# Generalization of Bieberbach's second theorem

Let $$F_0$$ and $$F_1$$ be compact flat manifolds of dimensions $$k$$ and $$m$$, respectively, where $$k \geq m$$. Suppose $$f : \pi_1(F_0) \to \pi_1(F_1)$$ is a surjective homomorphism. Consider the covering maps $$\mathbb{R}^k \overset{\phi_0}{\longrightarrow} \mathbb{T}^k \overset{\phi_1}{\longrightarrow} F_0$$ and $$\mathbb{R}^m \overset{\psi_0}{\longrightarrow} \mathbb{T}^m \overset{\psi_1}{\longrightarrow} F_1$$. Without loss of generality, suppose that $$\phi_0$$ and $$\psi_0$$ are the quotient maps by the standard integer lattices $$\mathbb{Z}^k$$ and $$\mathbb{Z}^m$$. Denote by $$\{ e_1,\ldots,e_k \}$$ and $$\{ e^1,\ldots,e^m \}$$ the standard bases for $$\mathbb{R}^k$$ and $$\mathbb{R}^m$$, respectively, and by $$\{ [s_1],\ldots,[s_k] \}$$ and $$\{ [s^1],\ldots,[s^m] \}$$ the standard generators for $$\pi_1(\mathbb{T}^k)$$ and $$\pi_1(\mathbb{T}^m)$$.

There exist integers $$c_{ij}$$ such that $$f([s_i]) = \sum_j c_{ij} [s^j]$$. Let $$T : \mathbb{R}^k \to \mathbb{R}^m$$ be the linear map defined by $$T(e_i) = \sum_j c_{ij} e^j$$. It's clear that $$T$$ descends to a map $$\tilde{T} : \mathbb{T}^k \to \mathbb{T}^m$$ such that the diagram $$\require{AMScd}$$ $$\begin{CD} \mathbb{R}^k @>T>> \mathbb{R}^m\\ @V \phi_0 V V @VV \psi_0 V\\ \mathbb{T}^k @>>\tilde{T}> \mathbb{T}^m \end{CD}$$ commutes. My question is whether there exists a map $$\hat{T} : F_0 \to F_1$$ such that the diagram $$\begin{CD} \mathbb{R}^k @>T>> \mathbb{R}^m\\ @V \phi_0 V V @VV \psi_0 V\\ \mathbb{T}^k @>>\tilde{T}> \mathbb{T}^m\\ @V \phi_1 V V @VV \psi_1 V\\ F_0 @>>\hat{T}> F_1 \end{CD}$$ commutes.

What I know: Bieberbach's second theorem states that, if $$G$$ and $$H$$ are Bieberbach subgroups of the isometry group $$\mathscr{I}(\mathbb{R}^k)$$, then for any isomorphism $$X : G \to H$$ there exists an affine bijection $$S : \mathbb{R}^k \to \mathbb{R}^k$$ such that $$X(\beta) = S \circ \beta \circ S^{-1}$$ for all $$\beta \in G$$. It's not difficult to show that $$k = m$$ if and only if $$f$$ is an isomorphism. In this case, it follows that $$S = T$$ and, therefore, $$T$$ descends to the desired map $$\hat{T}$$.

So, more specifically, my question is whether there is a generalization of Bieberbach's second theorem that implies $$T$$ descends when $$m < k$$.

Charles Frohman has given a simple argument for this: Since the universal cover of $$F_1$$ is contractible, $$f$$ is the induced homomorphism of a map $$F : F_0 \to F_1$$. (See also Realizing homomorphisms between fundamental groups.) As suggested in my edit to the original question, the result now follows, as it's a theorem of Eells–Sampson that $$F$$ is homotopic to a totally geodesic map.
There is also perhaps a more "hands-on" approach: each of the maps $$\phi_1\colon T^k\to F_0$$ and $$\psi_1\colon T^m\to F_1$$ is precisely the projection map corresponding to the quotient by the (free) action of the holonomy group of the base flat manifold on the covering torus (for details, see e.g. Sec 2 in https://arxiv.org/pdf/1705.08431.pdf or Charlap's book, referenced in the former). Thus, a map $$T^k \to T^m$$ descends to a map $$F_0\to F_1$$ if and only if it is equivariant with respect to these actions. This is the case in your example, by construction, using the short exact sequences for $$\pi_1(F_i)$$.
• At the risk of sounding incompetent: I've spent a good deal of time trying to write an elementary proof of the equivariance of $\tilde{T}$, but I don't see it. – James Dibble Nov 14 '18 at 3:53