[proofplan] All three parts rest on the integral form of the [Bochner–Weitzenböck Formula](/theorems/2759): for a harmonic $1$-form $\alpha$ on a closed oriented Riemannian manifold $(M, g)$, \begin{align*} 0 = \|\nabla\alpha\|_{L^2}^2 + \int_M \operatorname{Ric}(\alpha, \alpha)\,\omega_g. \end{align*} For Part (1), strict positivity of $\operatorname{Ric}$ together with $\alpha \neq 0$ makes the right side strictly positive, contradicting the left, so $\mathcal{H}^1(M) = 0$ and hence $H^1_{\mathrm{dR}}(M) = 0$ via the [Hodge Decomposition](/theorems/2745). For Part (2), non-negativity of $\operatorname{Ric}$ forces both $\nabla\alpha = 0$ and $\operatorname{Ric}(\alpha, \alpha) \equiv 0$; a parallel section of $T^*M$ is determined by its value at any point, so $\dim \mathcal{H}^1(M) \leq \dim T^*_{x_0} M = n$, hence $b^1(M) \leq n$. For Part (3), $b^1(M) = n$ produces $n$ linearly independent parallel $1$-forms, and dualising gives a parallel global frame; computing the curvature tensor in this frame using parallelism shows $R \equiv 0$, so $g$ is flat. [/proofplan]

[step:Set up the integral Bochner identity for harmonic $1$-forms] Let $(M, g)$ be a compact connected oriented Riemannian manifold of dimension $n$ with Riemannian volume form $\omega_g$. By the [Hodge Decomposition](/theorems/2745) and the [Harmonic Representatives](/theorems/2747) theorem, every de Rham cohomology class in $H^1_{\mathrm{dR}}(M)$ has a unique harmonic representative, so \begin{align*} H^1_{\mathrm{dR}}(M) \cong \mathcal{H}^1(M) := \{\alpha \in \Omega^1(M) : \Delta\alpha = 0\}. \end{align*} This identification is bijective and linear, so $b^1(M) = \dim H^1_{\mathrm{dR}}(M) = \dim \mathcal{H}^1(M)$. By the [Bochner–Weitzenböck Formula](/theorems/2759), for any $\alpha \in \Omega^1(M)$, \begin{align*} \Delta\alpha = \nabla^*\nabla\alpha + \operatorname{Ric}(\alpha). \end{align*} Pair both sides in the global $L^2$ inner product against $\alpha$: \begin{align*} \langle\langle \Delta\alpha, \alpha\rangle\rangle = \langle\langle \nabla^*\nabla\alpha, \alpha\rangle\rangle + \int_M \operatorname{Ric}(\alpha)(X_\alpha)\,\omega_g. \end{align*} By the definition of the formal adjoint $\nabla^*$ on the closed manifold $M$, $\langle\langle \nabla^*\nabla\alpha, \alpha\rangle\rangle = \langle\langle \nabla\alpha, \nabla\alpha\rangle\rangle = \|\nabla\alpha\|_{L^2}^2 \geq 0$ (this is the [Adjoint Formula for Covariant Derivative](/theorems/2758) integrated against $\alpha$). And $\operatorname{Ric}(\alpha)(X_\alpha) = \operatorname{Ric}(X_\alpha, X_\alpha) = \operatorname{Ric}(\alpha, \alpha)$ pointwise, where $X_\alpha$ is the metric dual of $\alpha$ and $\operatorname{Ric}(\alpha, \alpha) := \operatorname{Ric}(X_\alpha, X_\alpha)$. For a harmonic $\alpha$, $\Delta\alpha = 0$ implies $\langle\langle \Delta\alpha, \alpha\rangle\rangle = 0$, so \begin{align*} 0 = \|\nabla\alpha\|_{L^2}^2 + \int_M \operatorname{Ric}(\alpha, \alpha)\,\omega_g. \tag{IB} \end{align*} Identity (IB) is the engine of all three parts. [/step]

