text:
\section{Hypocoercivity for geometric Langevin dynamics}
\label{sec:Langevin}

In this section we apply the abstract Hilbert space hypocoercivity method
to the Langevin equation with some Riemannian manifold as position space,
see~\hyperref[thm:Langevin:main]{Theorem~\ref{thm:Langevin:main}} below.
That is the direct generalisation of the situation of~\cite{HypocoercMFAT}.
The techniques of proving we learned from~\cite{HypocoercJFA}.
Our interest arose from industrial fibre lay-down applications
and qualitative analysis of the nonwoven.
In this context, the position manifold~\posfold could e.\,g.\
reflect a sagging conveyor belt
or a belt moving over a cylindrical roller.
As we mentioned in the introduction,
Langevin dynamics have wide ranging applications
and our approach offers the freedom to include any `smooth' side condition on the position variable.

Consider the following Stratonovich~SDE in~\tanbundle\posfold:
\begin{equation}
\label{eq:geomLangevin}
    \diffd \eta
    =
    \spray{\riemannmetric*}
    \ \diffd t
    +
    \vertlift_\eta(-\grad{\riemannmetric*}\Psi)
    \ \diffd t
    +
    \sigma \pdot \vertlift_\eta\left(
        \sum_{j=1}^\posdim \frac{\partial}{\partial{x_\eta^j}}
        \right)
    \stratonovich\diffd W_t
    -
    \alpha \pdot \canonfield
    \ \diffd t
,
\end{equation}
where $\eta\colon\,\indexset\to\tanbundle\posfold$ is a curve with time interval~\indexset
and $\parentheses*{x_\eta^1,x_\eta^2,\ldots,x_\eta^\posdim}$ is a chart at~$\tanproj(\eta)$
providing normal coordinates.
Recall that~\canonfield denotes the canonical vector field.
Bellow we specify certain assumptions on the potential $\Psi\colon\,\posfold\to\Rnum$.
The nonnegative model parameters~$\alpha$ and~$\sigma$ are related by
$\sigma = \sqrt{\sfrac{2\alpha}\beta}$,
where~$\beta$ is a nonnegative rescaling of the potential as $\Phi=\beta\Psi$.
Recall that
the horizontal motion in \hyperref[eq:geomLangevin]{Equation~\eqref{eq:geomLangevin}}, i.\,e.\
$
    \horproj(\diffd \eta)
    =
    \spray{\riemannmetric*}
    \ \diffd t
$,
reflects the natural requirement of
$
    (\tanproj \bincirc \eta)^\prime
    =
    \eta
$
which is the form that any integral curve of~\spray{\riemannmetric*} attains by definition.
We call~\hyperref[eq:geomLangevin]{Equation~\eqref{eq:geomLangevin}}
the \emph{Langevin equation on~\posfold}
or just \emph{geometric Langevin equation}.

Using~\cite[Theorem~V.1.2]{IkedaWatanabe}
the Kolmogorov generator~$L$ is given as
\begin{equation}
\label{eq:Langevin-generator}
    L
    =
    \spray{\riemannmetric*}
    -\frac1\beta\,
    \grad{\vertmetric*}\hlfunc{\Phi}
    +
    \frac\alpha\beta\,\laplace{\vertmetric*}
    -
    \alpha\canonfield
.
\end{equation}
This operator is defined for all smooth functions on~\tanbundle\posfold,
but we consider the domain $D\vdef\smoothcompact{\tanbundle\posfold}$ of smooth test functions
with compact support.
Note that $(L,D)$ is densely defined
by~\hyperref[rem:L2dense-testfunctions]{Remark~\ref{rem:L2dense-testfunctions}}.
We usually call~$L$ as in~\hyperref[eq:Langevin-generator]{Equation~\eqref{eq:Langevin-generator}}
the \emph{Langevin generator}.
Compare this generator to~\cite[Equation~(1.2)]{HypocoercMFAT}
taking~\hyperref[ex:spray-acting]{Example~\ref{ex:spray-acting}} into account.
As we will see in~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}},
the Langevin generator basically decomposes into two components:
the vertical diffusion
and the (not entirely horizontal) component liaising the appropriate notion of second order differential equations.
Compare this decomposition to so-called hypoelliptic Laplacians;
via this concept J.-M.~Bismut links Brownian motion on manifolds and geodesic flow
in order to find a Langevin process in~\cite{Bismut}.

Before we start checking the data and hypocoercivity conditions,
we shall fix conditions on the potential in the geometric Langevin equation,
and thus on the base weight
$
    \rho_{\posfold}
    =
    \exp(-\Phi)
    =
    \exp(-\beta\Psi)
$.

\begin{condition}[Potential conditions~\ref{cond:named:Langevin:P}]
\label{cond:Langevin:P}
\namedlabel{cond:named:Langevin:P}{(P)}
\hfill

\begin{enumerate}[label={(P\arabic*)}, ref={(P\arabic*)}]
    \item\label{cond:Langevin:P1}
        \emph{General regularity and boundedness:}\
        Let $\Phi=\beta\Psi$ a loc\=/Lipschitzian potential
        which is bounded from below
        and such that
        $
            \Leb{\wriemannmetric*}
            =
            \rho_{\posfold}\,\Leb{\riemannmetric*}
            =
            \exp(-\Phi)\,\Leb{\riemannmetric*}
        $
        is a probability measure on $(\posfold,\borel{\posfold})$.
    \item\label{cond:Langevin:P2}
        \emph{Poincar\'e inequality:}\
        The weighted Riemannian measure~\Leb{\wriemannmetric*}
        satisfies the Poincar\'e inequality
        \begin{equation}
        \label{eq:Langevin:Poincare}
            \norm{
                \grad{\riemannmetric*} f_0
            }{\bigL2{\posfold\to\tanbundle\posfold}{\wriemannmetric*}}^2
            \geq
            \Lambda
            \norm{
                f_0-\scalarprod{f_0}{1}{\bigL2{\posfold}{\wriemannmetric*}}
            }{\bigL2{\posfold}{\wriemannmetric*}}^2
        \end{equation}
        for all $f_0\in\smoothcompact{\posfold}$
        and some $\Lambda\in(0,\infty)$.
    \item\label{cond:Langevin:P3}
        \emph{Hessian dominated by gradient:}\
        Assume $\Phi\in\contidiff2{\posfold}$.
        There is a constant $c\in(0,\infty)$ such that
        \begin{displaymath}
            \abs{ \operatorname{Hess}_{\riemannmetric*}(\Phi)(x) }
            \leq
            c\, (1+\abs{\grad{\riemannmetric*}\Phi(x)}{\riemannmetric*})
            \qquad\text{holds for all } x\in\posfold
        .
        \end{displaymath}
        Here, `$\operatorname{Hess}_{\riemannmetric*}(\cdot)$'
        denotes the Hessian wrt.\ the given Riemannian metric
        and the norm `$\abs{\operatorname{Hess}_{\riemannmetric*}(\cdot)}$'
        of the Hessian is the Frobenius tensor norm induced by the Riemannian metric.
\vspace*{-3ex}
\end{enumerate}
\end{condition}

As explained in~\cite[Remark~3.16]{HypocoercJFA}
the condition~\ref{cond:Langevin:P3} as above can be weakened.
Clearly, the~\hyperref[eq:Langevin:Poincare]{Poincar\'e inequality~\eqref{eq:Langevin:Poincare}}
is the most restrictive condition wrt.\
the geometry of the weighted position manifold~$(\posfold,\wriemannmetric*)$.
It's satisfied for compact manifolds, see~\cite[Theorems~2.10, 2.11]{Hebey}.
For the noncompact case we refer to~\cite[Lemma~3.1]{Hebey},
where the necessary geometric assumption is that
the Ricci curvature is bounded from below by some multiple of~\wriemannmetric*.
Furthermore, we point out that
in~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}}
the nasty assumption of a weakly harmonic potential appears.
In view of Weyl's theorem it would be quite a restrictive assumption.
That's why
in~\hyperref[prop:Langevin:SAD]{Proposition~\ref{prop:Langevin:SAD}}
we remove this assumption
for the general setting of this section.
Note that by~\hyperref[lem:weighted-M-complete]{Lemma~\ref{lem:weighted-M-complete}}
the Riemannian manifold $(\posfold,\wriemannmetric*)$ weighted by a potential as above
is complete again.
\smallskip

We formulate the main theorem of this section:
\begin{theorem}[Hypocoercivity of the geometric Langevin dynamic]\hfill
\label{thm:Langevin:main}

    Let $\alpha,\beta\in(0,\infty)$
    and $(\posfold,\riemannmetric*)$ be a Riemannian manifold satisfying~\ref{cond:M}.
    We assume that the potential $\Phi\colon\,\posfold\to\Rnum$
    fulfils the conditions~\ref{cond:named:Langevin:P} above.
    Denote by~$\nu$ the zero-mean Gaussian measure with covariance matrix~$\beta^{-1}\ID$
    and define
    $
        \mu
        \vdef
        \Leb{\wriemannmetric*}\locprod\nu
    $.

    Then, the Langevin operator
    \begin{displaymath}
        \parentheses*{
            L,
            \smoothcompact{\tanbundle{\posfold}}
        }
        =
        \parentheses*{
            \frac\alpha\beta\,\laplace{\vertmetric*}
            -
            \alpha\canonfield
            +
            \spray{\riemannmetric*}
            -
            \frac1\beta\,
            \grad{\vertmetric*}\hlfunc{\Phi},
            \smoothcompact{\tanbundle{\posfold}}
        }
    \end{displaymath}
    is closable in $H\vdef\bigL2{\tanbundle\posfold}{\mu}$.
    Moreover, its closure~\operator{L} generates a strongly continuous contraction semigroup $(T_t)_{t\in[0,\infty)}$.
    Finally, there are constants $\kappa_1,\kappa_2\in(0,\infty)$
    computable in terms of $\alpha$, $\beta$, $\Lambda$ and $c$
    such that for all $g\in H$ and times $t\in[0,\infty)$ holds
    \begin{displaymath}
        \norm{ T_tg-\scalarprod{g}{1}{H} }{H}
        \leq
        \kappa_1 \euler^{-\kappa_2 t}\norm{ g-\scalarprod{g}{1}{H} }{H}
    .
    \end{displaymath}
\end{theorem}

Clearly, we are going to prove it by applying the~\ref{thm:hypocoercivity-thm-named}.

\subsection{Data conditions}
\label{subsec:Langevin:D}

\begin{definition}[model Hilbert space~\ref{cond:D1}]
\label{def:Langevin:D1}
    Consider the probability space
    \begin{displaymath}
        (E,\mathfrak{E},\mu)
        =
        \parentheses*{
            \tanbundle{\posfold},
            \borel{\tanbundle{\posfold}},
            \Leb{\wsasakimetric*}
        }
    ,
    \end{displaymath}
    where $\Leb{\wsasakimetric*} = \Leb{\wriemannmetric*} \locprod \nu$
    is the weighted Sasaki volume measure with
    $\wriemannmetric*$ is weighted by
    $\rho_{\posfold}\vdef \exp(-\Phi)= \exp(-\beta\Psi)$ with $\beta\in(0,\infty)$ such that
    \Leb{\wriemannmetric*} is a probability measure on~$(\posfold,\borel{\posfold})$,
    and $\nu = \normaldist{0}{\beta^{-1}\ID_\posdim}$
    is the zero-mean normal distribution on the fibre~$F=\Rnum^\posdim$
    with covariance matrix~$\beta^{-1}\ID_\posdim$.
    In other words,
    \Leb{\wsasakimetric*} has the loc\=/density
    $\exp(-\Phi) \locprod \prod_{j=1}^\posdim\,\varphi_{0,\beta^{-1}}\bincirc\proj_j$,
    where~$\varphi_{0,\beta^{-1}}$ denotes the density of a one-dimensional normal distribution with variance~$\beta^{-1}$.
    The model Hilbert space is
    $H \vdef \bigL2{E}{\mu} = \bigL2{\tanbundle{\posfold}}{\wsasakimetric*}$,
    cf.~\hyperref[not:Lp-measure]{Notation~\ref{not:Lp-measure}}.
\end{definition}

\begin{lemma}[SAD-decomposition~\ref{cond:D3},~\ref{cond:D4},~\ref{cond:D6}]
\label{lem:Langevin:SAD}
    Let condition~\ref{cond:Langevin:P1} hold and let~$\Phi$ be weakly harmonic.
    Consider the SAD-decomposition $L=S-A$ on~$D$ with
    \begin{align*}
        Sf
        &\vdef
        \frac\alpha\beta\,\laplace{\wvertmetric*}f
        =
        \frac\alpha\beta\,\laplace{\vertmetric*}f
        -
        \alpha\canonfield f
        \\
        \quad\text{and}\quad
        Af
        &=
        -\spray{\wriemannmetric*}f
        \vdef
        -\spray{\riemannmetric*}f
        +
        \frac1\beta\,
        \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}(f)
    \end{align*}
    for all $f\in D$.
    Then, the following assertions hold:
    \begin{enumerate}[label={(\roman*)}]
        \item
        \label{itm:Langevin:SAD:S}
        $(S,D)$ is symmetric and negative semidefinite.
        \item
        \label{itm:Langevin:SAD:A}
        $(A,D)$ is antisymmetric.
        \item
        \label{itm:Langevin:SAD:L}
        For all $f\in D$ we have that
        $Lf \in \bigL1{\tanbundle{\posfold}}{\mu}$ and
        $\int_{\tanbundle{\posfold}} Lf\ \diffd\mu = 0$.
    \end{enumerate}
\end{lemma}
\begin{proof}
\hfill

