[proofplan] We apply the [Bochner formula for harmonic maps](/theorems/5688) to the energy density of $u$. Because the domain has nonnegative Ricci curvature and the target has nonpositive sectional curvature, every term on the right-hand side of the Bochner identity has the favourable sign. Integrating over the compact boundaryless manifold forces the full right-hand side to vanish pointwise. The positive definiteness of $\operatorname{Ric}^M$ at one point then forces $du$ to vanish at that point, and parallelness of $du$ propagates this vanishing to all of $M$; connectedness then implies that $u$ is constant. [/proofplan]

[step:Apply the Bochner identity to the energy density of the harmonic map]Let $e(u):M\to [0,\infty)$ denote the energy density map of $u$, defined by \begin{align*} e(u)(p)=\frac{1}{2}|du_p|_{g,h}^2 \end{align*} for $p\in M$, where $|du_p|_{g,h}$ is the Hilbert-Schmidt norm of the linear map $du_p:T_pM\to T_{u(p)}N$ computed using $g_p$ and $h_{u(p)}$. Let $\nabla du$ denote the covariant derivative of $du$ with respect to the Levi-Civita connections of $(M,g)$ and $(N,h)$. Let $\Delta_g:C^\infty(M)\to C^\infty(M)$ denote the Laplace-Beltrami operator associated to the Riemannian metric $g$. Since $u$ is harmonic, the external [Bochner formula for harmonic maps](/page/Harmonic%20Map) gives, at every $p\in M$, \begin{align*} \Delta_g e(u)(p) = |\nabla du|_{g,h}^2(p) + \sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr) - \sum_{i,j=1}^{m} h_{u(p)}\bigl(R^N(du_p(E_i),du_p(E_j))du_p(E_j),du_p(E_i)\bigr), \end{align*} where $m=\dim M$, $(E_1,\dots,E_m)$ is any $g_p$-[orthonormal basis](/page/Orthonormal%20Basis) of $T_pM$, $\operatorname{Ric}^{M,\sharp}_p:T_pM\to T_pM$ is the $g_p$-self-adjoint [linear map](/page/Linear%20Map) defined by \begin{align*} g_p(\operatorname{Ric}^{M,\sharp}_p X,Y)=\operatorname{Ric}^M_p(X,Y) \end{align*} for $X,Y\in T_pM$, and $R^N$ is the curvature tensor of $(N,h)$. Here we are citing a result not yet in the wiki: Bochner formula for harmonic maps.[/step]

[guided]The object to which the curvature assumptions apply is the energy density. Define the energy density map $e(u):M\to [0,\infty)$ by \begin{align*} e(u)(p)=\frac{1}{2}|du_p|_{g,h}^2 \end{align*} for $p\in M$. This is a smooth function because $u$ is smooth and the metrics $g$ and $h$ are smooth. The Bochner formula for harmonic maps is the bridge between curvature and rigidity. Since $u$ is harmonic, its tension field vanishes, and the formula states that, for every $p\in M$ and every $g_p$-orthonormal basis $(E_1,\dots,E_m)$ of $T_pM$, \begin{align*} \Delta_g e(u)(p) = |\nabla du|_{g,h}^2(p) + \sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr) - \sum_{i,j=1}^{m} h_{u(p)}\bigl(R^N(du_p(E_i),du_p(E_j))du_p(E_j),du_p(E_i)\bigr). \end{align*} The map $\operatorname{Ric}^{M,\sharp}_p:T_pM\to T_pM$ is the $g_p$-self-adjoint representative of the Ricci tensor, defined by \begin{align*} g_p(\operatorname{Ric}^{M,\sharp}_p X,Y)=\operatorname{Ric}^M_p(X,Y) \end{align*} for all $X,Y\in T_pM$. Let $\Delta_g:C^\infty(M)\to C^\infty(M)$ denote the Laplace-Beltrami operator associated to $g$. This identity is the key point: it separates the Laplacian of $e(u)$ into a square term, a Ricci term from the domain, and a sectional-curvature term from the target. We are citing the external [Bochner formula for harmonic maps](/page/Harmonic%20Map); no dedicated theorem page for this formula is currently available in the wiki.[/guided]

