The sublaplacian is defined by $\mathcal{L}=-\left(X_{0}^{2}+X_{1}^{2}+X_{2}^{2}+X_{3}^{2}\right)$, which is independent of the choice of the orthonormal basis of $\mathbb{H}$. It is well known that the sublaplacian $\mathcal{L}$ is positive and essentially self-adjoint. Let $\mathcal{L}=\int_{0}^{\infty} \lambda d E(\lambda)$ be the spectral decomposition of $\mathcal{L}$. Then the restriction operator $P_{\lambda}$ for $\mathcal{L}$ is defined by
$$ P_{\lambda} f=\lim _{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} \int_{\lambda}^{\lambda+\varepsilon} d E(\lambda) f $$
for $f \in \mathscr{S}(\mathcal{N})$. Thus $f=\int_{0}^{\infty} P_{\lambda} f d \lambda$ where $P_{\lambda} f$ is an eigenfunction of $\mathcal{L}$ with eigenvalue $\lambda$.
I have read the above in a paper but I have some questions:
- Why the operator $\mathcal{L}$ is independent of the choice of the orthonormal basis?
- why $P_{\lambda} f$ is an eigenfunction of $\mathcal{L}$ with eigenvalue $\lambda$? Thank you very much for your help in advance.