    \begin{enumerate}[label={(\roman*)}]
        \item
        Combining the form of~$\grad{\sasakimetric*}\rho$
        from~\hyperref[rem:grad-loc-density]{Remark~\ref{rem:grad-loc-density}}
        with the result on the weighted Laplace-Beltrami,
        see~\hyperref[lem:weighted-Laplace]{Lemma~\ref{lem:weighted-Laplace}},
        we know that
        we have to look at $\frac1{\rho_F}\grad{\vertmetric*}\rho_F$.
        Condensing notation a bit
        we calculate that
        \begin{align*}
            \frac1{\rho_F}\grad{\vertmetric*}\rho_F(f)
            &=
            \grad{\mathrm{euc}}\parentheses*{
                -\frac\beta2\,\abs{\ID_F}{\mathrm{euc}}^2
            }(f)
            =
            -\beta\dualpair{\ID_{\tanbundle\posfold}}{df_0}
            =
            -\beta\pdot\canonfield f
        \end{align*}
        holds for all $f=\hlfunc{f_0}\in\kappa^\ast\smoothcompact{\posfold}$.
        Hence, it follows
        \begin{displaymath}
            \scalarprod{Sf}{g}{H}
            =
            -\int_{\tanbundle\posfold}
                \vertmetric{\grad{\vertmetric*}f}{\grad{\vertmetric*}g}
            \ \diffd\mu
            \qquad\text{for all }f,g\in D_0
        ,
        \end{displaymath}
        and therefore $(S,D_0)$ is symmetric and nonpositive definite.
        Since~$S$ is well-defined on~$D$ and~$D_0$ is dense in~$D$ wrt.\ graph norm,
        part~\ref{itm:Langevin:SAD:S} follows directly.
        \item
        By~\ref{thm:named:Liouville-thm}
        and similar reasoning as in~\hyperref[rem:adjoint-sec]{Remark~\ref{rem:adjoint-sec}}
        we know the adjoint of the Riemannian semispray wrt.~\bigL2{\tanbundle\posfold}{\wsasakimetric*}\=/scalar product:
        \begin{align*}
            \adjoint{\spray{\riemannmetric*}}
            &=
            -\spray{\riemannmetric*}
            - 0
            - \frac1\rho\, \spray{\riemannmetric*}\rho
            =
            -\spray{\riemannmetric*}
            + \frac1{\rho_F}\, \beta\rho_F \pdot \spray{\riemannmetric*}\vlfunc{\Psi}
            \\
            &=
            -\spray{\riemannmetric*}
            + \beta \pdot \spray{\riemannmetric*}\vlfunc{\Psi}
        .
        \end{align*}
        Furthermore, we compute the adjoint wrt.~\bigL2{\tanbundle\posfold}{\wsasakimetric*}\=/scalar product
        of $\grad{\vertmetric*}\hlfunc{\Psi} = \frac1\beta\,\grad{\vertmetric*}\hlfunc{\Phi}$
        using that~$\Psi$ is weakly harmonic
        and adapting~\hyperref[rem:grad-loc-density]{Remark~\ref{rem:grad-loc-density}} accordingly:
        \begin{align*}
            \adjoint*{\grad{\vertmetric*}\hlfunc{\Psi}}
            &=
            -
            \grad{\vertmetric*}\hlfunc{\Psi}
            -
            \underbrace{
                \divergence{\sasakimetric*}\parentheses*{
                    \grad{\vertmetric*}\hlfunc{\Psi}
                    }
                }_{=0}
            -
            \frac1\rho\, \parentheses*{ \grad{\vertmetric*}\hlfunc{\Psi} }(\rho)
            \\
            &=
            -\grad{\vertmetric*}\hlfunc{\Psi}
            -\frac1\rho\,
            \vlfunc{\rho_{\posfold}}
            \pdot
            \underbrace{
                \parentheses*{ \grad{\vertmetric*}\hlfunc{\Psi} }(\rho_F)
            }_{
                =\,\vertmetric
                {\grad{\vertmetric*}\hlfunc{\Psi}}
                {\grad{\vertmetric*}\rho_F}
            }
            \\
            &=
            -\grad{\vertmetric*}\hlfunc{\Psi}
            -\underbrace{
                \vertmetric
                {\grad{\vertmetric*}\hlfunc{\Psi}}
                {-\beta\canonfield}
            }_{
                =\,\hormetric
                {\grad{\hormetric*}\vlfunc{\Psi}}
                {-\beta\spray{\riemannmetric*}}
            }
            =
            -\grad{\vertmetric*}\hlfunc{\Psi}
            +\beta \pdot \spray{\riemannmetric*}\vlfunc{\Psi}
        .
        \end{align*}
        Hence, $(A,D)$ is antisymmetric.
        Since $(A,D)$ is densely defined on~$H$,
        it is closable.
        \item
        The integrability statement is clear:
        Let $h\in D$ arbitrary,
        then we conclude that
        $\int_{\tanbundle{\posfold}} Sh\,\diffd\mu = 0$
        by part~\ref{itm:Langevin:SAD:S}
        and
        $\int_{\tanbundle{\posfold}} Ah\,\diffd\mu = 0$
        by part~\ref{itm:Langevin:SAD:A}.
        We finish the proof pointing out
        that $(L,D)$ is densely defined on~$H$ and dissipative,
        thus it is closable.
    \end{enumerate}
\end{proof}

\begin{notation}
    Due~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}}
    $(S,D)$, $(A,D)$ and $(L,D)$ are closable.
    The closures we denote by~\operator{S}, \operator{A} and~\operator{L} respectively.
\end{notation}

From the easiest example of Euclidean space
we learn that one actually doesn't need a weakly harmonic potential.
To see this, we make use of Poisson manifolds.

\begin{example}
    Let $\posfold\vdef\Rnum_x^\posdim$.
    Then, the inverse~$\begin{pmatrix} 0 & -\ID{\posdim} \\ \ID{\posdim} & 0\end{pmatrix}$
    of `the' symplectic matrix
    yields an almost complex structure~$\mathbb{J}$ on
    $\tanbundle{\Rnum_x^\posdim}\simeq\Rnum_x^\posdim\times\Rnum_v^\posdim$.
    It's the same as the one constructed
    in~\cite[Paragraph~5]{Dombrowski} or later in~\cite{TachibanaOkumura},
    by our convention of listing the vertical component first and the horizontal one second.
    Moreover, it is \emph{compatible} with the Euclidean metric
    and the canonical symplectic form~$\Omega$ on
    $\tanbundle*{\Rnum_x^\posdim}\simeq\Rnum_x^\posdim\times\parentheses*{\Rnum^\posdim}^\ast$
    in the sense that
    \begin{displaymath}
        \scalarprod
            {v}
            {w}
            {\mathrm{euc}}
        =
        \Omega(v,\mathbb{J}w)
        \quad\text{and}\quad
        \Omega(v,w)
        =
        \scalarprod
            {\mathbb{J}v}
            {w}
            {\mathrm{euc}}
        \qquad\text{for all } v,w\in\Rnum^{2\posdim}
    ,
    \end{displaymath}
    cf.~\cite[Exercise~2.2.1]{MarsdenRatiu}.
    The symplectic form gives rise to a Poisson bracket~\Poissonbracket{\cdot}{\cdot}
    such that the correpsonding Poisson tensor reads as
    \begin{displaymath}
        \Poissonbracket{f}{g}(x,v)
        =
        -
        \scalarprod
            {\grad{\mathrm{euc}}f(x,v)}
            {\mathbb{J}\grad{\mathrm{euc}}g(x,v)}
            {\mathrm{euc}}
        =
        \Omega_{(x,v)}(df,dg)
    \end{displaymath}
    for all $f,g\in\smoothfunc{\Rnum^{2\posdim}}$ and $(x,v)\in\Rnum_x^\posdim\times\Rnum_v^\posdim$.

    In this terminology, the proof of~\cite[Lemma~3.4 part~ii)]{HypocoercMFAT} relies on
    linking the operator~$(A,D)$ to the antisymmetric bilinear form~$(\mathcal{A},D)$
    of integrating minus the Poisson bracket wrt.~$\mu$:
    $
        \mathcal{A}(f,g)
        \vdef
        \int_{\Rnum^{2\posdim}}
            - \Poissonbracket{f}{g}
        \ \diffd\mu
    $
    for all $f,g\in D=\smoothcompact{\Rnum^{2\posdim}}$.
    Via integration by parts one can show that
    $
        \scalarprod
            {Af}
            {g}
            {\bigL2{\mu}}
        =
        \mathcal{A}(f,g)
    $ holds for all $f,g\in\smoothcompact{\Rnum^{2\posdim}}$.
    This can be done \emph{without the assumption} of a weakly harmonic potential,
    since the vector field action can be represented in terms of the Hamiltonian vector fields.
    Indeed:
    Denote by~\Hamilton{f} the \emph{Hamiltonian vector field} of~$f\in\smoothfunc{\Rnum^{2\posdim}}$, i.\,e.\
    it fulfils $\Hamilton{f}(g)=\Poissonbracket{f}{g}$
    for all $g\in\smoothfunc{\Rnum^{2\posdim}}$.
    We get the explicit formula
    $
        -\Hamilton{f}
        =
        \scalarprod
            {\grad{v}f}
            {\grad{x}}
            {\mathrm{euc}}
        -
        \scalarprod
            {\grad{x}f}
            {\grad{v}}
            {\mathrm{euc}}
    $
    and conclude that~\Hamilton{f} is solenoidal by Schwarz's theorem.
    Hence, the following equality is true:
    \begin{displaymath}
        \mathcal{A}(f,g)
        =
        \int_{\Rnum^{2\posdim}}
            \Hamilton{f}(\rho)
            \pdot g
        \ \diffd\Leb
        =
        \scalarprod
            {Af}
            {g}
            {\bigL2{\mu}}
        \qquad\text{for all }f,g\in D
    ,
    \end{displaymath}
    where $\rho\vdef\frac{d\mu}{d\Leb}$ is the (loc-)density of~$\mu$.
\end{example}

\begin{proposition}[SAD-decomposition (2nd version)]
\label{prop:Langevin:SAD}
    The assertions of~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}}
    are true without the assumption of~$\Psi$ being weakly harmonic.
\end{proposition}
\begin{proof}
    We enhance the proof of~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}} part~\ref{itm:Langevin:SAD:A}
    via the technique discussed in the previous example
    which works due to the particular choices of
    the Gaussian fibre measure and the Sasaki metric wrt.\ the most natural Ehresmann connection on~\tantanbundle\posfold.

    Let's abbreviate by
    $
        \rho
        \vdef
        \frac{d\mu}{d\Leb{\sasakimetric*}}
        =
        \frac{d\Leb{\wsasakimetric*}}{d\Leb{\sasakimetric*}}
    $
    the (loc-)density of~$\mu$.
    Denote by~$\mathbb{J}$ minus the almost complex structure on~\tanbundle\posfold
    constructed in~\cite{Dombrowski},
    and let~$\Omega$ be the canonical symplectic form on~\tanbundle*\posfold.
    By construction, the Sasaki metric, $\mathbb{J}$ and~$\Omega$ are compatible,
    cf.~\cite[page~341]{MarsdenRatiu}.
    Hence, they define the same Poisson bracket \Poissonbracket{\cdot}{\cdot} on~\tanbundle\posfold via the assignments
    \begin{displaymath}
        \Omega_{v}(df,dg)
        \fedv
        \Poissonbracket{f}{g}(v)
        \vdef
        -
        \sasakimetric
            {\grad{\sasakimetric*}f(v)}
            {\mathbb{J}\grad{\sasakimetric*}g(v)}
            {v}
    \end{displaymath}
    for all $f,g\in\smoothfunc{\tanbundle\posfold}$ and $v\in\tanbundle\posfold$.
    Thus, for any fixed $f\in\smoothfunc{\tanbundle\posfold}$
    there is a unique Hamilton vector field~\Hamilton{f}
    by~\cite[Proposition~10.2.1]{MarsdenRatiu}.
    Let a chart $(v^j)_{j=1}^{2\posdim}$
    that gives normal coordinates and respects the Ehresmann connection
    such that $(\partial v^i)_{i=1}^{\posdim}$ provides local basis for vertical vector fields
    and $(\partial v^{k+\posdim})_{k=1}^{\posdim}$ provides a basis for the horizontal vector fields;
    we find that in these coordinates the Hamiltonian vector fields attain the form
    \begin{displaymath}
        -\Hamilton{f}
        =
        \sum_{k=1}^{\posdim}
            \partial_{v^k}f \pdot \partial_{v^{k+\posdim}}
        -
        \sum_{i=1}^{\posdim}
            \partial_{v^{i+\posdim}}f \pdot \partial_{v^i}
    .
    \end{displaymath}
    In any such coordinates we easily compute that
    $\divergence{\sasakimetric*}(\Hamilton{f})=0$ using Schwarz's Theorem,
    in other words all~\Hamilton{f} are solenoidal.

    Define the antisymmetric bilinear form $(\mathcal{A},D)$ by
    $
        \mathcal{A}(f,g)
        \vdef
        \int_{\tanbundle\posfold}
            - \Poissonbracket{f}{g}
        \ \diffd\mu
    $
    for all $f,g\in D=\smoothcompact{\tanbundle\posfold}$.
    From the Divergence Theorem it follows that
    \begin{displaymath}
        \mathcal{A}(f,g)
        =
        \int_{\tanbundle\posfold}
            -\Hamilton{f}(g)
        \ \diffd\Leb{\wsasakimetric*}
        =
        \int_{\tanbundle\posfold}
            \Hamilton{f}(\rho)
            \pdot g
        \ \diffd\Leb{\sasakimetric*}
    .
    \end{displaymath}
    Using our comments on the Sasaki gradient of a loc\=/density,
    see~\hyperref[rem:grad-loc-density]{Remark~\ref{rem:grad-loc-density}},
    we can infer that
    $\Hamilton{f}(\rho) = -\rho\pdot\spray{\wriemannmetric*}f$.
    In a nutshell,
    we localise in the support of~$f$
    via a partition of unity argument
    where the corresponding open cover is formed by
    charts $(v^j)_{j=1}^{2\posdim}$
    that are respecting the Ehresmann connection
    and also are restricted to domains of local trivialisation;
    therein, we can use the local coordinate form of~\Hamilton{f}
    and that the loc\=/density~$\rho$ trivialises to a product of exponential-type densities.
    Hence, we gain that
    $
        \scalarprod
            {Af}
            {g}
            {H}
        =
        \mathcal{A}(f,g)
    $ for all $f,g\in D$
    which finishes the proof.
\end{proof}

We are going to construct the projection~$P$ mentioned in~\ref{cond:D5}.
For every $f\in H$ we call the mapping
\begin{displaymath}
    \expect[\nu]{f}\colon\
    \posfold\to\Rnum,\
    x \longmapsto \int_{\tanbundle{\posfold}{x}} f \ \diffd\nu
\end{displaymath}
the \emph{fibrewise average of~$f$}.
Clearly, the operator $\expect[\nu]$ acts trivially on vertically lifted functions,
since they are fibrewise constant:
$\expect[\nu]{\vlfunc{f_0}} = f_0$
for all $f_0\in\bigL2{\posfold}{\wriemannmetric*}$.
We define $P_Sf\vdef\expect[\nu]{f}\bincirc\tanproj$ assigning
to an element of the tangent space the average of~$f$ over the fibre corresponding to this element.
Thinking in a local trivialisation
the vertical lift of the fibrewise average erases the dependency of~$f$ on the velocity component,
since~$f$ can be thought as a bivariate function of position and velocity in this localisation.
The range of~$P_S$ is precisely
$P_S(H) = \tanproj^\ast\parentheses*{ \bigL2{\posfold}{\wriemannmetric*} }$,
the set of vertically lifted \bigL2\=/functions on the weighted position manifold,
and~$P_S$ is a projection.

