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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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another way of seeing the existence of such a 1-form is the following: $S^2$ is orientable, hence it has a non-degenerated 2-form $\omega$ (a volume form). Hence, once you vector field $X$ vanishes only in one point, the 1-form $i_X\omega$ vanishes only at this very point. – matgaio May 13 '12 at 3:56
another way of seeing this is the following: $S^2$ admits a riemannian metric $g$. Then, $g(X,\cdot)$ is a 1-form wich vanishes exactly where $X$ vanishes. – matgaio May 13 '12 at 4:04
You can do this without coordinates. Try showing that if $X$ is a smooth vector field, $g(X,\cdot)$ (as @matgaio suggests) is a smooth one-form. – Neal May 13 '12 at 4:28
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. – Jason DeVito May 25 '12 at 0:43
up vote 4 down vote accepted

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 $S^2$:


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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In fact, we all are using a riemannian metric implicitly, because when we do $i^*\alpha$, we restrict the metric structure of $\mathbb{R}^3$ to the submanifold $S^2$, wich is the canonical way of defining a metric on $S^2$. – matgaio May 14 '12 at 18:10
Dependos on what course you are taking, hehe. But using the inner product of $\mathbb{R}^3$ is a really canonical technique. I don't think it could be a problem using it. – matgaio May 14 '12 at 19:00
I'm quite sure it s not a problem using the metric structure of $\mathbb{R}^3$ in roder to prove things here. – matgaio May 14 '12 at 19:45

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