What are the linear maps which preserve the time-like cone? I'm looking at the set of time-like vectors:
$$\mathcal{T}_+ = \{ x \in \mathbb{R}^4 \mbox{ s.t. } x^T \eta x \geq 0 \:, x^0\geq 0\} ,$$ where $\eta = \mbox{diag}(1, -1, -1, -1)$.
I want to be able to characterize the set of linear operators that preserve this set:
$$\mathcal{V} = \{L\in M(4,\mathbb{R}) \mbox{ s.t. } L x \in \cal T_+\,, \forall x \in \cal T_+\}.$$
Clearly $L$ must satisfy
$$x^T L^T \eta L x \geq 0\:,  (Lx)^0\geq 0 \:\:\forall x \in \mathcal{T}_+\:.$$
Is $\mathcal{V}$ convex? It's clear that the orthochronous Lorentz transformations preserve $\mathcal{T}_+$, since $x^T \Lambda \eta \Lambda^T x = x^T \eta x$. Do these form the boundary of $\mathcal{V}$?
A clue: I know that $\mathcal{T}_+$ is convex.
 A: Replacing $GL(4, \mathbb R)$ with $M(4, \mathbb R)$ we have
$${\cal V}:= \{M \in M(4, \mathbb R)\:|\: \mbox{if $x\in {\cal T}_+$, then $Mx \in {\cal T}_+$}\}\tag{1}\:,$$
where
$${\cal T}_+ := \{x \in \mathbb R^4 \:|\: x^T\eta x \geq 0\:, x^0 \geq 0\}$$
so that we can equivalently restate the given definition as
$${\cal V}:=$$ $$\{M \in M(4, \mathbb R)\:|\: \mbox{ $x^T\eta x \geq 0$,  $x^0 \geq 0$ $\Rightarrow$ $x^TM^T\eta Mx \geq 0$, $(Mx)^0 \geq 0$}\}\tag{2}\:.$$
Proposition.
With the given definition of $\cal V$, it turns out to be a closed convex cone of the real vector space $M(4,\mathbb R)$.
PROOF.
The set is trivially a cone, because, if $M \in \cal V$, then $\lambda M \in \cal V$ in view of the given definition for $\lambda \geq 0$.
Let us establish that it is convex, too.
Let  $M, M' \in \cal V$ and $p,q \in [0,1]$ with $p+q=1$. If $x\in {\cal T}_+$, then $Mx$, $M'x \in {\cal T}_+$ by (1). Moreover, as  ${\cal T}_+$ is convex, it also holds $pMx+ qM'x \in  {\cal T}_+$. Using (2), this fact can equivalently be re-written as $$(pMx+ qM'x)^T \eta (pMx+ qM'x) \geq 0\:,$$
that is
$$x^T(pM+ qM')^T \eta (pM+ qM')x \geq 0$$
Similarly, because the left-hand side is sum of two non-negative numbers, $$((pM+qM')x)^0\geq 0\:.$$
Since the found identity hold for every $x\in {\cal T}_+$, looking at (2), we have proved that if $M,M' \in \cal V$ and $p,q \in [0,1]$ with $p+q=1$, then
$pM+qM' \in \cal V$. In other words $\cal V$ is convex.
Let us finally prove that $\cal V$ is closed in the natural topology of $M(4,\mathbb R) \subset \mathbb R^{16}$. If ${\cal V} \ni M_n \to M \in M(4,\mathbb R)$ with respect to the said topology for $n\to +\infty$, and $x^T\eta x \geq 0$, then  $$0\leq x^TM_n^T\eta M_nx \to x^TM^T\eta Mx\geq 0$$ since all involved operations are continuous and $[0,+\infty)$ is closed.
The condition on $x^0$ survives the limit procedure similarly.
(2) implies that $M \in \cal V$. Since $\cal V$ includes all of its limit points, it must be closed.
QED
ADDENDUM
Regarding the boundary of $\cal V$, we have $O(3,1)_+ \subset \partial \cal V$. In fact, if $\Lambda \in O(3,1)_+$, then (a)  it belongs to $\cal V$ and (b) there is a continuous curve joining $\Lambda$ and some point in $M(4,\mathbb R) \setminus \cal V$. It is $$[1/2,1]\ni \Lambda_\lambda := \lambda \mapsto \mbox{diag}(\lambda, 1, 1, 1) \Lambda\:.$$
For every $\lambda \in [1/2,1)$, $\Lambda_\lambda \not \in \cal V$ because it transforms future-oriented  lightlike vectors into spacelike ones. Summing up, $\Lambda \in \cal V$, but
$\Lambda \not \in Int(\cal V)$. It must be $\Lambda \in \partial \cal V$.
Nevertheless, the same reasoning holds if replacing $\Lambda$ with a pure dilatation transformation $D_z u := z u$ with $z>0$ fixed, for all $u \in \mathbb R^4$. Therfore, as $D_z \not \in O(3,1)_+$, we have that $O(3,1)_+ \neq \partial \cal V$.
Compositions $D_z \Lambda$ certainly belong to $\partial \cal V$. I suspect they are the only elements of that frontier and that (see Qmechanic's answer) the convex hull of the set of those elements coincides to $\cal V$ itself.
A: OP is asking about 3+1 dimensions, but let us here work out the corresponding construction in 1+1 dimensions. The 1+1 dimensional result may be used as a toy model to gain some intuition of what might (or might not) hold in higher dimensions. We use light-cone coordinates $x^{\pm}$.

*

*The future light-cone is
$$ {\cal T}_+~=~\{(x^+,x^-)\in \mathbb{R}^2 \mid x^{\pm}\geq 0\}.\tag{1}$$
The set
$$\begin{align}{\cal V} ~:=~& \{M\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid M({\cal T}_+)\subseteq {\cal T}_+\}\cr 
~=~&\left\{ \left.\begin{pmatrix}a &b\cr c &d \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{R}) \right| a,b,c,d\geq 0 \right\} \end{align}\tag{2}$$
is the set of non-negative matrices.


*The restricted Lorentz group is
$$ SO^+(1,1)~=~\left\{ \left.\begin{pmatrix}a &0\cr 0 &a^{-1} \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{R}) \right| a>0 \right\}.\tag{3} $$
The convex hull of $SO^+(1,1)$ is
$$ {\rm Conv}(SO^+(1,1))~=~\left\{ \left.\begin{pmatrix}a &0\cr 0 &d \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{R}) \right| a,d>0, ad\geq 1 \right\}.\tag{4}$$
The closed convex cone of $SO^+(1,1)$ is the set
$$ \overline{{\rm Convcone}(SO^+(1,1))}~=~\left\{ \left.\begin{pmatrix}a &0\cr 0 &d \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{R}) \right| a,d\geq0,  \right\}\tag{5}$$
of diagonal matrices with non-negative eigenvalues.


*The orthochronous Lorentz group is
$$ O^+(1,1)~=~SO^+(1,1) \cup \left\{ \left.\begin{pmatrix}0 &b\cr b^{-1} &0 \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{R}) \right| b>0 \right\}.\tag{6} $$
The closed convex cone
$$ \overline{{\rm Convcone}(O^+(1,1))}~=~{\cal V} \tag{7}$$
of $O^+(1,1)$ is OP's set (2).
