Let $G$ be a real, connected, compact Lie group equipped with the bi-invariant Haar measure satisfying $\mu(G)=1$. Let $(V_1,\dots,V_d)$ be a basis for its Lie algebra, viewed as left-invariant vector fields and hence linear maps $C^\infty(G)\to C^{\infty}(G)$. And define the (differential) operator $$L= \left(\sum_1^dV_i^2\right)^2:C^\infty(G)\to C^\infty(G).$$
My question Let $u\in C^\infty(G)$ satisfy $\int u\;d\mu=0$, then is it the case that $$\int u^2d\mu\leq c\int Lu\cdot ud\mu$$ for some $c>0$ independent of $u$? And, if not, what sort of further conditions would one need to impose on $G$ to make it true?
Ideas so far If $G= S^1$ (the circle, although the same argument would hold for the $n$-dim torus), then $L=\Delta^2$. Here, the trigonometric polynomials, $(e_n)$ form an orthonormal basis of $L^2(G)$ and are eigenvectors for $L$: $Le_n=cn^4e_n$ (some $c>0$ depending on $\pi$, etc). Hence $L$ has a spectral gap and the result follows easily. Here, I used crucially the fact that I could identify an orthonormal basis of eigenvectors for $L$ - something that I don't know how to do in the general case above.
Yes.
One key observation is that each left-invariant vector field $V$ is skew-adjoint. To see this, let $\Omega$ be the smooth $d$-form on $G$ associated with the measure $\mu$. Note that the flow of $V$ is given by right multiplication: $$ \theta_t(x) = x\cdot \exp(tV). $$ Thus since $\Omega$ is right-invariant, it follows that it is invariant under the flow of $V$ and thus $\mathscr L_V\Omega=0$. Using Cartan's magic formula, $$ 0 = \mathscr L_V(\Omega) = d(i_V \Omega) + i_V(d\Omega) = d(i_V\Omega) $$ because $d\Omega$ is an $(d+1)$-form on an $d$-manifold. A computation in local coordinates shows that for any smooth function $u$, $$ du \wedge (i_V\Omega) = (Vu)\Omega, $$ and therefore by Stokes's theorem, \begin{align*} 0 &= \int_G d( uv\cdot i_V\Omega)\\ &= \int_G d(uv)\wedge (i_V\Omega) + uv \,d(i_V \Omega) \\ &= \int_G V(uv)\Omega\\ &= \int_G uVv\,\Omega+ \int_G v Vu\,\Omega. \end{align*} This proves that $V$ is skew-adjoint.
Now set $$ \Delta = \sum_{i=1}^d V_i^2, $$ so $\Delta$ is self-adjoint and $L=\Delta^2$. Given $u$ satisfying $\int_G u\,\Omega=0$, one version of the ordinary Poincaré inequality reads \begin{align*} \|u\|_{L^2}^2 &\le c \sum_{i=1}^d\int_G (V_iu)^2\Omega\\ &=- c \sum_{i=1}^d\int_G u\cdot (V_i^2 u)\, \Omega\\ &= - c\int_G u\cdot \Delta u\,\Omega\\ &\le c\|u\|_{L^2}\, \|\Delta u\|_{L^2}. \end{align*} Dividing by $\|u\|_{L^2}$ and squaring both sides, we obtain \begin{align*} \int_G u^2\,\Omega \le c^2 \int_G (\Delta u)^2\,\Omega = c^2 \int_G u\cdot (\Delta^2 u)\,\Omega = c^2 \int_G u \cdot Lu\,\Omega. \end{align*}
Added in response to comment:
The standard Poincaré inequality for compact manifolds is proved, for example, in Hebey's Sobolev Spaces on Riemannian Manifolds (Lemma 3.8 on page 24), and probably many other places as well. Note that the usual statement of the theorem on a compact manifold $M$ involves a Riemannian metric $g$, as follows: for $u\in C^\infty(M)$ with average value zero, $$ \int_M u^2\, dV_g \le c \int_M |\nabla u|_g^2\, dV_g, $$ where $dV_g$ is the Riemannian volume form and $\nabla u$ is the total covariant derivative of $u$, which in this case is just the differential $du$. You can translate your case into this notation by letting $g$ be the left-invariant Riemannian metric on $G$ for which $(V_1,\dots,V_d)$ is a global orthonormal frame, and $dV_g$ is a constant multiple of $\Omega$.
This inequality is false if $G$ is noncompact or disconnected. For a disconnected counterexample, let $G=O(n)$, and let $u(A) = \det A$, which is equal to $1$ on one component and $-1$ on the other. (Note that Hebey's statement of Lemma 3.8 doesn't mention connectedness, but he adopts a blanket assumption of connectedness on page 3.)
For a noncompact counterexample, let $G=\mathbb R$ with its usual additive group structure, so $V=d/dx$ is left-invariant; and let $u$ be any nontrivial compactly supported odd function ($u(-x)=-u(x)$), so $\int_{\mathbb R}u(x)\,dx=0$. For $\delta>0$, let $u_\delta(x) = u(\delta x)$. Then a simple change of variables shows $$ \int_{\mathbb R} u_\delta(x)^2\,dx = \frac{1}{\delta}\int_{\mathbb R}u(x)^2\,dx, \qquad \int_{\mathbb R} u_\delta'(x)^2\,dx = \delta\int_{\mathbb R}u'(x)^2\,dx, $$ so the ratio $\left.\int_{\mathbb R} u_\delta(x)^2\,dx \middle/ \int_{\mathbb R} u_\delta'(x)^2\,dx\right.$ is unbounded as $\delta \to 0$.