Prove that the set of polynomials is a vector space
Proving this is a subspace.
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0I 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. – 2012-04-01
2 Answers
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!)
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0Yes, but the OP was hav$i$ng trouble gett$i$ng 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 – 2012-03-26
What you need is the inverse function theorem: http://en.wikipedia.org/wiki/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.
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2You'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 $\phi(u,v,x_3,x_4,x_5)=\cos(u),\cos(2v),\sin(2v)+x_3,\sin(u)\cos(v)+x_4,\sin(u) \sin (v)+x_5)$ – 2012-03-26