Sign of first eigenvalue of conformal Laplacian Let $(M^n,g)$ be some manifold of dimension $n \geq 3$. The conformal Laplacian is given by
$L=-4 \frac{n-1}{n-2} \Delta+ R$, where $R$ is the scalar curvature of $M$ and $\Delta= -\operatorname{div}\circ \operatorname{grad}$.

Now assume that $g$ is a Riemannian metric, $M$ is orientable, compact and has no boundary.
  Show that the sign of the first eigenvalue of $L$ is a conformal invariant.

Why is that so?
I was able to show some transformation law for $L$.
Namely, if $\tilde{g}=v^{4/(n-2)}g$, then for some function $\varphi >0$ we have
$L(\varphi)=v^{\frac{n+2}{n-2}} \tilde{L} (v^{-1} \varphi)$.
(Here $\tilde{L}$ is the conformal Laplacian with respect to the metric $\tilde{g}$) I originaly hoped I could conclude something similar to:
$(\varphi, \lambda)$ eigenpair for $L$ $\Leftrightarrow$ $(v^{-1} \varphi, \lambda \cdot \max v)$ eigenpair for $\tilde{L}$.
However I did not have any luck with that so far.
(And obviously the above line is NOT true. I just wanted to give an impression what kind of statement I was looking for)
How could I go about this exercise?
 A: Observe that the $L^2$ pairing satisfies 
$$ \langle L(\phi),\phi\rangle_g = \int L(\phi)\phi \mathrm{d}v_g = \int \tilde{L}(\nu^{-1} \phi) \nu^{-1} \phi \nu^{\frac{2n}{n-2}} \mathrm{d}v_g = \int\tilde{L}(\nu^{-1}\phi) \nu^{-1}\phi \mathrm{d}v_{\tilde{g}} = \langle \tilde{L}(\nu^{-1}\phi) , \nu^{-1}\phi\rangle_{\tilde{g}}$$


*

*The first eigenvalue of $L$ is negative if and only if there exists $\phi$ such that $\langle L(\phi),\phi\rangle < 0$. But then $\nu^{-1}\phi$ is such that $\langle \tilde{L}(\nu^{-1}\phi),\nu^{-1}\phi\rangle_{\tilde{g}}< 0$, showing that the first eigenvalue of $\tilde{L}$ is negative. 

*The first eigenvalue of $L$ is nonnegative if and only if $\forall \phi, \langle L(\phi),\phi\rangle_g \geq 0$. Given the equation above we see that this is also conformally invariant. 

*Furthermore, the first eigenvalue of $L$ is zero if and only if additionally there exists $\phi$ such that $L(\phi) = 0$. But then $\tilde{L}(\nu^{-1}\phi) = 0$ showing that the first eigenvalue of $\tilde{L}$ is also zero. 

