[proofplan] We verify the formula by checking that the right-hand side acts as the formal $L^2$-adjoint of $\nabla : \Omega^0(T^*M) \to \Omega^1(T^*M)$. Concretely, for any $\alpha \in \Omega^0(T^*M) = \Omega^1(M)$, we show \begin{align*} \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g = -\int_M \big\langle \alpha,\, \sum_i i(e_i)\nabla_{e_i}\beta \big\rangle\,\omega_g. \end{align*} The strategy is a "Leibniz plus divergence theorem" integration by parts: introduce a smooth vector field $W$ such that $\operatorname{div}(W) = \langle\nabla\alpha, \beta\rangle + \langle\alpha, \sum_i i(e_i)\nabla_{e_i}\beta\rangle$, then integrate over $M$ and use the [Divergence Theorem](/theorems/2754) to kill $\int_M \operatorname{div}(W)\,\omega_g$. The vector field is built from the natural pairing of $\alpha$ with $\beta$, and the divergence Leibniz rule plus metric compatibility produce exactly the desired identity. The fundamental lemma of the calculus of variations then converts the integral identity into the pointwise statement. [/proofplan] [step:Set up the formal adjoint to be verified] Let $(M, g)$ be an oriented closed Riemannian manifold of dimension $n$, $\nabla$ the Levi-Civita connection, $\omega_g$ the Riemannian volume form, and $\{e_1, \ldots, e_n\}$ a local orthonormal frame field on an open set $U \subseteq M$ (so $g(e_i, e_j) = \delta_{ij}$ on $U$). Let $\beta \in \Omega^1(T^*M) = \Gamma(T^*M \otimes T^*M)$ be a smooth $(0,2)$-tensor field on $M$. The covariant derivative \begin{align*} \nabla : \Gamma(T^*M) &\to \Gamma(T^*M \otimes T^*M), \\ \alpha &\mapsto \nabla\alpha, \end{align*} extends $\nabla$ from $\Omega^1(M) = \Gamma(T^*M)$ to its first jet bundle, with $(\nabla\alpha)(X, Y) := (\nabla_X \alpha)(Y)$. The codifferential-style operator \begin{align*} \nabla^* : \Gamma(T^*M \otimes T^*M) \to \Gamma(T^*M) \end{align*} is defined as the formal $L^2$-adjoint of $\nabla$ with respect to the metric-induced inner products on $\Gamma(T^*M)$ and $\Gamma(T^*M \otimes T^*M)$ — namely, \begin{align*} \int_M \langle \nabla\alpha, \beta \rangle\,\omega_g = \int_M \langle \alpha, \nabla^*\beta\rangle\,\omega_g \qquad \text{for every } \alpha \in \Omega^1(M). \tag{Adj} \end{align*} Here the inner products are \begin{align*} \langle \alpha, \alpha'\rangle &= g(\alpha, \alpha') = \sum_{i=1}^n \alpha(e_i)\alpha'(e_i) \quad \text{on } \Gamma(T^*M), \\ \langle \beta, \beta'\rangle &= \sum_{i, j = 1}^n \beta(e_i, e_j)\beta'(e_i, e_j) \quad \text{on } \Gamma(T^*M \otimes T^*M), \end{align*} both orthonormal-frame expressions of the metric trace. We will prove identity (Adj) holds with $\nabla^*\beta := -\sum_{i} i(e_i)\nabla_{e_i}\beta$, where the interior product $i(e_i)$ acts in the **first** slot of $\beta$: \begin{align*} \big(i(e_i)\beta\big)(Y) := \beta(e_i, Y). \end{align*} The fundamental lemma will then give pointwise equality. [/step] [step:Expand $\langle \nabla\alpha, \beta\rangle$ in the orthonormal frame] Fix $\alpha \in \Omega^1(M)$. In the orthonormal frame on $U$, the inner product on $\Gamma(T^*M \otimes T^*M)$ between $\nabla\alpha$ and $\beta$ unfolds to \begin{align*} \langle \nabla\alpha, \beta\rangle = \sum_{i, j = 1}^n (\nabla\alpha)(e_i, e_j)\,\beta(e_i, e_j) = \sum_{i, j = 1}^n (\nabla_{e_i}\alpha)(e_j)\,\beta(e_i, e_j). \end{align*} For each fixed $i$, the inner sum over $j$ is the natural pairing of the $1$-form $\nabla_{e_i}\alpha$ with the $1$-form $i(e_i)\beta$ — both viewed as elements of $T^*M$ — via the metric. Concretely, \begin{align*} \sum_{j=1}^n (\nabla_{e_i}\alpha)(e_j)\,\beta(e_i, e_j) = \sum_{j=1}^n (\nabla_{e_i}\alpha)(e_j)\,(i(e_i)\beta)(e_j) = g\big(\nabla_{e_i}\alpha,\, i(e_i)\beta\big). \end{align*} Hence \begin{align*} \langle \nabla\alpha, \beta\rangle = \sum_{i=1}^n g\big(\nabla_{e_i}\alpha,\, i(e_i)\beta\big). \tag{E} \end{align*} [/step] [step:Construct an auxiliary vector field $W$ for integration by parts] For each $i$, define the smooth function on $U$, \begin{align*} h_i := g\big(\alpha,\, i(e_i)\beta\big) = \sum_{j=1}^n \alpha(e_j)\,\beta(e_i, e_j) \in C^\infty(U), \end{align*} and set \begin{align*} W := \sum_{i=1}^n h_i\, e_i \in \mathfrak{X}(U). \end{align*} On a globally framed manifold, $W$ extends to a global smooth vector field on $M$. On a general $M$, the assertion is local — we work in $U$ and pair against $\alpha$ supported in $U$ via a partition of unity; the resulting integral identity is then patched. For clarity, we first complete the computation assuming the frame is global; the partition-of-unity reduction is standard. Differentiate $h_i$ in the direction $e_i$ using metric compatibility of $\nabla$: \begin{align*} e_i(h_i) = e_i\big(g(\alpha, i(e_i)\beta)\big) = g\big(\nabla_{e_i}\alpha,\, i(e_i)\beta\big) + g\big(\alpha,\, \nabla_{e_i}(i(e_i)\beta)\big). \end{align*} The covariant derivative satisfies the Leibniz rule across the interior product: \begin{align*} \nabla_{e_i}(i(e_i)\beta) = i(\nabla_{e_i} e_i)\beta + i(e_i)(\nabla_{e_i}\beta). \end{align*} Substituting, \begin{align*} e_i(h_i) = g\big(\nabla_{e_i}\alpha,\, i(e_i)\beta\big) + g\big(\alpha, i(\nabla_{e_i} e_i)\beta\big) + g\big(\alpha,\, i(e_i)(\nabla_{e_i}\beta)\big). \end{align*} Sum over $i$: \begin{align*} \sum_{i=1}^n e_i(h_i) = \sum_i g(\nabla_{e_i}\alpha, i(e_i)\beta) + \sum_i g\big(\alpha, i(\nabla_{e_i} e_i)\beta\big) + \sum_i g\big(\alpha, i(e_i)(\nabla_{e_i}\beta)\big). \tag{S} \end{align*} [/step] [step:Identify $\sum_i e_i(h_i)$ with $\operatorname{div}(W)$ modulo a frame term that cancels] By the [Divergence Leibniz Rule](/theorems/2752), \begin{align*} \operatorname{div}(h_i\, e_i) = h_i\,\operatorname{div}(e_i) + \langle dh_i, e_i\rangle = h_i\,\operatorname{div}(e_i) + e_i(h_i). \end{align*} Summing, \begin{align*} \operatorname{div}(W) = \sum_i \operatorname{div}(h_i\, e_i) = \sum_i h_i\,\operatorname{div}(e_i) + \sum_i e_i(h_i), \end{align*} so \begin{align*} \sum_i e_i(h_i) = \operatorname{div}(W) - \sum_i h_i\,\operatorname{div}(e_i). \tag{D} \end{align*} To compute $\operatorname{div}(e_i) = \sum_j g(\nabla_{e_j} e_i, e_j)$, use metric compatibility on $g(e_i, e_j) = \delta_{ij}$ differentiated in the direction $e_j$: \begin{align*} 0 = e_j(\delta_{ij}) = e_j(g(e_i, e_j)) = g(\nabla_{e_j} e_i, e_j) + g(e_i, \nabla_{e_j} e_j), \end{align*} giving $g(\nabla_{e_j} e_i, e_j) = -g(e_i, \nabla_{e_j} e_j)$. Summing, \begin{align*} \operatorname{div}(e_i) = -\sum_j g(e_i, \nabla_{e_j} e_j) = -g(e_i, V), \qquad