[step:Use the curvature hypotheses to make every Bochner term nonnegative]Fix $p\in M$ and choose a $g_p$-orthonormal basis $(E_1,\dots,E_m)$ of $T_pM$ diagonalising the self-adjoint map $\operatorname{Ric}^{M,\sharp}_p$. Let $\lambda_i(p)\geq 0$ be the corresponding eigenvalues, so that $\operatorname{Ric}^{M,\sharp}_pE_i=\lambda_i(p)E_i$. Then \begin{align*} \sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr) = \sum_{i=1}^{m}\lambda_i(p)|du_p(E_i)|_{h}^2 \geq 0. \end{align*} For the target term, each summand satisfies \begin{align*} h_{u(p)}\bigl(R^N(A,B)B,A\bigr) = K_N(\operatorname{span}(A,B))\bigl(|A|_h^2|B|_h^2-h(A,B)^2\bigr) \leq 0 \end{align*} when $A=du_p(E_i)$ and $B=du_p(E_j)$, with the value interpreted as $0$ when $A$ and $B$ are linearly dependent. Hence \begin{align*} - \sum_{i,j=1}^{m} h_{u(p)}\bigl(R^N(du_p(E_i),du_p(E_j))du_p(E_j),du_p(E_i)\bigr) \geq 0. \end{align*} Therefore \begin{align*} \Delta_g e(u)(p)\geq 0 \end{align*} for every $p\in M$.[/step]

[guided]We now check exactly where the two curvature hypotheses enter. Since $\operatorname{Ric}^M_p$ is nonnegative definite, the associated $g_p$-self-adjoint map $\operatorname{Ric}^{M,\sharp}_p$ has nonnegative eigenvalues. Choose a $g_p$-orthonormal eigenbasis $(E_1,\dots,E_m)$ and write \begin{align*} \operatorname{Ric}^{M,\sharp}_pE_i=\lambda_i(p)E_i, \qquad \lambda_i(p)\geq 0. \end{align*} Substituting this into the Ricci contribution gives \begin{align*} \sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr) = \sum_{i=1}^{m}\lambda_i(p)|du_p(E_i)|_{h}^2 \geq 0. \end{align*} Thus the domain curvature term cannot decrease $\Delta_g e(u)$. For the target curvature term, set \begin{align*} A_i:=du_p(E_i)\in T_{u(p)}N. \end{align*} If $A_i$ and $A_j$ are linearly independent, the definition of sectional curvature gives \begin{align*} h_{u(p)}\bigl(R^N(A_i,A_j)A_j,A_i\bigr) = K_N(\operatorname{span}(A_i,A_j)) \bigl(|A_i|_h^2|A_j|_h^2-h(A_i,A_j)^2\bigr). \end{align*} The Gram determinant \begin{align*} |A_i|_h^2|A_j|_h^2-h(A_i,A_j)^2 \end{align*} is nonnegative by the [Cauchy-Schwarz inequality](/theorems/432) for the [inner product](/page/Inner%20Product) $h_{u(p)}$, and $K_N\leq 0$ by hypothesis. Hence the displayed curvature expression is nonpositive. If $A_i$ and $A_j$ are linearly dependent, the alternating property of $R^N$ gives the same conclusion with value $0$. Therefore \begin{align*} - \sum_{i,j=1}^{m} h_{u(p)}\bigl(R^N(du_p(E_i),du_p(E_j))du_p(E_j),du_p(E_i)\bigr) \geq 0. \end{align*} Together with $|\nabla du|_{g,h}^2(p)\geq 0$, the Bochner formula gives \begin{align*} \Delta_g e(u)(p)\geq 0. \end{align*}[/guided]

[step:Integrate the Bochner identity to force $\nabla du$ and the Ricci contribution to vanish]Let $\mu_g$ denote the Riemannian volume measure on $M$. Since $M$ is compact and has no boundary, [integration by parts](/theorems/2098) gives \begin{align*} \int_M \Delta_g e(u)\,d\mu_g(p)=0. \end{align*} For each $p\in M$, define the Ricci contribution $\mathcal R:M\to \mathbb R$ and the target-curvature contribution $\mathcal S:M\to \mathbb R$ by \begin{align*} \mathcal R(p)=\sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr) \end{align*} and \begin{align*} \mathcal S(p)=-\sum_{i,j=1}^{m}h_{u(p)}\bigl(R^N(du_p(E_i),du_p(E_j))du_p(E_j),du_p(E_i)\bigr), \end{align*} where $(E_1,\dots,E_m)$ is any $g_p$-orthonormal basis of $T_pM$. These quantities are independent of the chosen orthonormal basis because they are the two trace contractions appearing in the Bochner identity. Integrating the Bochner identity over $M$ therefore yields \begin{align*} 0=\int_M \bigl(|\nabla du|_{g,h}^2(p)+\mathcal R(p)+\mathcal S(p)\bigr)\,d\mu_g(p). \end{align*} Each summand in the integrand is continuous and nonnegative by the previous step. If their sum were positive at some point, continuity would make it positive on an open neighbourhood, and every nonempty open subset of a Riemannian manifold has positive Riemannian volume; this would contradict the displayed zero integral. Hence the sum vanishes at every point of $M$. In particular, \begin{align*} |\nabla du|_{g,h}^2(p)=0 \end{align*} for every $p\in M$, and \begin{align*} \sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr)=0 \end{align*} for every $p\in M$.[/step]