Moreover, we have that
\begin{align*}
    \int_{\tanbundle{\posfold}} (P_Sf)^2 \ \diffd\mu
    &=
    \int_{\tanbundle{\posfold}}
        (\expect[\nu]{f}\bincirc\tanproj)^2
    \ \diffd\Leb{\wriemannmetric*}\locprod\nu
    =
    \int_{\tanbundle{\posfold}}
        (\expect[\nu]{f})^2\bincirc\tanproj
    \ \diffd\Leb{\wriemannmetric*}\locprod\nu
    \\
    &=
    \int_{\tanbundle{\posfold}}
        (\expect[\nu]{f})^2\bincirc\tanproj
    \ \diffd\tanproj^\ast\Leb{\wriemannmetric*}
    =
    \int_{\posfold}
        \expect[\nu]{f}^2
    \ \diffd\Leb{\wriemannmetric*}
\end{align*}
for all $f\in H$.
This implies
$
    \norm{P_Sf}{H}
    =
    \norm{ \expect[\nu]{f} }{\bigL2{\posfold}{\wriemannmetric*}}
    \leq
    \norm{ f }{H}
$
meaning that~$P_S$ is continuous with norm~1.
Last but not least, $P_S$ also is an orthogonal projection,
cf.~\cite[Proposition~3.3]{Conway}.
Finally, we define $P \vdef P_S - \scalarprod{\cdot}{1}{H}$.

\begin{lemma}[Properties of projection~$P$ and semigroup conservativity~\ref{cond:D5}, \ref{cond:D7}]
\label{lem:Langevin:D5D7}
    Let condition~\ref{cond:Langevin:P1} hold.
    Then, we have
    $P(H)\subseteq\opdomain{S}$, $SP=0$,
    $P(D)\subseteq\opdomain{A}$ and $AP(D)\subseteq\opdomain{A}$.
    Furthermore, $1\in\opdomain{L}$ and $L1 = 0$.
\end{lemma}
\begin{proof}
    We follow the lines of the proof of~\cite[Lemma~3.4~(iv)]{HypocoercMFAT}
    for the Euclidean case.

    First of all, we start with the statements related to~$S$.
    Let $\varphi\in\smoothcompact{\posfold;[0,1]}$ be a cut-off function\footnote{
        An explicit choice of~$\varphi$ is given as follows:
        Let~$h$ be a chart at~$o$ --
        wlog.\ the ball~\ball{o}{2} is contained in the chart domain.
        Define the auxiliary `mountain' function
        \begin{displaymath}
            m\colon\ \Rnum\to\Rnum,\
            t\longmapsto \exp\parentheses*{-\frac1{t(1-t)}}\pdot\indicator{(0,1)}(t)
        .
        \end{displaymath}
        We transform the `mountain' to a `table mountain'
        by the assignment
        \begin{displaymath}
            \tau\colon\
            \Rnum^\posdim\to[0,1],\,
            y\longmapsto\frac{\int_0^{2-\norm{y}{2}} m(t)\ \diffd t}{\int_0^1 m(t)\ \diffd t}
        .
        \end{displaymath}
        Then, the choice
        $\varphi\vdef\tau\bincirc h\pdot\indicator{\ball{o}{2}}$
        yields a smooth function with the desired properties.
    }
    such that $\varphi=1$ on~\ball{o}{1}
    and $\varphi=0$ on~\ball{o}{2}.
    Define $\varphi_n\vdef\varphi(\sfrac{\ID}{n})$ for all $n\in\Nnum\setminus\{0\}$.
    Note that there is some constant $c\in(0,\infty)$ such that
    \begin{displaymath}
        \abs{\grad{\riemannmetric*}\varphi_n(x)}{\riemannmetric*}
        \leq
        \frac{c}n\,

        \quad\text{and}\quad
        \abs{\operatorname{Hess}_{\riemannmetric*}\varphi_n(x)}{\infty}
        \leq
        \frac{c}{n^2}\,

    \end{displaymath}
    for all $x\in\posfold$ and $n\in\Nnum\setminus\{0\}$,
    cf.~\cite[Definition~3.3]{HypocoercMFAT}.
    Clearly, $(\varphi_n)_{n\in\Nnum}$ pointwisely converges to the constant function~1 as $n\to\infty$.
    Let $f_0\in\smoothcompact{\posfold}$
    and define
    \begin{equation}
    \label{eq:Langevin:approxvertical1}
        f_n
        \vdef
        \vlfunc{f_0}\tensorprod\hlfunc{\varphi_n}
        =
        \vlfunc{f_0}\pdot\hlfunc{\varphi_n}
        \qquad\text{for all }n\in\Nnum\setminus\{0\}
    .
    \end{equation}
    This yields a sequence in~$D$ converging to~$\vlfunc{f_0}$ in~$H$.
    From dominated convergence we can conclude that
    \begin{displaymath}
        Sf_n
        =
        \frac\alpha\beta\,\vlfunc{f_0}
        \pdot
        \laplace{\vertmetric*}\hlfunc{\varphi_n}
        -
        \alpha \vlfunc{f_0}\pdot\canonfield\hlfunc{\varphi_n}
        \longrightarrow
        0
        \qquad\text{in }H\text{ as } n\to\infty
    ,
    \end{displaymath}
    where we use
    that the function
    $
        \tanbundle\posfold\to\Rnum,\,
        v \mapsto
        \abs{\canonfield(v)}{\sasakimetric*}
        =
        \abs{v}{\riemannmetric*}
    $
    is in~$H$
    as on each fibre $\tanbundle{\posfold}{x}\simeq\Rnum^\posdim$
    the function
    $
        \abs{\ID_{\Rnum^\posdim}}{\riemannmetric*}
    $
    is in~\bigL2{\Rnum^\posdim}{\nu}.
    Since \operator{S} is closed,
    we have shown that $\vlfunc{f_0}\in\opdomain{S}$, and more specifically $S\vlfunc{f_0}=0$.
    Next, we prove that~$P$ maps into the null space of \operator{S}.
    As the range of~$P$ is contained in~$\tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}$,
    we pick an $g_0\in\bigL2{\posfold}{\wriemannmetric*}$
    and show that $\vlfunc{g_0}\in\opdomain{S}$ and $S\vlfunc{g_0}=0$.
    The space~\smoothcompact{\posfold} is dense in~\bigL2{\posfold}{\wriemannmetric*},
    so there is a sequence $(g_n)_{n\in\Nnum\setminus\{0\}}$ in~\smoothcompact{\posfold}
    approximating~$g_0$ in \bigL2{\posfold}{\wriemannmetric*}.
    We have seen just before
    that $(\vlfunc{g_n})_{n\in\Nnum\setminus\{0\}}$ is a sequence in~\opdomain{S}
    and $S\vlfunc{g_n}=0$ for all $n\in\Nnum$.
    Again, with closedness of \operator{S}
    it follows that $\vlfunc{g_0}\in\opdomain{S}$ and $S\vlfunc{g_0}=0$.

    Now, we turn to the statements involving~$A$.
    Let $f_0\in\smoothcompact{\posfold}$
    and define the sequence $(f_n)_{n\in\Nnum\setminus\{0\}}$ as in~\eqref{eq:Langevin:approxvertical1}
    approximating $\vlfunc{f_0}$ in~$H$.
    Using
    $
        \abs{\spray{\riemannmetric*}}{\sasakimetric*}
        =
        \abs{\ID_{\tanbundle\posfold}}{\riemannmetric*}\in H
    $
    as above and that
    $\grad{\riemannmetric*}\Phi\in\bigL2{\posfold\to\tanbundle\posfold}{\wriemannmetric*}$
    we can infer via dominated convergence that
    \begin{align*}
        Af_n
        &=
        -\hlfunc{\varphi_n}\pdot\spray{\riemannmetric*}\vlfunc{f_0}
        +
        \frac1\beta\vlfunc{f_0}
        \pdot
        \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}\hlfunc{\varphi_n}
        =
        -\hlfunc{\varphi_n}\pdot\spray{\riemannmetric*}\vlfunc{f_0}
        +
        \frac1\beta\vlfunc{f_0}
        \pdot
        \riemannmetric{\grad{\riemannmetric*}\Phi}{\grad{\riemannmetric*}\varphi_n}
        \\
        &\longrightarrow
        -\spray{\riemannmetric*}\vlfunc{f_0}
        +
        \frac1\beta\vlfunc{f_0}
        \pdot
        0
        \qquad\text{in } H\text{ as }n\to\infty
    .
    \end{align*}
    Closedness of \operator{A} implies $\vlfunc{f_0}\in\opdomain{A}$ as well as
    $
        A\vlfunc{f_0}
        =
        -\spray{\riemannmetric*}\vlfunc{f_0}
    $.
    For the inclusion $P(D)\subseteq\opdomain{A}$
    it's enough to show that $1\in\opdomain{A}$ and $A1=0$.
    Similar as before the function
    $
        \tanbundle\posfold\to\Rnum,\,
        v \mapsto
        \abs{\spray{\riemannmetric*}(v)}{\sasakimetric*}
        =
        \abs{v}{\riemannmetric*}
    $
    is in~$H$,
    and thus dominated convergence gives us
    \begin{displaymath}
        A\vlfunc{\varphi_n}
        =
        -\spray{\riemannmetric*}\vlfunc{\varphi_n}
        \longrightarrow
        0
        \qquad\text{in } H\text{ as }n\to\infty
    .
    \end{displaymath}
    In order to prove now that $AP(D)\subseteq\opdomain{A}$
    we adhere to our approximation strategy and define
    $
        h_n
        \vdef
        \hlfunc{\varphi_n}\pdot\spray{\riemannmetric*}\vlfunc{f_0}
    $
    for all $n\in\Nnum\setminus\{0\}$.
    The sequence $(h_n)_{n\in\Nnum}$ converges to~$\spray{\riemannmetric*}\vlfunc{f_0}$
    both pointwisely and in~$H$.
    We note that the function
    $
        \spray{\riemannmetric*}^2\vlfunc{f_0}
        =
        \spray{\riemannmetric*}(\spray{\riemannmetric*}\vlfunc{f_0})
    $
    is dominated by
    \begin{displaymath}
        \norm{\spray{\riemannmetric*}}{\bigL2{\tanbundle\posfold\to\tantanbundle\posfold}{\mu}}
        \pdot
        \norm{\spray{\riemannmetric*}\vlfunc{f_0}}{H}
        \leq
        \norm{\spray{\riemannmetric*}}{\bigL2{\tanbundle\posfold\to\tantanbundle\posfold}{\mu}}^2
        \pdot
        \norm{\vlfunc{f_0}}{H}
        =
        \norm{\abs{\ID_{\tanbundle\posfold}}{\riemannmetric*}}{H}^2
        \pdot
        \norm{\vlfunc{f_0}}{H}
    \end{displaymath}
    due to the Cauchy-Bunyakovsky-Schwarz inequality applied twice.
    This dominating function is in~$H$
    as on each fibre $\tanbundle{\posfold}{x}\simeq\Rnum^\posdim$
    the function
    $
        \abs{\ID_{\Rnum^\posdim}}^2
    $
    is in~\bigL2{\Rnum^\posdim}{\nu}.
    Together with afore-mentioned facts that
    $\abs{\ID_{\tanbundle\posfold}}{\riemannmetric*}\in H$
    and
    $\grad{\riemannmetric*}\Phi\in\bigL2{\posfold\to\tanbundle\posfold}{\wriemannmetric*}$
    this yields
    \begin{align*}
        Ah_n
        &=
        -\hlfunc{\varphi_n}\pdot\spray{\riemannmetric*}^2\vlfunc{f_0}
        +
        \frac1\beta
        \pdot
        \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}h_n
        \\
        &=
        -\hlfunc{\varphi_n}\pdot\spray{\riemannmetric*}^2\vlfunc{f_0}
        +
        \frac1\beta
        \pdot
        \parentheses*{
            \parentheses*{\spray{\riemannmetric*}\vlfunc{f_0}}
            \pdot
            \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}\hlfunc{\varphi_n}
            +
            \hlfunc{\varphi_n}
            \pdot
            \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}(\spray{\riemannmetric*}\vlfunc{f_0})
        }
        \\
        &\longrightarrow
        -\spray{\riemannmetric*}^2\vlfunc{f_0}
        +
        0
        +
        \frac1\beta
        \pdot
        \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}(\spray{\riemannmetric*}\vlfunc{f_0})
        \qquad\text{in } H\text{ as }n\to\infty
    \end{align*}
    by dominated convergence.
    Since \operator{A} is closed,
    the function $\spray{\riemannmetric*}\vlfunc{f_0}$ is an element of~\opdomain{A}.

