[proofplan] The proof is a pointwise computation in a normal orthonormal frame on $M$. We differentiate the identity $e(u)=\frac12\sum_i |du(e_i)|^2$ twice, using metric compatibility to produce the positive term $|\nabla du|^2$ and a remaining rough-Laplacian contraction. We then commute the covariant derivatives of $du$; the domain curvature contracts to the Ricci term, the target curvature gives the negative sectional-curvature term, and the derivative of the tension field vanishes because $u$ is harmonic. Since every term is tensorial, the pointwise normal-frame calculation proves the formula globally. [/proofplan]

[step:Declare the induced connection and reduce to a normal-frame computation] Let $E:=u^*TN$ be the [pullback vector bundle](/page/Pullback%20Bundle) over $M$. The connection induced by the Levi-Civita connection $\nabla^N$ on $E$ is denoted by $\nabla^E$, and the induced tensor-product connection on $T^*M\otimes E$ is denoted by $\nabla$. Thus $du$ is a smooth section \begin{align*} du \in \Gamma(T^*M\otimes E), \end{align*} and its covariant derivative is the section \begin{align*} \nabla du \in \Gamma(T^*M\otimes T^*M\otimes E) \end{align*} defined by \begin{align*} (\nabla du)(X,Y) := \nabla^E_X(du(Y))-du(\nabla^M_XY) \end{align*} for smooth vector fields $X,Y\in\mathfrak X(M)$. The tension field of $u$ is the section \begin{align*} \tau(u)\in\Gamma(E), \qquad \tau(u):=\sum_{i=1}^m(\nabla du)(e_i,e_i), \end{align*} where $e_1,\dots,e_m$ is any local $g$-orthonormal frame. The harmonicity assumption means \begin{align*} \tau(u)=0. \end{align*} Fix a point $p\in M$. Choose a local $g$-orthonormal frame $e_1,\dots,e_m$ near $p$ such that \begin{align*} \nabla^M_{e_i}e_j(p)=0 \end{align*} for all $1\leq i,j\leq m$. Because $\nabla^M$ is torsion-free, this also gives \begin{align*} [e_i,e_j](p)=0. \end{align*} The desired formula is tensorial in the frame and in the point $p$, so it suffices to prove it at $p$ in this frame. [/step]

[step:Differentiate the energy density twice]For $1\leq j\leq m$, define the local section $A_j\in \Gamma(E)$ by \begin{align*} A_j:=du(e_j). \end{align*} For $1\leq i,j\leq m$, define the local section $B_{ij}\in\Gamma(E)$ by \begin{align*} B_{ij}:=(\nabla du)(e_i,e_j). \end{align*} By the definition of the induced connection on $T^*M\otimes E$, \begin{align*} B_{ij}=\nabla^E_{e_i}A_j-du(\nabla^M_{e_i}e_j). \end{align*} At the point $p$, the normal-frame condition gives \begin{align*} B_{ij}(p)=\nabla^E_{e_i}A_j(p). \end{align*} For $1\leq i,j,k\leq m$, define the smooth real-valued connection coefficient $\Gamma_{ij}^k$ by \begin{align*} \nabla^M_{e_i}e_j=\sum_{k=1}^m \Gamma_{ij}^k e_k. \end{align*} Metric compatibility of $\nabla^M$ and the $g$-orthonormality of the frame give \begin{align*} 0=e_i g(e_j,e_k)=\Gamma_{ij}^k+\Gamma_{ik}^j, \end{align*} so $\Gamma_{ij}^k=-\Gamma_{ik}^j$. Since \begin{align*} e(u)=\frac12\sum_{j=1}^m h(A_j,A_j), \end{align*} metric compatibility of $\nabla^E$ gives, on