One has

\begin{align*} \label{eq:second-fund-form} \langle \tilde\nabla_i\xi_\sigma, f_* e_j\rangle &= \langle\xi_\sigma, \tilde\nabla_i (f_* e_j)\rangle = \langle \xi_\sigma,\tilde\nabla_i(f^\gamma_l dx^l\otimes \tilde e_\gamma) e_j + f_* \nabla_i e_j \rangle\\ &= \langle \xi_\sigma, (f^\gamma_{il} dx^l\tilde e_\gamma + f^\gamma_l dx^l \tilde\nabla_i\tilde e_\gamma) e_j \rangle\\ &= \langle \xi_\sigma, f^\gamma_{ij} \tilde e_\gamma + f^\gamma_j A^\alpha_{\gamma_i}\tilde e_\alpha \rangle = \left\langle \xi_\sigma, \left(f^\gamma_{ij} + \Gamma'^\gamma_{\alpha\beta} f^\alpha_i f^\beta_j \right)\tilde e_\gamma \right\rangle \\ &= \langle\xi_\sigma,\tilde\nabla_i(f_*).e_j \rangle = \langle \xi_\sigma,\beta(f)_{ij} \rangle \end{align*}

where we used \( \xi_\sigma \perp f_* e_j \) in the first line and \( \xi_\sigma \perp f_*([e_i, e_j]) \) in the second line. Hence \( \sff_{ij} \equiv -\beta(f)_{ij} \) modulo an element in \( TM \). It remains to see that \( \beta(f)_{ij}\perp M \) in order to conclude \( \sff = -\beta(f) \). By definition, one has \( \beta(f)_{ij} = \tilde \nabla_i (f_*). e_j \) and

\begin{align*} \langle \beta(f)_{ij}, f_* e_k \rangle &= \langle\tilde \nabla_i (f_*). e_j, f_* e_k \rangle = \tilde\nabla_i \langle f_* e_j, f_* e_k \rangle - \langle \nabla_i e_j, e_k \rangle - \langle f_*e_j, \tilde\nabla_i(f_* e_k) \rangle \\ &= \nabla_i \langle e_j,e_k \rangle - \langle \nabla_i e_j, e_k \rangle -\langle \beta(f)_{ik}, f_* e_j \rangle - \langle e_j, \nabla_i e_k \rangle\\ &= -\langle \beta(f)_{ik}, f_* e_j \rangle \end{align*}

Then using the symmetric of \( \beta(f)_{ij} \), one has \( \langle \beta(f)_{ij}, f_* e_k \rangle=0 \).

Finally, if \( \beta(f)=0 \) and \( X \) is a geodesic vector field of \( M \), one needs to prove that \( f_*X \) is a geodesic vector field of \( M' \). In fact \[ \tilde\nabla_{X}(f_* X) = (\tilde\nabla_X f_*) X + f_*\nabla_X X = \beta(f)(X,X) = 0. \] Hence \( f_* X \) is a geodesic field of \( M' \).

The first part is due to Ehresmann, take a small geodesic ball center at \( P \), and connect every point \( Q \) to \( P \) by a curve \( \gamma \). Map every point \(\hat \gamma(0)\in M_P \) to the point \( \hat\gamma(1)\in M_Q \) where \( \hat\gamma \) is the lift of \( \gamma \) starting from \( \hat\gamma(0) \). One has a diffeomorphism \( \theta_\gamma: M_{\gamma(0)} \longrightarrow M_{\gamma(1)} \).

The second part, due to Hermann, can be established in 2 steps:

First, by direct computation, one proves that if any geodesic field \( X \) on \( B \) lifts to a horizontal vector field \( \hat X \) then \( \hat X \) is a geodesic vector field. In fact, denote by \( \nabla \) and \( \tilde\nabla \) the Levi-Civita connection on \( M \) and \( B \) respectively and \( \mathcal{V}, \mathcal{H} \) the vertical and horizontal projection of tangent vectors of \( M \). Then \(\nabla_{\hat X}\hat X = \mathcal{V}\nabla_{\hat X}\hat X + \mathcal{H}\nabla_{\hat X}\hat X\) in which \( \mathcal{H}\nabla_{\hat X}\hat X \) is actually the horizontal lift of \( \tilde\nabla_X X \) therefore vanishes. We claim that \( \mathcal{V}\nabla_{Y}Y = \frac{1}{2}\mathcal{V}[Y,Y]\) hence also vanishes for every basic vector field \( Y \). In fact let \( U \) be any vertical vector field then \[ \langle U, \mathcal{V}\nabla_{Y}Y \rangle = \langle U,\nabla_Y Y \rangle = - \langle \nabla_X U, X \rangle= \langle \nabla_U X, X \rangle = \frac{1}{2}\nabla_U \langle X,X \rangle=0 \] where we used the fact that \( \)\( \nabla_X U -\nabla_U X = [X,U] = \widehat{[\pi_* X,\pi_* U]} = 0 \) and \( \langle X, X \rangle \) is constant on each fibre (being \(\langle \pi_* X,\pi_* X \rangle \)), hence in every vertical direction \( U \) (Remark: this corresponds to the fact that \( g' \) only depend on \( b \)).

