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Let $M$ a manifold of dimension $2$.

Q1) I know that we can have charts $U$ that gives coordinates $(x_1,x_2)$ and charts $V$ that gives coordinates $(y_1,y_1)$. Now are polar coordinates is given by a new charts $W$ or from each charts $U$ and $V$ I can get polar coordinate defined as $$(x_1,x_2)=(r\cos\theta ,r\sin\theta )$$ and $$(y_1,y_2)=(\rho\cos\varphi,\rho\sin\varphi) \ \ ?$$

Q2) For a function $f:M\to \mathbb R^2$ for example. Do we have that in $U$ that $\nabla f=(\frac{\partial f}{\partial x-1},\frac{\partial f}{\partial x_2})$ and in $V$ that $\nabla f=(\frac{\partial f}{\partial y_1},\frac{\partial f}{\partial y_2})$ ? Or the gradient is defined in a specific charts only ?

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The answer to Q1) for $U$ depends on the image of the chart in the $x_1,x_2$ plane. Let me use the notation $f : U \to \mathbb R^2$ for the coordinate map, where $f(p)=(x_1(p),x_2(p))$ for each $p \in U$. Also, for the image of $U$ under this coordinate map, let me use the notation $$\text{image}(f) = \{(x_1(p),x_2(p)) \mid p \in U\} \subset \mathbb R^2 $$ The problem is that polar coordinates are not defined for all subsets of the plane. For instance, polar coordinates are not defined for any subset that contains the origin $(0,0)$. Also, polar coordinates are not defined for any subset that contains a circle around the origin, because the angle coordinate $\theta$ is not well-defined (going all the way around the circle increases $\theta$ by $2\pi$). However, polar coordinates are defined for any simply connected subset that does not contain the origin. So, what you can conclude is that if $\text{image}(f)$ is simply connected and does not contain the origin, or even if $\text{image}(f)$ is contained in a simply connected set that does not contain the origin, then yes, polar coordinates can be defined on $U$. For a very specific example, if $U$ is homeomorphic to the open 2-disc then $\text{image}(f)$ is also homeomorphic to the open 2-disc and is therefore simply connected; hence, polar coordinates are defined as long as $\text{image}(f)$ does not contain the origin.

The answer to Q2) is that the gradient is not well-defined independent of the chart, because although you can certainly define it in each chart, where two charts overlap the result can be different. Generally speaking you need more structure on your manifold in order to get a well-defined gradient of a function. For example, if you have a Riemannian metric on your manifold then the gradient of a function is well-defined. These issues are covered in any course or book on differential geometry.

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