\subsection*{{1 -- Introduction}}
\end{center}


\begin{center}
 \textsc{1.1 -- A well-known result}
\end{center}

In the article \cite{GluckZiller}, H.~Gluck and W.~Ziller applied the theory of calibrations to determine the unit vector fields of minimal volume defined on the unit sphere $\mathbb{S}^3$.

Given a Riemannian manifold $(M,g)$ of dimension $m$, the volume of a unit-norm vector field $X\in{\mathfrak{X}}_M$ is defined as ${\mathrm{vol}}(X)=\int_M\sqrt{\deter(1+({\nabla} X)^t{\nabla} X)}\,{\mathrm{vol}}_M$. Using the Sasaki metric $g^S$, we have that
\begin{equation}  \label{Definition_volume}
 \begin{split}
&{\mathrm{vol}}(X) ={\mathrm{vol}}(M,X^*g^S)=\int_M\biggl(1+\sum_{j=1}^m\|{\nabla}_{e_j}X\|^2 \biggr. \\
&\ \  \biggl.+\sum_{j_1<j_2}\|{\nabla}_{e_{j_1}}X\wedge{\nabla}_{e_{j_2}}X\|^2+\cdots +\sum_{j_1<\cdots<j_{m-1}}\|{\nabla}_{e_{j_1}}X\wedge\cdots\wedge{\nabla}_{e_{j_{m-1}}}X\|^2\biggr)^{\frac{1}{2}}\,{\mathrm{vol}}
\end{split}
\end{equation}
where $\{e_1,\ldots,e_m\}$ is an orthonormal local frame on $M$, cf. \cite{BritoChaconNaveira, GilMedranoLlinaresFuster,GluckZiller}.


The celebrated article gives a clever construction of a calibration 3-form $\varphi$ on the total space of the unit tangent sphere bundle $\pi:T^1\mathbb{S}^3\longrightarrow \mathbb{S}^3$, over the 3-sphere of radius 1. Sections of this fibre bundle correspond to embedded submanifolds and it is observed that non-vanishing vector fields define a unique homology class. A Hopf vector field turns out to be a calibrated submanifold for $\varphi$ and therefore an optimal volume vector field in its homology class (\cite{HarLaw}).


Let us describe the essential steps taken in \cite{GluckZiller} for the construction of the 3-form $\varphi$. First, one recalls the circle bundle
\begin{equation}
\begin{split}
 & \mathrm{V}_2(\R^4)=T^1\mathbb{S}^3\longrightarrow\mathrm{Gr}_2(\R^4) \\
 &\qquad\qquad (x,y)\longmapsto x\wedge y
\end{split}
\end{equation}
mapping the Stiefel manifold of pairs of orthogonal units $(x,y)$ to the Grassmannian of oriented planes. The latter is the quadric $z_1^2+\cdots+z_4^2=0$ in $\C\Proj^3$, a K\"ahler surface embedded as $z_i=x_i+\sqrt{-1}y_i$. The vertical direction of the bundle, spanned by the standard fibre $S^1$, induces a 1-form $\theta$ dual to a horizontal direction $e_0$ of the fibration $T^1\mathbb{S}^3\rightarrow\mathbb{S}^3$. We shall see next that this corresponds with the velocity of the geodesic flow and thus integrates precisely to the Hopf fibration.


We see in the original paper that $\theta$ is a factor of the 3-form $\varphi$. It is also recalled that $\mathrm{Gr}_2(\R^4)=\mathbb{S}^2\times \mathbb{S}^2$ isometrically, so
taking the pullback of the two K\"ahler forms from $\mathbb{S}^2$, one finds two closed linearly independent 2-forms -- essential in the definition of $\varphi$.

The contact form and the two orthogonal and closed 2-forms on the tangent sphere bundle, which finally add to $\varphi$, are invariant under the isotropy subgroup ${\mathrm{SO}}(2)\times{\mathrm{SO}}(2)$ of the whole action group ${\mathrm{SO}}(4)\times{\mathrm{SO}}(2)$ on the 5-dimensional space. However, one must agree the decomposition of the 4-dimensional $e_0^\perp=\ker\theta$ into two 2-planes is not the decomposition into `horizontal plus vertical' of the Sasaki metric, induced by the Levi-Civita connection of $\mathbb{S}^3$.




\begin{center}
 \textsc{1.2 -- New motivation}
\end{center}

The setting above seems to be very particular of the 3-sphere. First, already Gluck and Ziller prove their method does not work for $\mathbb{S}^5$ and above,
even though it is observed  in \cite{GluckZiller} that a calibration on any $T^1\mathbb{S}^{2n+1}$ could also be defined, this time invariant under ${\mathrm{SO}}(2n+2)\times{\mathrm{SO}}(2)$. Secondly, besides the presence of a rank 1 symmetric space, implying 2-point homogeneous action, it is known the geodesic spray of a given Riemannian manifold acts by isometries if and only if the manifold has constant sectional curvature 1, \cite[p.345]{Chavel}.


Furthermore, sectional curvature 1 is equivalent to the general Riemannian fibration induced by the geodesic spray and Sasaki metric being of K\"ahler base, a result of Y.~Tashiro, cf. \cite{Alb2019b,Alb2020a}. In other words, equivalent to the canonical tangent almost contact structure on $T^1\mathbb{S}^m$ verifying the conditions of a Sasakian manifold.

The question of radius $r$ on the base vs radius 1 on the fibre is only considered in \cite{BorGil2006}. Certain stability results are deduced depending on the curvature of the base, although again just with $\dim\geq 5$.

As it is well known, the action of ${\mathrm{SO}}(4)$ on $\mathbb{S}^3$ by isometries is also transitive and isometric when lifted to the rank 1 symmetric space $(T^1\mathbb{S}^3,g^S)$ as the differential map. For $(x,y)\in V_2(\R^4)$ and $g\in{\mathrm{SO}}(4)$, we have simply $g(x,y)=(gx,gy)$; clearly this commutes with the geodesic flow, the ${\mathrm{SO}}(2)$ action of the great circles, as referred in \cite{GluckZiller}:
\begin{equation}\label{geodesic_flow_on_Sm}
 g_t(x,y)=(x\cos t+y\sin t,-x\sin t+y\cos t)  .
\end{equation}
This last result is true in any dimension and we take the opportunity later to prove it in a more general context, namely, for any constant $c\neq0$ sectional curvature.

There is indeed a similar description of the geodesic flow for hyperbolic metric, which we have not found in the literature:
\begin{equation}
  g_t(x,y)=(x\cosh t+y\sinh t,x\sinh t+y\cosh t) .
\end{equation}
In Theorem \ref{Theorem_velocity_proof} we prove that, both on ${T^1(\mathbb{S}^m(r))}$ and on $T^1(\mathbb{H}^m(r))$, the velocities of those curves $g_t(x,y)$ agree with a multiple by $r$ of the geodesic spray vector field $e_0$.

Finally, since the theory of calibrations is also valid with a fixed boundary volume, there should exist some \textit{hope} for the question of minimality of vector fields on a hyperbolic domain of finite volume.


Now, we have been studying a fundamental differential system of global $n$-forms on the tangent sphere bundle of any given oriented Riemannian $(n+1)$-dimensional manifold (\cite{Alb2019b}). In the present article we are thus interested in describing all invariant calibrations for 3-di\-men\-sional space form geometry. Our purposes are the same as above: finding minimal volume through calibrated geometry and observe any further developments. As we have done first elsewhere with the 2-dimensional case and due topological restrictions.

The 1-form $\theta$ from above is the same for \textit{our} differential system. Also the differential system is isometry invariant too by construction. In spite of similarities, the above picture of K\"ahler and 3- and 2-sphere-geometry is not a particular case. Our new proof of Gluck and Ziller's celebrated result for $\mathbb{S}^3$ is quite independent from theirs.