the neighbourhood where the frame is defined, \begin{align*} e_i e(u)=\sum_{j=1}^m h(\nabla^E_{e_i}A_j,A_j). \end{align*} Using $\nabla^E_{e_i}A_j=B_{ij}+du(\nabla^M_{e_i}e_j)$, this becomes \begin{align*} e_i e(u)=\sum_{j=1}^m h(B_{ij},A_j)+\sum_{j,k=1}^m \Gamma_{ij}^k h(A_k,A_j). \end{align*} The last sum vanishes identically for each $i$, because $\Gamma_{ij}^k$ is skew-symmetric in $j,k$ while $h(A_k,A_j)$ is symmetric in $j,k$. Hence \begin{align*} e_i e(u)=\sum_{j=1}^m h(B_{ij},A_j) \end{align*} as an identity of smooth functions on the frame neighbourhood. Since $\Delta_g f(p)=\sum_{i=1}^m e_i e_i f(p)$ for every smooth real-valued function $f:M\to\mathbb R$ in this normal frame, differentiating this identity and using metric compatibility of $\nabla^E$ gives \begin{align*} \Delta_g e(u)(p)=\sum_{i,j=1}^m h(\nabla^E_{e_i}B_{ij},A_j)(p)+\sum_{i,j=1}^m h(B_{ij},\nabla^E_{e_i}A_j)(p). \end{align*} At $p$, the normal-frame condition gives $\nabla^E_{e_i}A_j(p)=B_{ij}(p)$, so \begin{align*} \Delta_g e(u)(p) = \sum_{i,j=1}^m h(\nabla^E_{e_i}B_{ij},A_j)(p) + \sum_{i,j=1}^m h(B_{ij},B_{ij})(p). \end{align*} The second sum is the pointwise tensor norm $|\nabla du|^2(p)$ in the orthonormal frame. Thus \begin{align*} \Delta_g e(u)(p) = \sum_{j=1}^m h(C_j,A_j)(p)+|\nabla du|^2(p), \end{align*} where the vector $C_j\in E_p$ is defined by \begin{align*} C_j:=\sum_{i=1}^m\nabla^E_{e_i}B_{ij}(p). \end{align*}[/step]

[guided]We now compute the Laplacian of the scalar function $e(u)$ without differentiating the normal-frame identity $B_{ij}(p)=\nabla^E_{e_i}A_j(p)$ as if it held to first order. The reason for introducing $B_{ij}$ is precisely to keep the calculation covariant: $B_{ij}$ is the component of the tensor $\nabla du$, while $\nabla^E_{e_i}A_j$ alone would miss derivatives of the moving frame. For each $j$, define $A_j:=du(e_j)\in\Gamma(E)$, and for each pair $i,j$, define $B_{ij}:=(\nabla du)(e_i,e_j)\in\Gamma(E)$. By the definition of the induced connection on $T^*M\otimes E$, \begin{align*} B_{ij}=\nabla^E_{e_i}(du(e_j))-du(\nabla^M_{e_i}e_j). \end{align*} At the fixed point $p$, the normal-frame condition gives $\nabla^M_{e_i}e_j(p)=0$, so \begin{align*} B_{ij}(p)=\nabla^E_{e_i}A_j(p). \end{align*} This equality is used only at the point $p$, not differentiated as an identity of sections. The energy density is \begin{align*} e(u)=\frac12|du|^2=\frac12\sum_{j=1}^m h(A_j,A_j). \end{align*} The subtle point is that $B_{ij}(p)=\nabla^E_{e_i}A_j(p)$ is only a pointwise equality. Away from $p$ there is an extra moving-frame term. To account for it, define smooth real-valued functions $\Gamma_{ij}^k$ by \begin{align*} \nabla^M_{e_i}e_j=\sum_{k=1}^m \Gamma_{ij}^k e_k. \end{align*} Because the frame is $g$-orthonormal and $\nabla^M$ is metric-compatible, \begin{align*} 0=e_i g(e_j,e_k)=g(\nabla^M_{e_i}e_j,e_k)+g(e_j,\nabla^M_{e_i}e_k)=\Gamma_{ij}^k+\Gamma_{ik}^j. \end{align*} Thus $\Gamma_{ij}^k=-\Gamma_{ik}^j$: for fixed $i$, the connection