[step:Prove Part (1) — strict positivity of Ricci kills $H^1$] Assume $\operatorname{Ric}(g) > 0$ at every point of $M$, meaning that the symmetric bilinear form $\operatorname{Ric}_x$ is positive-definite on $T_x M$ for every $x$. Suppose for contradiction that $\alpha \in \mathcal{H}^1(M)$ with $\alpha \neq 0$. Since $\alpha \neq 0$ as a smooth $1$-form on $M$, the dual vector field $X_\alpha$ is non-zero on a non-empty open set $V \subseteq M$ (the complement of the closed zero set of $\alpha$). On $V$, $\operatorname{Ric}(X_\alpha, X_\alpha) > 0$ pointwise by strict positive definiteness of $\operatorname{Ric}$. By continuity of $\operatorname{Ric}(\alpha, \alpha)$ and openness of $V$, there is a non-empty open set $V' \subseteq V$ on which $\operatorname{Ric}(\alpha, \alpha) \geq c > 0$ for some constant $c$. Therefore \begin{align*} \int_M \operatorname{Ric}(\alpha, \alpha)\,\omega_g \geq \int_{V'} c\,\omega_g = c\,\omega_g(V') > 0, \end{align*} strictly positive because $V'$ has positive volume. Combined with $\|\nabla\alpha\|_{L^2}^2 \geq 0$, the right side of (IB) is strictly positive, contradicting (IB). Hence $\mathcal{H}^1(M) = \{0\}$, and via the harmonic representative isomorphism $H^1_{\mathrm{dR}}(M) = 0$. [/step]

[step:Prove Part (2) — non-negative Ricci bounds $b^1$ by $n$] Assume $\operatorname{Ric}(g) \geq 0$ at every point: the symmetric bilinear form $\operatorname{Ric}_x$ is positive-semidefinite on $T_x M$ for every $x$. Let $\alpha \in \mathcal{H}^1(M)$. Both terms on the right of (IB) are non-negative ($\|\nabla\alpha\|_{L^2}^2 \geq 0$ since it is a squared norm, and $\operatorname{Ric}(\alpha, \alpha) \geq 0$ pointwise by hypothesis), so their sum being zero forces each separately to vanish: \begin{align*} \|\nabla\alpha\|_{L^2}^2 = 0 \quad \text{and} \quad \int_M \operatorname{Ric}(\alpha, \alpha)\,\omega_g = 0. \end{align*} The first identity gives $\nabla\alpha = 0$ as an element of $L^2(\Gamma(T^*M \otimes T^*M))$, and since $\nabla\alpha$ is a continuous section, $\nabla\alpha \equiv 0$ pointwise. So $\alpha$ is **parallel**: $\nabla_Y\alpha = 0$ for every $Y \in \mathfrak{X}(M)$. Now bound $\dim \mathcal{H}^1(M)$. Fix $x_0 \in M$ and consider the linear evaluation map \begin{align*} \operatorname{ev}_{x_0} : \mathcal{H}^1(M) &\to T_{x_0}^* M, \\ \alpha &\mapsto \alpha_{x_0}. \end{align*} This is well-defined because every $\alpha \in \mathcal{H}^1(M)$ is a smooth $1$-form. We claim $\operatorname{ev}_{x_0}$ is **injective**. Suppose $\alpha \in \mathcal{H}^1(M)$ with $\alpha_{x_0} = 0$. Since $\alpha$ is parallel and $M$ is connected, parallel transport determines $\alpha$ globally from its value at $x_0$: for any $x \in M$, choose a smooth path $\gamma: [0, 1] \to M$ with $\gamma(0) = x_0$, $\gamma(1) = x$. Parallel transport $P_\gamma^{0, t}: T_{\gamma(0)}^* M \to T_{\gamma(t)}^* M$ along $\gamma$ in the cotangent bundle satisfies $\nabla_{\dot\gamma}\alpha = 0$ along $\gamma$, so $\alpha_{\gamma(t)} = P_\gamma^{0, t}(\alpha_{x_0})$. With $\alpha_{x_0} = 0$, $P_\gamma^{0, t}$ being linear gives $\alpha_{\gamma(t)} = 0$ for all $t$, in particular $\alpha_x = 0$. Since $x$ was arbitrary, $\alpha \equiv 0$. Hence $\operatorname{ev}_{x_0}$ is injective. Linearity and injectivity of $\operatorname{ev}_{x_0}: \mathcal{H}^1(M) \to T_{x_0}^* M$ give \begin{align*} b^1(M) = \dim \mathcal{H}^1(M) \leq \dim T_{x_0}^* M = n. \end{align*} [/step]