    Finally, the statements on~$L$ follow the very same way:
    Let $f_0\in\smoothcompact{\posfold}$
    and define the sequence $(f_n)_{n\in\Nnum\setminus\{0\}}$ as in~\eqref{eq:Langevin:approxvertical1};
    repeat the previous steps and conclude from closedness of~\operator{L}
    that $f=\vlfunc{f_0}\in\opdomain{L}$ and $L\vlfunc{f_0} = - A\vlfunc{f_0}$.
    In particular, the sequence
    $\parentheses*{L\vlfunc{\varphi_n}}_{n\in\Nnum}$
    converges in~$H$ to~0 as $n\to\infty$,
    and by closedness it follows
    $1\in\opdomain{L}$ with $L1=0$.
\end{proof}

\begin{remark}
\label{rem:Langevin:APf-Eq2}
    As we deduced $Pf\in\opdomain{A}$ for all $f\in D$,
    we can calculate that
    \begin{align}
    \label{eq:Langevin:APf-Eq1}
    \begin{aligned}
        APf
        &=
        -\spray{\wriemannmetric*}\parentheses*{\vlfunc{\expect[\nu]{f}}}
        =
        -\dualpair{
            \spray{\riemannmetric*}
        }{
            d\expect[\nu]{f} \bincirc d\tanproj
        }
        \\
        &=
        -\dualpair{ \ID_{\tanbundle\posfold} }{ d\expect[\nu]{f} }
        =
        -d\expect[\nu]{f}
    .
    \end{aligned}
    \end{align}
    However, one can give this formula a more intuitive form:
    \begin{align}
    \label{eq:Langevin:APf-Eq2}
    \begin{aligned}
        APf
        &=
        -\spray{\wriemannmetric*}\parentheses*{Pf}
        =
        -\spray{\riemannmetric*}\parentheses*{\vlfunc{\expect[\nu]{f}}}
        =
        -\hormetric{
            \spray{\riemannmetric*}
        }{
            \grad{\hormetric*}(P_Sf)
        }
        \\
        &=
        -\hormetric{
            \spray{\riemannmetric*}
        }{
            \horlift\parentheses*{ \grad{\riemannmetric*}\expect[\nu]{f} }
        }
        =
        -\riemannmetric{
            \ID_{\tanbundle\posfold}
        }{
            \grad{\riemannmetric*}\expect[\nu]{f} \bincirc \tanproj
        }{\tanproj}
    .
    \end{aligned}
    \end{align}
    Unlike \hyperref[eq:Langevin:APf-Eq1]{Equation~\eqref{eq:Langevin:APf-Eq1}},
    we recognise that~\hyperref[eq:Langevin:APf-Eq2]{Equation~\eqref{eq:Langevin:APf-Eq2}}
    is in perfect correspondence to~\cite[Equation~(3.12)]{HypocoercMFAT}.
\end{remark}

Turning to the remaining condition~\ref{cond:D2}, i.\,e.\
the question whether~\operator{L} generates a strongly continuous semigroup,
we first look into the case of a smooth potential $\Psi\in\smoothfunc{\posfold}$.
The proof relies on methods from~\cite{HelfferNier}
as explained in~\cite[Section~4]{HypocoercJFA}.
We briefly quote a consequence of the Hörmander~Theorem,
namely~\cite[Proposition~A.1]{HelfferNier}.
Let~$T$ be a second order differential operator
on a Riemannian manifold~$(\basefold,\basemetric*)$
of the form $T = c + \Xfield_0 + \sum_{k=1}^\ell \Xfield_k$
with $c\in\smoothfunc{\basefold}$
and $\Xfield_k\in\smoothsec{\tanbundle\basefold}$ for all $k\in\{0,\ldots,\ell\}$.
We say that~$T$ satisfies the \emph{Hörmander condition}
if at any point $b\in\basefold$ holds
\begin{displaymath}
    \dim(\Liespan{\Xfield_0,\ldots,\Xfield_\ell}{b})
    =
    \dim(\tanbundle{\basefold}{b})
    =
    \dim(\basefold)
    ,
\end{displaymath}
where \Liespan{\Xfield_0,\ldots,\Xfield_\ell} denotes the generated Lie~algebra.

\begin{proposition}
    Let~$T$ satisfy the Hörmander condition
    and let $f\in\Lloc1{\basefold}{\basemetric*}$ such that
    \begin{displaymath}
        \int_{\basefold} f \pdot T\psi \ \diffd\Leb{\basemetric*}
        =
        0
        \qquad\text{for all }
        \psi\in\smoothcompact{\basefold}
    .
    \end{displaymath}
    Then, $f$ has a smooth representative.
\end{proposition}
\begin{proof}
    The proof of this proposition works as for~\cite[Proposition~A.1]{HypocoercJFA}:
    It is done in chart domains,
    and within these domains it's perfectly fine to consider
    $\Xfield_j = \partial x^j = \frac{\partial}{\partial x^j}$,
    where $\parentheses*{x^j}_{j=1}^{\dim(\basefold)}$ denotes local coordinates provided by the chart.
    Then, we have smooth representatives in chart domains
    serving as starting point for a partition of unity argument.
    Thus, even if~$T$ is just available in local coordinate form,
    we can apply the previous proposition
    as soon as the respective chart domains form an open cover.
\end{proof}

\begin{lemma}[Hörmander condition for the Langevin generator]
\label{lem:Langevin:generator-Hörmander}
    Consider a smooth potential $\Psi\in\smoothfunc{\posfold}$.
    Then, the Langevin generator satisfies the Hörmander condition.
    More precisely, let $v\in\tanbundle\posfold$
    and $\parentheses*{x^j}_{j=1}^{\posdim}$ be local coordinates corresponding to a chart at~$\tanproj(v)$.
    Then, we have that
    \begin{align*}
        \dim\parentheses*{\Liespan{
            \spray{\riemannmetric*},
            \vertlift\parentheses*{\grad_{\riemannmetric*}\Psi},
            \vertlift\parentheses*{\partial x^1},
            \ldots,
            \vertlift\parentheses*{\partial x^\posdim},
            \canonfield
        }{v}}
        &=
        \dim(\tantanbundle{\posfold}{v})
        \\
        &=
        \dim(\tanbundle\posfold)
        =
        2\posdim
    .
    \end{align*}
\end{lemma}
\begin{proof}
    Obviously, neither $\vertlift\parentheses*{\grad_{\riemannmetric*}\Psi}$ nor~\canonfield
    contribute anything to the generated Lie algebra.
    From~\cite[Proposition~5.1]{GudmundssonKappos} we know explicit forms
    for the Lie brackets of vertically and horizontally lifted vector fields in any combination.
    With that said, we have the equations
    \begin{displaymath}
        \Liebracket{\vertlift(\Xfield)}{\vertlift(\Yfield)}
        =
        0
        \qquad\text{and}\qquad
        \Liebracket{\horlift(\Yfield)}{\vertlift(\Xfield)}
        =
        \vertlift\parentheses*{\connection{\Yfield}{\riemannmetric*}\Xfield}
    \end{displaymath}
    for all $\Xfield,\Yfield\in\smoothsec{\tanbundle\posfold}$.
    Thus, the only nontrivial pairing of the seeding vector fields
    $
        \Xfield,\Yfield
        \in
        \{
            \partial x^1,
            \ldots,
            \partial x^\posdim
        \}
    $
    is
    $
        \Liebracket{\horlift(\Yfield)}{\vertlift(\Xfield)}
        =
        \vertlift\parentheses*{\connection{\Yfield}{\riemannmetric*}\Xfield}
    $,
    but clearly all vector fields of this form are linear dependent of the vertical vector fields
    $
        \left\{
            \vertlift\parentheses*{\partial x^1},
            \ldots,
            \vertlift\parentheses*{\partial x^{\posdim}}
        \right\}
    $
    generating the Lie algebra.
    This just shows that the collection
    $
        \{
            \partial x^1,
            \ldots,
            \partial x^{\posdim}
        \}
    $
    might not serve our purpose
    even if we would build our generator with both kinds of liftings.

    The statement concerning Lie brackets of one vertically and one horizontally lifted vector fields
    does not apply to~\spray{\riemannmetric*},
    as a semispray can not arise as horizontal lift of a vector field.
    However, in~\hyperref[lem:app:bracket-with-spray]{Lemma~\ref{lem:app:bracket-with-spray}}
    we prove that
    $
        \Liebracket{
            \spray{\riemannmetric*}
        }{
            \vertlift\parentheses*{\partial x^j}
        }
        =
        \horlift\parentheses*{\partial x^j}
        -
        \sum_{i=1}^{\posdim}
            N_j^i \pdot \vertlift\parentheses*{\partial x^i}
    $
    for certain functions~$N_j^i$.
    This yields \posdim~many linear independent horizontal vector fields in the generated Lie~algebra
    which finishes the proof.
\end{proof}

\begin{theorem}[\ref{cond:D2}~for smooth potentials using a hypoellipticity startegy]
\label{thm:Langevin:D2-smooth-case}
    Let $\Psi\in\smoothfunc{\posfold}$ be a smooth potential.
    Then, $(L,D)$~is essentially m-dissipative.
    Thus, its closure~\operator{L} generates a strongly continuous semigroup.
\end{theorem}
\begin{proof}
    From~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}} we know that
    $(L,D)$ is dissipative.
    In view of the Lumer-Philips~Theorem,
    we have to show that
    the range $(\ID_H-L)(D)$ is dense in~$H$.
    Let $f\in H$ fixed such that
    \begin{equation}
    \label{eq:aux:hypoellip-D2}
        \scalarprod{(\ID_H-L)u}{f}{H}
        =
        0
        \qquad\text{for all }u\in D
    .
    \end{equation}
    We claim that $f=0$.

    Due to the choice of~$f$ we have that
    $\exp(-\vlfunc{\Phi})f\in\Lloc1{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}$
    and by~\hyperref[lem:Langevin:generator-Hörmander]{Lemma~\ref{lem:Langevin:generator-Hörmander}}
    we can assume that $\exp(-\vlfunc{\Phi})f$ is smooth.
    Let $(\varphi_n)_{n\in\Nnum}$ be a sequence of cut-off functions in \smoothcompact{\posfold;[0,1]}
    with $\varphi_n\rightarrow1$ pointwise as $n\to\infty$
    and such that
    $
        \norm{ \grad{\riemannmetric*}\varphi_n }{\bigL\infty{\posfold\to\tanbundle\posfold}{\wriemannmetric*}}^2
        \leq
        \frac1n\,C
    $
    holds for all $n\in\Nnum\setminus\{0\}$ and some $C\in(0,\infty)$.
    Now, define $u_n\vdef\parentheses*{\vlfunc{\varphi_n}}^2f$ for all $n\in\Nnum$.
    It's clear that
    \begin{displaymath}
        \scalarprod{u_n}{f}{H}
        \overset{\eqref{eq:aux:hypoellip-D2}}{=}
        \scalarprod{Lu_n}{f}{H}
        =
        \scalarprod{Su_n}{f}{H}
        -
        \scalarprod{Au_n}{f}{H}
    .
    \end{displaymath}
    Due to the fact that $(S,D)$ is nonpositive definite and multiplying with $\varphi$ commutes with the action of~$S$
    we see that
    $
        \scalarprod{Su_n}{f}{H}
        =
        \scalarprod{S(\vlfunc{\varphi_n}f)}{\vlfunc{\varphi_n}f}{H}
        \leq
        0
    $.
    Similarly, using antisymmetry we get that
    \begin{displaymath}
        \scalarprod{Au_n}{f}{H}
        =
        \scalarprod{
            A\parentheses*{\vlfunc{\varphi_n}}\,\vlfunc{\varphi_n}f
        }{f}{H}
        +
        \scalarprod{A(\vlfunc{\varphi_n}f)}{\vlfunc{\varphi_n}f}{H}
        =
        \scalarprod{
            A\parentheses*{\vlfunc{\varphi_n}}\,\vlfunc{\varphi_n}f
        }{f}{H}
    .
    \end{displaymath}
    Altogether, this yields the estimate
    \begin{displaymath}
        \scalarprod{u_n}{f}{H}
        =
        \int_{\tanbundle\posfold} \parentheses*{\vlfunc{\varphi_n}}^2f^2 \ \diffd\mu
        \leq
        \frac1n\,C
        \int_{\tanbundle\posfold} \vlfunc{\varphi_n} f^2 \ \diffd\mu
        \leq
        \frac1n\,C
        \norm{ f }{H}
    \end{displaymath}
    using~\eqref{eq:Langevin:APf-Eq2}.
    By dominated convergence this implies that
    $\norm{ f }{H}^2 \leq 0$,
    thus $f=0$.
\end{proof}

For the case of a loc-Lipschitzian potential,
we leave the base weight aside for a moment,
but keep the fibre weight.
In other words,
we could think of~\tanbundle\posfold endowed with bundle weight $1\locprod\rho_F$.
Let
$\Psi\in\Lloc1{\posfold}{\riemannmetric*}$,
and
$L_0 \vdef \frac\alpha\beta\,\laplace{\wvertmetric*} +\, \spray{\riemannmetric*}$
be defined on the very same set~$D_0$ as before --
it should be clear at this point that
$L_0$ has to consist of the weighted diffusion and the \emph{non}-corrected semispray
instead of~\spray{\wriemannmetric*}.

\begin{lemma}
\label{lem:Langevin:L0D0-ess-m-dissipative}
    $(L_0,D_0)$ is essentially m-dissipative on~\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}.
\end{lemma}
\begin{proof}
    From~\autoref{thm:Langevin:D2-smooth-case} applied to the smooth case of the zero potential,
    we know that
    $(L_0,D)$ is essentially m-dissipative
    on~$\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}$.
    We have to show that
    $(L_0,D)$ is contained in the closure of $(L_0,D_0)$.

    For any $f\in D$ there is an approximating sequence $(f_n)_{n\in\Nnum}$ in~$D_0$
    wrt.\ usual locally convex topology implying uniform convergence of all derivatives on compacts
    cf.~\hyperref[lem:D0isdense]{Lemma~\ref{lem:D0isdense}}.
    Furthermore, there is a common compact set in~\tanbundle{\posfold}  large enough
    containing~$\supp(f)$ and~$\supp(f_n)$ for all $n\in\Nnum$.
    Therefore, we have that
    $\sup_{v\in\tanbundle\posfold} \abs{ L_0 f_n(v) - L_0 f(v) } \rightarrow0$
    as $n\to\infty$.
    Thus, $f_n\rightarrow f$ and $L_0 f_n \rightarrow L_0 f$
    in~\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu} as $n\to\infty$.
\end{proof}

Denote by~$\diriboundsobolev1\infty{\posfold}$
the closure of~\smoothcompact{\posfold} wrt.\ \sobolev1\infty\=/norm.
Define
\begin{displaymath}
    D_1
    \vdef
    \tanproj^\ast\diriboundsobolev1\infty{\posfold}
    \tensorprod
    \kappa^\ast\smoothcompact{\posfold}
.
\end{displaymath}
Considering the operator $(L,D_1)$ we realise that
\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}} still does apply --
the proof has to be adapted just slightly.
Further on, we define the unitary isomorphism
\begin{displaymath}
    U\colon\
    H=\bigL2{\tanbundle\posfold}{\Leb{\wriemannmetric*}\locprod\nu}
    \to
    \bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu},\
    f \longmapsto \vlfunc{\exp\parentheses*{-\sfrac\Phi2}}\pdot f
.
\end{displaymath}
Note that $U(D_1) = D_1$.
Thus, we define the operator $\widetilde{L}\vdef ULU^{-1}$ on~$D_1$.
One directly observes that
\begin{displaymath}
    \widetilde{L}
    =
    \widetilde{S} - \widetilde{A}
    \vdef
    U\laplace{\wvertmetric*}U^{-1} - UAU^{-1}
    \qquad\text{holds on } D_1
\end{displaymath}
with operators $(\widetilde{S},D_1)$ symmetric, negative semidefinite
and $(\widetilde{A},D_1)$ antisymmetric
both on~\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu},
as well as
\begin{align*}
    \widetilde{A}\vlfunc{f_0}
    &=
    -U
    \parentheses*{
        \vlfunc{f_0}
        \pdot
        \riemannmetric
            {\ID_{\tanbundle\posfold}}
            {\grad{\riemannmetric*}\exp(\sfrac\Phi2)\bincirc\tanproj}
            {\tanproj}
        +
        \vlfunc{\exp(\sfrac\Phi2)}
        \pdot
        \riemannmetric
            {\ID_{\tanbundle\posfold}}
            {\grad{\riemannmetric*}f_0\bincirc\tanproj}
            {\tanproj}
    }
    \\
    &=
    -\frac12\,\vlfunc{f_0}
    \pdot
    \riemannmetric
        {\ID_{\tanbundle\posfold}}
        {\grad{\riemannmetric*}\Phi\bincirc\tanproj}
        {\tanproj}
    -
    \riemannmetric
        {\ID_{\tanbundle\posfold}}
        {\grad{\riemannmetric*}f_0\bincirc\tanproj}
        {\tanproj}
    \\
    &=
    -\frac12\,\vlfunc{f_0}
    \pdot
    \spray{\riemannmetric*}\vlfunc{\Phi}
    -
    \spray{\riemannmetric*}\vlfunc{f_0}
    \qquad\text{for all } f_0\in\diriboundsobolev1\infty{\posfold}
.
\end{align*}
For~$f=\vlfunc{f_0}\tensorprod\hlfunc{g_0} \in D_1$ we get that
\begin{align*}
    \widetilde{A}f
    &=
    -\frac12\,\vlfunc{f_0}\hlfunc{g_0}
    \pdot
    \spray{\riemannmetric*}\vlfunc{\Phi}
    -
    \hlfunc{g_0} \pdot \spray{\riemannmetric*}\vlfunc{f_0}
    +
    \vlfunc{f_0}\pdot
    \frac1\beta\,
    \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}} \parentheses*{\hlfunc{g_0}}
    \\
    &=
    -\frac12\,f
    \pdot
    \spray{\riemannmetric*}\vlfunc{\Phi}
    -
    \spray{\wriemannmetric*}f
    +
    \frac1\beta\,
    \parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}} (f)
.
\end{align*}