Now if \( M \) is complete then for every geodesic curve \( \gamma \) in \( B \), let \( X \) be the velocity field of \( \gamma \) and \( \hat X \) be the horizontal lift of \( X \), which is now a horizontal, geodesic field of \( M \), whose integral curves are lifts of \(\gamma \). Therefore \( B \) is complete and every geodesic curve of \( B \) lifts horizontally to \( M \).

For the general curve \( \gamma \) of \( M \), the idea will be to approximate it by geodesics and lift part by part.

1. One has \( e(f) = \frac{1}{2}\langle g, f^* g' \rangle = \frac{1}{2}\langle g, (f'\circ f)^* g'' \rangle = e(f'\circ f)\).
2. One has \( \tau(f'\circ f)^a = \tau(f)^a + g^{ij}\beta(f')^a_{\alpha\beta} f^\alpha_i f^\beta_j\) by \eqref{eq:tension-field-composition}. The conclusion follows since the second term is normal to \( M' \).

One can suppose that \( M' \) is a Riemannian product of \( M'' \) and its fibre, and \( f' \) is the projection to \( M'' \), since this is true locally and the proposition is local. Then the conclusion is that the tension field of the projection is the projection of the tension field, or equivalently the tension field of \( f= f_1\times f_2: M \longrightarrow M''\times F \) is \( \tau(f) = (\tau f_1,\tau f_2) \). This follows from the explicit formula of \( \tau(f) \), noting that the Christoffel symbols \( \Gamma^\alpha_{\beta\gamma} \) vanish except when the indice \( \alpha,\beta,\gamma \) belong to the same tangent space (\( TM'' \) or \( TF \)).

Let \( \iota \) be the isometry of \( T \) locally given by \( (y,d)\mapsto (y,-d) \) for \( (y,d)\in M'\times D \equiv T \) and pose \( G_t= \iota F_t \) then \( G_t \) and \( F_t \) coincide initially since \( M' \) is fixed by \( \iota \). Moreover \[ \frac{d G_t}{d t} = d\iota . \frac{d F_t}{d t} = d\iota (\tau_T(F_t)) = \tau_T(\iota F_t)=\tau_T(G_t) \] We conclude that \( F_t = G_t = \iota F_t \), hence \( F_t \) remains in \( M' \) for all relevant \( t \), using the following uniqueness of nonlinear heat equation.

It is sufficient to prove the theorem for \( \omega \) very close to \( \alpha \), therefore by compactness of \( M \), we can suppose that there exists a finite atlas \( M = \bigcup_i U_i \) with \( f_1 (U_i\times [\alpha,\omega]) \) and \( f_2 (U_i,[\alpha,\omega]) \) being in the same chart \( V_i \) of \( M' \). We consider the distance function \( \sigma(a,b) = \frac{1}{2}d_{M'}(a,b)^2 \) for \( a,b\in M' \) to measure the difference between \( f_1 \) and \( f_2 \) by \[ \rho(x,t) = \sigma(f_1(x,t), f_2(x,t)) \] The strategy is to prove that there exists \( C>0 \) such that \( \frac{d \rho}{d t} \leq -\Delta \rho + C\rho \), then by Maximum principle, one has \( \rho = 0 \).

Fix a chart \( U_i \) of \( M \) and the corresponding \( V_i \) of \( M' \), one has by straightforward calculation:

\begin{align} \frac{d\rho}{d t} &= -\Delta \rho -g^{ij}\left( \frac{\partial^2 \sigma}{\partial f_1^\beta \partial f_1^\gamma} - \frac{\partial \sigma}{\partial f_1^\alpha} {\Gamma'} ^{\alpha}_{\beta\gamma}(f_1) \right) {f_1}^\beta_i {f_1}^\gamma_j \nonumber \\ &- g^{ij}\left( \frac{\partial^2 \sigma}{\partial f_2^\beta \partial f_2^\gamma} - \frac{\partial \sigma}{\partial f_2^\alpha} {\Gamma'} ^{\alpha}_{\beta\gamma}(f_2) \right) {f_2}^\beta_i {f_2}^\gamma_j - 2g^{ij}\frac{\partial^2 \sigma}{\partial f_1^\beta \partial f_2^\gamma} {f_1}^\beta_i {f_2}^\gamma_j \label{eq:drho-dt} \end{align}

where \( g^{ij} \) is the metric on \( M \) and \( {\Gamma'}^\alpha_{\beta\gamma} \) are Christoffel symbols of \( M' \).

