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What is immersion and submersion at the intuitive level. What can be visually done in each case?

Watson
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Koushik
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    Both locally look like standard maps $\mathbb{R}^m \to \mathbb{R}^n$ – Yuri Vyatkin Feb 09 '13 at 08:56
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    they're conditions on induced maps of tangent spaces...so in an immersion $X\rightarrow Y$, $\dim X\leq\dim Y$, you're just allowing self-intersections of $X$ inside of $Y$, but you're not allowed to "compress" any of $X$ in such a way that the tangent space of a point goes to 0 (because the map must be injective on tangent spaces). For a submersion, you want the induced map to be surjective, so intuitively you're "crumpling up" $X$ to fit into $Y$ in such a way as to hit every possible curve in $Y$ with a curve in $X$. – gmoss Feb 09 '13 at 21:17
  • Related: http://math.stackexchange.com/questions/1214630/ – Watson Mar 05 '17 at 11:27

2 Answers2

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First of all, note that if $f : M \to N$ is a submersion, then $\dim M \geq \dim N$, and if $f$ is an immersion, $\dim M \leq \dim N$.

The Rank Theorem may provide some insight into these concepts. The following statement of the theorem is taken from Lee's Introduction to Smooth Manifolds (second edition); see Theorem $4.12$.

Suppose $M$ and $N$ are smooth manifolds of dimensions $m$ and $n$, respectively, and $F : M \to N$ is a smooth map with constant rank $r$. For each $p \in M$ there exist smooth charts $(U, \varphi)$ for $M$ centered at $p$ and $(V, \psi)$ for $N$ centered at $F(p)$ such that $F(U) \subseteq V$, in which $F$ has a coordinate representation of the form $$\hat{F}(x^1, \dots, x^r, x^{r+1}, \dots, x^m) = (x^1, \dots, x^r, 0, \dots, 0).$$ In particular, if $F$ is a smooth submersion, this becomes $$\hat{F}(x^1, \dots, x^n, x^{n+1}, \dots, x^m) = (x^1, \dots, x^n),$$ and if $F$ is a smooth immersion, it is $$\hat{F}(x^1, \dots, x^m) = (x^1, \dots, x^m, 0, \dots, 0).$$

So a submersion locally looks like a projection $\mathbb{R}^n\times\mathbb{R}^{m-n} \to \mathbb{R}^n$, while an immersion locally looks like an inclusion $\mathbb{R}^m \to \mathbb{R}^m\times\mathbb{R}^{n-m}$.

Michael Albanese
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  • What conditions are sufficient to ensure an immersion has a submersion as an inverse\dual? I'm considering principle G-bundles P over a (pseudo) Riemannian manifold M. We usually think of this case as a submersion, but is there a away to generate an immersion $M\rightarrow P$ since the bundle is generated by M itself? It would seem like you get 1 to 1 when you consider a G-valued (gauge) field everywhere, but I'm new to this – R. Rankin Oct 14 '22 at 06:23
  • @R.Rankin: If an immersion has an inverse, then it is a bijection, and hence a diffeomorphism (see this question). I don't know what you mean by "the bundle is generated by $M$ itself". Are you thinking of the map $M \to P$ as a section of $P \to M$? Such a thing exists if and only if the bundle is trivial. If my comments don't address your questions sufficiently, you should ask them in a separate post. – Michael Albanese Oct 15 '22 at 03:20
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My answer builds upon Michael Albanese's answer. I'm referring to the same textbook and use the same notation:

It might be worth noting, that in these same charts U and V in which the local representation $ \hat{F} $ of a map $F$ of constant rank $r$ has the form given by

$ \hat{F}(x^1,\dots,x^r,x^{r+1},\dots,x^m)=(x^1,\dots,x^r,\underbrace{0,\dots,0}_{n-r \, times}) $,

its total derivative $ D\hat{F}(p) $ at a point $ p \in U $ has exactly the same form

$ D\hat{F}(p)(x^1,\dots,x^r,x^{r+1},\dots,x^m)=(x^1,\dots,x^r,\underbrace{0,\dots,0}_{n-r \, times}) $,

as can be easily seen (this is a generalization of the fact that the derivative of the identity map is the identity map). Thus, the local representation $ D\hat{F}(p) $ of the differential $ dF_p $ is exactly the same as the local representation $ \hat{F} $ of $F$. By the way, the maps are linear and represent the canonical form of a linear map given in Theorem B.20 on p.626 of John Lee's book.

The same holds for the two special cases of an immersion and a submersion, respectively. Both locally look exactly like their differentials.

Roland Salz
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