In straight analogy to~\cite[Section~4]{HypocoercJFA},
the proof of the next lemma,
which deals with the globally Lipschitzian case,
is based on a perturbation theorem for essentially m-dissipative operators.
We present it here for sake of completeness.

\begin{theorem}[Kato perturbation of an essentially m-dissipative operator]
\label{thm:perturbation-thm}
    Let an essentially m-dissipative operator~$Z$
    and a dissipative operator~$T$
    have common domain in some given Hilbert space with norm
    $\norm{ \cdot }\vdef\sqrt{\scalarprod{\cdot}{\cdot}}$.
    Assume that there are constants $c_1\in\Rnum$ and $c_2\in(0,\infty)$ such that
    \begin{displaymath}
        \norm{ Tf }^2
        \leq
        c_1\scalarprod{Zf}{f}
        +
        c_2\norm{ f }^2
    \end{displaymath}
    holds for all~$f$ from the common domain.
    Then, the perturbation~$Z+T$ of~$Z$ by~$T$ defined on the common domain
    is essentially m-dissipative.
\end{theorem}
\begin{proof}
    See~\cite[Corollary~3.8, Lemma~3.9 and Problem~3.10]{Davies}.
\end{proof}

\begin{lemma}[Essential m-dissipativity in case of globally Lipschitzian potentials]
\label{lem:Langevin:D2-global-Lipschitz}
    Assume that $\Psi$ is globally Lipschitzian.
    Then, $(\widetilde{L},D_1)$ is essentially m-dissipative
    on~\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}.
    Hence, $(L,D_1)$ is essentially m-dissipative
    on the space~$H=\bigL2{\tanbundle\posfold}{\Leb{\wriemannmetric*}\locprod\nu}$.
\end{lemma}
\begin{proof}
    Define $Z\vdef L_0$ on~$D_1$.
    Then, $(Z,D_1)$ is a dissipative extension of $(Z,D_0)$.
    Thus, $(Z,D_1)$ is essentially m-dissipative
    on~\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}
    by~\hyperref[lem:Langevin:L0D0-ess-m-dissipative]{Lemma~\ref{lem:Langevin:L0D0-ess-m-dissipative}}.
    Define the perturbation
    \begin{displaymath}
        Tf\vdef
        -\frac1\beta\,\parentheses*{\grad{\vertmetric*}\hlfunc{\Phi}}(f)
        +
        \frac12\, f \pdot\,\spray{\riemannmetric*}\vlfunc{\Phi}
    \end{displaymath}
    for all $f=\vlfunc{f_0}\tensorprod\hlfunc{g_0}\in D_1$.
    Since by~\ref{thm:named:Liouville-thm}
    $(\spray{\riemannmetric*},D_1)$ is antisymmetric
    and also $(\widetilde{A},D_1)$ is antisymmetric
    in~\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu},
    $(T,D_1)$ is antisymmetric as well.
    Thus, $(T,D_1)$ is dissipative.

    Choose~$g$ such that $f=Ug$.
    Using the Cauchy-Bunyakovsky-Schwarz inequality we get that
    \begin{align*}
        &\int_{\tanbundle\posfold}
            \parentheses*{
                \grad{\vertmetric*}\hlfunc{\Phi}(f)
            }^2
        \ \diffd\Leb{\riemannmetric*}\locprod\nu
        \\
        &=
        \int_{\tanbundle\posfold}
            \abs{
                \vertmetric
                    {\grad{\vertmetric*}\hlfunc{\Phi}}
                    {\grad{\vertmetric*}f}
            }^2
        \ \diffd\Leb{\riemannmetric*}\locprod\nu
        =
        \int_{\tanbundle\posfold}
            \abs{
                \vertmetric
                    {\grad{\vertmetric*}\hlfunc{\Phi}}
                    {\grad{\vertmetric*}g}
            }^2
        \ \diffd\Leb{\wriemannmetric*}\locprod\nu
        \\
        &\leq
        \norm{
            \grad{\vertmetric*}\hlfunc{\Phi}
        }{\bigL2{\tanbundle\posfold\to\tantanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}}^2
        \pdot
        \int_{\tanbundle\posfold}
            \abs{ \grad{\vertmetric*}g }{\vertmetric*}^2
        \ \diffd\Leb{\wriemannmetric*}\locprod\nu
        \\
        &=
        \norm{
            \grad{\riemannmetric*}\Phi
        }{\bigL2{\posfold\to\tanbundle\posfold}{\riemannmetric*}}^2
        \pdot
        \scalarprod
            {-\laplace{\wvertmetric*}g}
            {g}
            {\bigL2{\tanbundle\posfold}{\Leb{\wriemannmetric*}\locprod\nu}}
        \\
        &=
        \norm{
            \grad{\riemannmetric*}\Phi
        }{\bigL2{\posfold\to\tanbundle\posfold}{\riemannmetric*}}^2
        \pdot
        \scalarprod
            {-\widetilde{S}f}
            {f}
            {\bigL2{\tanbundle\posfold}{\Leb{\riemannmetric*}\locprod\nu}}
    .
    \end{align*}
    by the integration by parts formula~\eqref{eq:weighted-ibp}.
    Abbreviate
    $
        C_{\Phi}
        \vdef
        \norm{ \grad{\riemannmetric*}\Phi }{\bigL\infty{\posfold\to\tanbundle\posfold}{\riemannmetric*}}^2
    $.
    Then, we immediately conclude
    \begin{align*}
        \norm{ Tf }{\bigL2{\Leb{\riemannmetric*}\locprod\nu}}^2
        &\leq
        \frac1{\beta^2}\,
        C_\Phi
        \pdot
        \scalarprod
            {-\widetilde{S}f}
            {f}
            {\bigL2{\Leb{\riemannmetric*}\locprod\nu}}
        +
        \frac14\,C_\Phi
        \pdot
        \norm{ f }{\bigL2{\Leb{\riemannmetric*}\locprod\nu}}^2
        \\
        &=
        c_1
        \scalarprod
            {-Zf}
            {f}
            {\bigL2{\Leb{\riemannmetric*}\locprod\nu}}
        +
        c_2
        \norm{ f }{\bigL2{\Leb{\riemannmetric*}\locprod\nu}}^2
    \end{align*}
    with
    $
        c_1
        \vdef
        \frac2{\alpha\beta}\, C_\Phi
    $
        and
    $
        c_2
        \vdef
        \frac14\,C_\Phi
    $, since we know
    $
        \scalarprod
            {\spray{\riemannmetric*}f}
            {f}
            {\bigL2{\Leb{\riemannmetric*}\locprod\nu}}
        =
        0
    $.
    Finally, the claim follows applying~\autoref{thm:perturbation-thm} to $(Z+T,D_1)$.
\end{proof}

\begin{corollary}[\ref{cond:D2} for globally Lipschitzian potentials]
\label{cor:Langevin:D2-global-Lipschitz}
    Assume that $\Psi$ is a globally Lipschitzian potential.
    Then, $(L,D)$ is essentially m-dissipative on~$H$.
\end{corollary}
\begin{proof}
    Note that $(L,D)$ is a dissipative extension of $(L,D_0)$.
    Thus, we show that $(L,D_1)$ is contained in the closure of $(L,D_0)$
    and then apply~\hyperref[lem:Langevin:D2-global-Lipschitz]{Lemma~\ref{lem:Langevin:D2-global-Lipschitz}}.

    Let $f = \vlfunc{f_0}\in\tanproj^\ast\diriboundsobolev1\infty{\posfold}$,
    $g\in\kappa^\ast\smoothcompact{\posfold}$ and
    a sequence $(f_n)_{n\in\Nnum\setminus\{0\}}$ in~\smoothcompact{\posfold} such that
    its vertical lifting approximates~$f$ in \sobolev12\=/sense, i.\,e.\
    \begin{enumerate}
        \item
        $f_n\longrightarrow f_0$ as $n\to\infty$
        in $\bigL2{\posfold}{\riemannmetric*}$\=/sense and
        \item
        $\frac{\partial f_n}{\partial x^j} \longrightarrow \frac{\partial f_0}{\partial x^j} $ as $n\to\infty$
        in $\bigL2{\posfold}{\riemannmetric*}$\=/sense
        for any chart $x=(x^j)_{j=1}^{\posdim}$.
    \end{enumerate}
    This convergence is maintained under passing to
    $\tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}$, i.\,e.\
    weighting the manifold.
    Finally, we conclude that
    \begin{displaymath}
        L\parentheses*{\vlfunc{f_n}\tensorprod g}
        \longrightarrow
        L\parentheses*{\vlfunc{f_0}\tensorprod g}
        =
        L\parentheses*{f\tensorprod g}
        \qquad\text{in } H \text{ as } n\to\infty
    .
    \end{displaymath}
\end{proof}

The final prove of this section basically is the same as in~\cite[Theorem~4.7]{HypocoercJFA}
as we have taken geometric effects into account before.

\begin{theorem}[\ref{cond:D2} for locally Lipschitzian potentials]
\label{thm:Langevin:D2-loc-Lipschitz}
    Let $\Psi$ be a loc-Lipschitzian potential bounded from below.
    Then, $(L,D)$ is essentially m-dissipative on~$H$.
\end{theorem}
\begin{proof}
    Wlog.\ we assume that $\Psi\geq0$.
    Let $\varepsilon\in(0,\infty)$ and fix some $g\in D\setminus\{0\}$.
    Choose $\varphi,\psi\in D$ such that
    \begin{displaymath}
        \varphi\vert_{\supp(g)}
        =
        1
        \text{,}\quad
        \psi\vert_{\supp(\varphi)}
        =
        1
        \quad\text{and}\quad
        0
        \leq
        \varphi
        \leq
        \psi
        \leq
        1
    .
    \end{displaymath}
    Let $f\in D$ arbitrary.
    Throughout the proof,
    we add to the generators and invariant measures a subscript to indicate the corresponding potential,
    e.\,g.\ $\mu_0 = \Leb{\sasakimetric*}$ in case of the zero potential.
    By construction
    and using dissipativity of $(L_{\psi\Psi},D)$ on~\bigL2{\tanbundle\posfold}{\mu_{\psi\Psi}}
    we get that
    \begin{align*}
        &\norm{
            (\ID-L_{\Psi})(\varphi f) -g
        }{\bigL2{\mu_\Psi}}
        \\
        &\leq
        \norm{
            \varphi\parentheses*{(\ID-L_{\psi\Psi})f -g}
        }{\bigL2{\mu_{\psi\Psi}}}
        +
        \norm{ f }{\bigL2{\mu_{\psi\Psi}}}
        \pdot
        \norm{
            \grad{\riemannmetric*}\varphi
        }{\bigL\infty{\Leb{\wriemannmetric*}}}
        \\
        &\leq
        \norm{ (\ID-L_{\psi\Psi})f -g }{\bigL2{\mu_{\psi\Psi}}}
        +
        \norm{ (\ID-L_{\psi\Psi})f }{\bigL2{\mu_{\psi\Psi}}}
        \pdot
        \norm{ \grad{\riemannmetric*}\varphi }{\bigL\infty{\Leb{\wriemannmetric*}}}
    .
    \end{align*}
    Now, we tighten the requirements on~$\varphi$ via additionally demanding that
    \begin{displaymath}
        \norm{
            \grad{\riemannmetric*}\varphi
        }{\bigL\infty{\posfold\to\tanbundle\posfold}{\wriemannmetric*}}
        =
        \norm{
            \grad{\riemannmetric*}\varphi
        }{\bigL\infty{\wriemannmetric*}}
        <
        \frac\varepsilon4 \pdot \norm{ g }{\bigL2{\mu_0}}^{-1}
        =
        \frac\varepsilon4 \pdot \norm{ g }{\bigL2{\sasakimetric*}}^{-1}
    .
    \end{displaymath}
    Due to~\hyperref[cor:Langevin:D2-global-Lipschitz]{Corollary~\ref{cor:Langevin:D2-global-Lipschitz}}
    $(L_{\psi\Psi},D)$ is essentially m-dissipative on~\bigL2{\tanbundle\posfold}{\mu_{\psi\Psi}},
    hence as a consequence of the Lumer-Philips Theorem,
    there is $f\in D$ such that simultaneously hold
    \begin{align*}
        \norm{
            (\ID-L_{\psi\Psi})f -g
        }{\bigL2{\mu_{\psi\Psi}}}
        \leq
        \frac\varepsilon2
        \quad\text{and}\quad
        \norm{
            (\ID-L_{\psi\Psi})f
        }{\bigL2{\mu_{\psi\Psi}}}
        \leq
        2
        \norm{
            g
        }{\bigL2{\mu_{\psi\Psi}}}
    .
    \end{align*}
    For such an~$f$ we end up with
    $
        \norm{
            (\ID-L_{\Psi})(\varphi f) -g
        }{\bigL2{\mu_{\psi\Psi}}}
        <
        \varepsilon
    $.
    In conclusion, we proved that
    $(\ID-L_\Psi)(D)$ is dense in $H=\bigL2{\tanbundle\posfold}{\mu_\Psi}$.
\end{proof}