Let \( c \) be a point in the chart \( V_i \) and choose the normal coordinates of \( M' \) at \( c \). Then for \( a,b\in M' \) near \( c \), one has, since \( \sigma(a,b) = \sigma(b,a) \) and \( \sigma(a,b)=0 \) if \( b^\gamma = k a^\gamma \) (the Euclidean straight line from \( a \) to \( ka \) viewed on \( M' \) is a geodesic): \[ \sigma(a,b) = \frac{1}{2}d_{M'}(a,b)^2 = \frac{1}{2}d_E(a,b)^2 + \lambda_{\beta\gamma,\delta}(a^\beta a^\gamma b^\delta + b^\beta b^\gamma a^\delta) \] where \( d_E \) is the Euclidean distance, with \( \lambda_{\beta\gamma,\delta} = \lambda_{\gamma\beta,\delta} \) and \( \lambda_{\beta\gamma,\delta} +\lambda_{\gamma\delta,\beta} + \lambda_{\beta\delta,\gamma}= 0 \). We then have the series development of \( \sigma \) at \( (0,0) \):

\begin{equation} \label{eq:dev-sig-unique} \sigma(a,b) = \frac{1}{2}\delta_{\beta\gamma} (a^\beta - b^\beta)(a^\gamma - b^\gamma) + \lambda_{\beta\gamma,\delta} (a^\beta a^\gamma b^\delta + b^\beta b^\gamma a^\delta) + O(|a|+|b|)^4 \end{equation}

and the development of its derivatives

\begin{align*} \frac{\partial^2 \sigma}{\partial a^\beta \partial b^\gamma}(a,b) &= -\delta_{\beta\gamma} + O(|a|+|b|)^2\\ \frac{\partial^2 \sigma}{\partial a^\beta \partial a^\gamma}(a,b) &= \delta_{\beta\gamma} + \lambda_{\beta\gamma,\delta}b^\delta + O(|a|+|b|)^2\\ \frac{\partial^2 \sigma}{\partial b^\beta \partial b^\gamma}(a,b) &= \delta_{\beta\gamma} + \lambda_{\beta\gamma,\delta}a^\delta + O(|a|+|b|)^2\\ \frac{\partial\sigma}{\partial a^\alpha}(a,b) &= O(|a|+|b|),\quad {\Gamma'}^\alpha_{\beta\gamma}(a) = O(|a|) \end{align*}

So choose \( c \) to be the midpoint of \( f_1(x,t) \) and \( f_2(x,t) \) and \( (f_1(x,t), f_2(x,t)) = (w,-w)\) in the chart, one has:

\begin{align} \frac{d\rho}{d t} &= -\Delta\rho - \left( \delta_{\beta\gamma} -\lambda_{\beta\gamma,\delta}w^\delta + O(|w|^2) \right) {f_1}^\beta_i {f_1}^\gamma_j g^{ij} - \left( \delta_{\beta\gamma} + \lambda_{\beta\gamma,\delta} w^\delta + O(|w|^2)\right) {f_2}^\beta_i {f_2}^\gamma_j g^{ij} \label{eq:sym-red-beta-before}\\ & - 2\left( -\delta_{\beta\gamma} + O(|w|^2) \right) {f_1}^\beta_i {f_2}^\gamma_j g^{ij} \label{eq:sym-red-beta}\\ & = -\Delta \rho - |df_1 -df_2|^2 - w^\delta \lambda_{\beta\gamma,\delta}g^{ij}\left( {f_2}^\beta_i {f_2}^\gamma_j - {f_1}^\beta_i {f_1}^\gamma_j \right) + O(|\omega|^2) \label{eq:first-order-w} \end{align}

The last term of \eqref{eq:first-order-w} can be bounded as follows:

\begin{align*} \left | w^\delta \lambda_{\beta\gamma,\delta}\left({f_2}^\beta_i {f_2}^\gamma_j - {f_1}^\beta_i {f_1}^\gamma_j\right) g^{ij} \right| &= \left | w^\delta \lambda_{\beta\gamma,\delta}\left({f_2}^\beta_i ({f_2}^\gamma_j - {f_1}^\gamma_j) + {f_1}^\gamma_j({f_2}^\beta_i - {f_1}^\beta_i) \right) g^{ij} \right| \\ &\leq 2 |w^\delta \lambda_{\beta\gamma,\delta}| |df_2 -df_1| (|df_1| + |df_2|)\\ &\leq |df_1 - df_2|^2 + O(|w|^2) \end{align*}

where for the last inequality, we use \( 2uv \leq u^2 + v^2 \) and the fact that \( |df_1| \) and \( |df_2| \) are bounded on \( M \). The estimate \eqref{eq:first-order-w} can be continued: \[ \frac{d \rho}{d t}\leq -\Delta\rho + C(x,t) |w|^2 \leq -\Delta \rho + C \rho \] where \( C >0 \) is a constant chosen to dominate all \( C(x,t) \) for \( x\in M \) in all charts and \( t\in [\alpha,\omega] \).