V := \sum_j \nabla_{e_j} e_j \in \mathfrak{X}(U). \end{align*} Therefore \begin{align*} \sum_i h_i\,\operatorname{div}(e_i) = -\sum_i h_i\, g(e_i, V) = -g\Big(\sum_i h_i\, e_i,\, V\Big) = -g(W, V). \end{align*} And the frame term in (S) computes: \begin{align*} \sum_i g\big(\alpha, i(\nabla_{e_i} e_i)\beta\big) = \sum_i \sum_j \alpha(e_j)\,\beta(\nabla_{e_i} e_i, e_j) = \sum_j \alpha(e_j)\,\beta(V, e_j) = g(\alpha, i(V)\beta). \end{align*} On the other hand, using $h_i = g(\alpha, i(e_i)\beta) = \sum_j \alpha(e_j)\beta(e_i, e_j)$ and bilinearity of $i(\cdot)\beta$, \begin{align*} g(W, V) = \sum_i h_i\, g(e_i, V) = \sum_{i, j} \alpha(e_j)\beta(e_i, e_j)\,g(e_i, V) = \sum_j \alpha(e_j)\,\beta(V, e_j) = g(\alpha, i(V)\beta), \end{align*} where the third equality used $\sum_i \beta(e_i, e_j)g(e_i, V) = \beta\big(\sum_i g(e_i, V) e_i,\, e_j\big) = \beta(V, e_j)$ (expanding $V$ in the orthonormal frame). Therefore $\sum_i h_i\,\operatorname{div}(e_i) = -g(W, V) = -g(\alpha, i(V)\beta) = -\sum_i g(\alpha, i(\nabla_{e_i} e_i)\beta)$. Substituting into (D), \begin{align*} \sum_i e_i(h_i) = \operatorname{div}(W) + \sum_i g\big(\alpha, i(\nabla_{e_i} e_i)\beta\big). \end{align*} Combining with (S) and cancelling the frame term $\sum_i g(\alpha, i(\nabla_{e_i} e_i)\beta)$ from both sides: \begin{align*} \operatorname{div}(W) = \sum_i g\big(\nabla_{e_i}\alpha, i(e_i)\beta\big) + \sum_i g\big(\alpha, i(e_i)(\nabla_{e_i}\beta)\big). \end{align*} By (E), the first sum on the right is $\langle \nabla\alpha, \beta\rangle$, so \begin{align*} \operatorname{div}(W) = \langle \nabla\alpha, \beta\rangle + \sum_i g\big(\alpha, i(e_i)(\nabla_{e_i}\beta)\big). \tag{IBP} \end{align*} [/step] [step:Integrate (IBP) and invoke the Divergence Theorem and the fundamental lemma] Integrate (IBP) over $M$ against $\omega_g$. Both sides are smooth on the compact $M$, hence integrable. The hypotheses of the [Divergence Theorem](/theorems/2754) — $M$ oriented, compact, and without boundary — hold by hypothesis, so $\int_M \operatorname{div}(W)\,\omega_g = 0$. Therefore \begin{align*} 0 = \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g + \int_M \sum_{i=1}^n g\big(\alpha, i(e_i)(\nabla_{e_i}\beta)\big)\,\omega_g, \end{align*} i.e., \begin{align*} \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g = -\int_M \Big\langle \alpha,\, \sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta)\Big\rangle\,\omega_g \qquad \text{for every } \alpha \in \Omega^1(M). \tag{T} \end{align*} Comparing (T) with the defining property (Adj) of $\nabla^*\beta$, \begin{align*} \int_M \langle \alpha, \nabla^*\beta\rangle\,\omega_g = -\int_M \Big\langle \alpha,\, \sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta)\Big\rangle\,\omega_g \qquad \text{for every } \alpha \in \Omega^1(M). \end{align*} The smooth $1$-form $\nabla^*\beta + \sum_i i(e_i)\nabla_{e_i}\beta \in \Omega^1(M)$ pairs to zero against every smooth $\alpha \in \Omega^1(M)$ in the global $L^2$ inner product. By the fundamental lemma of the calculus of variations on the closed Riemannian manifold $M$ — applied here in its $1$-form version, which follows by testing against $\alpha = f\,\xi$ for $f \in C^\infty(M)$ and $\xi$ ranging over a local frame — \begin{align*} \nabla^*\beta = -\sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta) \qquad \text{pointwise on } M. \end{align*} This is the asserted formula. [guided] We have arrived at the local pointwise identity (IBP): \begin{align*} \operatorname{div}(W) = \langle \nabla\alpha, \beta\rangle + \sum_{i=1}^n g\big(\alpha, i(e_i)(\nabla_{e_i}\beta)\big). \end{align*} The strategy now is to integrate this identity over $M$ and use the Divergence Theorem to kill the divergence term, leaving an integral identity that matches the defining property (Adj) of $\nabla^*\beta$. The fundamental lemma then upgrades the integral identity into a pointwise formula. Let us execute these moves carefully and verify every hypothesis. **Integrating (IBP) over $M$.** Because $M$ is compact and every term in (IBP) is a smooth function on $M$, all integrals against the volume form $\omega_g$ are finite, so we may integrate (IBP) termwise: \begin{align*} \int_M \operatorname{div}(W)\,\omega_g = \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g + \int_M \sum_{i=1}^n g\big(\alpha,\, i(e_i)(\nabla_{e_i}\beta)\big)\,\omega_g. \end{align*} **Why the left-hand side vanishes.** We invoke the [Divergence Theorem](/theorems/2754) to assert $\int_M \operatorname{div}(W)\,\omega_g = 0$. The Divergence Theorem on a Riemannian manifold without boundary states that for any smooth, compactly supported vector field $W$, \begin{align*} \int_M \operatorname{div}(W)\,\omega_g = \int_{\partial M} g(W, \nu)\,d\mathcal{H}^{n-1} = 0, \end{align*} where the second equality uses $\partial M = \emptyset$. We must check the hypotheses: (i) $M$ is oriented — given in the theorem hypotheses; (ii) $M$ is compact — given (we say "closed", meaning compact without boundary); (iii) $\partial M = \emptyset$ — given; (iv) $W$ is smooth — true since $\alpha$, $\beta$, the frame vectors $e_i$, and the metric $g$ are smooth, so $h_i = g(\alpha, i(e_i)\beta)$ is smooth and $W = \sum_i h_i e_i$ is smooth on $U$ and patches globally via partition of unity; (v) $W$ has compact support — automatic on the compact $M$. All hypotheses verified. Therefore \begin{align*} \int_M \operatorname{div}(W)\,\omega_g = 0. \end{align*} **The resulting integral identity.** Substituting back, the integrated form of (IBP) becomes \begin{align*} 0 = \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g + \int_M \sum_{i=1}^n g\big(\alpha,\, i(e_i)(\nabla_{e_i}\beta)\big)\,\omega_g. \end{align*} Recognizing the inner sum as the inner product on $\Gamma(T^*M)$ pairing $\alpha$ with the $1$-form $\sum_i i(e_i)(\nabla_{e_i}\beta)$, we rearrange: \begin{align*} \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g = -\int_M \Big\langle \alpha,\, \sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta)\Big\rangle\,\omega_g \qquad \text{for every } \alpha \in \Omega^1(M). \tag{T} \end{align*} This is the key integral identity. It says: **on the closed manifold $M$, the operator $-\sum_i i(e_i)\nabla_{e_i}$ is an $L^2$-adjoint of $\nabla$.** The minus sign — which is the only signed object in the identity — comes entirely from the cancellation of $\int_M \operatorname{div}(W)\,\omega_g = 0$ moving across the equality. **Comparison with the defining property (Adj).