\subsection{Hypocoercivity conditions}
\label{subsec:Langevin:H}

\begin{lemma}[algebraic relation~\ref{cond:H1}]
\label{lem:Langevin:H1}
    Let condition~\ref{cond:Langevin:P1} hold.
    Then, we have $PAP\vert_D = 0$.
\end{lemma}
\begin{proof}
    Recall~\hyperref[eq:Langevin:APf-Eq2]{Equation~\eqref{eq:Langevin:APf-Eq2}}
    from~\hyperref[rem:Langevin:APf-Eq2]{Remark~\ref{rem:Langevin:APf-Eq2}}.
    Furthermore, we are going to apply the formula for Gaussian integrals from~\cite[Lemma~3.1]{HypocoercJFA}:
    Let $f\in D$
    and consider polar coordinates in the fibre at $x\in\posfold$
    using~\cite[Satz~14.8]{Forster3}
    which is an application of the transformation formula and Fubini.
    We get that
    \begin{align*}
        &\int_{\tanbundle{\posfold}{x}} APf \ \diffd\nu
        \\
        &=
        \int_{(0,\infty)}\int_{\sphere{\posdim-1}}
            -\riemannmetric
                {v(r,u)}
                {\grad{\riemannmetric*}\expect[\nu]{f}(x)}
                {x}
            \pdot
            r^{\posdim-1}\,
            \frac
                {d \nu(v(r,u))}
                {d \Leb\otimes\spheresurfmeasure}
        \quad \diffd\spheresurfmeasure(u)\,\diffd r
        \\
        &=
        \int_{(0,\infty)}
            0
            \pdot
            r^{\posdim-1}\,
            \frac
                {d \nu(v(r,u_0))}
                {d \Leb\otimes\spheresurfmeasure}
        \ \diffd r
        =
        0
    ,
    \end{align*}
    where $u_0\in\sphere{\posdim-1}$ is arbitrary,
    and further \spheresurfmeasure denotes the surface measure of the sphere~\sphere{\posdim-1} and
    \begin{displaymath}
        v\colon\
        (0,\infty)\times\sphere{\posdim-1}
        \to
        v((0,\infty)\times\sphere{\posdim-1}) \subseteq \tanbundle{\posfold}{x},\
        (r,u) \longmapsto v(r,u)
    \end{displaymath}
    is the diffeomorphism corresponding to
    $
        (0,\infty)\times\sphere{\posdim-1} \to F=\Rnum^{\posdim},\,
        (r,u) \mapsto ru
    $.
    We point out that
    this argument just works
    as~$\nu$ is invariant wrt.\ rotations.

    We have seen that $P_SAP$ is trivial an~$D$,
    so we can use orthogonality of~$P_S$ to obtain
    \begin{displaymath}
        0
        =
        \scalarprod{P_SAPf}{1}{\bigL2{\posfold}{\wriemannmetric*}}
        =
        \scalarprod{APf}{1}{H}
        \qquad\text{for all }f\in D
    .
    \end{displaymath}
    Thus, $PAP\vert_D=0$.
\end{proof}

\begin{lemma}[microscopic hypocoercivity~\ref{cond:H2}]
\label{lem:Langevin:H2}
    Let condition~\ref{cond:Langevin:P1} hold.

    Then, condition~\ref{cond:H2} is fulfilled with
    $\Lambda_m = \alpha$.
\end{lemma}
\begin{proof}
    Let $f\in D$.
    Using the Poincar\'e inequality for Gaussian measures, see~\cite{Beckner89},
    we deduce that
    \begin{align*}
        \scalarprod{-Sf}{f}{H}
        &=
        \frac\alpha\beta\,
        \scalarprod
            {\grad{\wvertmetric*}f}
            {\grad{\wvertmetric*}f}
            {\bigL2{\tanbundle\posfold\to\tantanbundle\posfold}{\wsasakimetric*}}
        =
        \frac\alpha\beta\,
        \norm{
            \grad{\vertmetric*}f
        }{\bigL2{\tanbundle\posfold\to\tantanbundle\posfold}{\wsasakimetric*}}^2
        \\
        &\geq
        \alpha
        \norm{
            f
            -
            \scalarprod
                {f}
                {1}
                {\bigL2{\tanbundle\posfold}{\wvertmetric*}}
        }{\bigL2{\tanbundle\posfold}{\wsasakimetric*}}^2
        =
        \alpha
        \norm{
            (\ID_H-P_S)f
        }{H}^2
    \end{align*}
    and the claim follows.
\end{proof}

The strategy for proving condition~\ref{cond:H3} relies on~\cite[Corollary~2.13]{HypocoercJFA}.
Most importantly, we have to prove that
$(\ID_H-PA^2P,D)$ is essentially m-dissipative.
To do so, we characterise $\ID_H-PA^2P$ on~$D$
starting from~\hyperref[eq:Langevin:APf-Eq2]{Equation~\eqref{eq:Langevin:APf-Eq2}}
and show that the range $(\ID_H-PA^2P)(D)$ is dense in~$H$.

Let $s\colon\,(-\delta,\delta)\to\tanbundle\posfold$ be a curve such that
$s(0)=v$ and $s^\prime(0)=\spray{\riemannmetric*}(v)$
for $v\in\tanbundle\posfold$ fixed and some small $\delta\in(0,\infty)$.
Let $x\vdef\tanproj(v)$.
The following computation relies on
$\tanproj\bincirc s$ being a geodesic of~$\spray{\riemannmetric*}$
and the characterisation of the directional derivative
in terms of parallel transport `$\paralleltransport$' along~$s$ given by the Levi-Civita connection:
\begin{align*}
\begin{aligned}
    -\spray{\riemannmetric*}APf(v)
    &\overset{\eqref{eq:Langevin:APf-Eq2}}{=}
    -\spray{\riemannmetric*}
    \parentheses*{
        -\riemannmetric
            {\ID_{\tanbundle\posfold}}
            {\grad{\riemannmetric*}\expect*[\nu]{f} \bincirc\tanproj}
            {\tanproj}
    }(v)
    \\
    &=
    \sasakimetric
        {\lim_{t\to0}\frac1t\, \parentheses*{
            \paralleltransport_{s(t)}^{s(t)}(
                \ID_{\tanbundle\posfold}\bincirc s(t)
            )
            -
            \paralleltransport_{s(0)}^{s(t)}(
                \ID_{\tanbundle\posfold}\bincirc s(0)
            )
        }}
        {\grad{\hormetric*}P_S f(v)}
        {v}
    \\
    &\qquad\qquad+
    \hormetric
        {\spray{\riemannmetric*}(v)}
        {\connection{\spray{\riemannmetric*}}{\sasakimetric*}
            \parentheses*{\grad{\hormetric*}P_Sf}(v)
        }
        {v}
    \\
    &=
    \hormetric
        {\spray{\riemannmetric*}(v)}
        {\expect*[\nu]{\grad{\hormetric*}f}(v)}
        {v}
    +
    \riemannmetric
        {v}
        {\connection{v}{\riemannmetric*}
            \parentheses*{\grad{\riemannmetric*}\expect[\nu]{f}}(x)
        }
        {x}
    \\
    &=
    \riemannmetric
        {v}
        {\grad{\riemannmetric*}\expect*[\nu]{f}(x)}
        {x}
    +
    \riemannmetric
        {v}
        {\connection{v}{\riemannmetric*}
            \parentheses*{\grad{\riemannmetric*}\expect[\nu]{f}}(x)
        }
        {x}
.
\end{aligned}
\end{align*}
For the second equality we also used the metric compatibility of the Levi-Civita connection~\connection{}{\sasakimetric*}.
The last but one line is obtained
using~\hyperref[lem:app:covaderivative-with-spray]{Lemma~\ref{lem:app:covaderivative-with-spray}}
for the second summand.
Now, we transform into polar coordinates in the fibre~$\tanbundle{\posfold}{x}$
similar as in the proof of~\hyperref[lem:Langevin:H1]{Lemma~\ref{lem:Langevin:H1}}.
With this ansatz we calculate
applying~\cite[Lemma~3.1]{HypocoercJFA} twice that
\begin{align}
\label{eq:Langevin:A2P-part1}
\begin{aligned}
    &\int_{\tanbundle{\posfold}{x}}
        -\spray{\riemannmetric*}
        \parentheses*{
            -\riemannmetric
                {\ID_{\tanbundle\posfold}}
                {\grad{\riemannmetric*}\expect*[\nu]{f} \bincirc\tanproj}
                {\tanproj}
        }
    \ \diffd\nu
    \\
    ={}&
    \int_{\tanbundle{\posfold}{x}}
        \riemannmetric
            {v}
            {\grad{\riemannmetric*}\expect*[\nu]{f}(x)}
            {x}
    \ \diffd\nu(v)
    +
    \int_{\tanbundle{\posfold}{x}}
        \riemannmetric
            {v}
            {\connection{v}{\riemannmetric*}
                \parentheses*{\grad{\riemannmetric*}\expect[\nu]{f}}(x)
            }
            {x}
    \ \diffd\nu(v)
    \\
    ={}&
    \int_{(0,\infty)}\int_{\sphere{\posdim-1}}
        \riemannmetric
            {v(r,u)}
            {\connection{v(r,u)}{\riemannmetric*}
                \parentheses*{\grad{\riemannmetric*}\expect[\nu]{f}}(x)
            }
            {x}
    \\
    &\hphantom{\int_{(0,\infty)}\int_{\sphere{\posdim-1}}}\qquad\qquad
        \pdot
        r^{\posdim-1}\,
        \frac{d\nu(v(r,u))}{d\Leb\otimes\spheresurfmeasure}
    \quad \diffd\spheresurfmeasure(u)\,\diffd r
    \\
    ={}&
    \frac1{\posdim}
    \laplace{\riemannmetric*}\expect[\nu]{f}(x)
    \quad\pdot
    \int_{(0,\infty)}
        r^2\pdot
        \parentheses*{
            \frac1{\spheresurfmeasure(\sphere{\posdim-1})}\,
            \int_{\sphere{\posdim-1}} 1\ \diffd\spheresurfmeasure
        }
        \pdot
        r^{\posdim-1}\,
        \frac{d\nu(v(r,u_0))}{d\Leb\otimes\spheresurfmeasure}
    \ \diffd r
    \\
    ={}&
    \frac1{\beta}
    \laplace{\riemannmetric*}\expect[\nu]{f}(x)
    \quad\pdot1
,
\end{aligned}
\end{align}
where we have taken $u_0\in\sphere{\posdim-1}$ arbitrary,
since~$\nu$ is invariant wrt.\ rotations.
In order to arrive at the last but one line,
we consider some chart $\parentheses*{x^j}_{j=1}^{\posdim}$ at~$x\in\posfold$ providing normal coordinates;
in such coordinates the Levi-Civita connection is understood in terms of directional derivatives as
\begin{displaymath}
    \connection{\Yfield}\Xfield(x)
    =
    \sum_{i,j\in\{1,\ldots,\posdim\}}
        \Yfield^i(x)\, \frac{\partial\Xfield^j(x)}{\partial x^i}\, x^j
\end{displaymath}
for all $\Xfield,\Yfield\in\smoothsec{\tanbundle\posfold}$ with local coordinate expression
$\Xfield = \sum_{j=1}^{\posdim} \Xfield^j\pdot\partial x^j$
and
$\Yfield = \sum_{j=1}^{\posdim} \Yfield^j\pdot\partial x^j$.
Thus, we can understand the mapping
\begin{displaymath}
    \sphere{\posdim-1}\to\sphere{\posdim-1},\
    u \longmapsto \parentheses*{
        \connection{u}{\riemannmetric*}\grad{\riemannmetric*}\expect[\nu]{f}
    }(x)
\end{displaymath}
as the matrix in~\cite[Lemma~3.1]{HypocoercJFA}.
The last step of~\hyperref[eq:Langevin:A2P-part1]{Equation~\eqref{eq:Langevin:A2P-part1}} is due to the fact that
the mean of a chi-squared distribution equals the number of degrees of freedom, i.\,e.~\posdim in the present case.

With the proof of part~\ref{itm:Langevin:SAD:A} of~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}}
and with~\cite[Lemma~3.1]{HypocoercJFA}
we similarly get that
for every $v\in\tanbundle\posfold$ with $x\vdef\tanproj(v)$ holds
\begin{align}
\label{eq:Langevin:A2P-part2}
\begin{aligned}
    &
    P_S\parentheses*{
        \frac1\beta\,\grad{\vertmetric*}\hlfunc{\Phi}(APf)
        }
    (v)
    =
    \int_{\tanbundle{\posfold}{x}}
        \spray{\riemannmetric*}\vlfunc{\Phi}
        \pdot
        APf
    \ \diffd\nu
    \\
    \overset{\eqref{eq:Langevin:APf-Eq2}}{=}{}&
    -\int_{\tanbundle{\posfold}{x}}
        \hormetric
            {\spray{\riemannmetric*}}
            {\grad{\hormetric*}\vlfunc{\Phi}}
        \pdot
        \hormetric
            {\spray{\riemannmetric*}}
            {\grad{\hormetric*}(P_Sf)}
    \ \diffd\nu
    \\
    ={}&
    -\int_{\tanbundle{\posfold}{x}}
        \riemannmetric
            {\ID_{\tanbundle\posfold}}
            {\grad{\riemannmetric*}\Phi(x)}
            {x}
        \pdot
        \riemannmetric
            {\ID_{\tanbundle\posfold}}
            {\grad{\riemannmetric*}\expect[\nu]{f}(x)}
            {x}
    \ \diffd\nu
    \\
    ={}&
    -\frac1{\beta}\,
    \riemannmetric
        {\grad{\riemannmetric*}\Phi(x)}
        {\grad{\riemannmetric*}\expect[\nu]{f}(x)}
        {x}
    =
    -\frac1{\beta}\,
    \connection{\grad{\riemannmetric*}\Phi}(\expect[\nu]{f})(x)
.
\end{aligned}
\end{align}

Hence, we proved combining~\hyperref[eq:Langevin:A2P-part1]{Equation~\eqref{eq:Langevin:A2P-part1}}
and~\hyperref[eq:Langevin:A2P-part2]{Equation~\eqref{eq:Langevin:A2P-part2}} that
\begin{align}
\label{eq:Langevin:PA2P}
\begin{aligned}
    PA^2Pf
    &=
    P_SA^2Pf
    =
    \frac1{\beta}
    \pdot
    \parentheses*{
        \laplace{\riemannmetric*}\expect[\nu]{f}\bincirc\tanproj
        -
        \connection{\grad{\riemannmetric*}\Psi}\expect[\nu]{f}\bincirc\tanproj
    }
    \\
    &=
    \frac1{\beta}
    \pdot
    \laplace{\whormetric*}\parentheses*{\vlfunc{\expect[\nu]{f}}}
    =
    \frac1{\beta}
    \pdot
    \laplace{\whormetric*}(P_Sf)
\end{aligned}
\end{align}
for all $f\in D$.
Compare our result to~\cite[Equation~(3.16)]{HypocoercMFAT}.
These preparations give shape to the following corollary,
cf.~\cite[Proposition~3.9]{HypocoercMFAT}.

\begin{corollary}[$PA^2P$ is essentially m-dissipative]
\label{cor:Langevin:PA2P-m-dissipative}
    Let condition~\ref{cond:Langevin:P1} hold.
    Then, the range $(\ID_H-PA^2P)(D)$ is dense in~$H$,
    thus $PA^2P$ is essentially m-dissipative on~$D$.
\end{corollary}
\begin{proof}
    Right away, we know that
    $
        (PA^2P,D)
        \overset{\eqref{eq:Langevin:PA2P}}{=}
        \parentheses*{ \sfrac1\beta\laplace{\whormetric*}\bincirc P_S,D) }
    $
    is essentially m-dissipative on
    $
        P_S(H)
        =
        \tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}
    $,
    as $P_S(D) = \tanproj^\ast\smoothcompact{\posfold}$
    and~$(\laplace{\whormetric*},D)$
    is essentially self-adjoint in~$H$.
    The later is true,
    since~$(\laplace{\whormetric*},D_0)$ is essentially self-adjoint in~$H$ --
    as so is the Laplace-Beltrami on~$(\tanbundle\posfold,\whormetric*)$,
    cf.\ the beginning of~\autoref{sec:preliminaries} --
    and the fact that~$D_0$ is dense in~$D$.