[guided]Let $\mu_g$ denote the Riemannian volume measure on $M$. The goal is to turn the pointwise inequality from the Bochner formula into pointwise vanishing. Because $M$ is compact and has no boundary, [integration by parts](/theorems/210) for the Laplace-Beltrami operator gives \begin{align*} \int_M \Delta_g e(u)\,d\mu_g(p)=0. \end{align*} For each $p\in M$, define the Ricci contribution $\mathcal R:M\to \mathbb R$ and the target-curvature contribution $\mathcal S:M\to \mathbb R$ by \begin{align*} \mathcal R(p)=\sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr) \end{align*} and \begin{align*} \mathcal S(p)=-\sum_{i,j=1}^{m}h_{u(p)}\bigl(R^N(du_p(E_i),du_p(E_j))du_p(E_j),du_p(E_i)\bigr), \end{align*} where $(E_1,\dots,E_m)$ is any $g_p$-orthonormal basis of $T_pM$. These definitions do not depend on the chosen orthonormal basis because they are trace contractions of tensorial quantities appearing in the Bochner identity. Integrating the Bochner identity gives \begin{align*} 0=\int_M \bigl(|\nabla du|_{g,h}^2(p)+\mathcal R(p)+\mathcal S(p)\bigr)\,d\mu_g(p). \end{align*} The previous curvature-sign step proves that each of the three summands is nonnegative at every point, and smoothness of $u$, $g$, and $h$ makes the summands continuous. If the sum were positive at some point $p\in M$, continuity would give an open neighbourhood $U\subset M$ of $p$ on which the sum is still positive. Every nonempty open subset of a Riemannian manifold has positive Riemannian volume, so the integral over $U$ would be positive, contradicting the displayed equality. Therefore the full sum vanishes at every point. Since each summand is nonnegative, a pointwise sum equal to zero forces each summand to be zero. In particular, \begin{align*} |\nabla du|_{g,h}^2(p)=0 \end{align*} for every $p\in M$, and \begin{align*} \sum_{i=1}^{m} h_{u(p)}\bigl(du_p(\operatorname{Ric}^{M,\sharp}_p E_i),du_p(E_i)\bigr)=0 \end{align*} for every $p\in M$.[/guided]

[step:Use positive Ricci curvature at one point to show $du$ vanishes there]Let $p_0\in M$ be a point such that $\operatorname{Ric}^M_{p_0}$ is positive definite. Choose a $g_{p_0}$-orthonormal basis $(F_1,\dots,F_m)$ of $T_{p_0}M$ diagonalising $\operatorname{Ric}^{M,\sharp}_{p_0}$, and write \begin{align*} \operatorname{Ric}^{M,\sharp}_{p_0}F_i=\lambda_i(p_0)F_i. \end{align*} Positive definiteness gives $\lambda_i(p_0)>0$ for every $i$. The vanishing of the Ricci contribution at $p_0$ gives \begin{align*} 0 = \sum_{i=1}^{m}\lambda_i(p_0)|du_{p_0}(F_i)|_h^2. \end{align*} Every summand is nonnegative and every coefficient $\lambda_i(p_0)$ is strictly positive, so \begin{align*} du_{p_0}(F_i)=0 \end{align*} for every $i$. Since $(F_1,\dots,F_m)$ is a basis of $T_{p_0}M$, this proves \begin{align*} du_{p_0}=0:T_{p_0}M\to T_{u(p_0)}N. \end{align*}[/step]

[guided]Let $p_0\in M$ be a point at which $\operatorname{Ric}^M_{p_0}$ is positive definite. We use this single point to obtain one actual zero of the differential. Choose a $g_{p_0}$-orthonormal basis $(F_1,\dots,F_m)$ of $T_{p_0}M$ diagonalising the $g_{p_0}$-self-adjoint map $\operatorname{Ric}^{M,\sharp}_{p_0}:T_{p_0}M\to T_{p_0}M$, and write \begin{align*} \operatorname{Ric}^{M,\sharp}_{p_0}F_i=\lambda_i(p_0)F_i. \end{align*} Because $\operatorname{Ric}^M_{p_0}$ is positive definite, every eigenvalue satisfies $\lambda_i(p_0)>0$. From the previous step, the Ricci contribution vanishes at every point, hence at $p_0$: \begin{align*} 0 = \sum_{i=1}^{m}\lambda_i(p_0)|du_{p_0}(F_i)|_h^2. \end{align*} Each term in this finite sum is nonnegative: the coefficient $\lambda_i(p_0)$ is positive and the norm-square $|du_{p_0}(F_i)|_h^2$ is nonnegative. A finite sum of nonnegative [real numbers](/page/Real%20Numbers) can equal zero only when every summand is zero. Therefore \begin{align*} du_{p_0}(F_i)=0 \end{align*} for every $i\in\{1,\dots,m\}$. Since $(F_1,\dots,F_m)$ is a basis of $T_{p_0}M$, the linear map $du_{p_0}:T_{p_0}M\to T_{u(p_0)}N$ vanishes on a basis and hence is the zero map: \begin{align*} du_{p_0}=0:T_{p_0}M\to T_{u(p_0)}N. \end{align*}[/guided]