The assertion ``there is, up to constant multiple, just one isometry-invariant closed 3-form'', found in \cite[p.178]{GluckZiller}, is a singularity, a coincidence justified by sectional curvature 1, as referred.

Our method works well with any constant sectional curvature manifold. Most of all, we look forward to further applications of the differential system on 3-manifolds.






\begin{center}

\subsection*{2 -- A different differential system in 3D}

\end{center}


\begin{center}
\textsc{2.1 -- A fundamental differential system}
\end{center}


Let $(M,g)$ be an oriented Riemannian 3-manifold. We have the well-known Sasaki metric and the natural contact structure $\theta=e^0$ on the total space of $\pi:T^1M\longrightarrow M$. In the nomenclature of \cite{KatokHasel}, one may say $T^1M=\{u\in TM:\ \|u\|^2=1\}$ is a submanifold of contact type of the symplectic manifold $TM$ with its canonical structure induced from $T^*M$.

Let us recall the fundamental differential system from \cite{Alb2018a,Alb2019b} in dimension 3. 

The system is composed of four 2-forms on $T^1M$. Let $e_0,\ e_1,\ e_2,\ e_3=Be_1,\ e_4=Be_2$ be an adapted frame of $TT^1M$. This means $e_0$ is the geodesic spray, horizontal, --- indeed the unique horizontal such that ${\mathrm{d}}\pi_u(e_0)=u\in T_{\pi(u)}M$, for all $u\in TM$. It means $e_1,e_2$ span the horizontal plane $H^{\nabla}\cap e_0^\perp$ and their mirror $e_3,e_4$ by the map $B$ span the vertical distribution, tangent to the fibres $\mathbb{S}^2$ of $\pi$. The mirror map $B\in\End{TTM}$ is canonical.

One proves $TT^1M=\xi^\perp_{|T^1M}$, where $\xi=Be_0$ is the tautological vertical vector field over $TM$. And the above confirms the reduction of the structure group of $(T^1M,g^S)$ to ${\mathrm{SO}}(2)$.

The orientation on $T^1M$ is given by ${\mathrm{vol}}_{T^1M}=e^{01\cdots4}$, invariant of the orientation on $M$. The following 2-forms are globally defined:
\begin{equation} \label{thealphaiwithadaptedframe}
 {\mathrm{d}}\theta=e^{31}+e^{42},\qquad\alpha_0=e^{12}, \qquad\alpha_1=e^{14}-e^{23},\qquad\alpha_2=e^{34} .
\end{equation}
It follows that, $\forall i=0,1,2$,
\begin{equation}
\begin{split}
 & \quad\ \alpha_i\wedge{\mathrm{d}}\theta=\alpha_0\wedge\alpha_1=\alpha_2\wedge\alpha_1=0,\qquad\alpha_1\wedge\alpha_1=-2\alpha_0\wedge\alpha_2=({\mathrm{d}}\theta)^2 , \\
 &  *\alpha_0=\theta\wedge\alpha_2,\qquad *\alpha_1=-\theta\wedge\alpha_1,\qquad *\alpha_2=\theta\wedge\alpha_0,  \qquad *{\mathrm{d}}\theta=-\theta\wedge{\mathrm{d}}\theta .
\end{split}
\end{equation}
Such are the basic structural equations of the differential system associated to any given oriented Riemannian 3-manifold.



Let us now assume that $M=M_c$ has constant sectional curvature $c\in\R$. The structural equations
\begin{equation}
 {\mathrm{d}}\alpha_0=\theta\wedge\alpha_1,\qquad{\mathrm{d}}\alpha_1=2\theta\wedge\alpha_2-2c\,\theta\wedge\alpha_0,\qquad{\mathrm{d}}\alpha_2=-c\,\theta\wedge\alpha_1
\end{equation}
have become well-known.





\begin{center}
\textsc{2.2 -- Invariant calibrations}
\end{center}

By \cite[Lemma 2.1]{Alb2020a} or by straightforward computation, a 2-form arising from the invariant differential system with constant coefficients is closed if and only if there exist $Q,Q_1\in\R$ such that
\begin{equation} \label{omega_c}
 \omega_c:=cQ\alpha_0+Q\alpha_2+Q_1{\mathrm{d}}\theta .
\end{equation}


We are now interested in a calibration 3-form $\varphi$ on $T^1M$ which is a multiple of the geodesic flow (we only have $\theta$ and the invariant 2-forms to work with), this is, the wedge of $\theta$ with a 2-form $\omega$ of comass 1 on $e_0^\perp$. The definition of comass 1, for any form, is that $\varphi\leq{\mathrm{vol}}$ and that equality is attained, {in limit}, on the simple  3-vectors ${\mathfrak{u}}\wedge{\mathfrak{v}}\wedge\mathfrak{w}$, formed of unit tangent directions at the points of $T^1M$, cf. \cite[p.57]{HarLaw}.



Now, by \cite[Lemma 2.2]{Alb2020a} or again by straightforward computation, we find that $\varphi=\theta\wedge\omega$ is closed if and only if it is of the kind
\begin{equation}
\varphi= \theta\wedge(b_0\alpha_0+b_1\alpha_1+b_2\alpha_2)
\end{equation}
with any $b_0,b_1,b_2\in\R$; we have assumed a priori these coefficients to be constant. We remark there are closed \textit{invariant} 3-forms in the case of the base space with any sectional curvature; simply, they must also satisfy $b_2=0$.



The requirement that $\omega=b_0\alpha_0+b_1\alpha_1+b_2\alpha_2+b_3{\mathrm{d}}\theta$ is a symplectic form is not obligatory in order to have a calibration, but seems to lead to the less restrictive conditions on the $\varphi$-calibrated submanifolds.


Let us first suppose the orientations on $e_0^\perp$ induced by $({\mathrm{d}}\theta)^2={\mathrm{d}}\theta\wedge{\mathrm{d}}\theta=-2e^{1234}$ and $\omega\wedge\omega$ give the same volume. We find that $\omega$ must satisfy
\begin{equation}
 b_0b_2-{b_1}^2-{b_3}^2=-1.
\end{equation}
\begin{prop} \label{circleofcalibrations}
 An invariant 3-form $\varphi=\theta\wedge\omega$ as above defines a calibration on $T^1M$ if and only if
 \begin{equation}   \label{invariantthreeformcalibration}
    b_0+b_2=0,\qquad {b_0}^2+{b_1}^2=1,\qquad b_3=0 .
 \end{equation}
\end{prop}
\begin{proof}
 As mentioned, ${\mathrm{d}}\varphi=0$ if and only if $b_3=0$. Now we must have ${b_1}^2-b_0b_2=1$. The calibration $\varphi$ has norm at most $\sqrt{2}$, the maximum being attained where $\omega$ is symplectic (due to Wirtinger's inequality). This is equivalent to $\omega\wedge*\omega\leq2{\mathrm{vol}}_{T^1M}$. From
 \begin{equation*}
 \begin{split}
 \langle\omega,\omega\rangle{\mathrm{vol}}_{T^1M} &= (b_0\alpha_0+b_1\alpha_1+b_2\alpha_2)\wedge\theta\wedge (b_2\alpha_0-b_1\alpha_1+b_0\alpha_2) \\
 & = ({b_0}^2+{b_2}^2+2{b_1}^2){\mathrm{vol}}_{T^1M}
   \end{split}
 \end{equation*}
 we find ${b_0}^2+{b_2}^2+2{b_1}^2\leq2$. Hence $(b_0+b_2)^2\leq0$ and therefore ${b_0}^2+{b_1}^2=1$. 
\end{proof}