    Let $g\in H$ such that
    \begin{displaymath}
        \scalarprod
            {(\ID_H-PA^2P)f}
            {g}
            {H}
        =
        0
    \end{displaymath}
    for all $f\in D$
    and we claim that $g=0$.
    Our assumption immediately implies that
    \begin{displaymath}
        0
        =
        \scalarprod
            {(\ID_H-PA^2P)\vlfunc{f_0}}
            {P_Sg}
            {H}
        =
        \scalarprod
            {\vlfunc{f_0} - \frac1\beta\,\laplace{\whormetric*}\vlfunc{f_0}}
            {P_Sg}
            {\tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}}
    \end{displaymath}
    for all $f_0\in\smoothcompact{\posfold}$.
    Thus, $P_Sg=0$,
    since the range
    $
        \parentheses*{
            \ID_H-\sfrac1{\beta}\,\laplace{\whormetric*}
        }(\tanproj^\ast\smoothcompact{\posfold})
    $
    is dense in
    $
        \tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}
    $.
    Ultimately, this means that
    \begin{displaymath}
        \scalarprod{f}{g}{H}
        =
        \scalarprod{PA^2Pf}{g}{H}
        =
        \scalarprod
            {\sfrac1\beta\laplace{\whormetric*}(P_Sf)}
            {P_Sg}
            {\tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}}
        =
        0
        \qquad\text{for all }f\in D
    ,
    \end{displaymath}
    which implies
    $\scalarprod{f}{g}{H} = 0$
    for all $f\in D$,
    hence $g=0$ as claimed.
\end{proof}

\begin{proposition}[macroscopic hypocoercivity~\ref{cond:H3}]
    Let the conditions~\ref{cond:Langevin:P1} and~\ref{cond:Langevin:P2} hold.
    Then, condition~\ref{cond:H3} is fulfilled
    with $\Lambda_M = \frac1\beta\,\Lambda$.
\end{proposition}
\begin{proof}
    Let $f\in D$.
    Since
    $
        \parentheses*{PA^2P,D}
        =
        \parentheses*{\sfrac1\beta\,\laplace{\whormetric*}P_S,D}
    $
    pregenerates a variant of the weighted horizontal gradient form in the sense that
    \begin{displaymath}
        \scalarprod{PA^2Pf}{g}{H}
        =
        -\frac1\beta
        \int_{\tanbundle\posfold}
            \hormetric
                {\grad{\hormetric*}P_Sf}
                {\grad{\hormetric*}P_Sg}
        \ \diffd\mu
        \qquad\text{for all }f,g\in D
    ,
    \end{displaymath}
    we easily compute that
    \begin{align*}
        \norm{ APf }{H}^2
        &=
        \frac1\beta
        \int_{\tanbundle\posfold}
            \abs{
                \grad{\hormetric*}P_Sf(v)
            }{\hormetric*}^2
        \quad \diffd\mu(v)
        =
        \frac1\beta
        \int_{\posfold}
            \abs{
                \grad{\riemannmetric*}\expect[\nu]{f}(x)
            }{\riemannmetric*}^2
        \quad \diffd\Leb{\wriemannmetric*}(x)
        \\
        &=
        \frac1\beta\,
        \norm{
            \grad{\riemannmetric*}\expect[\nu]{f}
        }{\bigL2{\posfold\to\tanbundle\posfold}{\wriemannmetric*}}^2
        \geq
        \frac1\beta\,\Lambda
        \norm{
            \expect[\nu]{f}
            -
            \scalarprod{\expect[\nu]{f}}{1}
                {\bigL2{\posfold}{\wriemannmetric*}}
        }{\bigL2{\posfold}{\wriemannmetric*}}^2
    \end{align*}
    by Poincar\'e inequality.
    Combining this estimates with the previous corollary,
    then \cite[Corollary~2.13]{HypocoercJFA} finishes the proof.
\end{proof}

The remaining hypocoercivity condition~\ref{cond:H4} is checked
via a standard procedure relying on~\cite[Lemma~2.14]{HypocoercJFA} and~\cite[Proposition~2.15]{HypocoercJFA}
cf.\ also~\cite[Proposition~3.11]{HypocoercMFAT}.

\begin{lemma}[boundedness of~$(BS,D)$, first part of~\ref{cond:H4}]
\label{lem:Langevin:H4-1}
    Let condition~\ref{cond:Langevin:P1} hold.
    Then, with $c_1\vdef\frac12\,\alpha$ it holds that
    \begin{displaymath}
        \norm{ BSf }{H}
        \leq
        c_1 \norm{ (\ID_H-P_j)f }{H}
        \qquad\text{for all }f\in D
    \end{displaymath}
    and $P_j\in\{P,P_S\}$, $j\in\{1,2\}$.
\end{lemma}
\begin{proof}
    First, we show that
    $
        SAP
        =
        \alpha
        AP
    $
    on~$D$.
    For the first time, it will become important here that
    \spray{\riemannmetric*} is not just a semispray,
    but actually a \emph{spray}, i.\,e.\
    additionally we have that
    $
        \Liebracket{\canonfield}{\spray{\riemannmetric*}}
        =
        \spray{\riemannmetric*}
    $.
    This is due to the fact that
    \spray{\riemannmetric*}~was chosen in correspondence to the Levi-Civita connection
    which is an affine connection.
    See~\cite{AmbrosePalaisSinger}.
    Let $f\in D$ be fixed.
    Then, we immediately get that
    \begin{displaymath}
        SAPf
        =
        SAP_Sf
        =
        -\alpha\canonfield(-\spray{\wriemannmetric*}(P_Sf))
        =
        \alpha
        \canonfield\parentheses*{
            \spray{\riemannmetric*}\vlfunc*{\expect[\nu]{f}}
        }
    ,
    \end{displaymath}
    since we know from the proof of~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}}
    that $\scalarprod{Ah}{1}{H} = 0$ for all $h\in D$.
    Using the Koszul formula we calculate that
    for all $\Xfield\in\smoothsec{\tanbundle\posfold}$ holds
    \begin{displaymath}
        \canonfield
        \hormetric
            {\spray{\riemannmetric*}}
            {\horlift\Xfield}
        =
        0
        -
        \hormetric
            {\Liebracket
                {\canonfield}
                {\spray{\riemannmetric*}}
            }
            {\horlift\Xfield}
        +
        \hormetric
            {\Liebracket
                {\canonfield}
                {\horlift\Xfield}
            }
            {\spray{\riemannmetric*}}
    .
    \end{displaymath}
    Similar to~\cite[Proposition~5.1]{GudmundssonKappos}
    mentioned before in~\hyperref[lem:Langevin:generator-Hörmander]{Lemma~\ref{lem:Langevin:generator-Hörmander}},
    one could use local coordinates for~\canonfield
    in order to show that
    \Liebracket
        {\canonfield}
        {\horlift\Xfield}
    is purely vertical.
    As~\spray{\riemannmetric*} even is a spray,
    we gain that
    \begin{align*}
        SAPf
        &=
        \alpha
        \canonfield
        \hormetric
            {\spray{\riemannmetric*}}
            {\grad{\hormetric*}\vlfunc*{\expect[\nu]{f}}}
        =
        -\alpha
        \hormetric
            {\spray{\riemannmetric*}}
            {\grad{\hormetric*}\vlfunc*{\expect[\nu]{f}}}
        \\
        &=
        -\alpha
        \spray{\riemannmetric*} \vlfunc*{\expect[\nu]{f}}
        =
        \alpha(
        -\spray{\wriemannmetric*} \vlfunc*{\expect[\nu]{f}}
        )
        =
        \alpha APf
    \end{align*}
    Setting $c_1\vdef\frac\alpha2$
    the claim follows with~\cite[Lemma~2.14]{HypocoercJFA}.
\end{proof}

\begin{lemma}[boundedness of~$(BA(\ID_H-P),D)$, second part of~\ref{cond:H4})]
\label{lem:Langevin:H4-2}
    Let the potential conditions~\ref{cond:named:Langevin:P} hold.
    Then, there exists a constant $c_2\in(0,\infty)$ such that
    \begin{displaymath}
        \norm{ BA(\ID_H-P)f }{H}
        \leq
        c_2 \norm{ (\ID_H-P_j)f }{H}
        \qquad\text{for all }f\in D
    \end{displaymath}
    and $P_j\in\{P,P_S\}$, $j\in\{1,2\}$.
\end{lemma}
\begin{proof}
    Let $f\in D$ and $g\vdef(\ID_H-PA^2P)f$.
    We know that
    $g\in D(\adjoint{(BA)})$
    with $\adjoint{(BA)}g = - A^2Pf$,
    cf.~\cite[Proposition~2.15]{HypocoercJFA}.
    Using our knowledge from the proof of part~\ref{itm:Langevin:SAD:A} of~\hyperref[lem:Langevin:SAD]{Lemma~\ref{lem:Langevin:SAD}},
    furthermore~\hyperref[eq:Langevin:A2P-part1]{Equation~\eqref{eq:Langevin:APf-Eq2}}
    and the Cauchy-Bunyakovsky-Schwarz inequality
    we estimate that
    \begin{align*}
        \norm{ \adjoint{(BA)}g }{H}
        &\leq
        \norm{
            v\longmapsto
            \spray{\riemannmetric*}^2\parentheses*{\vlfunc{\expect[\nu]{f}}}(v)
        }{H}
        +
        \norm*{
            v\longmapsto \frac1\beta\,\grad{\vertmetric*}\hlfunc{\Phi}(APf)(v)
        }{H}
        \\
        &\leq
        \norm{
            \abs{ \spray{\riemannmetric*}^2\parentheses*{\vlfunc{\expect[\nu]{f}}} }
        }{H}
        +
        \norm{
            \abs{ \spray{\riemannmetric*}\vlfunc{\Phi} }{\hormetric*}
            \pdot
            \abs{ \spray{\riemannmetric*}(\vlfunc{\expect[\nu]{f}}) }{\hormetric*}
        }{H}
        \\
        &\leq
        \norm{ v\mapsto\abs{v}{\riemannmetric*}^2 }{H}
        \pdot
        \parentheses*{
            \norm{
                \abs{
                    \operatorname{Hess}_{\sasakimetric*}\parentheses*{\vlfunc{\expect[\nu]{f}}}
                }
            }{H}
            +
            \norm{
                \abs{ \grad{\hormetric*}\vlfunc{\Phi} }{\hormetric*}
                \pdot
                \abs{ \grad{\hormetric*}\vlfunc{\expect[\nu]{f}} }{\hormetric*}
            }{H}
        }
        \\
        &=
        \frac\posdim\beta
        \parentheses*{
            \norm{
                \vlfunc{\abs{
                    \operatorname{Hess}_{\riemannmetric*}\parentheses*{\expect[\nu]{f}}
                }}
            }{H}
            +
            \norm{
                \vlfunc{\abs{ \grad{\riemannmetric*}\Phi }{\riemannmetric*}}
                \pdot
                \vlfunc{\abs{ \grad{\riemannmetric*}\expect[\nu]{f} }{\riemannmetric*}}
            }{H}
        }
    .
    \end{align*}
    Due to the form of~$PA^2P$ we derived in~\hyperref[eq:Langevin:PA2P]{Equation~\eqref{eq:Langevin:PA2P}},
    we know that~$u \vdef Pf$ solves the elliptic equation
    \begin{align*}
        &
        u
        -
        \frac1{\beta}\,
        \laplace{\whormetric*}P_Su
        =
        g
        \\
        \text{in }\qquad&
        \left\{
            u\in\tanproj^\ast\bigL2{\posfold}{\wriemannmetric*}
        \ \middle\vert\
            \exists\, f_0\in\smoothcompact{\posfold}\colon\
            u
            =
            \vlfunc{f_0}
            -
            \scalarprod{\vlfunc{f_0}}{1}{H}
        \right\}
    .
    \end{align*}
    As we assumed the necessary potential conditions,
    the a priori estimates of Dolbeault, Mouhot and Schmeiser,
    cf.~\cite[Appendix]{HypocoercJFA},
    yield existence of a constant~$c_2\in(0,\infty)$ independent of~$Pf$ and~$g$
    such that
    \begin{displaymath}
        \norm{ \adjoint{(BA)}g }{H}
        \leq
        c_2
        \pdot
        \norm{ Pg }{H}
        \leq
        c_2
        \pdot
        \norm{ g }{H}
    .
    \end{displaymath}
    Now, \cite[Propositions~2.15]{HypocoercJFA} does apply which finishes the proof.
\end{proof}

Collecting the individual results of~\autoref{subsec:Langevin:D} and~\autoref{subsec:Langevin:H}
we can infer~\hyperref[thm:Langevin:main]{Theorem~\ref{thm:Langevin:main}}
using the~\ref{thm:hypocoercivity-thm-named}.


