Does every embedding of one unipotent group (over R) in another extend to an embedding of the respective upper triangular matrix groups?

Question

Let $T^*$ denote upper triangular matrices (of the appropriate size) with positive diagonal entries and $\mathrm{UT}$ upper triangular matrices with all diagonal entries equal to 1.

Does every (abstract group) embedding $\varphi:\mathrm{UT}(n,\mathbb{R})\to\mathrm{UT}(m,\mathbb{R})$ extend to $\bar{\varphi}:T^*(n,\mathbb{R})\to T^*(m,\mathbb{R})$?

(In the other direction, any $\bar{\varphi}:T^*(n,\mathbb{R})\to T^*(m,\mathbb{R})$ restricts to a homomorphism $\mathrm{UT}(n,\mathbb{R})\to\mathrm{UT}(m,\mathbb{R})$, since $\mathrm{UT}$ is the derived subgroup of $T^*$.)

An affirmative answer to this question implies a relatively easy affirmative answer to this question. I've recently answered the latter independently, but wonder if there's a general result that could be used, and which I should know about. I asked the current question on Math.stackexchange but without success.

EDIT: Florian Eisele has shown that as stated above the answer to the original question is no. This makes me wonder if there's a reasonably natural reformulation for which the answer is yes. For the sake of asking a concrete question, let me hazard the following.

Does every embedding $\varphi:\mathrm{UT}(n,\mathbb{Q})\to\mathrm{UT}(m,\mathbb{Q})$ extend to $\bar{\varphi}:T^*(n,\mathbb{Q})\to T^*(m,\mathbb{Q})$?

Answer 1

Here's a counterexample ($m=n=3$) which is a continuous homomorphism and actually probably also works for your second question with $\mathbf{Q}$.

In short: most automorphisms of $UT(3)$ do not extend to $T^*(3)$.

Since I deal with continuous automorphisms, it boils down to a Lie algebra problem. An automorphism of $UT(3)$ induces an automorphism of its abelianized subgroup, giving a $2\times 2$ matrix and every invertible matrix occurs this way. But my claim is that if the automorphism extends to $T^*(3)$ then (some power of) this matrix has to be diagonal.

Sketch of argument: I work upside down by considering an automorphism of $T^1(3)$, the group of determinant 1 matrices in $T^*(3)$ (it's enough and more convenient to deal with it) and describe its restriction to $UT(3)$.

First, because $UT(3)$ is the derived subgroup, it is stable. Write $T^1(3)=D.UT(3)$ where $D$ is the group of diagonal matrices. Since $D$ is a Cartan subgroup in $T^1(3)$ (that is, its Lie algebra is nilpotent and self-normalized, see Bourbaki, Groups and Lie algebras) and is unique up to conjugacy, it is mapped to a conjugate subgroup. So after composing by a conjugation, we can suppose $D$ mapped into itself. Now any automorphism of $T^1(3)$ induces a finite order automorphism of $D$: indeed, any one-parameter group of automorphisms of $T^1(3)$ induces the identity on $D$, as we see using the fact that the derived subgroup of the semidirect product $\mathbf{R}\ltimes T^1(3)$ is nilpotent (hence contained in the nilpotent radical $UT(3)$ of $T^1(3)$). So after conjugation, some power of the automorphism acts on $D$ as the identity and therefore preserves the weight decomposition of the Lie algebra of $UT(3)$ induced by the action of $D$, and thus induces a diagonal automorphism of the derived subgroup of $UT(3)$.

If you don't follow the argument you can also compute the group of continuous automorphisms of $T^*(3)$ and check that no one maps a matrix $\pmatrix{1 & x & * \cr 0 & 1 & y \cr 0 & 0 & 1\cr}$ to a matrix of the form $\pmatrix{1 & x+y & * \cr 0 & 1 & y \cr 0 & 0 & 1\cr}$.

NB: it's unclear from your question if you mean "extends as an automorphism" or "extends as an endomorphism". However, it is not hard to show that an endomorphism of $T^1(3)$ that is injective on $UT(3)$ remains injective on $T^1(3)$, and it easily follows that an automorphism of $UT(3)$ extends as an endomorphism of $T^*(3)$ iff it extends as an automorphism.

Answer 2

No (at least if you really mean abstract group embeddings). Choose $m=n=2$. Note that $$UT(2,\mathbb R) \cong (\mathbb R,+)\quad \textrm{ and }\quad T^*(2,\mathbb R)\cong UT(2,\mathbb R)\rtimes (\mathbb R_+,\cdot)^2 $$

Note secondly that the image of the conjugation action of $T^*(2,\mathbb R)$ on $UT(2,\mathbb R)$ is equal to $(\mathbb R_+,\cdot)\leq Aut((\mathbb R,+))$. In particular, $UT(2,\mathbb R)$ splits up into only three orbits under this action. I claim there is an injective homomorphism from $UT(2,\mathbb R)$ into itself such that its image $G$ splits up into many more orbits under the action of the normalizer of $G$ in $T^*(2,\mathbb R)$. Such an embedding clearly cannot be extended to all of $T^*(2,\mathbb R)$.

As a $\mathbb Q$-vector space (and hence also as an abelian group), $\mathbb R\cong \bigoplus_I \mathbb Q$ for some index set $I$ of the same cardinality as $\mathbb R$. Let $S$ be a transcendence basis of $\mathbb R$ over $\mathbb Q$ (so the cardinality of $S$ will be the same as that of $I$). Let $G$ be the $\mathbb Q$-span of $S$. So there is an isomorphism from $(\mathbb R, +)$ to $(G,+)\subsetneq (\mathbb R,+)$. Now if $s_1\neq s_2\in S$, then the unique element in $\mathbb R-\{0\}$ sending $s_1$ to $s_2$ is $r=s_2/s_1$. But if $r\cdot s_2 = s_2^2/s_1$ could be written as a $\mathbb Q$-linear combination $\sum_i c_i s_i$ of elements in $S$, then we would get an algebraic relation $$ s_1(\sum_i c_i s_i)-s_2^2 $$ between the elements in the transcendence basis $S$. Therefore $r\cdot s_2 \notin G$, i.e. $r\cdot G \nsubseteq G$, thus proving that all elements of $S$ lie in different orbits under the action of the normalizer of $G$ in $T^*(2,\mathbb R)$.