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An example in my book says that $f:\mathbb{R}\to S^1$ defined by $f(t)=(\cos(t),\sin(t))$ is a local but not global diffeomorphism.

By the inverse function theorem, $f$ is a local diffeomorphism if the determinant of $df_x$ is nonzero.

I must be doing something wrong. Isn't $$ df_x=\begin{bmatrix} -\sin x \\ \cos x\end{bmatrix}? $$

I was expecting a square matrix in order to take the determinant.

PSK
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    You've written $df_x$ for a map from $\mathbb{R}\rightarrow\mathbb{R}^2$, not $\mathbb{R}\rightarrow S^1$. You'll need to use 1-dimensional coordinates on $S^1$. – MartianInvader Sep 27 '13 at 20:17
  • @MartianInvader Does that mean something like $f(t)=e^{it}$, so $df_x=ie^{ix}\neq 0$? for all $x$? – PSK Sep 27 '13 at 20:20
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    It almost seems that to get $1$-dimensional coordinates on $S^1$, you already have the construction of your local homeomorphism. – Lubin Sep 27 '13 at 20:26
  • If you think it's ok to use $i$ here, I'd go back and double-check the definition of a differentiable manifold. – MartianInvader Sep 27 '13 at 20:38
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    If you've established that $S^1$ is a submanifold of $\mathbb R^2$, then you have local slice coordinates $(r,\theta)$ so that $S^1 = {r = 1}$ and $\theta$ is a chart on $S^1$. In this local coordinate system $f$ has a very simple expression. – Anthony Carapetis Sep 27 '13 at 21:03
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    You should probably tell us your book's definition of the smooth structure on $S^1$. –  Sep 27 '13 at 22:32

1 Answers1

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The map $f$ is smooth and the derivative at a point is an injective linear map. Restricting the codomain of derivative to the tangent space to a point on the circle, the derivative becomes a linear isomorphism.

Inverse function theorem of smooth manifold applies and you obtain that $f$ is really a local diffeomorphism.

Edit: To answer @Idonotknow's questions all at once.

  1. The map $f$ is not a global diffeomorphism for an obvious reason: the map is not injective as $f(x)=f(x+2\pi n)$ for all integers $n$ and all real numbers $x$.

  2. If you view the map as $f:\Bbb R\to\Bbb R^2$, the derivative at $x$ would be given by the linear transformation represented by the Jacobian matrix $$\begin{bmatrix}-\sin x\\\cos x\end{bmatrix}$$ just as what OP found.

  3. You check the above linear map's injectivity by letting it act on a nonzero vector $c\in\Bbb R$ (view it as a vector in the vector space $\Bbb R$), giving the nonzero vector $$c\begin{bmatrix}-\sin x\\\cos x\end{bmatrix}.$$ You check that is vector is nonzero, by calculating its inner product with itself: $$c\begin{bmatrix}-\sin x&\cos x\end{bmatrix}c\begin{bmatrix}-\sin x\\\cos x\end{bmatrix}=c^2\not=0.$$

  4. Viewing $f$ as a maps $f:\Bbb R\to\Bbb R^2$, the derivative at a point $x\in\Bbb R$ is a map between the tangent spaces $df_x:T_x\Bbb R\to T_{f(x)}\Bbb R^2$. What I mean by restricting the codomain, is that we choose the subspace $T_{f(x)}S^1$ of $T_{f(x)}\Bbb R^2$, and restrict the codomain to $T_{f(x)}S^1$. You can do so, because the image of $f$, by definition, is on $S^1$ only.

edm
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