Notice that $\varphi=\theta\wedge(b_0\alpha_0+b_1\alpha_1-b_0\alpha_2)=-*\omega$. The following result is more surprising. Recall $\omega_c$ from \eqref{omega_c}.
\begin{prop} \label{omegaclosedbadorientation}
 The 2-form $\omega$ is closed if and only if $c=-1$, $b_0=-b_2=\pm1$ and $b_1=0$.
\end{prop}
\begin{proof}
 $\omega=\omega_c\neq0$ if and only if $b_1=0$ and $cQ=b_0=-Q$; hence $b_0=\pm1$ and $c=-1$.
\end{proof}

Now keeping the orientation on $T^1M$ and choosing that on $e_0^\perp$ to be the opposite of $({\mathrm{d}}\theta)^2$, we have the following results analogous to the above.
\begin{prop} \label{omega_goodorientation}
(i)\  \,$\omega\wedge\omega=2e^{1234}$ if and only if $b_0b_2-{b_1}^2
 -{b_3}^2=1$\\
(ii)\  such $\varphi=\theta\wedge\omega$ is a calibration if and only if
$b_0=b_2=\pm1,\ b_1=b_3=0$\\
 (iii)\ moreover $\omega$ is closed if and only if $c=1$.
\end{prop}
\begin{proof}
 Notice that comass 1 implies all $|b_i|\leq1$. Now, in case (ii), we have $b_0b_2={b_1}^2+1$. The result follows.
\end{proof}

\begin{Rema}[Remarks.]
  1)\ As referred above, we do have two symplectic forms $\omega$ with either orientation on $e_0^\perp$. In case of \eqref{invariantthreeformcalibration} it is written in an orthonormal frame as $ \omega=e^1\wedge(b_0e^2+b_1e^4)+e^3\wedge(b_1e^2-b_0e^4)$. And in the case of the opposite orientation, $\pm\omega=e^{12}+e^{34}$. \,2)\
  In the last case, again, we have $\varphi=\theta\wedge\omega=-*\omega$ on the 5-dimensional space. \,3)\  Notice the uniqueness of invariant $\varphi$ is assured only when we ask that $\omega$ too is closed.  \,4)\ We stress that there are no calibrations of the kind given by Propositions \ref{omegaclosedbadorientation} or \ref{omega_goodorientation}(iii) on a flat manifold.
\end{Rema}


Notice that $\theta\wedge{\mathrm{d}}\theta$ cannot contribute to a closed invariant 3-form. Therefore, checking the ``Table of invariant forms and their exterior derivatives'' from \cite[p.182]{GluckZiller}, which is specific to the 3-sphere, case $c=1$, we conclude our differential system must stand perpendicularly to theirs, since we do not obtain the \textit{two} closed 2-forms on $T^1M$ which add to give a calibration.


\begin{center}
\textsc{2.3 -- Exact and cohomologous calibrations}
\end{center}


Here we find out when do two invariant calibrations on $T^1M$ define the same invariant cohomology class ($\sim$). By invariance of the structure one can restrict to a cohomology with constant coefficients. So we consider the closed 3-forms
\begin{equation}
\varphi_t=\theta\wedge(b_{0t}\alpha_0+b_{1t}\alpha_1+b_{2t}\alpha_2)\quad\mbox{and}\quad
\varphi=\theta\wedge(b_{0}\alpha_0+b_{1}\alpha_1+b_{2}\alpha_2).
\end{equation}

\begin{prop} \label{cohomologous3forms}
 \,\ $\varphi_t\sim\varphi\ \Longleftrightarrow\ b_{0t}-b_{0}=-c(b_{2t}-b_2)$.
\end{prop}
\begin{proof}
We have $\varphi_t=\varphi+{\mathrm{d}}(\sum_ip_i\alpha_i)$ for some constant $p_0,p_1,p_2$ if and only if
\begin{equation*}
 \varphi_t-\varphi
   = \theta\wedge(p_0\alpha_1+2p_1\alpha_2-2cp_1\alpha_0-p_2c\alpha_1) .
\end{equation*}
This is equivalent to the system
$b_{0t}-b_{0}=-2cp_1,\ \, b_{1t}-b_{1}=p_0-2p_2c,\ \, b_{2t}-b_{2}=2p_1$, whose solutions exist under the referred condition.
\end{proof}

We restrict to the two types of calibration found before:
\begin{equation}
 \varphi_t=\theta\wedge(\cos{t}\,\alpha_0+\sin t\,\alpha_1-\cos t\,\alpha_2) \qquad\mbox{and}\qquad \varphi_+=\theta\wedge(\alpha_0+\alpha_2) .
\end{equation}
Notice that $\varphi_+=\theta\wedge\omega_{1}$ and $\varphi_-=\varphi_0=\theta\wedge\omega_{-1}$, recalling $\omega_c$ from \eqref{omega_c} with $Q=\pm1,Q_1=0$.

The theory of exact calibrations is relevant on its own, hence the following results. By Proposition \ref{cohomologous3forms}, we have $\varphi_t\sim \varphi_0=\theta\wedge(\alpha_0-\alpha_2)$ iff $\cos t-1=c(\cos t-1)$. And $\varphi_t\sim\varphi_+$ iff $\cos t-1=c(\cos t+1)$.

\begin{prop} \label{cohomologous3formsc=1}
If $c=1$, then all $\varphi_t\sim\varphi_0\sim 0$ and  $\varphi_+\nsim 0$.
\end{prop}

\begin{prop}  \label{cohomologous3formscneq1}
If $c\neq1$, then:\\
(i)\ \,$\varphi_t\sim\varphi_{s}$ iff $\cos t=\cos s$\\
(ii)\,\ $\varphi_t\sim0$ iff  $t={\frac{\pi}{2}}\mod\Z\pi$\\
(iii)\,\ $\varphi_t\sim\varphi_+$ iff $\cos t=\frac{1+c}{1-c}=-1+\frac{2}{1-c}$; in particular, $\varphi_t\nsim\varphi_+$ for $c>0$\\
(iv)\ \,$\varphi_+\sim0$ iff $c=-1$.
\end{prop}

Once again we observe the particular case of $c=\pm1$.





\begin{center}

\subsection*{3 -- Unit vector field and calibrations}

\end{center}



\begin{center}
\textsc{3.1 -- Solutions with minimal volume}
\end{center}


Inspired by \cite{GluckZiller}, we also apply the theory of calibrations to the search for the minimal volume vector fields on constant sectional curvature 3-manifolds.

Let $X\in{\mathfrak{X}}_M$ be a unit norm vector field on a given Riemannian $(n+1)$-manifold. Within the homology class of $X(M)\subset T^1M$, the absolute minimal volume unit vector fields are those for which $\varphi={\mathrm{vol}}$ when restricted to the submanifold. In other words, $X$ has minimal volume when $X(M)$ is a calibrated or $\varphi$-submanifold of the pair $(T^1M,\varphi)$. Recalling ${\mathrm{vol}}_X$ from \eqref{Definition_volume}, the equation is equivalent to
\begin{equation}
X^*\varphi={\mathrm{vol}}_X.
\end{equation}

Certainly, every vector field induces a monomorphism $H_{n+1}(M)\hookrightarrow H_{n+1}(T^1M)$. And indeed we have a well-known fundamental relation, cf. \cite{HarLaw}: for any unit $X'\in{\mathfrak{X}}_M$ in the same homology class as $X$,
\begin{equation}
   \int_M{\mathrm{vol}}_X =\int_{X(M)}\varphi=\int_{X'(M)}\varphi \leq
   \int_M{\mathrm{vol}}_{X'}.
\end{equation}

The theory of calibrations holds for submanifolds with boundary. One may choose a fixed open subset, a domain $\Omega\subset M$ perhaps with non-empty boundary, and seek for an immersion $X:\Omega\longrightarrow T^1M$ giving a $\varphi$-submanifold. Prescribing values for $X$ on a given compact boundary $\partial\Omega$ implies that certain ``moment'' conditions are satisfied, cf. \cite[(6.9)]{HarLaw}.

