# A smooth vector field on $S^2$ vanishing at a single point

I need to find a vector field as described in the title.

I was given a couple hints, and this is what I have so far. Let $\varphi:S^2\setminus\{N\}\to\mathbb{R}^2$ be stereographic projection ($N$ is the north pole). Let $Y$ be a smooth, non-zero vector field on $\mathbb{R}^2$ that dies off to $0$ "at infinity." Since $\varphi$ is a diffeomorphism, we can push $Y$ forward along $\varphi^{-1}$ to get a smooth vector field $(\varphi^{-1})_*Y_{\varphi(p)}$ on $S^2$. But this vector field is not zero at the north pole... it's undefined. So can I just define my vector field on $S^2$, call it Z, to be the pushforward of $Y$ (as above) away from $N$ but $0$ at $N$? And if so, do we still have smoothness?

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I think that's the plan, yes. You will have to do some calculation involving the $\varphi$ to find out if your $Z$ is still smooth at the pole. If it isn't, then you may need to make $Y$ die off faster. Letting $Y$ be a something bounded times $e^{-(x^2+y^2)}$ ought to do the trick (but may be overkill). – Henning Makholm Oct 11 '11 at 1:39
@Henning: Well, I did end up multiplying by $e^{-(x^2+y^2)}$, but I'm confused on actually doing the calculation to show that I still have smoothness at N. – Bey Oct 14 '11 at 21:16
Select an "arctic" chart on $S^2$ that covers the north pole -- something simple like $(x,y)\mapsto(x,y,\sqrt{1-x^2-y^2})$ on a neighborhood of $(0,0)$ will probably do. The vector field is smooth at the north pole iff it is smooth by coordinates in the arctic chart. Thus, derive the transition map from the stereographic chart to the arctic chart, and express your vector field in the latter. Then the remaining argument ought to be reasonably similar to the standard proof of smoothness for $e^{1/x^2}$. – Henning Makholm Oct 14 '11 at 23:46

Short Version:

There is no need to use "complicated" functions to kill the field at "infinity". Standard coordinate fields will do the trick, thanks to the stereographic coordinates structure.

Let $(u,v)$ be the stereographic projection coordinates on $\mathbb{S}^2-\{N\}$. Then consider the vector field $\partial_u$ (or also $\partial_v$). A priori, it is defined only on $\mathbb{S}^2-\{N\}$

Now try to compute the local expression of this vector field in stereographic coordinatates but from the south pole. You will notice that the field vanishes at zero (i.e. at the south pole $S$).

Therefore, you can extend the field $\partial_u$ to a well defined vector field on $\mathbb{S}^2$ which vanishes exactly at one point (i.e. the south pole $S$).

Long Version:

More precisely, let us denote $(u,v)$ the stereographic coordinates relative to the projection from the north pole, that is the map $\phi_N : \mathbb{S}^2-\{N\} \to \mathbb{R}^2$. Let us denote $(\overline{u},\overline{v})$ the stereographic coordinates relative to the projection from the south pole, which is the map $\phi_S : \mathbb{S}^2-\{S\} \to \mathbb{R}^2$.

Now, consider the vector field $\partial_u$ (in coordinates), defined on $\mathbb{S}^2-\{N\}$. On the intersection of the two charts, $\mathbb{S}^2-\{N,S\}$, we can compute the expression of $\partial_u$ in $(\overline{u},\overline{v})$ coordinates. The result is:

$\partial_u = (\overline{v}^2-\overline{u}^2)\partial_\overline{u} - 2\overline{u}\overline{v}\partial_\overline{v} \qquad (1)$

Where, of course, $(\overline{u},\overline{v}) = \phi_S \circ \phi_N^{-1} (u,v)$. You can easily see that this vector field can be extended at the north pole, by formula (1).

Therefore, a field with the property you required is:

$X_p = \begin{cases} \left(\phi_N^{-1}\right)_* \left(\frac{\partial}{\partial u}\right) & p \in \mathbb{S}-\{N\} \\ \left(\phi_S^{-1}\right)_* \left((\overline{v}^2-\overline{u}^2)\frac{\partial}{\partial_\overline{u}} - 2\overline{u}\overline{v}\frac{\partial}{\partial_\overline{v}}\right) & p \in \mathbb{S}-\{S\} \end{cases}$

$X_p$ is a well defined vector field on the whole $\mathbb{S}^2$. It is also obviousvly smooth, since it is smooth in local coordinates. Moreover $X_N =0$, and $X_p = \partial_u \neq 0$ on $\mathbb{S}-\{N\}$, as required.

Explicit calculation for the change of coordinates

The change of coordinates map (and its inverse) can be computed explicitly:

$(u,v) = \phi_N \circ \phi_S^{-1}(\overline{u},\overline{v}) = \frac{(\overline{u},\overline{v})}{\overline{u}^2+\overline{v}^2}$ $(\overline{u},\overline{v}) = \phi_S \circ \phi_N^{-1}(u,v) = \frac{(u,v)}{u^2+v^2}$

By using these explicit expression it's easy to express $\partial_u$ in terms of barred coordinates:

$\frac{\partial}{\partial u} = \frac{\partial \overline{u}}{\partial u}\frac{\partial}{\partial \overline{u}} + \frac{\partial \overline{v}}{\partial u}\frac{\partial}{\partial \overline{v}}$

Now, I leave to you the last step (i.e. to explicitly compute the derivatives). REMEMBER, when computing the derivatives of the barred coordinates with respect to $u$, to express them in terms of BARRED coordinates.

