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.
Algebraist's definition of the tangent space of a manifold
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differential-topology
manifolds
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2I 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_? – 2012-08-05
1 Answers
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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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0Yes, $[f]_{(t_0,\sqrt{1-t_0^2})} \mapsto {d \over dt}_{t=t_0} f(t,\sqrt{1-t^2})$ is the derivation you are looking for – 2012-04-03