I was reading a math paper and got stuck on a fact that was considered to be elementary by the author.
The title is pretty much all about my question. More detailed question is as the following:
Let $A$ be a finite dimensional affine space, and let $g$ be an affine transformation on $A$ which acts freely, i.e. it does not fix any element of $A$. Then the linear part of $A$, which is a linear transformation on the vector space associated to $A$, has 1 as an eigenvalue.
I tried to verify this statement, but I do not have any idea to start with. I will appreciate any helps or hints.
Thanks in advance.
$\newcommand\mysetminus{\raise.3ex\smallsetminus}$ I don't know if a replay to your post is still useful to you, as it has been a long time. But, in doubt, I post it anyway.
A simple proof is in https://it.wikipedia.org/wiki/Trasformazione_affine#Punti_fissi (in italian, sorry). This proof use coordinate systems, matrices,... I want to propose here a proof that does not make use of coordinates, let us say synthetic; a little less easy, to be honest.
Let $\mathcal A(E)$ be an $n$-dimensional affine space over the vector space $E$, and
$$ g\;:\; \mathcal A(E) \longrightarrow \mathcal A(E) $$
an affine function without fixed points. Suppose by absurd that the linear part $dg$ of $g$ does not admit the eigenvalue $1$. This means that $dg(x) \neq x$ for all $x\in E\mysetminus\{0\}$, i.e.
$$ (dg-1_E)(x) \neq 0\qquad \forall x\in E\mysetminus\{0\}, $$
where $1_E$ is the identity function of $E$. It follows that the kernel of $dg-1_E$ is $\{0\}$, so the linear function $dg-1_E\;:\;E \to E$ is invertible [remember the rank-nullity Theorem].
Let us now consider the function
$$ g-1_{\mathcal A(E)}\;:\;\mathcal A(E) \longrightarrow E^{\text{aff}}(E) $$ $$ (g-1_{\mathcal A(E)})(P) := g(P)-1_{\mathcal A(E)}(P) = g(P)-P $$
where $E^{\text{aff}}$ is the vector space $E$ equipped with the canonical structure of affine space (the points are the vectors,...) over the vector space $E$, and $1_{\mathcal A(E)}$ is the identity function of $\mathcal A(E)$. It is an affine function because:
\begin{align} \big(g-1_{\mathcal A(E)}\big)(&P)-\big(g-1_{\mathcal A(E)}\big)(Q) = \big(g(P)-P\big)-\big(g(Q)-Q\big)\\[2ex] &= g(P)-g(Q)-(P-Q)\\[2ex] &= dg(P-Q)-1_E(P-Q)\\[2ex] &= (dg-1_E)(P-Q), \end{align}
and this also proves that
$$ d(g-1_{\mathcal A(E)}) = dg-1_E $$
and therefore that $g-1_{\mathcal A(E)}$ is invertible, being invertible its linear part. Taking then
$$ Q := (g-1_{\mathcal A(E)})^{-1}(\vec 0), $$
we deduce that
$$ g(Q)-Q = (g-1_{\mathcal A(E)})(Q) = \big(g-1_{\mathcal A(E)}\big)\big((g-1_{\mathcal A(E)})^{-1}(\vec 0)\big) = \vec 0, $$
so $g(Q)=Q$, a contradiction.