This observation also tells that we may take just the invariant $\varphi$ when considering a homogeneous Riemannian manifold. Since the minimal solutions will be invariant too.






Now, following  e.g. notation from \cite{Alb2018a,Alb2019b}, we use $\pi^*X,\pi^{\boldsymbol{\star}} X$ for the canonical lifts of a vector field on $M$. We write the well-known `horizontal plus vertical' decomposition in $TTM$
\begin{equation}
 {\mathrm{d}} X(Y)=\pi^*Y+\pi^{\boldsymbol{\star}}({\nabla}_YX).
\end{equation}


Since $T^1M=\{u\in TM:\ \|u\|^2=1\}$ is also characterized as the locus of $\|\xi\|^2=1$, where $\xi$ is the tautological vertical tangent vector field over $TM$, this is, $\xi_u=u$, and since $(\pi^{\boldsymbol{\star}}{\nabla})_Y\xi=Y^v$, it follows that $TT^1M=\xi^\perp$ for the Sasaki metric. Moreover, if $X$ is a unit norm vector field on $M$, then ${\mathrm{d}} X(Y)\in TT^1M$.

Now let us recall the adapted oriented orthonormal frames $\{e_0,e_1,\ldots,e_n,e_{n+1},\ldots,e_{2n}\}$ on $T^1M$. The geodesic spray ${e_0}$ equals $\pi^*X$ over the submanifold $X(\Omega)$, for any $X$. In other words, $X^*\theta=X^\flat$. The $e_i,\ i\leq n$, are the horizontal vectors and thus project at each $u\in T^1M$ to give a frame $e_i\in TM$. Hence we may write
\begin{equation}
{\mathrm{d}} X(e_i)=e_i+\sum_{j=1}^n A_{ij}e_{j+n} ,
\end{equation}
for all $i=0,1\ldots,n$, where $A_{ij}=\langle{\nabla}_{e_i}X,e_j\rangle$.


We now resume with the study in dimension three.

It is not at all clear how the choice of a 3-form $\varphi=\theta\wedge\omega$ as in Section 2 may influence the existence of a solution vector field as calibrated submanifold, if it ever does.

\begin{lemma} \label{Lemma_varphicalibratedvectorfield}
 Let $\varphi=\theta\wedge(b_0\alpha_0+b_1\alpha_1+b_2\alpha_2)$. Then $X^*\varphi={\mathrm{vol}}_X$ if and only if
 \begin{equation}  \label{varphicalibratedvectorfield}
 b_0+b_1(A_{11}+A_{22})+b_2(A_{11}A_{22}-A_{12}A_{21})=\sqrt{1+\sum_{i,j=0}^2{A_{ij}}^2+\sum_{j=0}^2{A_{(j0)}}^2}
 \end{equation}
 where the $A_{(j0)}$ are defined below.
\end{lemma}
\begin{proof}
  First, by definition of Riemannian immersion, we have the formula from \eqref{Definition_volume} as well as the helpful identity developed in \cite{BritoChaconNaveira} for all dimensions: ${\mathrm{vol}}_X=\|X_*(e_0\wedge e_1\wedge\cdots\wedge e_n)\|\,{\mathrm{vol}}_M$. Thus we compute
 \begin{eqnarray*}
  \lefteqn{X_*(e_0\wedge e_1\wedge e_2) =}  \\
  & &= (e_0+A_{01}e_3+A_{02}e_4)\wedge(e_1+A_{11}e_3+A_{12}e_4)\wedge(e_2+A_{21}e_3+A_{22}e_4) \\
  & & = e_{012}+A_{21}e_{013}+A_{11}e_{032}+A_{22}e_{014}+A_{12}e_{042} +A_{01}e_{312}+A_{02}e_{412}+\\
  & & \ \ \ \ \underbrace{(A_{11}A_{22}-A_{21}A_{12})}_{A_{(00)}}e_{034}+\underbrace{(A_{02}A_{21}-A_{01}A_{22})}_{-A_{(10)}}e_{134}+\underbrace{(A_{01}A_{12}-A_{02}A_{11})}_{A_{(20)}}e_{234}
 \end{eqnarray*}
 where $A_{(ij)}$ denote the minors of $(A_{ij})_{i,j=0,1,2}$. Then, since all $A_{i0}=0$,
 \begin{equation*}
  \|X_*(e_0\wedge e_1\wedge e_2)\|^2=1+\sum_{i,j=0}^2{A_{ij}}^2+\sum_{j=0}^2{A_{(j0)}}^2.
 \end{equation*}

 Considering $\varphi=\theta\wedge(b_0\alpha_0+b_1\alpha_1+b_2\alpha_2)$ and recalling \eqref{thealphaiwithadaptedframe} it is easy to deduce that $X^*\varphi(e_0,e_1,e_2)=b_0+b_1(A_{11}+A_{22})+b_2(A_{11}A_{22}-A_{12}A_{21})$.

 One may also compute $X^*\varphi$ directly from $X^*\theta=X^\flat=e^0$ and, moreover, $X^*e^i=e^i$, for $0\leq i\leq2$, and $X^*e^3=A_{01}e^0+A_{11}e^1+A_{21}e^2$ and $X^*e^4=A_{02}e^0+A_{12}e^1+A_{22}e^2$.
\end{proof}






We let
\begin{equation}\label{veryclosedcalibration}
 \varphi=\varphi_\pm=\theta\wedge(\alpha_0\pm\alpha_2).
\end{equation}


\begin{prop} \label{Propfundamental}
 Let the oriented Riemannian manifold $(M,g)$ have constant sectional curvature (any constant). Then a vector field $X$ on the manifold $M$ corresponds to a $\varphi$-submanifold if and only if (i) ${\nabla}_XX=0$ and
 (ii)  $\langle{\nabla}_YX,Y\rangle=\pm\langle{\nabla}_ZX,Z\rangle$ and $\langle{\nabla}_YX,Z\rangle=\mp\langle{\nabla}_ZX,Y\rangle$, $\forall Y,Z$ such that $X,Y,Z$ is any orthonormal frame on $M$.
\end{prop}
\begin{proof}
Applying the lemma, if $\|X^*\varphi\|^2=(X^*\varphi(e_0,e_1,e_2))^2$ equals $\|X_*(e_0\wedge e_1\wedge e_2)\|^2$, then
 \begin{equation}  \label{normofvarphivectorfield}
      1\pm2A_{(00)}+{A_{(00)}}^2 = 1+\sum_{i,j=0}^2{A_{ij}}^2+\sum_{j=0}^2{A_{(j0)}}^2,
\end{equation}
this is,
\begin{equation*}
 {A_{01}}^2+{A_{02}}^2+{A_{11}}^2+{A_{12}}^2+{A_{21}}^2+{A_{22}}^2+ {A_{(10)}}^2+{A_{(20)}}^2\mp2A_{(00)}=0.
\end{equation*}
 Since $A_{(00)}=A_{11}A_{22}-A_{21}A_{12}$, we find
 \begin{equation*}
  {A_{01}}^2+{A_{02}}^2+(A_{11}\mp A_{22})^2+(A_{12}\pm A_{21})^2+
  {A_{(10)}}^2+{A_{(20)}}^2=0 .
 \end{equation*}
 Giving immediately the three identities in (i-ii). The reciprocal, the three identities implying \eqref{varphicalibratedvectorfield}, is easily checked.
 \end{proof}

