special relativity - Homogeneity of space implies linearity of Lorentz transformations

In the derivation of Lorentz transformations, the Wikipedia article mentions a couple of times that the linearity comes from the homogeneity of space. I am looking for a thorough explanation on this.

Answer

I claim that if the transformation between frames is homogeneous and differentiable, then it is affine (homogeneity is not strictly speaking sufficient for linearity since the full transformation between frames is actually a Poincare transformation which is affine, not linear)

For a mathematically precise proof, we need a mathematical definition of homogeneity. To arrive at such a definition, we note that the basic idea is that we can pick our origin wherever we choose, and it won't "affect the measurement results of different observers." In particular, this applies to measurements of the differences between the coordinates of two events. Let's put this in mathematical terms.

Let $L:\mathbb R^4\to\mathbb R^4$ be a transformation. We say that $L$ is homogeneous provided \begin{align} L(x+\epsilon) - L(y+\epsilon) = L(x) - L(y) \end{align} for all $\epsilon\in\mathbb R^4$ and for all $x,y\in\mathbb R^4$.

We can now precisely state and prove the desired result. Note that I also assume that the transformation is differentiable. I haven't thought very hard about if or how one can weaken and/or motivate this assumption.

Proposition. If $L$ is homogeneous and differentiable, then $L$ is affine.

Proof. The definition of homogeneity implies that , \begin{align} L(x+\epsilon)-L(x) = L(y+\epsilon) - L(y) \tag{1} \end{align} for all $\epsilon, x, y$. Now we note that the derivative $L'(x)$ of $L$ at a point $x$ is a linear operator on $\mathbb R^4$ that satisfies \begin{align} L(x+\epsilon) - L(x) = L'(x)\cdot \epsilon +o(|\epsilon|) \end{align} and plugging this into $(1)$ gives \begin{align} (L'(x)-L'(y))\cdot\epsilon = o(|\epsilon|) \end{align} for all $\epsilon,x,y$, where $|\cdot|$ is the euclidean norm. Now simply choose $\epsilon = |\epsilon|e_j$ with $|\epsilon|\neq 0$ where $e_0, \dots e_3$ are the standard, ordered basis elements on $\mathbb R^4$, multiply both sides on the left by $(e_i)^t$ where $^t$ means transpose, divide both sides by $|\epsilon|$, and take the limit $|\epsilon|\to 0$ to show that all matrix elements of $L'(x)-L'(y)$ are zero. If follows immediately that \begin{align} L'(x) = L'(y) \end{align} In other words, the derivative of $L$ is constant. It follows pretty much immediately that $L$ is affine, namely that there exists a linear operator $\Lambda$ on $\mathbb R^4$, and a vector $a\in\mathbb R^4$ such that \begin{align} L(x) = \Lambda x + a \end{align} for all $x\in\mathbb R^4$. $\blacksquare$