Poincare' s inequality for vectorfields on the sphere Let $\mathbb{S}^2$ be the standard 2-sphere, and let $V$ be $\mathcal{C}^1$ vectorfield on it.
I'd like to understand if it is true that there exists $C > 0$ such that, for all such $V$, we have
$$
\int_{\mathbb{S}^2}|\nabla V|^2 \mbox{d}\sigma_{\mathbb{S}^2}\geq C \int_{\mathbb{S}^2}|V|^2 \mbox{d}\sigma_{\mathbb{S}^2},
$$
where $\nabla$ is the Levi-Civita connection associated to the standard metric, and $\sigma_{\mathbb{S}^2}$ is the standard volume form.
My heuristic reasoning was the following: usually, for a Poincare' estimate on functions, you need either some condition on the support or on the integral mean of the function. Here, by the Hairy Ball Theorem, $V$ is "pinned" to be zero at one point, hence I wonder if this is sufficient to say something.
In case it worked, I would also be interested in understanding if such an inequality holds for some subset of $H^1$-vectorfields on the sphere.
 A: In this case it will be easier to work with one form. Let $\alpha$ be the one form dual to $V$, defined by
$$\alpha(X) = g (V, X). $$
Then 
$$\int_S |\nabla V|^2 ds = \int_S |\nabla \alpha|^2 ds = \int_S g(\alpha, \nabla^* \nabla \alpha) ds$$
where $\nabla^*$ is the adjoint of $\nabla$, locally given by 
$$(\nabla^* \beta)_i = - g^{jk}\nabla_j \beta_{ki} $$
By the Bochner identity, 
$$\Delta_d \alpha = (dd^* +d^*d)\alpha = \nabla^*\nabla \alpha + g^{jk}R_{ij}\alpha_k, $$
In our case, $R_{ij} = g_{ij}$ and so $g^{jk}R_{ij}\alpha_k = \alpha_i$ and 
$$g(\alpha, \nabla^*\nabla \alpha) = g (\alpha, (\Delta_d - 1)\alpha).$$
So we want to know if $\Delta_d - 1$ is positive. Using the fact in:
https://mathoverflow.net/questions/57749/laplace-derham-operator-for-1-forms-on-the-sphere
The spectrum on one form is the same as the spectrum on functions (except the constant function as $df = 0$), and it is known that $\lambda_1 = 2$. Thus we have 
$$\int_S |\nabla V|^2 ds \geq \int_S |V|^2 ds . $$
Remark The Bochner identity that I was using can be found in p.221 of this book:
http://www.amazon.ca/Riemannian-Geometry-Sylvestre-Gallot/dp/3540204938
(The calculation is eignevalue for Laplacian on spheres can also be found in this book). 
The set of eigenfunctions actually forms an orthonormal basis for $L^2(M)$ when $M$ is compact. The reason is that the smallest eigenvalue can be found by minimizing
$$E(f) = \frac{\int_M |\nabla f|^2 dS}{\int_M|f|^2 dS}$$
among all nonzero $f\in W^{1, 2}(M)$ and in general the $n$-th eigenvalue is the infimum of $E(f)$ where now $f$ is the set of all $W^{1, 2}(M)$ orthogonal to the first $(n-1)$-th eigenfunctions. From this construction it is clear that the eigenfunctions are dense in $W^{1, 2}(M)$,thus also dense in $L^2(M)$. 
Another remark Let's me briefly explain the proof in the first link. First of all, as $\Delta * = *\Delta$, $f$ is an eigenfunction for the scalar $\Delta$ if and only if $*f$ is an eigenform for $\Delta$ (on two forms). Since $** = \pm id$, the eigenvectors for $\Delta$ on two forms are the same as that of the scalar $\Delta$. Now we use the Hodge decomposition:
$$\Omega^1(\mathbb S^2) = \ker \Delta \oplus Im (d) \oplus Im (d^*) =  Im (d) \oplus Im (d^*) \  \ \ \ \text{as  }H^1(\mathbb S^2) =0,$$
Thus all one forms $\alpha$ are of the form $\alpha = df + d^*(* g)= df \pm * dg$ for some functions $f, g$. From here one see that $\Omega^1(\mathbb S^2$ is also generated by eigenform of the form $df$ and $*dg$. 
(It is also in general true that for any compact $(M, g)$, the eigenforms of $\Delta$ forms a complete orthonormal basis for the space of $H^1$ eigenforms)