 Recall the notions of closed vector field, ${\mathrm{d}} X^\flat=0$, of divergence-free or co-closed vector field, $\mathrm{tr}_g{{\nabla} X}=0$, and that of harmonic vector field (as 1-form, not as a map).
 \begin{prop}
 Let $X$ be a solution of equations (i-ii) in Proposition \ref{Propfundamental}. We have that:
\begin{itemize}
 \item in the case $\varphi_+$, then $X$ is a Killing vector field and therefore co-closed. It is not closed and thus not a harmonic vector field.
 \item (cf. Proposition \ref{prop_nonexistenceminimalvolumevectorfield} below) in the case $\varphi_-$, then $X$ is closed and co-closed. Hence harmonic. $X$ is \emph{not} Killing. On the other hand, the distribution $X^\perp\subset TM$ is integrable.
\end{itemize}
 \end{prop}
\begin{proof}
We have ${\nabla}_XX=0$ and $\langle{\nabla}_\cdot X,X\rangle=0$. The matrix of $A$ restricted to $X^\perp$ is skew-symmetric and symmetric in the respective cases $\varphi_\pm$. In the positive case, there exists an orthonormal frame $Y,Z$ such that ${A=\scriptstyle{\left(\begin{array}{cc} 0 &\lambda \\ -\lambda &0 \end{array}\right)}}$. In the negative case, then $A$ has eigenfunctions $\lambda,-\lambda\neq0$. For both we may use the formula
 \[ {\mathrm{d}} X^\flat= X^\flat\wedge{\nabla}_XX^\flat+   Y^\flat\wedge{\nabla}_YX^\flat+Z^\flat\wedge{\nabla}_ZX^\flat  .  \]
 To see if the orthogonal distribution to $X$ is integrable we are bound to check, for every $Y,Z$ orthonormal frame, that $\langle [Y,Z],X\rangle=-\langle{\nabla}_YX,Z\rangle+ \langle{\nabla}_ZX,Y\rangle=-(1\pm1)\langle{\nabla}_YX,Z\rangle$ vanishes. Which is obvious in one case.
 \end{proof}


 The following result generalizes to any curvature the well-known result from \cite{GluckZiller} for the 3-sphere. It has been proved by completely different means in \cite{GilMedranoLlinaresFuster,GilMedrano2022} and in \cite{DavilaVanhecke1,DavilaVanhecke2}.

\begin{teo}
A Hopf vector fields on $\Omega\subset\mathbb{S}^3(r)$ has minimal volume $(1+\frac{1}{r^2}){\mathrm{vol}}(\Omega)$ in its homology class. Such solution is unique in any given $\Omega$.

A global Hopf vector field has minimal volume $2\pi^2(r+r^3)$.
\end{teo}
\begin{proof}
 We start by uniqueness. Equations (i-ii) in Proposition  \ref{Propfundamental} yield $A_{0j}=0$, $A_{11}=A_{22}$, $A_{12}=-A_{21}$.
 Applying formula \eqref{normofvarphivectorfield} with our choice of $\varphi_+$, we find the 3-form $X^*\varphi_+$ is $(1+{A_{11}}^2+{A_{12}}^2){\mathrm{vol}}_M$. Uniqueness comes from invariance (not from K\"ahler geometry as in the case $r=1$). Indeed, $A_{ij}$ must be constant in order to have minimality everywhere, ie. every $\Omega\subset\mathbb{S}^3(r)$ with free boundary conditions.



 Those equations are satisfied by a Hopf vector field. Indeed we deduce in Section 4 that $A_{0j}=0$, $A_{11}=A_{22}=0$, $A_{12}=-A_{21}=\frac{1}{r}=\sqrt{c}$. These are the only possible values and every such vector field $X$ is a Hopf vector field.

 For the last assertion, recall ${\mathrm{vol}}(\mathbb{S}^3(r))=2\pi^2r^3$.
\end{proof}

Hopf vector fields are induced by linear complex structures $J_0$ on $\R^4$ compatible with the metric and orientation. On the 3-sphere, $X_x=J_0x$.

Let us recall $\mathrm{Isom}_+(\mathbb{S}^3)={\mathrm{SO}}(4)$ and raise the \textit{usual} question for the quotient manifolds of the sphere. This has also deserved the attention of \cite{GilMedrano2022}.
\begin{coro}
 Given a subgroup $\Gamma\subset{\mathrm{SO}}(4)$ acting freely on the 3-sphere, the manifold $\mathbb{S}^3/\Gamma$ has as minimal vector fields the Hopf vector fields for which $[\Gamma,J_0]=0$.
\end{coro}
The proof is clear from the same theory of calibrations, since the canonical connection on $M$, the Sasaki metric and the fundamental differential system on $T^1M$ are all invariant by isometries, for any oriented Riemannian manifold $M$.

The author suspects that not even every finite $\Gamma$ lies in some $\mathrm{U}(2)$, cf. \cite{Hong,Wolf}, somewhat in contradiction with \cite{GilMedrano2022}.





Proposition \ref{Propfundamental} gives also a solution to the flat case (obviously the parallel vector fields). Indeed, notice we may use both $\varphi_-\sim\varphi_+\nsim0$ in this case, cf. Proposition \ref{cohomologous3formscneq1} (iii).



\begin{center}
\textsc{3.2 -- No solutions}
\end{center}



The above result of a solution of the minimal volume vector field on $\mathbb{S}^3$ as the geodesic flow of the 2-sphere, as well as the appearance in the hyperbolic case of a system of differential equations (i-ii) leading to strong similarity with Anosov flows, which furthermore require a \textit{hyperbolic set for the flow}, cf. \cite[Definition 17.14.1]{KatokHasel}, and yield precisely that a geodesic flow on hyperbolic metric is Anosov, cf. \cite[p. 346]{Chavel}\cite{KatokHasel}, leads us to ask if, for the case $c=-1$, a local solution $X$ of minimal volume on $\mathbb{H}^3$ exists as the geodesic flow of some unknown surface yet to be discovered. However, no such analogy is true.

\begin{prop}  \label{prop_nonexistenceminimalvolumevectorfield}
There exists no vector field $X$ on a given domain of a negative constant sectional curvature manifold which is a solution of (i-ii) from Proposition \ref{Propfundamental}.
\end{prop}
\begin{Rema}
 Equations (ii) are invariant under rotation of $Y,Z$ in both cases $\pm$.
\end{Rema}
\begin{proof}[A proof for case $\varphi_-$]
 Suppose a solution exists on a non-empty open subset. Notice $X$ is locally a gradient, since $X^\flat$ is closed, and moreover $X$ is harmonic. Then a well-known result of Cartan, owing to the search for isoparametric hypersurfaces, states that there do not exist harmonic functions with gradient of constant norm on hyperbolic space, cf. \cite{Plaue} and the references therein.
\end{proof}
\begin{proof}
 Let us see the case $\varphi_-$ first.
 Suppose $X$ exists on an open subset $\neq\emptyset$. 
 Recall ${\nabla}_XX=0$ and $\langle{\nabla}_\cdot X,X\rangle=0$. The matrix of $A$ restricted to $X^\perp$ is symmetric and has eigenfunctions $\lambda,-\lambda\neq0$. For two local orthonormal vector fields $Y,Z$ in $X^\perp$, we have ${\nabla}_YX=\lambda Y$, ${\nabla}_ZX=-\lambda Z$ (this is the kind of phenomena observed with an Anosov flow, cf. \cite[Chapter 17.4]{KatokHasel}). Then it follows that ${\nabla}_YZ=\mu Y$ and ${\nabla}_ZY=\nu Z$ for some functions $\mu,\nu$, and therefore ${\nabla}_YY=-\mu Z-\lambda X$ and ${\nabla}_ZZ=\lambda X-\nu Y$. In the same way we find ${\nabla}_XY=pZ$, ${\nabla}_XZ=-pY$ for some function $p$. Regarding the Lie brackets, we get
 \[  [X,Y]=pZ-\lambda Y,\quad [X,Z]=-pY+\lambda Z,\quad [Y,Z]=\mu Y-\nu Z. \]
 Then $R(X,Y)X={\mathrm{d}}\lambda(X)Y+2\lambda pZ+\lambda^2 Y$ and, from constant $-1$ sectional curvature tensor, it follows that $p=0$ and ${\mathrm{d}}\lambda(X)+\lambda^2=1$. From $R(X,Y)Z={\mathrm{d}}\mu(X)Y+\lambda\mu Y=0$, follows ${\mathrm{d}}\mu(X)+\lambda\mu=0$. As with $Y$, we have $R(X,Z)X=-{\mathrm{d}}\lambda(X)Z+\lambda^2Z$, and therefore we find now $-{\mathrm{d}}\lambda(X)+\lambda^2=1$. We conclude $\lambda^2=1$. Continuing, from $R(X,Z)Y={\mathrm{d}}\nu(X)Z-\lambda\nu Z$, follows ${\mathrm{d}}\nu(X)-\lambda\nu=0$. From $R(Y,Z)X=-2\mu\lambda Y-2\nu\lambda Z=0$, comes that $\mu=\nu=0$. Finally, a last deduction $R(Y,Z)Y=-\lambda^2Z$ implies that $\lambda^2=-1$.

