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. 
 A: This answer is essentially the same as JoshPhysics's Answer but with the following points:


*

*We use a more "off-the-shelf" mathematical result to get rid of the differentiability assumption JoshPhysics used in his answer and instead we simply need to assume that the Lorentz transformation is only continuous;

*We show that the continuity assumption is a necessary and the minimum necessary assumption further to the OP's homogeneity assumption i.e. the OP's assertion that linearity comes from the homogeneity of spacetime alone is wrong.



JoshPhysics's equation (1) implies:
$$L(X+Y)-L(Y) = L(X)-L(0);\;\forall X,\,Y\in \mathbb{R}^{1+3}\tag{1}$$
Now we define $h:\mathbb{R}^{1+3}\to\mathbb{R}^{1+3}$ by $h(Z)=L(Z)-L(0)$; then it follows from (1) alone that:
$$h(X+Y)=h(X)+h(Y);\quad\forall\,X,\,Y\in\mathbb{R}^{1+3}\tag{2}$$
But this is the famous Cauchy functional equation generalized to $3+1$ dimensions. For one, real dimension, the only continuous solution is $h(X)\propto X$; there are other solutions, but they are everywhere discontinuous, as shown in:
E. Hewitt & K. R. Stromberg, "Real and Abstract Analysis" (Graduate Texts in Mathematics), Springer-Verlag, Berlin, 1965. Chapter 1, section 5 
It is easy to broaden the Hewitt-Stromberg argument to any number of dimensions, so that, given an assumption of continuity of $L:\mathbb{R}^{1+3}\to\mathbb{R}^{1+3}$, we must have:
$$L(X) = \Lambda\,X + \Delta\tag{3}$$
where $\Lambda$ is a linear operator - a $4\times4$ matrix and $\Delta$ a spacetime offset.
Note that we must invoke the continuity assumption; otherwise, following the reasoning in Hewitt and Stromberg, our $h$ function could be one of the everywhere discontinuous Cauchy equation solutions, and we could then, by reversing the step from my equation (1) to (2), construct everywhere discontinuous, nonlinear functions that fulfill JoshPhysics's homogeneity postulate. So unless we require exactly continuity, we shan't "choose" the right solution of the Cauchy equation. Thus continuity of transformation as well as homogeneity are the minimum assumptions needed to imply linearity.
A: 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 $x\in\mathbb R^4$. $\blacksquare$
A: Intuitively, this is fairly easy to understand. This isn't a proof, 
but suppose Bob is traveling at a constant velocity relative to you in 
such a way that: 


*

*3 minutes on Bob's clock equal 15 minutes on your clock (time dilation) 