coefficients are skew-symmetric in the frame indices $j,k$. Metric compatibility of $\nabla^E$ gives \begin{align*} e_i e(u)=\sum_{j=1}^m h(\nabla^E_{e_i}A_j,A_j). \end{align*} Using the identity $\nabla^E_{e_i}A_j=B_{ij}+du(\nabla^M_{e_i}e_j)$, we obtain \begin{align*} e_i e(u)=\sum_{j=1}^m h(B_{ij}+du(\nabla^M_{e_i}e_j),A_j). \end{align*} Substituting $\nabla^M_{e_i}e_j=\sum_{k=1}^m\Gamma_{ij}^k e_k$ gives \begin{align*} e_i e(u)=\sum_{j=1}^m h(B_{ij},A_j)+\sum_{j,k=1}^m \Gamma_{ij}^k h(A_k,A_j). \end{align*} Why does the second sum disappear? The coefficient $\Gamma_{ij}^k$ changes sign when $j$ and $k$ are interchanged, while $h(A_k,A_j)=h(A_j,A_k)$ is symmetric. Pairing the $(j,k)$ and $(k,j)$ terms therefore cancels the whole sum, including the diagonal terms because $\Gamma_{ij}^j=0$. Hence, as an identity of functions on the frame neighbourhood, \begin{align*} e_i e(u)=\sum_{j=1}^m h(B_{ij},A_j). \end{align*} This is the identity we may safely differentiate. At $p$, the scalar Laplacian in a normal orthonormal frame is \begin{align*} \Delta_g f(p)=\sum_{i=1}^m e_i e_i f(p) \end{align*} for every smooth real-valued function $f:M\to\mathbb R$. Differentiating the preceding identity gives \begin{align*} \Delta_g e(u)(p)=\sum_{i,j=1}^m e_i h(B_{ij},A_j)(p). \end{align*} Using metric compatibility of $\nabla^E$ on each summand, this becomes \begin{align*} \Delta_g e(u)(p)=\sum_{i,j=1}^m h(\nabla^E_{e_i}B_{ij},A_j)(p)+\sum_{i,j=1}^m h(B_{ij},\nabla^E_{e_i}A_j)(p). \end{align*} Only now do we use the normal-frame equality at the point $p$: since $\nabla^M_{e_i}e_j(p)=0$, we have $\nabla^E_{e_i}A_j(p)=B_{ij}(p)$. Therefore \begin{align*} \Delta_g e(u)(p)=\sum_{j=1}^m h\left(\sum_{i=1}^m\nabla^E_{e_i}B_{ij},A_j\right)(p)+\sum_{i,j=1}^m h(B_{ij},B_{ij})(p). \end{align*} The last sum is $|\nabla du|^2(p)$ by the definition of the pointwise tensor norm in the orthonormal frame. Therefore \begin{align*} \Delta_g e(u)(p)=\sum_{j=1}^m h(C_j,A_j)(p)+|\nabla du|^2(p), \end{align*} where $C_j:=\sum_{i=1}^m\nabla^E_{e_i}B_{ij}(p)\in E_p$. The remaining task is to identify $C_j$ by commuting the covariant derivatives of the tensor $du$.[/guided]

[step:Commute the covariant derivatives of $du$] At $p$, write \begin{align*} C_j = \sum_{i=1}^m\nabla^E_{e_i}B_{ij}(p). \end{align*} We first justify the symmetry of $\nabla du$. For smooth vector fields $X,Y\in\mathfrak X(M)$, the torsion-free property of $\nabla^M$ gives $[X,Y]=\nabla^M_XY-\nabla^M_YX$, while the torsion-free property of $\nabla^N$ gives \begin{align*} \nabla^E_X(du(Y))-\nabla^E_Y(du(X))=du([X,Y]). \end{align*} Substituting the definition of $\nabla du$ therefore gives \begin{align*} (\nabla du)(X,Y)=(\nabla du)(Y,X). \end{align*} Thus $B_{ij}=B_{ji}$, and hence \begin{align*} C_j = \sum_{i=1}^m\nabla^E_{e_i}B_{ji}(p). \end{align*} Now compare $\nabla^E_{e_i}B_{ji}$ with $\nabla^E_{e_j}B_{ii}$. For a section $\alpha\in\Gamma(T^*M\otimes E)$, the curvature of the [tensor-product