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

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.

share|cite|improve this question
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

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 $(x+h,y) \notin S^1$

share|cite|improve this answer
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
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

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.