F$.

An standard example where $F(N)$ is not a manifold is the curve $$F:\mathbb{R}\to S^1\times S^1,\quad F(x)=(e^{2\pi i x},e^{2\pi i \alpha x})$$ where $S^1$ is the unit circle and $\alpha\in\mathbb{R}$ is irrational. The image $F(\mathbb{R})$ is a dense set in $S^1\times S^1$ which is not locally Euclidean. See here.

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Spenser
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