I have seen that in the definition of a smooth function $f: M \to \mathbb{R}$, we firstly take $M$ to be a smooth manifold but i am not getting why do we need to take smooth manifold? The definition is as follows:

Let $M$ be a smooth manifold. A function $f: M \to \mathbb{R}$ is said to be smooth if $f \circ \phi^{-1}: \phi(U) \to \mathbb{R}$ is smooth for all local coordinate charts $(U,\phi(U))$ of $M$.

  • $\begingroup$ I'm not sure exactly what your question is but without the smoothness of the transition maps in the atlas of coordinate charts, the definition of "smooth function" (that you have indicated) is not well-defined. $\endgroup$ – Amitesh Datta Apr 19 '14 at 3:18

The point of that definition is that without it there is no definition of differentiation on the set $M$. In general, $M$ is abstract and there is no intrinsic notion of how one should differentiate. On the other hand, given a coordinate chart on $M$ the answer is simply to differentiate the coordinate representative. The derivative of the representative is a derivative in $\mathbb{R}^n$ which is a bit less abstract.

Maybe this is an answer to your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.