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Dear Luca, thanks to your answer I have realized that there was no need to introduce a rapidly decreasing function because $(D\rho)(\rho(x))$ is a homogeneous polynomial in $x$. Thank you. – Giuseppe Oct 16 '11 at 19:59
@Luca: I understand the argument, but I can't recreate the calculation that $\partial_u=(\overline{v}^2-\overline{u}^2)\partial_{\overline{u}}-2\overline{u}‌​\overline{v}\partial_{\overline{v}}$. I know what $\overline{u}$ and $\overline{v}$ look like, at least, but I can't seem to make the change of coordinate match what you have – Bey Oct 17 '11 at 5:35
Also, did you want the vector field $\partial_u$ to be defined on $\mathbb{R}^2$? This seems to be the case, since you are later pushing it forward under the inverse of stereographic projection (from N) to get a vector field on $S^2$. – Bey Oct 17 '11 at 5:44
@Bey: Yes, as usual I identified $\partial_u$ with the field $\left(\phi_N^{-1}\right)_*(\partial_u)$ defined on $\mathbb{S}-\{N\}. In the last formula for$X_p$, however, I explicitly written out everything, without any identification, just to be clear. About the calculation, I'll add it explicitly to the answer in a second. – Raziel Oct 17 '11 at 7:11 @Luca: Thanks so much for your help! – Bey Oct 17 '11 at 11:49 Let$N$be the northern pole in the sphere$S^2$. The planes passing through$N$and perpendicular to the$yz$plane cut the sphere in a circle. Consider the field$Y$of unit vectors tangent to those circles: this is defined in the complement of$N$. Now multiply$Y$by the function$1-z$, to obtained a field$X$. Now notice that$X$extends to the whole of$S^2$, and vanishes only at$N$. - I don't quite understand. Each of the circles you reference contains$N$, so why would$Y$not be defined at$N$? Is it because the different tangent vectors we get as we move "from plane to plane" don't match up? Also, I assume that we are choosing a point$(x,y,z)\in S^2$when we multiply by$1-z$? – Bey Oct 11 '11 at 2:37 We can use extension lemma for vector fields (Introduction to Smooth Manifolds by John M. Lee, Exercise 4.2) to see that X extends to the whole$\mathbb{S^2}$. – HbCwiRoJDp Dec 3 '11 at 14:37 The Sphere$S^n$has an atlas with two charts$(U_1=S^n\setminus\{N\},\phi_1)$and$(U_2=S^n\setminus\{S\},\phi_2)$, where$\phi_1$and$\phi_2$are the stereographic projections respectively from the north and the south pole. The transition map$\phi_2\circ\phi_1^{-1}:\phi_1(S^n\setminus\{S,N\})=\mathbb{R}^n\setminus\{0\}\to\phi_2(S^n\setminus\{S,N\})=\mathbb{R}^n\setminus\{0\}$is given by the inversion$\rho$through the unit sphere in$\mathbb{R}^n$, i.e.$\phi_2\circ\phi_1^{-1}(x)=\rho(x)=|x|^{-2}x$. Consequently the tangent space$TS^n$has an atlas consisting of the two charts$(U_1\times\mathbb{R}^n,T\phi_1)$and$(U_2\times\mathbb{R}^n,T\phi_2)$, and the transition map$(T\phi_1)\circ(T\phi_1)^{-1}$is$T\rho$. So the vector field$v$on$S^n$is given a ordered pair of maps$v_1,v_2:\mathbb{R}^n\to\mathbb{R}^n$such that$v_2(\rho(x))=(T_x\rho)v_1(x)$for all$x\in\mathbb{R}^n\setminus\{0\}$. What you want is a smooth map$v_1:\mathbb{R}^n\to\mathbb{R}^n\setminus\{0\}$such that: 1.$x\in\mathbb{R}^n\setminus\{0\}\to (T_{\rho(x)}\rho)v_1(\rho(x))\in\mathbb{R}\setminus\{0\}$extends to a smooth map defined on all$\mathbb{R}^n$and vanishing at$0$. Let us take$v_1(x)=f(|x|)v^0$where$v^0$is a nonzero vector and$f:\mathbb{R}\to\mathbb{R}_+$a smooth function. The condition (1) becomes: 1.$x\in\mathbb{R}_+\to f(x^{-1})(T_{\rho(x)}\rho)v^0)\in\mathbb{R}\setminus\{0\}$extends to a smooth map vanishing at$0\in\mathbb{R}$.}$

Because $x\in\mathbb{R}_+\to x^{-1}\in\mathbb{R}_+$ together with all its derivatives diverges polynomially as $x$ goes to $0$ it is necessary for us that $f\in\mathcal{S}(\mathbb{R})$, i.e. $f$ rapidly decreases at infinity. For example we could take $f(x)=e^{-x^2}$

Edit after Luca's Answer The introduction of $f$ is redundant.
By the definition $\rho(x)=|x|^{-2}x$ we get the components of $T_{\rho(x)}\rho\equiv(D\rho)(\rho(x))$ are homogeneous polynomial of second degree in the componens of $x$.

So the condition 1 is satisfied in particular taking $v^0$ constant non zero as in the answer of Luca.

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