TITLE: Finding the vertex of an elliptic cone
QUESTION [0 upvotes]: I have a elliptic cone
$$x^2 + (y - 4)^2 = (z - 4)^2$$
I know to solve for the vertex by setting each variable equal to zero, which gives $(0, 4, 4)$. I can see from the graph that this point is the vertex. But I only know this because I've seen an example of it being done. Can someone give a proper explanation on why this is the vertex?

REPLY [2 votes]: mostly because there are (distinct) straight lines within the cone that all pass through the  vertex. These are lines parametrized by a new variable $t$
$$ (0, 4+t, 4+t)  $$
$$ (0, 4-t, 4+t)  $$
$$ (t, 4, 4+t)  $$
$$ (-t, 4, 4+t)  $$
For each one, at $t=0$  it goes through the vertex. The lines do not meet elsewhere.   To prove that for a pair, change the variable $t$ to $s$  in one of them, for example  $ (t, 4, 4+t) , \; \; (-s, 4, 4+s)  . \; \;   $  For the coordinates to match you need $t=-s$  but you also need $4+t=4+s. \;$ That happens only when $s=t=0$
using Pythagorean triples, we can find other lines (within the cone) without trig functions.
$$ (4t, 4+3t, 4+5t)  $$
$$ (5t, 4+12t, 4+13t)  $$
$$ (8t, 4+15t, 4+17t)  $$
$$ (20t, 4+21t, 4+29t)  $$


TITLE: What kinds of sets can replace lattices in the Shannon sampling theorem?
QUESTION [19 upvotes]: Background: The Shannon sampling theorem states that a bandlimited function (an $L^2$ function whose Fourier transform has compact support) can be determined uniquely from sampling it an integer lattice. More precisely, if the Fourier transform* $\hat{f}$ is supported in $[-\Omega/2, \Omega/2]$ (which we express by writing $f \in B_{\Omega}$) and $\tau>0$ satisfies $\Omega \tau \leq 1$, then $f$ can be reconstructed from the family $\{f(k\tau)\}_{k\in \mathbb Z}$.   This is proved more or less by periodizing and reducing to the theory of Fourier series.   A higher-dimensional analog using lattices in higher dimensions can be proved similarly.
Question:  What sets can these uniformly spaced lattices be replaced by?  Is there a general criterion, or some characterization of the collection of such sets (e.g. measure-theoretic)?

If $\Omega$ is fixed, then for what sets $E$ (say, countable), is the map $B_{\Omega} \to \mathbb{R}^E$ injective?

Motivation:  I overheard a conversation between a graduate student and a faculty member; the student wished to determine if one had a ``deformed lattice'' $\{ k\tau  + X_k \}$ where $\tau$ was a very small mesh relative to $\Omega$  and the $X_k$ was a sequence of reals, for what sequences $\{X_k\}$ would the samples $\{f(k\tau + X_k)\}$ determine $f \in B_{\Omega}$?  We could think of the $X_k$ as independent bounded random variables with a small variance, for instance.  He suspected (and wanted to know) whether the set of singular parameters (where this information was insufficient) would be a set of measure zero.  Intuitively, the small mesh size suggests something like this should be the case.
It was suggested that one might use some sort of finite-dimensional approximations to the full space $B_{\Omega}$ and taking finite subsets of the whole deformed lattice, and that the set of admissible $X_k$ would be some algebraic subvariety of proper codimension, therefore of measure zero, but it wasn't clear how to do this.  This led to my more general question boxed above.
Another reason this question may be of interest is that the sampling used in practical applications (e.g. CAT-scans) does not sample based upon a lattice pattern as in the Shannon theorem.
*I am using the normalization with the factor $2\pi$ for the Fourier transform.

REPLY [8 votes]: One sufficient condition is that, if $|X_k| <= L$ for some $L < 1/4$ then the sampling map would be an injection,  as well as a bounded map from $B$ to $l^2$.  This is known as Kadec's theorem.
See
Nonuniform Sampling and Reconstruction in Shift-Invariant Spaces,
Akram Aldroubi and Karlheinz Gröchenig,
SIAM Review, Vol. 43, No. 4 (Dec., 2001), pp. 585-620
for a survey and further references.


TITLE: How does a mathematician choose on which problem to work?
QUESTION [52 upvotes]: Main question:

How does a mathematician choose on which problem to work?

An example approach to framing one's answer:

What is a mathematical problem - big or small - that you solved or are working on, how did you choose this problem, and why did/do you continue to work on it?

An earlier incarnation of this question asked also for problem sources, and about finding mathematics questions (and engaging in mathematical research) more generally; hopefully this revised version focuses it a bit more (though already posted answers should be understood as responses to the previous version).
As one example of what a retrospective look at a problem (problems) might look like, see the note (How the Upper Bound Conjecture Was Proved) provided by Richard Stanley in a comment below.

REPLY [22 votes]: I've found that "problem creation by analogy" can be very helpful. In grad school I learned about elliptic curves from many great sources (courses of Mazur and Serre, grad student friends too numerous to enumerate, survey articles by Cassels and by Tate, books by Lang,...) and started working on an elliptic curve problem posed by Lang. And every few weeks I'd go to the library and skim the titles and abstracts of lots of journals. (Nowadays, the ArXiv can serve as a similar source.) And I noticed an article with a new improvement on something called Lehmer's conjecture, which I'd never heard of, but it had something to do with heights of algebraic numbers. So I thought, well, algebraic numbers (more properly, the multiplicative group $\bar{\mathbb Q}^*$) are analogous to points on elliptic curves. So I translated Lehmer's conjecture to elliptic curves and proved a result. (Admittedly, it was rather weak, and Masser and other people had stronger results via different methods; but over the years, I've returned to the problem and have papers with Marc Hindry and with Matt Baker.) Fast-forward a few years, I was at a conference at Union College, where the inimitable John Milnor gave a beautiful colloquium-level talk on complex dynamics. I knew nothing about the subject, but for the first half, which he devoted to a survey of the classical theory (Fatou, Julia, etc.), almost every concept that he mentioned seemed to have an elliptic curve (or arithmetic geometry) analog. Thus orbits of points via iteration of rational maps looked analogous to the Mordell-Weil group of an elliptic curve, points with finite orbits were the torsion points, one could look at integer points in orbits as being analogous to integer points on curves (Siegel's theorem), etc. Pursuing that analogy has lead me, and many other people, to a host of fascinating problems, including 10 PhD theses that my students have written in this relatively new field of arithmetic dynamics.


TITLE: Why do the following expressions make no sense?
QUESTION [2 upvotes]: I have an exercise that says:
If $u, v$ and $w$ are vectors in $R^n, n\geq 2$, and $c$ is a scalar explain why the following expressions make no sense:
$A: u \cdot(v\cdot w)$
$B: c\cdot(u+w)$
So in $A$, $v\cdot w$ equals a number that is multiplied by $u$ that is a vector, so that should be a scalar multiple.
In $B$, $u+w $ equals a vector multiplied by $c$ that is a scalar should make another scalar multiple.
What am I assuming wrong?

REPLY [1 votes]: You can't take the dot product of a vector with a scalar. In the first, $u$ is a vector and $v\cdot w$ is a scalar. In the second, $u+w$ is a vector and $c$ is a scalar. The dot product is only defined for vectors.


TITLE: Dielectric material between two plates
QUESTION [0 upvotes]: I wonder if the capacitance is the same if we have two plates with a dielectric material between. I mean, is it the same when the dielectric material only touches one of the plates, compared with when it touches both plates?

REPLY [2 votes]: No it isn't the same. After all, the electric field in the dielectric will be different (reduced). By placing an infinitesimally small metallic slab along the bottom face of the dielectric, you can model the capacitor as two capacitors in series, like in this diagram:
Hope this helps.


TITLE: Establishing a 'weak' union without knowing elements
QUESTION [0 upvotes]: I am wondering if there is a canonical name for the following operation/condition. I imagine it as a kind of generalized Union operation for cases where element labels are unknown.
Suppose I know that $X \subset Z$ ($X$ is a subset of $Z$ - all of $X$'s elements are also elements of $Z$ - but I don't know which elements of $Z$), and that $Y \subset Z.$
I also know that, for any other subset that shares elements with $Z$ (any $S\sim Z$), that subset also shares elements with $X$ and/or $Y.$
This condition is weaker than the union operation, but any $X,Y,Z$ where $X \cup Y = Z$ will satisfy it.
What do we call this? I'd like to learn more about how it shows up in esp. topology.
Thanks.

REPLY [1 votes]: This condition is not weaker than $X\cup Y = Z$.
Suppose $X$, $Y$, $Z$ satisfy your condition. Since $X\subseteq Z$ and $Y\subseteq Z$, we have $X\cup Y\subseteq Z$. So suppose for contradiction that $X\cup Y\neq Z$. Then there is some $z\in Z$ which is not in $X$ or $Y$. The set $S = \{z\}$ shares elements with $Z$, but it does not share elements with either $X$ or $Y$, contradicting your condition.
Your language "for cases where element labels are unknown" and "all of $X$'s elements are also elements of $Z$ - but I don't know which elements of $Z$" suggests to me that you're confused about the basics of set theory. The operation of union is not defined by manipulating element labels, and the subset relation is not defined by listing elements.


TITLE: Maximum Principle fails when u∉C²(Ω)? Can't find example.
QUESTION [3 upvotes]: I would like an example where the maximum principle fails in a bounded smooth domain $\Omega$ where one has a solution which is not $C^2(\Omega)$ to $Lu=0$ where $L$ is elliptic and linear. This obviously must rely on the coefficients being discontiuous for the elliptic operator since otherwise one can do interior regularity estimates. All of the examples I have tried to come up with turn out to not actually be weak solutions so I'm stuck on this. Perhaps maximum principles extend to the non-smooth setting? Any insight on this would be appreciated.

REPLY [3 votes]: Maximum principle, in general, can be applied in the viscosity solutions setting (in this case, the viscosity solution is only continuous). You can have a look at the excellent paper by Crandall-Ishii-Lions as following:
http://arxiv.org/abs/math/9207212


TITLE: Is the reciprocal of a Satake parameter a Satake parameter?
QUESTION [0 upvotes]: Let $F$ be an automorphic L-function for $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$. Is it known whether for all but finitely many primes $p$ the set of Satake parameters of $F$ at $p$ is invariant under the map $x\mapsto x^{-1}$?

REPLY [1 votes]: No, this surely is false. First off, you shouldn't write $F$ an automorphic $L$-function; this is saying that $F(s) = L(s,\pi)$, which is terrible notation. Just write "let $\pi$ be an automorphic representation of $\mathrm{GL}_n(\mathbb{A_Q})$".
Anyway, your question is almost correct in one specific case: $n = 2$ and $\omega_{\pi} = 1$, where $\omega_{\pi}$ is the central character of $\pi$. This corresponds to the case of the $L$-function $L(s,f)$ of a newform $f$ on $\Gamma_0(q) \backslash \mathbb{H}$ of conductor $q$ with principal nebentypus $\chi = \chi_{0(q)}$. Then the $L$-function is $L(s,f) = \prod_p L_p(s,f)$ with
\[L_p(s,f) = \frac{1}{1 - \lambda_f(p) p^{-s} + \chi_{0(q)}(p) p^{-2s}} = \frac{1}{(1 - \alpha_{f,1}(p) p^{-s}) (1 - \alpha_{f,2}(p) p^{-s})},\]
where $\lambda_f(p) = \alpha_{f,1}(p) + \alpha_{f,2}(p)$ is the $p$-th Hecke eigenvalue of $f$ and $\alpha_{f,1}(p), \alpha_{f,2}(p)$ are the two Satake parameters at $p$. Then one sees that
\[\alpha_{f,1}(p) \alpha_{f,2}(p) = \chi_{0(p)}(p),\]
which is $1$ if $p \nmid q$ and $0$ if $p \mid q$, so indeed the Satake parameters do satisfy the invariance map for all but the primes dividing the conductor $q$.
In general, however, this is not the case. Let $\pi$ be an automorphic representation of $\mathrm{GL}_n(\mathbb{A_Q})$ with central character $\omega_{\pi}$, and let $\alpha_{\pi,1}(p),\ldots,\alpha_{\pi,n}(p)$ denote the Satake parameters at a prime $p$. Then it is the case that
\[\alpha_{\pi,1}(p) \cdots \alpha_{\pi,n}(p) = \omega_{\pi}(p).\]
But in general, one shouldn't expect that $\alpha_{\pi,j}(p)^{-1}$ is equal to $\alpha_{\pi,k}(p)$ for some $k$.


TITLE: If $T$ is closed and densely-defined, then so is $T^\ast T$?
QUESTION [1 upvotes]: We suppose $H$ is a Hilbert space, and that $T : H \to H$ is a closed and densely-defined operator.
Under what conditions is $T^\ast T: H \rightarrow H$ a closed and densely-defined operator?

REPLY [2 votes]: The product space $H\times H$ with natural inner product has an orthogonal decomposition
$$
           H \times H = \mathcal{G}(T)\oplus J\mathcal{G}(T^*)
$$
where $J(x,y)=(-y,x)$. Recall that this is how the proof of the existence of adjoint proceeds for a closed densely-defined $T$. The above orthogonality gives $\langle (x,Tx),(-T^*y,y)\rangle=0$ which is the adjoint relation $(Tx,y)=(x,T^*y)$ for all $x\in\mathcal{D}(T)$, $y\in\mathcal{D}(T^*)$.
Surjectivity: Because of the orthogonal decomposition, for any $z\in H$ there exists $x\in\mathcal{D}(T)$ and $y\in\mathcal{D}(T^*)$ such that
$$
              (z,0) = (x,Tx)+(-T^*y,y) \\
          \implies y=-Tx, \;\; z=x-T^*y=x+T^*Tx.
$$
So $(I+T^*T)$ is surjective because every $z\in H$ is in the range of $I+T^*T$.
Density of Domain: $T^*T$ is densely-defined because, if $y\perp\mathcal{D}(T^*T)$, then $y=(I+T^*T)x$ for some $x$ and
$$
              0=(y,x)=\langle (I+T^*T)x,x\rangle =\|x\|^2+\|Tx\|^2 \\
             \implies x=0 \implies y=(I+T^*T)x=0.
$$
Closed Graph: Because $\langle (I+T^*T)x,x\rangle \ge \|x\|^2$ for $x\in\mathcal{D}(T^*T)$, then $(I+T^*T)^{-1}$ is a bounded operator on $H$ and, hence, closed. It follows that $I+T^*T$ is also closed because its graph is the transpose of the graph of $(I+T^*T)^{-1}$. So $T^*T$ is closed because $(T^*T+I)-I$ is a bounded perturbation of a closed operator.