# 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.

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.