[step:Prove Part (3) — equality $b^1(M) = n$ forces flatness]Assume $\operatorname{Ric}(g) \geq 0$ and $b^1(M) = n$. From Part (2), $\dim \mathcal{H}^1(M) = n$ and the evaluation map $\operatorname{ev}_{x_0}: \mathcal{H}^1(M) \to T_{x_0}^* M$ is injective; since both spaces have dimension $n$, it is a linear isomorphism. Choose a basis $\alpha_1, \ldots, \alpha_n$ of $\mathcal{H}^1(M)$. By Part (2), each $\alpha_i$ is parallel: $\nabla\alpha_i = 0$. By the isomorphism $\operatorname{ev}_{x_0}$, the values $(\alpha_i)_{x_0}$ form a basis of $T_{x_0}^* M$. Parallel transport is a linear isomorphism on each cotangent fibre, so $(\alpha_i)_x$ form a basis of $T_x^* M$ for every $x \in M$ — that is, $\{\alpha_1, \ldots, \alpha_n\}$ is a global parallel coframe. Define the dual frame $X_1, \ldots, X_n \in \mathfrak{X}(M)$ by $\alpha_j(X_i) = \delta_{ij}$ at every point. We show $\nabla X_i = 0$. Differentiate the constant scalar function $\alpha_j(X_i) = \delta_{ij}$ in any direction $Y \in \mathfrak{X}(M)$: \begin{align*} 0 = Y\big(\alpha_j(X_i)\big) = (\nabla_Y \alpha_j)(X_i) + \alpha_j(\nabla_Y X_i) = 0 + \alpha_j(\nabla_Y X_i), \end{align*} since $\nabla\alpha_j = 0$. Hence $\alpha_j(\nabla_Y X_i) = 0$ for every $j$. Because $\{\alpha_j\}$ spans $T^*M$ pointwise, this forces $\nabla_Y X_i = 0$ for every $Y$, i.e., $\nabla X_i = 0$. So $\{X_1, \ldots, X_n\}$ is a global parallel frame on $TM$. Now compute the Riemann curvature on this frame. By [Curvature as Commutator of Covariant Derivatives](/theorems/2703) under the chapter sign convention $R(X, Y) = -[\nabla_X, \nabla_Y] + \nabla_{[X, Y]}$: \begin{align*} R(X_i, X_j) X_k = -[\nabla_{X_i}, \nabla_{X_j}] X_k + \nabla_{[X_i, X_j]} X_k. \end{align*} Each $X_\ell$ is parallel, so $\nabla_{X_i} X_k = 0$, hence $\nabla_{X_j}\nabla_{X_i}X_k = 0$ and similarly $\nabla_{X_i}\nabla_{X_j}X_k = 0$, giving $[\nabla_{X_i}, \nabla_{X_j}] X_k = 0$. For the second term: since $\nabla$ is torsion-free, \begin{align*} [X_i, X_j] = \nabla_{X_i} X_j - \nabla_{X_j} X_i = 0 - 0 = 0, \end{align*} so $\nabla_{[X_i, X_j]} X_k = 0$. Therefore $R(X_i, X_j) X_k = 0$ for all $i, j, k$. At every point $x$, $\{X_i|_x\}$ is a basis of $T_x M$, so by trilinearity $R(Y, Z) W = 0$ for all $Y, Z, W \in T_x M$, hence $R \equiv 0$ on $M$. By definition, this is the statement that $g$ is **flat**.[/step]

[guided]We assume $\operatorname{Ric}(g) \geq 0$ and $b^1(M) = n$. The strategy is to manufacture a global parallel frame of $TM$ from harmonic $1$-forms, and then read off $R \equiv 0$ from parallelism. **Step 1: upgrade the bound to an isomorphism.** From Part (2), the evaluation map \begin{align*} \operatorname{ev}_{x_0} : \mathcal{H}^1(M) &\to T_{x_0}^* M, \\ \alpha &\mapsto \alpha_{x_0} \end{align*} is injective, and we proved $\dim \mathcal{H}^1(M) \leq n$. The hypothesis $b^1(M) = n$ gives $\dim \mathcal{H}^1(M) = b^1(M) = n = \dim T_{x_0}^* M$. A linear injection between finite-dimensional spaces of the same dimension is automatically a linear isomorphism, so $\operatorname{ev}_{x_0}$ is bijective. **Step 2: produce a global parallel coframe.** Choose a basis $\alpha_1, \ldots, \alpha_n$ of $\mathcal{H}^1(M)$. By Part (2) (which used identity (IB) with $\operatorname{Ric} \geq 0$ to force the squared-norm term and the Ricci term to vanish separately), each $\alpha_i$ is parallel: $\nabla\alpha_i = 0$. Why are the values $(\alpha_i)_x$ a basis of $T_x^* M$ at *every* point $x$, not just at $x_0$? Because parallel transport along any path from $x_0$ to $x$ is a linear isomorphism between the cotangent fibres, and parallelism of $\alpha_i$ means \begin{align*} (\alpha_i)_{\gamma(t)} = P_\gamma^{0,t}\big((\alpha_i)_{x_0}\big) \end{align*} for any path $\gamma$ from $x_0$ to $\gamma(t)$. A linear isomorphism sends a basis to a basis, so $(\alpha_i)_x$ form a basis of $T_x^* M$ for every $x$. Thus $\{\alpha_1, \ldots, \alpha_n\}$ is a **global parallel coframe**. **Step 3: dualise to a global parallel frame on $TM$.