Set of homomorphisms from discrete upper triangular group into continuous u.t. group Let $G$ be the group
$$
\begin{pmatrix}
1 & a_{12} & a_{13} & a_{14}\\
0 & 1 & a_{23} & a_{24}\\
0 & 0 & 1 & a_{34}\\
0 & 0 & 0 & 1
\end{pmatrix}
$$
where $a_{ij}\in\mathbb R$ and $\Gamma=G\cap\mathrm{GL}_4\,\mathbb Z$. 
What is the set of injective homomorphisms $\Gamma\to G$?
 A: There a neat answer to the question of describing all homomorphisms $\Gamma\to G$. Malcev indeed proved (see the book by Raghunathan) a general "superrigidity" result for general unipotent groups, implying in particular that every homomorphism $\Gamma\to G_1$ has a unique extension as a continuous homomorphism $G\to G_1$ ($G_1$ any real unipotent group). Since $G$ is a simply connected Lie group, this shows that the set of continuous homomorphisms $G\to G$ can be identitied to the set of endomorphisms of its Lie algebra, which is more explicitly computable, but is not yet a linear object.
However assuming the homomorphism injective in restriction to $\Gamma$ here simplifies a bit. Namely, the Lie algebra $\mathfrak{g}$ of $G$ has the property that its center is 1-dimensional, so any non-injective homomorphism from $G$ to any Lie group will kill the center and hence be non-injective in restriction to $\Gamma$.
This shows that the set of injective homomorphisms $\Gamma\to G$ can be identified to the group $H$ of automorphisms of the Lie algebra $\mathfrak{g}$. This is a finite dimensional Lie group with finitely many components (as the group of real points of an algebraic group), whose Lie algebra $\mathfrak{h}$ can be identified to the space of derivations of the Lie algebra $\mathfrak{g}$; the latter object being even more easily computable (because unlike the set of endomorphisms, this is a linear object).
I haven't done the computation of $\mathfrak{h}$ (do it!); just note that $H$ is at least 8-dimensional, because it contains the group $T$ of conjugations by elements of the upper triangular group (which is 10-dimensional but scalar matrices as well as $(1,4)$-matrices act trivially). It also contains the involution $j:Z\mapsto Qt(Z^{-1})Q^{-1}$, where $t$ denotes transpose and $Q$ is the antidiagonal matrix with antidiagonal entries equal to 1; the latter normalizes $T$, so the group generated by $T$ and $j$ contains $T$ with index 2.
