Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that the set of polynomials is a vector space

share|cite|improve this question
Are you trying to argue that $M$ is diffeomorphic to a $2$-manifold, or that $M$ is a two-dimensional submanifold of $\mathbb{R}^5$? If the former, you just need to produce charts around each $p\in M$, which should be provided by periodicity of $\phi$. If the latter, you need to produce submanifold charts about each $p\in M$. You would probably want to think about using the implicit function theorem for that. – Neal Mar 26 '12 at 2:18
Sorry, I'm not sure. Our definition is as follows: M is a k-manifold if for each $p \in M$, there exists an open set $V$ containing p and an open set $U \subset \mathbb R^k$ and a map $\phi: U \rightarrow V$ such that $\phi$ is $C^r$, $\phi$ is a continuous bijection with continuous inverse and $D\phi$ has rank k at each point in $U$ – AnalysisHelp Mar 26 '12 at 2:28
I think we're in the same class based on the questions you're asking (this one and the cartesian product one). I've been working on this assignment too and am also trying to finish up this problem. You want to get together? We could meet tomorrow evening in the classroom. – user28058 Apr 1 '12 at 2:08
up vote 1 down vote accepted

Your map is periodic of period $2\pi$ in both variables, so you can factor it through the quotient $\mathbb R^2\to\mathbb R^2/\mathbb Z^2$ to obtain a map $\phi:\mathbb R^2/\mathbb Z^2\to\mathbb R^5$ with exactly the same image.

Now this map $\phi$ is an immersion (because, as you observed, the original map has differential with full rank) and it is not hard to see that it is injective. Since its domain is compact, the image is a submanifold.

(Of course, one has to check this last claim!)

share|cite|improve this answer
This is all true and relevant, but the OP asks to show that the map is indeed a local diffeomorphism, which is assumed in the definition of immersion. To assume the concept of immersions is to assume the answer that the OP is looking for. – Brett Frankel Mar 26 '12 at 3:18
The question is «how do I prove the image of this map is a manifold,» rather—not that the map is a local diffeo. By immersion I mean simply that the differential is injective, as usual, and this is something the OP already knows. (Notice that the image of an injective immersion from a non-compact manifold need not be a submanifold... so one has to deal with non-compactness somehow...) – Mariano Suárez-Alvarez Mar 26 '12 at 3:21
Yes, but the OP was having trouble getting from injective differential to local diffeomorphism (existence of inverse function). To be fair, I do think that the OP assumed that the local existence of an inverse function is enough to prove that the image is manifold, an issue which you certainly did address. +1 – Brett Frankel Mar 26 '12 at 3:30

What you need is the inverse function theorem:

To satisfy the conditions of the theorem, you just need to check that the total derivative of your map never has rank equal to less than 2, which follows from the fact that $\det D\phi^T D\phi$ is always nonzero.

share|cite|improve this answer
Doesn't the inverse function theorem only apply to maps between $\mathbb R^n \rightarrow \mathbb R^n$ (between same dimensional spaces)? – AnalysisHelp Mar 26 '12 at 3:06
You can add a few extra dimensions onto your domain, for example by defining the function $\phi+x_3+x_4+x_5$ from $\mathbb{R}^5$ to $\mathbb{R}^5$. Now use the inverse function theorem to get a local diffeomorphism, and then once you have that you can restrict your attention to $\mathbb{R}^2$ and its image. – Brett Frankel Mar 26 '12 at 3:10
Sorry, but I still don't understand. Even if I somehow tack on $x_3, x_4, x_5$, I'll still have $det(D\phi) = 0$ The inverse function theorem we learned didn't go into cases where the rank was $\le n$ – AnalysisHelp Mar 26 '12 at 4:51
$\det D\phi=0$, but $\det D(\phi +x_3+x_4+x_5)\neq0$ – Brett Frankel Mar 26 '12 at 16:35
You're right, I could have said that better. My goal was to put a $3\times 3$ identity block in the bottom right of the derivative matrix. The explicit map I meant was $\phi(u,v,x_3,x_4,x_5)=\cos(u),\cos(2v),\sin(2v)+x_3,\sin(u)\cos(v)+x_4,\sin(u) \sin (v)+x_5)$ – Brett Frankel Mar 26 '12 at 18:08

Your Answer


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.