Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm learning the concepts of immersions at the moment. However I'm a bit confused when they define an immersion as a function $f: X\rightarrow Y$ where $X$ and $Y$ are manifolds with dim$X <$ dim$Y$ such that $df_x: T_x(X)\rightarrow T_y(Y)$ is injective.

I was wondering why don't we let $f$ be injective and say that's the best case we can get for the condition dim$X <$ dim$Y$(since under this condition we can't apply the inverse function theorem)?

Also does injectivity of $df_x$ inply the injectivity of $f$ (it seems that I can't prove it)?

How should we picture immersion as (something like the tangent space of $X$ always "immerses" into the tangent space of $Y$)?

Thanks for everyone's help!

share|improve this question
3  
"Does injectivity of $\mathrm{d}f_x$ imply the injectivity of $f$?" Consider the standard covering map $f:\mathbb{R}^1 \to \mathbb{S}^1$... –  Willie Wong May 27 '13 at 11:01
    
"How should we picture...?" Locally immersions are imbeddings. Near a point $p$ in the domain and near $f(p)$ in the image, there are coordinates under which the (restriction of the) immersion is the standard imbedding of Euclidean space into a Euclidean space of equal or higher dimension: $(x_1,..., x_m) \mapsto (x_1,...,x_m, 0,...,0)$. There is an important subtelty here: the image of this restriction may not be open in the image of the whole domain. This is what distinguishes imbeddings from injective immersions. –  Tim kinsella May 27 '13 at 11:53

2 Answers 2

up vote 2 down vote accepted

Think of a particle moving around a figure 8 with nowhere zero speed. This parametric curve gives you an immersion $f\colon\mathbb R \to\mathbb R^2$ that is not injective. If you restrict the domain to make it a bijection (which you can do), the image is not a submanifold but is called an immersed submanifold.

share|improve this answer
    
Hi Prof. Shifrin! Thanks a lot for your great answer! –  Evariste Jun 1 '13 at 5:51

I think a glimpse on the wikipedia's article helps. Immersions usually are not injective because the image can appear "knotted" in the target space.

I do not think you need $\dim X<\dim Y$ in general. You can define it for $\dim X=\dim Y$, it is only because we need $df_{p}$ to have rank equal to $\dim X$ that made you "need" $\dim X\le \dim Y$.

I think you can find in classical differential topology/manifold books (like Boothby's ) that an immersion is locally an injection map $$\mathbb{R}^{n}\times \{0\}\rightarrow \mathbb{R}^{m+n}$$You can attempt to prove this via inverse function theorem or implicit function theorem. The proof is quite standard.

share|improve this answer
    
Thanks a lot! Great answer! –  Evariste Jun 1 '13 at 5:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.