I know the following to be true:
If $(M,g)$ is a compact flat Riemannian manifold whose first Betti number ($= \dim H_{dR}^1(M)$) is equal to the dimension, then it is (isometric to) a flat torus.
In the notes I am following it is stated that this follows from Bieberbach's Theorem (I guess it is the one about flat manifolds being finitely covered by tori), yet I have not found the way to link it to the first Betti number.
Can someone show me how to prove the statement using Bieberbach's Theorem?
I'm not sure of the theorem you quote but here is my understanding (apologies to Mods if this is seen as not an answer to the OP).
Let $(M^n, g)$ be Riemannian manifold and $T^2$ a torus. Let $b_1$ be the first Betti number of our manifold $M$. Using an integral basis of harmonic 1-forms we can define, by integration, the Jacobi map $\pi$; this in turn gives a map onto a torus $T^{b_1}$, on which we put the usual flat metric.
In our case, as all harmonic $1$-forms are of constant length, we are allowed to choose this basis to be pointwise orthonormal. Therefore $\pi$ is a Riemannian submersion and besides we must have $b_1 \leq n$.
Proposition
Let $(M^n, g)$ be a compact Riemannian manifold and $b_1$ be its first Betti number. Then all harmonic $1$-forms are of constant length if and only if $(M^n, g)$ is a locally trivial fibre bundle, with minimal fibres, over a $b_1$-dimensional flat torus and such that $b_1 \leq n$. Moreover under the sequence
$$F^{n-b_1} \hookrightarrow M^n \xrightarrow{\pi}T^{b_1}$$
we see that if $b_1 = n$, then $\pi$ is a Riemannian isometry and hence $(M^n, g)$ is a flat torus.