 Now the case $\varphi_+$. We suppose $X$ exists such that ${\nabla}_XX=0$ and that for a local orthonormal frame $Y,Z$ of $X^\perp$ we have now ${\nabla}_YX=\lambda Z,\ {\nabla}_ZX=-\lambda Y$ for some function $\lambda$. We also see ${\nabla}_XZ=\mu Y$ and ${\nabla}_XY=-\mu Z$ for some function $\mu$. Hence $[X,Z]=\mu Y+\lambda Y$ and it follows that $R(X,Z)X=-{\mathrm{d}}\lambda(X)Y-\lambda^2Z$ and therefore $\lambda^2=-1$, a same contradiction as above.
\end{proof}
It is not proved above that there do not exist minimal volume vector fields e.g. on a constant hyperbolic manifold of finite volume, but it seems reasonable to conjecture so, with the infimum $(1-c){\mathrm{vol}}(M)$, which is the lower bound for compact $M$ found in \cite[Corollary 4]{BritoChaconNaveira} (cf. also \cite[Proposition 13]{GilMedranoLlinaresFuster}).

If the circle ${b_0}^2+{b_1}^2=1$ of calibrations from Proposition \ref{circleofcalibrations} may still yield the right equations for the problem of minimal volume on hyperbolic ambient, then we should probably notice that always some $\varphi_t\sim\varphi_+$, for some $t$, and that $\varphi_+$ has no solutions.
Case $c=-1$ is specially skew, since $\varphi_{\frac{\pi}{2}}=\theta\wedge\alpha_1={\mathrm{d}}\alpha_0\sim0\sim\varphi_+$.

Although this case $b_0=0,\ \pm b_1=1$ leads to complicated conditions on $X$ in general, we shall give a simple solution in Section 5.







\begin{center}
\subsection*{4 -- On the geodesic flow and Hopf vector fields}
\end{center}



We think of a Riemannian space form as a hypersurface of the Euclidean or Lorentzian space. Defining
\begin{equation}
 \langle\ ,\ \rangle_\pm=\pm({\mathrm{d}} x^1)^2+({\mathrm{d}} x^2)^2+\cdots+({\mathrm{d}} x^{m+1})^2
\end{equation}
we let $M_+=\mathbb{S}^m(r)=\{x:\ \langle x,x\rangle_+=r^2\}$ and $M_-=\mathbb{H}^m(r)=\{x:\ \langle x,x\rangle_-=-r^2,\ x^1>0\}$, where $r>0$. These metrics restrict to positive definite metrics in both subspaces. They are also given as the invariant metrics. We remark $M_-={\mathrm{SO}}_0(1,m)/{\mathrm{SO}}(m)$.

Next we prove a result which is asserted in the case of spheres. The proof would seem quite immediate perhaps through the exponential map, but we wish to see it in detail.

Let $x\in M_\pm$ and $(x,y)\in T^1_xM_{\pm}$. Hence $y$ is a unit vector, $\langle y,y\rangle_\pm=1$, orthogonal to $x$.

\begin{teo} \label{Theorem_velocity_proof}
The geodesic spray vector field $e_0$ of $M_+$ or $M_-$ along the curves in $T^1M_+$ and $T^1M_-$, respectively,
\begin{equation}  \label{twogeodesicflowcurves}
\begin{split}
 &  g_t(x,y)=(x\cos t+yr\sin t,-x\frac{\sin t}{r}+y\cos t), \\
 &  g_t(x,y)=(x\cosh t+yr\sinh t,x\frac{\sinh t}{r}+y\cosh t) ,
\end{split}
\end{equation}
agrees with $\frac{1}{r}$ times the velocity of these same curves.
\end{teo}
\begin{proof}
We have $\pi:TM_\pm\longrightarrow M_\pm$, $\pi(x,y)=x$, where $\langle x,x\rangle_\pm=\pm r^2$. Differentiating this equation, it follows that $x\perp_\pm y$. Next we let drop the $\pm$ where there is no fear of confusion.

In the same way, it is easy to see
\begin{equation*}
 T(TM)=\bigl\{(x,y,u,v)\in(\R^{m+1})^4:\ \langle x,x\rangle=\pm r^2,\ \langle x,y\rangle=\langle u,x\rangle=\langle u,y\rangle+\langle v,x\rangle=0\bigr\}.
\end{equation*}
Vertical vectors are those for which $u=0$. The tautological vector field $\xi$ is clearly given by $\xi_{(x,y)}=(x,y,0,y)$. The mirror map $B\in\End{TTM}$ is simply the map $B(x,y,u,v)=(x,y,0,u)$.


The Levi-Civita connections of the given hypersurfaces are found to be, respectively, ${\nabla}_XY={\mathrm{d}} Y(X)\pm\langle X,Y\rangle \frac{x}{r^2}$ at a point $x\in M$. Indeed well-defined, it is the metric torsion-free connection. Invariant under the respective ${\mathrm{SO}}(m+1)$ and ${\mathrm{SO}}_0(1,m)$.

Notice that we are starting to ommit the base point in $TM$; this will be even \textit{worse} with $TTM$.


A simple computation yields $R^{\nabla}(X,Y)Z=\pm\frac{1}{r^2}(\langle Y,Z\rangle X-\langle  X,Z\rangle Y)$, and hence the conclusion that $M_\pm$ has constant sectional curvature $c=\pm\frac{1}{r^2}$.


Regarding the tangents just to $T^1M_\pm$ we have, moreover, $\langle y,y\rangle_\pm=\|y\|^2=1$, $\langle y,v\rangle=0$. We then lift the Levi-Civita connection to a metric connection over $(TM,g^S)$, by taking the pullback connection to $\pi^\star TM=\ker{\mathrm{d}}\pi$, the vertical subspace. For any vector field ${\mathfrak{u}}$ over $TM$, we have
\begin{equation*}
 {\mathfrak{u}}_{(x,y)}=(x,y,U_1,U_2) \quad\mbox{and}\quad {\mathrm{d}}\pi({\mathfrak{u}}_{(x,y)})=(x,U_1)
\end{equation*}
and thus
\begin{equation*}
 {\nabla}^*_{\mathfrak{u}}\xi=(x,y,0,U_2\pm\langle U_1,y\rangle \frac{x}{r^2}),
\end{equation*}
giving the horizontal subspace as the kernel:
\begin{equation}\label{eqhorizontal}
 {\mathfrak{u}}\in H^{\nabla}\quad\mbox{if and only if}\quad r^2U_2\pm\langle U_1,y\rangle x=0.
\end{equation}
It is  easy to find the horizontal and vertical lifts of $X=(x,U)$ tangent to $M$ at $x$. For instance, the horizontal lift of $X$ is $\pi^*X=(x,y,U,\mp\langle U,y\rangle\frac{x}{r^2})$. As expected, $B\pi^*X=\pi^{\boldsymbol{\star}} X$.

