[proofplan] We compute the second derivative of the energy along the variation $U$. The geodesic-homotopy hypothesis says that the variation field $V=\partial_t U$ is parallel in the $t$-direction, so the only terms in the second variation are a nonnegative square term and a curvature term. The curvature term has sign controlled by $K_N \leq 0$, and therefore the second derivative of the energy is nonnegative for every $t \in [0,1]$. [/proofplan] [step:Introduce the variation field and the pullback connection] Let $\nabla^N$ denote the Levi-Civita connection of $(N,h)$. Let $\nabla^U$ denote the pullback connection on the vector bundle $U^*TN \to M \times [0,1]$ induced by $\nabla^N$. Let $R^N$ denote the curvature tensor of $\nabla^N$ with the sign convention $R^N(A,B)C := \nabla^N_A\nabla^N_B C - \nabla^N_B\nabla^N_A C - \nabla^N_{[A,B]}C$. Define the variation field $V: M \times [0,1] \to U^*TN$ by $V(p,t) := \partial_t U(p,t) \in T_{U(p,t)}N$. For a vector field $X \in \mathfrak{X}(M)$, viewed as a vector field on $M \times [0,1]$ by making it independent of $t$, define $X_t(p) := du_t|_p(X_p) \in T_{u_t(p)}N$. Since $\gamma_p(t)=U(p,t)$ is a geodesic for every $p \in M$, the geodesic equation gives \begin{align*} \nabla^U_{\partial_t} V = 0. \end{align*} [/step] [step:Differentiate the energy once] Fix $t \in [0,1]$. Let $(e_1,\dots,e_m)$ be a local $g$-orthonormal frame on an [open set](/page/Open%20Set) in $M$, where $m=\dim M$. Define the local expression \begin{align*} e(u_t) = \frac{1}{2}\sum_{i=1}^m h((e_i)_t,(e_i)_t). \end{align*} If $(\tilde e_1,\dots,\tilde e_m)$ is another local $g$-orthonormal frame, the two frames differ pointwise by an orthogonal matrix and the above sum is the squared Hilbert-Schmidt norm of $du_t:T_pM\to T_{u_t(p)}N$. Hence this local formula defines a global smooth scalar function on $M$, so it may be integrated against $d\operatorname{vol}_g$. Because $M$ is compact and $U$ is smooth on $M \times [0,1]$, differentiation under the integral sign is valid. The metric compatibility of $\nabla^N$ gives \begin{align*} \frac{d}{dt}E(u_t) &= \int_M \sum_{i=1}^m h(\nabla^U_{\partial_t}(e_i)_t,(e_i)_t)\,d\operatorname{vol}_g(p). \end{align*} The Levi-Civita connection is torsion-free, and $[\partial_t,e_i]=0$ on $M \times [0,1]$, hence \begin{align*} \nabla^U_{\partial_t}(e_i)_t = \nabla^U_{e_i}V. \end{align*} Therefore \begin{align*} \frac{d}{dt}E(u_t) &= \int_M \sum_{i=1}^m h(\nabla^U_{e_i}V,(e_i)_t)\,d\operatorname{vol}_g(p). \end{align*} [guided] The first derivative records how the spatial differential $du_t$ changes as $t$ varies. We use a local $g$-orthonormal frame $(e_1,\dots,e_m)$ on $M$, where $m=\dim M$. Although the frame is only local, the sum over a $g$-orthonormal frame is independent of the chosen frame, so the scalar quantities obtained below patch to global functions on $M$. With such a frame, the squared Hilbert-Schmidt norm of $du_t$ is \begin{align*} |du_t|_{g,h}^2 = \sum_{i=1}^m h((e_i)_t,(e_i)_t). \end{align*} Thus \begin{align*} E(u_t)=\frac{1}{2}\int_M \sum_{i=1}^m h((e_i)_t,(e_i)_t)\,d\operatorname{vol}_g(p). \end{align*} Since $U$ is smooth and $M$ is compact, the integrand and its $t$-derivative are continuous and bounded on $M \times [0,1]$. Hence we may differentiate under the integral sign. Metric compatibility of $\nabla^N$ gives \begin{align*} \frac{d}{dt} h((e_i)_t,(e_i)_t) = 2h(\nabla^U_{\partial_t}(e_i)_t,(e_i)_t). \end{align*} After the factor $\frac{1}{2}$ in the definition of energy cancels the factor $2$, we obtain \begin{align*} \frac{d}{dt}E(u_t) &= \int_M \sum_{i=1}^m