If $w$ is a $1$-form on $S^2$ such that $A^*w = w$ for all $A \in SO(3)$, then $w = 0$.

Question

Assume that $w$ is a $1$-form on the $2$-sphere $S^{2}$ so that $A^{*}w = w$ for all $A \in SO(3)$. Show that $w = 0$.

I have tried to apply the definition of pullback and special orthogonal group, but I don't know how to combine these to get the result. Any hint is appreciated!

Answer 1

Consider first the following linear algebra proposition:

Let $(V, \left< \cdot, \cdot \right>)$ be a finite dimensional inner product space with $\dim V \geq 2$ and let $\varphi \colon V \rightarrow \mathbb{R}$ be a linear functional. Assume that $\varphi \circ S = \varphi$ for all orientation preserving orthogonal transformations $S \colon V \rightarrow V$. Then $\varphi = 0$. To see this, let $0 \neq w \in \ker(\varphi)$ and let $0 \neq v \in V$. Then there exists an orientation preserving orthogonal transformation $S \colon V \rightarrow V$ with $S \left( \frac{w}{||w||} \right) = \left( \frac{v}{||v||} \right)$. But then

$$ 0 = \varphi \left( \frac{w}{||w||} \right) = \varphi \left( S \left( \frac{w}{||w||} \right) \right) = \varphi \left( \frac{v}{||v||} \right) = \frac{\varphi(v)}{||v||} $$

which implies that $\varphi(v) = 0$.

Now, given a one-form $\omega$ on $S^2$, note that given any point $p \in S^2$ and orientation preserving orthogonal map $S \colon T_p(S^2) \rightarrow T_p(S^2)$, you can construct an element $A \in SO(3)$ such that $A(p) = p$ and $dA|_p = S$. Applying the proposition above, you can conclude that $\omega_p = 0$.

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To be even more explicit and precise, consider the rotation matrices

$$ A_\theta = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{pmatrix}. $$

Denote by $e_i$ the standard basis vectors of $\mathbb{R}^3$. Note that $A \in SO(3)$, $A(e_1) = e_1$ and after identifying the vectors $(e_2,e_3)$ with tangent vectors in $T_{e_1}(S^2)$, we have

$$ [d(A_\theta)|_{e_1}]_{(e_1,e_2)} = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} $$

so the differential $d(A_\theta)_{e_1}$ acts on $T_{e_1}(S^2)$ by rotation. Since $\omega_{e_1} = A_{\theta}^{*}(\omega)_{e_1} = \omega_{e_1} \circ d(A_\theta)|_{e_1}$ for all $\theta$, we must have $\omega_{e_1} = 0$. Next, given $p \in S^2$, construct a matrix $A \in SO(3)$ satisfying $Ap = e_1$ and conclude that $\omega_p = \omega_{e_1} \circ dA|_{p}$ and so $\omega_p = 0$ for all $p \in S^2$.

Answer 2

Suppose $w_q = 0$. As $SO(3)$ acts transitively on $S^2$, for any $p \in S^2$ there is $A \in SO(3)$ such that $A(p) = q$. As $(A^*w)_p(X, Y) = w_{A(p)}(A_*X, A_*Y) = w_q(A_*X, A_*Y) = 0$, we see that $w = 0$. That is, if $w$ has a zero, then it is identically zero.

Suppose that $w$ has no zeroes and let $g$ be a Riemmanian metric on $S^2$. As $g$ is non-degenerate, there is a vector field $V$ such that $w(Z) = g(V, Z)$ for all $Z$. As $w$ is nowhere zero, $V$ is nowhere zero. In particular, $V$ has isolated zeroes, so by the Poincaré-Hopf Theorem, $\chi(S^2)$ is the number of zeroes of $V$ counted with sign. This is a contradiction as $Z$ has no zeroes while $\chi(S^2) = 2 \neq 0$.

Therefore $w = 0$.

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More generally, suppose $M$ is a closed smooth manifold and there is a smooth transitive group action by $G$ on $M$, i.e. $M$ is a homogeneous space for $G$. If $\chi(M) \neq 0$, then the only $G$-invariant $1$-form on $M$ is $0$.