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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?

Alessio Di Lorenzo
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1 Answers1

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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.

  • Thank you for your answer! Unfortunately, I already knew (more or less) how to prove it using the Jacobi map $\pi$. Anyway, I think that this answer is useful anyway; can you specify, though, what $F$ is? In the notes I am following it is stated that this theorem follows directy from Bieberbach's Theorem about the structure of compact flat manifolds. I will edit so that it's clearer. – Alessio Di Lorenzo Sep 13 '17 at 13:52
  • @AlessioDiLorenzo Ah I see, perhaps I should have stated clearer that $F$ is a fibration.It allows for the following long exact sequence on the homotopy groups $$\ldots \rightarrow π_n(F^2) \rightarrow \pi_n(M^n) \rightarrow \pi_n(T^{n−2}) \rightarrow \pi_{n−1}(F^2) \rightarrow \pi_{n−1}(M^n) \rightarrow \ldots$$ –  Sep 13 '17 at 14:06