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Let $S^2$ be the 2-sphere in $\mathbb{R}^3$, given with vector field $X=x \partial_x+ y \partial_y$ on the stereographic projection from north pole chart. What is the global extension of $X$ to $S^2$ as a vector field on $\mathbb{R}^3$?

Edit: The same vector field $X=x \partial_x+ y \partial_y$ on south pole chart(I checked that on overlap these define same vector fields) gives a globally defined vector field on sphere. I think this problem wants to ask how can $X$ be expressed in terms of $x,y,z$ in $\mathbb{R}^3$. (Since $x,y$ in $X$ is not the same $x,y$ in $\mathbb{R}^3$.)

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You've defined $X$ at every point of $\mathbb{S}^2$ except the north pole. What value does $X$ need to take in $T_{\mbox{north pole}}\mathbb{S}^2$ so that it is smooth there? –  Neal Dec 12 '12 at 15:37
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$\mathbb{R}^3$ is completely redundant here. –  Rhys Dec 12 '12 at 16:05
    
I made some edits to clarify things. –  Gobi Dec 12 '12 at 16:13
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Dear Gobi,if I correctly interpret what you say, there is a change of sign in the expression of the field in the south pole chart: see my answer. –  Georges Elencwajg Dec 23 '12 at 11:49
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2 Answers

As far as I know, there is no canonical way to extend vector fields from a submanifold. The field can be extended in many ways, for example by using a smooth radial cut-off function $\eta$ that is equal to $1$ in a neighborhood of $S^2$ and to $0$ at some distance from $S^2$. For example, if the original field was $\sin \phi \frac{d}{d\phi}$ in spherical coordinates $(\theta,\phi)$ on $S^2$, then $\eta(\rho)\sin \phi \frac{d}{d\phi}$ would be its extension to $\mathbb R^3$ in spherical coordinates $(\rho, \theta,\phi)$.

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The change of coordinates from north pole coordinates to south pole oordinates is $$\phi:\mathbb R^2\setminus \lbrace (0,0)\rbrace \to \mathbb R^2\setminus \lbrace (0,0)\rbrace:\begin{pmatrix} x \\\\ y\end{pmatrix}\mapsto \begin{pmatrix} u \\\\ v\end{pmatrix}=\frac {1}{(x^2+y^2)}\begin{pmatrix} x \\\\ y\end{pmatrix}$$ and the jacobian of that map is given by the matrix

$$Jac(\phi)(x,y)=\frac {1}{(x^2+y^2)^2}\begin{pmatrix} -x^2+y^2&-2xy \\\\ -2xy&x^2-y^2 \end{pmatrix}$$

This implies that your vector field has as expression in the south pole chart(beware the minus sign!) $$Y(u,v)=-(u\partial_u+v\partial_v)\quad \text{for} \begin{pmatrix} u \\\\ v\end{pmatrix}\in \mathbb R^2\setminus \lbrace (0,0)\rbrace $$ which can be smoothly extended to the origin by $Y(0,0)=0\partial_u+0\partial_v$.

Conclusion
The initial vector field $X$ on the sphere minus the north pole smoothly extends to a global vector field on the whole sphere by defining it as zero at the north pole.

Edit
Slightly more generally we can define a global vector field (vanishing at the two poles and only there if the real numbers $a,b$ are not both zeroon the sphere by the following expressions in the two charts given by the two stereographic projections: $$(ax+by)\partial_x+(-bx+ay) \partial_y=(-au+bv)\partial_u+(-bu-av)\partial_v$$ If the real numbers $a,b $ are not both zero, that vector field vanishes at the two poles and only there.

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