** Define $X_1, \ldots, X_n \in \mathfrak{X}(M)$ to be the pointwise dual basis: $\alpha_j(X_i) = \delta_{ij}$ at every point. Since $\{(\alpha_j)_x\}$ is a basis of $T^*_x M$ at every $x$, the dual is well-defined and smooth. The non-trivial claim is that $\nabla X_i = 0$. Why should this be true? The natural strategy is to differentiate the constancy relation $\alpha_j(X_i) = \delta_{ij}$ and use the metric compatibility of the dual pairing — that is, the connection's compatibility with the natural pairing $T^*M \otimes TM \to \mathbb{R}$. For any $Y \in \mathfrak{X}(M)$, applying $Y$ to the scalar constant $\delta_{ij}$: \begin{align*} 0 = Y\big(\alpha_j(X_i)\big) = (\nabla_Y \alpha_j)(X_i) + \alpha_j(\nabla_Y X_i). \end{align*} The first term on the right vanishes since $\nabla\alpha_j = 0$, leaving $\alpha_j(\nabla_Y X_i) = 0$ for every $j$. Because $\{\alpha_j\}_{j=1}^n$ spans $T^*_x M$ pointwise — every cotangent vector is a linear combination of the $(\alpha_j)_x$ — the only $v \in T_x M$ on which every $\alpha_j$ vanishes is $v = 0$. Hence $\nabla_Y X_i = 0$ for every $Y$, i.e., $\nabla X_i = 0$. So $\{X_1, \ldots, X_n\}$ is a **global parallel frame** on $TM$. **Step 4: compute $R$ in the parallel frame.** Once we have a parallel frame, the curvature collapses by direct computation. We use the convention $R(X, Y) = -[\nabla_X, \nabla_Y] + \nabla_{[X, Y]}$ from [Curvature as Commutator of Covariant Derivatives](/theorems/2703) (this is the chapter sign convention; the opposite convention would change the sign on the commutator but not the conclusion): \begin{align*} R(X_i, X_j) X_k = -[\nabla_{X_i}, \nabla_{X_j}] X_k + \nabla_{[X_i, X_j]} X_k. \end{align*} We claim both terms on the right vanish. *Commutator term.* Each $X_\ell$ is parallel, so $\nabla_{X_i} X_k = 0$. Differentiating zero gives zero: $\nabla_{X_j}(\nabla_{X_i} X_k) = \nabla_{X_j}(0) = 0$, and symmetrically $\nabla_{X_i}(\nabla_{X_j} X_k) = 0$. Therefore \begin{align*} [\nabla_{X_i}, \nabla_{X_j}] X_k = \nabla_{X_i}\nabla_{X_j} X_k - \nabla_{X_j}\nabla_{X_i} X_k = 0 - 0 = 0. \end{align*} *Lie bracket term.* The Levi-Civita connection $\nabla$ is torsion-free, so $T(X_i, X_j) := \nabla_{X_i} X_j - \nabla_{X_j} X_i - [X_i, X_j] = 0$, which rearranges to \begin{align*} [X_i, X_j] = \nabla_{X_i} X_j - \nabla_{X_j} X_i = 0 - 0 = 0, \end{align*} using parallelism of $X_i$ and $X_j$ in the second equality. Hence $\nabla_{[X_i, X_j]} X_k = \nabla_0 X_k = 0$. Combining the two: \begin{align*} R(X_i, X_j) X_k = -0 + 0 = 0 \quad \text{for all } i, j, k \in \{1, \ldots, n\}. \end{align*} **Step 5: extend to all of $TM$ by trilinearity.** Curvature vanishing on a frame extends to curvature vanishing everywhere. Fix $x \in M$. The triple $(Y, Z, W) \mapsto R_x(Y, Z) W$ is trilinear in $T_x M$, and $\{X_i|_x\}_{i=1}^n$ is a basis of $T_x M$. Writing $Y = \sum_i a_i X_i|_x$, $Z = \sum_j b_j X_j|_x$, $W = \sum_k c_k X_k|_x$ and expanding by trilinearity: \begin{align*} R_x(Y, Z) W = \sum_{i, j, k} a_i b_j c_k\, R_x(X_i, X_j) X_k = 0. \end{align*} Since $x \in M$ and $Y, Z, W \in T_x M$ were arbitrary, $R \equiv 0$ on $M$. By definition, a Riemannian manifold whose Riemann curvature tensor vanishes identically is **flat**. Hence $g$ is flat, completing Part (3).[/guided]