Orthogonal decomposition of $L^2(SM)$ I have been stuck on the following problem for a long time but I could not get the answer. Would you please help me? I was reading one paper [Paternian Salo Uhlmann: Tensor tomography on the surface] in that authors directly write that The space $L^{2}(S M)$ decomposes orthogonally as a direct sum
$$
L^{2}(S M)=\bigoplus_{k \in \mathbb{Z}} H_{k}
$$
where $H_{k}$ is the eigenspace of $-i V$ corresponding to the eigenvalue $k$. A function $u \in L^{2}(S M)$ has a Fourier series expansion
$$
u=\sum_{k=-\infty}^{\infty} u_{k}
$$
I wanted to justify that. Would you please help me? What I know is the following:
Theory: The standard Hilbert space theory that the complex Hilbert space $H$ is said to be a direct-sum of the subspaces $H_{m}, m \in \mathbb{Z}^{+}$and we write $H=\oplus_{m=0}^{\infty} H_{m}$, if
1.The subspaces $H_{m}$ are closed for every $m$.
2.The subspaces $H_{m}$ are pairwise orthogonal, i.e., if $h \in H_{m}$ and $g \in H_{m^{\prime}}$, then $\langle f, g\rangle=0$ whenever $m \neq m^{\prime}$
3.Every element $f \in H$ has a representation of the form
$$
f=h_{0}+h_{1}+\cdots+h_{m}+\cdots
$$
where $h_{m} \in H_{m}\left(m \in \mathbb{Z}^{+}\right)$and the sum converges in the norm of $H$.
The unit sphere bundle is defined as $S M:=\left\{(x, v) ; x \in M, v \in T_{x} M,|v|_{g}=1\right\}$.
The vertical vector field $V$ which differentiate with respect to direction variable $v$ i.e.
$$
V u(x, v)=\left.\frac{d}{d s} u\left(\rho_{s}(x, v)\right)\right|_{s=0}
$$
where $\rho_{s}(x, v)=\left(x, e^{i s} v\right)$.
For any $k \in \mathbb{Z}$ define
$$
H_{k}=\left\{u \in L^{2}(S M):-i V u=k u\right\}
$$
Any help or hint or reference will be highly appreciated. Thank you so much.
 A: I asume $M$ is a surface and $SM$ is the  unit circle bundle.  The bundle $SM\to M$  is also a principal $S^1$-bundle and the obvious $S^1$ action on $SM$ that preserves the natural metric on $SM$.    $\newcommand{\ii}{\boldsymbol{i}}$ For $p\in SM$ and $e^{\ii t}\in S^1$ denote by $e^{\ii t}p$ the action of $e^{\ii t}$ on $p$.   For $\newcommand{\bR}{\mathbb{R}}$ $t\in \bR$  define
$$
U_t: L^2(SM)\to L^2(SM),\;\;U_t f(p) =f(e^{\ii t}p).
$$
Since the $S^1$-action  preserves the metric on $SM$ the operator $U_t$ is  unitary.  We now have a one-parameter group of unitary operators. Clearly if  $f$  is continuous, then
$$
\lim_{t\to 0} \Vert U_t f-f\Vert=0,\;\;\Vert-\Vert:=\Vert-\Vert_{L^2}.
$$
If $f$ is in $L^2$, then for any $\newcommand{\ve}{{\varepsilon}}$ $\ve>0$ we can find a continuous function $f_\ve$ such that
$$
\Vert f-f_\ve\Vert<\ve.
$$
We have
$$
\Vert U_t  f-U_tf_\ve\Vert= \Vert  f-f_\ve\Vert<\ve,\;\;\forall t.
$$
Hence
$$
\Vert U_t f-f\Vert \leq \Vert  U_t f-U_t f_\ve\Vert+\Vert U_t f_\ve-f_\ve\Vert<\ve +\Vert U_t f_\ve-f_\ve\Vert
$$
We deduce that
$$
\limsup_{t\to 0}\Vert U_t f-f\Vert\leq \ve,\;\;\forall \ve>0,
$$
and thus
$$
\lim_{t\to 0} \Vert U_t f-f\Vert=0,\;\;\forall f\in L^2.
$$
This shows that $(U_t)_{t\in \bR}$ is a strongly continuous  one-parameter group of unitary operators.    Invoking Stone's theorem  we deduce that $U_t$ has the form
$$ 
U_t=e^{\ii t A},
$$
where the generator $A$ is a closed selfadjoint operator.   In this case the generator is $-\ii V$ so $U_t= e^{t V}$. Note that $U_{2\pi}=1$. Now use the spectral theorem for  (unbounded) selfadjoint operators  to prove the claims in the paper you mentioned.