*15 meters of Bob's distance is equal to 3 meters of your distance 
(Lorentz contraction) 
Note that I'm making no assumption of linearity. I don't know how long 
4 minutes on Bob's clock will be on my clock. I'm only going to use 
the two specific observations above to show linearity (intuitively). 
Suppose Bob starts a 3 minute egg timer (hourglass), and, the moment 3 
minutes elapse, he turns it around to measure another 3 minutes. 
Since Bob is in an inertial (constant velocity) reference frame, his 3 
minutes plus 3 minutes add to 6 minutes. 
In your reference frame, the first 3 minutes took 15 minutes (by our 
observation above) and the second 3 minutes also took 15 minutes, 
since Bob's velocity relative to us remains constant. Thus, Bob's 6 
minutes took 15 + 15 minutes, or 30 minutes. 
Of course, you can apply this observation to any amount of time, thus 
showing linearity. 
The argument for distance is similar. If Bob walks 15 meters, pauses 
(for a length of time that will differ for the two of you), and then 
walks another 15 meters, he has walked a total of 30 meters, since 
distances add. 
You don't know how long 30 of Bob's meters are for you, but you do 
know the first 15 meters translates to 3 meters, as does the second 15 
meters. Since distance adds for you as well, you now know that 30 
meters of Bob's distance equals 6 meters of your distance. 
In other words, time and distance add in all intertial reference frames. 
Why is this not a proof? 
I assume that 3 minutes on Bob's clock is always equal to 15 minutes 
on your clock, since Bob is traveling at a constant velocity relative 
to you. 
However, it's at least theoretically possible that the speed of Bob's 
clock depends on his distance from you. Perhaps 3 minutes on Bob's 
clock equals 15 minutes on your clock the instant he passes you, but, 
when he's half a light year away, 3 minutes on his clock is now an 
hour on your clock. 
So, this isn't a proof, but if you intuitively accept that the 
difference in time and distance between two observers depends solely 
on their relative velocity, this may be helpful. 
A: The following proof only requires the continuity of the Lorentz transformations $L$ but it also requires that the two observers start measuring time at the same instant and in the the same point of space, thus $L(0)=0$.
As already mentioned by joshphysics homogeneity of the space translates into the following property:
\begin{equation}
 L(y+\varepsilon)-L(x+\varepsilon) = L(y)-L(x)\quad \forall\, x,\,y,\,\varepsilon\,. 
\end{equation}
Let now $\varepsilon = -x $ so that
$$ L(y-x) = L(y) - L(x) + L(0)\,, $$
then we have the following
$$ L(y+x) = L(y-(-x)) = L(y) - L(-x) + L(0) $$
and
$$ L(-x) = L(0-x) = L(0)-L(x)+L(0) = -L(x) + 2\,L(0)\quad .$$
Combining the last two equations we get
$$L(y+x) = L(y)+L(x)-L(0)\quad.$$
If we assume that $L(0) = 0$ then
$$ (1)\quad\begin{cases}L(y+x) = L(y)+L(x) \\ \\ L(-x) = - L(x)\end{cases}$$
It is easy to check from $(1)$ that $L(z\,y) = z\,L(y)\,$ $\forall z\in\mathbb{Z}\,.$
Consider now $q\in \mathbb{Q}$ and let $a\in\mathbb{Z}\,,$ $b\in\mathbb{N}$ such that $q=\dfrac{a}{b}\,.$ Then
$$ L(y) = L\left(\dfrac{b}{b}\,y\right) = b\,L\left(\dfrac{1}{b}y\right) \qquad \Rightarrow \qquad L\left(\dfrac{1}{b}y\right) = \dfrac{1}{b}\,L(y)$$
so that
$$L(q\,y) = L\left(\dfrac{a}{b}\,y\right) = a L\left(\dfrac{1}{b}\,y\right) = \dfrac{a}{b}\,L(y) = q\,L(y)\quad.$$
Consider now $\alpha\in\mathbb{R}$, since $\mathbb{Q}$ is dense in $\mathbb{R}$ there exists a sequence $\{q_n\}_{n=0}^\infty$ of rational numbers such that $q_n \rightarrow \alpha $ as $n\rightarrow \infty$. From the continuity of $L$ we have that
$$L(\alpha\,y ) = \lim_{n\rightarrow \infty} L(q_n\,y)  = \lim_{n\rightarrow \infty}q_n\, L(y) = \alpha\,L(y)\qquad .$$
Finally given any two real numbers $\alpha$ and $\beta$, and given any two events $x$ and $y$, we have linearity of Lorentz transformations:
$$ L(\alpha\,y+\beta\,x) = L(\alpha\,y)+L(\beta\,x) = \alpha\,L(y) + \beta\,L(x)\quad.    $$
Note that if $L(0)\neq 0$ instead of $(1)$ we have
$$ (2)\quad\begin{cases}L(y+x) = L(y)+L(x)-L(0) \\ \\ L(-x) = - L(x)+2\,L(0)\end{cases}$$
Letting $\Lambda(x) = L(x)-L(0)$ we can rewrite $(2)$ as follows
$$ (1')\quad\begin{cases}\Lambda(y+x) = \Lambda(y)+\Lambda(x) \\ \\ \Lambda(-x) = - \Lambda(x)\end{cases}$$
Since $\Lambda$ is also continuous we can repeat the previous steps showing its linearity. In the end if the Lorentz transformation is just continuous the homogeneity of the space implies that it's affine.