Finally the geodesic spray $e_0=B^\mathrm{t}\xi$ equals $e_0=(x,y,y,\mp\frac{x}{r^2})$.
Notice the defining conditions are satisfied: $e_0$ is horizontal and ${\mathrm{d}}\pi({e_0}_{X_x})=X_x$, for any $X$ tangent to $M$ at $x$.

Recalling \eqref{geodesic_flow_on_Sm}, which exists in any dimension,
we may easily deduce, that at any point $g_t(x,y)=(x\cos t+yr\sin t,-x\frac{\sin t}{r}+y\cos t)$ lies in $T^1\mathbb{S}^m(r)$. And secondly, that
\begin{eqnarray*}
 \papa{ }{t}g_t(x,y)&=
 &(-x\sin t+yr\cos t,-x\frac{\cos t}{r}-y\sin t) \\
 &=&  r\bigl(-x\frac{\sin t}{r}+y\cos t, -\frac{1}{r^2}(x\cos t+yr\sin t)\bigr)
 \\ &=& r e_0\,_{|_{g_t(x,y)}}
 \end{eqnarray*}

The hyperbolic case is just as easy: $g_t(x,y)$ lies in $T^1\mathbb{H}^m(r)$ and $\papa{ }{t}g_t(x,y)= r e_0\,_{|_{g_t(x,y)}}$.
\end{proof}

We have again the well-defined and non-singular projections
\begin{equation}
   V_{2,\pm}(\R^{m+1})=T^1M \longrightarrow \mathrm{Gr}_2(\R^{m+1}), \quad (x,y)\longmapsto x\wedge y,
\end{equation}
with kernel the curves in \eqref{twogeodesicflowcurves}.


We have the ${\mathrm{SO}}(2)$ action on $T^1M_+$ arising from the geodesic flow. This action is by isometries, as pointed in \cite{GluckZiller}, for the case $r=1$. Moreover,  it is asserted in \cite{Chavel} that the action of the general geodesic flow is by isometries if and only if we are in the case of sectional curvature 1. We shall reconfirm this independently.

Notice that the geodesic flow of hyperbolic metric is not always complete, since it must lie in $\{x^1>0\}$.

Following the definition in Theorem \ref{Theorem_velocity_proof} we still observe the (pseudo) Lie group action of $\R$ on $T^1M_-$. Indeed the identities $g_0=1,\ g_{t_1+t_2}=g_{t_1}\circ g_{t_2}$ hold true for any $t_1,t_2\in\R$.

\begin{prop}
 The Lie groups ${\mathrm{SO}}(m+1)$ and ${\mathrm{SO}}_0(m,1)$ act by isometries on $T^1\mathbb{S}^m(r)$ and $T^1\mathbb{H}^m(r)$, respectively, leaving the fundamental differential system from \cite{Alb2019b} invariant.

 The actions of ${\mathrm{SO}}(2)$ and $\R$ as geodesic flows commute with the previous respective symmetric space structures. These actions are by isometries if and only if we are in the case of $\mathbb{S}^m(1)$.
\end{prop}
\begin{proof}
 The Levi-Civita connection and therefore the Sasaki metric are invariant by isometry. This yields quite immediately the first part. We are left with the proof of the last assertion. Since $g_t$ acts linearly, at a point $(x,y)$ we have ${\mathrm{d}} g_t(u,v)=g_t(u,v) \in T^1_{g_t(x,y)}M$. With $X=(x,U),\ x\perp U$, we have
 \[ {\mathrm{d}} g_t(\pi^*X)=g_t(U,\mp\langle U,y\rangle\frac{x}{r^2})=
  \bigl(Uc\mp\langle U,y\rangle\frac{x}{r}s,\mp U\frac{s}{r}\mp\langle U,y\rangle   \frac{x}{r^2}c\bigr) , \]
 where $c=\cos t,\ s=\sin t$ in the elliptic case and $c=\cosh t,\ s=\sinh t$ in the hyperbolic case. At point $g_t(x,y)=(xc+yrs,\mp x\frac{s}{r}+yc)$, the vertical part is, cf. \eqref{eqhorizontal}:
 \begin{eqnarray*}
  \lefteqn{ \mp U\frac{s}{r}\mp\langle U,y\rangle
  \frac{x}{r^2}c\pm\frac{1}{r^2}\Bigl\langle Uc\mp\langle U,y\rangle\frac{x}{r}s,\mp x\frac{s}{r}+yc\Bigr\rangle(xc+yrs) \ \ }\\
  &=& \mp U\frac{s}{r}\mp\langle U,y\rangle\frac{c}{r^2}x\pm\frac{c^2}{r^2} \langle U,y\rangle(xc+yrs)+\frac{s^2}{r^2}\langle U,y\rangle(xc+yrs) \\
  &=& \mp\frac{s}{r}(U-\langle U,y\rangle y).
 \end{eqnarray*}
Now for verticals $(x,y,0,U_2),\ U_2\perp x,y$, we have ${\mathrm{d}} g_t(0,U_2)=(U_2rs,U_2c)$, which has vertical part
\[  U_2c\pm\frac{1}{r^2}\langle U_2rs,\mp x\frac{s}{r}+yc\rangle(xc+yrs)=U_2c  \]

Finally, with any $U,U_1\perp x$ giving horizontal vector fields and any $U_2,U_3$ giving verticals:
 \begin{equation*}
  \langle {\mathrm{d}} g_t(U,\mp\langle U,y\rangle\frac{x}{r^2}),{\mathrm{d}} g_t(U_1,\mp\langle U_1,y\rangle\frac{x}{r^2}) \rangle = (c^2+\frac{s^2}{r^2})\langle U,U_1\rangle+ (\pm s^2-\frac{s^2}{r^2})\langle U,y\rangle\langle U_1,y\rangle,
 \end{equation*}
\[   \langle{\mathrm{d}} g_t(0,U_2),{\mathrm{d}} g_t(0,U_3)\rangle = (r^2s^2+c^2)\langle U_2,U_3\rangle \]
and
 \begin{equation*}
  \langle{\mathrm{d}} g_t(U,\mp\langle U,y\rangle\frac{x}{r^2}),{\mathrm{d}} g_t(0,U_2)\rangle = (crs\mp\frac{cs}{r})\langle U,U_2\rangle .
 \end{equation*}
 $g_t$ is an isometry for the Sasaki metric if and only if the three formulas above are, respectively, equal to $\langle U,U_1\rangle,\ \langle U_2,U_3\rangle$ and $0$. The result follows.
\end{proof}



When $m=2k+1$ is odd we have further understanding of the geodesic flow.

We see it through the means of Hopf vector fields. Let $J_0\in\End{\R^{2k+2}}$ be a linear complex structure compatible with the metric. It gives rise to a Hopf vector field, defined, at each $x\in\mathbb{S}^m(r)$, by $X=\frac{1}{r}J_0x$. We have omitted the base point on the target manifold. Clearly
\begin{equation}\label{identitiesforHopfvf}
 {\nabla}_{W}X={\mathrm{d}} X(W)+\langle X,W\rangle \frac{x}{r^2}=\frac{1}{r} J_0W+\langle X,W\rangle \frac{x}{r^2}
\end{equation}
and so ${\nabla}_XX=0$. Finally, for any $Y$ such that $Y,J_0Y\perp \{x,X\}$, we have successively
\begin{equation}\label{identitiesforHopfvf2}
 {\nabla}_YX=\frac{1}{r} J_0Y,\qquad \langle{\nabla}_YX,Y\rangle=0,\qquad\langle{\nabla}_YX,J_0Y\rangle=\frac{1}{r}\|Y\|^2=-\langle{\nabla}_{J_0Y}X,Y\rangle.
\end{equation}
These equations are very useful in regard to Proposition \ref{Propfundamental}.

