Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\Gamma\to G$?

share|cite|improve this question
Where do the $a_{ij}$ live? The reals? – JSchlather Jul 24 '12 at 17:49
Yes, sorry, added. – Earthliŋ Jul 25 '12 at 1:22
Hm, so what have you tried? You certainly get the natural inclusion and then you can look at the conjugates of that subgroup. – JSchlather Jul 25 '12 at 13:03
But will conjugates alone give all injective homomorphisms? For example, in dimension 3, the set of injective homomorphisms is the 6-dimensional space $GL_2\mathbb R\times\mathbb R^2$ with $G$ acting transitively on the $\mathbb R^2$ factor, but trivially on $GL_2\mathbb R$. Including and conjugating would thus only yield $\{\text{pt}\}\times\mathbb R^2$ and not the whole set... – Earthliŋ Jul 25 '12 at 15:07

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.