Apply Lemma lem:calculs-general to \( s = d f_t \) and the Riemannian-connected bundle \( F^* TM' \) over \( M\times [\alpha,\omega] \) where \( F(\cdot,t) = f_t \), the curvature terms cancel out and it remains to see that \( \frac{d e(f_t)}{dt}= - \langle df_t, \Delta df_t \rangle \), meaning that \( \tilde \nabla_{\partial t} df_t = -\Delta df_t \). This can be easily justified: \[ \tilde \nabla_{\partial t} df_t = \tilde \nabla_{\partial t} \tilde\nabla^M F= \tilde\nabla^M \tilde\nabla_{\partial t} F= \tilde\nabla^M\ \tau (f_t) = -D\delta (df_t) = -\Delta df_t \] where the last "=" is due to \( D df_t = 0 \). Note that \( D \) and \( \delta \) are the exterior derivative and its adjoint of the bundle \( (f_t)^*TM' \) on \( M \), where \( t \) can be fixed after the third "=" sign.

Let \( F:\ I\times M \longrightarrow M' \) be the total function with \( F(t,\cdot) = f_t\) for \( t\in I =[\alpha,\omega] \) and \( E = F^* TM' \) is a Riemannian-connected bundle on \( I\times M \) with curvature form \( \Theta \), then

\begin{equation} \label{eq:den-kin-1} \tilde \nabla_{\partial t} \tilde\nabla_v (dF.v) = \tilde\nabla_v \tilde\nabla_{\partial t} (dF.v) + \Theta(\partial t, v) dF.v \end{equation}

where \( dF \) is the exterior derivative of \( f_t \) on \( M \). Note that \( \tilde \nabla_v \tilde \nabla_{\partial t} (dF.v) = \tilde \nabla_v (\tilde \nabla_{\partial t} dF).v = \tilde \nabla_v (\tilde\nabla^M \frac{\partial f_t}{\partial t}).v \) since \( \tilde\nabla^M \frac{\partial f_t}{\partial t}= \tilde\nabla_{\partial t}^{I\times M} dF = \tilde\nabla^{I}_{\partial_t}dF \) because \( \tilde \nabla \) is torsionless on \( M' \). Plugging this in \eqref{eq:den-kin-1} and taking contraction in \( v \), one has

\begin{equation} \label{eq:den-kin-2} \tilde \nabla_{\partial t}\ \tau(f_t) = -\tilde \Delta \frac{\partial f_t}{\partial t} + \tr \left( v\mapsto \Theta (\partial t,v) dF.v\right) \end{equation}

But \( \Theta^\beta_\alpha = R'^\beta_{\alpha\nu\mu} F^\mu_i F^\nu_j dx^i\otimes dx^j \) where \( R' \) denotes the Riemannian curvature of \( M' \) and the indices \( i,j \) can be \( 0 \), with \( x^0 \equiv t \). Hence \[ \Theta (\partial t, v) dF.v = R'^\beta_{\alpha \nu \mu} \frac{\partial f^\mu_t}{\partial t}\frac{\partial f_t^\nu}{\partial v} \frac{\partial f_t^\alpha}{\partial v} \tilde e_\beta = \Riem(M')\left(\nabla_v f_t, \frac{\partial f_t}{\partial t}\right)\nabla_v f_t \] Plugging in \eqref{eq:den-kin-2} and taking inner product with \( \frac{\partial f_t}{\partial t} \), one has

\begin{align*} \frac{\partial k(f_t)}{\partial t} &= \left\langle \tilde \nabla_{\partial t}\ \tau(f_t), \frac{\partial f_t}{\partial t} \right \rangle = - \left\langle \tilde \Delta \frac{\partial f_t}{\partial t}, \frac{\partial f_t}{\partial t} \right\rangle + \left \langle \Riem(M') (\nabla_v f_t, \frac{\partial f_t}{\partial t}) \nabla_v f_t, \frac{\partial f_t}{\partial t} \right \rangle \\ &= - \Delta \left(\frac{1}{2} \left|\frac{\partial f_t}{\partial t}\right|^2\right) - \left| \tilde \nabla \frac{\partial f_t}{\partial t}\right|^2 + \left \langle \Riem(M') (\nabla_v f_t, \frac{\partial f_t}{\partial t}) \nabla_v f_t, \frac{\partial f_t}{\partial t} \right \rangle \end{align*}