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Covector field on the sphere $S^2$ vanishing?

There exists a smooth vector field $X$ on $S^2$ that vanishes at exactly one point, for example at the north pole. My idea is the following: Let $\beta:=\{Y_1:=X, Y_2, Y_3\}$ be a basis for $\mathbb{R}^3$. Now, take $\beta^*:=\{\phi^1,\phi^2,\phi^3\}$, the dual basis of $\beta$.

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    More generally, a choice of Riemannian metric gives rise to a bundle isomorphism between $TM$ and $T^\ast M$ for any manifold $M$. Hence, any vector field property is equivalent to a covector field property and vice versa.2012-05-25

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Yes, your 1-form will be smooth.

To see this withouth using coordinate systems, we can think this way: $S^2$ is a smooth surface of $\mathbb{R}^3-\{0\}$. Define in $\mathbb{R}^3-\{0\}$ a 1-form $\alpha$ in the following way: given $x\in\mathbb{R}^3-\{0\}$ and a vector field $V$ in $\mathbb{R}^3-\{0\}$,

$\alpha_x(V(x))=\left\langle X\left(\frac{x}{\|x\|}\right),V(x)\right\rangle$

where $\langle\cdot,\cdot\rangle$ is the canonical inner product on $\mathbb{R}^3$.

Claim: $\alpha$ is a smooth 1-form.

Proof: Normalization $\frac{x}{\|x\|}$ is smooth in $\mathbb{R}^3-\{0\}$ and $X$ is, by hypothesis smooth. Hence, $\alpha$ is smooth.

Now, just notice that your 1-form is the restriction of this 1-form to $S^2$:

$\omega=i^*\alpha$

where $i\colon S^2\rightarrow\mathbb{R}^3-\{0\}$ is the inclusion. It follows from the fact that you are taking dual basis and I'm representing a vector of this basis with inner product. Hence, your 1-form is smooth.

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    I'm quite sure it s not a problem using the metric structure of $\mathbb{R}^3$ in roder to prove things here.2012-05-14