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Let $S^2 = \{p \in \mathbb{R}^3\mid ||p||_2 = 1\}$. Given that the partial derivatives of $\frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}$ form a basis for the tangent bundle of $S^2$, I was wondering whether the vector field $X_p = (1 - z)\left(\frac{\partial}{\partial x^1} + \frac{\partial}{\partial x^2}\right)|_p$ would work as a $\mathcal{C}^\infty$ vector field on $S^2$ vanishing only at a single point? I came across this particular post, A smooth vector field on $S^2$ vanishing at a single point, and given that the discussion in it is quite lengthy, I have a feeling that I am missing something. So, does the aforementioned $X_p$ suffice as the stated vector field? It not, why?

Epsilon Away
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  • IIRC, each coordinate vector field itself extends to a smooth (in fact, holomorphic on the Riemann sphere) vector field vanishing only at the north pole. (The proposed vector field also does, but is not holomorphic.) – Andrew D. Hwang Jan 29 '22 at 17:19

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What do you mean precisely by $x^1$ and $x^2$? If you are working in the ambient space $\mathbb R^3=(x^1,x^2,z)$ then the vector field you have defined does not actually lie in the tangent space for all $p\in S^2$, since it is usually not pointing in the normal direction to the surface. If, however, you are using $x^1$ and $x^2$ as local coordinates for $S^2$ in a chart, then the issue is you have not given a complete specification of your vector field, since you will need to tell us how to glue the different local coordinates together at the intersections of the charts (note that you must use more than one chart since $S^2$ has nontrivial topology).

pre-kidney
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    By $x^1, x^2$ I mean the local coordinates of $S^2$. What would be this complete specification? If we take, say, the stereographic projection $\phi: S^2\setminus {(0, 0, -1)}\to \mathbb{R}^2, (x, y, z) \mapsto \left(\frac{x}{1 + z}, \frac{y}{1 + z}\right)$, then what would I need to do to tell how to glue together the intersection of different charts? Should I consider the transition function $\phi\circ \psi^{-1}$ of $\phi$ with another projection, say $\psi:S^2\setminus {(0, 0, 1)}\to\mathbb{R}^2, (x, y, z) \mapsto \left(\frac{x}{1 - z}, \frac{y}{1 - z}\right)$? – Epsilon Away Jan 29 '22 at 15:07
  • Yes, this is along the right lines, if you are going for a local description (which I'm not claiming is the most elegant or simplest way to understand the situation). Again, I stress that there is no precise meaning when you say "the" local coordinates, as they are only defined in a local chart (which you clearly specify in advance) and then $x^1,x^2$ will have different meanings relative to the two charts where they overlap, hence the need for a transition function. Also, to be technically complete, you would express $z$ in terms of local coords (getting different functions on each chart). – pre-kidney Jan 29 '22 at 15:12
  • So $z$ is related in $S^2$ by the fact that $||p|| = 1$, so that $z = \pm\sqrt{1 - x^2 - y^2}, (x, y, z) \in S^2$. The inverse of $\phi$ is given by $\psi^{-1}(a, b) = \frac{1}{\sqrt{1 + u^2 + v^2}}\left(u, v, \sqrt{1+ u^2 + v^2} - 1\right)$, and $\phi\circ\psi^{-1}(u, v) = \frac{1}{2\sqrt{1 + u^2+ v^2}}(u, v)$. Are we done by writing out the transition function $\psi\circ \phi^{-1}$ and plugging in $\pm\sqrt{1 - x^2 - y^2}$ for $z$? – Epsilon Away Jan 29 '22 at 15:28
  • No, because you have not said how $x^1$ and $x^2$ are related to $x$ and $y$. Or $a,b$ or $u,v$. Which textbook are you learning differential geometry from? – pre-kidney Jan 29 '22 at 15:30
  • @pre-kidny From Loring's An Introduction to Differential Geometry. Well, given that $z$ is a function of $x$ and $y$, can we just declare that $x^1 = x, x^2 = y$? – Epsilon Away Jan 29 '22 at 16:10