connection](/page/Connection) on $T^*M\otimes E$ is \begin{align*} ((R^{(T^*M)\otimes E}(X,Y)\alpha)(Z))=R^E(X,Y)(\alpha(Z))-\alpha(R^M(X,Y)Z). \end{align*} Indeed, the first term is the curvature action on the $E$ factor, and the second term is the dual curvature action on the covector factor. Applying this identity to $\alpha=du$, $X=e_i$, $Y=e_j$, and $Z=e_i$, we evaluate the left-hand side at $p$. The terms coming from differentiating the tensor slot vanish there because $\nabla^M_{e_i}e_k(p)=0$ for every $k$, and the Lie-bracket term in the curvature commutator vanishes because $[e_i,e_j](p)=0$. Hence \begin{align*} \nabla^E_{e_i}B_{ji}-\nabla^E_{e_j}B_{ii}=R^N(A_i,A_j)A_i-du(R^M(e_i,e_j)e_i). \end{align*} Therefore \begin{align*} C_j=\sum_{i=1}^m\nabla^E_{e_j}B_{ii}(p)+\sum_{i=1}^m R^N(A_i,A_j)A_i(p)-\sum_{i=1}^m du(R^M(e_i,e_j)e_i)(p). \end{align*} Since $\nabla^M_{e_j}e_i(p)=0$, the first sum is $\nabla^E_{e_j}\tau(u)(p)$. Therefore \begin{align*} C_j=\nabla^E_{e_j}\tau(u)(p)+\sum_{i=1}^m R^N(A_i,A_j)A_i(p)-du\left(\sum_{i=1}^m R^M(e_i,e_j)e_i\right)(p). \end{align*} Since $u$ is harmonic, $\tau(u)=0$, so $\nabla^E_{e_j}\tau(u)(p)=0$. With the stated curvature convention, \begin{align*} \sum_{i=1}^m R^M(e_i,e_j)e_i=-\operatorname{Ric}^M(e_j). \end{align*} Thus \begin{align*} C_j = du(\operatorname{Ric}^M(e_j))(p) + \sum_{i=1}^m R^N(A_i,A_j)A_i(p). \end{align*} [/step]

[step:Convert the curvature contraction to the stated sign] Substituting the expression for $C_j$ into the second-derivative identity gives \begin{align*} \Delta_g e(u)(p)=|\nabla du|^2(p)+\sum_{j=1}^m h(du(\operatorname{Ric}^M(e_j)),A_j)(p)+\sum_{i,j=1}^m h(R^N(A_i,A_j)A_i,A_j)(p). \end{align*} Using metric compatibility of the [Levi-Civita connection](/page/Levi-Civita%20Connection) on $N$, the curvature endomorphism $R^N(X,Y):T_{u(p)}N\to T_{u(p)}N$ is skew-adjoint with respect to $h$, so \begin{align*} h(R^N(X,Y)Z,W)=-h(R^N(X,Y)W,Z). \end{align*} We apply this identity with $X=A_i$, $Y=A_j$, $Z=A_i$, and $W=A_j$. This gives \begin{align*} h(R^N(A_i,A_j)A_i,A_j) = -h(R^N(A_i,A_j)A_j,A_i). \end{align*} Since $A_i=du(e_i)$ and $A_j=du(e_j)$, this becomes \begin{align*} \Delta_g e(u)(p)=|\nabla du|^2(p)+\sum_{j=1}^m h(du(\operatorname{Ric}^M(e_j)),du(e_j))(p)-\sum_{i,j=1}^m h(R^N(du(e_i),du(e_j))du(e_j),du(e_i))(p). \end{align*} This is the desired identity at $p$. [/step]

[step:Use tensoriality to obtain the global formula] The point $p\in M$ was arbitrary. Each term in the identity is independent of the chosen normal frame: $\Delta_g e(u)$ is a scalar function, $|\nabla du|^2$ is the pointwise norm of a tensor, the Ricci term is the trace of the tensor pairing between $du\circ\operatorname{Ric}^M$ and $du$, and the target-curvature term is the full orthonormal-frame contraction of the curvature tensor of $N$ against $du$. Hence the identity proved at $p$ holds at every point of $M$ and in every local $g$-orthonormal frame. This proves the Bochner formula for harmonic maps. [/step]