Reciprocally, a vector field $X$ on the sphere which satisfies \eqref{identitiesforHopfvf2} must be a Hopf vector field.


In the case of hyperbolic space, we \textit{could} follow exactly the same steps taking $J_0$ compatible with the metric. Again $X=J_0x\perp x\in\mathbb{H}^{2k+1}$. And ${\nabla}_{W}X=J_0W-\langle X,W\rangle x$. And all the identities in \eqref{identitiesforHopfvf2} would be accordingly satisfied. However, such a $J_0$ does not exist in ${\mathrm{SO}}(1,2k+1)$.






\begin{center}
\subsection*{5 -- Towards minimal vector fields on hyperbolic metric}
\end{center}




Let us now assume that $M$ is any 3-dimensional oriented Riemannian manifold, not necessarily of constant sectional curvature. The 1st order structure equations are as follows, \cite{Alb2020a},
\begin{equation}\label{equacoesestruturaisgerais}
 {\mathrm{d}}\alpha_0=\theta\wedge\alpha_1,\qquad{\mathrm{d}}\alpha_1=2\theta\wedge\alpha_2-r\,\theta\wedge\alpha_0,\qquad{\mathrm{d}}\alpha_2=\theta\wedge\gamma-\frac{r}{2}\,\theta\wedge\alpha_1+\alpha_0\wedge\rho,
\end{equation}
where $\rho$ is the vertical pullback
\begin{equation}
 \rho=\xi\lrcorner\pi^{\star}{\mathrm{Ric}\,}=R_{1012}e^4-R_{2012}e^3
\end{equation}
and $r$ is the function on $T^1M$ defined by
\begin{equation}
 r(u)={\mathrm{Ric}\,}(u,u),\ \ \ u\in T^1M.
\end{equation}
Regarding $\gamma$, we refer to \cite{Alb2018a,Alb2019b} for details and other results, where the above identities have all been proved by two different means.


Proposition \ref{Propfundamental} together with \eqref{equacoesestruturaisgerais} yields a remarkable result.
\begin{teo}
 For every oriented Riemannian 3-manifold $M$, the total space $T^1M$ carries two natural degree 3 calibrations (up to a $\pm$ sign):
\begin{equation}
 \theta\wedge\alpha_0\qquad\mbox{and}\qquad{\mathrm{d}}\alpha_0=\theta\wedge\alpha_1 .
\end{equation}
\end{teo}
Notice $\pm\theta\wedge\alpha_1$ is of the kind found in Proposition \ref{circleofcalibrations}.

From Lemma \ref{varphicalibratedvectorfield}, a vector field is a calibrated submanifold of $\theta\wedge\alpha_0$ if and only if it is parallel. More important now is the solution of the optimal vector field problem on hyperbolic space, with which we almost finish the article.

 Consider the following model of hyperbolic metric on $M=\R^2\times\R_+$, with $a>0$ fixed, and constant sectional curvature $c=-a$:
 \[  \langle\ ,\ \rangle=\frac{1}{at^2}\bigl(({\mathrm{d}} x^1)^2+({\mathrm{d}} x^2)^2+({\mathrm{d}} t)^2\bigr) .\]


\begin{teo}
For $c=-1$, the vector field $X\in{\mathfrak{X}}^1_\Omega$ defined on a domain $\Omega\subset M$ by $X=t\partial_t$, has minimal volume
\[  {\mathrm{vol}}(X)=2{\mathrm{vol}}(\Omega)=-\int_{\partial\Omega}t^4\,{\mathrm{d}} x^1\wedge{\mathrm{d}} x^2  \]
among all vector fields in its homology class.
\end{teo}
\begin{proof}
 It is easy to see the Levi-Civita connection is given by
 \[  {\nabla}_{\partial_i}\partial_j=\frac{1}{t}\delta_{ij}\partial_t ,\quad\quad
 {\nabla}_{\partial_i}\partial_t={\nabla}_{\partial_t}\partial_i=-\frac{1}{t}\partial_i ,\quad\quad {\nabla}_{\partial_t}\partial_t=-\frac{1}{t}\partial_t , \qquad\forall i,j=1,2,\]
 and even easier to check the metric has constant sectional curvature $-a$.

 Now we use the exact calibration $\varphi=-\theta\wedge\alpha_1$ and recall Lemma \ref{Lemma_varphicalibratedvectorfield} with $X=X_0=\sqrt{a}t\partial_t$. For this, we let $X_1=\sqrt{a}t\partial_1$, $X_2=\sqrt{a}t\partial_2$, composing the orthonormal frame. We find ${\nabla}_{\partial_t}t\partial_t=\partial_t-t\frac{1}{t}\partial_t=0$ and ${\nabla}_{\partial_i}t\partial_t=-t\frac{1}{t}\partial_i=-\partial_i$ ($i=1,2$). Hence, for the $A_{ij}=\langle{\nabla}_{X_i}X,X_j\rangle,\ i,j=0,1,2$, we obtain
 \[  A_{i0}=0,\qquad A_{0j}=\langle{\nabla}_XX,X_i\rangle=0,\qquad A_{ij}=\langle{\nabla}_{X_i}X,X_j\rangle=0,\ \ i\neq j ,\]
\[  A_{ii}=\langle{\nabla}_{X_i}X,X_i\rangle= a^{\frac{3}{2}}t^2\langle{\nabla}_{\partial_i}t\partial_t,\partial_i\rangle=-\sqrt{a},\ \ i=1,2. \]
Finally, equation \eqref{varphicalibratedvectorfield} reads
\[  -(-2\sqrt{a})=\sqrt{1+a+a+a^2} ,  \]
which has the desired solution with $a=1$. Moreover, in this case,
 \[ {\mathrm{vol}}(X)=\int_\Omega X^*\varphi=-\int_{\partial\Omega}X^*\alpha_0=
  -\int_{\partial\Omega}X\lrcorner{\mathrm{vol}}_M=-\int_{\partial\Omega}t^4\,{\mathrm{d}} x^1\wedge{\mathrm{d}} x^2,
 \]
since the volume form of $M$ is $t^3\,{\mathrm{d}} x^1{\mathrm{d}} x^2 {\mathrm{d}} t$.
\end{proof}

Recalling the Introduction, we seem to have a \textit{one-time-deal} too with $c=-1$, as in \cite{GluckZiller} with $c=1$. The volume of $X$ is twice the volume of $\Omega$, as expected in the end of Section 3.2.


The last result is quite curious since $X_1$ or $X_2$ have less volume than $X_0$. In order to see this, let us rephrase the notation and write
\[   X=t\partial_1, \qquad Y=t\partial_2,\qquad Z=t\partial_t  .\]
Then $ {\nabla}_XX=t^2{\nabla}_{\partial_1}\partial_1=t\partial_t=Z,\
  {\nabla}_YX=t^2{\nabla}_{\partial_2}\partial_1=0,\
 {\nabla}_ZX=t{\nabla}_{\partial_t}t\partial_1=t\partial_1+t^2{\nabla}_{\partial_t}\partial_1= 0$.
And therefore, with the usual ordering,
$A_{00}=A_{01}=A_{1j}=A_{2j}=0,\ \forall j$, and only $A_{02}=1$. Whence ${\mathrm{vol}}(X)=\sqrt{2}{\mathrm{vol}}(\Omega)$. We note that no calibration studied in this article can deal with this vector field.

The conclusion is that the minimal vector field $t\partial_t$ and $t\partial_1$ are not in the same homology class, for every domain.

We remark we have not found any other concrete solution for hyperbolic metric in the literature. And that it should be an interesting problem concerning 3-manifold geometry.
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