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By the "algebraist's definition" of the tangent space of manifolds, can we say that the partial derivative $d/dx$ belongs to the the tangent space of $S^1$? It feels strange, but I can't see why it shouldn't be true.

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I don't understand what you mean by the tangent space of a manifold. Do you mean the tangent space at a point, or do you mean the tangent bundle? –  Qiaochu Yuan Aug 5 '12 at 4:39

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If $f:S^1\rightarrow \mathbb R$ is a smooth function, then how can you differentiate w.r.t. the variable $x$? This is impossible! $$ {f(x+h,y)-f(x,y) \over h} $$ makes no sense, since $(x+h,y) \notin S^1$

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Right, I can see a problem indeed. What would a proper derivation (in the sense of the tangent space) then be on $TS^1$? We have the same problem with $(x\frac{d}{dy}-y\frac{d}{dx})$ don't we? (thanks for the help I'm a beginner:) –  Kim Apr 3 '12 at 10:48
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Yep. You should employ a chart $x$ for $S^1$, write $f\circ x^{-1}$ as a function of one real variable. The tangent space is a one dimensional real vectorspace with basis $\partial_\varphi$ –  Blah Apr 3 '12 at 10:53
    
You should compose with a curve from an interval to $S^1$ and then differentiate! –  checkmath Apr 3 '12 at 11:01
    
Just to be sure. If I take a chart $x: S^1\to \mathbb{R}$, then you suggest to use $$[f \to \tfrac{d}{dt}(f \circ x^{-1})(t)]$$ as a derivation (with $f: S^1\to\mathbb{R}$)? For example $x: (x_1,x_2) \to x_1$ gives $x^{-1}: t \to (t,\sqrt{1-t^2})$ and then $$[f \to \tfrac{d}{dt}(f(t,\sqrt{1-t^2}))(t)]$$ gives the derivation? –  Kim Apr 3 '12 at 11:30
    
Yes, $[f]_{(t_0,\sqrt{1-t_0^2})} \mapsto {d \over dt}_{t=t_0} f(t,\sqrt{1-t^2})$ is the derivation you are looking for –  Blah Apr 3 '12 at 16:49

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