[step:Propagate the vanishing of $du$ from one point to all of $M$]From the integral argument we already know that \begin{align*} \nabla du=0 \end{align*} on $M$. Thus $du$ is a parallel section of the vector bundle $T^*M\otimes u^*TN\to M$ with respect to the induced connection. Let $q\in M$. Since $M$ is connected, there exists a piecewise smooth curve \begin{align*} \gamma:[0,1]\to M \end{align*} with $\gamma(0)=p_0$ and $\gamma(1)=q$. Parallelness of $du$ means that $du_{\gamma(t)}$ is obtained from $du_{p_0}$ by parallel transport along $\gamma$ in the bundle $T^*M\otimes u^*TN$. Since parallel transport is linear and $du_{p_0}=0$, it follows that \begin{align*} du_q=0. \end{align*} As $q\in M$ was arbitrary, $du=0$ everywhere on $M$.[/step]

[guided]The equation $\nabla du=0$ says more than pointwise vanishing of a norm: it says that $du$ is parallel as a section of the bundle \begin{align*} T^*M\otimes u^*TN\to M. \end{align*} This is the correct bundle because, at each point $p\in M$, the differential $du_p$ is a linear map $T_pM\to T_{u(p)}N$, equivalently an element of $T_p^*M\otimes T_{u(p)}N$. Fix an arbitrary point $q\in M$. Since $M$ is connected, and smooth connected manifolds are path-connected, choose a piecewise smooth curve \begin{align*} \gamma:[0,1]\to M \end{align*} such that $\gamma(0)=p_0$ and $\gamma(1)=q$. Along this curve, the induced connection on $T^*M\otimes u^*TN$ defines a parallel transport map from the fibre over $p_0$ to the fibre over $q$. Because $\nabla du=0$, the section $du$ is transported into itself along $\gamma$. Therefore $du_q$ is the parallel transport of $du_{p_0}$. But we proved that \begin{align*} du_{p_0}=0. \end{align*} Parallel transport is linear, so it sends the zero element of the fibre over $p_0$ to the zero element of the fibre over $q$. Hence \begin{align*} du_q=0. \end{align*} Since $q$ was arbitrary, $du=0$ at every point of $M$.[/guided]

[step:Conclude that the harmonic map is constant]Let $p,q\in M$. Since $M$ is connected, choose a piecewise smooth curve \begin{align*} \gamma:[0,1]\to M \end{align*} with $\gamma(0)=p$ and $\gamma(1)=q$. The composition \begin{align*} u\circ\gamma:[0,1]\to N \end{align*} satisfies \begin{align*} \frac{d}{dt}(u\circ\gamma)(t)=du_{\gamma(t)}(\dot{\gamma}(t))=0 \end{align*} on each smooth subinterval of $\gamma$. Hence $u\circ\gamma$ is constant on $[0,1]$, so $u(p)=u(q)$. Since $p$ and $q$ were arbitrary, $u$ is constant on $M$.[/step]

[guided]We now translate the differential statement $du=0$ into the global statement that $u$ is constant. Let $p,q\in M$ be arbitrary. Since $M$ is connected, and smooth connected manifolds are path-connected, choose a piecewise smooth curve \begin{align*} \gamma:[0,1]\to M \end{align*} with $\gamma(0)=p$ and $\gamma(1)=q$. Consider the composition \begin{align*} u\circ\gamma:[0,1]\to N. \end{align*} On each smooth subinterval of $\gamma$, the chain rule gives \begin{align*} \frac{d}{dt}(u\circ\gamma)(t)=du_{\gamma(t)}(\dot{\gamma}(t)). \end{align*} But the previous step proved $du_{\gamma(t)}=0:T_{\gamma(t)}M\to T_{u(\gamma(t))}N$ for every $t$, so \begin{align*} \frac{d}{dt}(u\circ\gamma)(t)=0 \end{align*} on each smooth subinterval. Hence $u\circ\gamma$ is constant on every smooth subinterval, and continuity across the finitely many breakpoints of the piecewise smooth curve makes it constant on all of $[0,1]$. Therefore \begin{align*} u(p)=u(q). \end{align*} Since $p$ and $q$ were arbitrary points of $M$, the map $u:M\to N$ is constant.[/guided]