h(\nabla^U_{\partial_t}(e_i)_t,(e_i)_t)\,d\operatorname{vol}_g(p). \end{align*} The torsion-free property of the Levi-Civita connection identifies the $t$-derivative of the spatial differential with the spatial derivative of the variation field. Since each $e_i$ is extended to $M \times [0,1]$ independently of $t$, we have $[\partial_t,e_i]=0$, and therefore \begin{align*} \nabla^U_{\partial_t}(e_i)_t = \nabla^U_{e_i}V. \end{align*} Substituting this identity gives the [first variation formula](/theorems/2728) \begin{align*} \frac{d}{dt}E(u_t) &= \int_M \sum_{i=1}^m h(\nabla^U_{e_i}V,(e_i)_t)\,d\operatorname{vol}_g(p). \end{align*} [/guided] [/step] [step:Differentiate again and use the geodesic equation] Differentiate the first variation formula. Again using compactness of $M$ and smoothness of $U$ to justify differentiation under the integral sign, we obtain \begin{align*} \frac{d^2}{dt^2}E(u_t) &= \int_M \sum_{i=1}^m \left[ h(\nabla^U_{\partial_t}\nabla^U_{e_i}V,(e_i)_t) + h(\nabla^U_{e_i}V,\nabla^U_{\partial_t}(e_i)_t) \right]\,d\operatorname{vol}_g(p). \end{align*} Using $\nabla^U_{\partial_t}(e_i)_t=\nabla^U_{e_i}V$, the second term becomes $|\nabla^U_{e_i}V|_h^2$. For the first term, the curvature identity for the pullback connection gives \begin{align*} \nabla^U_{\partial_t}\nabla^U_{e_i}V - \nabla^U_{e_i}\nabla^U_{\partial_t}V = R^N(V,(e_i)_t)V. \end{align*} Since $\nabla^U_{\partial_t}V=0$, this reduces to \begin{align*} \nabla^U_{\partial_t}\nabla^U_{e_i}V = R^N(V,(e_i)_t)V. \end{align*} Therefore \begin{align*} \frac{d^2}{dt^2}E(u_t) = \int_M \sum_{i=1}^m \left[ |\nabla^U_{e_i}V|_h^2 + h(R^N(V,(e_i)_t)V,(e_i)_t) \right]\,d\operatorname{vol}_g(p). \end{align*} Using the standard curvature symmetry $h(R^N(A,B)C,D)=-h(R^N(A,B)D,C)$ gives the equivalent formula \begin{align*} \frac{d^2}{dt^2}E(u_t) = \int_M \sum_{i=1}^m \left[ |\nabla^U_{e_i}V|_h^2 - h(R^N(V,(e_i)_t)(e_i)_t,V) \right]\,d\operatorname{vol}_g(p), \end{align*} where $A,B,C,D \in T_qN$ for an arbitrary point $q \in N$. [/step] [step:Use nonpositive sectional curvature to make the curvature term nonnegative] Fix $p \in M$, $t \in [0,1]$, and $i \in \{1,\dots,m\}$. Let \begin{align*} A := V(p,t) \in T_{U(p,t)}N, \qquad B := (e_i)_t(p) \in T_{U(p,t)}N. \end{align*} If $A$ and $B$ are linearly dependent, then \begin{align*} h(R^N(A,B)B,A)=0 \end{align*} by the alternating property of the curvature tensor in its first two entries. If $A$ and $B$ are linearly independent, the definition of sectional curvature gives \begin{align*} h(R^N(A,B)B,A) = K_N(\operatorname{span}\{A,B\}) \left(|A|_h^2|B|_h^2-h(A,B)^2\right). \end{align*} Since $K_N \leq 0$ and the Gram determinant \begin{align*} |A|_h^2|B|_h^2-h(A,B)^2 \end{align*} is positive for linearly independent $A$ and $B$, we have \begin{align*} h(R^N(A,B)B,A) \leq 0. \end{align*} Thus, in all cases, \begin{align*} -h(R^N(V,(e_i)_t)(e_i)_t,V) \geq 0. \end{align*} [/step] [step:Conclude that the energy is convex] Combining the [second variation formula](/theorems/2729) with the sign of the curvature term gives \begin{align*} \frac{d^2}{dt^2}E(u_t) = \int_M \sum_{i=1}^m \left[ |\nabla^U_{e_i}V|_h^2 - h(R^N(V,(e_i)_t)(e_i)_t,V) \right]\,d\operatorname{vol}_g(p). \end{align*} Each summand is nonnegative, so \begin{align*} \frac{d^2}{dt^2}E(u_t) \geq 0. \end{align*} Hence the smooth function $F: [0,1] \to \mathbb{R}$ defined by $F(t) := E(u_t)$ has nonnegative ordinary second derivative on the interior $(0,1)$. Since $F$ is continuous on $[0,1]$ and smooth on $(0,1)$, this implies that $F$ is convex on $[0,1]$. Therefore $t \mapsto E(u_t)$ is convex. [/step]