= Solution
Use the nonnegative Hodge-Laplacian convention from Question 4. The <Bochner-Weitzenbock formula for one-forms> is
$$
\boxed{\Delta\alpha=\nabla^*\nabla\alpha+\operatorname{Ric}\cdot\alpha.}
$$
Here $\nabla$ is the <Levi-Civita connection> induced on the <cotangent bundle>, and the <rough Laplacian> in a local orthonormal frame is
$$
\nabla^*\nabla\alpha=-\sum_i\bigl(\nabla_{e_i}\nabla_{e_i}\alpha-\nabla_{\nabla_{e_i}e_i}\alpha\bigr).
$$
It is the composition of <covariant derivative> with its <formal adjoint>. The Ricci endomorphism is defined by $g(\operatorname{Ric}^{\sharp}X,Y)=\operatorname{Ric}(X,Y)$ and acts on one-forms by $(\operatorname{Ric}\cdot\alpha)(X)=\alpha(\operatorname{Ric}^{\sharp}X)$. The <musical isomorphism> defines $\alpha^{\sharp}$ by $g(\alpha^{\sharp},X)=\alpha(X)$. The associated scalar formula is
$$
\tfrac12\Delta|\alpha|^2=\langle\Delta\alpha,\alpha\rangle-|\nabla\alpha|^2-\operatorname{Ric}(\alpha^{\sharp},\alpha^{\sharp}).
$$
These signs make the integrated rough-Laplacian term $\|\nabla\alpha\|^2$ on a <closed manifold>.
For a connected manifold, the full <Riemannian holonomy group> at $p$ is the subgroup of $O(T_pM,g_p)$ consisting of parallel transports around all piecewise smooth loops based at $p$. Its natural action on $T_pM$ is the <holonomy representation>; it induces actions on cotangent spaces and all tensor spaces. The <holonomy representation> is an <irreducible representation> when it has no nonzero proper <invariant subspace>.
The <fundamental principle of Riemannian holonomy> identifies parallel tensor fields with tensors at $p$ fixed by full holonomy. A parallel field returns to its value under every loop. Conversely, transport a fixed tensor along a path from $p$ to each point. Any two paths differ by a loop, so the result is independent of the path; local smooth <parallel transport> yields a smooth parallel field. Evaluation and construction are inverse. Full holonomy, not merely the contractible-loop subgroup, is required for this global correspondence.
On a compact manifold without boundary and with nonnegative <Ricci curvature>, a <harmonic one-form> satisfies the integrated Bochner identity
$$
0=\int_M\langle\Delta\alpha,\alpha\rangle\,d\mathrm{vol}_g
=\int_M\bigl(|\nabla\alpha|^2+\operatorname{Ric}(\alpha^{\sharp},\alpha^{\sharp})\bigr)\,d\mathrm{vol}_g.
$$
Both terms are nonnegative; therefore $\nabla\alpha=0$. This proves that <harmonic one-forms are parallel under nonnegative Ricci curvature>. Integration uses the Riemannian volume density and does not require an <orientation>.
If $n=\dim M\ge2$, a nonzero such form would give a nonzero fixed tangent vector via metric duality and hence a proper invariant line, contradicting irreducibility. This is precisely the <irreducible holonomy in dimension at least two has no parallel one-form> criterion. Since the question permits harmonic representatives of all classes, it gives $b_1(M)=0$, where $b_1$ is the first <Betti number>.
The covering $\pi:X\times T^k\to M$ is finite: its fibre over a point is a closed discrete subset of the compact total space and hence finite. Use the explicitly permitted equality of the Betti numbers of $M$ and its finite cover, in particular $b_1(X\times T^k)=b_1(M)=0$. This equality is a permission of this question, not a general theorem about finite covers.
The $k$ angular one-forms on $T^k$, pulled back to $X\times T^k$, represent linearly independent de Rham classes: their periods on the $k$ coordinate circles are the standard basis vectors. <exact differential forms> have zero periods. Thus $b_1(X\times T^k)\ge k$, forcing $k=0$. Now $\pi:X\to M$ is a finite <universal covering map> because $X$ is simply connected. The <fundamental group> acts freely and transitively on a fibre, or equivalently loop-lifting identifies its elements with the finite set of possible endpoints upstairs. Consequently the intended conclusion is
$$
\boxed{n\ge2\quad\Longrightarrow\quad k=0,\qquad |\pi_1(M)|=\deg\pi<\infty.}
$$
The printed statement needs the dimension qualification under the usual definition of irreducibility. Take the standard circle $M=S^1$, with $X$ a point, $k=1$, and the identity covering $X\times T^1\to S^1$. Its <Ricci curvature> is zero. <parallel transport> fixes its global unit tangent, so its holonomy representation is the trivial representation on a one-dimensional real vector space, which is irreducible. The allowed Betti-number equality holds for this identity cover, yet
$$
\boxed{\pi_1(S^1)=\mathbb Z\text{ is infinite}.}
$$
Thus there is no proof of the unqualified literal assertion in dimension one. If “irreducible holonomy” is instead intended to exclude the one-dimensional trivial representation, the preceding intended proof applies. A connected zero-dimensional manifold is a point and has trivial <fundamental group>.
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