** The codifferential $\nabla^*$ is defined (in (Adj) of Step 1) by \begin{align*} \int_M \langle \alpha, \nabla^*\beta\rangle\,\omega_g = \int_M \langle \nabla\alpha, \beta\rangle\,\omega_g \qquad \text{for every } \alpha \in \Omega^1(M). \end{align*} Combined with (T), this gives \begin{align*} \int_M \langle \alpha, \nabla^*\beta\rangle\,\omega_g = -\int_M \Big\langle \alpha,\, \sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta)\Big\rangle\,\omega_g \qquad \text{for every } \alpha \in \Omega^1(M). \end{align*} Subtracting the right-hand side from both sides and using bilinearity of the inner product: \begin{align*} \int_M \Big\langle \alpha,\, \nabla^*\beta + \sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta)\Big\rangle\,\omega_g = 0 \qquad \text{for every } \alpha \in \Omega^1(M). \end{align*} **From integral identity to pointwise identity via the fundamental lemma.** Set $\gamma := \nabla^*\beta + \sum_i i(e_i)(\nabla_{e_i}\beta) \in \Omega^1(M)$. We have shown $\int_M \langle \alpha, \gamma\rangle\,\omega_g = 0$ for every $\alpha \in \Omega^1(M)$, and we want to conclude $\gamma = 0$ pointwise. This is the fundamental lemma of the calculus of variations applied to $1$-forms. Let us verify it. Suppose $\gamma(p) \neq 0$ at some $p \in M$. Choose a coordinate neighborhood $V \ni p$ on which $\gamma$ is expressed in a local frame $\{\xi_1, \ldots, \xi_n\}$ for $T^*M$, and pick the component $\gamma_k := \gamma(\xi_k^\sharp) \neq 0$ at $p$. By continuity, $\gamma_k$ has constant sign on a smaller neighborhood $V' \ni p$. Choose a non-negative bump function $f \in C^\infty(M)$ supported in $V'$ with $f(p) > 0$, and set $\alpha := f\,\xi_k$. Then \begin{align*} \int_M \langle \alpha, \gamma\rangle\,\omega_g = \int_{V'} f(x)\,\gamma_k(x)\,\omega_g(x) \neq 0, \end{align*} since the integrand has constant sign and is strictly nonzero at $p$ — a contradiction. Therefore $\gamma \equiv 0$, i.e., \begin{align*} \nabla^*\beta = -\sum_{i=1}^n i(e_i)(\nabla_{e_i}\beta) \qquad \text{pointwise on } M. \end{align*} **Why this is independent of the chosen frame.** A subtle point: we computed everything in a *local* orthonormal frame on $U$. Why is the conclusion a *global* identity on $M$? Because the right-hand side $\sum_i i(e_i)\nabla_{e_i}\beta$, computed in any orthonormal frame, is the metric trace of $\nabla\beta$ in the first two slots — this is a tensorial, frame-independent expression. So the local formula in any orthonormal frame defines the same global tensor on $M$, and the identity holds globally. **Why this formula matters.** This is the foundation of the **Bochner technique**: the operator $\nabla^*\nabla$ — the **connection Laplacian** — is the composition of $\nabla : \Omega^1 \to \Gamma(T^*M \otimes T^*M)$ with $\nabla^*$ in the reverse direction, and the formula we just proved gives the explicit local expression $\nabla^*\nabla\alpha = -\sum_i (\nabla_{e_i}\nabla_{e_i}\alpha - \nabla_{\nabla_{e_i} e_i}\alpha)$ in any orthonormal frame — the standard "trace of Hessian minus frame correction". The Bochner–Weitzenböck identity $\Delta = \nabla^*\nabla + \mathcal{R}$ (Hodge Laplacian = connection Laplacian + curvature term) takes off from here. [/guided] [/step]